Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

ATKINS’

PHYSICAL
CHEMISTRY

This page intentionally left blank

ATKINS’

PHYSICAL
CHEMISTRY
Eighth Edition
Peter Atkins
Professor of Chemistry,
University of Oxford,
and Fellow of Lincoln College, Oxford

Julio de Paula
Professor and Dean of the College of Arts and Sciences
Lewis and Clark College,
Portland, Oregon

W
...
Freeman and Company
New York

Library of Congress Control Number: 2005936591
Physical Chemistry, Eighth Edition
© 2006 by Peter Atkins and Julio de Paula
All rights reserved
ISBN: 0-7167-8759-8
EAN: 9780716787594
Published in Great Britain by Oxford University Press
This edition has been authorized by Oxford University Press for sale in the
United States and Canada only and not for export therefrom
...
H
...
whfreeman
...
The bulk of textbooks is a perennial concern: we
have sought to tighten the presentation in this edition
...

The most striking change in presentation is the use of colour
...
The text is still divided into three parts, but material has been moved
between chapters and the chapters have been reorganized
...
We no longer make a distinction between ‘concepts’
and ‘machinery’, and as a result have provided a more compact presentation of thermodynamics with less artificial divisions between the approaches
...

In Part 2 (Structure) the principal changes are within the chapters, where we have
sought to bring into the discussion contemporary techniques of spectroscopy and
approaches to computational chemistry
...
Moreover, we have
introduced concepts of nanoscience throughout much of Part 2
...
We
regard this material as highly important in a contemporary context, but as a final
chapter it rarely received the attention it deserves
...

We have discarded the Boxes of earlier editions
...
By liberating these topics from
their boxes, we believe they are more likely to be used and read; there are end-ofchapter problems on most of the material in these sections
...
That discussion continues
...
The strategic aim of this revision
is to make it possible to work through the text in a variety of orders and at the end of
this Preface we once again include two suggested road maps
...
Thus, we give more help with the development of equations, motivate

vi

PREFACE

them, justify them, and comment on the steps
...

We are, of course, alert to the developments in electronic resources and have made
a special effort in this edition to encourage the use of the resources on our Web site (at
www
...
com/pchem8) where you can also access the eBook
...
To do so, wherever we
call out a Living graph (by an icon attached to a graph in the text), we include an
Exploration in the figure legend, suggesting how to explore the consequences of
changing parameters
...

Oxford
Portland

P
...
A
...
de P
...
One of the problems that make the
subject daunting is the sheer amount of information: we have introduced several
devices for organizing the material: see Organizing the information
...
Problem solving—especially, ‘where do I start?’—is often a challenge, and
we have done our best to help overcome this first hurdle: see Problem solving
...
The following paragraphs explain the features in
more detail
...
A gas is a form of matter that fills any container it occupies
...
An equation of state interrelates pressure, volume,
temperature, and amount of substance: p = f(T,V,n)
...
The pressure is the force divided by the area to which the force
is applied
...

4
...

5
...

6
...
An adiabatic boundary is a boundary that
prevents the passage of energy as heat
...
Thermal equilibrium is a condition in which no change of
state occurs when two objects A and B are in contact through
a diathermic boundary
...
The Zeroth Law of thermodynamics states that, if A is in
thermal equilibrium with B, and B is in thermal equilibrium
with C, then C is also in thermal equilibrium with A
...
The Celsius and thermodynamic temperature scales are
related by T/K = θ/°C + 273
...

10
...
The partial pressure of any gas i
xJ = nJ/n is its mole fraction in a
pressure
...
In real gases, molecular interact
state; the true equation of state i
coefficients B, C,
...
We suggest checking off the box that precedes each
entry when you feel confident about the topic
...
The vapour pressure is the press
with its condensed phase
...
The critical point is the point at
end of the horizontal part of the
a single point
...

16
...

17
...
A reduced variable is the actual
corresponding critical constant

IMPACT ON NANOSCIENCE

I20
...
1, I9
...
3) that research on nanometre-sized materials is motivated by the possibility that they will form the basis for
cheaper and smaller electronic devices
...
An important type of nanowire is based on carbon nanotubes, which,
like graphite, can conduct electrons through delocalized π molecular orbitals that
form from unhybridized 2p orbitals on carbon
...
The SWNT in Fig
...
45 is a semiconductor
...

Carbon nanotubes are promising building blocks not only because they have useful
electrical properties but also because they have unusual mechanical properties
...

Silicon nanowires can be made by focusing a pulsed laser beam on to a solid target
composed of silicon and iron
...

The Impact sections show how the principles developed in
the chapter are currently being applied in a variety of modern
contexts
...
This scale is absolute, and the lowest
temperature is 0 regardless of the size of the divisions on the scale (just as we write
p = 0 for zero pressure, regardless of the size of the units we adopt, such as bar or
pascal)
...


The material on regular solutions presented in Section 5
...
The
starting point is the expression for the Gibbs energy of mixing for a regular solution
(eqn 5
...
We show in the following Justification that eqn 5
...
We have used this feature to help encourage the use
of the language and procedures of science in conformity to
international practice and to help avoid common mistakes
...
8 The activities of regular solutions

2
ln γA = βxB

ix

(5
...

Justification 5
...
However, mathematical development is an intrinsic part of physical chemistry, and it is
important to see how a particular expression is obtained
...


∆mixG = nRT{xA ln aA + xB ln aB}
This relation follows from the derivation of eqn 5
...
If each activity is replaced by γ x, this expression becomes
∆mixG = nRT{xA ln xA + xB ln xB + xA ln γA + xB ln γB}
Now we introduce the two expressions in eqn 5
...
31
...


Molecular interpretation sections
Molecular interpretation 5
...
If it is not an enthalpy
effect, it must be an entropy effect
...
Its vapour pressure reflects the tendency of the solution towards greater entropy, which can be achieved if the liquid vaporizes to
form a gas
...
Because the entropy of the liquid is
already higher than that of the pure liquid, there is a weaker tendency to form the
gas (Fig
...
22)
...

Similarly, the enhanced molecular randomness of the solution opposes the
tendency to freeze
...
Hence, the freezing point is
lowered
...
The Molecular
interpretation sections enhance and enrich coverage of that
material by explaining how it can be understood in terms of
the behaviour of atoms and molecules
...
1 The Debye–Hückel theory of ionic
solutions

Imagine a solution in which all the ions have their actual positions,
but in which their Coulombic interactions have been turned off
...
For a salt M p Xq, we write

where rD is called the Debye length
...


1
...
8

Potential, f/(Z /rD)

ideal
ideal
= p(µ+ − µ + ) + q(µ− − µ − )

From eqn 5
...
73)

Zi

Zi =

r

zte

φi =

966

r

0
...
3

0

0

0
...
74)

4πε

The ionic atmosphere causes the potential to decay with distance
more sharply than this expression implies
...
6

0
...

The Coulomb potential at a distance r from an isolated ion of
charge zie in a medium of permittivity ε is

φi =

In some cases, we have judged that a derivation is too long,
too detailed, or too different in level for it to be included
in the text
...


−r/rD

e

(5
...
5
...


Exploration Write an expression f
unshielded and shielded Coulom
Then plot this expression against rD and
interpretation for the shape of the plot
...
6 Partial derivatives
A partial derivative of a function of more than one variable
of the function with respect to one of the variables, all the
constant (see Fig
...
Although a partial derivative show
when one variable changes, it may be used to determine
when more than one variable changes by an infinitesimal a
tion of x and y, then when x and y change by dx and dy, res
df =

Physical chemistry draws on a lot of background material, especially in mathematics and physics
...


A ∂f D
A ∂f D
dx +
dy
C ∂x F y
C ∂y F x

where the symbol ∂ is used (instead of d) to denote a parti
df is also called the differential of f
...
8 Expansion coefficients, α, and isothermal
compressibilities, κT
a/(10 − 4 K−1 )

Table 2
...
4

92
...
4

90
...
2

76
...
82

38
...
1

49
...
501

Tf /K

603
723

83
...
7s

40
202

14
...
6

Methane
0
...
030

0
...
354

0
...
861

2
...

Data: AIP(α), KL(κT)
...
6

Neon

231

24
...
3

Oxygen

764

54
...

Data: AIP, JL, and M
...
Zemansky, Heat and
New York (1957)
...

We provide a lot of data in the Data section at the end of the
text and short extracts in the Synoptic tables in the text itself to
give an idea of the typical values of the physical quantities we
are introducing
...
5
Comment 1
...


e
e

The partial-differential operation
(∂z/∂x)y consists of taking the first
derivative of z(x,y) with respect to x,
treating y as a constant
...


A ∂z D A ∂[x 2y] D
dx 2
B E =B
E =y
= 2yx
C ∂x F y C ∂x F y
dx
Partial derivatives are reviewed in
Appendix 2
...
These appendices do not go
into great detail, but should be enough to act as reminders of
topics learned in other courses
...
3 The trajectory in terms of the energy
The velocity, V, of a particle is the rate of change of its po
V=
py

dr
dt

The velocity is a vector, with both direction and magnit
velocity is the speed, v
...

A3
...
A3
...
In terms of the linear momentu
ticle is
2

Problem solving
Illustrations
Illustration 5
...
9 × 104 kPa kg mol−1

= 2
...
29 mmol kg−1
...
99709 kg dm−3
...
29 mmol kg−1 × 0
...
29 mmol dm−3
A note on good practice The number of significant figures in the result of a calcu-

lation should not exceed the number in the data (only two in this case)
...
5 Calculate the molar solubility of nitrogen in water exposed to air at

25°C; partial pressures were calculated in Example 1
...


[0
...
In particular, we show how to use data and
how to manipulate units correctly
...
1 Calculating the number of photons

Calculate the number of photons emitted by a 100 W yellow lamp in 1
...
Take the
wavelength of yellow light as 560 nm and assume 100 per cent efficiency
...
To use this equation, we need to know the frequency
of the radiation (from ν = c/λ) and the total energy emitted by the lamp
...

Answer The number of photons is

N=

E
hν

=

P∆t
h(c/λ)

=

A Worked example is a much more structured form of
Illustration, often involving a more elaborate procedure
...
Then there is the
worked-out Answer
...
60 × 10−7 m) × (100 J s−1) × (1
...
626 × 10−34 J s) × (2
...
8 × 1020

Note that it would take nearly 40 min to produce 1 mol of these photons
...
Moreover, an analytical result may be used for other data
without having to repeat the entire calculation
...
1 How many photons does a monochromatic (single frequency)
infrared rangefinder of power 1 mW and wavelength 1000 nm emit in 0
...
12 Calculate the change in Gm for ice at −10°C, with density 917 kg m−3,

[+2
...
0 bar to 2
...


Discussion questions

Discussion questions
1
...

1
...

1
...


Each Worked example, and many of the Illustrations, has a Selftest, with the answer provided as a check that the procedure has
been mastered
...
Think of Self-tests as in-chapter Exercises designed to
help monitor your progress
...
4 What is the significance of the critical co
1
...
6 Explain how the van der Waals equation

behaviour
...


ABOUT THE BOOK
Exercises and Problems

Exercises
Molar absorption coefficient, e

+
14
...
What is the total

spin and total orbital angular momentum of the molecule? Show that the term
symbol agrees with the electron configuration that would be predicted using
the building-up principle
...
1b One of the excited states of the C2 molecule has the valence electron
2
2
3
configuration 1σ g 1σ u1π u1π 1
...

g
14
...
Calculate the percentage
reduction in intensity when light of that wavelength passes through 2
...
25 mmol dm−3
...
2b The molar absorption coefficient of a substance dissolved in hexane is
known to be 327 dm3 mol−1 cm−1 at 300 nm
...
50 mm
of a solution of concentration 2
...

14
...
00 cm transmits 20
...
If the concentration of the component is
0
...
3b When light of wavelength 400 nm passes through 3
...
667 mmol dm−3, the
transmission is 65
...
Calculate the molar absorption coefficient of the
solute at this wavelength and express the answer in cm2 mol−1
...
14
...
7b The following data were obtained for th

in methylbenzene using a 2
...
Calcu
coefficient of the dye at the wavelength emplo
[dye]/(mol dm−3)

0
...
0050

0
...
2

ll

fill d

h

l

Problems
Assume all gases are perfect unless stated otherwise
...
013 25 bar
...
15 K
...
1 A sample consisting of 1 mol of perfect gas atoms (for which
3
CV,m = – R) is taken through the cycle shown in Fig
...
34
...
(b) Calculate q, w, ∆U, and ∆H for each
step and for the overall cycle
...


1
...
2
...

2
...
9 The standard enthalpy of formation of t

bis(benzene)chromium was measured in a c
reaction Cr(C6H6)2(s) → Cr(s) + 2 C6H6(g) t
Find the corresponding reaction enthalpy an
of formation of the compound at 583 K
...
1 J K−1 mol−1
81
...


2

1

in a calorimeter and then ignited in the prese
temperature rose by 0
...
In a separate ex
the combustion of 0
...
10‡ From the enthalpy of combustion dat

3

0
...
44

44
...
2
...
2 A sample consisting of 1
...
The heating was carried out in a container fitted with a piston
that was initially resting on the solid
...
0 atm
...
11 It is possible to investigate the thermoc

hydrocarbons with molecular modelling me
software to predict ∆cH 7 values for the alkan
calculate ∆cH 7 values, estimate the standard
CnH2(n+1)(g) by performing semi-empirical c
or PM3 methods) and use experimental stan
values for CO2(g) and H2O(l)
...
5) and
the molecular modelling method
...
The Exercises are straightforward numerical tests that give practice with manipulating
numerical data
...
They are divided into ‘numerical’, where the emphasis is on the manipulation of data, and ‘theoretical’, where the emphasis is on the
manipulation of equations before (in some cases) using numerical data
...


About the Web site
The Web site to accompany Physical Chemistry is available at:
www
...
com/pchem8

It includes the following features:
Living graphs

A Living graph is indicated in the text by the icon
attached
to a graph
...
To encourage
the use of this resource (and the more extensive Explorations in
Physical Chemistry) we have added a question to each figure
where a Living graph is called out
...
16 The boundary surfaces of d orbitals
...
The dark and light areas
denote regions of opposite sign of the
wavefunction
...

l

y
x
dx 2- y 2

dz 2

dyz
dxy

dz

ABOUT THE WEB SITE
Artwork

An instructor may wish to use the illustrations from this text
in a lecture
...
This edition is in full
colour: we have aimed to use colour systematically and helpfully, not just to make the page prettier
...


xv

integrating all student media resources and adds features unique to the eBook
...
Access to the eBook is included
with purchase of the special package of the text (0-7167-85862), through use of an activation code card
...
whfreeman
...

Key features of the eBook include:
• Easy access from any Internet-connected computer via a
standard Web browser
...


Web links

• Integration of all Living Graph animations
...

Also, a piece of information may be needed that we have not
included in the text
...


• Text highlighting, down to the level of individual phrases
...

• A powerful Notes feature that allows students or instructors to add notes to any page
...


Interactive calculators, plotters and a periodic table for the
study of chemistry
...

• Automatic saving of all notes, highlighting, and bookmarks
...


Explorations in Physical Chemistry
Now from W
...
Freeman & Company, the new edition of the
popular Explorations in Physical Chemistry is available on-line
at www
...
com/explorations, using the activation
code card included with Physical Chemistry 8e
...
They motivate
students to simulate physical, chemical, and biochemical
phenomena with their personal computers
...
and
Excel® by Microsoft Corporation, students can manipulate
over 75 graphics, alter simulation parameters, and solve equations to gain deeper insight into physical chemistry
...


The Physical Chemistry, Eighth Edition eBook
A complete online version of the textbook
...

• Instructor notes: Lecturers can choose to create an annotated version of the eBook with their notes on any page
...

• Custom content: Lecturer notes can include text, web
links, and even images, allowing lecturers to place any
content they choose exactly where they want it
...
The chapters from Physical Chemistry, 8e that appear in
each volume are as follows:

Volume 1: Thermodynamics and Kinetics
(0-7167-8567-6)
1
...
The first law

xvi
3
...

5
...

7
...

22
...

24
...
Quantum theory: introduction and principles
9
...

11
...

13
...

15
...

17
...
A Student’s Solutions Manual (0-71676206-4) provides full solutions to the ‘a’ exercises and the
odd-numbered problems
...


About the authors

Julio de Paula is Professor of Chemistry and Dean of the College of Arts & Sciences at
Lewis & Clark College
...
A
...
D
...
His research activities encompass the areas of
molecular spectroscopy, biophysical chemistry, and nanoscience
...


Peter Atkins is Professor of Chemistry at Oxford University, a fellow of Lincoln
College, and the author of more than fifty books for students and a general audience
...
A frequent lecturer in the United States
and throughout the world, he has held visiting prefessorships in France, Israel, Japan,
China, and New Zealand
...


Acknowledgements
A book as extensive as this could not have been written without
significant input from many individuals
...
Our warm thanks go Charles Trapp, Carmen Giunta,
and Marshall Cady who have produced the Solutions manuals that
accompany this book
...
We therefore wish to thank the following colleagues most
warmly:
Joe Addison, Governors State University
Joseph Alia, University of Minnesota Morris
David Andrews, University of East Anglia
Mike Ashfold, University of Bristol
Daniel E
...
M
...
Braiman, Syracuse University
Alex Brown, University of Alberta
David E
...
Michael Duncan, Cornell University
Christer Elvingson, Uppsala University
Cherice M
...
Frantzen, Stephen F
...
Garza-López, Pomona College
Robert J
...
Halpern, Indiana State University
Tom Halstead, University of York
Todd M
...
Harbison, University Nebraska at Lincoln
Ulf Henriksson, Royal Institute of Technology, Sweden
Mike Hey, University of Nottingham
Paul Hodgkinson, University of Durham
Robert E
...
Kapoor, University of Delhi
Peter Karadakov, University of York

Miklos Kertesz, Georgetown University
Neil R
...
Madura, Duquesne University
Andrew Masters, University of Manchester
Paul May, University of Bristol
Mitchell D
...
Micha, University of Florida
Sergey Mikhalovsky, University of Brighton
Jonathan Mitschele, Saint Joseph’s College
Vicki D
...
Perona, CSU Stanislaus
Nils-Ola Persson, Linköping University
Richard Pethrick, University of Strathclyde
John A
...
Prasad, University of Hyderabad
Steve Price, University College London
S
...
Ramaraj, Madurai Kamaraj University
David Ritter, Southeast Missouri State University
Bent Ronsholdt, Aalborg University
Stephen Roser, University of Bath
Kathryn Rowberg, Purdue University Calumet
S
...
Safron, Florida State University
Kari Salmi, Espoo-Vantaa Institute of Technology
Stephan Sauer, University of Copenhagen
Nicholas Schlotter, Hamline University
Roseanne J
...
J
...
Siders, University of Minnesota, Duluth
Harjinder Singh, Panjab University
Steen Skaarup, Technical University of Denmark
David Smith, University of Exeter
Patricia A
...
Warner, University of Southern Denmark

ACKNOWLEDGEMENTS
Richard Wells, University of Aberdeen
Ben Whitaker, University of Leeds
Christopher Whitehead, University of Manchester
Mark Wilson, University College London
Kazushige Yokoyama, State University of New York at Geneseo
Nigel Young, University of Hull
Sidney H
...
We would also like to thank our two publishers,
Oxford University Press and W
...
Freeman & Co
...
Authors could
not wish for a more congenial publishing environment
...
1
1
...
1

The states of gases
The gas laws
Impact on environmental science: The gas laws
and the weather

Real gases
1
...
4
1
...
1
2
...
3
2
...
5
I2
...
6

Work, heat, and energy
The internal energy
Expansion work
Heat transactions
Enthalpy
Impact on biochemistry and materials science:
Differential scanning calorimetry
Adiabatic changes

Thermochemistry
2
...
2
2
...
9

Standard enthalpy changes
Impact on biology: Food and energy reserves
Standard enthalpies of formation
The temperature-dependence of reaction enthalpies

State functions and exact differentials
2
...
11
2
...
1: Adiabatic processes
Further information 2
...
1
3
...
1
3
...
4

The dispersal of energy
Entropy
Impact on engineering: Refrigeration
Entropy changes accompanying specific processes
The Third Law of thermodynamics

Concentrating on the system
3
...
6

The Helmholtz and Gibbs energies
Standard reaction Gibbs energies

Combining the First and Second Laws

49

49
52
54
56

94

95
100
102

102
103
105

Checklist of key ideas
Further reading
Further information 3
...
2: Real gases: the fugacity
Discussion questions
Exercises
Problems

109
110
110
111
112
113
114

Phase diagrams
4
...
2
I4
...
3

The stabilities of phases
Phase boundaries
Impact on engineering and technology:
Supercritical fluids
Three typical phase diagrams

57

Phase stability and phase transitions

57
59
63

4
...
5
4
...
7

67
68
69
69

77
78
85
87
92

The fundamental equation
Properties of the internal energy
Properties of the Gibbs energy

3
...
8
3
...
1
5
...
3
I5
...
4
5
...
2

Liquid mixtures
Colligative properties
Impact on biology: Osmosis in physiology
and biochemistry

Activities

136
136

The response of equilibria to the conditions

136
141
143

7
...
4
I7
...
1: The Debye–Hückel theory
of ionic solutions
Discussion questions
Exercises
Problems

166
167

5
...
7
5
...
9

The Gibbs energy minimum
The description of equilibrium

7
...
2

167
169
169
171

How equilibria respond to pressure
The response of equilibria to temperature
Impact on engineering: The extraction
of metals from their oxides

Equilibrium electrochemistry
7
...
6
7
...
8
7
...
2

Half-reactions and electrodes
Varieties of cells
The electromotive force
Standard potentials
Applications of standard potentials
Impact on biochemistry: Energy conversion
in biological cells

Checklist of key ideas
Further reading
Discussion questions
Exercises
Problems

PART 2 Structure
8 Quantum theory: introduction and principles

6 Phase diagrams

The failures of classical physics
Wave–particle duality
Impact on biology: Electron microscopy

Definitions
The phase rule

174
176

8
...
2
I8
...
3
8
...
1
6
...
3
6
...
5
6
...
1
I6
...
5
8
...
7

193
194
194
195
197

Checklist of key ideas
Further reading
Discussion questions
Exercises
Problems

243

244
249
253
254

254
256
260

260
269
272
273
273
274
274
275

CONTENTS
9 Quantum theory: techniques and applications
Translational motion
9
...
2
9
...
1

A particle in a box
Motion in two and more dimensions
Tunnelling
Impact on nanoscience: Scanning
probe microscopy

Vibrational motion
9
...
5

The energy levels
The wavefunctions

Rotational motion
9
...
7
I9
...
8

Rotation in two dimensions: a particle on a ring
Rotation in three dimensions: the particle on a
sphere
Impact on nanoscience: Quantum dots
Spin

Techniques of approximation
9
...
10

Time-independent perturbation theory
Time-dependent perturbation theory

Checklist of key ideas
Further reading
Further information 9
...
2: Perturbation theory
Discussion questions
Exercises
Problems

10 Atomic structure and atomic spectra

277

11 Molecular structure
The Born–Oppenheimer approximation

362

278
283
286

Valence-bond theory

363

288

Molecular orbital theory

290

291
292

11
...
2

11
...
4
11
...
1

297

297
301
306
308
310

310
311
312
313
313
313
316
316
317

320

Homonuclear diatomic molecules
Polyatomic molecules

The hydrogen molecule-ion
Homonuclear diatomic molecules
Heteronuclear diatomic molecules
Impact on biochemistry: The
biochemical reactivity of O2, N2, and NO

Molecular orbitals for polyatomic systems
11
...
7
11
...
1
12
...
3

Operations and symmetry elements
The symmetry classification of molecules
Some immediate consequences of symmetry

Applications to molecular orbital theory and
spectroscopy

321
326
335

12
...
5
12
...
1
10
...
3

10
...
5

The structure of hydrogenic atoms
Atomic orbitals and their energies
Spectroscopic transitions and selection rules

The spectra of complex atoms
I10
...
6
10
...
8
10
...
1: The separation of motion
Discussion questions
Exercises
Problems

362

277

320

The structure and spectra of hydrogenic atoms

xxv

345

346
346
347
348
352
356
357
357
358
358
359

Character tables and symmetry labels
Vanishing integrals and orbital overlap
Vanishing integrals and selection rules

363
365
368

368
373
379
385
386

387
392
396
398
399
399
399
400

404
404

405
406
411

413

413
419
423
425
426
426
426
427

13 Molecular spectroscopy 1: rotational and

vibrational spectra

430

General features of spectroscopy

431

13
...
2
13
...
1

Experimental techniques
The intensities of spectral lines
Linewidths
Impact on astrophysics: Rotational and
vibrational spectroscopy of interstellar space

431
432
436
438

xxvi

CONTENTS
Pure rotation spectra
13
...
5
13
...
7
13
...
9
13
...
11
13
...
13

Molecular vibrations
Selection rules
Anharmonicity
Vibration–rotation spectra
Vibrational Raman spectra of diatomic
molecules

441

441
443
446
449
450

15 Molecular spectroscopy 3: magnetic resonance

508
509
510

513

452

The effect of magnetic fields on electrons and nuclei

513

452
454
455
457

15
...
2
15
...
14 Normal modes
13
...
2 Impact on environmental science: Global
warming
13
...
3 Impact on biochemistry: Vibrational microscopy
13
...
1: Spectrometers
Further information 13
...
4
15
...
6
15
...
8
15
...
1
15
...
11
15
...
13

The magnetization vector
Spin relaxation
Impact on medicine: Magnetic resonance imaging
Spin decoupling
The nuclear Overhauser effect
Two-dimensional NMR
Solid-state NMR

Electron paramagnetic resonance
473
476
476
478

481

The characteristics of electronic transitions

481

14
...
2 The electronic spectra of polyatomic molecules
I14
...
1: Fourier transformation of the
FID curve
Discussion questions
Exercises
Problems

554
555

16 Statistical thermodynamics 1: the concepts

560

15
...
15
15
...
2

492

14
...
2 Impact on biochemistry: Fluorescence

517

14
...
5
14
...
1: Examples of practical lasers

496

496
500
505
506
506

555
556
556
557

The distribution of molecular states

561

16
...
2 The molecular partition function
I16
...
3
16
...
7 Independent molecules

577

Checklist of key ideas
Further reading
Further information 16
...
2: The Boltzmann formula
Further information 16
...
5
16
...
1
17
...
3
17
...
5
17
...
7
17
...
1
18
...
3

Electric dipole moments
Polarizabilities
Relative permittivities

578
579

589

589
591
599

599
601
604
606
609
610
615
615
617
617
618

620
620

620
624
627

Interactions between molecules

629
637

19 Materials 1: macromolecules and aggregates

646
646
646
647
648
648
649

652

Determination of size and shape

652

Mean molar masses
Mass spectrometry
Laser light scattering
Ultracentrifugation
Electrophoresis
Impact on biochemistry: Gel electrophoresis in
genomics and proteomics
19
...
1
19
...
3
19
...
5
I19
...
7
19
...
9
I19
...
10
19
...
12

The different levels of structure
Random coils
The structure and stability of synthetic polymers
Impact on technology: Conducting polymers
The structure of proteins
The structure of nucleic acids
The stability of proteins and nucleic acids

Self-assembly
19
...
14
19
...
3

Colloids
Micelles and biological membranes
Surface films
Impact on nanoscience: Nanofabrication with
self-assembled monolayers

Checklist of key ideas
Further reading
Further information 19
...
4 Interactions between dipoles
18
...
1 Impact on medicine: Molecular recognition

Checklist of key ideas
Further reading
Further information 18
...
2: The basic principles of
molecular beams
Discussion questions
Exercises
Problems

xxvii

and drug design
Gases and liquids
18
...
7
18
...
4 Neutron and electron diffraction

697
700
702

20
...
2
20
...
1

711
713

xxviii

CONTENTS

Crystal structure
20
...
6
20
...
8
20
...
2
20
...
11
20
...
1 The kinetic model of gases
I21
...
1
22
...
3
22
...
5

Experimental techniques
The rates of reactions
Integrated rate laws
Reactions approaching equilibrium
The temperature dependence of reaction rates

791
791

792
794
798
804
807

Accounting for the rate laws

809

22
...
7 Consecutive elementary reactions
I22
...
8

Checklist of key ideas
Further reading
Further information 22
...
2 Collision with walls and surfaces
21
...
4 Transport properties of a perfect gas

22 The rates of chemical reactions

21
...
6
21
...
8
I21
...
11 Diffusion probabilities
21
...
3
23
...
1: The transport characteristics of
a perfect gas
Discussion questions
Exercises
Problems

783
783
784
785
786
788

Stepwise polymerization
Chain polymerization

Homogeneous catalysis
23
...
6

Features of homogeneous catalysis
Enzymes

Photochemistry
23
...
1
I23
...
9 The thermodynamic view
21
...
3 Impact on biochemistry: Transport of non-

23
...
2

23
...
3

Kinetics of photophysical and photochemical
processes
Impact on environmental science: The chemistry
of stratospheric ozone
Impact on biochemistry: Harvesting of light
during plant photosynthesis
Complex photochemical processes
Impact on medicine: Photodynamic therapy

Checklist of key ideas
Further reading
Further information 23
...
1
24
...
3

Collision theory
Diffusion-controlled reactions
The material balance equation

Transition state theory

869
869

870
876
879
880

The Eyring equation
Thermodynamic aspects

880
883

The dynamics of molecular collisions

885

24
...
5

24
...
7
24
...
9

Reactive collisions
Potential energy surfaces
Some results from experiments and calculations
The investigation of reaction dynamics with
ultrafast laser techniques

Electron transfer in homogeneous systems
24
...
11
24
...
1

The rates of electron transfer processes
Theory of electron transfer processes
Experimental results
Impact on biochemistry: Electron transfer in and
between proteins

Checklist of key ideas
Further reading
Further information 24
...
1
25
...
3
25
...
5
I25
...
1 Logarithms and exponentials
A2
...
3 Vectors

963
963
964

Calculus

965

Differentiation and integration
Power series and Taylor expansions
Partial derivatives
Functionals and functional derivatives
Undetermined multipliers
Differential equations

965
967
968
969
969
971

910
911

A2
...
5
A2
...
7
A2
...
9

916

Statistics and probability

973

916
917
922
925

A2
...
11 Some results of probability theory

973
974

Matrix algebra

975

A2
...
13 Simultaneous equations
A2
...
6 Mechanisms of heterogeneous catalysis
25
...
2 Impact on technology: Catalysis in the

25
...
9
25
...
11

Checklist of key ideas
Further reading
Further information 25
...
12
I25
...
13
I25
...
1 Kinetic and potential energy
A3
...
3 The trajectory in terms of the

energy
A3
...
5 Rotational motion
A3
...
11
A3
...
13
A3
...
7
A3
...
9
A3
...
1
I2
...
2
I3
...
1
I5
...
2
I6
...
2
I7
...
2
I8
...
1
I9
...
1
I11
...
1
I13
...
3
I14
...
2
I15
...
2
I16
...
1
I19
...
2
I19
...
1
I20
...
1
I21
...
3
I22
...
1
I23
...
3
I24
...
1
I25
...
3
I25
...
Equilibria include physical change, such as fusion and
vaporization, and chemical change, including electrochemistry
...

We see that we can obtain a unified view of equilibrium and the direction of
spontaneous change in terms of the chemical potentials of substances
...


1 The properties of gases
2 The First Law
3 The Second Law
4 Physical transformations of pure substances
5 Simple mixtures
6 Phase diagrams
7 Chemical equilibrium

This page intentionally left blank

1

The properties
of gases
This chapter establishes the properties of gases that will be used throughout the text
...
We then see how the properties of real
gases differ from those of a perfect gas, and construct an equation of state that describes
their properties
...
Initially we consider only pure gases, but later in the chapter we see that the
same ideas and equations apply to mixtures of gases too
...
1 The states of gases
1
...
1 Impact on environmental

science: The gas laws and the
weather
Real gases
1
...
4 The van der Waals equation

The perfect gas

1
...

A gas differs from a liquid in that, except during collisions, the molecules of a gas are
widely separated from one another and move in paths that are largely unaffected by
intermolecular forces
...
1 The states of gases
The physical state of a sample of a substance, its physical condition, is defined by its
physical properties
...
The state of a pure gas, for example, is specified by giving its
volume, V, amount of substance (number of moles), n, pressure, p, and temperature,
T
...
That is, it is an experimental fact that each substance is described by an equation of state, an equation that
interrelates these four variables
...
1)

This equation tells us that, if we know the values of T, V, and n for a particular substance, then the pressure has a fixed value
...
One very important example is the equation of state of a ‘perfect gas’, which has
the form p = nRT/V, where R is a constant
...


4

1 THE PROPERTIES OF GASES
Table 1
...
325 kPa

torr

1 Torr

(101 325/760) Pa = 133
...
Pa

millimetres of mercury

1 mmHg

133
...
Pa

pound per square inch

1 psi

6
...
kPa

(a) Pressure

Comment 1
...

Movable
wall
High
Low
pressure
pressure

Pressure is defined as force divided by the area to which the force is applied
...
The origin of the force
exerted by a gas is the incessant battering of the molecules on the walls of its container
...

The SI unit of pressure, the pascal (Pa), is defined as 1 newton per metre-squared:
1 Pa = 1 N m−2

[1
...
2b]

Several other units are still widely used (Table 1
...
013 25 × 105 Pa exactly) and bar (1 bar = 105 Pa)
...


(a)
Motion

Equal pressures

Self-test 1
...
0 kg pressing through the point of a pin of area 1
...
Hint
...

[0
...
7 × 103 atm]

(b)
Low
pressure

High
pressure

(c)

When a region of high pressure is
separated from a region of low pressure by
a movable wall, the wall will be pushed into
one region or the other, as in (a) and (c)
...
The latter
condition is one of mechanical equilibrium
between the two regions
...
1
...
1
...
The pressure of the high-pressure gas will fall as it expands
and that of the low-pressure gas will rise as it is compressed
...
This
condition of equality of pressure on either side of a movable wall (a ‘piston’) is a state
of mechanical equilibrium between the two gases
...

(b) The measurement of pressure

The pressure exerted by the atmosphere is measured with a barometer
...
When the column of mercury is in
mechanical equilibrium with the atmosphere, the pressure at its base is equal to that

1
...
It follows that the height of the mercury column is proportional to the external pressure
...
1 Calculating the pressure exerted by a column of liquid

Derive an equation for the pressure at the base of a column of liquid of mass
density ρ (rho) and height h at the surface of the Earth
...
To calculate F we need to know the mass m of the column of liquid,
which is its mass density, ρ, multiplied by its volume, V: m = ρV
...


1

Answer Let the column have cross-sectional area A; then its volume is Ah and its

mass is m = ρAh
...
3)

Note that the pressure is independent of the shape and cross-sectional area of the
column
...


(a)

Self-test 1
...


Low
temperature

Energy as heat

[p = ρgl cos θ]

The pressure of a sample of gas inside a container is measured by using a pressure
gauge, which is a device with electrical properties that depend on the pressure
...
In a capacitance manometer, the deflection of a diaphragm relative to a fixed electrode is monitored through its effect on the capacitance of the arrangement
...


Equal temperatures

(b)
Low
temperature

High
temperature

(c) Temperature

The concept of temperature springs from the observation that a change in physical
state (for example, a change of volume) can occur when two objects are in contact
with one another, as when a red-hot metal is plunged into water
...
1)
we shall see that the change in state can be interpreted as arising from a flow of energy
as heat from one object to another
...
If
energy flows from A to B when they are in contact, then we say that A has a higher
temperature than B (Fig
...
2)
...
A boundary is diathermic (thermally conducting) if a change of state is
observed when two objects at different temperatures are brought into contact
...


(c)

Energy flows as heat from a region
at a higher temperature to one at a lower
temperature if the two are in contact
through a diathermic wall, as in (a) and
(c)
...
The latter condition corresponds
to the two regions being at thermal
equilibrium
...
1
...


Fig
...
3

metal container has diathermic walls
...
A
vacuum flask is an approximation to an adiabatic container
...
Thermal
equilibrium is established if no change of state occurs when two objects A to B are in
contact through a diathermic boundary
...
Then it
has been found experimentally that A and C will also be in thermal equilibrium when
they are put in contact (Fig
...
3)
...

The Zeroth Law justifies the concept of temperature and the use of a thermometer,
a device for measuring the temperature
...
Then, when A is in contact with B, the mercury column in the latter has a
certain length
...
Moreover, we can use the length of the mercury column as a measure of the temperatures
of A and C
...
This procedure led to the Celsius scale of temperature
...
However,
because different liquids expand to different extents, and do not always expand
uniformly over a given range, thermometers constructed from different materials
showed different numerical values of the temperature between their fixed points
...
The perfect-gas scale turns out to be
identical to the thermodynamic temperature scale to be introduced in Section 3
...
On
the thermodynamic temperature scale, temperatures are denoted T and are normally
reported in kelvins, K (not °K)
...
15

(1
...
15, is the current definition of the Celsius
scale in terms of the more fundamental Kelvin scale
...

A note on good practice We write T = 0, not T = 0 K for the zero temperature
on the thermodynamic temperature scale
...
However, we write 0°C because the Celsius scale is not absolute
...
2 THE GAS LAWS

7

Illustration 1
...
00°C as a temperature in kelvins, we use eqn 1
...
00°C)/°C + 273
...
00 + 273
...
15
Increasing
temperature, T

Pressure, p

Note how the units (in this case, °C) are cancelled like numbers
...
00) and a unit (1°C)
...
15 K
...
00°C)/°C = 25
...
Units may be multiplied and cancelled just like numbers
...
2 The gas laws
The equation of state of a gas at low pressure was established by combining a series of
empirical laws
...
Each curve is a hyperbola
(pV = constant) and is called an isotherm
...
1
...
5)°

Charles’s law: V = constant × T, at constant n, p

(1
...
5 mol CO2(g) varies
with volume as it is compressed at
(a) 273 K, (b) 373 K from 30 dm3 to
15 dm3
...
6b)°

Avogadro’s principle: V = constant × n at constant p, T
2

(1
...
Equations valid in this limiting sense will be
signalled by a ° on the equation number, as in these expressions
...
In this form, it is increasingly
true as p → 0
...

Figure 1
...
Each of the curves in the graph corresponds to a single temperature and
hence is called an isotherm
...
An alternative depiction, a plot of pressure against 1/volume, is shown in
Fig
...
5
...
1
...
The lines in this illustration are examples of isobars, or
lines showing the variation of properties at constant pressure
...
7 illustrates the
linear variation of pressure with temperature
...

2
Avogadro’s principle is a principle rather than a law (a summary of experience) because it depends on
the validity of a model, in this case the existence of molecules
...

3
To solve this and other Explorations, use either mathematical software or the Living graphs from the
text’s web site
...
2

A hyperbola is a curve obtained by
plotting y against x with xy = constant
...


Fig
...
5

Exploration Repeat Exploration 1
...


Extrapolation

Extrapolation
0

1/V

Decreasing
volume, V

0
0

Temperature, T

The variation of the volume of a
fixed amount of gas with the temperature
at constant pressure
...


Fig
...
6

Exploration Explore how the volume
of 1
...
00 bar, (b) 0
...


0

Temperature, T

The pressure also varies linearly
with the temperature at constant volume,
and extrapolates to zero at T = 0 (−273°C)
...
1
...
5 mol CO2(g) in a container of
volume (a) 30 dm3, (b) 15 dm3 varies with
temperature as it is cooled from 373 K to
273 K
...


The empirical observations summarized by eqns 1
...
The
constant of proportionality, which is found experimentally to be the same for all
gases, is denoted R and called the gas constant
...
8)°

is the perfect gas equation
...
A gas that obeys
eqn 1
...
A real gas,
an actual gas, behaves more like a perfect gas the lower the pressure, and is described
exactly by eqn 1
...
The gas constant R can be determined by
evaluating R = pV/nT for a gas in the limit of zero pressure (to guarantee that it is

1
...
However, a more accurate value can be obtained by measuring
the speed of sound in a low-pressure gas (argon is used in practice) and extrapolating
its value to zero pressure
...
2 lists the values of R in a variety of units
...
2 The gas constant
R
J K−1 mol−1

8
...
1 The kinetic model of gases

8
...
As a result, the average force exerted on
the walls is doubled
...
Boyle’s law applies to all gases regardless of their
chemical identity (provided the pressure is low) because at low pressures the average separation of molecules is so great that they exert no influence on one another
and hence travel independently
...
The molecules collide with the walls more frequently and with greater
impact
...

These qualitative concepts are expressed quantitatively in terms of the kinetic
model of gases, which is described more fully in Chapter 21
...
314 47 × 10

−2

dm3 bar K−1 mol−1

8
...
364

dm3 Torr K−1 mol−1

1
...
The gas consists of molecules of mass m in ceaseless random motion
...
The size of the molecules is negligible, in the sense that their diameters are
much smaller than the average distance travelled between collisions
...
The molecules interact only through brief, infrequent, and elastic collisions
...
From the very economical assumptions of the kinetic
model, it can be deduced (as we shall show in detail in Chapter 21) that the pressure and volume of the gas are related by
pV = 1 nMc 2
3

(1
...
10)

We see that, if the root mean square speed of the molecules depends only on the
temperature, then at constant temperature
pV = constant
which is the content of Boyle’s law
...
9 to be the equation of
state of a perfect gas, its right-hand side must be equal to nRT
...
11)°

We can conclude that the root mean square speed of the molecules of a gas is proportional to the square root of the temperature and inversely proportional to the square
root of the molar mass
...
The root mean square speed of N2 molecules, for
instance, is found from eqn 1
...


Comment 1
...

2
The potential energy, EP or V, of an
object is the energy arising from its
position (not speed)
...


1 THE PROPERTIES OF GASES
p µ 1/V
isotherm
VµT
isobar

A region of the p,V,T surface of a
fixed amount of perfect gas
...


Fig
...
8

T
Volu
me,
V

Tem
pera
ture
,

Tem
pera
ture
,T

Volu
me,
V

Pressure, p

pµT
isochore

Surface
of possible
states

Pressure, p

10

Sections through the surface shown
in Fig
...
8 at constant temperature give the
isotherms shown in Fig
...
4 and the isobars
shown in Fig
...
6
...
1
...
1
...
8
...
The graphs in Figs
...
4 and 1
...
1
...

Example 1
...

If it enters the vessel at 100 atm and 300 K, what pressure would it exert at the
working temperature if it behaved as a perfect gas?
Method We expect the pressure to be greater on account of the increase in tem-

perature
...
12)°

The known and unknown data are summarized in (2)
...
2 THE GAS LAWS
Substitution of the data then gives
p2 =

500 K
× (100 atm) = 167 atm
300 K

Experiment shows that the pressure is actually 183 atm under these conditions, so
the assumption that the gas is perfect leads to a 10 per cent error
...
3 What temperature would result in the same sample exerting a pressure

of 300 atm?

[900 K]

The perfect gas equation is of the greatest importance in physical chemistry because
it is used to derive a wide range of relations that are used throughout thermodynamics
...
For instance, the molar volume, Vm = V/n, of a perfect gas under the conditions called standard ambient temperature and pressure
(SATP), which means 298
...
789 dm3 mol−1
...
414 dm3 mol−1
...
8 can be used to discuss processes
in the atmosphere that give rise to the weather
...
1 The gas laws and the weather

The biggest sample of gas readily accessible to us is the atmosphere, a mixture of gases
with the composition summarized in Table 1
...
The composition is maintained moderately constant by diffusion and convection (winds, particularly the local turbulence
called eddies) but the pressure and temperature vary with altitude and with the local
conditions, particularly in the troposphere (the ‘sphere of change’), the layer extending up to about 11 km
...
3 The composition of dry air at sea level
Percentage
Component

By volume

By mass

Nitrogen, N2

78
...
53

Oxygen, O2

20
...
14

Argon, Ar

0
...
28

Carbon dioxide, CO2

0
...
047

Hydrogen, H2

5
...
0 × 10 −4

Neon, Ne

1
...
3 × 10 −3

5
...
2 × 10 −5

2
...
1 × 10 −4

1
...
2 × 10 −4

Nitric oxide, NO

5
...
7 × 10 −6

Xenon, Xe

8
...
2 × 10 −5

7
...
2 × 10 −5

2
...
3 × 10 −6

Helium, He
Methane, CH4
Krypton, Kr

Ozone, O3: summer
winter

11

12

1 THE PROPERTIES OF GASES
In the troposphere the average temperature is 15°C at sea level, falling to –57°C at
the bottom of the tropopause at 11 km
...
If we
suppose that the temperature has its average value all the way up to the tropopause,
then the pressure varies with altitude, h, according to the barometric formula:

30

Altitude, h/km

20
15

p = p0e−h/H
10
6
0
0

Pressure, p

p0

Fig
...
10 The variation of atmospheric
pressure with altitude, as predicted by the
barometric formula and as suggested by the
‘US Standard Atmosphere’, which takes
into account the variation of temperature
with altitude
...


H
H
H
H

L

Fig
...
11 A typical weather map; in this case,
for the United States on 1 January 2000
...

More specifically, H = RT/Mg, where M is the average molar mass of air and T is the
temperature
...
1
...
It implies that the
pressure of the air and its density fall to half their sea-level value at h = H ln 2, or 6 km
...
A small region of air is termed a parcel
...
As a parcel rises, it
expands adiabatically (that is, without transfer of heat from its surroundings), so it
cools
...
Cloudy skies can therefore be associated with rising air and
clear skies are often associated with descending air
...
The former result in the formation
of regions of high pressure (‘highs’ or anticyclones) and the latter result in regions of
low pressure (‘lows’, depressions, or cyclones)
...
1
...
The lines of constant pressure—differing
by 4 mbar (400 Pa, about 3 Torr)—marked on it are called isobars
...

In meteorology, large-scale vertical movement is called convection
...
1
...
Winds
coming from the north in the Northern hemisphere and from the south in the
Southern hemisphere are deflected towards the west as they migrate from a region
where the Earth is rotating slowly (at the poles) to where it is rotating most rapidly (at
the equator)
...
At the
surface, where wind speeds are lower, the winds tend to travel perpendicular to the
isobars from high to low pressure
...

The air lost from regions of high pressure is restored as an influx of air converges
into the region and descends
...
It also becomes warmer by compression as it descends, so regions of high pressure are associated with high surface temperatures
...
Geographical conditions may also trap cool air, as
in Los Angeles, and the photochemical pollutants we know as smog may be trapped
under the warm layer
...
1
...


When dealing with gaseous mixtures, we often need to know the contribution that
each component makes to the total pressure of the sample
...
13]

1
...
14]

When no J molecules are present, xJ = 0; when only J molecules are present, xJ = 1
...
15)

This relation is true for both real and perfect gases
...
13 is also
the pressure that each gas would occupy if it occupied the same container alone at
the same temperature
...
That
identification was the basis of the original formulation of Dalton’s law:
The pressure exerted by a mixture of gases is the sum of the pressures that each one
would exist if it occupied the container alone
...
13) and total
pressure (as given by eqn 1
...

Example 1
...
5;
O2: 23
...
3
...
00 atm?
Method We expect species with a high mole fraction to have a proportionally high

partial pressure
...
13
...
To calculate mole fractions, which are
defined by eqn 1
...
The mole fractions are independent of the
total mass of the sample, so we can choose the latter to be 100 g (which makes
the conversion from mass percentages very easy)
...
5 per cent of 100 g, which is 75
...

Answer The amounts of each type of molecule present in 100 g of air, in which the

masses of N2, O2, and Ar are 75
...
2 g, and 1
...
5 g
75
...
02
28
...
2 g
23
...
00 g mol −1 32
...
3 g
1
...
95 g mol −1 39
...
69 mol, 0
...
033 mol, respectively,
for a total of 3
...
The mole fractions are obtained by dividing each of the

13

14

1 THE PROPERTIES OF GASES
above amounts by 3
...
00 atm):
Mole fraction:
Partial pressure/atm:

N2
0
...
780

O2
0
...
210

Ar
0
...
0096

We have not had to assume that the gases are perfect: partial pressures are defined
as pJ = xJ p for any kind of gas
...
4 When carbon dioxide is taken into account, the mass percentages are
75
...
15 (O2), 1
...
046 (CO2)
...
900 atm?
[0
...
189, 0
...
00027 atm]

Real gases
Real gases do not obey the perfect gas law exactly
...


0

Repulsions dominant

Potential energy

Contact

1
...
1
...
High positive potential energy
(at very small separations) indicates that
the interactions between them are strongly
repulsive at these distances
...
At large separations
(on the right) the potential energy is zero
and there is no interaction between the
molecules
...
Repulsive forces between molecules assist expansion and attractive forces
assist compression
...
1
...

Because they are short-range interactions, repulsions can be expected to be important
only when the average separation of the molecules is small
...
On the other hand, attractive
intermolecular forces have a relatively long range and are effective over several molecular diameters
...
1
...
Attractive forces
are ineffective when the molecules are far apart (well to the right in Fig
...
13)
...

At low pressures, when the sample occupies a large volume, the molecules are so far
apart for most of the time that the intermolecular forces play no significant role, and
the gas behaves virtually perfectly
...
In this case, the gas can be expected to be more compressible than
a perfect gas because the forces help to draw the molecules together
...

(a) The compression factor

The compression factor, Z, of a gas is the ratio of its measured molar volume, Vm =
o
V/n, to the molar volume of a perfect gas, Vm, at the same pressure and temperature:

1
...
16]

Because the molar volume of a perfect gas is equal to RT/p, an equivalent expression
o
is Z = RT/pV m, which we can write as
pVm = RTZ

(1
...

Some experimental values of Z are plotted in Fig
...
14
...
At high pressures, all the
gases have Z > 1, signifying that they have a larger molar volume than a perfect gas
...
At intermediate pressures, most gases have Z < 1,
indicating that the attractive forces are reducing the molar volume relative to that of a
perfect gas
...
15 shows the experimental isotherms for carbon dioxide
...
The small differences suggest that the perfect gas law is in fact
the first term in an expression of the form
pVm = RT(1 + B′p + C′p2 + · · · )

(1
...
04°C (Tc)

*

20°C

60

CH4

E D C
B

1
...
98
NH3
0
0

40°C

p/atm

Compression factor, Z

H2

0
...
1
...
A perfect gas has Z = 1 at all pressures
...


0

0
...
4
Vm /(dm3 mol -1)

0
...
1
...
The
‘critical isotherm’, the isotherm at the
critical temperature, is at 31
...
The
critical point is marked with a star
...
A more convenient expansion for many applications is

Comment 1
...


⎛
⎞
B
C
pVm = RT ⎜1 +
+ 2 + ⋅ ⋅ ⋅⎟
Vm V m
⎝
⎠

Synoptic Table 1
...
7

11
...
7

−12
...
5

21
...
7

−19
...
19)

These two expressions are two versions of the virial equation of state
...
17 we see that the term in parentheses can be identified with
the compression factor, Z
...
, which depend on the temperature, are the second, third,

...
4); the first virial coefficient is 1
...

<
We can use the virial equation to demonstrate the important point that, although
the equation of state of a real gas may coincide with the perfect gas law as p → 0, not
all its properties necessarily coincide with those of a perfect gas in that limit
...
For a perfect gas dZ/dp = 0 (because Z = 1 at all pressures), but for a real gas
from eqn 1
...


as

p→0

(1
...
1
...
Because several
physical properties of gases depend on derivatives, the properties of real gases do not
always coincide with the perfect gas values at low pressures
...
20b)

Because the virial coefficients depend on the temperature, there may be a temperature at which Z → 1 with zero slope at low pressure or high molar volume (Fig
...
16)
...
According to eqn 1
...

It then follows from eqn 1
...
For helium
TB = 22
...
8 K; more values are given in Table 1
...


Perfect
gas
Lower
temperature
0

Synoptic Table 1
...
0

75
...
7

0
...
5

CO2

72
...
8

Pressure, p

The compression factor, Z,
approaches 1 at low pressures, but does so
with different slopes
...
At the Boyle
temperature, the slope is zero and the gas
behaves perfectly over a wider range of
conditions than at other temperatures
...
1
...
0

304
...
274

He

2
...
8

5
...
305

O2

50
...
0

154
...
308

22
...
9

* More values are given in the Data section
...
The coefficients are sometimes denoted B2, B3,
...
4 THE VAN DER WAALS EQUATION
(c) Condensation

Now consider what happens when we compress a sample of gas initially in the state
marked A in Fig
...
15 at constant temperature (by pushing in a piston)
...
Serious deviations from that law begin to appear when the volume has been reduced to B
...
Examination of the contents
of the vessel shows that just to the left of C a liquid appears, and there are two phases
separated by a sharply defined surface
...
There is no additional resistance to the piston
because the gas can respond by condensing
...

At E, the sample is entirely liquid and the piston rests on its surface
...
Even a small reduction of volume from E to F
requires a great increase in pressure
...
19 K, or 31
...
An isotherm slightly below Tc behaves as we have
already described: at a certain pressure, a liquid condenses from the gas and is distinguishable from it by the presence of a visible surface
...
The temperature, pressure, and molar volume
at the critical point are called the critical temperature, Tc, critical pressure, pc, and
critical molar volume, Vc, of the substance
...
5)
...
Such a phase is, by definition, a gas
...
The critical temperature of
oxygen, for instance, signifies that it is impossible to produce liquid oxygen by compression alone if its temperature is greater than 155 K: to liquefy oxygen—to obtain a
fluid phase that does not occupy the entire volume—the temperature must first be
lowered to below 155 K, and then the gas compressed isothermally
...

1
...
It is often useful to have a broader, if less precise, view of all
gases
...
D
...
This equation is an excellent example of an expression that can
be obtained by thinking scientifically about a mathematically complicated but physically simple problem, that is, it is a good example of ‘model building’
...
21a)

Comment 1
...


17

18

1 THE PROPERTIES OF GASES

Synoptic Table 1
...
337

3
...
610

4
...
0341
4
...
16

p=

RT
a
− 2
Vm − b V m

(1
...
They are characteristic of each gas but independent of the temperature (Table 1
...


2
...
1
...


Justification 1
...
The nonzero volume of the molecules implies that instead of moving in a volume V they are
restricted to a smaller volume V − nb, where nb is approximately the total volume
taken up by the molecules themselves
...
The closest distance of two hard-sphere molecules
4
4
of radius r, and volume Vmolecule = 3 πr 3, is 2r, so the volume excluded is 3 π(2r)3, or
8Vmolecule
...

The pressure depends on both the frequency of collisions with the walls and
the force of each collision
...
Therefore, because both
the frequency and the force of the collisions are reduced by the attractive forces,
the pressure is reduced in proportion to the square of this concentration
...
21
...
The equation can be
derived in other ways, but the present method has the advantage that it shows how
to derive the form of an equation out of general ideas
...


Example 1
...

Method To express eqn 1
...
4 THE VAN DER WAALS EQUATION

19

Although closed expressions for the roots of a cubic equation can be given, they
are very complicated
...

Answer According to Table 1
...
592 dm6 atm mol−2 and b = 4
...
Under the stated conditions, RT/p = 0
...
The coefficients
in the equation for Vm are therefore
b + RT/p = 0
...
61 × 10−2 (dm3 mol−1)2
ab/p = 1
...
453x 2 + (3
...
55 × 10−3) = 0
The acceptable root is x = 0
...
366 dm3 mol−1
...
410 dm3 mol−1
...
5 Calculate the molar volume of argon at 100°C and 100 atm on the

[0
...


(a) The reliability of the equation

We now examine to what extent the van der Waals equation predicts the behaviour
of real gases
...
The advantage of the van der Waals equation, however, is
that it is analytical (that is, expressed symbolically) and allows us to draw some general conclusions about real gases
...
7), invent
a new one, or go back to the virial equation
...
1
...
Some
Table 1
...
5
...
5
1
...
5

1
...
8

Volum
e

1

1
...
5

0
...

Compare this surface with that shown in
Fig
...
8
...
1
...
1

1
Reduced volume, V/Vc

10

Van der Waals isotherms at several values of T/Tc
...
1
...
The van der Waals loops are normally replaced by horizontal straight lines
...


Fig
...
18

Exploration Calculate the molar volume of chlorine gas on the basis of the van der Waals
equation of state at 250 K and 150 kPa and calculate the percentage difference from the
value predicted by the perfect gas equation
...
1
...
18
...

The oscillations, the van der Waals loops, are unrealistic because they suggest that
under some conditions an increase of pressure results in an increase of volume
...
The
van der Waals coefficients, such as those in Table 1
...

(b) The features of the equation

The principal features of the van der Waals equation can be summarized as follows
...

When the temperature is high, RT may be so large that the first term in eqn 1
...
Furthermore, if the molar volume is large in the sense
Vm > b, then the denominator Vm − b ≈ Vm
...

(2) Liquids and gases coexist when cohesive and dispersing effects are in balance
...
21b have similar magnitudes
...

(3) The critical constants are related to the van der Waals coefficients
...
5 THE PRINCIPLE OF CORRESPONDING STATES
For T < Tc, the calculated isotherms oscillate, and each one passes through a minimum followed by a maximum
...
From the properties of
curves, we know that an inflexion of this type occurs when both the first and second
derivatives are zero
...
The solutions of these two equations (and using eqn 1
...
22)

These relations provide an alternative route to the determination of a and b from the
values of the critical constants
...
23)

for all gases
...
5 that, although Zc < 3 = 0
...
3) and the discrepancy is reasonably small
...
5 The principle of corresponding states
An important general technique in science for comparing the properties of objects is
to choose a related fundamental property of the same kind and to set up a relative
scale on that basis
...
We
therefore introduce the dimensionless reduced variables of a gas by dividing the
actual variable by the corresponding critical constant:
pr =

p
pc

Vr =

Vm
Vc

Tr =

T
Tc

[1
...
Van der Waals, who first
tried this procedure, hoped that gases confined to the same reduced volume, Vr, at the
same reduced temperature, Tr, would exert the same reduced pressure, pr
...
1
...
The illustration shows the dependence of the compression factor on the reduced pressure for a variety of gases at various reduced temperatures
...
1
...
The observation that real gases at the same reduced volume and reduced temperature exert the
same reduced pressure is called the principle of corresponding states
...
It works best for gases composed of spherical molecules;
it fails, sometimes badly, when the molecules are non-spherical or polar
...
First, we express
eqn 1
...
0

2
...
8

Compression factor, Z

22

1
...
6
1
...
4
Nitrogen
Methane
Propane

0
...
1
...
1
...
The curves are labelled with the reduced temperature Tr = T/Tc
...


Exploration Is there a set of conditions at which the compression factor of a van der
Waals gas passes through a minimum? If so, how does the location and value of the
minimum value of Z depend on the coefficients a and b?

Then we express the critical constants in terms of a and b by using eqn 1
...
25)

This equation has the same form as the original, but the coefficients a and b, which
differ from gas to gas, have disappeared
...
1
...
This is precisely the
content of the principle of corresponding states, so the van der Waals equation is
compatible with it
...
7)
...
The observation that real gases obey the principle
approximately amounts to saying that the effects of the attractive and repulsive interactions can each be approximated in terms of a single parameter
...
1
...
1
...


DISCUSSION QUESTIONS

23

Checklist of key ideas
1
...

2
...

3
...
The standard pressure is p7 = 1 bar (105 Pa)
...
Mechanical equilibrium is the condition of equality of
pressure on either side of a movable wall
...
Temperature is the property that indicates the direction of the
flow of energy through a thermally conducting, rigid wall
...
A diathermic boundary is a boundary that permits the passage
of energy as heat
...

7
...

8
...

9
...
15
...
A perfect gas obeys the perfect gas equation, pV = nRT, exactly
under all conditions
...
Dalton’s law states that the pressure exerted by a mixture of
gases is the sum of the partial pressures of the gases
...
The partial pressure of any gas is defined as pJ = xJ p, where
xJ = nJ/n is its mole fraction in a mixture and p is the total
pressure
...
In real gases, molecular interactions affect the equation of
state; the true equation of state is expressed in terms of virial
2
coefficients B, C,
...

14
...

15
...
The critical constants pc, Vc, and Tc are the
pressure, molar volume, and temperature, respectively, at the
critical point
...
A supercritical fluid is a dense fluid phase above its critical
temperature and pressure
...
The van der Waals equation of state is an approximation to
the true equation of state in which attractions are represented
by a parameter a and repulsions are represented by a
parameter b: p = nRT/(V − nb) − a(n/V)2
...
A reduced variable is the actual variable divided by the
corresponding critical constant
...
According to the principle of corresponding states, real gases
at the same reduced volume and reduced temperature exert
the same reduced pressure
...
L
...
H
...
J
...
Educ
...

M
...
In Encyclopedia of applied physics
(ed
...
L
...
VCH, New York (1993)
...
J
...
Oxford University Press
(1983)
...
P
...

Oxford University Press (2000)
...
H
...
B
...
Oxford University Press (1980)
...
D
...
Wilkinson, Compendium of chemical
terminology
...


Discussion questions
1
...

1
...

1
...


1
...
5 Describe the formulation of the van der Waals equation and suggest a

rationale for one other equation of state in Table 1
...

1
...


24

1 THE PROPERTIES OF GASES

Exercises
1
...
0 dm3 exert a
pressure of 20 atm at 25°C if it behaved as a perfect gas? If not, what pressure
would it exert? (b) What pressure would it exert if it behaved as a van der
Waals gas?

1
...
15 K
...
1(b) (a) Could 25 g of argon gas in a vessel of volume 1
...
9649

ρ/(g dm−3)

1
...
0 bar at 30°C if it behaved as a perfect gas? If not, what pressure
would it exert? (b) What pressure would it exert if it behaved as a van der
Waals gas?
1
...
20 dm3
...
04 bar
and 4
...
Calculate the original pressure of the gas in (a) bar,
(b) atm
...

p/atm

0
...
500 000
44
...
714110

0
...
6384
0
...
8(a) At 500°C and 93
...
710 kg m−3
...
8(b) At 100°C and 1
...
6388 kg m−3
...
9(a) Calculate the mass of water vapour present in a room of volume 400 m3

1
...
80 dm3
...
97 bar and
2
...
Calculate the original pressure of the gas in (a) bar,
(b) Torr
...
3(a) A car tyre (i
...
an automobile tire) was inflated to a pressure of 24 lb in

(1
...
7 lb in−2) on a winter’s day when the temperature was –5°C
...
3(b) A sample of hydrogen gas was found to have a pressure of 125 kPa

when the temperature was 23°C
...

1
...

1
...
987 bar and 27°C is 1
...
0 mole per cent Ar
...
10(b) A gas mixture consists of 320 mg of methane, 175 mg of argon, and
225 mg of neon
...
87 kPa
...

1
...
23 kg m−3 at

1
...
00 dm at 122 K
...
What is the molar mass of the compound?

gas law to calculate the pressure of the gas
...
11(b) In an experiment to measure the molar mass of a gas, 250 cm3 of the

1
...
00 × 103 m3 of natural gas in a year to heat a

gas was confined in a glass vessel
...
5 mg
...
Assume that natural gas is all methane, CH4, and that methane is a
perfect gas for the conditions of this problem, which are 1
...

What is the mass of gas used?
1
...
0 m3 when on the deck of a boat
...
025 g cm−3 and assume
that the temperature is the same as on the surface
...
5(b) What pressure difference must be generated across the length of a

15 cm vertical drinking straw in order to drink a water-like liquid of density
1
...
6(a) A manometer consists of a U-shaped tube containing a liquid
...
The
pressure inside the apparatus is then determined from the difference in
heights of the liquid
...
0 cm lower than the side connected to the
apparatus
...
997 07 g cm−3
...
6(b) A manometer like that described in Exercise 1
...
Suppose the external pressure is 760 Torr, and the open side is
10
...
What is the pressure
in the apparatus? (The density of mercury at 25°C is 13
...
)
1
...
000 dm filled with 0
...
402 cm of water in a
manometer at 25°C
...
(The density of
water at 25°C is 0
...
6a
...
12(a) The densities of air at −85°C, 0°C, and 100°C are 1
...
294 g dm−3, and 0
...
From these data, and assuming
that air obeys Charles’s law, determine a value for the absolute zero of
temperature in degrees Celsius
...
12(b) A certain sample of a gas has a volume of 20
...
000 atm
...
0741 dm3 (°C)−1
...

1
...
0 mol C2H6 behaving as (a) a
perfect gas, (b) a van der Waals gas when it is confined under the following
conditions: (i) at 273
...
414 dm3, (ii) at 1000 K in 100 cm3
...
6
...
13(b) Calculate the pressure exerted by 1
...
15 K in 22
...
Use the data
in Table 1
...

1
...
751 atm dm6 mol−2 and

b = 0
...


1
...
32 atm dm6 mol−2 and

b = 0
...


1
...
Calculate (a) the compression
factor under these conditions and (b) the molar volume of the gas
...
15(b) A gas at 350 K and 12 atm has a molar volume 12 per cent larger than
that calculated from the perfect gas law
...
Which are
dominating in the sample, the attractive or the repulsive forces?

1
...
6 atm,
Vc = 98
...
6 K
...


1
...
4 K
...


volume of 1
...
The gas enters the container at 300 K and 100 atm
...
4 kg
...

For nitrogen, a = 1
...
0387 dm3 mol−1
...
16(b) Cylinders of compressed gas are typically filled to a pressure of
200 bar
...

For oxygen, a = 1
...
19 × 10−2 dm3 mol−1
...
17(a) Suppose that 10
...
860 dm3 at 27°C
...
Calculate the compression factor
based on these calculations
...
507 dm6 atm mol−2,
b = 0
...

1
...
86
...
2 mmol of the gas under these conditions and
(b) an approximate value of the second virial coefficient B at 300 K
...
18(a) A vessel of volume 22
...
0 mol H2 and 1
...
15 K
...

1
...
4 dm3 contains 1
...
5 mol N2 at

273
...
Calculate (a) the mole fractions of each component, (b) their partial
pressures, and (c) their total pressure
...
19(b) The critical constants of ethane are pc = 48
...
20(a) Use the van der Waals parameters for chlorine to calculate

approximate values of (a) the Boyle temperature of chlorine and (b) the radius
of a Cl2 molecule regarded as a sphere
...
20(b) Use the van der Waals parameters for hydrogen sulfide to calculate
approximate values of (a) the Boyle temperature of the gas and (b) the
radius of a H2S molecule regarded as a sphere (a = 4
...
0434 dm3 mol−1)
...
21(a) Suggest the pressure and temperature at which 1
...
0 mol H2 at 1
...

1
...
0 mol of (a) H2S,
(b) CO2, (c) Ar will be in states that correspond to 1
...
0 atm and
25°C
...
22(a) A certain gas obeys the van der Waals equation with a = 0
...

Its volume is found to be 5
...
0 MPa
...
What is the
compression factor for this gas at the prevailing temperature and pressure?
1
...
76 m6 Pa mol−2
...
00 × 10−4 m3 mol−1 at 288 K and 4
...
From
this information calculate the van der Waals constant b
...
4 The molar mass of a newly synthesized fluorocarbon was measured in a

1
...
Further communications have revealed that the Neptunians know
about perfect gas behaviour and they find that, in the limit of zero pressure,
the value of pV is 28 dm3 atm at 0°N and 40 dm3 atm at 100°N
...
2 Deduce the relation between the pressure and mass density, ρ, of a perfect
gas of molar mass M
...


p/kPa

12
...
20

36
...
37

85
...
3

ρ/(kg m−3)

0
...
456

0
...
062

1
...
734

1
...
The following values for α have been reported for nitrogen at 0°C:
p/Torr

749
...
6

333
...
6

103α /(°C)−1

3
...
6697

3
...
6643

For these data calculate the best value for the absolute zero of temperature on
the Celsius scale
...
This device consists of a glass bulb forming one end of a
beam, the whole surrounded by a closed container
...
In one experiment, the balance
point was reached when the fluorocarbon pressure was 327
...
014 g mol−1)
was introduced at 423
...
A repeat of the experiment with a different
setting of the pivot required a pressure of 293
...
22 Torr of the CHF3
...

1
...
69

kPa at the triple point temperature of water (273
...
(a) What change of
pressure indicates a change of 1
...
00°C? (c) What change of pressure indicates a
change of 1
...
6 A vessel of volume 22
...
0 mol H2 and 1
...
15 K initially
...
Calculate
the partial pressures and the total pressure of the final mixture
...
7 Calculate the molar volume of chlorine gas at 350 K and 2
...
Use the answer to
(a) to calculate a first approximation to the correction term for attraction and
then use successive approximations to obtain a numerical answer for part (b)
...


26

1 THE PROPERTIES OF GASES

1
...
7 cm3 mol−1 and
C = 1200 cm6 mol−2, where B and C are the second and third virial coefficients
in the expansion of Z in powers of 1/Vm
...
From your result, estimate the molar volume of argon under these
conditions
...
9 Calculate the volume occupied by 1
...
Assume that
the pressure is 10 atm throughout
...
3 K, a = 1
...
0387 dm3 mol−1
...
10‡ The second virial coefficient of methane can be approximated
2
by the empirical equation B′(T) = a + be−c/T , where a = −0
...
2002 bar , and c = 1131 K with 300 K < T < 600 K
...
11 The mass density of water vapour at 327
...
4 K is 133
...

Given that for water Tc = 647
...
3 atm, a = 5
...
03049 dm3 mol−1, and M = 18
...
Then calculate the compression factor (b) from the data,
(c) from the virial expansion of the van der Waals equation
...
12 The critical volume and critical pressure of a certain gas are
160 cm3 mol−1 and 40 atm, respectively
...
Estimate the radii
of the gas molecules on the assumption that they are spheres
...
13 Estimate the coefficients a and b in the Dieterici equation of state from

the critical constants of xenon
...
0 mol Xe
when it is confined to 1
...


Theoretical problems
1
...

1
...
The expansion you will need is (1 − x)−1 = 1 + x + x2 + · · ·
...
7 cm3 mol−1 and C = 1200 cm6 mol−2
for the virial coefficients at 273 K
...
16‡ Derive the relation between the critical constants and the Dieterici
equation parameters
...
Compare the van der Waals and Dieterici
predictions of the critical compression factor
...
17 A scientist proposed the following equation of state:

p=

RT
Vm

−

B
2
Vm

+

C
3
Vm

Show that the equation leads to critical behaviour
...

1
...
18 and 1
...
Find

the relation between B, C and B′, C′
...
19 The second virial coefficient B′ can be obtained from measurements of

the density ρ of a gas at a series of pressures
...
Use the data
on dimethyl ether in Problem 1
...


1
...
Find (∂V/∂T)p
...
21 The following equations of state are occasionally used for approximate
calculations on gases: (gas A) pVm = RT(1 + b/Vm), (gas B) p(Vm – b) = RT
...

1
...
If the pressure
and temperature are such that Vm = 10b, what is the numerical value of the
compression factor?
1
...
Rayleigh
prepared some samples of nitrogen by chemical reaction of nitrogencontaining compounds; under his standard conditions, a glass globe filled
with this ‘chemical nitrogen’ had a mass of 2
...
He prepared other
samples by removing oxygen, carbon dioxide, and water vapour from
atmospheric air; under the same conditions, this ‘atmospheric nitrogen’ had a
mass of 2
...

With the hindsight of knowing accurate values for the molar masses of
nitrogen and argon, compute the mole fraction of argon in the latter sample
on the assumption that the former was pure nitrogen and the latter a mixture
of nitrogen and argon
...
24‡ A substance as elementary and well known as argon still receives
research attention
...
B
...
T
...
Phys
...
Ref
...


p/MPa

0
...
5000

0
...
8000

1
...
2208

4
...
1423

3
...
4795

p/MPa

1
...
000

2
...
000

4
...
6483

1
...
98357

0
...
60998

3

−1

Vm/(dm mol )

(a) Compute the second virial coefficient, B, at this temperature
...


Applications: to environmental science
1
...

Not all pollution, however, is from industrial sources
...
The Kilauea volcano in Hawaii emits
200–300 t of SO2 per day
...
0 atm, what
volume of gas is emitted?
1
...
One Dobson unit is the
thickness, in thousandths of a centimetre, of a column of gas if it were
collected as a pure gas at 1
...
What amount of O3 (in moles) is
found in a column of atmosphere with a cross-sectional area of 1
...
00 dm2
area? Most atmospheric ozone is found between 10 and 50 km above the
surface of the earth
...
27 The barometric formula relates the pressure of a gas of molar mass M at
an altitude h to its pressure p0 at sea level
...
Remember that ρ depends on the pressure
...
0 atm
...
29‡ The preceding problem is most readily solved (see the Solutions

1
...
It is possible to investigate
some of the technicalities of ballooning by using the perfect gas law
...
0 m and that it is spherical
...
0 atm in an ambient temperature of
25°C at sea level? (b) What mass can the balloon lift at sea level, where the
density of air is 1
...
30 ‡ Chlorofluorocarbons such as CCl3F and CCl2F2 have been linked

manual) with the use of the Archimedes principle, which states that the lifting
force is equal to the difference between the weight of the displaced air and the
weight of the balloon
...
Hint
...

to ozone depletion in Antarctica
...
Compute the molar
concentration of these gases under conditions typical of (a) the mid-latitude
troposphere (10°C and 1
...
050 atm)
...
1 Work, heat, and energy
2
...
3 Expansion work
2
...
5 Enthalpy
I2
...
6 Adiabatic changes

The First Law
This chapter introduces some of the basic concepts of thermodynamics
...

Much of this chapter examines the means by which a system can exchange energy with its
surroundings in terms of the work it may do or the heat that it may produce
...
We also begin to unfold some of the power of thermodynamics by
showing how to establish relations between different properties of a system
...
The relations we derive also enable
us to discuss the liquefaction of gases and to establish the relation between the heat
capacities of a substance under different conditions
...
7 Standard enthalpy changes
I2
...
8 Standard enthalpies of

formation
2
...
In chemistry, we
encounter reactions that can be harnessed to provide heat and work, reactions that
liberate energy which is squandered (often to the detriment of the environment) but
which give products we require, and reactions that constitute the processes of life
...


State functions and exact
differentials
2
...
11 Changes in internal energy
2
...
1: Adiabatic
processes
Further information 2
...
The system is the part of the world in which we have a special interest
...
The surroundings comprise the region outside the system and are
where we make our measurements
...
2
...
If matter can be
transferred through the boundary between the system and its surroundings the system is classified as open
...
Both open and closed systems can exchange energy with their surroundings
...


2
...

2
...
Doing work is equivalent to raising a weight somewhere
in the surroundings
...
A chemical reaction that drives an electric current
through a resistance also does work, because the same current could be driven
through a motor and used to raise a weight
...
When work is done on an otherwise isolated system (for instance, by compressing a gas or winding a spring), the capacity of the system to do work is increased; in other words, the energy of the system
is increased
...

Experiments have shown that the energy of a system may be changed by means
other than work itself
...
When a heater is immersed in a beaker of water (the system), the
capacity of the system to do work increases because hot water can be used to do more
work than the same amount of cold water
...

An exothermic process is a process that releases energy as heat into its surroundings
...
An endothermic process is a process in which energy is acquired from its surroundings as heat
...
To avoid a lot of awkward circumlocution, we say that in an exothermic process energy is transferred ‘as heat’ to the
surroundings and in an endothermic process energy is transferred ‘as heat’ from
the surroundings into the system
...

An endothermic process in a diathermic container results in energy flowing into the
system as heat
...
When an endothermic process takes
place in an adiabatic container, it results in a lowering of temperature of the system;
an exothermic process results in a rise of temperature
...
2
...

Molecular interpretation 2
...
The disorderly motion of molecules is called thermal motion
...
When a system heats its surroundings, molecules of
the system stimulate the thermal motion of the molecules in the surroundings
(Fig
...
3)
...
2
...
When a weight is raised or lowered, its atoms move in an organized way
(up or down)
...

(b) A closed system can exchange energy
with its surroundings, but it cannot
exchange matter
...


Fig
...
1

29

30

2 THE FIRST LAW

When energy is transferred to the
surroundings as heat, the transfer
stimulates random motion of the atoms in
the surroundings
...


Fig
...
3

(a) When an endothermic process
occurs in an adiabatic system, the
temperature falls; (b) if the process is
exothermic, then the temperature rises
...
(d) If the
process is exothermic, then energy leaves as
heat, and the process is isothermal
...
2
...
For instance, the atoms
shown here may be part of a weight that is
being raised
...


Fig
...
4

electrons in an electric current move in an orderly direction when it flows
...
Likewise, when work is done on a system, molecules in the surroundings are used to transfer energy to it in an organized way, as the atoms in a
weight are lowered or a current of electrons is passed
...
The fact
that a falling weight may stimulate thermal motion in the system is irrelevant to the
distinction between heat and work: work is identified as energy transfer making
use of the organized motion of atoms in the surroundings, and heat is identified as
energy transfer making use of thermal motion in the surroundings
...
Because collisions between molecules
quickly randomize their directions, the orderly motion of the atoms of the weight
is in effect stimulating thermal motion in the gas
...


2
...
The
internal energy is the total kinetic and potential energy of the molecules in the system
(see Comment 1
...
1 We denote by
∆U the change in internal energy when a system changes from an initial state i with
internal energy Ui to a final state f of internal energy Uf :
∆U = Uf − Ui

[2
...


2
...

In other words, it is a function of the properties that determine the current state of
the system
...
The internal energy is an extensive property
...
10
...
The
joule, which is named after the nineteenth-century scientist J
...
Joule, is defined as
1 J = 1 kg m2 s−2
A joule is quite a small unit of energy: for instance, each beat of the human heart consumes about 1 J
...
Certain other energy units are also used, but are more
common in fields other than thermodynamics
...
16 aJ (where 1 aJ = 10−18 J)
...
Thus, the energy to remove an electron from a sodium atom is close to
5 eV
...
The current definition
of the calorie in terms of joules is
1 cal = 4
...

Molecular interpretation 2
...
Many physical and
chemical properties depend on the energy associated with each of these modes of
motion
...

The equipartition theorem of classical mechanics is a useful guide to the average
energy associated with each degree of freedom when the sample is at a temperature
T
...
For example, the kinetic energy an atom of mass m as it moves
through space is
1
1
1
2
2
2
EK = –mv x + –mv y + –mv z
2
2
2

and there are three quadratic contributions to its energy
...
381 × 10−23 J K −1)
...
In practice, it can be used for molecular translation and rotation but not
1
vibration
...

2
According to the equipartition theorem, the average energy of each term in the
1
3
expression above is –kT
...
1

An extensive property is a property that
depends on the amount of substance in
the sample
...
Two
examples of extensive properties are
mass and volume
...


32

2 THE FIRST LAW
3
total energy of the gas (there being no potential energy contribution) is –NkT, or
2
3
– nRT (because N = nNA and R = NAk)
...
(a) A linear molecule can
rotate about two axes perpendicular to the
line of the atoms
...


Fig
...
5

where Um(0) is the molar internal energy at T = 0, when all translational motion
has ceased and the sole contribution to the internal energy arises from the internal
structure of the atoms
...
At 25°C, –RT = 3
...

When the gas consists of polyatomic molecules, we need to take into account the
effect of rotation and vibration
...
2
...

2
Therefore, the mean rotational energy is kT and the rotational contribution to the
molar internal energy is RT
...

2
3
Therefore, the mean rotational energy is –kT and there is a rotational contribution
2
3
of –RT to the molar internal energy of the molecule
...

The internal energy of interacting molecules in condensed phases also has a
contribution from the potential energy of their interaction
...
Nevertheless, the crucial molecular
point is that, as the temperature of a system is raised, the internal energy increases
as the various modes of motion become more highly excited
...
Whereas we may know
how the energy transfer has occurred (because we can see if a weight has been raised
or lowered in the surroundings, indicating transfer of energy by doing work, or if ice
has melted in the surroundings, indicating transfer of energy as heat), the system is
blind to the mode employed
...
A system is like a bank: it accepts deposits in either currency, but stores
its reserves as internal energy
...
This
summary of observations is now known as the First Law of thermodynamics and
expressed as follows:
The internal energy of an isolated system is constant
...

The evidence for this property is that no ‘perpetual motion machine’ (a machine that

2
...

These remarks may be summarized as follows
...
2)

Equation 2
...
The equation states that the change in
internal energy of a closed system is equal to the energy that passes through its boundary as heat or work
...
In other words, we view the flow of energy as work
or heat from the system’s perspective
...
1 The sign convention in thermodynamics

If an electric motor produced 15 kJ of energy each second as mechanical work and
lost 2 kJ as heat to the surroundings, then the change in the internal energy of the
motor each second is
∆U = −2 kJ − 15 kJ = −17 kJ
Suppose that, when a spring was wound, 100 J of work was done on it but 15 J
escaped to the surroundings as heat
...
3 Expansion work
The way can now be opened to powerful methods of calculation by switching attention to infinitesimal changes of state (such as infinitesimal change in temperature)
and infinitesimal changes in the internal energy dU
...
2 we have
dU = dq + dw

(2
...

We begin by discussing expansion work, the work arising from a change in volume
...
Many chemical reactions result in the generation or consumption of
gases (for instance, the thermal decomposition of calcium carbonate or the combustion of octane), and the thermodynamic characteristics of a reaction depend on the
work it can do
...

(a) The general expression for work

The calculation of expansion work starts from the definition used in physics, which
states that the work required to move an object a distance dz against an opposing force
of magnitude F is
dw = −Fdz

[2
...
Now consider the
arrangement shown in Fig
...
6, in which one wall of a system is a massless, frictionless,
rigid, perfectly fitting piston of area A
...
When the system expands
through a distance dz against an external pressure pex, it follows that the work done is
dw = −pex Adz
...

Therefore, the work done when the system expands by dV against a pressure pex is
dw = −pexdV

(2
...
The external pressure pex
is equivalent to a weight pressing on the
piston, and the force opposing expansion is
F = pex A
...
2
...
6)

Vi

The force acting on the piston, pex A, is equivalent to a weight that is raised as the system expands
...
6 can still be used, but now Vf < Vi
...
This somewhat perplexing conclusion seems to be inconsistent with the fact that the gas inside
the container is opposing the compression
...

Other types of work (for example, electrical work), which we shall call either nonexpansion work or additional work, have analogous expressions, with each one the
product of an intensive factor (the pressure, for instance) and an extensive factor (the
change in volume)
...
1
...
5 and 2
...

(b) Free expansion

By free expansion we mean expansion against zero opposing force
...
According to eqn 2
...
Hence, overall:
Free expansion:

w=0

(2
...
1 Varieties of work*
Type of work

dw

Comments

Units†

Expansion

−pexdV

pex is the external pressure
dV is the change in volume

Pa
m3

Surface expansion

γ dσ

γ is the surface tension
dσ is the change in area

N m−1
m2

Extension

fdl

f is the tension
dl is the change in length

N
m

Electrical

φ dQ

φ is the electric potential
dQ is the change in charge

V
C

* In general, the work done on a system can be expressed in the form dw = −Fdz, where F is a ‘generalized force’
and dz is a ‘generalized displacement’
...
Note that 1 N m = 1 J and 1 V C = 1 J
...
3 EXPANSION WORK

35

That is, no work is done when a system expands freely
...

(c) Expansion against constant pressure

Now suppose that the external pressure is constant throughout the expansion
...
A chemical example of this condition is the expansion
of a gas formed in a chemical reaction
...
6 by taking the constant
pex outside the integral:

Ύ

w = −pex

Vf

dV = −pex(Vf − Vi)

Vi

Therefore, if we write the change in volume as ∆V = Vf − Vi,
w = −pex ∆V

(2
...
2
...
The magnitude of w, denoted |w|, is equal to the
area beneath the horizontal line at p = pex lying between the initial and final volumes
...


The work done by a gas when it
expands against a constant external
pressure, pex, is equal to the shaded area in
this example of an indicator diagram
...
2
...
2

(d) Reversible expansion

The value of the integral

A reversible change in thermodynamics is a change that can be reversed by an
infinitesimal modification of a variable
...
We
say that a system is in equilibrium with its surroundings if an infinitesimal change
in the conditions in opposite directions results in opposite changes in its state
...
The transfer of energy as heat between the
two is reversible because, if the temperature of either system is lowered infinitesimally, then energy flows into the system with the lower temperature
...

Suppose a gas is confined by a piston and that the external pressure, pex, is set equal
to the pressure, p, of the confined gas
...
1) because an infinitesimal change in the
external pressure in either direction causes changes in volume in opposite directions
...
If
the external pressure is increased infinitesimally, then the gas contracts slightly
...
If, on the other hand,
the external pressure differs measurably from the internal pressure, then changing pex
infinitesimally will not decrease it below the pressure of the gas, so will not change the
direction of the process
...

To achieve reversible expansion we set pex equal to p at each stage of the expansion
...
When we set pex = p, eqn 2
...
9)rev

(Equations valid only for reversible processes are labelled with a subscript rev
...
For instance, the
area under the curve f(x) = x 2 shown in
the illustration that lies between x = 1
and 3 is

Ύ x dx = (–x + constant)
3

2

1

1 3
3

3
1

1
26
–
= –(33 − 13) = – ≈ 8
...
The total work
of reversible expansion is therefore

Ύ

w=−

Vf

pdV

(2
...
Equation 2
...

(e) Isothermal reversible expansion

Comment 2
...
The expansion is made
isothermal by keeping the system in thermal contact with its surroundings (which
may be a constant-temperature bath)
...

The temperature T is constant in an isothermal expansion, so (together with n and R)
it may be taken outside the integral
...
The work done
during the irreversible expansion against
the same final pressure is equal to the
rectangular area shown slightly darker
...


Fig
...
8

Exploration Calculate the work of

isothermal reversible expansion of
1
...
0 m3 to
3
...


dV
V

= −nRT ln

Vf
Vi

(2
...
11 is positive and hence w < 0
...
2 The equations also show that more work is done for a given change of volume
when the temperature is increased
...

We can express the result of the calculation as an indicator diagram, for the magnitude of the work done is equal to the area under the isotherm p = nRT/V (Fig
...
8)
...
More work is obtained when the expansion is reversible (the
area is greater) because matching the external pressure to the internal pressure at each
stage of the process ensures that none of the system’s pushing power is wasted
...
We may infer from
this discussion that, because some pushing power is wasted when p > pex, the maximum work available from a system operating between specified initial and final states
and passing along a specified path is obtained when the change takes place reversibly
...
Later (in Section 3
...

Example 2
...

2
We shall see later that there is a compensating influx of energy as heat, so overall the internal energy is
constant for the isothermal expansion of a perfect gas
...
4 HEAT TRANSACTIONS
Method We need to judge the magnitude of the volume change and then to decide

how the process occurs
...
If the system expands against a constant external
pressure, the work can be calculated from eqn 2
...
A general feature of processes in
which a condensed phase changes into a gas is that the volume of the former may
usually be neglected relative to that of the gas it forms
...
In (b) the gas drives back the atmosphere and therefore w = −pex ∆V
...
Therefore,
w = −pex ∆V ≈ −pex ×

nRT
pex

= −nRT

Because the reaction is Fe(s) + 2 HCl(aq) → FeCl2(aq) + H2(g), we know that 1 mol
H2 is generated when 1 mol Fe is consumed, and n can be taken as the amount of
Fe atoms that react
...
85 g mol−1, it follows that
w≈−

50 g
55
...
3145 J K−1 mol−1) × (298 K)

≈ −2
...
2 kJ of work driving back the atmosphere
...

Self-test 2
...

[−10 kJ]

2
...
12)

where dwe is work in addition (e for ‘extra’) to the expansion work, dwexp
...
A
system kept at constant volume can do no expansion work, so dwexp = 0
...
Under these
circumstances:
dU = dq

(at constant volume, no additional work)

(2
...
For a measurable change,
∆U = qV

(2
...


37

38

2 THE FIRST LAW
(a) Calorimetry

Calorimetry is the study of heat transfer during physical and chemical processes
...
The most common
device for measuring ∆U is an adiabatic bomb calorimeter (Fig
...
9)
...
The bomb is immersed in a stirred water bath, and the
whole device is the calorimeter
...
The water in the calorimeter and of the outer bath are both monitored and
adjusted to the same temperature
...

The change in temperature, ∆T, of the calorimeter is proportional to the heat
that the reaction releases or absorbs
...
The conversion of ∆T to qV is best achieved by calibrating the
calorimeter using a process of known energy output and determining the calorimeter
constant, the constant C in the relation
q = C∆T

(2
...
The ‘bomb’ is the central
vessel, which is strong enough to withstand
high pressures
...
To ensure
adiabaticity, the calorimeter is immersed
in a water bath with a temperature
continuously readjusted to that of the
calorimeter at each stage of the
combustion
...
2
...
14b)

Alternatively, C may be determined by burning a known mass of substance (benzoic
acid is often used) that has a known heat output
...

Illustration 2
...
0 A from a 12 V supply for 300 s, then from eqn 2
...
0 A) × (12 V) × (300 s) = 3
...
If the observed rise in temperature is 5
...
5 K) = 6
...


Comment 2
...
The motion of charge gives
rise to an electric current, I, measured in
coulombs per second, or amperes, A,
where 1 A = 1 C s−1
...
We write the electrical
power, P, as
P = (energy supplied)/(time interval)
= IV t/t = IV

(b) Heat capacity

The internal energy of a substance increases when its temperature is raised
...
For example,
the sample may be a gas in a container of fixed volume
...
2
...
The slope of
the tangent to the curve at any temperature is called the heat capacity of the system at
that temperature
...
All applications in this chapter refer to a single substance, so this complication
can be ignored
...
4 HEAT TRANSACTIONS

Fig
...
10 The internal energy of a system
increases as the temperature is raised; this
graph shows its variation as the system is
heated at constant volume
...
Note that, for the system
illustrated, the heat capacity is greater at B
than at A
...
2
...
The
variation of the internal energy with
temperature at one particular constant
volume is illustrated by the curve drawn
parallel to T
...


A ∂U D
C ∂T F V

Comment 2
...
15]

In this case, the internal energy varies with the temperature and the volume of the
sample, but we are interested only in its variation with the temperature, the volume
being held constant (Fig
...
11)
...
3 Estimating a constant-volume heat capacity

The heat capacity of a monatomic perfect gas can be calculated by inserting the
expression for the internal energy derived in Molecular interpretation 2
...
There we
3
saw that Um = Um(0) + –RT, so from eqn 2
...
47 J K−1 mol−1
...
The molar heat capacity at constant
volume, CV,m = CV /n, is the heat capacity per mole of material, and is an intensive
property (all molar quantities are intensive)
...
For certain applications it is useful to know the
specific heat capacity (more informally, the ‘specific heat’) of a substance, which is
the heat capacity of the sample divided by the mass, usually in grams: CV,s = CV /m
...
In general,

The partial-differential operation
(∂z/∂x)y consists of taking the first
derivative of z(x,y) with respect to x,
treating y as a constant
...


39

40

2 THE FIRST LAW
heat capacities depend on the temperature and decrease at low temperatures
...

The heat capacity is used to relate a change in internal energy to a change in temperature of a constant-volume system
...
15 that
dU = CV dT

(at constant volume)

(2
...
If
the heat capacity is independent of temperature over the range of temperatures of
interest, a measurable change of temperature, ∆T, brings about a measurable increase
in internal energy, ∆U, where
∆U = CV ∆T

(at constant volume)

(2
...
13b), the last equation can be written
qV = CV ∆T

(2
...
The ratio of the energy transferred as heat to the temperature rise it causes (qV /∆T) is the constant-volume heat
capacity of the sample
...
An infinite heat capacity implies that there will be no increase in temperature
however much energy is supplied as heat
...
Therefore, at the temperature of
a phase transition, the heat capacity of a sample is infinite
...
7
...
5 Enthalpy
The change in internal energy is not equal to the energy transferred as heat when the
system is free to change its volume
...
2
...
However, we shall now show that in this case the
energy supplied as heat at constant pressure is equal to the change in another
thermodynamic property of the system, the enthalpy
...
In such a case, the
change in internal energy is smaller than
the energy supplied as heat
...
2
...
18]

where p is the pressure of the system and V is its volume
...
As is true of any state function, the
change in enthalpy, ∆H, between any pair of initial and final states is independent of
the path between them
...
5 ENTHALPY
Although the definition of enthalpy may appear arbitrary, it has important implications for thermochemisty
...
18 implies that the change in enthalpy is equal to the energy supplied as heat at constant
pressure (provided the system does no additional work):
dH = dq

(at constant pressure, no additional work)

(2
...
19b)

Justification 2
...
18,
H changes from U + pV to
H + dH = (U + dU ) + (p + dp)(V + dV)
= U + dU + pV + pdV + Vdp + dpdV
The last term is the product of two infinitesimally small quantities and can therefore
be neglected
...
Then
dH = dq

(at constant pressure, no additional work)

as in eqn 2
...


The result expressed in eqn 2
...
For example, if we supply 36 kJ of energy through an electric
heater immersed in an open beaker of water, then the enthalpy of the water increases
by 36 kJ and we write ∆H = +36 kJ
...
A calorimeter for studying processes at constant pressure is called an isobaric
calorimeter
...
For a combustion reaction an adiabatic flame calorimeter may be
used to measure ∆T when a given amount of substance burns in a supply of oxygen
(Fig
...
13)
...
Because solids and liquids have
small molar volumes, for them pVm is so small that the molar enthalpy and molar
internal energy are almost identical (Hm = Um + pVm ≈ Um)
...

Physically, such processes are accompanied by a very small change in volume, the
system does negligible work on the surroundings when the process occurs, so the
energy supplied as heat stays entirely within the system
...
Changes in enthalpy and internal energy may also be measured by noncalorimetric methods (see Chapter 7)
...
2 Relating ∆H and ∆U

The internal energy change when 1
...
21 kJ
...
0 bar given that the densities of the
solids are 2
...
93 g cm−3, respectively
...

Combustion occurs as a known amount of
reactant is passed through to fuel the flame,
and the rise of temperature is monitored
...
2
...
18)
...

Answer The change in enthalpy when the transition occurs is

∆H = H(aragonite) − H(calcite)
= {U(a) + pV(a)} − {U(c) + pV(c)}
= ∆U + p{V(a) − V(c)} = ∆U + p∆V
The volume of 1
...
0 mol
CaCO3 as calcite is 37 cm3
...
0 × 105 Pa) × (34 − 37) × 10− 6 m3 = −0
...
Hence,
∆H − ∆U = −0
...
1 per cent of the value of ∆U
...

Self-test 2
...
0 mol Sn(s, grey)
of density 5
...
31 g cm−3 at 10
...
At
298 K, ∆H = +2
...

[∆H − ∆U = −4
...
20)°

2
...
21)°

where ∆ng is the change in the amount of gas molecules in the reaction
...
4 The relation between ∆H and ∆U for gas-phase reactions

In the reaction 2 H2(g) + O2(g) → 2 H2O(l), 3 mol of gas-phase molecules is
replaced by 2 mol of liquid-phase molecules, so ∆ng = −3 mol
...
5 kJ mol−1, the enthalpy and internal energy changes taking place in
the system are related by
∆H − ∆U = (−3 mol) × RT ≈ −7
...
2
...


Example 2
...
0 atm
...
50 A from a 12 V supply is passed for 300 s through a resistance in thermal contact with it, it is found that 0
...
Calculate the molar internal energy and enthalpy changes at the boiling point (373
...

Method Because the vaporization occurs at constant pressure, the enthalpy change

is equal to the heat supplied by the heater
...
To convert from enthalpy change to internal energy
change, we assume that the vapour is a perfect gas and use eqn 2
...

Answer The enthalpy change is

∆H = qp = (0
...
50 × 12 × 300) J
Here we have used 1 A V s = 1 J (see Comment 2
...
Because 0
...
798 g)/(18
...
798/18
...
50 × 12 × 300 J
(0
...
02) mol

= +41 kJ mol−1

In the process H2O(l) → H2O(g) the change in the amount of gas molecules is
∆ng = +1 mol, so
∆Um = ∆Hm − RT = +38 kJ mol−1
The plus sign is added to positive quantities to emphasize that they represent an
increase in internal energy or enthalpy
...


43

44

2 THE FIRST LAW
Self-test 2
...
25 K) is 30
...
What is the molar internal energy change? For how
long would the same 12 V source need to supply a 0
...
9 kJ mol−1, 660 s]

(c) The variation of enthalpy with temperature

Fig
...
14 The slope of the tangent to a curve
of the enthalpy of a system subjected to a
constant pressure plotted against
temperature is the constant-pressure heat
capacity
...
Thus, the
heat capacities at A and B are different
...


The enthalpy of a substance increases as its temperature is raised
...
The most important
condition is constant pressure, and the slope of the tangent to a plot of enthalpy
against temperature at constant pressure is called the heat capacity at constant pressure, Cp, at a given temperature (Fig
...
14)
...
22]

The heat capacity at constant pressure is the analogue of the heat capacity at constant
volume, and is an extensive property
...

The heat capacity at constant pressure is used to relate the change in enthalpy to a
change in temperature
...
23a)

If the heat capacity is constant over the range of temperatures of interest, then for a
measurable increase in temperature
∆H = Cp ∆T

(at constant pressure)

(2
...
24)

This expression shows us how to measure the heat capacity of a sample: a measured
quantity of energy is supplied as heat under conditions of constant pressure (as in a
sample exposed to the atmosphere and free to expand), and the temperature rise is
monitored
...
However, when it is
necessary to take the variation into account, a convenient approximate empirical
expression is
Cp,m = a + bT +

c
T2

(2
...
2)
...
All applications in this chapter refer to pure substances, so this
complication can be ignored
...
5 ENTHALPY

45

Synoptic Table 2
...
86

4
...
54

CO2(g)

44
...
79

−8
...
29

0

N2(g)

28
...
77

0
−0
...


Example 2
...
2
...
23b (which assumes that the heat capacity of the substance is constant)
...
23a, substitute eqn 2
...

Answer For convenience, we denote the two temperatures T1 (298 K) and T2 (373 K)
...
Now we use the integrals

Ύdx = x + constant Ύx dx = –x + constant Ύ x
1 2
2

dx
2

=−

1
x

+ constant

to obtain
1
2
2
H(T2) − H(T1) = a(T2 − T1) + –b(T 2 − T 1) − c
2

A 1
1D
−
C T2 T1 F

Substitution of the numerical data results in
H(373 K) = H(298 K) + 2
...
14 J K−1 mol−1 (the value given
by eqn 2
...
19 kJ mol−1
...
4 At very low temperatures the heat capacity of a solid is proportional
to T 3, and we can write Cp = aT 3
...
Such systems do work on
the surroundings and therefore some of the energy supplied to them as heat escapes

Comment 2
...


46

2 THE FIRST LAW
back to the surroundings
...
A smaller increase in temperature
implies a larger heat capacity, so we conclude that in most cases the heat capacity at
constant pressure of a system is larger than its heat capacity at constant volume
...
11) that there is a simple relation between the two heat
capacities of a perfect gas:
Cp − CV = nR

(2
...
Because the heat capacity at constant volume of a monatomic gas is about 12 J K−1 mol−1, the difference is highly significant
and must be taken into account
...
1 Differential scanning calorimetry
A differential scanning calorimeter
...
The output is the difference in power
needed to maintain the heat sinks at equal
temperatures as the temperature rises
...
2
...
The term
‘differential’ refers to the fact that the behaviour of the sample is compared to that of
a reference material which does not undergo a physical or chemical change during the
analysis
...

A DSC consists of two small compartments that are heated electrically at a constant
rate
...
A
computer controls the electrical power output in order to maintain the same temperature in the sample and reference compartments throughout the analysis (see Fig
...
15)
...
To maintain the same temperature in both
compartments, excess energy is transferred as heat to or from the sample during the
process
...

If no physical or chemical change occurs in the sample at temperature T, we write
the heat transferred to the sample as qp = Cp ∆T, where ∆T = T − T0 and we have
assumed that Cp is independent of temperature
...
We interpret qp,ex in terms of an
apparent change in the heat capacity at constant pressure of the sample, Cp, during the
temperature scan
...
2
...
45
...
(Adapted from B
...
LeHarne, J
...
Educ
...
)

Cp,ex =

qp,ex
∆T

=

qp,ex

αt

=

Pex

α

where Pex = qp,ex /t is the excess electrical power necessary to equalize the temperature
of the sample and reference compartments
...
2
...
Broad peaks in the thermogram indicate processes requiring transfer of
energy as heat
...
23a, the enthalpy change associated with the process is

2
...
This relation shows that the enthalpy change is then the area under the curve of
Cp,ex against T
...
5 mg, which is a significant advantage over bomb or flame
calorimeters, which require several grams of material
...
Large molecules, such as synthetic or biological polymers,
attain complex three-dimensional structures due to intra- and intermolecular interactions, such as hydrogen bonding and hydrophobic interactions (Chapter 18)
...
For example, the thermogram shown in the illustration indicated that the protein ubiquitin retains its native structure up to about 45°C
...
The same principles also apply to the study of
structural integrity and stability of synthetic polymers, such as plastics
...
6 Adiabatic changes
We are now equipped to deal with the changes that occur when a perfect gas expands
adiabatically
...
In molecular terms, the kinetic energy of the molecules falls
as work is done, so their average speed decreases, and hence the temperature falls
...
2
...
In the first step, only the volume changes and the temperature is
held constant at its initial value
...
Provided the heat capacity is independent of temperature, this change is
∆U = CV (Tf − Ti) = CV ∆T
Because the expansion is adiabatic, we know that q = 0; because ∆U = q + w, it then
follows that ∆U = wad
...
Therefore,
by equating the two values we have obtained for ∆U, we obtain
wad = CV ∆T

(2
...
That is exactly what
we expect on molecular grounds, because the mean kinetic energy is proportional
to T, so a change in internal energy arising from temperature alone is also expected
to be proportional to ∆T
...
1 we show that the initial and
final temperatures of a perfect gas that undergoes reversible adiabatic expansion
(reversible expansion in a thermally insulated container) can be calculated from

A Vi D
Tf = Ti
C Vf F

1/c

(2
...
2
...
In the first step, the system expands
at constant temperature; there is no change
in internal energy if the system consists of a
perfect gas
...
The overall change in
internal energy is the sum of the changes
for the two steps
...
28b)°
rev

This result is often summarized in the form VT = constant
...
5 Work of adiabatic expansion

Consider the adiabatic, reversible expansion of 0
...
50 dm3 to 1
...
The molar heat capacity of argon at constant volume is
12
...
501
...
28a,

A 0
...
00 dm3 F

1/1
...
27, that
w = {(0
...
48 J K−1 mol−1)} × (−110 K) = −27 J
Note that temperature change is independent of the amount of gas but the work
is not
...
5 Calculate the final temperature, the work done, and the change of
internal energy when ammonia is used in a reversible adiabatic expansion from
0
...
00 dm3, the other initial conditions being the same
...
1 that the pressure of a perfect gas that
undergoes reversible adiabatic expansion from a volume Vi to a volume Vf is related
to its initial pressure by
pfV γ = piV γ
f
i

(2
...
2
...
(a) An adiabat for a perfect
gas undergoing reversible expansion
...


Exploration Explore how the

parameter γ affects the dependence
of the pressure on the volume
...
This result is summarized in the form pV = constant
...
3), and from eqn 2
...
For a gas of nonlinear polyatomic molecules (which can rotate as well as
3
4
translate), CV,m = 3R, so γ = –
...
2
...

Because γ > 1, an adiabat falls more steeply (p ∝ 1/V γ ) than the corresponding
isotherm (p ∝ 1/V)
...

Illustration 2
...


2
...
Thermochemistry is a branch of thermodynamics because
a reaction vessel and its contents form a system, and chemical reactions result in the
exchange of energy between the system and the surroundings
...
Conversely, if
we know ∆U or ∆H for a reaction, we can predict the energy (transferred as heat) the
reaction can produce
...
Because the release of energy by heating the
surroundings signifies a decrease in the enthalpy of a system (at constant pressure), we
can now see that an exothermic process at constant pressure is one for which ∆H < 0
...

2
...
In most of our discussions we shall consider the standard
enthalpy change, ∆H 7, the change in enthalpy for a process in which the initial and
final substances are in their standard states:
The standard state of a substance at a specified temperature is its pure form at
1 bar
...
The standard enthalpy change for a reaction or a physical process is the difference between the products in their standard states and the reactants in their standard
states, all at the same specified temperature
...
66 kJ mol−1

As implied by the examples, standard enthalpies may be reported for any temperature
...
15 K (corresponding to 25
...
Unless otherwise mentioned, all thermodynamic data in this text will refer to this conventional temperature
...
However, the older convention,
∆Hvap, is still widely used
...


5
The definition of standard state is more sophisticated for a real gas (Further information 3
...
6 and 5
...


49

50

2 THE FIRST LAW
Synoptic Table 2
...
81

C6H6
H2O

1
...
61
273
...
5

He

10
...
29

Vaporization
6
...
2

30
...
008

373
...
656 (44
...
021

4
...
084

* More values are given in the Data section
...
3)
...
Another is the standard enthalpy
of fusion, ∆fusH 7, the standard enthalpy change accompanying the conversion of a
solid to a liquid, as in
H2O(s) → H2O(l)

∆fusH 7(273 K) = +6
...

Because enthalpy is a state function, a change in enthalpy is independent of the path
between the two states
...
For example, we can picture the conversion of a solid to a vapour either as occurring by sublimation (the direct conversion
from solid to vapour),
H2O(s) → H2O(g)

∆subH 7

or as occurring in two steps, first fusion (melting) and then vaporization of the resulting liquid:
H2O(s) → H2O(l)

∆ fusH 7

H2O(l) → H2O(g)

∆ vapH 7

Overall: H2O(s) → H2O(g)

∆fusH 7 + ∆ vapH 7

Because the overall result of the indirect path is the same as that of the direct path, the
overall enthalpy change is the same in each case (1), and we can conclude that (for
processes occurring at the same temperature)
∆subH 7 = ∆ fusH 7 + ∆ vapH 7

(2
...

Another consequence of H being a state function is that the standard enthalpy
changes of a forward process and its reverse differ in sign (2):
∆H 7(A → B) = −∆H 7(B → A)

(2
...


2
...
4 Enthalpies of transition
Transition

Process

Symbol*

Transition

Phase α → phase β

∆trsH

Fusion

s→l

∆fusH

Vaporization

l→g

∆vapH

Sublimation

s→g

∆subH

Mixing

Pure → mixture

∆mixH
∆solH

Solution

Solute → solution

Hydration

X±(g) → X±(aq)

∆hydH

Atomization

Species(s, l, g) → atoms(g)

∆atH

Ionization

X(g) → X+(g) + e−(g)

∆ionH

−

−

Electron gain

X(g) + e (g) → X (g)

Reaction

Reactants → products

∆egH
∆rH

Combustion

Compounds(s, l, g) + O2(g) → CO2(g), H2O(l, g)

∆cH

Formation

Elements → compound

∆fH

Activation

Reactants → activated complex

∆‡H

* IUPAC recommendations
...


The different types of enthalpies encountered in thermochemistry are summarized
in Table 2
...
We shall meet them again in various locations throughout the text
...
There are two
ways of reporting the change in enthalpy that accompanies a chemical reaction
...
For the combustion of methane, the standard value refers to the reaction in which 1 mol CH4 in the form of pure methane gas at 1 bar reacts completely
with 2 mol O2 in the form of pure oxygen gas to produce 1 mol CO2 as pure carbon
dioxide at 1 bar and 2 mol H2O as pure liquid water at 1 bar; the numerical value is for
the reaction at 298 K
...
Thus, for the combustion of reaction, we write
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
For the reaction
2A+B→3C+D

∆ r H 7 = −890 kJ mol−1

51

52

2 THE FIRST LAW
Synoptic Table 2
...
0

−3268

Ethane, C2H6(g)

−84
...
8

Methanol, CH3OH(l)

−890

−238
...


the standard reaction enthalpy is
7
7
7
7
∆ r H 7 = {3Hm(C) + Hm(D)} − {2Hm(A) + Hm(B)}
7
where Hm(J) is the standard molar enthalpy of species J at the temperature of interest
...
We interpret the ‘per mole’ by noting the stoichiometic
coefficients in the chemical equation
...
In general,

∆r H 7 =

7
7
∑ νH m − ∑ νH m

Products

(2
...
6
Some standard reaction enthalpies have special names and a particular significance
...
An
example is the combustion of glucose:
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)

∆c H 7 = −2808 kJ mol−1

The value quoted shows that 2808 kJ of heat is released when 1 mol C6H12O6 burns
under standard conditions (at 298 K)
...
5
...
2 Food and energy reserves

The thermochemical properties of fuels Table 2
...

Thus, if the standard enthalpy of combustion is ∆c H 7 and the molar mass of the compound is M, then the specific enthalpy is ∆c H 7/M
...
6 lists the specific enthalpies
of several fuels
...
If the entire consumption were in the form of glucose
(3; which has a specific enthalpy of 16 kJ g−1), that would require the consumption of
750 g of glucose for a man and 560 g for a woman
...
For a more sophisticated way
of writing eqn 2
...
2
...
7 STANDARD ENTHALPY CHANGES
Table 2
...
8 × 104

−726

23

1
...

The specific enthalpy of fats, which are long-chain esters like tristearin (beef fat), is
much greater than that of carbohydrates, at around 38 kJ g−1, slightly less than the value
for the hydrocarbon oils used as fuel (48 kJ g−1)
...
In Arctic species, the stored fat also acts as a layer of insulation; in desert
species (such as the camel), the fat is also a source of water, one of its oxidation products
...
When proteins are oxidized (to urea, CO(NH2)2), the equivalent
enthalpy density is comparable to that of carbohydrates
...
6–37
...
A variety of
mechanisms contribute to this aspect of homeostasis, the ability of an organism to
counteract environmental changes with physiological responses
...
When heat needs to be dissipated rapidly, warm blood is allowed to flow
through the capillaries of the skin, so producing flushing
...
Evaporation removes about 2
...

When vigorous exercise promotes sweating (through the influence of heat selectors
on the hypothalamus), 1–2 dm3 of perspired water can be produced per hour, corresponding to a heat loss of 2
...
0 MJ h−1
...
This application of the First Law is called Hess’s law:
The standard enthalpy of an overall reaction is the sum of the standard enthalpies
of the individual reactions into which a reaction may be divided
...

The thermodynamic basis of the law is the path-independence of the value of ∆r H 7
and the implication that we may take the specified reactants, pass through any (possibly hypothetical) set of reactions to the specified products, and overall obtain the
same change of enthalpy
...

Example 2
...
The standard reaction enthalpy for the combustion of propane,
CH3CH2CH3(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
is −2220 kJ mol−1
...

Method The skill to develop is the ability to assemble a given thermochemical

equation from others
...
Then add or subtract the
reaction enthalpies in the same way
...
5
...
7* Standard
enthalpies of formation of inorganic
compounds at 298 K

−124
−2220
+286
−2058

Self-test 2
...


[−205 kJ mol−1]

∆f H 7/(kJ mol−1)
H2O(l)

−285
...
78

NH3(g)

−46
...
63

NO2(g)

33
...
16

NaCl(s)

−411
...
75

* More values are given in the Data section
...
7

The NIST WebBook listed in the web
site for this book links to online
databases of thermochemical data
...
8 Standard enthalpies of formation
The standard enthalpy of formation, ∆f H 7, of a substance is the standard reaction
enthalpy for the formation of the compound from its elements in their reference
states
...
For example, at 298 K the reference state of nitrogen is a gas of N2
molecules, that of mercury is liquid mercury, that of carbon is graphite, and that of tin
is the white (metallic) form
...
Standard enthalpies of formation are expressed as enthalpies per mole
of molecules or (for ionic substances) formula units of the compound
...
0 kJ mol−1
...
Some enthalpies of formation are listed in Tables 2
...
7
...
8 STANDARD ENTHALPIES OF FORMATION
The standard enthalpy of formation of ions in solution poses a special problem
because it is impossible to prepare a solution of cations alone or of anions alone
...
33]

Thus, if the enthalpy of formation of HBr(aq) is found to be −122 kJ mol−1, then the
whole of that value is ascribed to the formation of Br−(aq), and we write ∆ f H 7(Br−, aq)
= −122 kJ mol−1
...
In essence,
this definition adjusts the actual values of the enthalpies of formation of ions by a fixed
amount, which is chosen so that the standard value for one of them, H+(aq), has the
value zero
...
The value of
∆ r H 7 for the overall reaction is the sum of these ‘unforming’ and forming enthalpies
...
Hence, in the enthalpies of formation of
substances, we have enough information to calculate the enthalpy of any reaction by
using
∆rH 7 =

∑ν∆f H 7 − ∑ν∆f H 7

Products

(2
...

Illustration 2
...
78 + 4(0)} kJ mol−1 − {2(264
...
25)} kJ mol−1
= −896
...
The question that now arises is whether we can construct
standard enthalpies of formation from a knowledge of the chemical constitution
of the species
...
In the past, approximate procedures based on mean bond enthalpies,
∆H(A-B), the average enthalpy change associated with the breaking of a specific
A-B bond,
A-B(g) → A(g) + B(g)

∆H(A-B)

55

56

2 THE FIRST LAW
have been used
...
Nor does the
approach distinguish between geometrical isomers, where the same atoms and bonds
may be present but experimentally the enthalpies of formation might be significantly
different
...
Commercial software packages use the principles developed in Chapter 11
to calculate the standard enthalpy of formation of a molecule drawn on the computer
screen
...
In the case of methylcyclohexane, for instance, the calculated conformational energy difference ranges from 5
...
9 kJ mol−1, with the equatorial conformer
having the lower standard enthalpy of formation
...
5 kJ mol−1
...
Computational methods almost
always predict correctly which conformer is more stable but do not always predict the
correct magnitude of the conformational energy difference
...
When the temperature is
increased, the enthalpy of the products and
the reactants both increase, but may do so
to different extents
...
The change in
reaction enthalpy reflects the difference in
the changes of the enthalpies
...
2
...
9 The temperature-dependence of reaction enthalpies
The standard enthalpies of many important reactions have been measured at different temperatures
...
2
...
In many cases heat capacity
data are more accurate that reaction enthalpies so, providing the information is available, the procedure we are about to describe is more accurate that a direct measurement of a reaction enthalpy at an elevated temperature
...
23a that, when a substance is heated from T1 to T2, its enthalpy changes from H(T1) to
H(T2) = H(T1) +

Ύ

T2

CpdT

(2
...
) Because this equation applies to each substance in the reaction, the standard
reaction enthalpy changes from ∆rH 7(T1) to
∆ r H 7(T2) = ∆r H 7(T1) +

Ύ

T2

∆rC 7dT
p

(2
...
37]

Reactants

Equation 2
...
It is normally a good approximation to
assume that ∆rCp is independent of the temperature, at least over reasonably limited
ranges, as illustrated in the following example
...
In some cases the temperature
dependence of heat capacities is taken into account by using eqn 2
...


2
...
6 Using Kirchhoff’s law

The standard enthalpy of formation of gaseous H2O at 298 K is −241
...

Estimate its value at 100°C given the following values of the molar heat capacities
at constant pressure: H2O(g): 33
...
84 J K−1 mol−1; O2(g):
29
...
Assume that the heat capacities are independent of temperature
...
36 evaluates to (T2 − T1)∆rC 7
...

p
1
Answer The reaction is H2(g) + – O2(g) → H2O(g), so
2
1 7
7
7
∆rC 7 = C p,m(H2O, g) − {C p,m(H2, g) + –C p,m(O2, g)} = −9
...
82 kJ mol−1 + (75 K) × (−9
...
6 kJ mol−1
Self-test 2
...
5
...
2 that a ‘state function’ is a property that is independent of how a
sample is prepared
...
The internal energy
and enthalpy are examples of state functions, for they depend on the current state of
the system and are independent of its previous history
...
Examples of path functions are the
work and heating that are done when preparing a state
...
In each case, the energy transferred as
work or heat relates to the path being taken between states, not the current state itself
...
The practical importance of these results is
that we can combine measurements of different properties to obtain the value of a
property we require
...
10 Exact and inexact differentials
Consider a system undergoing the changes depicted in Fig
...
20
...
Work is done by the system as it
expands adiabatically to a state f
...
Notice our
use of language: U is a property of the state; w is a property of the path
...
The internal energy of both the

Fig
...
20 As the volume and temperature of
a system are changed, the internal energy
changes
...


58

2 THE FIRST LAW
initial and the final states are the same as before (because U is a state function)
...
The work and the heat are path functions
...
2, the change in altitude (a state function) is
independent of the path, but the distance travelled (a path function) does depend on
the path taken between the fixed endpoints
...
38)

i

The value of ∆U depends on the initial and final states of the system but is independent of the path between them
...
In general, an exact differential is an
infinitesimal quantity that, when integrated, gives a result that is independent of the
path between the initial and final states
...
39)

i, path

Notice the difference between this equation and eqn 2
...
First, we do not write ∆q,
because q is not a state function and the energy supplied as heat cannot be expressed
as qf − qi
...
This path-dependence is expressed
by saying that dq is an ‘inexact differential’
...
Often dq is written pq to emphasize that it is inexact and requires the specification of a path
...
It follows that
dw is an inexact differential
...

Example 2
...
Let the initial state be
T, Vi and the final state be T, Vf
...
Calculate w, q, and ∆U for each process
...
We saw in Molecular interpretation 2
...
We also know that
in general ∆U = q + w
...
11 CHANGES IN INTERNAL ENERGY
expressions
...

Answer Because ∆U = 0 for both paths and ∆U = q + w, in each case q = −w
...
3b); so in Path 1, w = 0 and q = 0
...
11, so w = −nRT ln(Vf /Vi) and consequently
q = nRT ln(Vf /Vi)
...

Self-test 2
...

[q = pex ∆V, w = −pex ∆V, ∆U = 0]

2
...
The internal energy U can be regarded as a function of V, T, and p,
but, because there is an equation of state, stating the values of two of the variables fixes
the value of the third
...
Expressing U as a function of volume and
temperature fits the purpose of our discussion
...
2
...
If, instead, T changes to T +
dT at constant volume (Fig
...
22), then the internal energy changes to

Fig
...
21 The partial derivative (∂U/∂V)T is
the slope of U with respect to V with the
temperature T held constant
...
2
...


59

60

2 THE FIRST LAW
U′ = U +

A ∂U D
dT
C ∂T F V

Now suppose that V and T both change infinitesimally (Fig
...
23)
...
Therefore, from
the last equation we obtain the very important result that
An overall change in U, which is
denoted dU, arises when both V and T
are allowed to change
...

Fig
...
23

dU =

The internal pressure, πT , is the
slope of U with respect to V with the
temperature T held constant
...
40)

The interpretation of this equation is that, in a closed system of constant composition,
any infinitesimal change in the internal energy is proportional to the infinitesimal
changes of volume and temperature, the coefficients of proportionality being the two
partial derivatives
...
In the present case, we have already met (∂U/∂T)V in eqn 2
...
The other coefficient, (∂U/∂V)T,
plays a major role in thermodynamics because it is a measure of the variation of
the internal energy of a substance as its volume is changed at constant temperature
(Fig
...
24)
...
2
...
41]

In terms of the notation CV and πT, eqn 2
...
42)

(b) The Joule experiment

When there are no interactions between the molecules, the internal energy is independent of their separation and hence independent of the volume of the sample (see
Molecular interpretation 2
...
Therefore, for a perfect gas we can write πT = 0
...
If the internal energy increases (dU > 0)
as the volume of the sample expands isothermally (dV > 0), which is the case when
there are attractive forces between the particles, then a plot of internal energy against
volume slopes upwards and π T > 0 (Fig
...
25)
...
He used two metal vessels
immersed in a water bath (Fig
...
26)
...
He then tried to measure the change in temperature of the water
of the bath when a stopcock was opened and the air expanded into a vacuum
...


2
...
2
...
If attractions are
dominant in a real gas, the internal energy
increases with volume because the
molecules become farther apart on average
...


Fig
...
26 A schematic diagram of the
apparatus used by Joule in an attempt to
measure the change in internal energy
when a gas expands isothermally
...


The thermodynamic implications of the experiment are as follows
...
No energy entered or left the system
(the gas) as heat because the temperature of the bath did not change, so q = 0
...
It follows that U does
not change much when a gas expands isothermally and therefore that π T = 0
...
In particular, the heat capacity of the apparatus
was so large that the temperature change that gases do in fact cause was too small to
measure
...

(c) Changes in internal energy at constant pressure

Partial derivatives have many useful properties and some that we shall draw on
frequently are reviewed in Appendix 2
...

As an example, suppose we want to find out how the internal energy varies with
temperature when the pressure of the system is kept constant
...
42 by dT and impose the condition of constant pressure on the resulting
differentials, so that dU/dT on the left becomes (∂U/∂T)p, we obtain

A ∂U D
A ∂V D
= πT
+ CV
C ∂T F p
C ∂T F p
It is usually sensible in thermodynamics to inspect the output of a manipulation like
this to see if it contains any recognizable physical quantity
...
8* Expansion
coefficients (α) and isothermal
compressibilities (κT) at 298 K
α /(10−4 K−1)
Benzene

90
...
4

constant pressure)
...
030

0
...
861

2
...
1

49
...
43]

V C ∂T F p

and physically is the fractional change in volume that accompanies a rise in temperature
...
Table 2
...


1 A ∂V D

[2
...

Example 2
...

Method The expansion coefficient is defined in eqn 2
...
To use this expression,

substitute the expression for V in terms of T obtained from the equation of state
for the gas
...
43, the pressure, p, is treated as a
constant
...

Self-test 2
...


[κT
...
45)

This equation is entirely general (provided the system is closed and its composition is
constant)
...
For a perfect
gas, π T = 0, so then

A ∂U D
= CV
C ∂T F
p
7

(2
...
Throughout this chapter, we deal only with pure substances, so this complication can
be disregarded
...
12 THE JOULE–THOMSON EFFECT
That is, although the constant-volume heat capacity of a perfect gas is defined as the
slope of a plot of internal energy against temperature at constant volume, for a perfect
gas CV is also the slope at constant pressure
...
46 provides an easy way to derive the relation between Cp and CV for a
perfect gas expressed in eqn 2
...
Thus, we can use it to express both heat capacities in
terms of derivatives at constant pressure:
Cp − CV =

A ∂H D A ∂U D
−
C ∂T F p C ∂T F p

(2
...
48)°

which is eqn 2
...
We show in Further information 2
...
49)

Equation 2
...
It reduces to
eqn 2
...
Because expansion
coefficients α of liquids and solids are small, it is tempting to deduce from eqn 2
...
But this is not always so, because the compressibility κT might
also be small, so α 2/κT might be large
...
As an illustration, for water at 25°C, eqn
2
...
3 J K−1 mol−1 compared with CV,m = 74
...
In some
cases, the two heat capacities differ by as much as 30 per cent
...
12 The Joule–Thomson effect
We can carry out a similar set of operations on the enthalpy, H = U + pV
...
It turns out that H is a useful thermodynamic function when the pressure
is under our control: we saw a sign of that in the relation ∆H = qp (eqn 2
...
We shall
therefore regard H as a function of p and T, and adapt the argument in Section 2
...
As set
out in Justification 2
...
50)

where the Joule–Thomson coefficient, µ (mu), is defined as

µ=

A ∂T D
C ∂p F H

[2
...


Justification 2
...
40 but with H regarded as a function of p and
T we can write
A ∂H D
A ∂H D
E dp + B
dH = B
E dT
C ∂p F T
C ∂T F p

(2
...
The chain relation (see Further information 2
...
2) twice:
A ∂H D
(∂T/∂p)H A ∂T D A ∂H D
B
E =−
E B
E = −µCp
=B
∂p F T
(∂T/∂H)p C ∂p F H C ∂T F p
C

(2
...
51)
...
50 now follows directly
...
2
...
The gas expands
through the porous barrier, which acts as a
throttle, and the whole apparatus is
thermally insulated
...
Whether the expansion
results in a heating or a cooling of the gas
depends on the conditions
...
We need to be able to interpret it physically and to measure it
...
They let a gas expand
through a porous barrier from one constant pressure to another, and monitored
the difference of temperature that arose from the expansion (Fig
...
27)
...
They observed a lower temperature on the low pressure side, the difference in temperature being proportional to
the pressure difference they maintained
...


Justification 2
...
Because all changes to the gas occur adiabatically,
q = 0, which implies ∆U = w
Consider the work done as the gas passes through the barrier
...
2
...
The gas emerges on
the low pressure side, where the same amount of gas has a pressure pf, a temperature
Tf, and occupies a volume Vf
...
The relevant pressure is pi and the volume changes
from Vi to 0; therefore, the work done on the gas is
w1 = −pi(0 − Vi) = piVi
The gas expands isothermally on the right of the barrier (but possibly at a different
constant temperature) against the pressure pf provided by the downstream gas acting as a piston to be driven out
...
12 THE JOULE–THOMSON EFFECT

65

It follows that the change of internal energy of the gas as it moves adiabatically from
one side of the barrier to the other is
Uf − Ui = w = piVi − pfVf
Reorganization of this expression gives
Uf + pfVf = Ui + piVi, or Hf = Hi
Therefore, the expansion occurs without change of enthalpy
...
Adding the constraint of constant enthalpy and taking
the limit of small ∆p implies that the thermodynamic quantity measured is (∂T/∂p)H,
which is the Joule–Thomson coefficient, µ
...

The modern method of measuring µ is indirect, and involves measuring the
isothermal Joule–Thomson coefficient, the quantity

µT =

A ∂H D
C ∂p F T

[2
...
2
...
Comparing eqns 2
...
54, we see that the two coefficients are related
by:

µT = −Cp µ

(2
...
The steep pressure drop is measured,
and the cooling effect is exactly offset by an electric heater placed immediately after
the plug (Fig
...
30)
...
Because the
energy transferred as heat can be identified with the value of ∆H for the gas (because

Fig
...
29 The isothermal Joule–Thomson
coefficient is the slope of the enthalpy with
respect to changing pressure, the
temperature being held constant
...
2
...

The electrical heating required to offset
the cooling arising from expansion is
interpreted as ∆H and used to calculate
(∂H/∂p)T, which is then converted to µ as
explained in the text
...
2
...
The pistons
represent the upstream and downstream
gases, which maintain constant pressures
either side of the throttle
...


66

2 THE FIRST LAW

Synoptic Table 2
...
8

Tb/K

µ /(K bar−1)

87
...
10

1500

194
...
2

−0
...
4

+0
...
3

* More values are given in the Data section
...
2
...

Inside the boundary, the shaded area, it is
positive and outside it is negative
...
For a given pressure, the
temperature must be below a certain value
if cooling is required but, if it becomes too
low, the boundary is crossed again and
heating occurs
...
The inversion
temperature curve runs through the points
of the isenthalps where their slope changes
from negative to positive
...
Table 2
...

Real gases have nonzero Joule–Thomson coefficients
...
1), and the temperature, the sign of
the coefficient may be either positive or negative (Fig
...
31)
...

Gases that show a heating effect (µ < 0) at one temperature show a cooling effect
(µ > 0) when the temperature is below their upper inversion temperature, TI
(Table 2
...
2
...
As indicated in Fig
...
32, a gas typically has two inversion temperatures, one at high temperature and the other at low
...
2
...
The gas at high pressure is allowed to expand through a throttle; it cools
and is circulated past the incoming gas
...
There comes a stage when the circulating gas becomes so cold
that it condenses to a liquid
...
8 This characteristic points clearly to the involvement
of intermolecular forces in determining the size of the effect
...

The coefficient behaves like the properties discussed in Section 1
...


Fig
...
32 The inversion temperatures for
three real gases, nitrogen, hydrogen, and
helium
...
2
...
The
gas is recirculated, and so long as it is
beneath its inversion temperature it cools
on expansion through the throttle
...

Eventually liquefied gas drips from the
throttle
...
6
...
3 Molecular interactions and the Joule–Thomson effect

The kinetic model of gases (Molecular interpretation 1
...
2) imply that the mean kinetic energy of
molecules in a gas is proportional to the temperature
...
If the speed of the
molecules can be reduced to the point that neighbours can capture each other by
their intermolecular attractions, then the cooled gas will condense to a liquid
...
We
saw in Section 1
...
It follows that, if we can cause the
molecules to move apart from each other, like a ball rising from a planet, then they
should slow
...

To cool a gas, therefore, we allow it to expand without allowing any energy to enter
from outside as heat
...

Because some kinetic energy must be converted into potential energy to reach
greater separations, the molecules travel more slowly as their separation increases
...
The cooling effect, which corresponds to
µ > 0, is observed under conditions when attractive interactions are dominant
(Z < 1, eqn 1
...
For molecules under conditions
when repulsions are dominant (Z > 1), the Joule–Thomson effect results in the gas
becoming warmer, or µ < 0
...
Thermodynamics is the study of the transformations of
energy
...
The system is the part of the world in which we have a special
interest
...

3
...
A closed system has a boundary through which
matter cannot be transferred
...

4
...
The internal energy is the
total energy of a system
...
Work is the transfer of energy by motion against an opposing
force, dw = −Fdz
...


6
...
An endothermic process absorbs energy as heat
from the surroundings
...
A state function is a property that depends only on the current
state of the system and is independent of how that state has
been prepared
...
The First Law of thermodynamics states that the internal
energy of an isolated system is constant, ∆U = q + w
...
Expansion work is the work of expansion (or compression) of
a system, dw = −pexdV
...

The work of expansion against a constant external pressure is
w = −pex ∆V
...

10
...

11
...


68

2 THE FIRST LAW
12
...

13
...
The heat capacity at constant pressure is
Cp = (∂H/∂T)p
...

14
...
The enthalpy change is
the energy transferred as heat at constant pressure, ∆H = qp
...
During a reversible adiabatic change, the temperature of a
perfect gas varies according to Tf = Ti(Vi/Vf)1/c, c = CV,m/R
...

16
...
The standard state is the pure substance at
1 bar
...
Enthalpy changes are additive, as in ∆subH 7 = ∆fus H 7 + ∆ vap H 7
...
The enthalpy change for a process and its reverse are related
by ∆forwardH 7 = −∆reverseH 7
...
The standard enthalpy of combustion is the standard reaction
enthalpy for the complete oxidation of an organic compound
to CO2 gas and liquid H2O if the compound contains C, H,
and O, and to N2 gas if N is also present
...
Hess’s law states that the standard enthalpy of an overall
reaction is the sum of the standard enthalpies of the individual
reactions into which a reaction may be divided
...
The standard enthalpy of formation (∆ f H 7) is the standard
reaction enthalpy for the formation of the compound from its
elements in their reference states
...

22
...

23
...

T2

r

7
p

T1

24
...
An inexact differential
is an infinitesimal quantity that, when integrated, gives a
result that depends on the path between the initial and final
states
...
The internal pressure is defined as πT = (∂U/∂V)T
...

26
...

27
...

The isothermal Joule–Thomson coefficient is defined as
µT = (∂H/∂p)T = −Cp µ
...
The inversion temperature is the temperature at which the
Joule–Thomson coefficient changes sign
...
W
...
C
...

W
...
Freeman, New York (2005)
...
W
...
(ed
...
Published
as J
...
Chem
...
Data, Monograph no
...
American Institute of
Physics, New York (1998)
...
A
...
Yang, and B
...
Dasgupta, Thermodynamic partial
derivatives and experimentally measurable quantities
...
Chem
...
66, 890 (1989)
...
M
...
M
...
Wiley–Interscience, New York (2000)
...
N
...
Randall, Thermodynamics
...
S
...
Brewer
...

J
...
J
...
Educ
...


J
...
Cox, D
...
Wagman, and V
...
Medvedev, CODATA key values
for thermodynamics
...
, New York
(1989)
...
B
...
H
...
B
...
H
...
Halow,
S
...
Bailey, K
...
Churney, and R
...
Nuttall, The NBS tables of
chemical thermodynamic properties
...
Phys
...
Ref
...

R
...
Weast (ed
...
81
...

M
...
Ruzicka Jr
...
Majer, and E
...
Domalski
...
Published as J
...
Chem
...
Data,
Monograph no
...
American Institute of Physics, New York
(1996)
...
1 Adiabatic processes

Consider a stage in a reversible adiabatic expansion when the
pressure inside and out is p
...

Therefore, because for an adiabatic change (dq = 0) dU = dw + dq =
dw, we can equate these two expressions for dU and write
CV dT = −pdV
We are dealing with a perfect gas, so we can replace p by nRT/V and
obtain
CV dT

=−

T

nRdV
V

To integrate this expression we note that T is equal to Ti when V is
equal to Vi, and is equal to Tf when V is equal to Vf at the end of the
expansion
...
) Then, because
∫dx/x = ln x + constant, we obtain
CV ln

Tf
Ti

= −nR ln

Vf
Vi

Because ln(x /y) = −ln(y/x), this expression rearranges to
CV
nR

ln

Tf
Ti

= ln

Vi
Vf

With c = CV /nR we obtain (because ln x a = a ln x)
c

A Tf D
A Vi D
ln B E = ln B E
C Ti F
C Vf F
which implies that (Tf /Ti)c = (Vi /Vf) and, upon rearrangement,
eqn 2
...

The initial and final states of a perfect gas satisfy the perfect gas law
regardless of how the change of state takes place, so we can use
pV = nRT to write
piVi
pfVf

=

Ti
Tf

However, we have just shown that
A Vf D
=B E
Tf C Vi F
Ti

1/c

A Vf D
=B E
C Vi F

γ −1

where we use the definition of the heat capacity ratio where
γ = Cp,m/CV,m and the fact that, for a perfect gas, Cp,m – CV,m = R (the
molar version of eqn 2
...
Then we combine the two expressions, to
obtain
A Vf D
= ×B E
pf Vi C Vi F
pi

Vf

γ −1

A Vf D
E
=B
C Vi F

γ

which rearranges to piV γ = pfV γ , which is eqn 2
...

i
f

Further information 2
...
In the present problem we do this twice, first by
expressing Cp and CV in terms of their definitions and then by
inserting the definition H = U + pV:
A ∂H D
A ∂U D
E −B
E
Cp − CV = B
C ∂T F p C ∂T F V
A ∂U D
A ∂(pV) D
A ∂U D
E
E +B
E −B
=B
C ∂T F p C ∂T F p C ∂T F V
We have already calculated the difference of the first and third terms
on the right, and eqn 2
...
The
factor αV gives the change in volume when the temperature is raised,
and πT = (∂U/∂V)T converts this change in volume into a change in
internal energy
...

Collecting the two contributions gives
Cp − CV = α(p + πT)V

(2
...

At this point we can go further by using the result we prove in
Section 3
...
57)

We now transform the remaining partial derivative
...
8

The Euler chain relation states that, for a differentiable function
z = z(x,y),
A ∂y D A ∂x D A ∂z D
B E B E B E = −1
C ∂x F z C ∂z F y C ∂y F x
For instance, if z(x,y) = x2y,

70

2 THE FIRST LAW
A ∂y D
A ∂(z/x 2)D
d(1/x 2)
2z
=− 3
B E =B
E =z
dx
x
C ∂x F z C ∂x F z
A ∂(z/y)1/2 D
A ∂x D
1 dz1/2
1
=
E = 1/2
B E =B
dz
2(yz)1/2
C ∂z F y C ∂z F y y
A ∂z D
A ∂(x 2y) D
dy
B E =B
E = x2 = x2
dy
C ∂y F x C ∂y F x

Multiplication of the three terms together gives the result −1
...
However, the
‘reciprocal identity’ allows us to invert partial derivatives and to
write

Comment 2
...


Insertion of this relation into eqn 2
...
49
...
1 Provide mechanical and molecular definitions of work and heat
...
2 Consider the reversible expansion of a perfect gas
...
5 Explain the significance of the Joule and Joule–Thomson experiments
...


2
...
3 Explain the difference between the change in internal energy and the

2
...
2
...


baseline below T1 is at a different level from that above T2
...


2
...


compile a list of as many state functions as you can identify
...
Unless otherwise stated,
thermochemical data are for 298
...

2
...
0 m
on the surface of (a) the Earth and (b) the Moon (g = 1
...

2
...

2
...
As a result of the reaction, a piston is pushed out through 10 cm
against an external pressure of 1
...
Calculate the work done by the system
...
2(b) A chemical reaction takes place in a container of cross-sectional area

50
...
As a result of the reaction, a piston is pushed out through 15 cm
against an external pressure of 121 kPa
...

2
...
00 mol Ar is expanded isothermally at 0°C
from 22
...
8 dm3 (a) reversibly, (b) against a constant external
pressure equal to the final pressure of the gas, and (c) freely (against zero
external pressure)
...


2
...
00 mol He is expanded isothermally at 22°C

from 22
...
7 dm3 (a) reversibly, (b) against a constant external
pressure equal to the final pressure of the gas, and (c) freely (against zero
external pressure)
...

2
...
00 mol of perfect gas atoms, for which
3
CV,m = –R, initially at p1 = 1
...
Calculate the final pressure, ∆U, q, and w
...
4(b) A sample consisting of 2
...
Calculate the final pressure, ∆U, q, and w
...
5(a) A sample of 4
...
7 dm3 at 310 K
...
3 dm3
...

2
...
56 g occupies 18
...


(a) Calculate the work done when the gas expands isothermally against a

EXERCISES
constant external pressure of 7
...
5 dm3
...

2
...
00 mol H2O(g) is condensed isothermally and reversibly

to liquid water at 100°C
...
656 kJ mol−1
...

2
...
00 mol CH3OH(g) is condensed isothermally and

reversibly to liquid at 64°C
...
3 kJ mol−1
...

2
...
Calculate the work done by the system as a result of the
reaction
...
0 atm and the temperature 25°C
...
7(b) A piece of zinc of mass 5
...
Calculate the work done by the system as a result of the
reaction
...
1 atm and the temperature 23°C
...
8(a) The constant-pressure heat capacity of a sample of a perfect gas was
found to vary with temperature according to the expression Cp /(J K−1) = 20
...
3665(T/K)
...

2
...
17
+ 0
...
Calculate q, w, ∆U, and ∆H when the temperature is raised
from 0°C to 100°C (a) at constant pressure, (b) at constant volume
...
9(a) Calculate the final temperature of a sample of argon of mass 12
...
0 dm3 at 273
...
0 dm3
...
9(b) Calculate the final temperature of a sample of carbon dioxide of mass
16
...
15 K
to 2
...

2
...
45 g at 27
...
00 dm3
...
10(b) A sample of nitrogen of mass 3
...
0°C is allowed to expand
reversibly and adiabatically from 400 cm3 to 2
...
What is the work done
by the gas?
2
...
4 kPa and 1
...
0 dm3
...
4
...
11(b) Calculate the final pressure of a sample of water vapour that expands
reversibly and adiabatically from 87
...
0 dm3
...
3
...
12(a) When 229 J of energy is supplied as heat to 3
...
55 K
...

2
...
9 mol of gas molecules,

the temperature of the sample increases by 1
...
Calculate the molar heat
capacities at constant volume and constant pressure of the gas
...
13(a) When 3
...
25 atm, its
temperature increases from 260 K to 285 K
...
4 J K−1 mol−1, calculate q, ∆H, and ∆U
...
13(b) When 2
...
25 atm, its

temperature increases from 250 K to 277 K
...
11 J K−1 mol−1, calculate q, ∆H, and ∆U
...
14(a) A sample of 4
...
0
...
(The final pressure of the gas is not necessarily 600 Torr
...
14(b) A sample of 5
...
5 kPa
until the volume has increased by a factor of 4
...
Calculate q, w, ∆T, ∆U, and
∆H
...
5 kPa
...
15(a) A sample consisting of 1
...
8 J K−1 is initially at 3
...
It undergoes reversible adiabatic
expansion until its pressure reaches 2
...
Calculate the final volume and
temperature and the work done
...
15(b) A sample consisting of 1
...
8 J K−1 mol−1 is initially at 230 kPa and 315 K
...
Calculate the final
volume and temperature and the work done
...
16(a) A certain liquid has ∆ vapH 7 = 26
...
Calculate q, w, ∆H, and

∆U when 0
...


2
...
0 kJ mol−1
...
75 mol is vaporized at 260 K and 765 Torr
...
17(a) The standard enthalpy of formation of ethylbenzene is −12
...


Calculate its standard enthalpy of combustion
...
17(b) The standard enthalpy of formation of phenol is −165
...


Calculate its standard enthalpy of combustion
...
18(a) The standard enthalpy of combustion of cyclopropane is −2091 kJ

mol−1 at 25°C
...
The
enthalpy of formation of propene is +20
...
Calculate the enthalpy of
isomerization of cyclopropane to propene
...
18(b) From the following data, determine ∆f H 7 for diborane, B2H6(g), at

298 K:
(1) B2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g)
3
(2) 2 B(s) + – O2(g) → B2O3(s)
2
1
– O2(g) → H2O(g)
(3) H2(g) + 2

∆r H 7 = −1941 kJ mol−1
∆r H 7 = −2368 kJ mol−1
∆r H 7 = −241
...
19(a) When 120 mg of naphthalene, C10H8(s), was burned in a bomb

calorimeter the temperature rose by 3
...
Calculate the calorimeter
constant
...
19(b) When 2
...
35 K
...
By how much will the temperature rise when 135 mg of phenol,
C6H5OH(s), is burned in the calorimeter under the same conditions?
(∆cH 7(C14H10, s) = −7061 kJ mol−1
...
20(a) Calculate the standard enthalpy of solution of AgCl(s) in water from

the enthalpies of formation of the solid and the aqueous ions
...
20(b) Calculate the standard enthalpy of solution of AgBr(s) in water from
the enthalpies of formation of the solid and the aqueous ions
...
21(a) The standard enthalpy of decomposition of the yellow complex

H3NSO2 into NH3 and SO2 is +40 kJ mol−1
...

2
...
51 kJ mol−1 and that of diamond is −395
...

2
...

(1) H2(g) + Cl2(g) → 2 HCl(g)
(2) 2 H2(g) + O2(g) → 2 H2O(g)
(3) 4 HCl(g) + O2(g) → Cl2(g) + 2 H2O(g)

∆rH 7 = −184
...
64 kJ mol−1

72

2 THE FIRST LAW

2
...


(1) H2(g) + I2(s) → 2 HI(g)
(2) 2 H2(g) + O2(g) → 2 H2O(g)
(3) 4 HI(g) + O2(g) → 2 I2(s) + 2 H2O(g)

∆r H 7 = +52
...
64 kJ mol−1

2
...
Calculate ∆r H 7
...
23(b) For the reaction 2 C6H5COOH(s) + 13 O2(g) → 12 CO2(g) +
7

−1

7

6 H2O(g), ∆rU = −772
...
Calculate ∆r H
...
24(a) Calculate the standard enthalpies of formation of (a) KClO3(s) from

the enthalpy of formation of KCl, (b) NaHCO3(s) from the enthalpies of
formation of CO2 and NaOH together with the following information:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
NaOH(s) + CO2(g) → NaHCO3(s)

∆r H 7 = −89
...
5 kJ mol−1

2
...
5, together with the following
information:

2 NOCl(g) → 2 NO(g) + Cl2(g)

∆r H 7 = +75
...
25(a) Use the information in Table 2
...

2
...
5 to predict the standard reaction
enthalpy of 2 H2(g) + O2(g) → 2 H2O(l) at 100°C from its value at 25°C
...
26(a) From the data in Table 2
...
Assume
all heat capacities to be constant over the temperature range of interest
...
26(b) Calculate ∆r H 7 and ∆rU 7 at 298 K and ∆r H 7 at 348 K for the

hydrogenation of ethyne (acetylene) to ethene (ethylene) from the enthalpy of
combustion and heat capacity data in Tables 2
...
7
...

2
...
7 in the Data section
...
27(b) Calculate ∆r H 7 for the reaction NaCl(aq) + AgNO3(aq) → AgCl(s) +

NaNO3(aq) from the information in Table 2
...

2
...
2 kJ mol−1; first and second ionization enthalpies of Mg(g),
7
...
035 eV; dissociation enthalpy of Cl2(g), +241
...
78 eV; enthalpy of solution of MgCl2(s),
−150
...
7 kJ mol−1
...
28(b) Set up a thermodynamic cycle for determining the enthalpy of
hydration of Ca2+ ions using the following data: enthalpy of sublimation
of Ca(s), +178
...
7 kJ mol−1 and 1145 kJ mol−1; enthalpy of vaporization of bromine,
+30
...
9 kJ mol−1; electron
gain enthalpy of Br(g), −331
...
1 kJ mol−1; enthalpy of hydration of Br−(g), −337 kJ mol−1
...
29(a) When a certain freon used in refrigeration was expanded adiabatically

from an initial pressure of 32 atm and 0°C to a final pressure of 1
...
Calculate the Joule–Thomson coefficient, µ, at 0°C,
assuming it remains constant over this temperature range
...
29(b) A vapour at 22 atm and 5°C was allowed to expand adiabatically to a
final pressure of 1
...
Calculate the
Joule–Thomson coefficient, µ, at 5°C, assuming it remains constant over this
temperature range
...
30(a) For a van der Waals gas, πT = a/V 2
...
00 dm3 to 24
...
What are the values of q and w?
2
...
30(a) for argon, from an initial volume of 1
...
1 dm3 at 298 K
...
31(a) The volume of a certain liquid varies with temperature as

V = V′{0
...
9 × 10−4(T/K) + 1
...
Calculate its expansion coefficient, α, at 320 K
...
31(b) The volume of a certain liquid varies with temperature as

V = V′{0
...
7 × 10−4(T/K) + 1
...
Calculate its expansion coefficient, α, at 310 K
...
32(a) The isothermal compressibility of copper at 293 K is 7
...
Calculate the pressure that must be applied in order to increase its
density by 0
...

2
...
21 × 10−6 atm−1
...
08 per cent
...
33(a) Given that µ = 0
...
Calculate the energy that must be
supplied as heat to maintain constant temperature when 15
...

2
...
11 K atm−1 for carbon dioxide, calculate the value of
its isothermal Joule–Thomson coefficient
...
0 mol CO2 flows
through a throttle in an isothermal Joule–Thomson experiment and the
pressure drop is 55 atm
...
Note that 1 atm =
1
...
Unless otherwise stated, thermochemical data are for 298
...


Numerical problems
2
...
2
...
(a) Determine the
2
temperature at the points 1, 2, and 3
...
If a numerical answer cannot be obtained from
the information given, then write in +, −, 0, or ? as appropriate
...
2
...

2
...
727 g was placed
in a calorimeter and then ignited in the presence of excess oxygen
...
910 K
...
825 g of benzoic acid, for which the internal energy of
combustion is −3251 kJ mol−1, gave a temperature rise of 1
...
Calculate
the internal energy of combustion of d-ribose and its enthalpy of formation
...
9 The standard enthalpy of formation of the metallocene
bis(benzene)chromium was measured in a calorimeter
...
0 kJ mol−1
...
The constant-pressure molar
heat capacity of benzene is 136
...
67 J K−1 mol−1 as a gas
...
10‡ From the enthalpy of combustion data in Table 2
...

Predict ∆cH 7 for decane and compare to the known value
...
2
...
2 A sample consisting of 1
...
The heating was carried out in a container fitted with a piston
that was initially resting on the solid
...
0 atm
...
3 A sample consisting of 2
...
0 dm3

at 300 K
...
35 kJ of energy as heat its temperature
increases to 341 K
...

2
...
25 cm3 to 6
...
5 J
...
7 cm3 mol−1 to calculate w, q, and ∆H for this change of state
...
5 A sample of 1
...
00 atm
...

2
...
Account physically for the way in which the coefficients a
and b appear in the final expression
...
11 × 10−2 dm3 mol−1, and
(c) a = 4
...
The values selected exaggerate the
imperfections but give rise to significant effects on the indicator diagrams
...
0 dm3, n = 1
...

2
...
73 +
0
...
The corresponding expressions for C(s) and H2(g) are given in

2
...
(a) Use electronic structure
software to predict ∆cH 7 values for the alkanes methane through pentane
...
(b) Compare your estimated values with the
experimental values of ∆cH 7 (Table 2
...
(c) Test the extent to which the relation
∆cH 7 = k{(M/(g mol−1)}n holds and find the numerical values for k and n
...
12‡ When 1
...
0 cm3

of 0
...
397°C on account of the reaction:
H3O+(aq) + NaCH3CO2 · 3 H2O(s)
→ Na+(aq) + CH3COOH(aq) + 4 H2O(l)
...
0 J K−1 and the heat capacity density
of the acid solution is 4
...
Determine the standard enthalpy of
formation of the aqueous sodium cation
...

2
...
Kolesov et al
...
P
...
M
...
K
...
B
...
A
...
Chem
...
In one of their
runs, they found the standard specific internal energy of combustion to be
−36
...
15 K Compute ∆c H 7 and ∆ f H 7 of C60
...
14‡ A thermodynamic study of DyCl3 (E
...
P
...
S
...
Yu
...
Chem
...
0 m HCl)
(2) Dy(s) + 3 HCl(aq, 4
...
0 m HCl(aq)) + – H2(g)
2
1
1
– H2(g) + – Cl2(g) → HCl(aq, 4
...


* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady
...
06 kJ mol−1
∆r H 7 = −699
...
31 kJ mol−1

74

2 THE FIRST LAW

2
...
Moffat et al
...
K
...
F
...
W
...
Phys
...
95, 145 (1991)) report
∆f H 7(SiH2) = +274 kJ mol−1
...
3 kJ mol−1 and
∆ f H 7(Si2H6) = +80
...
16‡ Silanone (SiH2O) and silanol (SiH3OH) are species believed to be
important in the oxidation of silane (SiH4)
...
C
...
Darling and H
...
Schlegel
(J
...
Chem
...
3 kJ mol−1 and
∆f H 7(SiH3OH) = −282 kJ mol−1
...
3 kJ mol−1 (CRC Handbook (2004))
...
17 The constant-volume heat capacity of a gas can be measured by
observing the decrease in temperature when it expands adiabatically and
reversibly
...
A fluorocarbon gas was allowed to expand
reversibly and adiabatically to twice its volume; as a result, the temperature
fell from 298
...
44 K and its pressure fell from 202
...
840 kPa
...

2
...
00 mol of a van der Waals gas is compressed
3

3

from 20
...
0 dm at 300 K
...
2 kJ of work is done
on the gas
...
4 J K−1 mol−1,
a = 3
...
44 dm3 mol−1, calculate ∆H for the process
...
19 Take nitrogen to be a van der Waals gas with a = 1
...
0387 dm3 mol−1, and calculate ∆Hm when the pressure on the gas is
decreased from 500 atm to 1
...
For a van der Waals gas,
7
µ = {(2a/RT) − b}/Cp,m
...

2

Theoretical problems
2
...


2
...
(c) Let z = xy − y + ln x + 2
...


expressing (∂H/∂U)p as the ratio of two derivatives with respect to volume
and then using the definition of enthalpy
...
26 (a) Write expressions for dV and dp given that V is a function of p and

T and p is a function of V and T
...

2
...
19
...
0 mol Ar at 273 K (for data, see Table 1
...
Let the expansion be from 500 cm3 to 1000 cm3 in each case
...
28 Express the work of isothermal reversible expansion of a van der Waals
gas in reduced variables and find a definition of reduced work that makes the
overall expression independent of the identity of the gas
...

2
...
Will the temperature increase, decrease, or
remain the same?
2
2
...
(Hint
...
)

2
...
Calculate (∂T/∂p)V and confirm
that (∂T/∂p)V = 1/(∂p/∂T)V
...

2
...
Show, using Euler’s chain relation, that
κT R = α(Vm − b)
...
33 Given that µCp = T(∂V/∂T)p − V, derive an expression for µ in terms of

the van der Waals parameters a and b, and express it in terms of reduced
variables
...
0 atm, when the molar volume of the gas
is 24
...
Use the expression obtained to derive a formula for the
inversion temperature of a van der Waals gas in terms of reduced variables,
and evaluate it for the xenon sample
...
34 The thermodynamic equation of state (∂U/∂V)T = T(∂p/∂T)V − p was

quoted in the chapter
...

2
...
0 atm
...
22 (a) Express (∂CV /∂V)T as a second derivative of U and find its relation
to (∂U/∂V)T and (∂Cp /∂p)T as a second derivative of H and find its relation
to (∂H/∂p)T
...


heat capacities γ by cs = (γ RT/M)1/2
...
Calculate the speed of sound in argon at 25°C
...
23 (a) Derive the relation CV = −(∂U/∂V)T (∂V/∂T)U from the expression

2
...


2
...
Obtain expressions for (a) the
Joule–Thomson coefficient and (b) its constant-volume heat capacity
...
24 Starting from the expression Cp − CV = T(∂p/∂T)V (∂V/∂T)p, use the

appropriate relations between partial derivatives to show that
Cp − CV =

2
T(∂V/∂T)p

(∂V/∂T)T

Evaluate Cp − CV for a perfect gas
...
25 (a) By direct differentiation of H = U + pV, obtain a relation between

(∂H/∂U)p and (∂U/∂V)p
...
38 It is possible to see with the aid of a powerful microscope that a long
piece of double-stranded DNA is flexible, with the distance between the ends
of the chain adopting a wide range of values
...
It is convenient to visualize a long piece

PROBLEMS
of DNA as a freely jointed chain, a chain of N small, rigid units of length l
that are free to make any angle with respect to each other
...
You will now explore the work associated with extending a
DNA molecule
...
Systems showing this
behaviour are said to obey Hooke’s law
...
Draw a graph
of your conclusion
...
In this case, the restoring
force of a chain extended by x = nl is given by
F=

A1+νD
ln B
E
2l
C1−νF

kT

ν = n/N

where k = 1
...

(i) What are the limitations of this model? (ii) What is the magnitude of the
force that must be applied to extend a DNA molecule with N = 200 by 90 nm?
(iii) Plot the restoring force against ν, noting that ν can be either positive or
negative
...
(v) Calculate the work of extending a DNA molecule from
ν = 0 to ν = 1
...
Hint
...
The task can
be accomplished easily with mathematical software
...
See Appendix 2 for a review of series expansions of functions
...

2
...
Some nutritionists recommend diets that are largely devoid of
carbohydrates, with most of the energy needs being met by fats
...
A –-cup serving of pasta contains
4
40 g of carbohydrates
...
40 An average human produces about 10 MJ of heat each day through
metabolic activity
...
What mass of
water should be evaporated each day to maintain constant temperature?
2
...
Sucrose, or table sugar, is a complex sugar with molecular formula
C12H22O11 that consists of a glucose unit covalently bound to a fructose unit
(a water molecule is given off as a result of the reaction between glucose and
fructose to form sucrose)
...
5 g is burned in air
...
5 g
...
(d) To what height could you climb on the energy a cube
provides assuming 25 per cent of the energy is available for work?
2
...
Muscle cells
may be deprived of O2 during vigorous exercise and, in that case, one

75

molecule of glucose is converted to two molecules of lactic acid
(CH3CH (OH)COOH) by a process called anaerobic glycolysis (see
Impact I7
...
(a) When 0
...
793 K
...
(b) What is the biological advantage (in kilojoules
per mole of energy released as heat) of complete aerobic oxidation
compared with anaerobic glycolysis to lactic acid?
2
...
Describe how you would use differential scanning
calorimetry to determine the mole percentage composition of P in the
allegedly impure sample
...
44‡ Alkyl radicals are important intermediates in the combustion
and atmospheric chemistry of hydrocarbons
...
(P
...

Seakins, M
...
Pilling, J
...
Niiranen, D
...
N
...
Phys
...
96, 9847 (1992)) report ∆ f H 7 for a variety of alkyl radicals
in the gas phase, information that is applicable to studies of pyrolysis and
oxidation reactions of hydrocarbons
...
Use the following set of data to compute the standard reaction
enthalpies for three possible fates of the tert-butyl radical, namely,
(a) tert-C4H9 → sec-C4H9, (b) tert-C4H9 → C3H6 + CH3, (c) tert-C4H9 →
C2H4 + C2H5
...
0

+67
...
3

2
...
0–3
...
0°C its best estimate
...
0°C, 2
...
5°C given that the volume of the Earth’s oceans is 1
...

2
...
One such
alternative is 2,2-dichloro-1,1,1-trifluoroethane (refrigerant 123)
...
A
...
McLinden, J
...
Chem
...
Data 23, 7
(1994)), from which properties such as the Joule–Thomson coefficient µ
can be computed
...
00 bar and 50°C given that (∂H/∂p)T
= −3
...
0 J K−1 mol−1
...
0 mol
of this refrigerant from 1
...
5 bar at 50°C
...
47‡ Another alternative refrigerant (see preceding problem) is 1,1,1,2tetrafluoroethane (refrigerant HFC-134a)
...
TillnerRoth and H
...
Baehr, J
...
Chem
...
Data 23, 657 (1994)), from which
properties such as the Joule–Thomson coefficient µ can be computed
...
100 MPa and 300 K from the following data (all referring
to 300 K):

p/MPa

0
...
100

0
...
48

426
...
76

(The specific constant-pressure heat capacity is 0
...
)
(b) Compute µ at 1
...
80

1
...
2

Specific enthalpy/(kJ kg−1)

461
...
12

456
...
0392 kJ K−1 kg−1
...
1 The dispersal of energy
3
...
1 Impact on engineering:

Refrigeration
3
...
4 The Third Law of

thermodynamics

The Second Law
The purpose of this chapter is to explain the origin of the spontaneity of physical and chemical change
...
The chapter also
introduces a major subsidiary thermodynamic property, the Gibbs energy, which lets us express the spontaneity of a process in terms of the properties of a system
...
As we
began to see in Chapter 2, one application of thermodynamics is to find relations between
properties that might not be thought to be related
...
We also see
how to derive expressions for the variation of the Gibbs energy with temperature and pressure and how to formulate expressions that are valid for real gases
...


Concentrating on the system
3
...
6 Standard reaction Gibbs

energies
Combining the First and
Second Laws
3
...
8 Properties of the internal

energy
3
...
1: The Born
equation
Further information 3
...
A gas expands to fill the available
volume, a hot body cools to the temperature of its surroundings, and a chemical reaction runs in one direction rather than another
...
A gas can be confined to a smaller volume, an object
can be cooled by using a refrigerator, and some reactions can be driven in reverse
(as in the electrolysis of water)
...
An important point, though, is that
throughout this text ‘spontaneous’ must be interpreted as a natural tendency that may
or may not be realized in practice
...

The recognition of two classes of process, spontaneous and non-spontaneous, is
summarized by the Second Law of thermodynamics
...
One statement was formulated by Kelvin:
No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work
...
3
...
All real heat engines have both a hot source and a cold sink; some energy is
always discarded into the cold sink as heat and not converted into work
...
1 THE DISPERSAL OF ENERGY
statement is a generalization of another everyday observation, that a ball at rest on a
surface has never been observed to leap spontaneously upwards
...


Hot source

Flow of
energy

The direction of spontaneous change
What determines the direction of spontaneous change? It is not the total energy of the
isolated system
...

Is it perhaps the energy of the system that tends towards a minimum? Two arguments show that this cannot be so
...
Secondly, if the energy
of a system does happen to decrease during a spontaneous change, the energy of its
surroundings must increase by the same amount (by the First Law)
...

When a change occurs, the total energy of an isolated system remains constant but
it is parcelled out in different ways
...


77

Heat

Engine

Work

The Kelvin statement of the Second
Law denies the possibility of the process
illustrated here, in which heat is changed
completely into work, there being no other
change
...


Fig
...
1

3
...
The ball does not rise as
high after each bounce because there are inelastic losses in the materials of the ball and
floor
...
The direction of
spontaneous change is towards a state in which the ball is at rest with all its energy dispersed into random thermal motion of molecules in the air and of the atoms of the
virtually infinite floor (Fig
...
2)
...
For
bouncing to begin, something rather special would need to happen
...
This accumulation requires a spontaneous localization of
energy from the myriad vibrations of the atoms of the floor into the much smaller
number of atoms that constitute the ball (Fig
...
3)
...
The localization of random, disorderly motion as concerted, ordered
motion is so unlikely that we can dismiss it as virtually impossible
...
This
principle accounts for the direction of change of the bouncing ball, because its energy
is spread out as thermal motion of the atoms of the floor
...
A gas does not contract spontaneously because

1
Concerted motion, but on a much smaller scale, is observed as Brownian motion, the jittering motion of
small particles suspended in water
...
On
each bounce some of its energy is degraded
into the thermal motion of the atoms
of the floor, and that energy disperses
...


Fig
...
2

78

3 THE SECOND LAW

(a)

(b)

The molecular interpretation of the
irreversibility expressed by the Second Law
...
(b) For the ball to fly upwards,
some of the random vibrational motion
would have to change into coordinated,
directed motion
...


Fig
...
3

to do so the random motion of its molecules, which spreads out the distribution of
kinetic energy throughout the container, would have to take them all into the same
region of the container, thereby localizing the energy
...
An object does not spontaneously become warmer
than its surroundings because it is highly improbable that the jostling of randomly
vibrating atoms in the surroundings will lead to the localization of thermal motion
in the object
...

It may seem very puzzling that the spreading out of energy and matter, the collapse
into disorder, can lead to the formation of such ordered structures as crystals or proteins
...

3
...

The internal energy is a state function that lets us assess whether a change is permissible: only those changes may occur for which the internal energy of an isolated system
remains constant
...
We shall see that the entropy (which we shall define shortly,
but is a measure of the energy dispersed in a process) lets us assess whether one state
is accessible from another by a spontaneous change
...

The Second Law of thermodynamics can be expressed in terms of the entropy:
The entropy of an isolated system increases in the course of a spontaneous change:
∆S tot > 0
where Stot is the total entropy of the system and its surroundings
...

(a) The thermodynamic definition of entropy

The thermodynamic definition of entropy concentrates on the change in entropy,
dS, that occurs as a result of a physical or chemical change (in general, as a result of
a ‘process’)
...
As we
have remarked, heat stimulates random motion in the surroundings
...

The thermodynamic definition of entropy is based on the expression
dS =

dqrev
T

[3
...
2)

3
...

Example 3
...

Method The definition of entropy instructs us to find the energy supplied as heat

for a reversible path between the stated initial and final states regardless of the
actual manner in which the process takes place
...
2
...

The work of reversible isothermal expansion was calculated in Section 2
...

Answer Because the temperature is constant, eqn 3
...
11, we know that
qrev = −wrev = nRT ln

Vf
Vi

It follows that
∆S = nR ln

Vf
Vi

As an illustration of this formula, when the volume occupied by 1
...
00 mol) × (8
...
76 J K−1
A note on good practice According to eqn 3
...
Entropy is an extensive property
...
2 The molar entropy is an intensive property
...
1 Calculate the change in entropy when the pressure of a perfect gas is
changed isothermally from pi to pf
...
1 to formulate an expression for the change in
entropy of the surroundings, ∆Ssur
...
The surroundings consist of a reservoir of constant volume, so
the energy supplied to them by heating can be identified with the change in their
2

The units of entropy are the same as those of the gas constant, R, and molar heat capacities
...
3 The internal energy is a state function, and dUsur is an exact
differential
...
The same remarks therefore apply to dqsur, to which dUsur is
equal
...
1 to write
dSsur =

dqsur,rev
Tsur

=

dqsur

(3
...
3b)

Tsur

That is, regardless of how the change is brought about in the system, reversibly or
irreversibly, we can calculate the change of entropy of the surroundings by dividing
the heat transferred by the temperature at which the transfer takes place
...
3 makes it very simple to calculate the changes in entropy of the surroundings that accompany any process
...
4)

This expression is true however the change takes place, reversibly or irreversibly, provided no local hot spots are formed in the surroundings
...
If hot spots do form, then the localized
energy may subsequently disperse spontaneously and hence generate more entropy
...
1 Calculating the entropy change in the surroundings

To calculate the entropy change in the surroundings when 1
...
7
...
Therefore,
∆Ssur =

2
...

Self-test 3
...
00 mol

N2O4(g) is formed from 2
...

[−192 J K−1]

Molecular interpretation 3
...
The continuous thermal agitation that molecules
3
Alternatively, the surroundings can be regarded as being at constant pressure, in which case we could
equate dqsur to dHsur
...
One particular molecule may be in one low energy state at one
instant, and then be excited into a high energy state a moment later
...

Only the lowest energy state is occupied at T = 0
...
3
...
Nevertheless, whatever the temperature, there is always a higher population in a state of low energy than one of high
energy
...
These remarks were summarized quantitatively by the Austrian physicist Ludwig Boltzmann in the Boltzmann distribution:
Ni =

81

Energy

3
...
381 × 10−23 J K−1 and Ni is the number of molecules in a sample of
N molecules that will be found in a state with an energy Ei when it is part of a system in thermal equilibrium at a temperature T
...

Boltzmann also made the link between the distribution of molecules over energy
levels and the entropy
...
5)

where W is the number of microstates, the ways in which the molecules of a system
can be arranged while keeping the total energy constant
...
When we measure the properties of a system, we are measuring an average
taken over the many microstates the system can occupy under the conditions of
the experiment
...

Equation 3
...
We see that if W = 1, which
corresponds to one microstate (only one way of achieving a given energy, all
molecules in exactly the same state), then S = 0 because ln 1 = 0
...
But, if more
molecules can participate in the distribution of energy, then there are more
microstates for a given total energy and the entropy is greater than when the energy
is confined so a smaller number of molecules
...

The molecular interpretation of entropy advanced by Boltzmann also suggests
the thermodynamic definition given by eqn 3
...
To appreciate this point, consider
that molecules in a system at high temperature can occupy a large number of the
available energy levels, so a small additional transfer of energy as heat will lead to a
relatively small change in the number of accessible energy levels
...
(a) At low temperatures, only the
lowest states are significantly populated;
(b) at high temperatures, there is
significant population in high-energy states
as well as in low-energy states
...


Fig
...
4

3 THE SECOND LAW

Final
state

Pressure, p

82

Initial
state

number of microstates does not increase appreciably and neither does the entropy
of the system
...
Hence, the change
in entropy upon heating will be greater when the energy is transferred to a cold
body than when it is transferred to a hot body
...
1
...


(b) The entropy as a state function

Fig
...
5

Entropy is a state function
...
To do so, it is sufficient to prove that the integral of
eqn 3
...
3
...
That is, we need to show that

Ώ

dqrev
T

=0

(3
...
There are three steps in
the argument:
1
...
6 is true for a special cycle (a ‘Carnot cycle’) involving a
perfect gas
...
Then to show that the result is true whatever the working substance
...
Finally, to show that the result is true for any cycle
...
3
...
Reversible isothermal expansion from A to B at Th; the entropy change is qh/Th,
where qh is the energy supplied to the system as heat from the hot source
...
Reversible adiabatic expansion from B to C
...
In the course of this expansion, the temperature falls
from Th to Tc, the temperature of the cold sink
...
Reversible isothermal compression from C to D at Tc
...

4
...
No energy enters the system as
heat, so the change in entropy is zero
...

The total change in entropy around the cycle is

The basic structure of a Carnot
cycle
...

Step 2 is a reversible adiabatic expansion in
which the temperature falls from Th to Tc
...

Fig
...
6

ΏdS = T + T
qh

qc

h

c

However, we show in Justification 3
...
7)rev

Substitution of this relation into the preceding equation gives zero on the right, which
is what we wanted to prove
...
2 ENTROPY

83

Justification 3
...
7 lie on the
same adiabat in Fig
...
6
...
1, for a perfect gas:
VB

qh = nRTh ln

qc = nRTc ln

VA

VD
VC

From the relations between temperature and volume for reversible adiabatic processes (eqn 2
...
7
...
7 applies to any material, not just a
perfect gas (which is why, in anticipation, we have not labelled it with a °)
...
8]

|w |

The definition implies that, the greater the work output for a given supply of heat
from the hot reservoir, the greater is the efficiency of the engine
...
3
...
9)

Cold sink

(Remember that qc < 0
...
7 that

εrev = 1 −

Tc
Th

Suppose an energy qh (for example,
20 kJ) is supplied to the engine and qc is lost
from the engine (for example, qc = −15 kJ)
and discarded into the cold reservoir
...
The
efficiency is the work done divided by the
energy supplied as heat from the hot
source
...
3
...
10)rev

Now we are ready to generalize this conclusion
...

To see the truth of this statement, suppose two reversible engines are coupled together
and run between the same two reservoirs (Fig
...
8)
...
Initially, suppose that

3 THE SECOND LAW
Th

Hot source

Cancel

qh

q´
h

Survive

Pressure, p

84

B

A

qc

qc

w´

Volume, V

Tc

Cold sink

(a) The demonstration of the
equivalence of the efficiencies of all
reversible engines working between the
same thermal reservoirs is based on the
flow of energy represented in this diagram
...


Hot source

Fig
...
8

A general cycle can be divided into
small Carnot cycles
...

Paths cancel in the interior of the
collection, and only the perimeter, an
increasingly good approximation to the
true cycle as the number of cycles increases,
survives
...


Fig
...
9

(a)

q - q´
w´
(b)

engine A is more efficient than engine B and that we choose a setting of the controls
that causes engine B to acquire energy as heat qc from the cold reservoir and to release
a certain quantity of energy as heat into the hot reservoir
...
The net result is that the cold reservoir
is unchanged, work has been done, and the hot reservoir has lost a certain amount of
energy
...
In molecular terms, the random
thermal motion of the hot reservoir has been converted into ordered motion characteristic of work
...
It follows that
the relation between the heat transfers and the temperatures must also be independent of the working material, and therefore that eqn 3
...

For the final step in the argument, we note that any reversible cycle can be approximated as a collection of Carnot cycles and the cyclic integral around an arbitrary path
is the sum of the integrals around each of the Carnot cycles (Fig
...
9)
...

The entropy change around each individual cycle is zero (as demonstrated above), so
the sum of entropy changes for all the cycles is zero
...
Therefore, all the entropy changes cancel except
for those along the perimeter of the overall cycle
...
2 ENTROPY
In the limit of infinitesimal cycles, the non-cancelling edges of the Carnot cycles
match the overall cycle exactly, and the sum becomes an integral
...
6 then
follows immediately
...


Th
Hot sink

IMPACT ON ENGINEERING

Entropy
change

qc

I3
...
First, we consider the work required to cool an
object, and refer to Fig
...
10
...
7 to express this result in terms of the temperatures alone,
which is possible if the transfer is performed reversibly
...
For a refrigerator
withdrawing heat from ice-cold water (Tc = 273 K) in a typical environment (Th =
293 K), c = 14, so, to remove 10 kJ (enough to freeze 30 g of water), requires transfer
of at least 0
...
Practical refrigerators, of course, have a lower coefficient of
performance
...
No thermal insulation is perfect, so there is always a flow of energy as heat into
the sample at a rate proportional to the temperature difference
...

Because |qc | is removed from the cold source, and the work |w| is added to the
energy stream, the energy deposited as heat in the hot sink is |qh | = |qc | + |w|
...
To generate more entropy,
energy must be added to the stream that enters the warm sink
...
The outcome is expressed as the coefficient
of performance, c:
c=

85

× A(Th − Tc) = A ×

(Th − Tc)2
Tc

Entropy
change

w
qc
Tc
Cold source
(b)

Fig
...
10 (a) The flow of energy as heat
from a cold source to a hot sink is not
spontaneous
...

(b) The process becomes feasible if work is
provided to add to the energy stream
...


86

3 THE SECOND LAW
We see that the power increases as the square of the temperature difference we are trying to maintain
...

(c) The thermodynamic temperature

Suppose we have an engine that is working reversibly between a hot source at a temperature Th and a cold sink at a temperature T; then we know from eqn 3
...
1

The triple point of a substance
represents the set of conditions at which
the three phases coexist in equilibrium
...
16 K and 611 Pa
...
2 for
details
...
11)

This expression enabled Kelvin to define the thermodynamic temperature scale
in terms of the efficiency of a heat engine
...
The size of the unit is entirely arbitrary, but on the Kelvin scale is
defined by setting the temperature of the triple point of water as 273
...

Then, if the heat engine has a hot source at the triple point of water, the temperature
of the cold sink (the object we want to measure) is found by measuring the efficiency
of the engine
...

(d) The Clausius inequality

We now show that the definition of entropy is consistent with the Second Law
...
That is, −dwrev ≥ −dw, or dw − dwrev ≥ 0
...
Now we use the thermodynamic definition of the entropy (eqn 3
...
When the same quantity of
energy enters a cooler reservoir, the
entropy increases by a larger amount
...

Relative changes in entropy are indicated
by the sizes of the arrows
...
3
...
12)

T

This expression is the Clausius inequality
...
5
...
2 Spontaneous cooling

|dq|

Tc

dq

Consider the transfer of energy as heat from one system—the hot source—at a
temperature Th to another system—the cold sink—at a temperature Tc (Fig
...
11)
...
When |dq | enters the cold sink the Clausius inequality implies that
dS ≥ dqc /Tc (with dqc > 0)
...
Hence, cooling (the transfer of heat
from hot to cold) is spontaneous, as we know from experience
...
3 ENTROPY CHANGES ACCOMPANYING SPECIFIC PROCESSES
We now suppose that the system is isolated from its surroundings, so that dq = 0
...
3 Entropy changes accompanying specific processes

DS /nR

3

and we conclude that in an isolated system the entropy cannot decrease when a spontaneous change occurs
...


2

We now see how to calculate the entropy changes that accompany a variety of basic
processes
...
1 that the change in entropy of a perfect gas that expands
isothermally from Vi to Vf is
∆S = nR ln

Vf

°

(3
...
The logarithmic dependence of entropy on
volume is illustrated in Fig
...
12
...
For any process dqsur = −dq, and for a reversible change we use the expression
in Example 3
...
3b
∆Ssur =

qsur
T

=−

qrev
T

= −nR ln

Vf
Vi

°
(3
...
If the isothermal expansion occurs freely (w = 0) and irreversibly, then q = 0 (because ∆U = 0)
...
13 itself:
∆Stot = nR ln

Vf
Vi

(3
...

(b) Phase transition

The degree of dispersal of matter and energy changes when a substance freezes or boils
as a result of changes in the order with which the molecules pack together and the
extent to which the energy is localized or dispersed
...
For example, when a substance
vaporizes, a compact condensed phase changes into a widely dispersed gas and we can
expect the entropy of the substance to increase considerably
...

Consider a system and its surroundings at the normal transition temperature,
Ttrs, the temperature at which two phases are in equilibrium at 1 atm
...
At the transition temperature,
any transfer of energy as heat between the system and its surroundings is reversible

1

10

20

30

Vf /Vi
Fig
...
12 The logarithmic increase in
entropy of a perfect gas as it expands
isothermally
...
00 mol CO2(g) from
0
...
010 m3 at 298 K, treated
as a van der Waals gas
...
1* Standard entropies (and temperatures) of phase transitions,
∆ trsS 7/(J K−1 mol−1)
Fusion (at Tf)

Vaporization (at Tb)

Argon, Ar

14
...
8 K)

74
...
3 K)

Benzene, C6H6

38
...
19 (at 353 K)

Water, H2O

22
...
15 K)

Helium, He

109
...
15 K)

4
...
9 (at 4
...


Synoptic Table 3
...
8

θ b /°C
80
...
2

Carbon tetrachloride

30

76
...
8

Cyclohexane

30
...
7

85
...
7

Methane

8
...
7

−60
...
9

−161
...
2

100
...
1

* More values are given in the Data section
...
Because at constant pressure
q = ∆ trs H, the change in molar entropy of the system is4
∆ trsS =

∆ trsH
Ttrs

(3
...
This decrease in entropy is consistent with localization of
matter and energy that accompanies the formation of a solid from a liquid or a liquid
from a gas
...

Table 3
...
Table 3
...

An interesting feature of the data is that a wide range of liquids give approximately the
same standard entropy of vaporization (about 85 J K−1 mol−1): this empirical observation is called Trouton’s rule
...
2 Trouton’s rule

The explanation of Trouton’s rule is that a comparable change in volume occurs
(with an accompanying change in the number of accessible microstates) when any
liquid evaporates and becomes a gas
...

Recall from Section 2
...

4

3
...
As a result, there
is a greater dispersal of energy and matter when the liquid turns into a vapour than
would occur for a liquid in which molcular motion is less restricted
...
Hydrogen bonds tend to organize
the molecules in the liquid so that they are less random than, for example, the
molecules in liquid hydrogen sulfide (in which there is no hydrogen bonding)
...
A part of the reason
is that the entropy of the gas itself is slightly low (186 J K−1 mol−1 at 298 K); the
entropy of N2 under the same conditions is 192 J K−1 mol−1
...


Illustration 3
...
To predict the standard molar enthalpy of vaporization of
bromine given that it boils at 59
...
4 K) × (85 J K−1 mol−1) = +2
...
45 kJ mol−1
...
3 Predict the enthalpy of vaporization of ethane from its boiling point,

[16 kJ mol−1]

−88
...


(c) Heating

We can use eqn 3
...
17)

We shall be particularly interested in the entropy change when the system is subjected
to constant pressure (such as from the atmosphere) during the heating
...
22), dqrev = CpdT provided the
system is doing no non-expansion work
...
18)

The same expression applies at constant volume, but with Cp replaced by CV
...
19)

Ti

Calculate the entropy change when argon at 25°C and 1
...
500 dm3 is allowed to expand to 1
...


1

10

T

= S(Ti) + Cp ln

Example 3
...
The logarithmic dependence
of entropy on temperature is illustrated in Fig
...
13
...
3
...
Different curves
correspond to different values of the
constant-volume heat capacity (which is
assumed constant over the temperature
range) expressed as CV,m/R
...


Method Because S is a state function, we are free to choose the most convenient
path from the initial state
...
The entropy change in the first step is given by eqn 3
...
19 (with CV
in place of Cp)
...
The heat capacity at constant volume is given by the equipartition
3
theorem as –R
...
7, converting to the
value at constant volume by using the relation Cp,m − CV,m = R
...
13

∆S(Step 1) =

A piVi D
Vf piVi Vf
× R ln =
ln
C RTi F
Vi
Ti
Vi

The entropy change in the second step, from 298 K to 373 K at constant volume, is
∆S(Step 2) =

A piVi D 3
Tf piVi A Tf D
ln
× –R ln =
2
C RTi F
C Ti F
Ti
Ti

3/2

The overall entropy change, the sum of these two changes, is

A Tf D
∆S =
ln +
ln
C Ti F
Ti
Vi
Ti
piVi

Vf

piVi

3/2

1 V A T D 3/2 5
f
f
6
=
ln 2
Ti
Vi C Ti F 7
3
piVi

At this point we substitute the data and obtain (by using 1 Pa m3 = 1 J)
∆S =

(1
...
500 × 10−3 m3)
298 K

1 1
...
500 C 298F 7

= +0
...

Self-test 3
...
0500 dm3 and cooled to −25°C
...
44 J K−1]

The entropy of a system at a temperature T is related to its entropy at T = 0 by measuring its heat capacity Cp at different temperatures and evaluating the integral in eqn
3
...
For example, if a substance melts at Tf
and boils at Tb, then its entropy above its boiling temperature is given by
Tf

0
Tb

Tf

Cp(s)dT
T
Cp(1)dT
T

+
+

Cp /T

∆fusH
Tf
∆ vapH
Tb

Ύ

T

+

Liquid

Ύ
+Ύ

S(T) = S(0) +

Debye
approximation

(a)

Solid

Cp(g)dT

Tb

T

(3
...
The former
procedure is illustrated in Fig
...
14: the area under the curve of Cp /T against T is the
integral required
...

One problem with the determination of entropy is the difficulty of measuring heat
capacities near T = 0
...
1), and this dependence
is the basis of the Debye extrapolation
...
That fit
determines the value of a, and the expression Cp = aT 3 is assumed valid down to T = 0
...
4 Calculating a standard molar entropy

The standard molar entropy of nitrogen gas at 25°C has been calculated from the
following data:
Debye extrapolation
Integration, from 10 K to 35
...
61 K
Integration, from 35
...
14 K
Fusion at 63
...
14 K to 77
...
32 K
Integration, from 77
...
15 K
Correction for gas imperfection
Total

7
S m/(J K−1 mol−1)
1
...
25
6
...
38
11
...
41
72
...
20
0
...
06

Therefore,
Sm(298
...
1 J K−1 mol−1

Example 3
...
2 K is 0
...
What is its molar entropy at that temperature?

91

Boil

(d) The measurement of entropy

Melt

3
...
3
...
(a) The variation of
Cp /T with the temperature for a sample
...


Exploration Allow for the
temperature dependence of the heat
capacity by writing C = a + bT + c/T 2, and
plot the change in entropy for different
values of the three coefficients (including
negative values of c)
...
18 to calculate the
entropy at a temperature T in terms of the entropy at T = 0 and the constant a
...

Answer The integration required is

Ύ

T

S(T) = S(0) +

0

aT 3dT
T

Ύ T dT = S(0) + –aT
T

= S(0) + a

2

1
3

3

0

However, because aT 3 is the heat capacity at the temperature T,
1
S(T) = S(0) + – Cp(T)
3

from which it follows that
Sm(10 K) = Sm(0) + 0
...
5 For metals, there is also a contribution to the heat capacity from the
electrons which is linearly proportional to T when the temperature is low
...

[S(T) = S(0) + Cp(T)]

3
...
The localization of matter and the
absence of thermal motion suggest that such materials also have zero entropy
...

(a) The Nernst heat theorem

The experimental observation that turns out to be consistent with the view that the
entropy of a regular array of molecules is zero at T = 0 is summarized by the Nernst
heat theorem:
The entropy change accompanying any physical or chemical transformation
approaches zero as the temperature approaches zero: ∆S → 0 as T → 0 provided all
the substances involved are perfectly crystalline
...
5 Using the Nernst heat theorem

Consider the entropy of the transition between orthorhombic sulfur, S(α), and
monoclinic sulfur, S(β), which can be calculated from the transition enthalpy
(−402 J mol−1) at the transition temperature (369 K):
∆trsS = Sm(α) − Sm(β) =

(−402 J mol−1)
369 K

= −1
...
It is found that Sm(α) = Sm(α,0) + 37 J K−1 mol−1

3
...
These two values imply that at the transition
temperature
∆ trsS = Sm(α,0) − Sm(β,0) = −1 J K−1 mol−1
On comparing this value with the one above, we conclude that Sm(α,0) − Sm(β,0)
≈ 0, in accord with the theorem
...
This conclusion is summarized by the Third Law
of thermodynamics:
The entropy of all perfect crystalline substances is zero at T = 0
...
The molecular interpretation of entropy, however, justifies
the value S = 0 at T = 0
...
3 The statistical view of the Third Law of thermodynamics

We saw in Molecular interpretation 3
...
In most cases,
W = 1 at T = 0 because there is only one way of achieving the lowest total energy:
put all the molecules into the same, lowest state
...
In certain cases, though, W may differ
from 1 at T = 0
...
For instance, for a diatomic molecule AB
there may be almost no energy difference between the arrangements
...
and
...
, so W > 1 even at T = 0
...
Ice has a residual entropy of 3
...
It
stems from the arrangement of the hydrogen bonds between neighbouring water
molecules: a given O atom has two short O-H bonds and two long O···H bonds to
its neighbours, but there is a degree of randomness in which two bonds are short
and which two are long
...
3* Standard
Third-Law entropies at 298 K
7
S m /(J K−1 mol−1)

Solids
Graphite, C(s)
Diamond, C(s)

5
...
4

Sucrose, C12H22O11(s)
Iodine, I2(s)

(b) Third-Law entropies

Entropies reported on the basis that S(0) = 0 are called Third-Law entropies (and
often just ‘entropies’)
...
A list of values at 298 K is
given in Table 3
...

The standard reaction entropy, ∆rS 7, is defined, like the standard reaction enthalpy, as the difference between the molar entropies of the pure, separated products
and the pure, separated reactants, all substances being in their standard states at the
specified temperature:
∆rS 7 =

7
7
∑νS m − ∑νS m

Products

(3
...

Standard reaction entropies are likely to be positive if there is a net formation of gas in
a reaction, and are likely to be negative if there is a net consumption of gas
...
2
116
...
3

Water, H2O(l)

69
...
0

Gases
Methane, CH4(g)

186
...
7

Hydrogen, H2(g)

130
...
2

Ammonia, NH3(g)

126
...


94

3 THE SECOND LAW
Illustration 3
...
7 of the Data Section to write
1 7
7
7
∆rS 7 = S m(H2O, l) − {S m(H2, g) + – S m(O2, g)}
2
1
= 69
...
7 + – (205
...
4 J K mol

The negative value is consistent with the conversion of two gases to a compact liquid
...

Self-test 3
...


[−243 J K−1 mol−1]

Just as in the discussion of enthalpies in Section 2
...
22]
5

The values based on this choice are listed in Table 2
...
Because the
entropies of ions in water are values relative to the hydrogen ion in water, they may be
either positive or negative
...
For instance, the standard molar entropy of Cl−(aq) is +57
J K−1 mol−1 and that of Mg2+(aq) is −128 J K−1 mol−1
...
Small, highly charged ions induce local structure in the surrounding water, and the disorder of the solution is decreased more than
in the case of large, singly charged ions
...
The negative
value indicates that the proton induces order in the solvent
...
We have seen
that it is always very simple to calculate the entropy change in the surroundings, and
we shall now see that it is possible to devise a simple method for taking that contribution into account automatically
...
1, the entropies of ions in solution are actually partial molar entropies, for their values include the consequences of their presence on the organization of the
solvent molecules around them
...
5 THE HELMHOLTZ AND GIBBS ENERGIES
and simplifies discussions
...

3
...

When a change in the system occurs and there is a transfer of energy as heat between
the system and the surroundings, the Clausius inequality, eqn 3
...
23)

We can develop this inequality in two ways according to the conditions (of constant
volume or constant pressure) under which the process occurs
...
Then, in the absence of non-expansion
work, we can write dqV = dU; consequently
dS −

dU
T

≥0

(3
...
The inequality is easily rearranged to
TdS ≥ dU

(constant V, no additional work)6

(3
...
26)

where the subscripts indicate the constant conditions
...
26 expresses the criteria for spontaneous change in terms of properties
relating to the system
...
That statement is essentially the content of the Second Law
...
Do
not interpret this criterion as a tendency of the system to sink to lower energy
...

When energy is transferred as heat at constant pressure, and there is no work other
than expansion work, we can write dqp = dH and obtain
TdS ≥ dH

(constant p, no additional work)

(3
...
28)

The interpretations of these inequalities are similar to those of eqn 3
...
The entropy
of the system at constant pressure must increase if its enthalpy remains constant (for
6

Recall that ‘additional work’ is work other than expansion work
...
Alternatively, the
enthalpy must decrease if the entropy of the system is constant, for then it is essential
to have an increase in entropy of the surroundings
...
25 and 3
...
One is the Helmholtz energy, A, which is defined as
A = U − TS

[3
...
30]

All the symbols in these two definitions refer to the system
...
31)

When we introduce eqns 3
...
27, respectively, we obtain the criteria of spontaneous change as
(a) dAT,V ≤ 0

(b) dGT,p ≤ 0

(3
...
They are developed in subsequent sections and chapters
...

That is, a change under these conditions is spontaneous if it corresponds to a decrease
in the Helmholtz energy
...
The criterion of equilibrium, when neither the forward nor
reverse process has a tendency to occur, is
dAT,V = 0

(3
...
A
negative value of dA is favoured by a negative value of dU and a positive value of TdS
...

However, this interpretation is false (even though it is a good rule of thumb for
remembering the expression for dA) because the tendency to lower A is solely a tendency towards states of greater overall entropy
...
The form of dA may give the impression that
systems favour lower energy, but that is misleading: dS is the entropy change of the
system, −dU/T is the entropy change of the surroundings (when the volume of the
system is constant), and their total tends to a maximum
...
34)

As a result, A is sometimes called the ‘maximum work function’, or the ‘work function’
...


3
...
2 Maximum work

To demonstrate that maximum work can be expressed in terms of the changes in
Helmholtz energy, we combine the Clausius inequality dS ≥ dq/T in the form TdS ≥
dq with the First Law, dU = dq + dw, and obtain
dU ≤ TdS + dw
(dU is smaller than the term on the right because we are replacing dq by TdS, which
in general is larger
...
Because at constant temperature dA = dU − TdS, we conclude
that dwmax = dA
...
34 becomes
wmax = ∆A

(3
...
36)

This expression shows that in some cases, depending on the sign of T∆S, not all the
change in internal energy may be available for doing work
...
For the change to be spontaneous, some of the energy must escape as
heat in order to generate enough entropy in the surroundings to overcome the reduction in entropy in the system (Fig
...
15)
...
This is the origin of the alternative name
‘Helmholtz free energy’ for A, because ∆A is that part of the change in internal energy
that we are free to use to do work
...
4 Maximum work and the Helmholtz energy

Further insight into the relation between the work that a system can do and the
Helmholtz energy is obtained by recalling that work is energy transferred to the
surroundings as the uniform motion of atoms
...

Because energy stored in random thermal motion cannot be used to achieve
uniform motion in the surroundings, only the part of U that is not stored in that
way, the quantity U − TS, is available for conversion into work
...
In this case,
the maximum work that can be obtained from the system is greater than ∆U
...
3
...
Moreover, the process is
spontaneous if overall the entropy of
the global, isolated system increases
...
Therefore, less
work than ∆U can be obtained
...
Because the entropy of the system increases, we can
afford a reduction of the entropy of the surroundings yet still have, overall, a spontaneous process
...
3
...

Nature is now providing a tax refund
...
4 Calculating the maximum available work

DSsur < 0

Fig
...
16 In this process, the entropy of the
system increases; hence we can afford to
lose some entropy of the surroundings
...
This energy can be
returned to them as work
...


When 1
...
4 J K−1
mol−1 at 25°C
...
To do so, we suppose that
all the gases involved are perfect, and use eqn 2
...

For the maximum work available from the process we use eqn 3
...

Answer (a) Because ∆νg = 0, we know that ∆ r H 7 = ∆ rU 7 = −2808 kJ mol−1
...
(b) Because
T = 298 K, the value of ∆ r A7 is
∆ r A7 = ∆ rU 7 − T∆ rS 7 = −2862 kJ mol−1
Therefore, the combustion of 1
...
The maximum work available is greater than the change in internal energy on account of the positive entropy of reaction (which is partly due to the
generation of a large number of small molecules from one big one)
...

Self-test 3
...
000 mol CH4(g) under

the same conditions, using data from Table 2
...


[|qp | = 890 kJ, |wmax | = 813 kJ]

(d) Some remarks on the Gibbs energy

The Gibbs energy (the ‘free energy’) is more common in chemistry than the Helmholtz
energy because, at least in laboratory chemistry, we are usually more interested in
changes occurring at constant pressure than at constant volume
...

Therefore, if we want to know whether a reaction is spontaneous, the pressure and
temperature being constant, we assess the change in the Gibbs energy
...
If G increases, then the reverse reaction is spontaneous
...
In such reactions, H increases, the system rises spontaneously to states
of higher enthalpy, and dH > 0
...
Endothermic reactions are therefore driven by
the increase of entropy of the system, and this entropy change overcomes the reduction of entropy brought about in the surroundings by the inflow of heat into the system (dSsur = −dH/T at constant pressure)
...
5 THE HELMHOLTZ AND GIBBS ENERGIES
(e) Maximum non-expansion work

The analogue of the maximum work interpretation of ∆A, and the origin of the name
‘free energy’, can be found for ∆G
...
37)

The corresponding expression for a measurable change is
wadd,max = ∆G

(3
...

Justification 3
...
Therefore, with d(pV) = pdV + Vdp,
dG = (−pdV + dwadd,rev) + pdV + Vdp = dwadd,rev + Vdp
If the change occurs at constant pressure (as well as constant temperature), we can
set dp = 0 and obtain dG = dwadd,rev
...
However, because the process is reversible, the work done must
now have its maximum value, so eqn 3
...


Example 3
...
00 mol of glucose molecules under standard conditions at
37°C (blood temperature)? The standard entropy of reaction is +182
...

Method The non-expansion work available from the reaction is equal to the

change in standard Gibbs energy for the reaction (∆ rG 7, a quantity defined more
fully below)
...
5, and to substitute the data into ∆rG 7 = ∆r H 7 − T∆ r S 7
...
4 J K−1 mol−1) = −2865 kJ mol−1

99

100

3 THE SECOND LAW
Therefore, wadd,max = −2865 kJ for the combustion of 1 mol glucose molecules, and
the reaction can be used to do up to 2865 kJ of non-expansion work
...
1 kJ of work
to climb vertically through 3
...
13 g of glucose is needed to
complete the task (and in practice significantly more)
...
8 How much non-expansion work can be obtained from the com-

bustion of 1
...

[818 kJ]

3
...
39]

The standard Gibbs energy of reaction is the difference in standard molar Gibbs
energies of the products and reactants in their standard states at the temperature
specified for the reaction as written
...
8 Standard Gibbs energies of formation of the elements in their reference
states are zero, because their formation is a ‘null’ reaction
...
4
...
40)

Reactants

with each term weighted by the appropriate stoichiometric coefficient
...
7 Calculating a standard Gibbs energy of reaction
1
To calculate the standard Gibbs energy of the reaction CO(g) + – O2(g) → CO2(g)
2
at 25°C, we write
1
∆rG 7 = ∆ f G 7(CO2, g) − {∆ f G 7(CO, g) + – ∆ f G 7(O2, g)}
2
1
= −394
...
2) + – (0)} kJ mol−1
2
= −257
...
4* Standard Gibbs
energies of formation (at 298 K)
∆f G 7/(kJ mol−1)
Diamond, C(s)
Benzene, C6H6(l)
Methane, CH4(g)

[−818 kJ mol−1]

+124
...
7
−394
...
1

Sodium chloride, NaCl(s)

CH4(g) at 298 K
...
9

Carbon dioxide, CO2(g)
Ammonia, NH3(g)

Self-test 3
...
5

Just as we did in Section 2
...
1

* More values are given in the Data section
...
7
...
41]

3
...
Then for the reaction
1
1
– H2(g) + – Cl2(g) → H+(aq) + Cl−(aq)
2
2

∆rG 7 = −131
...
23 kJ mol−1
...

Illustration 3
...
12 kJ mol−1

which leads to ∆f G 7(Ag+, aq) = +77
...

The factors responsible for the magnitude of the Gibbs energy of formation of an
ion in solution can be identified by analysing it in terms of a thermodynamic cycle
...
We do so by treating the formation reaction
1
1
– H2(g) + – X2(g) → H+(aq) + X−(aq)
2
2

as the outcome of the sequence of steps shown in Fig
...
17 (with values taken from the
Data section)
...
2

H+(g) + I(g) + e_

-

H (aq) + Cl (aq)

° +
DsolvG – (H )

+218

1
1
- H2(g) + - I2(g)
2
2
– (H+, aq) + D G – (I-, aq)}
°
- {DfG°
f

+

-

H (aq) + I (aq)

(b)

Fig
...
17 The thermodynamic cycles for the discussion of the Gibbs energies of solvation
(hydration) and formation of (a) chloride ions, (b) iodide ions in aqueous solution
...


The standard Gibbs energies of
formation of the gas-phase ions are
unknown
...
The conclusions
from the cycles are therefore only
approximate
...

Gibbs energies of solvation of individual ions may be estimated from an equation
derived by Max Born, who identified ∆solvG 7 with the electrical work of transferring an
ion from a vacuum into the solvent treated as a continuous dielectric of relative permittivity εr
...
1, is
∆solvG 7 = −

z i2e2NA A
8πε0ri C

1−

1D

(3
...

Note that ∆solvG 7 < 0, and that ∆solvG 7 is strongly negative for small, highly charged
ions in media of high relative permittivity
...
86 × 104 kJ mol−1)

(3
...
9 Using the Born equation

To see how closely the Born equation reproduces the experimental data, we calculate the difference in the values of ∆f G 7 for Cl− and I− in water, for which εr = 78
...
3), respectively, is
∆solvG 7(Cl−) − ∆solvG 7(I−) = −

A 1
1 D
−
× (6
...

Self-test 3
...

[−26 kJ mol−1 experimental; −29 kJ mol−1 calculated]

Comment 3
...


Calorimetry (for ∆H directly, and for S via heat capacities) is only one of the ways
of determining Gibbs energies
...


Combining the First and Second Laws
The First and Second Laws of thermodynamics are both relevant to the behaviour of
matter, and we can bring the whole force of thermodynamics to bear on a problem by
setting up a formulation that combines them
...
7 The fundamental equation
We have seen that the First Law of thermodynamics may be written dU = dq + dw
...
8 PROPERTIES OF THE INTERNAL ENERGY

103

any additional (non-expansion) work, we may set dwrev = −pdV and (from the definition of entropy) dqrev = TdS, where p is the pressure of the system and T its temperature
...
43)

However, because dU is an exact differential, its value is independent of path
...
Consequently, eqn 3
...
We shall call this combination of the First and Second Laws the fundamental equation
...
The reason is that only in the case of a reversible
change may TdS be identified with dq and −pdV with dw
...
The sum of dw and dq
remains equal to the sum of TdS and −pdV, provided the composition is constant
...
8 Properties of the internal energy
Equation 3
...
These simple proportionalities
suggest that U should be regarded as a function of S and V
...

The mathematical consequence of U being a function of S and V is that we can
express an infinitesimal change dU in terms of changes dS and dV by
dU =

A ∂U D
A ∂U D
dS +
dV
C ∂S F V
C ∂V F S

(3
...

When this expression is compared to the thermodynamic relation, eqn 3
...
5 and are reviewed in
Appendix 2
...
44 was first obtained in Section 2
...


(3
...
We are beginning to generate relations between the properties of
a system and to discover the power of thermodynamics for establishing unexpected
relations
...
The mathematical criterion for df being an exact differential (in the sense that its integral is independent of path) is that

A ∂g D A ∂h D
=
C ∂y F x C ∂x F y

Comment 3
...
46)

Because the fundamental equation, eqn 3
...
Therefore,
it must be the case that

Comment 3
...
46, let’s test whether df = 2xydx + x 2dy
is an exact differential
...


104

3 THE SECOND LAW

Table 3
...
47)

We have generated a relation between quantities that, at first sight, would not seem to
be related
...
47 is an example of a Maxwell relation
...
Nevertheless, it does suggest that
there may be other similar relations that are more useful
...
The
argument to obtain them runs in the same way in each case: because H, G, and A are
state functions, the expressions for dH, dG, and dA satisfy relations like eqn 3
...
All
four relations are listed in Table 3
...

(b) The variation of internal energy with volume

The quantity πT = (∂U/∂V)T , which represents how the internal energy changes as the
volume of a system is changed isothermally, played a central role in the manipulation
of the First Law, and in Further information 2
...
48)

This relation is called a thermodynamic equation of state because it is an expression
for pressure in terms of a variety of thermodynamic properties of the system
...

Justification 3
...
43 by
dV, imposing the constraint of constant temperature, which gives
A ∂UD
A ∂UD A ∂S D
A ∂UD
B E =B E B E +B E
C ∂V F T C ∂S F V C ∂V F T C ∂V F S
Next, we introduce the two relations in eqn 3
...
5 turns (∂S/∂V)T into (∂p/∂T)V , which completes the proof of eqn 3
...


Example 3
...

Method Proving a result ‘thermodynamically’ means basing it entirely on general

thermodynamic relations and equations of state, without drawing on molecular
arguments (such as the existence of intermolecular forces)
...
48
...
7, and for the second part of the question it should be used in eqn 3
...

Answer For a perfect gas we write

3
...
48 becomes

πT =

nRT
V

−p=0

The equation of state of a van der Waals gas is
p=

nRT
V − nb

−a

n2
V2

Because a and b are independent of temperature,

A ∂p D
nR
=
C ∂T F V V − nb
Therefore, from eqn 3
...
A larger molar volume, corresponding to a greater average separation
between molecules, implies weaker mean intermolecular attractions, so the total
energy is greater
...
11 Calculate

(Table 1
...


π T for a gas that obeys the virial equation of state
2
[π T = RT 2(∂B/∂T)V /V m + · · · ]

3
...
They lead to expressions showing how G varies with pressure and temperature
that are important for discussing phase transitions and chemical reactions
...
As in Justification 2
...
49)

This expression, which shows that a change in G is proportional to a change in p or
T, suggests that G may be best regarded as a function of p and T
...
In other words, G carries around the combined
consequences of the First and Second Laws in a way that makes it particularly suitable
for chemical applications
...
45, when applied to the exact differential dG
= Vdp − SdT, now gives

A ∂GD
= −S
C ∂T F p

A ∂GD
=V
C ∂p F T

(3
...
3
...
The first implies that:
• Because S > 0 for all substances, G always decreases when the temperature is raised
(at constant pressure and composition)
...

Therefore, the Gibbs energy of the gaseous phase of a substance, which has a high
molar entropy, is more sensitive to temperature than its liquid and solid phases
(Fig
...
19)
...


Gas

Slope = -S

Slope = V

T

Pressu
re, p

Fig
...
18 The variation of the Gibbs energy
of a system with (a) temperature at
constant pressure and (b) pressure at
constant temperature
...


Gibbs energy, G

Gibbs
energy,
G

Te
m
pe
ra
tu
re
,

106

Liquid
Solid

Temperature, T
Fig
...
19 The variation of the Gibbs energy
with the temperature is determined by
the entropy
...


3
...


As we remarked in the introduction, because the equilibrium composition of a system
depends on the Gibbs energy, to discuss the response of the composition to temperature we need to know how G varies with temperature
...
50, (∂G/∂T)p = −S, is our starting point for this discussion
...
Then

A ∂GD G − H
=
C ∂T F p
T

(3
...
52)

This expression is called the Gibbs–Helmholtz equation
...


Justification 3
...
51 in the form
A ∂G D
G
H
B E − =−
T
C ∂T F p T
It follows that
A ∂ GD
1 1 H5
H
B
E = 2− 6 = − 2
T
C ∂T T F p T 3 T 7
which is eqn 3
...


The Gibbs–Helmholtz equation is most useful when it is applied to changes,
including changes of physical state and chemical reactions at constant pressure
...
2b we derive the result that the equilibrium constant for a reaction is related to its standard
reaction Gibbs energy by ∆rG 7/T = −R ln K
...
3
...
Because the volume
of the gaseous phase of a substance is
greater than that of the same amount of
liquid phase, and the entropy of the solid
phase is smallest (for most substances), the
Gibbs energy changes most steeply for the
gas phase, followed by the liquid phase, and
then the solid phase of the substance
...


Comment 3
...
3
...


107

d(x2eax)
dx

= x2

deax
dx
2 ax

+ eax

dx2
dx
ax

= ax e + 2xe

108

3 THE SECOND LAW

A ∂ ∆G D
∆H
=− 2
C ∂T T F p
T

(3
...
As we
shall see, this is a crucial piece of information in chemistry
...
49, which gives dG = Vdp, and
integrate:
pf

G(pf) = G(pi) +

Ύ V dp

(3
...
54b)

pi

Volume, V

This expression is applicable to any phase of matter, but to evaluate it we need to know
how the molar volume, Vm, depends on the pressure
...
3
...
55)

pi

pi

Pressure, p

pf

Self-test 3
...
0 bar to 2
...

Fig
...
21 The difference in Gibbs energy of a
solid or liquid at two pressures is equal to
the rectangular area shown
...


V = nRT/p

[+2
...

Hence, we may usually suppose that the Gibbs energies of solids and liquids are independent of pressure
...
If the pressures are so great that there are substantial volume changes
over the range of integration, then we must use the complete expression, eqn 3
...


Volume, V

Illustration 3
...
0 cm3 mol−1 independent of pressure
...
0 Mbar (3
...
0 bar (1
...
3
...


∆trsG(3 Mbar) = ∆trsG(1 bar) + (1
...
0 × 1011 Pa − 1
...
0 × 102 kJ mol−1
where we have used 1 Pa m3 = 1 J
...
Furthermore, because the volume also varies markedly with
the pressure, we cannot treat it as a constant in the integral in eqn 3
...
3
...


CHECKLIST OF KEY IDEAS

pf

Ύ

Gm(pf) = Gm(pi) + RT

pi

dp
p

= Gm(pi) + RT ln

pf
pi

Molar Gibbs energy, Gm

For a perfect gas we substitute Vm = RT/p into the integral, treat RT as a constant,
and find
(3
...
It also follows from this
equation that, if we set pi = p7 (the standard pressure of 1 bar), then the molar Gibbs
energy of a perfect gas at a pressure p (set pf = p) is related to its standard value by
7
Gm(p) = G m + RT ln

p
p7

109

Gm
°

(3
...
13 Calculate the change in the molar Gibbs energy of water vapour
(treated as a perfect gas) when the pressure is increased isothermally from 1
...
0 bar at 298 K
...
12) is a few joules per mole, the answer you should get for
a gas is of the order of kilojoules per mole
...
7 kJ mol−1]

Pressure, p

Fig
...
23 The molar Gibbs energy potential
of a perfect gas is proportional to ln p, and
the standard state is reached at p7
...


Exploration Show how the first
derivative of G, (∂G/∂p)T , varies
with pressure, and plot the resulting
expression over a pressure range
...
57 is illustrated in Fig
...
23
...
Further information 3
...


Checklist of key ideas
1
...

2
...


9
...

10
...


Ti

3
...

The statistical definition of entropy is given by the Boltzmann
formula, S = k ln W
...
The entropy of a substance is measured from the area under a
graph of Cp /T against T, using the Debye extrapolation at low
temperatures, Cp = aT 3 as T → 0
...
A Carnot cycle is a cycle composed of a sequence of
isothermal and adiabatic reversible expansions and
compressions
...
The efficiency of a heat engine is ε = |w| /qh
...


12
...


6
...
16 K
...
Third Law of thermodynamics: The entropy of all perfect
crystalline substances is zero at T = 0
...
The Clausius inequality is dS ≥ dq/T
...
The standard reaction entropy is calculated from
7
7
∆rS 7 = ∑ProductsνS m − ∑ReactantsνS m
...
The normal transition temperature, Ttrs, is the temperature at
which two phases are in equilibrium at 1 atm
...


15
...


110

3 THE SECOND LAW

16
...
The Gibbs energy is
G = H − TS
...
The standard Gibbs energies of formation of ions are reported
on a scale in which ∆ f G 7(H+, aq) = 0 at all temperatures
...
The criteria of spontaneity may be written as: (a) dSU,V ≥ 0
and dUS,V ≤ 0, or (b) dAT,V ≤ 0 and dGT,p ≤ 0
...
The fundamental equation is dU = TdS − pdV
...
The criterion of equilibrium at constant temperature and
volume, dAT,V = 0
...


26
...


19
...
The maximum additional (non-expansion) work
and the Gibbs energy are related by wadd,max = ∆G
...
The Gibbs energy is best described as a function of pressure
and temperature, dG = Vdp − SdT
...


20
...

21
...

22
...


25
...
5
...
The temperature dependence of the Gibbs energy is given by
the Gibbs–Helmholtz equation, (∂(G/T)/∂T)p = −H/T 2
...
For a condensed phase, the Gibbs energy varies with pressure
as G(pf) = G(pi) + Vm∆p
...


10

Further reading
Articles and texts

N
...
Craig, Entropy analyses of four familiar processes
...
Chem
...
65, 760 (1988)
...
B
...
W
...
Freeman and Co
...

F
...
Hale, Heat engines and refrigerators
...
G
...
Trigg), 7, 303
...

D
...
Prigogine, Modern thermodynamics: from heat
engines to dissipative structures
...


P
...
Nelson, Derivation of the Second Law of thermodynamics from
Boltzmann’s distribution law
...
Chem
...
65, 390 (1988)
...
W
...
(ed
...
Published
as J
...
Chem
...
Data, Monograph no
...
American Institute of
Physics, New York (1998)
...
C
...
), Handbook of chemistry and physics, Vol
...
CRC
Press, Boca Raton (2004)
...
1 The Born equation

The electrical concepts required in this derivation are reviewed in
Appendix 3
...
That work is calculated by taking the
difference of the work of charging an ion when it is in the solution
and the work of charging the same ion when it is in a vacuum
...
The permittivity of vacuum is
ε0 = 8
...
The relative permittivity (formerly
10

called the ‘dielectric constant’) of a substance is defined as εr = ε /ε0
...
See Chapter 18 for more details
...
It turns out that, when the charge of the sphere is
q, the electric potential, φ, at its surface is the same as the potential
due to a point charge at its centre, so we can use the last expression
and write

See Further reading in Chapter 2 for additional articles, texts, and sources of thermochemical data
...
Therefore,
the total work of charging the sphere from 0 to zie is

Ύ

φ dq =

0

zie

1
4πε ri

Ύ

0

q dq =

z 2e 2
i

Molar Gibbs energy, Gm

zie

w=

8πε ri

This electrical work of charging, when multiplied by Avogadro’s
constant, is the molar Gibbs energy for charging the ions
...
The corresponding value for
charging the ion in a medium is obtained by setting ε = εrε0, where εr
is the relative permittivity of the medium
...
42
...
2 Real gases: the fugacity

At various stages in the development of physical chemistry it is
necessary to switch from a consideration of idealized systems to real
systems
...
Then
deviations from the idealized behaviour can be expressed most
simply
...
3
...
To adapt
eqn 3
...
58]

The fugacity, a function of the pressure and temperature, is defined
so that this relation is exactly true
...
To develop this relation we write the
fugacity as
f = φp

[3
...

Equation 3
...

Expressing it in terms of the fugacity by using eqn 3
...
If the gas were perfect, we would
write
11

Repulsions
dominant
(f > p)

Attractions
dominant
(f < p)

4πε ri

111

Perfect
gas
°
G–
m

Real
gas

p°

Pressure, p

-¥

The molar Gibbs energy of a real gas
...
When attractive forces are dominant (at intermediate
pressures), the molar Gibbs energy is less than that of a perfect
gas and the molecules have a lower ‘escaping tendency’
...
Then the
‘escaping tendency’ is increased
...
3
...
Therefore, f ′/p′ → 1 as p′ → 0
...
For a real gas, Vm = RTZ/p, where
Z is the compression factor of the gas (Section 1
...
With these two
substitutions, we obtain

The name ‘fugacity’ comes from the Latin for ‘fleetness’ in the sense of ‘escaping tendency’; fugacity has the same dimensions as pressure
...
The curves are labelled
with the reduced temperature Tr = T/Tc
...
0

Exploration Evaluate the fugacity
coefficient as a function of the
reduced volume of a van der Waals gas and
plot the outcome for a selection of reduced
temperatures over the range 0
...


2
...
3
...
50

5

3
...
0
6

8

15

2
...
5

35

Fugacity coefficient, f = f /p

112

1
...
00

1
...
50
1
...
0

1
...
6* The fugacity of
nitrogen at 273 K

(3
...
59, to relate the fugacity to the
pressure of the gas
...
1
...
If Z < 1 throughout the
range of integration, then the integrand in eqn 3
...
This value implies that f < p (the molecules tend to stick
together) and that the molar Gibbs energy of the gas is less than that
of a perfect gas
...
The integral is then positive,
φ > 1, and f > p (the repulsive interactions are dominant and tend to
drive the particles apart)
...

Figure 3
...
999 55
9
...
03

1000

1839

* More values are given in the Data section
...
5)
...
6, the graphs can be used for
quick estimates of the fugacities of a wide range of gases
...
6
gives some explicit values for nitrogen
...
1 The evolution of life requires the organization of a very large number of

molecules into biological cells
...

3
...

This procedure is potentially very lucrative because, after an initial extraction
of energy from the ground, no fossil fuels would be required to keep the device
running indefinitely
...


and dGT,p ≤ 0
...

3
...
Discuss the origin, significance, and
applicability of each criterion
...
5 Discuss the physical interpretation of any one Maxwell relation
...
6 Account for the dependence of πT of a van der Waals gas in terms of the

significance of the parameters a and b
...
7 Suggest a physical interpretation of the dependence of the Gibbs energy

on the pressure
...
3 The following expressions have been used to establish criteria

3
...


EXERCISES

113

Exercises
Assume that all gases are perfect and that data refer to 298
...

3
...

3
...

3
...
22 J K−1 mol−1 at 298 K
...
2(b) Calculate the molar entropy of a constant-volume sample of argon at

250 K given that it is 154
...


3
...
8(b) Calculate the standard reaction entropy at 298 K of

(a) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
(b) C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l)
3
...
8a with the
reaction enthalpies, and calculate the standard reaction Gibbs energies at
298 K
...
9(b) Combine the reaction entropies calculated in Exercise 3
...
3(a) Calculate ∆S (for the system) when the state of 3
...
00 atm to 125°C and
2
5
...
How do you rationalize the sign of ∆S?

3
...
00 mol diatomic
7
perfect gas molecules, for which Cp,m = –R, is changed from 25°C and
2
1
...
00 atm
...
4(a) A sample consisting of 3
...
Given that CV,m = 27
...

3
...
00 mol of diatomic perfect gas molecules at

250 K is compressed reversibly and adiabatically until its temperature reaches
300 K
...
5 J K−1 mol−1, calculate q, w, ∆U, ∆H, and ∆S
...
5(a) Calculate ∆H and ∆Stot when two copper blocks, each of mass 10
...
The specific heat capacity of copper is 0
...

3
...
00 kg, one

at 200°C and the other at 25°C, are placed in contact in an isolated container
...
449 J K−1 g−1 and may be assumed
constant over the temperature range involved
...
6(a) Consider a system consisting of 2
...
0 cm
...
0 atm until the piston
has moved outwards through 20 cm
...
8 J K−1 mol−1 and calculate (a) q,
(b) w, (c) ∆U, (d) ∆T, (e) ∆S
...
6(b) Consider a system consisting of 1
...
0 atm and confined to a cylinder of cross-section 100
...
The sample is
allowed to expand adiabatically against an external pressure of 1
...
Assume that carbon dioxide
may be considered a perfect gas with CV,m = 28
...

3
...
4 kJ mol−1 at

reaction enthalpies, and calculate the standard reaction Gibbs energies at
298 K
...
10(a) Use standard Gibbs energies of formation to calculate the standard

reaction Gibbs energies at 298 K of the reactions in Exercise 3
...

3
...
8b
...
11(a) Calculate the standard Gibbs energy of the reaction 4 HCl(g) + O2(g)
→ 2 Cl2(g) + 2 H2O(l) at 298 K, from the standard entropies and enthalpies of
formation given in the Data section
...
11(b) Calculate the standard Gibbs energy of the reaction CO(g) +
CH3OH(l) → CH3COOH(l) at 298 K, from the standard entropies and
enthalpies of formation given in the Data section
...
12(a) The standard enthalpy of combustion of solid phenol (C6H5OH) is
−3054 kJ mol−1 at 298 K and its standard molar entropy is 144
...

Calculate the standard Gibbs energy of formation of phenol at 298 K
...
12(b) The standard enthalpy of combustion of solid urea (CO(NH2)2) is

−632 kJ mol−1 at 298 K and its standard molar entropy is 104
...

Calculate the standard Gibbs energy of formation of urea at 298 K
...
13(a) Calculate the change in the entropies of the system and the

surroundings, and the total change in entropy, when a sample of nitrogen gas
of mass 14 g at 298 K and 1
...

3
...
50 bar increases from 1
...
60 dm3 in (a) an isothermal reversible expansion, (b) an isothermal
irreversible expansion against pex = 0, and (c) an adiabatic reversible
expansion
...
14(a) Calculate the maximum non-expansion work per mole that may be

obtained from a fuel cell in which the chemical reaction is the combustion of
methane at 298 K
...
88 K
...


3
...


3
...
27 kJ mol−1 at its

3
...
(b) Repeat the
calculation for a modern steam turbine that operates with steam at 300°C
and discharges at 80°C
...
1°C
...


114

3 THE SECOND LAW

3
...
(a) What is
the maximum efficiency of the engine? (b) Calculate the maximum work that
can be done by for each 1
...
(c) How
much heat is discharged into the cold sink in a reversible process for each
1
...
16(a) Suppose that 3
...
Calculate ∆G for the process
...
16(b) Suppose that 2
...
Calculate ∆G for the process
...
17(a) The change in the Gibbs energy of a certain constant-pressure process

was found to fit the expression ∆G/J = −85
...
5(T/K)
...


3
...
8 atm to 29
...

3
...
0 kPa to 252
...

3
...
72
...

3
...
1 MPa is 0
...

Calculate the difference of its molar Gibbs energy from that of a perfect gas in
the same state
...
21(a) Estimate the change in the Gibbs energy of 1
...
0 atm to 100 atm
...
17(b) The change in the Gibbs energy of a certain constant-pressure process
was found to fit the expression ∆G/J = −73
...
8(T/K)
...


3
...
0 dm3 of water when the
pressure acting on it is increased from 100 kPa to 300 kPa
...
18(a) Calculate the change in Gibbs energy of 35 g of ethanol (mass density

3
...
789 g cm−3) when the pressure is increased isothermally from 1 atm to
3000 atm
...
0 atm to 100
...


3
...
791 g cm−3) when the pressure is increased isothermally from
100 kPa to 100 MPa
...
22(b) Calculate the change in the molar Gibbs energy of oxygen when its
pressure is increased isothermally from 50
...
0 kPa at 500 K
...


Numerical problems
3
...
00 atm
...
3 J K−1 mol−1 and −41
...
Distinguish between
the entropy changes of the sample, the surroundings, and the total system,
and discuss the spontaneity of the transitions at the two temperatures
...
2 The heat capacity of chloroform (trichloromethane, CHCl3) in the range
240 K to 330 K is given by Cp,m /(J K−1 mol−1) = 91
...
5 × 10−2 (T/K)
...
00 mol CHCl3 is heated from 273 K to 300 K
...

3
...
00 kg (Cp,m = 24
...
00 mol H2O(g) at 100°C and 1
...
(a) Assuming all the steam is
condensed to water, what will be the final temperature of the system, the heat
transferred from water to copper, and the entropy change of the water,
copper, and the total system? (b) In fact, some water vapour is present at
equilibrium
...

(Hint
...
)
3
...
All changes in B is
isothermal; that is, a thermostat surrounds B to keep its temperature constant
...
00 mol of the gas in each section
...
00 dm3
...
00 dm3
...
If numerical values cannot be obtained, indicate
whether the values should be positive, negative, or zero or are indeterminate
from the information given
...
)
3
...
00 mol of a monatomic perfect gas as the working

substance from an initial state of 10
...
It expands isothermally
to a pressure of 1
...
This expansion is followed by an isothermal compression
(Step 3), and then an adiabatic compression (Step 4) back to the initial state
...
Express your answer as a table of values
...
6 1
...
00 atm to a final pressure of 1
...
00 atm
...

3
...
45 J K−1 mol−1 at 298 K, and

its heat capacity is given by eqn 2
...
2
...

3
...
00 kΩ and negligible mass
...
00 A is passed for 15
...
Calculate the change in entropy of the copper,
taking Cp,m = 24
...
The experiment is then repeated with the
copper immersed in a stream of water that maintains its temperature at 293 K
...

3
...
Evaluate the

* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady
...
4 J K−1
mol−1, taking Th = 500 K and Tc = 250 K
...
17 Estimate the standard reaction Gibbs energy of N2(g) + 3 H2(g) →

3
...
00 mol molecules is described by the
equation of state pVm = RT(1 + Bp)
...
00 atm
...
21 K atm , B = −0
...


3
...
Evaluate the fugacity of oxygen at this temperature and
100 atm
...
8

T/K

70

Cp,m /(J K−1 mol−1)

23
...
8

7
...
1

16
...
4

150

24
...
3

250

25
...
2

26
...

3
...
Assume that the heat capacities are constant over
the temperature range involved
...
13 The heat capacity of anhydrous potassium hexacyanoferrate(II) varies

with temperature as follows:
T/K

Cp,m /(J K−1 mol−1)

T/K

2
...
6

20

14
...
8

30

36
...
6

40

62
...
3

87
...
0

180

131
...
0

80

149
...
3

90

165
...
33

20
...
15

44
...
81

12
...
18

32
...
86

66
...
90

140
...
59

225
...
99

298
...
05

121
...
4

163
...
2

196
...
492

Calculate the molar enthalpy relative to its value at T = 0 and the Third-Law
molar entropy of the compound at these temperatures
...
15‡ Given that S m = 29
...
G
...
Chem
...
Data 40,
1015 (1995)), compute the standard molar entropy of bismuth at 200 K
...
00

120
23
...
25

40
...
96956

0
...
00

100
...
7764

0
...
19 Represent the Carnot cycle on a temperature–entropy diagram and show
that the area enclosed by the cycle is equal to the work done
...
20 Prove that two reversible adiabatic paths can never cross
...
(Hint
...
Consider the changes
accompanying each stage of the cycle and show that they conflict with the
Kelvin statement of the Second Law
...
21 Prove that the perfect gas temperature scale and the thermodynamic
temperature scale based on the Second Law of thermodynamics differ from
each other by at most a constant numerical factor
...
25 Two of the four Maxwell relations were derived in the text, but two were
not
...


3
...
L
...
F
...
Chem
...

Faraday Trans
...
871 (1973))
...
14

10
...
24 Show that, for a perfect gas, (∂U/∂S)V = T and (∂U/∂V)S = −p
...


T/K

7
...
97880

(Table 1
...
For an isothermal expansion, for which kind of gas
(and a perfect gas) will ∆S be greatest? Explain your conclusion
...
1

70

4
...
98796

3
...
5

60

1
...
9971

3
...
Obtain the equation of
2
3
state of the gas
...
11 The molar heat capacity of lead varies with temperature as follows:

T/K

2 NH3(g) at (a) 500 K, (b) 1000 K from their values at 298 K
...
44

160
24
...
89

200
25
...
44 J K−1 mol−1 over this range
...
16 Calculate ∆rG 7(375 K) for the reaction 2 CO(g) + O2(g) → 2 CO2(g) from

the value of ∆rG 7(298 K), ∆rH 7(298 K), and the Gibbs–Helmholtz equation
...
26 Use the Maxwell relations to express the derivatives (a) (∂S/∂V)T and

(∂V/∂S)p and (b) (∂p/∂S)V and (∂V/∂S)p in terms of the heat capacities, the
expansion coefficient α, and the isothermal compressibility, κT
...
27 Use the Maxwell relations to show that the entropy of a perfect gas
depends on the volume as S ∝ R ln V
...
28 Derive the thermodynamic equation of state

A ∂H D
A ∂V D
B
E =V−TB
E
C ∂p F T
C ∂T F p
Derive an expression for (∂H/∂p)T for (a) a perfect gas and (b) a van der Waals
gas
...
0 mol Ar(g) at 298 K and 10 atm
...
29 Show that if B(T) is the second virial coefficient of a gas, and

∆B = B(T″) − B(T′), ∆T = T″ − T′, and T is the mean of T″ and T′, then
πT ≈ RT 2∆B/V 2 ∆T
...
0 cm3
m
mol−1 and B(300 K) = −15
...
0 atm, (b) 10
...

3
...
Show that

µJCV = p − α T/κT
...
31 Evaluate πT for a Dieterici gas (Table 1
...
Justify physically the form of

the expression obtained
...
32 The adiabatic compressibility, κS, is defined like κT (eqn 2
...
Show that for a perfect gas pγκS = 1 (where γ is the ratio of
heat capacities)
...
33 Suppose that S is regarded as a function of p and T
...
Hence, show that the energy transferred as heat when
the pressure on an incompressible liquid or solid is increased by ∆p is equal to
−αTV∆p
...
0 kbar
...
82 × 10−4 K−1
...
34 Suppose that (a) the attractive interactions between gas particles can be

neglected, (b) the attractive interaction is dominant in a van der Waals gas,
and the pressure is low enough to make the approximation 4ap/(RT)2 << 1
...
00 atm and 298
...

3
...
Use the resulting expression to
m
estimate the fugacity of argon at 1
...
13 cm3
−1
6
−2
mol and C = 1054 cm mol
...
0 K, assuming that
the relative humidity remains constant
...
0189 bar
...
40‡ Nitric acid hydrates have received much attention as possible catalysts
for heterogeneous reactions that bring about the Antarctic ozone hole
...
investigated the thermodynamic stability of these hydrates
under conditions typical of the polar winter stratosphere (D
...
Worsnop, L
...

Fox, M
...
Zahniser, and S
...
Wofsy, Science 259, 71 (1993))
...
Given ∆ r G 7 and ∆ r H 7 for these reactions at 220 K, use the
Gibbs–Helmholtz equation to compute ∆rG 7 at 190 K
...
36 The protein lysozyme unfolds at a transition temperature of 75
...
Calculate the entropy of
unfolding of lysozyme at 25
...
28 kJ K−1 mol−1 and can be
assumed to be independent of temperature
...
Imagine that the transition
at 25
...
0°C to
the transition temperature, (ii) unfolding at the transition temperature, and
(iii) cooling of the unfolded protein to 25
...
Because the entropy is a state
function, the entropy change at 25
...

3
...

Estimate the additional non-expansion work that may be obtained by raising
the temperature to blood temperature, 37°C
...
38 In biological cells, the energy released by the oxidation of foods (Impact
on Biology I2
...
The
essence of ATP’s action is its ability to lose its terminal phosphate group by
hydrolysis and to form adenosine diphosphate (ADP or ADP3−):
2−
ATP4−(aq) + H2O(l) → ADP3−(aq) + HPO4 (aq) + H3O+(aq)

At pH = 7
...
Under these conditions, the hydrolysis of 1 mol ATP4−(aq) results
in the extraction of up to 31 kJ of energy that can be used to do non-expansion
work, such as the synthesis of proteins from amino acids, muscular
contraction, and the activation of neuronal circuits in our brains
...
0 and
310 K
...
What is the power
density of the cell in watts per cubic metre (1 W = 1 J s−1)? A computer battery
delivers about 15 W and has a volume of 100 cm3
...
2 kJ mol−1 of energy input
...
How many moles of ATP must be hydrolysed to form 1 mol
glutamine?
3
...
0 –3
...
0°C its best estimate
...
Predict the relative increase in water

1

2
46
...
4

188

93
...
41‡ J
...
H
...
For such a chain, S(l) = −3kl2/2Na2 + C, where k is the
Boltzmann constant and C is a constant
...

3
...
What is the maximum height, neglecting all forms of friction, to
which a car of mass 1000 kg can be driven on 1
...
43 The cycle involved in the operation of an internal combustion engine is
called the Otto cycle
...
The cycle consists of the following steps:
(1) reversible adiabatic compression from A to B, (2) reversible constantvolume pressure increase from B to C due to the combustion of a small
amount of fuel, (3) reversible adiabatic expansion from C to D, and (4)
reversible and constant-volume pressure decrease back to state A
...
Evaluate the efficiency for a compression
ratio of 10:1
...
00 dm3, p = 1
...

2
3
...
(a) Find an expression for the work of cooling an
object from Ti to Tf when the refrigerator is in a room at a temperature Th
...
Write dw = dq/c(T), relate dq to dT through the heat capacity Cp,
and integrate the resulting expression
...
(b) Use the result in part
(a) to calculate the work needed to freeze 250 g of water in a refrigerator at
293 K
...
45 The expressions that apply to the treatment of refrigerators also describe
the behaviour of heat pumps, where warmth is obtained from the back of a
refrigerator while its front is being used to cool the outside world
...
Compare
heating of a room at 295 K by each of two methods: (a) direct conversion of
1
...
00 kJ of
electrical energy to run a reversible heat pump with the outside at 260 K
...


Physical
transformations of
pure substances
The discussion of the phase transitions of pure substances is among the simplest applications of thermodynamics to chemistry
...
First,
we describe the interpretation of empirically determined phase diagrams for a selection of
materials
...
The practical importance of the expressions we derive is that they show how the vapour pressure of a substance varies with temperature and how the melting point varies with pressure
...
This chapter also introduces the chemical
potential, a property that is at the centre of discussions of phase transitions and chemical
reactions
...
1 The stabilities of phases
4
...
1 Impact on engineering and

technology: Supercritical fluids
4
...
4 The thermodynamic criterion

of equilibrium

Vaporization, melting, and the conversion of graphite to diamond are all examples of
changes of phase without change of chemical composition
...


4
...
6 The location of phase

boundaries
4
...
We present the concept in this
section
...
1 The stabilities of phases
A phase of a substance is a form of matter that is uniform throughout in chemical
composition and physical state
...
A phase transition, the spontaneous conversion of one phase into
another phase, occurs at a characteristic temperature for a given pressure
...
This difference indicates that below 0°C the Gibbs energy decreases as liquid
water changes into ice and that above 0°C the Gibbs energy decreases as ice changes
into liquid water
...


118

4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES

Critical
point
Pressure, p

Solid
Liquid
Triple
point
Vapour

T3
Temperature, T

Tc

The general regions of pressure and
temperature where solid, liquid, or gas is
stable (that is, has minimum molar Gibbs
energy) are shown on this phase diagram
...
In the following paragraphs we
locate the precise boundaries between the
regions
...
A transition that is predicted from thermodynamics to be spontaneous may
occur too slowly to be significant in practice
...
However,
for this transition to take place, the C atoms must change their locations, which is an
immeasurably slow process in a solid except at high temperatures
...
In gases and liquids the mobilities of the molecules allow phase
transitions to occur rapidly, but in solids thermodynamic instability may be frozen in
...
Diamond is a metastable phase of carbon
under normal conditions
...
4
...
2 Phase boundaries
The phase diagram of a substance shows the regions of pressure and temperature at
which its various phases are thermodynamically stable (Fig
...
1)
...

Consider a liquid sample of a pure substance in a closed vessel
...
4
...
Therefore, the liquid–vapour phase boundary in a phase diagram shows
how the vapour pressure of the liquid varies with temperature
...
The vapour pressure of a substance
increases with temperature because at higher temperatures more molecules have
sufficient energy to escape from their neighbours
...


Fig
...
2

Comment 4
...


When a liquid is heated in an open vessel, the liquid vaporizes from its surface
...
The condition of free vaporization throughout the liquid
is called boiling
...
For the special
case of an external pressure of 1 atm, the boiling temperature is called the normal
boiling point, Tb
...
Because 1 bar is slightly less than 1 atm
(1
...
987 atm), the standard boiling point of a liquid is slightly lower than
its normal boiling point
...
0°C; its standard
boiling point is 99
...

Boiling does not occur when a liquid is heated in a rigid, closed vessel
...
4
...
At the same time, the density of the liquid decreases slightly as a result of its
expansion
...
The temperature
at which the surface disappears is the critical temperature, Tc, of the substance
...
3d
...
At and above the critical temperature, a
single uniform phase called a supercritical fluid fills the container and an interface no

4
...
That is, above the critical temperature, the liquid phase of the substance
does not exist
...
Because a substance melts at exactly the same temperature as it freezes, the melting temperature of
a substance is the same as its freezing temperature
...
The normal and standard freezing points are negligibly different for most purposes
...

There is a set of conditions under which three different phases of a substance
(typically solid, liquid, and vapour) all simultaneously coexist in equilibrium
...
The temperature at the triple point is denoted T3
...
The triple point of water lies at 273
...
11 mbar, 4
...
This invariance of the triple point is the basis of its use in the definition of
the thermodynamic temperature scale (Section 3
...

As we can see from Fig
...
1, the triple point marks the lowest pressure at which a
liquid phase of a substance can exist
...

IMPACT ON CHEMICAL ENGINEERING AND TECHNOLOGY

I4
...
The critical temperature of CO2, 304
...
0°C) and
its critical pressure, 72
...
The density of scCO2 at its critical point is 0
...
However, the transport
properties of any supercritical fluid depend strongly on its density, which in turn is
sensitive to the pressure and temperature
...
1 g cm−3 to a liquid-like 1
...
A useful rule of thumb is that the
solubility of a solute is an exponential function of the density of the supercritical fluid,
so small increases in pressure, particularly close to the critical point, can have very
large effects on solubility
...
It is used, for
instance, to remove caffeine from coffee
...

Supercritical CO2 has been used since the 1960s as a mobile phase in supercritical
fluid chromatography (SFC), but it fell out of favour when the more convenient technique of high-performance liquid chromatography (HPLC) was introduced
...

Samples as small as 1 pg can be analysed
...
(b) When a liquid is heated in a
sealed container, the density of the vapour
phase increases and that of the liquid
decreases slightly
...
This disappearance occurs at
the critical temperature
...


Fig
...
3

120

4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
liquids, so there is less resistance to the transfer of solutes through the column, with
the result that separations may be effected rapidly or with high resolution
...
Indeed, scCO2-based dry cleaning depends on the availability of cheap surfactants; so too does the use of scCO2 as a solvent for homogeneous catalysts, such
as metal complexes
...
One solution is to use fluorinated and siloxane-based polymeric stabilizers, which allow polymerization reactions to proceed in scCO2
...
An alternative
and much cheaper approach is poly(ether-carbonate) copolymers
...

The critical temperature of water is 374°C and its pressure is 218 atm
...
Thus, as the density of scH2O
decreases, the characteristics of a solution change from those of an aqueous solution
through those of a non-aqueous solution and eventually to those of a gaseous solution
...

4
...

(a) Carbon dioxide

The phase diagram for carbon dioxide is shown in Fig
...
4
...
Notice also that, as the triple point
lies above 1 atm, the liquid cannot exist at normal atmospheric pressures whatever the
temperature, and the solid sublimes when left in the open (hence the name ‘dry ice’)
...
11 atm
...
When
the gas squirts through the throttle it cools by the Joule–Thomson effect, so when
it emerges into a region where the pressure is only 1 atm, it condenses into a finely
divided snow-like solid
...
Note that, as the triple
point lies at pressures well above
atmospheric, liquid carbon dioxide does
not exist under normal conditions (a
pressure of at least 5
...


Fig
...
4

Figure 4
...
The liquid–vapour boundary in the phase
diagram summarizes how the vapour pressure of liquid water varies with temperature
...
The solid–liquid boundary shows how the melting temperature varies with the pressure
...
Notice that the line has a negative slope
up to 2 kbar, which means that the melting temperature falls as the pressure is raised
...
The decrease in volume is a result of the very

4
...
Each O atom is linked by two
covalent bonds to H atoms and by two
hydrogen bonds to a neighbouring O atom,
in a tetrahedral array
...
4
...
4
...


open molecular structure of ice: as shown in Fig 4
...

Figure 4
...
4
...
Some of these phases melt at high
temperatures
...
Note
that five more triple points occur in the diagram other than the one where vapour,
liquid, and ice I coexist
...
The solid phases of ice differ in the arrangement of the water
molecules: under the influence of very high pressures, hydrogen bonds buckle and the
H2O molecules adopt different arrangements
...

(c) Helium

Figure 4
...
Helium behaves unusually at low temperatures
...
Solid helium can be obtained, but only by holding the atoms together by applying pressure
...
Pure helium-4 has two liquid phases
...
1

0
...
17
4
...
20
(Tb) (Tc)
(Tl)
Temperature, T /K

The phase diagram for helium
(4He)
...
Helium-II is the superfluid
phase
...
The labels hcp and bcc denote
different solid phases in which the atoms
pack together differently: hcp denotes
hexagonal closed packing and bcc denotes
body-centred cubic (see Section 20
...

Fig
...
7

122

4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES

Same
chemical
potential

it is so called because it flows without viscosity
...
6, helium is the only known substance
with a liquid–liquid boundary, shown as the λ-line (lambda line) in Fig
...
7
...
Helium-3 is unusual in that the entropy of the liquid is
lower than that of the solid, and melting is exothermic
...


Fig
...
8

We shall now see how thermodynamic considerations can account for the features of
the phase diagrams we have just described
...
In fact,
this quantity will play such an important role in this chapter and the rest of the text
that we give it a special name and symbol, the chemical potential, µ (mu)
...
The name ‘chemical potential’ is also instructive: as we
develop the concept, we shall see that µ is a measure of the potential that a substance
has for undergoing change in a system
...
In Chapter 7 we shall see that µ is the potential of
a substance to undergo chemical change
...
4 The thermodynamic criterion of equilibrium
We base our discussion on the following consequence of the Second Law: at equilibrium,
the chemical potential of a substance is the same throughout a sample, regardless of how many
phases are present
...
4
...

To see the validity of this remark, consider a system in which the chemical potential of a substance is µ1 at one location and µ2 at another location
...
When an amount dn of the substance is transferred from one location to the other, the Gibbs energy of the system changes by
−µ1dn when material is removed from location 1, and it changes by +µ2dn when
that material is added to location 2
...

If the chemical potential at location 1 is higher than that at location 2, the transfer is
accompanied by a decrease in G, and so has a spontaneous tendency to occur
...
We conclude that the transition temperature, Ttrs, is the temperature at which the chemical
potentials of two phases are equal
...
5 The dependence of stability on the conditions
At low temperatures and provided the pressure is not too low, the solid phase of a
substance has the lowest chemical potential and is therefore the most stable phase
...

When that happens, a transition to the second phase is spontaneous and occurs if it is
kinetically feasible to do so
...


4
...
50 ((∂G/∂T)p = −S)
...
1)

This relation shows that, as the temperature is raised, the chemical potential of a pure
substance decreases: Sm > 0 for all substances, so the slope of a plot of µ against T is
negative
...
1 implies that the slope of a plot of µ against temperature is steeper for
gases than for liquids, because Sm(g) > Sm(l)
...
These features are illustrated in Fig
...
9
...
The chemical potential of the gas phase plunges steeply downwards as the
temperature is raised (because the molar entropy of the vapour is so high), and there
comes a temperature at which it lies lowest
...

(b) The response of melting to applied pressure

Most substances melt at a higher temperature when subjected to pressure
...
Exceptions to this behaviour include water, for which the liquid is denser than the solid
...
That is,
water freezes at a lower temperature when it is under pressure
...
The
variation of the chemical potential with pressure is expressed (from the second of
eqn 3
...
The
phase with the lowest chemical potential at
a specified temperature is the most stable
one at that temperature
...


Fig
...
9

(4
...
An increase in pressure raises the chemical
potential of any pure substance (because Vm > 0)
...
As shown in Fig
...
10a, the effect of pressure in such

High
pressure

Chemical potential, m

(a) The temperature dependence of phase stability

(b)

High
pressure
Low
pressure

Tf´ Tf
Temperature, T

Fig
...
10 The pressure dependence of the
chemical potential of a substance depends
on the molar volume of the phase
...
(a) In this case the molar
volume of the solid is smaller than that of
the liquid and µ(s) increases less than µ(l)
...

(b) Here the molar volume is greater for
the solid than the liquid (as for water), µ(s)
increases more strongly than µ(l), and the
freezing temperature is lowered
...
For water, however, Vm(l) < Vm(s),
and an increase in pressure increases the chemical potential of the solid more than
that of the liquid
...
4
...

Example 4
...
00 bar to 2
...
The density of ice is 0
...
999 g cm−3 under these conditions
...
2, we know that the change in chemical potential of an incom-

pressible substance when the pressure is changed by ∆p is ∆ µ = Vm∆p
...
These values are obtained from the mass density, ρ, and the molar mass, M,
by using Vm = M/ρ
...

Answer The molar mass of water is 18
...
802 × 10−2 kg mol−1); therefore,

∆µ(ice) =
∆µ(water) =

(1
...
00 × 105 Pa)
917 kg m−3
(1
...
00 × 105 Pa)
999 kg m−3

= +1
...
80 J mol−1

We interpret the numerical results as follows: the chemical potential of ice rises
more sharply than that of water, so if they are initially in equilibrium at 1 bar, then
there will be a tendency for the ice to melt at 2 bar
...
1 Calculate the effect of an increase in pressure of 1
...
0 g mol−1) in equilibrium
with densities 2
...
50 g cm−3, respectively
...
87 J mol−1, ∆µ(s) = +1
...
4
...

When pressure is applied, the vapour
pressure of the condensed phase increases
...
Pressure can be exerted
on the condensed phases mechanically or by subjecting it to the applied pressure of an
inert gas (Fig
...
11); in the latter case, the vapour pressure is the partial pressure of the
vapour in equilibrium with the condensed phase, and we speak of the partial vapour
pressure of the substance
...
Another complication is that the gas phase molecules might
attract molecules out of the liquid by the process of gas solvation, the attachment of
molecules to gas phase species
...
3)

This equation shows how the vapour pressure increases when the pressure acting on
the condensed phase is increased
...
5 THE DEPENDENCE OF STABILITY ON THE CONDITIONS
Justification 4
...
It follows that, for any change that preserves equilibrium, the resulting change
in µ(l) must be equal to the change in µ(g); therefore, we can write dµ(g) = dµ(l)
...
The chemical potential of the vapour changes by
dµ(g) = Vm(g)dp where dp is the change in the vapour pressure we are trying to find
...

When there is no additional pressure acting on the liquid, P (the pressure experienced by the liquid) is equal to the normal vapour pressure p*, so when P = p*, p =
p* too
...
Provided the effect
of pressure on the vapour pressure is small (as will turn out to be the case) a good
approximation is to replace the p in p + ∆P by p* itself, and to set the upper limit of
the integral to p* + ∆P
...
3 because eln x = x
...
1 The effect of applied pressure on the vapour pressure of liquid water

For water, which has density 0
...
1 cm3 mol−1, when the pressure is increased by 10 bar (that is, ∆P = 1
...
81 × 10−5 m3 mol−1) × (1
...
3145 J K mol ) × (298 K)

=

1
...
0 × 10
8
...
It follows that p = 1
...
73 per cent
...
2 Calculate the effect of an increase in pressure of 100 bar on the vapour
pressure of benzene at 25°C, which has density 0
...

[43 per cent]

125

126

4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
4
...
Therefore, where
the phases α and β are in equilibrium,

µα(p,T) = µβ(p,T)

(4
...

dT

(a) The slopes of the phase boundaries
Temperature, T
Fig
...
12 When pressure is applied to a
system in which two phases are in
equilibrium (at a), the equilibrium is
disturbed
...
It follows that there is a
relation between dp and dT that ensures
that the system remains in equilibrium as
either variable is changed
...
Let p and T be changed infinitesimally, but in such a way that the two phases α
and β remain in equilibrium
...
They remain equal when the conditions are
changed to another point on the phase boundary, where the two phases continue to
be in equilibrium (Fig
...
12)
...
Because, from eqn 3
...
Hence
(Vβ,m − Vα,m)dp = (Sβ,m − Sα,m)dT

(4
...
6)

In this expression ∆trsS = Sβ,m − Sα,m and ∆trsV = Vβ,m − Vα,m are the entropy and
volume of transition, respectively
...
It implies that we can use thermodynamic data to predict the appearance of
phase diagrams and to understand their form
...


Pressure, p

(b) The solid–liquid boundary
Solid

Melting (fusion) is accompanied by a molar enthalpy change ∆fusH and occurs at a
temperature T
...
3),
and the Clapeyron equation becomes
Liquid

dp
dT

Temperature, T
Fig
...
13 A typical solid–liquid phase
boundary slopes steeply upwards
...
Most
substances behave in this way
...
7)

where ∆fusV is the change in molar volume that occurs on melting
...
Consequently, the slope dp/dT is steep and usually positive
(Fig
...
13)
...
If the melting temperature is T* when the pressure is p*, and T
when the pressure is p, the integration required is

4
...
8)

T*

This equation was originally obtained by yet another Thomson—James, the brother
of William, Lord Kelvin
...
2

therefore,
p ≈ p* +

∆ fusH
T*∆ fusV

(T − T*)

(4
...
4
...

(c) The liquid–vapour boundary

The entropy of vaporization at a temperature T is equal to ∆vapH/T; the Clapeyron
equation for the liquid–vapour boundary is therefore
dp
dT

=

∆ vapH

(4
...
Therefore, dp/dT
is positive, but it is much smaller than for the solid–liquid boundary
...

Example 4
...

Method To use eqn 4
...
At the boiling

point, the term ∆vapH/T is Trouton’s constant (Section 3
...
Because the molar
volume of a gas is so much greater than the molar volume of a liquid, we can write
∆vapV = Vm(g) − Vm(l) ≈ Vm(g)
and take for Vm(g) the molar volume of a perfect gas (at low pressures, at least)
...
The molar volume of a

perfect gas is about 25 dm3 mol−1 at 1 atm and near but above room temperature
...
5 × 10−2 m3 mol−1

= 3
...
If x < 1, a good
<
2
3
approximation is ln(1 + x) ≈ x
...
This value corresponds to 0
...
Therefore, a change of pressure of +0
...


Pressure, p

Liquid

Self-test 4
...
4
...
The boundary can be regarded
as a plot of the vapour pressure against the
temperature
...
4
...

This phase boundary terminates at the
critical point (not shown)
...
2 and Vm(g) = RT/p
...
2)
...
These two approximations turn the exact Clapeyron equation into
dp
dT

∆vapH

=

T(RT/p)

which rearranges into the Clausius–Clapeyron equation for the variation of vapour
pressure with temperature:
d ln p
dT

=

∆vapH

(4
...
) Like the Clapeyron equation, the Clausius–Clapeyron
equation is important for understanding the appearance of phase diagrams, particularly the location and shape of the liquid–vapour and solid–vapour phase boundaries
...
For instance, if we also assume that the enthalpy of
vaporization is independent of temperature, this equation can be integrated as follows:

Ύ

ln p

d ln p =

ln p*

∆vapH
R

Ύ

T

T*

dT
T

2

=−

∆vapH A 1
R

CT

−

1D
T* F

where p* is the vapour pressure when the temperature is T* and p the vapour pressure
when the temperature is T
...
12)°

T* F

Equation 4
...
4
...
The line does not
extend beyond the critical temperature Tc, because above this temperature the liquid
does not exist
...
2 The effect of temperature on the vapour pressure of a liquid

Equation 4
...
00 atm (101 kPa)
...
3), ∆vapH 7 = 30
...
08 × 104 J mol−1 A

1

8
...
08 × 104 A 1
1 D
=
−
353 K F
8
...
7 THE EHRENFEST CLASSIFICATION OF PHASE TRANSITIONS
and substitute this value into eqn 4
...
The result is 12 kPa
...


practice to carry out numerical calculations like this without evaluating the intermediate steps and using rounded values
...
Because the enthalpy of sublimation is greater than the enthalpy of vaporization (∆subH = ∆fusH + ∆vapH), the
equation predicts a steeper slope for the sublimation curve than for the vaporization curve at similar temperatures, which is near where they meet at the triple point
(Fig
...
15)
...
7 The Ehrenfest classification of phase transitions
There are many different types of phase transition, including the familiar examples of
fusion and vaporization and the less familiar examples of solid–solid, conducting–
superconducting, and fluid–superfluid transitions
...
The classification
scheme was originally proposed by Paul Ehrenfest, and is known as the Ehrenfest
classification
...
These changes have implications for the slopes of the
chemical potentials of the phases at either side of the phase transition
...
13)

A ∂µβ D A ∂µα D
∆trsH
−
= −Sβ,m + Sα,m = ∆trsS =
C ∂T F p C ∂T F p
Ttrs
Because ∆trsV and ∆trsH are non-zero for melting and vaporization, it follows that for
such transitions the slopes of the chemical potential plotted against either pressure or
temperature are different on either side of the transition (Fig
...
16a)
...

A transition for which the first derivative of the chemical potential with respect to
temperature is discontinuous is classified as a first-order phase transition
...
At a first-order phase transition, H changes by a finite
amount for an infinitesimal change of temperature
...
The physical reason is that heating drives the transition rather
than raising the temperature
...


Temperature, T
Fig
...
15 Near the point where they coincide
(at the triple point), the solid–gas
boundary has a steeper slope than the
liquid–gas boundary because the enthalpy
of sublimation is greater than the enthalpy
of vaporization and the temperatures that
occur in the Clausius–Clapeyron equation
for the slope have similar values
...
4
...


A second-order phase transition in the Ehrenfest sense is one in which the first
derivative of µ with respect to temperature is continuous but its second derivative is
discontinuous
...
4
...
The heat capacity is discontinuous at the transition but does not become infinite there
...
2
The term λ-transition is applied to a phase transition that is not first-order yet
the heat capacity becomes infinite at the transition temperature
...
4
...
This type of transition includes order–disorder transitions in alloys, the
onset of ferromagnetism, and the fluid–superfluid transition of liquid helium
...
1 Second-order phase transitions and λ-transitions

One type of second-order transition is associated with a change in symmetry of
the crystal structure of a solid
...
4
...
Such a crystal structure is classified
as tetragonal (see Section 20
...
Moreover, suppose the two shorter dimensions
increase more than the long dimension when the temperature is raised
...
At that point the crystal has
cubic symmetry (Fig
...
18b), and at higher temperatures it will expand equally in
all three directions (because there is no longer any distinction between them)
...

Fig
...
17 The λ-curve for helium, where the
heat capacity rises to infinity
...


2

A metallic conductor is a substance with an electrical conductivity that decreases as the temperature increases
...
See Chapter 20 for more
details
...
4
...
There is no
rearrangement of atoms at the transition temperature, and hence no enthalpy of transition
...

The low-temperature phase is an orderly array of alternating Cu and Zn atoms
...
4
...
At T = 0 the
order is perfect, but islands of disorder appear as the temperature is raised
...
The islands grow in extent, and merge throughout the crystal at the transition
temperature (742 K)
...


(c)
Fig
...
19 An order–disorder transition
...
(b) As the temperature is
increased, atoms exchange locations and
islands of each kind of atom form in
regions of the solid
...
(c) At and above the
transition temperature, the islands occur at
random throughout the sample
...
A phase is a form of matter that is uniform throughout in
chemical composition and physical state
...
The boiling temperature is the temperature at which the
vapour pressure of a liquid is equal to the external pressure
...
A transition temperature is the temperature at which the two
phases are in equilibrium
...
The critical temperature is the temperature at which a liquid
surface disappears and above which a liquid does not exist
whatever the pressure
...


3
...

4
...

5
...


10
...

11
...


6
...


12
...


7
...


13
...


132

4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES

14
...


17
...


15
...


18
...


16
...


19
...


Further reading
Articles and texts

E
...
H
...
In Encyclopedia of applied
physics (ed
...
L
...
VCH, New York (1995)
...
M
...
In Encyclopedia of applied
physics (ed
...
L
...
VCH, New York (1995)
...
M
...
J
...
Educ
...


W
...
Callister, Jr
...

Wiley, New York (2000)
...
Boublik, V
...
Hála, The vapor pressures of pure
substances
...

R
...
Weast (ed
...
81
...


Discussion questions
4
...

4
...

4
...

4
...
4
...
0 atm and 298 K, is subjected to the
following cycle: (a) isobaric (constant–pressure) heating to 320 K,
(b) isothermal compression to 100 atm, (c) isobaric cooling to 210 K,
(d) isothermal decompression to 1
...


4
...
Consult
library and internet resources and prepare a discussion of the principles,
advantages, disadvantages, and current uses of supercritical fluid extraction
technology
...
6 Explain the significance of the Clapeyron equation and of the
Clausius–Clapeyron equation
...
7 Distinguish between a first-order phase transition, a second-order phase
transition, and a λ-transition at both molecular and macroscopic levels
...
1(a) The vapour pressure of dichloromethane at 24
...
3 kPa and its
enthalpy of vaporization is 28
...
Estimate the temperature at which
its vapour pressure is 70
...

4
...
0°C is 58
...
7 kJ mol−1
...
0 kPa
...
2(a) The molar volume of a certain solid is 161
...
00 atm and
350
...
The molar volume of the liquid at this
temperature and pressure is 163
...
At 100 atm the melting
temperature changes to 351
...
Calculate the enthalpy and entropy of
fusion of the solid
...


4
...
0 cm3 mol−1 at 1
...
15 K, its melting temperature
...
6 cm3 mol−1
...
2 MPa the melting
temperature changes to 429
...
Calculate the enthalpy and entropy of
fusion of the solid
...
3(a) The vapour pressure of a liquid in the temperature range 200 K to
260 K was found to fit the expression ln(p/Torr) = 16
...
8/(T/K)
...

4
...
361 – 3036
...

Calculate the enthalpy of vaporization of the liquid
...
4(a) The vapour pressure of benzene between 10°C and 30°C fits the
expression log(p/Torr) = 7
...
Calculate (a) the enthalpy of
vaporization and (b) the normal boiling point of benzene
...
4(b) The vapour pressure of a liquid between 15°C and 35°C fits the

expression log(p/Torr) = 8
...
Calculate (a) the enthalpy of
vaporization and (b) the normal boiling point of the liquid
...
5(a) When benzene freezes at 5
...
879 g cm−3 to

0
...
Its enthalpy of fusion is 10
...
Estimate the freezing
point of benzene at 1000 atm
...
5(b) When a certain liquid freezes at −3
...
789 g cm−3 to 0
...
Its enthalpy of fusion is 8
...
Estimate
the freezing point of the liquid at 100 MPa
...
6(a) In July in Los Angeles, the incident sunlight at ground level has a power

density of 1
...
A swimming pool of area 50 m2 is directly
exposed to the sun
...


4
...
87 kW m−2 at noon
...
0 ha? (1 ha = 104 m2
...


133

substance will be found in the air if there is no ventilation? (The vapour
pressures are (a) 3
...
1 kPa, (c) 0
...
)
4
...
30 kPa
...
8(a) Naphthalene, C10H8, melts at 80
...
If the vapour pressure of the

liquid is 1
...
8°C and 5
...
3°C, use the Clausius–Clapeyron
equation to calculate (a) the enthalpy of vaporization, (b) the normal boiling
point, and (c) the enthalpy of vaporization at the boiling point
...
8(b) The normal boiling point of hexane is 69
...
Estimate (a) its enthalpy
of vaporization and (b) its vapour pressure at 25°C and 60°C
...
9(a) Calculate the melting point of ice under a pressure of 50 bar
...
92 g cm−3 and
that of liquid water is 1
...

4
...
Assume
that the density of ice under these conditions is approximately 0
...
998 g cm−3
...
10(a) What fraction of the enthalpy of vaporization of water is spent on

expanding the water vapour?

4
...
10(b) What fraction of the enthalpy of vaporization of ethanol is spent on

a laboratory measuring 5
...
0 m × 3
...
What mass of each

expanding its vapour?

Problems*
Numerical problems
4
...
5916 − 1871
...
3186 − 1425
...
Estimate the temperature and pressure of the triple
point of sulfur dioxide
...
2 Prior to the discovery that freon-12 (CF2Cl2) was harmful to the Earth’s

boiling point of water
...
917 g cm−3
and 1
...
958 g
cm−3 and 0
...
By how much does the chemical potential
of water vapour exceed that of liquid water at 1
...
6 The enthalpy of fusion of mercury is 2
...
3 K with a change in molar volume of +0
...
At what temperature will the bottom of a column of mercury
(density 13
...
0 m be expected to freeze?

ozone layer, it was frequently used as the dispersing agent in spray cans for
hair spray, etc
...
2°C is 20
...
Estimate the pressure that a can of hair spray using
freon-12 had to withstand at 40°C, the temperature of a can that has been
standing in sunlight
...
2°C
...
Calculate the final
temperature
...
17 kPa throughout, and its heat capacity is 75
...
Assume that the
air is not heated or cooled and that water vapour is a perfect gas
...
3 The enthalpy of vaporization of a certain liquid is found to be 14
...
8 The vapour pressure, p, of nitric acid varies with temperature as follows:

mol−1 at 180 K, its normal boiling point
...
5 dm3 mol−1,
respectively
...


4
...
(c) By how much does the chemical
potential of water supercooled to −5
...
5 Calculate the difference in slope of the chemical potential against pressure
on either side of (a) the normal freezing point of water and (b) the normal

4
...
0 dm3 of dry air was slowly bubbled through a thermally insulated

θ/°C

0

20

p/kPa 1
...
38

40

50

70

80

90

100

17
...
7

62
...
3

124
...
9

What are (a) the normal boiling point and (b) the enthalpy of vaporization of
nitric acid?
4
...
2 g mol−1), a

component of oil of spearmint, is as follows:
θ/°C
p/Torr

57
...
00

100
...
0

157
...
5

227
...
0

40
...


134

4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES

4
...
50°C using the following data: ∆fusH = 10
...
8 kJ
mol−1, ρ(s) = 0
...
879 g cm−3
...
11‡ In an investigation of thermophysical properties of toluene (R
...

Goodwin J
...
Chem
...
Data 18, 1565 (1989)) presented expressions for
two coexistence curves (phase boundaries)
...
60 + 11
...
4362 µbar and T3 = 178
...
The liquid–vapour curve is given by:
ln(p/bar) = −10
...
157 − 15
...
015y 2 − 5
...
7224(1 − y)1
...
95 K)
...
(b) Estimate the standard melting point of toluene
...
(d) Compute the standard
enthalpy of vaporization of toluene, given that the molar volumes of the
liquid and vapour at the normal boiling point are 0
...
3 dm3 mol−1, respectively
...
12‡ In a study of the vapour pressure of chloromethane, A
...

Dupont-Pavlovsky (J
...
Eng
...
Some of that
data is shown below:

T/K

145
...
96

149
...
94

153
...
94

p/Pa

13
...
49

25
...
76

50
...
56

Estimate the standard enthalpy of sublimation of chloromethane at 150 K
...
)

Theoretical problems
4
...

4
...
The Clapeyron

equation relates dp and dT at equilibrium, and so in combination the two
equations can be used to find how the enthalpy changes along a phase
boundary as the temperature changes and the two phases remain in
equilibrium
...

4
...
Show that the vapour pressure, p, of the liquid
is related to its molar mass, M, by p = AmP/(1 + Am), where A = RT/MPV
...
2 g mol−1), which is a component of oil
of roses, was measured at 110°C
...
00 dm3 of nitrogen
at 760 Torr was passed slowly through the heated liquid, the loss of mass was
0
...
Calculate the vapour pressure of geraniol
...
16 Combine the barometric formula (stated in Impact I1
...
Take the mean ambient temperature
as 20°C and predict the boiling temperature of water at 3000 m
...
17 Figure 4
...
All have a negative slope, but it is unlikely that they are truly
straight lines as indicated in the illustration
...
Is there a restriction on the curvature of these lines? Which state of
matter shows the greatest curvature?

4
...
They are:

dp
dT

=

α 2 − α1

dp

κT,2 − κT,1

dT

=

Cp,m2 − Cp,m1
TVm(α2 − α1)

where α is the expansion coefficient, κT the isothermal compressibility, and
the subscripts 1 and 2 refer to two different phases
...
Why does the Clapeyron equation not apply to second-order
transitions?
4
...


Applications: to biology and engineering
4
...
However,
when certain conditions are changed, the compact structure of a polypeptide
chain may collapse into a random coil
...
A thermodynamic
treatment allows predictions to be made of the temperature Tm for the
unfolding of a helical polypeptide held together by hydrogen bonds into a
random coil
...
Because the first and last residues in the
chain are free to move, n − 2 residues form the compact helix and have
restricted motion
...
(a) Justify the form of the
equation for the Gibbs energy of unfolding
...
At what value of n does Tm change
by less than 1% when n increases by one?
4
...
The solubility parameter, δ, is defined as
(∆Ucohesive /Vm)1/2, where ∆Ucohesive is the cohesive energy of the solvent, the
energy per mole needed to increase the volume isothermally to an infinite
value
...
(a) Derive a practical
equation for the computation of the isotherms for the reduced internal energy
change, ∆Ur(Tr,Vr) defined as

∆Ur(Tr,Vr) =

Ur(Tr,Vr) − Ur(Tr,∞)
pcVc

(b) Draw a graph of ∆Ur against pr for the isotherms Tr = 1,1
...
5 in the
reduced pressure range for which 0
...
(c) Draw a graph of δ against pr
for the carbon dioxide isotherms Tr = 1 and 1
...
In what pressure range at Tf = 1 will carbon dioxide have

PROBLEMS
solvent properties similar to those of liquid carbon tetrachloride? Hint
...

4
...
Friend et al
...
G
...
F
...
Ingham, J
...
Chem
...
Data 18,
583 (1989)), which included the following data describing the liquid–vapour
phase boundary
...
034 0
...
088 0
...
122 0
...
368 0
...
041 1
...
329 4
...
(b) Estimate the standard boiling
point of methane
...
80 × 10−2 and 8
...


135

4
...
For these reasons, diamond is used widely
in industrial applications that require a strong abrasive
...
To illustrate this point, calculate the pressure
required to convert graphite into diamond at 25°C
...
Assume the specific volume, Vs, and κT are constant with
respect to pressure changes
...
8678

∆ f G /(kJ mol )

0

Vs /(cm3 g−1)

0
...
284

κT /kPa

3
...
187 × 10−8

5
The thermodynamic description
of mixtures
5
...
2 The thermodynamics of mixing
5
...
1 Impact on biology: Gas

solubility and breathing
The properties of solutions
5
...
5 Colligative properties
I5
...
6 The solvent activity
5
...
8 The activities of regular

Simple mixtures
This chapter begins by developing the concept of chemical potential to show that it is a particular case of a class of properties called partial molar quantities
...

The underlying principle to keep in mind is that at equilibrium the chemical potential of a
species is the same in every phase
...
With this result established, we can
calculate the effect of a solute on certain thermodynamic properties of a solution
...
Finally, we see
how to express the chemical potential of a substance in a real mixture in terms of a property known as the activity
...


Chemistry deals with mixtures, including mixtures of substances that can react
together
...
As a first step towards dealing with chemical
reactions (which are treated in Chapter 7), here we consider mixtures of substances
that do not react together
...
We shall therefore often be able to simplify
equations by making use of the relation xA + xB = 1
...
9 The activities of ions in

solution
Checklist of key ideas
Further reading
Further information 5
...
For
a more general description of the thermodynamics of mixtures we need to introduce
other analogous ‘partial’ properties
...
1 Partial molar quantities
The easiest partial molar property to visualize is the ‘partial molar volume’, the contribution that a component of a mixture makes to the total volume of a sample
...
When a further 1 mol H2O is added, the
volume increases by 18 cm3 and we can report that 18 cm3 mol−1 is the molar volume

5
...
2 0
...
6 0
...
Note the
different scales (water on the left, ethanol
on the right)
...
5
...
3)

Although we have envisaged the two integrations as being linked (in order to preserve
constant composition), because V is a state function the final result in eqn 5
...

Partial molar volumes can be measured in several ways
...
Once the function has been found, its slope
can be determined at any composition of interest by differentiation
...
For instance, the partial molar volume of NaCl in water could be
written {(NaCl, aq) to distinguish it from the volume of the solution, V(NaCl, aq)
...
2)

Provided the composition is held constant as the amounts of A and B are increased,
the final volume of a mixture can be calculated by integration
...
1]

where the subscript n′ signifies that the amounts of all other substances present are
constant
...
5
...
Its value depends on the composition, as we saw for
water and ethanol
...
1 implies that, when the composition of the
mixture is changed by the addition of dnA of A and dnB of B, then the total volume of
the mixture changes by
dV =

58
Ethanol

–1
3
Partial molar volume of ethanol, V (C2H5OH)/(cm mol )

of pure water
...
The reason for the different increase in volume
is that the volume occupied by a given number of water molecules depends on the
identity of the molecules that surround them
...
The quantity 14 cm3 mol−1 is the partial molar volume of water in pure
ethanol
...

The partial molar volumes of the components of a mixture vary with composition
because the environment of each type of molecule changes as the composition
changes from pure A to pure B
...

The partial molar volumes of water and ethanol across the full composition range at
25°C are shown in Fig
...
1
...
In general, partial molar
quantities vary with the composition, as
shown by the different slopes at the
compositions a and b
...


Fig
...
2

138

5 SIMPLE MIXTURES

Partial molar volume, VE/(cm3 mol 1)

56

Illustration 5
...
000 kg of water is
55

v = 1002
...
6664x − 0
...
028 256x 3
where v = V/cm3, x = nE/mol, and nE is the amount of CH3CH2OH present
...
1
...
1, determine the value
of b at which VE has a minimum value
...
6664 − 2(0
...
028256)x 2

we can conclude that

Fig
...
3

VE/(cm3 mol−1) = 54
...
72788x + 0
...
3 is a graph of this function
...
1 At 25°C, the density of a 50 per cent by mass ethanol/water solution
is 0
...
Given that the partial molar volume of water in the solution is
17
...
4 cm3 mol−1]

Molar volumes are always positive, but partial molar quantities need not be
...
4 cm3 mol−1, which means that the addition
of 1 mol MgSO4 to a large volume of water results in a decrease in volume of 1
...

The mixture contracts because the salt breaks up the open structure of water as the
ions become hydrated, and it collapses slightly
...
For a substance in a mixture, the chemical potential is defined as the partial molar
Gibbs energy:

µJ =
a
0

b
Amount of A, nA

The chemical potential of a
substance is the slope of the total Gibbs
energy of a mixture with respect to the
amount of substance of interest
...
In this case, both chemical
potentials are positive
...
5
...
4]

That is, the chemical potential is the slope of a plot of Gibbs energy against the amount
of the component J, with the pressure and temperature (and the amounts of the other
substances) held constant (Fig
...
4)
...
4 obtain µJ = GJ,m: in this case, the chemical potential is simply the molar
Gibbs energy of the substance, as we used in Chapter 4
...
3, it follows that the total Gibbs energy of a
binary mixture is
G = nA µA + nB µB

(5
...
That
is, the chemical potential of a substance in a mixture is the contribution of that

5
...
Because the chemical potentials
depend on composition (and the pressure and temperature), the Gibbs energy of a
mixture may change when these variables change, and for a system of components A,
B, etc
...
6)

This expression is the fundamental equation of chemical thermodynamics
...

At constant pressure and temperature, eqn 5
...
7)

We saw in Section 3
...
Therefore, at
constant temperature and pressure,
dwadd,max = µAdnA + µBdnB + · · ·

(5
...
For instance, in an electrochemical cell, the chemical reaction is arranged to
take place in two distinct sites (at the two electrodes)
...

(c) The wider significance of the chemical potential

The chemical potential does more than show how G varies with composition
...
43 (that dU = TdS − pdV) to systems in
which the composition may change
...
9)

and hence that

µJ =

A ∂U D
C ∂nJ F S,V,n′

(5
...
In the same way it is easy to deduce that
(a) µJ =

A ∂H D
C ∂nJ F S,p,n′

(b) µJ =

A ∂A D
C ∂nJ F V,T,n′

(5
...
This is why the chemical potential is so central
to chemistry
...
5 and the
chemical potentials depend on the composition, when the compositions are changed
infinitesimally we might expect G of a binary system to change by
dG = µAdnA + µ BdnB + nAd µA + nBd µB

139

140

5 SIMPLE MIXTURES
However, we have seen that at constant pressure and temperature a change in Gibbs
energy is given by eqn 5
...
Because G is a state function, these two equations must be
equal, which implies that at constant temperature and pressure
nAdµA + nBdµB = 0

(5
...
12b)

J

The significance of the Gibbs–Duhem equation is that the chemical potential of one
component of a mixture cannot change independently of the chemical potentials of
the other components
...
13)
nB
The same line of reasoning applies to all partial molar quantities
...
5
...
Moreover, as eqn 5
...
5
...
In practice, the Gibbs–Duhem equation is used to determine the partial molar
volume of one component of a binary mixture from measurements of the partial
molar volume of the second component
...
1

The molar concentration (colloquially,
the ‘molarity’, [J] or cJ) is the amount of
solute divided by the volume of the
solution and is usually expressed in
moles per cubic decimetre (mol dm−3)
...
The term
molality, b, is the amount of solute
divided by the mass of solvent and is
usually expressed in moles per kilogram
of solvent (mol kg−1)
...


Example 5
...
280 + 18
...
1)
...
The molar volume of pure
water at 298 K is 18
...

Method Let A denote H2O, the solvent, and B denote K2SO4, the solute
...
This relation implies that dvA = −(nB/nA)dvB, and therefore that
vA can be found by integration:
vA = vA −
*

Ύn

nB

dvB

A

where vA = VA/(cm3 mol−1) is the numerical value of the molar volume of pure A
...

Answer It follows from the information in the question that, with B = K2SO4,

dvB/dx = 9
...
Therefore, the integration required is
b/b 7

vB = vB − 9
...
2 THE THERMODYNAMICS OF MIXING
40

nB
(1 kg)/MA

=

nBMA
1 kg

= bMA = xb7MA

Ύ

x1/2dx = vA − – {9
...
802 × 10−2 kg mol−1, the
molar mass of water), that
VA/(cm3 mol−1) = 18
...
1094(b/b7)3/2

36

V(H2O)/(cm3 mol–1)

b/b7

vA = vA − 9
...
078

34

and hence
7

V(K2SO4)/(cm3 mol–1)

nA

=

18
...
076

The partial molar volumes are plotted in Fig
...
5
...
2 Repeat the calculation for a salt B for which VB/(cm3 mol−1) = 6
...
146b − 7
...


32

[VA/(cm3 mol−1) = 18
...
0464b2 + 0
...
05
b/(mol kg–1)

18
...
1

The partial molar volumes of the
components of an aqueous solution of
potassium sulfate
...
5
...
2 The thermodynamics of mixing
The dependence of the Gibbs energy of a mixture on its composition is given by
eqn 5
...
This is the link we need in order to apply thermodynamics to the
discussion of spontaneous changes of composition, as in the mixing of two substances
...
The mixing is spontaneous, so it must correspond
to a decrease in G
...

(a) The Gibbs energy of mixing of perfect gases

nA, T, p

Let the amounts of two perfect gases in the two containers be nA and nB; both are at
a temperature T and a pressure p (Fig
...
6)
...
57:

T, pA, pB with pA + pB = p

µ = µ 7 + RT ln

p
p7

(5
...
It will be much simpler notationally if we agree to let p denote the pressure
relative to p 7; that is, to replace p/p7 by p, for then we can write

µ = µ 7 + RT ln p

{5
...
; to use the
equations, we have to remember to replace p by p/p7 again
...
The Gibbs energy of the total system is
then given by eqn 5
...
15}°

The arrangement for calculating the
thermodynamic functions of mixing of two
perfect gases
...
5
...
The total
Gibbs energy changes to

0

7
7
Gf = nA(µ A + RT ln pA) + nB(µ B + RT ln pB)

DmixG /nRT

-0
...
4

pA
p

+ nB RT ln

pB
p

(5
...
2b) to write
pJ/p = xJ for each component, which gives

-0
...
8
0

{5
...
5
Mole fraction of A, xA

1

The Gibbs energy of mixing of two
perfect gases and (as discussed later) of two
liquids that form an ideal solution
...


(5
...
5
...
The conclusion that ∆mixG is negative for all compositions confirms that perfect gases mix spontaneously in all proportions
...


Fig
...
7

Exploration Draw graphs of ∆mixG
against xA at different temperatures
in the range 298 K to 500 K
...
0 mol H2

1
...
2 Calculating a Gibbs energy of mixing

A container is divided into two equal compartments (Fig
...
8)
...
0 mol H2(g) at 25°C; the other contains 1
...
Calculate the
Gibbs energy of mixing when the partition is removed
...

Method Equation 5
...
We proceed by calculating the initial Gibbs energy from the
chemical potentials
...
Write the pressure
of nitrogen as p; then the pressure of hydrogen as a multiple of p can be found from
the gas laws
...
The volume occupied by each gas doubles, so its initial partial pressure is
halved
...
0 mol){µ 7(H2) + RT ln 3p} + (1
...
0 mol H2

1
...

Fig
...
8

When the partition is removed and each gas occupies twice the original volume,
1
3
the partial pressure of nitrogen falls to – p and that of hydrogen falls to – p
...
0 mol){µ 7(H2) + RT ln – p} + (1
...
0 mol)RT ln

3
1
A –p D
A –p D
2
+ (1
...
0 mol)RT ln2 − (1
...
0 mol)RT ln2 = −6
...

When 3
...
0 mol N2 at the same pressure, with the volumes of
the vessels adjusted accordingly, the change of Gibbs energy is −5
...


5
...
8

Self-test 5
...
0 mol H2 at 2
...
0 mol N2 at

3
...
Calculate ∆mixG
...
7 kJ, −9
...
18 that, for a mixture of
perfect gases initially at the same pressure, the entropy of mixing, ∆mixS, is
∆mixS =

A ∂∆mixG D
C ∂T F p,n

= −nR(xA ln xA + xB ln xB)

(5
...
6

0
...
2

A,nB

Because ln x < 0, it follows that ∆mixS > 0 for all compositions (Fig
...
9)
...
This increase in entropy is what we expect when one
gas disperses into the other and the disorder increases
...
It follows from eqns 5
...
19 that
∆mix H = 0

(5
...
It follows that the
whole of the driving force for mixing comes from the increase in entropy of the system, because the entropy of the surroundings is unchanged
...
3 The chemical potentials of liquids
To discuss the equilibrium properties of liquid mixtures we need to know how the
Gibbs energy of a liquid varies with composition
...


0

0

0
...
The
entropy increases for all compositions
and temperatures, so perfect gases mix
spontaneously in all proportions
...

Hence, the graph also shows the total
entropy of the system plus the
surroundings when perfect gases mix
...
5
...
Because the vapour pressure of the pure liquid is pA it follows from
*,
7
eqn 5
...
These two chem*
ical potentials are equal at equilibrium (Fig
...
10), so we can write
7
µ A = µ A + RT ln pA
*
*

{5
...
The vapour and
solvent are still in equilibrium, so we can write
7
µA = µ A + RT ln pA

{5
...
To do so, we write eqn 5
...
22 to obtain

Equal at
equilibrium

mA(l)
A(l) + B(l)

Fig
...
10 At equilibrium, the chemical
potential of the gaseous form of a substance
A is equal to the chemical potential of its
condensed phase
...
Because the
chemical potential of A in the vapour
depends on its partial vapour pressure, it
follows that the chemical potential of liquid
A can be related to its partial vapour
pressure
...
5
...


Benzene

Methylbenzene

0
1
Mole fraction of
methylbenzene, x (C6H5CH3)

Fig
...
12 Two similar liquids, in this case
benzene and methylbenzene (toluene),
behave almost ideally, and the variation of
their vapour pressures with composition
resembles that for an ideal solution
...
23)

In the final step we draw on additional experimental information about the relation
between the ratio of vapour pressures and the composition of the liquid
...
That is, he established
what we now call Raoult’s law:

Blocked

pA = xA pA
*

(5
...
5
...
Some mixtures obey Raoult’s law very well, especially when the components are structurally similar (Fig
...
12)
...
When we write equations that are valid only for ideal solutions, we shall label
them with a superscript °, as in eqn 5
...

For an ideal solution, it follows from eqns 5
...
24 that

µA = µ A + RT ln xA
*

(5
...
It is in fact a better definition than
eqn 5
...

Molecular interpretation 5
...
The large
spheres represent solvent molecules at the
surface of a solution (the uppermost line of
spheres), and the small spheres are solute
molecules
...

Fig
...
13

The origin of Raoult’s law can be understood in molecular terms by considering
the rates at which molecules leave and return to the liquid
...
5
...


5
...
The rate at which molecules condense is
proportional to their concentration in the gas phase, which in turn is proportional
to their partial pressure:
rate of condensation = k′pA
At equilibrium, the rates of vaporization and condensation are equal, so k′pA = kxA
...
Equation 5
...

Some solutions depart significantly from Raoult’s law (Fig
...
14)
...
The law is therefore a good approximation for
the properties of the solvent if the solution is dilute
...
However, the
English chemist William Henry found experimentally that, for real solutions at low
concentrations, although the vapour pressure of the solute is proportional to its mole
fraction, the constant of proportionality is not the vapour pressure of the pure substance (Fig
...
15)
...
5
...


0

Mole fraction of B, xB

1

Fig
...
15 When a component (the solvent) is
nearly pure, it has a vapour pressure that is
proportional to mole fraction with a slope
pB (Raoult’s law)
...


145

146

5 SIMPLE MIXTURES
pB = xBKB

Fig
...
16 In a dilute solution, the solvent
molecules (the purple spheres) are in an
environment that differs only slightly from
that of the pure solvent
...


(5
...

Mixtures for which the solute obeys Henry’s law and the solvent obeys Raoult’s law
are called ideal-dilute solutions
...
The difference in behaviour of the
solute and solvent at low concentrations (as expressed by Henry’s and Raoult’s laws,
respectively) arises from the fact that in a dilute solution the solvent molecules are in
an environment very much like the one they have in the pure liquid (Fig
...
16)
...
Thus, the solvent behaves like a slightly
modified pure liquid, but the solute behaves entirely differently from its pure state
unless the solvent and solute molecules happen to be very similar
...

Example 5
...
3

0
...
7
33
...
40
11
23
...
60
18
...
3

0
...
7
4
...
4
0

Confirm that the mixture conforms to Raoult’s law for the component in large excess and to Henry’s law for the minor component
...

Method Both Raoult’s and Henry’s laws are statements about the form of the

0
0

Henry's
law

0
...
4

0
...
8 1
...
5
...
3
...


graph of partial vapour pressure against mole fraction
...
Raoult’s law is tested by comparing the
data with the straight line pJ = xJ p * for each component in the region in which it is
J
in excess (and acting as the solvent)
...

Answer The data are plotted in Fig
...
17 together with the Raoult’s law lines
...
3 kPa for propanone and K = 22
...
Notice how the system deviates from both Raoult’s and Henry’s
laws even for quite small departures from x = 1 and x = 0, respectively
...
5
...
4 The vapour pressure of chloromethane at various mole fractions in a
mixture at 25°C was found to be as follows:

x
p/kPa

0
...
3

0
...
4

Estimate Henry’s law constant
...
019
101

0
...
3 THE CHEMICAL POTENTIALS OF LIQUIDS
pB = bBKB
Some Henry’s law data for this convention are listed in Table 5
...
As well as providing
a link between the mole fraction of solute and its partial pressure, the data in the table
may also be used to calculate gas solubilities
...


Synoptic Table 5
...
01 × 103

H2

1
...
56 × 105

O2

7
...
2 Using Henry’s law
* More values are given in the Data section
...
9 × 104 kPa kg mol−1

= 2
...
29 mmol kg−1
...
99709 kg dm−3
...
29 mmol kg−1 × 0
...
29 mmol dm−3
A note on good practice The number of significant figures in the result of a calcu-

lation should not exceed the number in the data (only two in this case)
...
5 Calculate the molar solubility of nitrogen in water exposed to air at

25°C; partial pressures were calculated in Example 1
...


[0
...
1 Gas solubility and breathing

We inhale about 500 cm3 of air with each breath we take
...
Expiration occurs as the diaphragm rises and the chest contracts, and gives rise
to a differential pressure of about 100 Pa above atmospheric pressure
...
5 dm3
...

A knowledge of Henry’s law constants for gases in fats and lipids is important for
the discussion of respiration
...
Alveolar gas is in fact a mixture of newly inhaled air
and air about to be exhaled
...
3 kPa), whereas the partial pressure of freshly inhaled air is about 104 Torr (13
...
Arterial blood remains in
the capillary passing through the wall of an alveolus for about 0
...
25 s
...
Carbon dioxide moves in the opposite direction across the respiratory

147

Comment 5
...


148

5 SIMPLE MIXTURES
tissue, but the partial pressure gradient is much less, corresponding to about 5 Torr
(0
...
3 kPa) in air at equilibrium
...

A hyperbaric oxygen chamber, in which oxygen is at an elevated partial pressure,
is used to treat certain types of disease
...
Diseases that are caused by anaerobic
bacteria, such as gas gangrene and tetanus, can also be treated because the bacteria
cannot thrive in high oxygen concentrations
...

The latter increases by about 1 atm for each 10 m of descent
...
The result is nitrogen narcosis, with symptoms like intoxication
...
Many cases of scuba drowning appear to be consequences of arterial
embolisms (obstructions in arteries caused by gas bubbles) and loss of consciousness
as the air bubbles rise into the head
...
First, we consider
the simple case of mixtures of liquids that mix to form an ideal solution
...
The calculation provides a background for discussing the deviations from ideal behaviour exhibited by real solutions
...
4 Liquid mixtures
Thermodynamics can provide insight into the properties of liquid mixtures, and a few
simple ideas can bring the whole field of study together
...
2)
...
25 and the
total Gibbs energy is
Gf = nA{µ A + RT ln xA} + nB{µ B +RT ln xB}
*
*
Consequently, the Gibbs energy of mixing is
∆mixG = nRT{xA ln xA + xB ln xB}

(5
...
As for gases, it follows that the ideal entropy of mixing of two
liquids is
∆mixS = −nR{xA ln xA + xB ln xB}

(5
...
4 LIQUID MIXTURES
and, because ∆mix H = ∆mixG + T∆mixS = 0, the ideal enthalpy of mixing is zero
...
50 ((∂G/∂p)T = V) that ∆mixV = (∂∆mixG/∂p)T, but ∆mixG in
eqn 5
...

Equation 5
...
It should be noted,
however, that solution ideality means something different from gas perfection
...
In ideal solutions there are
interactions, but the average energy of A-B interactions in the mixture is the same as
the average energy of A-A and B-B interactions in the pure liquids
...
5
...
5
...

Real solutions are composed of particles for which A-A, A-B, and B-B interactions are all different
...
If the enthalpy change is large and positive or if the
entropy change is adverse (because of a reorganization of the molecules that results in
an orderly mixture), then the Gibbs energy might be positive for mixing
...
Alternatively, the liquids
might be partially miscible, which means that they are miscible only over a certain
range of compositions
...
The excess entropy, SE, for example, is
defined as
is given by eqn 5
...
The excess enthalpy and volume are both equal
where ∆mixS
to the observed enthalpy and volume of mixing, because the ideal values are zero in
each case
...
In this connection a useful model system is the regular solution, a
solution for which HE ≠ 0 but SE = 0
...
Figure 5
...

We can make this discussion more quantitative by supposing that the excess enthalpy depends on composition as
(5
...
The function given by eqn 5
...
5
...
5
...
If
β < 0, mixing is exothermic and the solute–solvent interactions are more favourable
than the solvent–solvent and solute–solute interactions
...
29]

ideal

H E = nβRTxAxB

400

It is on the basis of this distinction that the term ‘perfect gas’ is preferable to the more common ‘ideal gas’
...
5
x (C6H6)

1

8
4

V E/(mm3 mol–1)

SE = ∆mixS − ∆mixSideal

800

H E/(J mol–1)

(b) Excess functions and regular solutions

149

0
–4
–8

–12
(b) 0

0
...
5
...
(a) HE for benzene/cyclohexane;
this graph shows that the mixing is
endothermic (because ∆mix H = 0 for an
ideal solution)
...


5 SIMPLE MIXTURES
+0
...
1
3

0

1

2
...
1

H E/nRT

150

0

0

-1

0

0
...
5

-0
...
4

-2
-0
...
2

1

The excess enthalpy according to a
model in which it is proportional to βxAxB,
for different values of the parameter β
...
5
...
For
what value of xA does the excess enthalpy
depend on temperature most strongly?

-0
...
5
xA

1

Fig
...
20 The Gibbs energy of mixing for
different values of the parameter β
...
5 and vary the temperature
...
Because the entropy of mixing has its ideal value for a regular solution, the
excess Gibbs energy is equal to the excess enthalpy, and the Gibbs energy of mixing is
∆mixG = nRT{xA ln xA + xB ln xB + βxAxB}

(5
...
20 shows how ∆mixG varies with composition for different values of β
...
The implication of this observation is that, provided β > 2, then the system will
separate spontaneously into two phases with compositions corresponding to the two
minima, for that separation corresponds to a reduction in Gibbs energy
...
8 and 6
...

5
...
In dilute solutions these properties depend only on the number of solute particles present, not their identity
...

We assume throughout the following that the solute is not volatile, so it does not
contribute to the vapour
...
The latter
assumption is quite drastic, although it is true of many mixtures; it can be avoided at
the expense of more algebra, but that introduces no new principles
...
For an ideal-dilute solution, the

5
...
2 The lowering of vapour pressure of a solvent in a mixture

The molecular origin of the lowering of the chemical potential is not the energy of
interaction of the solute and solvent particles, because the lowering occurs even in
an ideal solution (for which the enthalpy of mixing is zero)
...

The pure liquid solvent has an entropy that reflects the number of microstates
available to its molecules
...
When a solute is present, there is an additional contribution to the
entropy of the liquid, even in an ideal solution
...
5
...
The effect of the solute appears as a lowered vapour pressure, and
hence a higher boiling point
...
Consequently, a lower temperature must be reached before
equilibrium between solid and solution is achieved
...


Pure liquid

Solid
Chemical potential, m

reduction is from µA for the pure solvent to µA + RT ln xA when a solute is present
*
*
(ln xA is negative because xA < 1)
...
As can be seen from Fig
...
21, the reduction in
chemical potential of the solvent implies that the liquid–vapour equilibrium occurs at
a higher temperature (the boiling point is raised) and the solid–liquid equilibrium
occurs at a lower temperature (the freezing point is lowered)
...
5
...
We denote the
solvent by A and the solute by B
...
32)°

(The pressure of 1 atm is the same throughout, and will not be written explicitly
...
33)°

T´
b

Boiling point
elevation

Fig
...
21 The chemical potential of a solvent
in the presence of a solute
...


p*
A

The strategy for the quantitative discussion of the elevation of boiling point and the
depression of freezing point is to look for the temperature at which, at 1 atm, one
phase (the pure solvent vapour or the pure solid solvent) has the same chemical potential as the solvent in the solution
...


151

(a)

pA

(b)

Fig
...
22 The vapour pressure of a pure
liquid represents a balance between the
increase in disorder arising from
vaporization and the decrease in disorder
of the surroundings
...

(b) When solute (the dark squares) is
present, the disorder of the condensed
phase is higher than that of the pure liquid,
and there is a decreased tendency to
acquire the disorder characteristic of the
vapour
...
1 The elevation of the boiling point of a solvent

Equation 5
...


RT

=

The series expansion of a natural
logarithm (see Appendix 2) is
1
1
ln(1 − x) = −x − –x 2 − – x 3 · · ·
2
3

provided that −1 < x < 1
...


RT

1 d(∆ vapG/T)
R

dT

=−

∆ vapH
RT 2

Now multiply both sides by dT and integrate from xA = 1, corresponding to ln xA = 0
(and when T = T*, the boiling point of pure A) to xA (when the boiling point is T):

Ύ

ln xA

d ln xA = −

RΎ

T

1

∆ vapH

T*

0

T2

dT

The left-hand side integrates to ln xA, which is equal to ln(1 − xB)
...

Thus, we obtain
ln(1 − xB) =

Comment 5
...
First, to find
the relation between a change in composition and the resulting change in boiling
temperature, we differentiate both sides with respect to temperature and use the
Gibbs–Helmholtz equation (eqn 3
...
5
...
We can
<
then write ln(1 − xB ) ≈ −xB and hence obtain
xB =

∆ vapH A 1
1D
B
− E
R C T* T F

Finally, because T ≈ T*, it also follows that
1
T*

−

1
T

=

T − T*
TT*

≈

∆T
T*2

with ∆T = T − T*
...
33
...
33 makes no reference to the identity of the solute, only to its mole
fraction, we conclude that the elevation of boiling point is a colligative property
...
3 For practical applications of eqn 5
...
34)

where Kb is the empirical boiling-point constant of the solvent (Table 5
...

(c) The depression of freezing point
Fig
...
24 The heterogeneous equilibrium
involved in the calculation of the lowering
of freezing point is between A in the pure
solid and A in the mixture, A being the
solvent and B a solute that is insoluble in
solid A
...
5
...
At the freezing point,
the chemical potentials of A in the two phases are equal:
3
By Trouton’s rule (Section 3
...
33 has the form ∆T ∝ T* and is
independent of ∆ vapH itself
...
5 COLLIGATIVE PROPERTIES

153

Synoptic Table 5
...
12

Camphor

Kb /(K kg mol−1)
2
...
27

3
...
86

0
...


µA = µA + RT ln xA
*(s)
*(l)

(5
...
Therefore we can write the result
directly from eqn 5
...
36)°

where ∆T is the freezing point depression, T* − T, and ∆fusH is the enthalpy of fusion
of the solvent
...
When the solution is dilute, the mole fraction is
proportional to the molality of the solute, b, and it is common to write the last equation as
∆T = Kf b

(5
...
2)
...

(d) Solubility

Although solubility is not strictly a colligative property (because solubility varies with
the identity of the solute), it may be estimated by the same techniques as we have been
using
...
Saturation is a state of equilibrium, with the undissolved solute in
equilibrium with the dissolved solute
...
5
...
Because the latter is

B(dissolved
in A)

mB(solution)
Equal at
equilibrium

µB = µB + RT ln xB
*(l)
we can write

B(s)

µB = µB + RT ln xB
*(s)
*(l)

(5
...
We now show in the following
Justification that
ln xB =

∆fusH A 1
R

C Tf

−

1D
TF

m*(s)
B

(5
...
5
...


154

5 SIMPLE MIXTURES
1

Justification 5
...


The starting point is the same as in Justification 5
...
In the
present case, we want to find the mole fraction of B in solution at equilibrium when
the temperature is T
...
38 into

Mole fraction of B, xB

0
...
1
0
...
3

0
...
2
10
0

RT

=−

∆fusG
RT

As in Justification 5
...
Then we
integrate from the melting temperature of B (when xB = 1 and ln xB = 0) to the lower
temperature of interest (when xB has a value between 0 and 1):

1
3

0

*(s)
*(l)
µB − µB

0
...
5
...

Individual curves are labelled with the
value of ∆fusH/RT*
...


Ύ

ln xB

0

d ln xB =

1
R

Ύ

T

Tf

∆fusH
T2

dT

If we suppose that the enthalpy of fusion of B is constant over the range of temperatures of interest, it can be taken outside the integral, and we obtain eqn 5
...


Equation 5
...
5
...
It shows that the solubility of B decreases exponentially as the temperature is lowered from its melting point
...
However, the detailed content of eqn 5
...
One aspect of its approximate character is
that it fails to predict that solutes will have different solubilities in different solvents,
for no solvent properties appear in the expression
...
5
...


The phenomenon of osmosis (from the Greek word for ‘push’) is the spontaneous
passage of a pure solvent into a solution separated from it by a semipermeable membrane, a membrane permeable to the solvent but not to the solute (Fig
...
27)
...
Important examples of osmosis include transport of fluids through
cell membranes, dialysis and osmometry, the determination of molar mass by the
measurement of osmotic pressure
...

In the simple arrangement shown in Fig
...
28, the opposing pressure arises from
the head of solution that the osmosis itself produces
...
The
complicating feature of this arrangement is that the entry of solvent into the solution
results in its dilution, and so it is more difficult to treat than the arrangement in
Fig
...
27, in which there is no flow and the concentrations remain unchanged
...

The chemical potential of the solvent is lowered by the solute, but is restored to its
‘pure’ value by the application of pressure
...


(5
...
5 COLLIGATIVE PROPERTIES
Justification 5
...
On the solution side, the chemical potential is lowered by the presence
of the solute, which reduces the mole fraction of the solvent from 1 to xA
...
At equilibrium the chemical potential of A is the same in
both compartments, and we can write

µA = µA(xA, p + Π )
*(p)
The presence of solute is taken into account in the normal way:

µA(xA, p + Π ) = µA + Π ) + RT ln xA
*(p
We saw in Section 3
...
54) how to take the effect of pressure into account:

µA + Π ) = µA +
*(p
*(p)

Ύ

p+Π

Vmdp

p

where Vm is the molar volume of the pure solvent A
...
For
dilute solutions, ln xA may be replaced by ln (1 − xB) ≈ −xB
...
That being so, Vm may be taken outside the integral, giving
RTxB = Π Vm
When the solution is dilute, xB ≈ nB/nA
...
40
...
As these huge molecules
dissolve to produce solutions that are far from ideal, it is assumed that the van ’t Hoff
equation is only the first term of a virial-like expansion:4

Π = [J]RT{1 + B[J] + · · · }

(5
...

Example 5
...
The pressures are expressed in terms of the heights of
solution (of mass density ρ = 0
...

Determine the molar mass of the polymer
...
00
h/cm
0
...
00
0
...
00
2
...
00
5
...
00
8
...


155

156

5 SIMPLE MIXTURES

(h/c)/(cm g-1 dm3)

0
...

We use eqn 5
...
The osmotic pressure is related to the hydrostatic pressure
by Π = ρgh (Example 1
...
81 m s−2
...
41
becomes

0
...
4

h

0
...
5
...
The molar mass is calculated
from the intercept at c = 0; in Chapter 19
we shall see that additional information
comes from the slope
...


=

RT A

ρgM C

1+

D
F

Bc

+··· =

M

RT

ρgM

+

A RTB D
c+···
C ρgM 2 F

Therefore, to find M, plot h/c against c, and expect a straight line with intercept
RT/ρgM at c = 0
...
00
−1
3
(h/c)/(cm g dm ) 0
...
00
0
...
00 7
...
00
0
...
729 0
...
5
...
The intercept is at 0
...
Therefore,
M=
=

RT

1

×

ρg 0
...
3145 J K−1 mol−1) × (298 K)
−3

−2

(980 kg m ) × (9
...
2 × 102 kg mol−1

×

1
−3

2
...
Molar masses of macromolecules are often
reported in daltons (Da), with 1 Da = 1 g mol−1
...
Modern osmometers give readings of
osmotic pressure in pascals, so the analysis of the data is more straightforward and
eqn 5
...
As we shall see in Chapter 19, the value obtained
from osmometry is the ‘number average molar mass’
...
5
...


Self-test 5
...
8 mK]
these solutions, taking Kf as about 10 K/(mol kg−1)
...
2 Osmosis in physiology and biochemistry

Osmosis helps biological cells maintain their structure
...
The difference in concentrations of
solutes inside and outside the cell gives rise to an osmotic pressure, and water passes
into the more concentrated solution in the interior of the cell, carrying small nutrient
molecules
...
These effects are important in everyday medical practice
...
If the injected solution is too
dilute, or hypotonic, the flow of solvent into the cells, required to equalize the osmotic
pressure, causes the cells to burst and die by a process called haemolysis
...


5
...
In a purification experiment, a solution of macromolecules
containing impurities, such as ions or small molecules (including small proteins or
nucleic acids), is placed in a bag made of a material that acts as a semipermeable
membrane and the filled bag is immersed in a solvent
...
In practice,
purification of the sample requires several changes of solvent to coax most of the
impurities out of the dialysis bag
...
Suppose the molar
concentration of the macromolecule M is [M] and the total concentration of the small
molecule A in the bag is [A]in
...
At
equilibrium, the chemical potential of free A in the macromolecule solution is equal
to the chemical potential of A in the solution on the other side of the membrane,
where its concentration is [A]out
...
7 that the equality µA,free =
µA,out implies that [A]free = [A]out, provided the activity coefficient of A is the same in
both solutions
...
The average number of A molecules bound to M molecules, ν, is
then the ratio

ν=

[A]bound
[M]

=

[A]in − [A]out
[M]

The bound and unbound A molecules are in equilibrium, M + A 5 MA, so their
concentrations are related by an equilibrium constant K, where
K=

[MA]
[M]free[A]free

=

[A]bound
([M] − [A]bound)[A]free

We have used [MA] = [A]bound and [M]free = [M] − [MA] = [M] − [A]bound
...
The average number of A molecules per site is ν/N, so the last equation becomes
K=

ν/N

A
νD
1−
[A]out
C
NF

It then follows that

ν
[A]out

= KN − Kν

157

158

5 SIMPLE MIXTURES
Intercept = KN

n/[A]out

Slope = -K

This expression is the Scatchard equation
...
5
...
From
these two quantities, we can find the equilibrium constant for binding and the number of binding sites on each macromolecule
...


Intercept = N

Activities
Number of binding sites, n

A Scatchard plot of ν/[A]out against
ν
...

Fig
...
30

Exploration The following tasks will
give you an idea of how graphical
analysis can distinguish between systems
with the same values of K or N
...
Then repeat the process, this time
varying N but fixing K
...
In Chapter 3 (specifically, Further information 3
...

Here we see how the expressions encountered in the treatment of ideal solutions can
also be preserved almost intact by introducing the concept of ‘activity’
...
3
...

5
...
23 (that µA = µA + RT ln(pA/pA where pA is the
vapour pressure of pure A and pA is the vapour pressure of A when it is a component
of a solution
...
25 (that is, as µA = µA +
*
RT ln xA)
...
42]

The quantity aA is the activity of A, a kind of ‘effective’ mole fraction, just as the
fugacity is an effective pressure
...
23 is true for both real and ideal solutions (the only approximation
being the use of pressures rather than fugacities), we can conclude by comparing it
with eqn 5
...
43)

pA
*

Table 5
...


5
...
43
...
3 Calculating the solvent activity

The vapour pressure of 0
...
95 kPa, so the activity of
water in the solution at this temperature is
aA =

99
...
325 kPa

= 0
...
44)

A convenient way of expressing this convergence is to introduce the activity
coefficient, γ , by the definition
aA = γAxA

γA → 1 as xA → 1

[5
...
The chemical potential of the solvent is then

µA = µA + RT ln xA + RT ln γA
*

(5
...

5
...
We shall show how to set up the definitions for a solute that
obeys Henry’s law exactly, and then show how to allow for deviations
...
In this case, the chemical potential of B is

µB = µB + RT ln
*

pB
pB
*

= µB + RT ln
*

KB
pB
*

+ RT ln xB

Both KB and pB are characteristics of the solute, so the second term may be combined
*
with the first to give a new standard chemical potential:
7
µB = µB + RT ln
*

KB
pB
*

[5
...
48)°

7
If the solution is ideal, KB = pB and eqn 5
...

*
*,

159

160

5 SIMPLE MIXTURES
(b) Real solutes

We now permit deviations from ideal-dilute, Henry’s law behaviour
...
48, and obtain
7
µB = µB + RT ln aB

[5
...
The value of the activity at any concentration
can be obtained in the same way as for the solvent, but in place of eqn 5
...
50)

KB

As for the solvent, it is sensible to introduce an activity coefficient through
aB = γBxB

[5
...
Because
the solute obeys Henry’s law as its concentration goes to zero, it follows that
aB → xB and γ B → 1

as xB → 0

(5
...
Deviations of the solute from ideality disappear as
zero concentration is approached
...
5 Measuring activity

Use the information in Example 5
...

For convenience, the data are repeated here:
0
0
46
...
20
4
...
3

0
...
3

0
...
9
12
...
80
26
...
9

1
36
...
For its activity as a solute (the Henry’s law activity),
form aC = pC /KC and γC = aC /xC
...
4 kPa and KC = 22
...
For instance, at xC = 0
...
7 kPa)/
(36
...
13 and γC = 0
...
20 = 0
...
7 kPa)/(22
...
21 and γC = 0
...
20 = 1
...

From Raoult’s law (chloroform regarded as the solvent):
aC
γC

0

0
...
65

0
...
75

0
...
87

0
...
91

1
...
00

From Henry’s law (chloroform regarded as the solute):
aC
γC

0
1

0
...
05

0
...
25

0
...
43

1
...
51

1
...
65

These values are plotted in Fig
...
31
...

Self-test 5
...

[At xA = 0
...
50; γR = 0
...
00, γH = 1
...
7 THE SOLUTE ACTIVITY
1

2
(a)

(b)

0
...
6

0
...
2

0
...
8

0
...
4

0

Fig
...
31 The variation of activity and
activity coefficient of chloroform
(trichloromethane) and acetone
(propanone) with composition according
to (a) Raoult’s law, (b) Henry’s law
...
2 0
...
6 0
...
2 0
...
6 0
...
In chemistry, compositions are often expressed as molalities, b, in place of mole fractions
...
53}°

7

where µ has a different value from the standard values introduced earlier
...
Note that as bB → 0, µB → −∞;
that is, as the solution becomes diluted, so the solute becomes increasingly stabilized
...

Now, as before, we incorporate deviations from ideality by introducing a dimensionless activity aB, a dimensionless activity coefficient γB, and writing
aB = γB

bB
b7

where γB → 1

as bB → 0

[5
...
The standard state remains unchanged in this last
stage and, as before, all the deviations from ideality are captured in the activity
coefficient γB
...
55)

(d) The biological standard state

One important illustration of the ability to choose a standard state to suit the circumstances arises in biological applications
...
Therefore, in biochemistry it is common to adopt the biological standard state, in which pH = 7 (an activity of 10−7, neutral solution) and to label the
corresponding standard thermodynamic functions as G⊕, H⊕, µ⊕, and S⊕ (some texts
use X 7 ′)
...


162

5 SIMPLE MIXTURES
To find the relation between the thermodynamic and biological standard values of
the chemical potential of hydrogen ions we need to note from eqn 5
...
56)
−1

At 298 K, 7RT ln 10 = 39
...

5
...
4 gives further insight into
the origin of deviations from Raoult’s law and its relation to activity coefficients
...
31)
...
31 implies that the activity coefficients are given by expressions of the form
2
ln γA = βxB

2
ln γB = βxA

(5
...

Justification 5
...
5

∆mixG = nRT{xA ln aA + xB ln aB}

3

This relation follows from the derivation of eqn 5
...
If each activity is replaced by γ x, this expression becomes

2
...
57, and use xA + xB = 1, which gives

*
pA /pA

2
1
...
5

2
2
∆mixG = nRT{xA ln xA + xB ln xB + βxAx B + βxBxA}
= nRT{xA ln xA + xB ln xB + βxAxB(xA + xB)}
= nRT{xA ln xA + xB ln xB + βxAxB}

0
-1

0

0

0
...
31
...

1

Fig
...
32 The vapour pressure of a mixture
based on a model in which the excess
enthalpy is proportional to βxAxB
...

Positive values of β give vapour pressures
higher than ideal
...


Exploration Plot pA/pA against xA
*

with β = 2
...
24 and
then eqn 5
...
Above what value of xA do
the values of pA/pA given by these equations
*
differ by more than 10 per cent?

At this point we can use the Margules equations to write the activity of A as
2

2

aA = γAxA = xAeβx B = xAeβ(1−xA)

(5
...
The activity of A, though, is just the ratio of the vapour
pressure of A in the solution to the vapour pressure of pure A (eqn 5
...
59)

This function is plotted in Fig
...
32
...
59
becomes pA = xA pA which is Raoult’s law)
...

Negative values of β (exothermic mixing, favourable solute–solvent interactions) give
a lower vapour pressure
...
59 approaches 1
...
59 approaches
<1,

5
...
60)
β

This expression has the form of Henry’s law once we identify K with e pA which is
*,
different for each solute–solvent system
...
9 The activities of ions in solution
Interactions between ions are so strong that the approximation of replacing activities
by molalities is valid only in very dilute solutions (less than 10−3 mol kg−1 in total ion
concentration) and in precise work activities themselves must be used
...

(a) Mean activity coefficients

If the chemical potential of a univalent cation M+ is denoted µ+ and that of a univalent
anion X− is denoted µ−, the total molar Gibbs energy of the ions in the electrically neutral solution is the sum of these partial molar quantities
...
61)°
+

−

However, for a real solution of M and X of the same molality,
ideal
ideal
Gm = µ+ + µ− = µ + + µ ideal + RT ln γ+ + RT ln γ− = G m + RT ln γ+γ−
−

(5
...

There is no experimental way of separating the product γ+γ− into contributions
from the cations and the anions
...
Therefore, for a 1,1electrolyte, we introduce the mean activity coefficient as the geometric mean of the
individual coefficients:

γ± = (γ+γ−)1/2

[5
...
64)

The sum of these two chemical potentials is the same as before, eqn 5
...

We can generalize this approach to the case of a compound Mp Xq that dissolves
to give a solution of p cations and q anions from each formula unit
...
65)

If we introduce the mean activity coefficient
p
γ± = (γ + γ q)1/s
−

s=p+q

[5
...
67)

we get the same expression as in eqn 5
...
68)

However, both types of ion now share equal responsibility for the nonideality
...
4

The geometric mean of x p and y q is
(x py q)1/(p+q)
...


164

5 SIMPLE MIXTURES

Fig
...
33 The picture underlying the
Debye–Hückel theory is of a tendency for
anions to be found around cations, and of
cations to be found around anions (one
such local clustering region is shown by the
circle)
...
The solutions to which the
theory applies are far less concentrated
than shown here
...
This domination is the
basis of the Debye–Hückel theory of ionic solutions, which was devised by Peter
Debye and Erich Hückel in 1923
...
The calculation itself, which is a profound example of how a
seemingly intractable problem can be formulated and then resolved by drawing on
physical insight, is described in Further information 5
...

Oppositely charged ions attract one another
...
5
...
Overall, the solution is
electrically neutral, but near any given ion there is an excess of counter ions (ions of
opposite charge)
...
This time-averaged, spherical haze around the central ion, in which
counter ions outnumber ions of the same charge as the central ion, has a net charge
equal in magnitude but opposite in sign to that on the central ion, and is called its
ionic atmosphere
...

This lowering of energy appears as the difference between the molar Gibbs energy Gm
ideal
and the ideal value G m of the solute, and hence can be identified with RT ln γ±
...

The model leads to the result that at very low concentrations the activity coefficient
can be calculated from the Debye–Hückel limiting law
log γ± = − | z+z− | AI1/2

(5
...
509 for an aqueous solution at 25°C and I is the dimensionless ionic
strength of the solution:
1
I=–
2

∑ z i2(bi /b7)

[5
...
4 Ionic strength and molality,
I = kb/b7
X−

k
M+

X2 −

X3 −

X4 −

1

3

6

10

2+

M

3

4

15

12

M3+

6

15

9

42

M4+

10

12

42

16

For example, the ionic strength of an M2X3
solution of molality b, which is understood to
give M3+ and X2− ions in solution is 15b/b7
...
The ionic strength occurs widely wherever ionic solutions are discussed, as we shall see
...
For solutions consisting of two types of ion at molalities b+ and b−,
1
2
I = – (b+z+ + b−z 2)/b7
−
2

(5
...
Table 5
...

Illustration 5
...
0 × 10−3 mol kg−1 KCl(aq) at 25°C is calculated by
writing
1
I = –(b+ + b−)/b7 = b/b7
2

where b is the molality of the solution (and b+ = b− = b)
...
69,
log γ± = −0
...
0 × 10−3)1/2 = −0
...
92
...
927
...
8 Calculate the ionic strength and the mean activity coefficient of

1
...


[3
...
880]

5
...
69 because ionic solutions of moderate
molalities may have activity coefficients that differ from the values given by this
expression, yet all solutions are expected to conform as b → 0
...
5 lists some
experimental values of activity coefficients for salts of various valence types
...
34 shows some of these values plotted against I1/2, and compares them with the
theoretical straight lines calculated from eqn 5
...
The agreement at very low molalities (less than about 1 mmol kg−1, depending on charge type) is impressive, and convincing evidence in support of the model
...


Synoptic Table 5
...
001

0
...
888

0
...
902

0
...
1

0
...
524

1
...
607

0
...


(c) The extended Debye–Hückel law

When the ionic strength of the solution is too high for the limiting law to be valid, the
activity coefficient may be estimated from the extended Debye–Hückel law:
log γ± = −

A | z+z− | I1/2
1 + BI1/2

+ CI

(5
...
Although B can be interpreted as a
measure of the closest approach of the ions, it (like C) is best regarded as an adjustable
empirical parameter
...
5
...
It is clear that
0

0

NaCl
(1,1)
-0
...
1

MgCl2

log g±

Extended law

-0
...
06

-0
...
08

0

4

8
1/2
100I

12

16

Fig
...
34 An experimental test of the
Debye–Hückel limiting law
...


0

4

8
12
1/2
100I

16

Fig
...
35 The extended Debye–Hückel law
gives agreement with experiment over a
wider range of molalities (as shown here
for a 1,1-electrolyte), but it fails at higher
molalities
...
50 and
C = 0 as a representation of experimental
data for a certain 1,1-electrolyte
...
72 accounts for some activity coefficients over a moderate range of dilute solutions (up to about 0
...

Current theories of activity coefficients for ionic solutes take an indirect route
...
12) to
estimate the activity coefficient of the solute
...
1 mol kg−1 and are valuable for the discussion of mixed salt solutions, such as sea water
...
The partial molar volume is the change in volume per mole of
A added to a large volume of the mixture: VJ = (∂V/∂n J)p,T,n′
...


15
...


2
...
The total Gibbs energy
of a mixture is G = nA µA + nB µB
...
The elevation of boiling point is given by ∆T = Kbb, where Kb
is the ebullioscopic constant
...


3
...


18
...


4
...

5
...

6
...

7
...

8
...

9
...

10
...

11
...


16
...


19
...

20
...

*
...
The activity is defined as aA = pA/pA
22
...
The activity may be written in terms of the
*
activity coefficient γA = aA/xA
...
The chemical potential of a solute in an ideal-dilute solution is
7
given by µ B = µ B + RT ln a B
...

24
...

H
25
...


12
...


26
...


13
...


27
...

i
2

14
...


28
...


FURTHER INFORMATION

167

Further reading
Articles and texts6

Sources of data and information

B
...
In Encyclopedia of applied physics (ed
...
L
...
VCH, New York (1995)
...
R
...
Dack, Solutions and solubilities
...
A
...
W
...
Wiley, New York
(1975)
...
N
...
D
...

Wiley–Interscience, New York (1994)
...
S
...
L
...

Butterworths, London (1982)
...
C
...
), Handbook of chemistry and physics, Vol
...
CRC
Press, Boca Raton (2004)
...
Sattar, Thermodynamics of mixing real gases
...
Chem
...
77,
1361 (2000)
...
1 The Debye–Hückel theory of ionic
solutions

Imagine a solution in which all the ions have their actual positions,
but in which their Coulombic interactions have been turned off
...
For a salt M p Xq, we write

1
...
When rD is large, the shielded
potential is virtually the same as the unshielded potential
...
5
...


ideal
ideal
we = (pµ+ + qµ −) − (pµ + + qµ − )
ideal
ideal
= p(µ+ − µ + ) + q(µ− − µ − )

Potential, f/(Z /rD)

0
...
64 we write
ideal
ideal
µ+ − µ + = µ− − µ − = RT ln γ±

So it follows that
ln γ± =

we

s=p+q

sRT

(5
...
75)

¥
1 3
0
...
74)

The ionic atmosphere causes the potential to decay with distance
more sharply than this expression implies
...
4

0
...

The Coulomb potential at a distance r from an isolated ion of
charge zie in a medium of permittivity ε is
Zi

0
...
5
Distance, r/rD

1

Fig
...
36 The variation of the shielded Coulomb potential with
distance for different values of the Debye length, rD/a
...
In each
case, a is an arbitrary unit of length
...

Then plot this expression against rD and provide a physical
interpretation for the shape of the plot
...


168

5 SIMPLE MIXTURES

To calculate rD, we need to know how the charge density, ρi, of the
ionic atmosphere, the charge in a small region divided by the volume
of the region, varies with distance from the ion
...
76)

ε

where ∇2 = (∂2/∂x 2 + ∂2/∂y 2 + ∂2/∂z 2) is called the laplacian
...
5

For systems with spherical symmetry, it is best to work in spherical
polar coordinates r, θ, and φ (see the illustration): x = r sin θ cos φ,
y = r sin θ sin φ, and z = r cos θ
...
The quantity eNA, the magnitude of the charge per mole
of electrons, is Faraday’s constant, F = 96
...
It follows that

ρ i = c+z+F + c−z−F = c + +Fe−z+eφi /kT + c ° z−Fe−z−eφi /kT
°z
−

(5
...
Because the average electrostatic
interaction energy is small compared with kT we may write eqn 5
...
6

Replacing e by F/NA and NAk by R results in the following expression:

z

ρ i = (c + + + c ° z−)F − (c + + + c ° z 2 )
°z
°z 2 − −
−

f

F 2φi
RT

+···

(5
...

The unwritten terms are assumed to be too small to be significant
...
70, by noting that in the dilute aqueous solutions we
are considering there is little difference between molality and molar
concentration, and c ≈ bρ, where ρ is the mass density of the solvent

r

x

cj
c°
j

The expansion of an exponential function used here is e−x = 1 − x
1
<
+ –x 2 − · · ·
...

2

1 ∂ A 2∂D
1
∂ A
∂D
1
∂2
Br
E+ 2
B sin θ E + 2 2
∇2 = 2
r ∂r C ∂r F r sin θ ∂θ C
∂θ F r sin θ ∂θ 2

q

Therefore, according to the Boltzmann distribution, the ratio of
the molar concentration, cj , of ions at a distance r and the molar
concentration in the bulk, c °, where the energy is zero, is:
j

y

°z 2 − −
c + + + c ° z 2 ≈ (b °z 2 + b° z 2 )ρ = 2Ib7ρ
+ +
− −
With these approximations, eqn 5
...
77)

To solve this equation we need to relate ρi and φi
...
1) to work out the probability
that an ion will be found at each distance
...
75,
results in

εφi

2ρF 2Ib7φi

We can now solve eqn 5
...
80)

To calculate the activity coefficient we need to find the electrical
work of charging the central ion when it is surrounded by its
atmosphere
...
This potential is the difference between the
total potential, given by eqn 5
...
If the charge of the central ion were q and
not zie, then the potential due to its atmosphere would be

φatmos(0) = −

Comment 5
...


ln γ± = −

q
4πεrD

ln γ± = −

dwe = φatmos(0)dq
zie

φatmos(0) dq = −

0

=−

NAz 2e 2
i
8πεrD

=−

NA
4πεrD

zie

Ύ

q dq

0

z 2F 2
i
8πεNArD

pwe,+ + qwe,−
sRT

=−

2
(pz 2 + qz −)F 2
+

| z+z− | F 2 A 2ρF 2Ib7 D
B
E
8πεNART C ε RT F

1/2

where we have grouped terms in such a way as to show that this
expression is beginning to take the form of eqn 5
...
Indeed,
conversion to common logarithms (by using ln x = ln 10 × log x) gives
1

A ρb7 D
B 3 3 3E
3 4πNA ln 10 C 2ε R T F

log γ± = −| z+z− | 2

where in the last step we have used F = NAe
...
73
that the mean activity coefficient of the ions is

ln γ± =

8πεNARTrD

1 F 3 A ρb7 D 1/2 5
B
E 6 I 1/2
= −| z+z− | 2
4πNA C 2ε 3R 3T 3 F 7
3

Therefore, the total molar work of fully charging the ions is

Ύ

| z+z− | F 2

Replacing rD with the expression in eqn 5
...
69 (log γ± = −| z+z− |AI 1/2) with

8πεsNARTrD

A=

However, for neutrality pz+ + qz− = 0; therefore

A ρb7 D
B
E
4πNA ln 10 C 2ε 3R 3T 3 F
F3

1/2

(5
...
1 State and justify the thermodynamic criterion for solution–vapour

equilibrium
...
2 How is Raoult’s law modified so as to describe the vapour pressure of real

solutions?

5
...

5
...

5
...


5
...


Exercises
5
...
4693
are 74
...
235 cm3 mol−1, respectively
...
000 kg?
5
...
3713 are 188
...
14 cm3 mol−1,
respectively
...
1 g mol−1 and 198
...

What is the volume of a solution of mass 1
...
2(a) At 25°C, the density of a 50 per cent by mass ethanol–water solution is

0
...
Given that the partial molar volume of water in the solution is
17
...


5
...
7 kg m−3
...
2 cm3 mol−1, calculate the partial molar volume of the water
...
3(a) At 300 K, the partial vapour pressures of HCl (that is, the partial
pressure of the HCl vapour) in liquid GeCl4 are as follows:

xHCl

0
...
012

0
...
0

76
...
8

Show that the solution obeys Henry’s law in this range of mole fractions, and
calculate Henry’s law constant at 300 K
...
3(b) At 310 K, the partial vapour pressures of a substance B dissolved in a
liquid A are as follows:

xB

0
...
015

0
...
0

122
...
1

Show that the solution obeys Henry’s law in this range of mole fractions, and
calculate Henry’s law constant at 310 K
...
4(a) Predict the partial vapour pressure of HCl above its solution in liquid
germanium tetrachloride of molality 0
...
For data, see Exercise 5
...


5
...
8 kJ mol−1 and its melting
point is 217°C
...


5
...
13(b) Predict the ideal solubility of lead in bismuth at 280°C given that its
melting point is 327°C and its enthalpy of fusion is 5
...


solution in A in Exercise 5
...
25 mol kg−1
...
1 g mol−1
...
5(a) The vapour pressure of benzene is 53
...
6°C, but it fell to

5
...
004 g cm−3:

51
...
0 g of an involatile organic compound was dissolved in
500 g of benzene
...


c/(g dm−3)

2
...
613

9
...
602

5
...
00 kPa at 338
...
592

1
...
750

3
...
62 kPa when 8
...
Calculate the molar mass of the compound
...


5
...
5 K
...


5
...
The following data were obtained:

c/(mg cm−3)

3
...
618

5
...
722

5
...
00 g of a compound to 250 g of naphthalene lowered

h/cm

5
...
238

9
...
990

the freezing point of the solvent by 0
...
Calculate the molar mass of the
compound
...


5
...


Calculate the freezing point of the solution
...
7(b) The osmotic pressure of an aqueous solution at 288 K is 99
...


Calculate the freezing point of the solution
...
15(a) Substances A and B are both volatile liquids with pA = 300 Torr,
*
pB = 250 Torr, and KB = 200 Torr (concentration expressed in mole fraction)
...
9, bB = 2
...
Calculate
the activities and activity coefficients of A and B
...


5
...
0 dm3 that is divided into two
compartments of equal size
...
0 atm and 25°C; in the right compartment there is hydrogen at the same
temperature and pressure
...
Assume that the gases are perfect
...
15(b) Given that p*(H2O) = 0
...
02239 atm in a
solution in which 0
...
920 kg water at 293 K, calculate the activity and activity
coefficient of water in the solution
...
8(b) Consider a container of volume 250 cm3 that is divided into two

5
...
In the left compartment there is argon at 100 kPa
and 0°C; in the right compartment there is neon at the same temperature and
pressure
...
Assume that the gases are perfect
...
The vapour pressure of pure CCl4 is 33
...

The Henry’s law constant when the concentration of Br2 is expressed as a mole
fraction is 122
...
Calculate the vapour pressure of each component,
the total pressure, and the composition of the vapour phase when the mole
fraction of Br2 is 0
...


5
...
4
...

5
...
00 mol C6H14 (hexane) is mixed with 1
...

5
...
10(b) What proportions of benzene and ethylbenzene should be mixed

(a) by mole fraction, (b) by mass in order to achieve the greatest entropy of
mixing?
5
...
1 to calculate the solubility (as

a molality) of CO2 in water at 25°C when its partial pressure is (a) 0
...
00 atm
...
11(b) The mole fractions of N2 and O2 in air at sea level are approximately

0
...
21
...

5
...
0 atm
...

5
...
0 atm
...


5
...
The boiling point of
pure benzene is 80
...
Calculate the chemical potential of benzene relative to
that of pure benzene when xbenzene = 0
...
If the activity
coefficient of benzene in this solution were actually 0
...
00,
what would be its vapour pressure?
5
...
2°C at 1
...
400 when yA = 0
...
Calculate the activities and activity coefficients
of both components in this solution on the Raoult’s law basis
...
5 kPa
...
)
5
...
00 atm, it was found that xA = 0
...
314
...
The vapour pressures
of the pure components at this temperature are: pA = 73
...
1 kPa
...
)
5
...
10 mol kg−1 in
KCl(aq) and 0
...

5
...
040 mol kg−1 in

K3[Fe(CN)6](aq), 0
...
050 mol kg−1 in NaBr(aq)
...
19(a) Calculate the masses of (a) Ca(NO3)2 and, separately, (b) NaCl to add

to a 0
...
250
...
19(b) Calculate the masses of (a) KNO3 and, separately, (b) Ba(NO3)2 to
add to a 0
...
00
...
20(a) Estimate the mean ionic activity coefficient and activity of a solution

that is 0
...
030 mol kg−1 NaF(aq)
...
21(a) The mean activity coefficients of HBr in three dilute aqueous
solutions at 25°C are 0
...
0 mmol kg−1), 0
...
0 mmol kg−1),
and 0
...
0 mmol kg−1)
...

5
...
927 (at 5
...
902 (at 10
...
816 (at 50
...
Estimate the value of B in the extended
Debye–Hückel law
...
20(b) Estimate the mean ionic activity coefficient and activity of a solution
that is 0
...
035 mol kg−1 Ca(NO3)2(aq)
...
1 The following table gives the mole fraction of methylbenzene (A) in liquid

and gaseous mixtures with butanone at equilibrium at 303
...
Take the vapour to be perfect and calculate the partial pressures of
the two components
...


Calculate the apparent molar mass of the solute and suggest an interpretation
...
4 kJ mol−1 and T * = 290 K
...
7 In a study of the properties of an aqueous solution of Th(NO3)4 (by A
...
Azoulay, and A
...
Chem
...
Faraday Trans
...
0703 K was observed for an aqueous
solution of molality 9
...
What is the apparent number of ions per
formula unit?

xA

0

0
...
2476

0
...
5194

0
...
0410

0
...
1762

0
...
3393

p/kPa

36
...
121

30
...
626

25
...
402

xA

0
...
8019

0
...
Find the activity coefficients of both
components on (a) the Raoult’s law basis, (b) the Henry’s law basis with
iodoethane as solute
...
4450

0
...
7284

1

xI

0

0
...
1095

0
...
2353

0
...
6984

18
...
496

12
...
73

7
...
7

14
...
72

pA/kPa

37
...
48

33
...
85

29
...
05

xI

0
...
6349

0
...
9093

1
...
44

31
...
58

43
...
12

pA/kPa

19
...
39

8
...
09

0

5
...
62x + 1
...
12x 2 where v = V/cm3, V is the volume of a
solution formed from 1
...
Calculate the partial
molar volume of the components in a solution of molality 0
...

5
...
000 kg of water fits the expression v = 1001
...
69(x − 0
...
Calculate the partial molar volumes of the salt and
the solvent when in a solution of molality 0
...


5
...
9 Plot the vapour pressure data for a mixture of benzene (B) and acetic acid

measured as set out below
...


(A) given below and plot the vapour pressure/composition curve for the
mixture at 50°C
...
Deduce the activities and activity coefficients of the
components on the Raoult’s law basis and then, taking B as the solute, its
activity and activity coefficient on a Henry’s law basis
...


m(CuSO4)/g

5

10

15

20

ρ /(g cm−3)

0
...
0835

0
...
1714

1
...
167

1
...
0160

1
...
484

0
...
535

1
...
45

pB/kPa

35
...
29

33
...
64

30
...
2973

0
...
5834

0
...
8437

0
...
31

3
...
84

5
...
76

7
...
16

26
...
42

18
...
0

0
...
4 The densities of aqueous solutions of copper(II) sulfate at 20°C were

where m(CuSO4) is the mass of CuSO4 dissolved in 100 g of solution
...
5 What proportions of ethanol and water should be mixed in order to

produce 100 cm3 of a mixture containing 50 per cent by mass of ethanol?
What change in volume is brought about by adding 1
...
5
...
)
have a number of unusual properties
...
Emsley, J
...
Soc
...
The following data were obtained:

5
...
examined mixtures of cyclohexane with various longchain alkanes (T
...
Aminabhavi, V
...
Patil, M
...
D
...
C
...
Chem
...
Data 41, 526 (1996))
...
15 K:

b/(mol kg−1)

0
...
037

0
...
295

0
...
6965

0
...
9004

∆T/K

0
...
295

0
...
381

2
...
7661

0
...
7697

5
...


172

5 SIMPLE MIXTURES

Compute the partial molar volume for each component in a mixture that has
a mole fraction cyclohexane of 0
...


Find an expression for the chemical potential of A in the mixture and sketch
its dependence on the composition
...
11‡ Comelli and Francesconi examined mixtures of propionic acid with
various other organic liquids at 313
...
Comelli and R
...
Chem
...
Data 41, 101 (1996))
...
4697 cm3 mol−1 and
a1 = 0
...
The density of propionic acid at this temperature is
0
...
86398 g cm−3
...

(b) Compute the partial molar volume for each component in an equimolar
mixture
...
18 Use the Gibbs–Duhem equation to derive the Gibbs–Duhem–Margules
equation

5
...
studied the liquid–vapour equilibria of

trichloromethane and 1,2-epoxybutane at several temperatures (R
...
Lunelli, and F
...
Chem
...
Data 41, 310 (1996))
...
15 K as a function of pressure
...
Use the relation to show that, when the fugacities are
replaced by pressures, if Raoult’s law applies to one component in a mixture
it must also apply to the other
...
19 Use the Gibbs–Duhem equation to show that the partial molar volume
(or any partial molar property) of a component B can be obtained if the
partial molar volume (or other property) of A is known for all compositions
up to the one of interest
...
40

21
...
25

18
...
15

20
...
50

26
...
129

0
...
353

0
...
700

0
...
500
...
065

0
...
285

0
...
805

0
...
194

0
...
559

0
...
889

1
...
99

75
...
50

77
...
08

79
...
67

Compute the activity coefficients of both components on the basis of Raoult’s
law
...
13‡ Chen and Lee studied the liquid–vapour equilibria of cyclohexanol

with several gases at elevated pressures (J
...
Chen and M
...
Lee, J
...

Eng
...
Among their data are the following measurements
of the mole fractions of cyclohexanol in the vapour phase (y) and the liquid
phase (x) at 393
...

p/bar

10
...
0

30
...
0

60
...
0

ycyc

0
...
0149

0
...
00947

0
...
00921

xcyc

0
...
9464

0
...
892

0
...
773

Determine the Henry’s law constant of CO2 in cyclohexanol, and compute
the activity coefficient of CO2
...
14‡ Equation 5
...
The data in the table below gives the solubility, S, of calcium
acetate in water as a function of temperature
...
4

34
...
7

32
...
7

Determine the extent to which the data fit the exponential S = S0eτ/T and
obtain values for S0 and τ
...

5
...
15 K was found to fit the expression

GE = RTx(1 − x){0
...
1077(2x − 1) + 0
...
Calculate the Gibbs
energy of mixing when a mixture of 1
...
00 mol of THF is
prepared
...
16 The mean activity coefficients for aqueous solutions of NaCl at 25°C are

given below
...

b/(mmol kg−1)

1
...
0

5
...
0

20
...
9649

0
...
9275

0
...
8712

Theoretical problems
5
...


5
...
Integrate d ln xA from xA = 0 to the value of interest, and
integrate the right–hand side from the transition temperature for the pure
liquid A to the value in the solution
...
33 and 5
...

5
...
By writing r

= xB/xA, and using the Gibbs–Duhem equation, show that we can calculate
the activity of B from the activities of A over a composition range by using
the formula
A aB D
ln B E
CrF

r

Ύ

= φ − φ (0) +

0

Aφ − 1 D
B
E dr
C r F

5
...
Go on to show that, provided the concentration of the solution is low,
this expression takes the form ΠV = φRT[B] and hence that the osmotic
coefficient, φ, (which is defined in Problem 5
...

5
...
Use the Debye–Hückel
limiting law to show that the osmotic coefficient (φ, Problem 5
...
303A and I = b/b7
...
24 Haemoglobin, the red blood protein responsible for oxygen transport,
binds about 1
...
Normal blood has a haemoglobin
concentration of 150 g dm−3
...


PROBLEMS
What volume of oxygen is given up by 100 cm3 of blood flowing from the
lungs in the capillary?
5
...

Breathing air at high pressures, such as in scuba diving, results in an
increased concentration of dissolved nitrogen
...
18 µg/(g H2O atm)
...
0 atm and 20°C? Compare
your answer to that for 100 g of water saturated with air at 1
...
(Air is
78
...
) If nitrogen is four times as soluble in fatty tissues as
in water, what is the increase in nitrogen concentration in fatty tissue in
going from 1 atm to 4 atm?
5
...
An equilibrium dialysis experiment was used to study the binding of
ethidium bromide (EB) to a short piece of DNA
...
00 µmol dm−3 aqueous
solution of the DNA sample was dialysed against an excess of EB
...
042

0
...
204

0
...
150

Side with DNA

0
...
590

1
...
531

4
...
Is the
identical and independent sites model for binding applicable?
5
...
2 applies only

when the macromolecule has identical and independent binding sites
...
(a) There are four independent sites on
an enzyme molecule and the intrinsic binding constant is K = 1
...

(b) There are a total of six sites per polymer
...
The binding constants for
the other two sites are 2 × 106
...
28 The addition of a small amount of a salt, such as (NH4)2SO4, to a
solution containing a charged protein increases the solubility of the protein
in water
...
However, the addition
of large amounts of salt can decrease the solubility of the protein to such an
extent that the protein precipitates from solution
...
Consider the equilibrium PX ν(s) 5 Pν+(aq) + νX−(aq), where Pν+
is a polycationic protein of charge +ν and X− is its counter ion
...

5
...

For example, in the determination of molar masses of polymers in solution
by osmometry, osmotic pressures are often reported in grams per square
centimetre (g cm−2) and concentrations in grams per cubic centimetre
(g cm−3)
...
Leonard and H
...
Polymer Sci
...
From these data, determine the molar mass of polyisobutene by
plotting Π/c against c
...
From your plot, how would you classify
chlorobenzene as a solvent for polyisobutene? Rationalize the result in terms
of the molecular structure of the polymer and solvent
...
(e) Experimentally, it is often found
that the virial expansion can be represented as

Π/c = RT/M (1 + B′c + gB′2c′2 + · · · )
and in good solvents, the parameter g is often about 0
...
With terms beyond
the second power ignored, obtain an equation for (Π /c)1/2 and plot this
quantity against c
...
Does this plot confirm the
assumed value of g?
10 −2(Π/c)/(g cm−2/g cm−3)

2
...
9

3
...
3

6
...
0

c/(g cm−3)

0
...
010

0
...
033

0
...
10

10 −2(Π/c)/(g cm−2/g cm−3)

19
...
0

38
...
145

0
...
245

0
...
29

5
...
Sato, F
...
Eirich, and J
...
Mark (J
...
, Polym
...
14,

619 (1976)) have reported the data in the table below for the osmotic
pressures of polychloroprene (ρ = 1
...
858 g cm−3)
at 30°C
...

c/(mg cm−3)
−2

Π /(N m )

1
...
10

4
...
18

9
...
1 Definitions
6
...
3 Vapour pressure diagrams

Phase diagrams
Phase diagrams for pure substances were introduced in Chapter 4
...
To set the stage, we introduce the famous phase rule of Gibbs,
which shows the extent to which various parameters can be varied yet the equilibrium
between phases preserved
...
The chapter then introduces systems of gradually increasing complexity
...


6
...
5 Liquid–liquid phase diagrams
6
...
1 Impact on materials science:

Liquid crystals
I6
...
In particular, we see how to use phase diagrams to judge whether two substances are mutually miscible, whether an equilibrium can exist over a range of conditions, or whether the system must be brought to a definite pressure, temperature,
and composition before equilibrium is established
...
They are also the basis of separation procedures in the
petroleum industry and of the formulation of foods and cosmetic preparations
...
W
...
We shall derive this rule first, and then apply it to a wide variety of systems
...

6
...
1 Thus we speak of the solid, liquid, and gas phases of a substance,
and of its various solid phases (as for black phosphorus and white phosphorus)
...
A gas, or a gaseous mixture, is a single
phase, a crystal is a single phase, and two totally miscible liquids form a single phase
...


6
...
Ice is a single phase (P = 1)
even though it might be chipped into small fragments
...
A system in which calcium carbonate undergoes thermal decomposition consists of two solid phases (one consisting of calcium carbonate and the other of calcium
oxide) and one gaseous phase (consisting of carbon dioxide)
...
This example shows that it is not
always easy to decide whether a system consists of one phase or of two
...
In a solution, atoms of A are surrounded by atoms of A and B, and
any sample cut from the sample, however small, is representative of the composition
of the whole
...
A small
sample could come entirely from one of the minute grains of pure A and would not be
representative of the whole (Fig
...
1)
...
The ability to control this
microstructure resulting from phase equilibria makes it possible to tailor the mechanical properties of the materials to a particular application
...
Thus, a mixture of ethanol and water has two constituents
...
The term constituent should be carefully distinguished from ‘component’, which has a more technical meaning
...

The number of components, C, in a system is the minimum number of independent
species necessary to define the composition of all the phases present in the system
...
Thus,
pure water is a one-component system (C = 1), because we need only the species
H2O to specify its composition
...
An aqueous solution of sodium chloride has two components because, by
charge balance, the number of Na+ ions must be the same as the number of Cl− ions
...
When a reaction can occur under the conditions
prevailing in the system, we need to decide the minimum number of species that, after
allowing for reactions in which one species is synthesized from others, can be used to
specify the composition of all the phases
...
To specify the composition of
the gas phase (Phase 3) we need the species CO2, and to specify the composition of
Phase 2 we need the species CaO
...

Hence, the system has only two components (C = 2)
...
6
...


176

6 PHASE DIAGRAMS
Example 6
...
Then
decide whether, under the conditions prevailing in the system, any of the constituents can be prepared from any of the other constituents
...
Finally, identify the
minimum number of these independent constituents that are needed to specify the
composition of all the phases
...
However, NH3
and HCl are formed in fixed stoichiometric proportions by the reaction
...
It follows that there is only one component in the system (C = 1)
...

Self-test 6
...

[(a) 1, (b) 2, (c) 2]

The variance, F, of a system is the number of intensive variables that can be
changed independently without disturbing the number of phases in equilibrium
...
We
say that such a system is bivariant, or that it has two degrees of freedom
...
That is,
the variance of the system has fallen to 1
...
1

Josiah Willard Gibbs spent most of his
working life at Yale, and may justly be
regarded as the originator of chemical
thermodynamics
...
He needed interpreters before
the power of his work was recognized
and before it could be applied to
industrial processes
...


6
...
W
...
1)

Justification 6
...
For two phases in equilibrium, we can write µJ(α) = µJ(β)
...
2 THE PHASE RULE

177

This is an equation relating p and T, so only one of these variables is independent
(just as the equation x + y = 2 is a relation for y in terms of x: y = 2 − x)
...
For three phases in mutual equilibrium,

µJ(α; p,T) = µJ(β; p,T) = µJ(γ; p,T)
This relation is actually two equations for two unknowns (µJ(α; p,T) = µJ(β; p,T)
and µJ(β; p,T) = µJ(γ; p,T)), and therefore has a solution only for a single value of p
and T (just as the pair of equations x + y = 2 and 3x − y = 4 has the single solution
3
1
x = – and y = –)
...
Four phases cannot be
2
2
in mutual equilibrium in a one-component system because the three equalities

µJ(β; p,T) = µJ(γ; p,T)

µJ(γ; p,T) = µJ(δ; p,T)

are three equations for two unknowns (p and T) and are not consistent (just as x + y
= 2, 3x − y = 4, and x + 4y = 6 have no solution)
...
We begin by counting the total number of
intensive variables
...
We can specify
the composition of a phase by giving the mole fractions of C − 1 components
...
Because there are P
phases, the total number of composition variables is P(C − 1)
...

At equilibrium, the chemical potential of a component J must be the same in
every phase (Section 4
...


Phase a

Pressure, p

µJ(α; p,T) = µJ(β; p,T)

F = P(C − 1) + 2 − C(P − 1) = C − P + 2
which is eqn 6
...


Phase b

F = 2,
one phase

Phase d
F = 0, three
phases in
equilibrium
F = 1, two
phases in
equilibrium

for P phases

That is, there are P − 1 equations of this kind to be satisfied for each component J
...
Each equation reduces our freedom to vary one of the P(C − 1) + 2 intensive variables
...
The lines
represent conditions under which the
two adjoining phases are in equilibrium
...
Four phases cannot
mutually coexist in equilibrium
...
6
...
When only one phase is
present, F = 2 and both p and T can be varied independently without changing the
number of phases
...
When two phases are in equilibrium F = 1, which implies that pressure is not
freely variable if the temperature is set; indeed, at a given temperature, a liquid has a
characteristic vapour pressure
...
Instead of selecting the temperature, we could
select the pressure, but having done so the two phases would be in equilibrium at a
single definite temperature
...

When three phases are in equilibrium, F = 0 and the system is invariant
...
The equilibrium of three phases
is therefore represented by a point, the triple point, on a phase diagram
...

These features are summarized in Fig
...
2
...
6
...
6
...
This diagram summarizes the changes that take place as
a sample, such as that at a, is cooled at constant pressure
...
4
...
The label T3
marks the temperature of the triple point,
Tb the normal boiling point, and Tf the
normal freezing point
...
6
...
Two phases are now in
equilibrium and F = 1
...
Lowering the temperature takes the system to c in the one-phase,
liquid region
...

(b) Experimental procedures

Detecting a phase change is not always as simple as seeing water boil in a kettle, so
special techniques have been developed
...
7), and differential scanning calorimetry (see Impact I2
...
They are useful for solid–solid transitions, where simple visual inspection of the sample may be
inadequate
...
At a first-order transition, heat is evolved and the cooling stops until the
transition is complete
...
6
...
6
...
The transition temperature is obvious, and is used to
mark point d on the phase diagram
...
Some of the highest
pressures currently attainable are produced in a diamond-anvil cell like that illustrated
in Fig
...
5
...
The advance in design this
represents is quite remarkable for, with a turn of the screw, pressures of up to about
1 Mbar can be reached that a few years ago could not be reached with equipment
weighing tons
...
One application of the technique is
to study the transition of covalent solids to metallic solids
...
6
...
The halt marked d corresponds
to the pause in the fall of temperature while
the first-order exothermic transition
(freezing) occurs
...

Fig
...
4

Diamond
anvils

Ultrahigh pressures (up to about
2 Mbar) can be achieved using a diamond
anvil
...

The principle of its action is like that of a
nutcracker: the pressure is exerted by
turning the screw by hand
...
6
...
3 VAPOUR PRESSURE DIAGRAMS
solid at around 210 kbar
...


179

p*
A

Pressure

Liquid

Two-component systems
When two components are present in a system, C = 2 and F = 4 − P
...
(The prime on F indicates that one of the degrees of freedom has been discarded,
in this case the temperature
...
Hence, one form of the phase diagram is a map of pressures and compositions at which each phase is stable
...


Vapour

p*
B
0

Mole fraction
of A, xA

1

The variation of the total vapour
pressure of a binary mixture with the mole
fraction of A in the liquid when Raoult’s
law is obeyed
...
6
...
3 Vapour pressure diagrams
The partial vapour pressures of the components of an ideal solution of two volatile liquids
are related to the composition of the liquid mixture by Raoult’s law (Section 5
...
2)°

where p* is the vapour pressure of pure A and p* that of pure B
...
6
...

B
A
(a) The composition of the vapour

The compositions of the liquid and vapour that are in mutual equilibrium are not
necessarily the same
...
This expectation can be confirmed as follows
...
2
...
4)

p

1

(6
...
5)°

Figure 6
...
We see that in all cases yA > xA, that is, the vapour
A B
is richer than the liquid in the more volatile component
...

Equation 6
...
Because we can relate the composition of the liquid to the
composition of the vapour through eqn 6
...
8

10
4

0
...
4

0
...
2 for pJ and eqn 6
...
2 0
...
6 0
...
5 for various values
of p* /p* (the label on each curve) with
A B
A more volatile than B
...

Fig
...
7

Exploration To reproduce the results
of Fig
...
7, first rearrange eqn 6
...
Then plot yA against xA for
A B
several values of pA/pB > 1
...
6)°

p* + (p* − p*)yA
A
B
A

This expression is plotted in Fig
...
8
...
8

(b) The interpretation of the diagrams

2

0
...
4

10
0
...
2

0
...
6

1000
0
...
6
...
6
...

A B

Fig
...
8

Exploration To reproduce the results
of Fig
...
8, first rearrange eqn 6
...
Then
A B
plot pA/p* against yA for several values of
A
p* /p* > 1
...
It is therefore sensible to combine Figs
...
7 and 6
...
6
...

The point a indicates the vapour pressure of a mixture of composition xA, and the
point b indicates the composition of the vapour that is in equilibrium with the liquid
at that pressure
...
That is, if the composition is specified (so using up the only remaining degree of freedom), the pressure at which the two phases are in equilibrium is fixed
...
If the horizontal axis
of the vapour pressure diagram is labelled with zA, then all the points down to the solid
diagonal line in the graph correspond to a system that is under such high pressure that
it contains only a liquid phase (the applied pressure is higher than the vapour pressure), so zA = xA, the composition of the liquid
...

Points that lie between the two lines correspond to a system in which there are two
phases present, one a liquid and the other a vapour
...
6
...
The lowering of pressure can be achieved by drawing out a piston
(Fig
...
11)
...
The
changes to the system do not affect the overall composition, so the state of the system
moves down the vertical line that passes through a
...
A point
between the two lines corresponds to both
liquid and vapour being present; outside
that region there is only one phase present
...


Fig
...
9

Vapour

1
0

Mole fraction of A, zA

1

Fig
...
10 The points of the pressure–
composition diagram discussed in the text
...


6
...
Until the point a1 is reached
(when the pressure has been reduced to p1), the sample consists of a single liquid
phase
...
As we have seen,
the composition of the vapour phase is given by point a1 A line joining two points
′
...
The composition of the liquid is
the same as initially (a1 lies on the isopleth through a), so we have to conclude that at
this pressure there is virtually no vapour present; however, the tiny amount of vapour
that is present has the composition a1
′
...

below the vapour pressure of the original liquid, so it vaporizes until the vapour
pressure of the remaining liquid falls to p2
...
Moreover, the composition of the vapour in equilibrium
with that liquid must be given by the point a2 at the other end of the tie line
...
6
...
If the pressure is reduced to p3, a
similar readjustment in composition takes place, and now the compositions of the
liquid and vapour are represented by the points a3 and a3 respectively
...
A
further decrease in pressure takes the system to the point a4; at this stage, only vapour
is present and its composition is the same as the initial overall composition of the
system (the composition of the original liquid)
...
To find the relative amounts of two phases α and β that are in
equilibrium, we measure the distances lα and lβ along the horizontal tie line, and then
use the lever rule (Fig
...
13):

na
la

nb

lb
b

One phase,
F=2

Composition

Composition
Fig
...
12 The general scheme of
interpretation of a pressure–composition
diagram (a vapour pressure diagram)
...
The distances lα and
lβ are used to find the proportions of the
amounts of phases α (such as vapour)
and β (for example, liquid) present at
equilibrium
...


Fig
...
13

(a)

(b)

181

(c)

Fig
...
11 (a) A liquid in a container exists
in equilibrium with its vapour
...
(b) When the pressure is changed by
drawing out a piston, the compositions of
the phases adjust as shown by the tie line in
the phase diagram
...


182

6 PHASE DIAGRAMS
nαlα = nβlβ

(6
...
In the case illustrated
in Fig
...
13, because lβ ≈ 2lα, the amount of phase α is about twice the amount of
phase β
...
2 The lever rule

To prove the lever rule we write n = nα + nβ and the overall amount of A as nzA
...
7
...
1 Using the lever rule

At p1 in Fig
...
10, the ratio lvap /lliq is almost infinite for this tie line, so nliq /nvap is also
almost infinite, and there is only a trace of vapour present
...
3, so nliq /nvap ≈ 0
...
3 times the amount of vapour
...


Temperature, T

Vapour
composition

a2

T2

a2¢

6
...
An example is shown
in Fig
...
14
...

(a) The distillation of mixtures

Fig
...
14 The temperature–composition
diagram corresponding to an ideal mixture
with the component A more volatile than
component B
...
The separation technique is called
fractional distillation
...
2

The textbook’s web site contains links to
online databases of phase diagrams
...
6
...
As usual,
the prime indicates that one degree of freedom has been discarded; in this case, the
pressure is being kept fixed, and hence at a given temperature the compositions of the
phases in equilibrium are fixed
...

Consider what happens when a liquid of composition a1 is heated
...
Then the liquid has composition a2 (the same as a1) and the
vapour (which is present only as a trace) has composition a2 The vapour is richer in
′
...
From
the location of a2, we can state the vapour’s composition at the boiling point, and
from the location of the tie line joining a2 and a2 we can read off the boiling tempera′
ture (T2) of the original liquid mixture
...
4 TEMPERATURE–COMPOSITION DIAGRAMS

(b) Azeotropes

Although many liquids have temperature–composition phase diagrams resembling
the ideal version in Fig
...
14, in a number of important cases there are marked deviations
...
6
...
In such cases
the excess Gibbs energy, GE (Section 5
...
Examples of this behaviour include trichloromethane/propanone and nitric
acid/water mixtures
...
6
...
For such mixtures GE is positive (less favourable to mixing than ideal),
and there may be contributions from both enthalpy and entropy effects
...

Vapour
composition

a4¢

a4
a3

a2
¢

a2

Temperature, T

Temperature, T

Boiling
temperature
of liquid

Vapour
composition

a3
¢

a2
¢
Boiling
temperature
of liquid
a3
¢
b

b
0

0

a

Mole fraction of A, zA

1

A high-boiling azeotrope
...

Fig
...
16

a4

a2

a3

Mole fraction of A, zA

a1
a
1

Fig
...
17 A low-boiling azeotrope
...


Temperature, T

1

2
3

A
(a)

Temperature, T

In a simple distillation, the vapour is withdrawn and condensed
...
In fractional distillation, the boiling and condensation cycle is repeated successively
...
We can follow the changes that occur by seeing
what happens when the first condensate of composition a3 is reheated
...
That vapour is drawn off, and
the first drop condenses to a liquid of composition a4
...

The efficiency of a fractionating column is expressed in terms of the number of
theoretical plates, the number of effective vaporization and condensation steps that
are required to achieve a condensate of given composition from a given distillate
...
6
...
To achieve the same separation
for the system shown in Fig
...
15b, in which the components have more similar
partial pressures, the fractionating column must be designed to correspond to five
theoretical plates
...
6
...
The two systems
shown correspond to (a) 3, (b) 5
theoretical plates
...
Consider a liquid of composition a on the right of the maximum
in Fig
...
16
...
If that
′)
vapour is removed (and condensed elsewhere), then the remaining liquid will move
to a composition that is richer in B, such as that represented by a3, and the vapour in
equilibrium with this mixture will have composition a3 As that vapour is removed,
′
...

remaining liquid shifts towards B as A is drawn off
...
When so much A has been evaporated that
the liquid has reached the composition b, the vapour has the same composition as
the liquid
...
The mixture is
said to form an azeotrope
...
One example of azeotrope formation is hydrochloric acid/water, which is azeotropic at 80 per cent by mass of water and boils
unchanged at 108
...

The system shown in Fig
...
17 is also azeotropic, but shows its azeotropy in a different way
...
The mixture
boils at a2 to give a vapour of composition a2 This vapour condenses in the column
′
...
That liquid reaches equilibrium
with its vapour at a3 which condenses higher up the tube to give a liquid of the same
′,
composition, which we now call a4
...
An example is ethanol/water, which boils
unchanged when the water content is 4 per cent by mass and the temperature is 78°C
...
6
...


Finally we consider the distillation of two immiscible liquids, such as octane and
water
...

(Fig
...
18a)
...
However, this boiling results in a vigorous agitation of
the mixture, so each component is kept saturated in the other component, and the
purging continues as the very dilute solutions are replenished
...
6
...
The presence of the saturated solutions
means that the ‘mixture’ boils at a lower temperature than either component would
alone because boiling begins when the total vapour pressure reaches 1 atm, not when
either vapour pressure reaches 1 atm
...
The only snag is that
the composition of the condensate is in proportion to the vapour pressures of the
components, so oils of low volatility distil in low abundance
...


6
...
5 Liquid–liquid phase diagrams
Tuc

P=1
P=2

a¢

a²
a

0

(a) Phase separation

Suppose a small amount of a liquid B is added to a sample of another liquid A at a
temperature T′
...

As more B is added, a stage comes at which no more dissolves
...

In the temperature–composition diagram drawn in Fig
...
19, the composition of the
former is represented by the point a′ and that of the latter by the point a″
...

When more B is added, A dissolves in it slightly
...

However, the amount of one phase increases at the expense of the other
...
The addition of more B now simply dilutes the solution, and from
then on it remains a single phase
...
For
hexane and nitrobenzene, raising the temperature increases their miscibility
...
We can construct the entire phase diagram by repeating the observations
at different temperatures and drawing the envelope of the two-phase region
...
An example is hexane and nitrobenzene
...
When P = 2, F′ = 1 (the prime denoting
the adoption of constant pressure), and the selection of a temperature implies that the
compositions of the immiscible liquid phases are fixed
...


185

Mole fraction of
nitrobenzene, xB

1

Fig
...
19 The temperature–composition
diagram for hexane and nitrobenzene at
1 atm
...
The upper critical
temperature, Tuc, is the temperature above
which the two liquids are miscible in all
proportions
...
2 Interpreting a liquid–liquid phase diagram

A mixture of 50 g of hexane (0
...
41 mol
C6H5NO2) was prepared at 290 K
...
Their
proportions are given by the lever rule (eqn 6
...
The temperature at which the
components are completely miscible is found by following the isopleth upwards
and noting the temperature at which it enters the one-phase region of the phase
diagram
...
6
...
6
...
The point xN = 0
...
The horizontal tie line cuts the phase

lb

P=2
273
0

0
...
4

0
...
8

x (C6H5NO2)
Fig
...
20 The temperature–composition
diagram for hexane and nitrobenzene at
1 atm again, with the points and lengths
discussed in the text
...
35 and xN = 0
...
According to the lever rule, the ratio of amounts of each phase is equal to
the ratio of the distances lα and lβ:
nα
nβ

=

lβ
lα

=

0
...
41
0
...
35

=

0
...
06

=7

That is, there is about 7 times more hexane-rich phase than nitrobenzene-rich
phase
...

Because the phase diagram has been constructed experimentally, these conclusions are not based on any assumptions about ideality
...

Self-test 6
...
09 and 0
...
3; 294 K]

273 K
...
5
Mole fraction of H, xH

1

Fig
...
21 The phase diagram for palladium
and palladium hydride, which has an upper
critical temperature at 300°C
...
3

This expression is an example of a
transcendental equation, an equation
that does not have a solution that can be
expressed in a closed form
...


Comment 6
...


The upper critical solution temperature, Tuc, is the highest temperature at which
phase separation occurs
...
This temperature exists because the greater thermal motion overcomes any potential energy advantage in molecules of one type being close together
...
6
...
An example of a
solid solution is the palladium/hydrogen system, which shows two phases, one a solid
solution of hydrogen in palladium and the other a palladium hydride, up to 300°C but
forms a single phase at higher temperatures (Fig
...
21)
...
We saw in
Section 5
...
5
...
Provided the parameter β that was introduced
in eqn 5
...
6
...
As a result, for β > 2 we can expect phase separation to occur
...
31 shows that we have to solve
ln

x
1−x

+ β(1 − 2x) = 0

The solutions are plotted in Fig
...
23
...

Some systems show a lower critical solution temperature, Tlc, below which they
mix in all proportions and above which they form two phases
...
6
...
In this case, at low temperatures the two components
are more miscible because they form a weak complex; at higher temperatures the
complexes break up and the two components are less miscible
...
They
occur because, after the weak complexes have been disrupted, leading to partial
miscibility, the thermal motion at higher temperatures homogenizes the mixture
again, just as in the case of ordinary partially miscible liquids
...
6
...


6
...
1
2

3

2

-0
...
5

-0
...
5
0

0
...
5

Composition
of second
phase

P=1

Tlc
1

Fig
...
22 The temperature variation of the
Gibbs energy of mixing of a system that is
partially miscible at low temperatures
...
This illustration is a duplicate
of Fig
...
20
...
31,
write an expression for Tmin, the
temperature at which ∆mixG has a
minimum, as a function of β and xA
...

Provide a physical interpretation for any
maxima or minima that you observe in
these plots
...
5
xA

1

The location of the phase
boundary as computed on the basis of
the β-parameter model introduced in
Section 5
...

Fig
...
23

Exploration Using mathematical
software or an electronic
spreadsheet, generate the plot of β against
xA by one of two methods: (a) solve the
transcendental equation ln {(x/(1− x)} +
β(1 − 2x) = 0 numerically, or (b) plot the
first term of the transcendental equation
against the second and identify the points
of intersection as β is changed
...

This combination is quite common because both properties reflect the tendency of
the two kinds of molecule to avoid each other
...

Figure 6
...
Distillation of a mixture of composition a1 leads to a vapour of composition b1, which condenses to the completely miscible single-phase solution at b2
...
This description applies only to the first drop of distillate
...
In the end,
when the whole sample has evaporated and condensed, the composition is back to a1
...
27 shows the second possibility, in which there is no upper critical solution
temperature
...
One phase has composition b3 and the other
′
has composition b3
″
...
6
...
A system at e1 forms two phases, which persist (but with changing
proportions) up to the boiling point at e2
...
2 0
...
6
0
...
6
...
This
system shows a lower critical temperature
at 292 K
...


Nicotine

H2O

Tuc

210

Temperature, q /°C

DmixG/nRT

-0
...
5

(C2H5)3N

Composition
of one
phase
P=2

0
-0
...
2

0
...
6
0
...
6
...

Note the high temperatures for the liquid
(especially the water): the diagram
corresponds to a sample under pressure
...
6
...
The
mixture forms a low-boiling azeotrope
...
6
...


composition as the liquid (the liquid is an azeotrope)
...
At a fixed
temperature, the mixture vaporizes and condenses like a single substance
...
3 Interpreting a phase diagram

State the changes that occur when a mixture of composition xB = 0
...
6
...

Method The area in which the point lies gives the number of phases; the composi-

tions of the phases are given by the points at the intersections of the horizontal tie
line with the phase boundaries; the relative abundances are given by the lever rule
(eqn 6
...

Answer The initial point is in the one-phase region
...
49

330
320
298
0

0
...
80

0
...
20

b3

0
...
66

b²
3

a1

0
...
6
...
6
...
3
...
66 (b1)
...
The boiling range of the liquid
is therefore 350 to 390 K
...
66
...
95
...
66 isopleth
...
87, the vapour xB = 0
...
At 320 K the sample consists of three phases: the
vapour and two liquids
...
30; the other
has composition xB = 0
...
62:1
...
20 and 0
...
82:1
...
When the last drop has been condensed the phase
composition is the same as at the beginning
...
3 Repeat the discussion, beginning at the point xB = 0
...


6
...
6 Liquid–solid phase diagrams

1
...
The system enters the two-phase region labelled ‘Liquid + B’
...

2
...
More of the solid forms, and the relative amounts of the solid and liquid
(which are in equilibrium) are given by the lever rule
...
The liquid phase is richer in A than before (its composition is
given by b3) because some B has been deposited
...
a3 → a4
...
This liquid now freezes to give a two-phase system of pure B and pure A
...
6
...
3 A liquid with the eutectic composition freezes at a single
temperature, without previously depositing solid A or B
...
Solutions of composition to the right of e deposit B as they cool, and solutions to the left deposit A: only the eutectic mixture (apart from pure A or pure B)
solidifies at a single definite temperature (F′ = 0 when C = 2 and P = 3) without gradually unloading one or other of the components from the liquid
...
The eutectic formed by
23 per cent NaCl and 77 per cent H2O by mass melts at −21
...
When salt is added to
ice under isothermal conditions (for example, when spread on an icy road) the mixture melts if the temperature is above −21
...
When salt is added to ice under adiabatic conditions (for example, when
added to ice in a vacuum flask) the ice melts, but in doing so it absorbs heat from the
rest of the mixture
...
Eutectic formation occurs in the
great majority of binary alloy systems, and is of great importance for the microstructure of solid materials
...
The two microcrystalline
phases can be distinguished by microscopy and structural techniques such as X-ray
diffraction (Chapter 20)
...
We can see how
it is used by considering the rate of cooling down the isopleth through a1 in Fig
...
29
...
6
...

Cooling is now slower because the solidification of B is exothermic and retards the
cooling
...
If the liquid has the eutectic composition e
initially, the liquid cools steadily down to the freezing temperature of the eutectic,

3

The name comes from the Greek words for ‘easily melted’
...
In this section, we shall consider systems where solid and
liquid phases may both be present at temperatures below the boiling point
...
6
...
The changes
that occur may be expressed as follows
...
6
...
Note the similarity to Fig
...
27
...


a1
a2

cooling e
Liquid
Temperature

Eutectic
g freezing
tatin
ecipi
B pr
ing
a4
ool
n
id c
sitio
Sol
Compo
Time
a5

a3

Fig
...
30 The cooling curves for the system
shown in Fig
...
29
...
There is a complete
halt at a4 while the eutectic solidifies
...

The eutectic halt shortens again for
compositions beyond e (richer in A)
...


190

6 PHASE DIAGRAMS
Liquid, P = 1

when there is a long eutectic halt as the entire sample solidifies (like the freezing of a
pure liquid)
...
The solid–liquid boundary is given by the
points at which the rate of cooling changes
...


a1

Temperature

a2
a3
a4 e
Solid,
P=2

A

Solid,
P=2

C
Composition

(b) Reacting systems

B

Fig
...
31 The phase diagram for a system in
which A and B react to form a compound
C = AB
...
6
...
The
constituent C is a true compound, not just
an equimolar mixture
...
Although three constituents are
present, there are only two components because GaAs is formed from the reaction
Ga + As 5 GaAs
...
6
...

A system prepared by mixing an excess of B with A consists of C and unreacted B
...
The principal change
from the eutectic phase diagram in Fig
...
29 is that the whole of the phase diagram
is squeezed into the range of compositions lying between equal amounts of A and B
(xB = 0
...
6
...
The interpretation of the information
in the diagram is obtained in the same way as for Fig
...
32
...
At temperatures below a4 there are
two solid phases, one consisting of C and the other of B
...

(c) Incongruent melting

In some cases the compound C is not stable as a liquid
...
6
...
Consider what happens as a liquid at a1 is
cooled:
1
...
Some solid Na is deposited, and the remaining liquid is richer in K
...
a2 → just below a3
...

b1

T1
Liquid
+ solid K
containing
some Na

Fig
...
32 The phase diagram for an actual
system (sodium and potassium) like
that shown in Fig
...
35, but with two
differences
...
The second is that the
compound exists only as the solid, not as
the liquid
...


Solid K
+ solid K
containing
some Na

a1
Liquid
+ solid Na
containing
some K

Liquid,
P=1

T2
¢
T2
T3

a2
Liquid +
solid Na2K

Solid Na2K
+ solid K
containing
some Na

b2
b3

T4

a3

Solid Na2K
+ solid Na
containing
some K

b4

P=2
K

Solid Na
+ solid Na
containing
some K

Na2K
Composition

P=2
Na

6
...
b1 → b2
...

2
...
Solid Na deposits, but at b3 a reaction occurs to form Na2K: this compound is formed by the K atoms diffusing into the solid Na
...
b3
...
The horizontal line representing this three-phase equilibrium is
called a peritectic line
...

4
...
As cooling continues, the amount of solid compound increases until at
b4 the liquid reaches its eutectic composition
...

If the solid is reheated, the sequence of events is reversed
...
This behaviour is an example of
incongruent melting, in which a compound melts into its components and does not
itself form a liquid phase
...
1 Liquid crystals

A mesophase is a phase intermediate between solid and liquid
...
A
mesophase may arise when molecules have highly non-spherical shapes, such as being
long and thin (1), or disk-like (2)
...
Calamitic liquid crystals (from the Greek word for reed) are made from long and
thin molecules, whereas discotic liquid crystals are made from disk-like molecules
...
6
...
In the cholesteric phase, the
stacking of layers continues to give a helical
arrangement of molecules
...
A lyotropic liquid crystal is a solution that undergoes a transition to the liquid crystalline phase as the composition is changed
...
6
...

Other materials, and some smectic liquid crystals at higher temperatures, lack the
layered structure but retain a parallel alignment; this mesophase is called a nematic
phase (from the Greek for thread, which refers to the observed defect structure of the
phase)
...
That is, they form helical
structures with a pitch that depends on the temperature
...
Disk-like
molecules such as (2) can form nematic and columnar mesophases
...
5 nm)
...
Nematic liquid crystals also respond in
special ways to electric fields
...
In a ‘twisted nematic’
LCD, the liquid crystal is held between two flat plates about 10 µm apart
...
The plates also have a surface that causes the liquid crystal to adopt
a particular orientation at its interface and are typically set at 90° to each other but
270° in a‘supertwist’ arrangement
...
The
incident light passes through the outer polarizer, then its plane of polarization is rotated
as it passes through the twisted nematic and, depending on the setting of the second
polarizer, will pass through (if that is how the second polarizer is arranged)
...

Although there are many liquid crystalline materials, some difficulty is often experienced in achieving a technologically useful temperature range for the existence of
the mesophase
...
An example of the
type of phase diagram that is then obtained is shown in Fig
...
34
...

IMPACT ON MATERIALS SCIENCE

I6
...
For example,
semiconductor devices consist of almost perfectly pure silicon or germanium doped to
a precisely controlled extent
...
4
In the technique of zone refining the sample is in the form of a narrow cylinder
...
The advancing liquid zone accumulates the impurities as it passes
...


CHECKLIST OF KEY IDEAS
A

Heating coil

140

100

(a)

Nematic

q/°C
120

Purified
material
Solid
solution

B

a1
a2
b1
b2 a3
b3

Isotropic

Collected
impurities

Solid
solution

0
...
6
...


a3
¢

Liquid

Solid

(b)
0

a2
¢

Temperature, T

160

193

The procedure for zone refining
...
(b) After a
molten zone is passed along the rod, the
impurities are more concentrated at the
right
...

Fig
...
35

a
0

Composition, xB

1

Fig
...
36 A binary temperature–
composition diagram can be used to
discuss zone refining, as explained in the
text
...
6
...
The zone at the end of the sample is the impurity dump: when
the heater has gone by, it cools to a dirty solid which can be discarded
...
It relies
on the impurities being more soluble in the molten sample than in the solid, and
sweeps them up by passing a molten zone repeatedly from one end to the other along
a sample
...
6
...
Consider
a liquid (this represents the molten zone) on the isopleth through a1, and let it cool
without the entire sample coming to overall equilibrium
...

′
heater has moved on) is at a2 Cooling that liquid down an isopleth passing through a2
deposits solid of composition b3 and leaves liquid at a3 The process continues until
′
...
There is plenty of
everyday evidence that impure liquids freeze in this way
...
It cannot escape from the interior of the cube, and so when
that freezes it occludes the air in a mist of tiny bubbles
...
It is used to introduce controlled
amounts of impurity (for example, of indium into germanium)
...
The zone is
then dragged repeatedly in alternate directions through the sample, where it deposits
a uniform distribution of the impurity
...
A phase is a state of matter that is uniform throughout, not
only in chemical composition but also in physical state
...
A constituent is a chemical species (an ion or a molecule)
...


3
...

4
...


194

6 PHASE DIAGRAMS
5
...

6
...
The composition of the vapour,
B
A
B
yA = xA p* /{p* + (p* − p*)xA}, yB = 1 − yA
...
The total vapour pressure of a mixture is given by
*p* A
p = pA B /{p* + (p* − p* )yA}
...
An isopleth is a line of constant composition in a phase
diagram
...

9
...


10
...


11
...

12
...

13
...
The lower critical solution temperature is the
temperature below which the components of a binary mixture
mix in all proportions and above which they form two phases
...
A eutectic is the mixture with the lowest melting point;
a liquid with the eutectic composition freezes at a single
temperature
...

15
...


Further reading
Articles and texts

J
...
Alper, The Gibbs phase rule revisited: interrelationships between
components and phases
...
Chem
...
76, 1567 (1999)
...
D
...
, Materials science and engineering, an introduction
...

P
...
Collings and M
...
Taylor & Francis, London (1997)
...
Hillert, Phase equilibria, phase diagrams and phase transformations:
a thermodynamic basis
...

H
...
Lee, Chemical thermodynamics for metals and materials
...


R
...
Stead and K
...

J
...
Educ
...

S
...
Sandler, Chemical and engineering thermodynamics
...

Sources of data and information

A
...
1, 2, and 3
...

J
...
Elsevier,
Amsterdam (1981–86)
...
1 Define the following terms: phase, constituent, component, and degree of

freedom
...
2 What factors determine the number of theoretical plates required to

achieve a desired degree of separation in fractional distillation?
6
...
Label the regions

and intersections of the diagrams, stating what materials (possibly compounds
or azeotropes) are present and whether they are solid liquid or gas
...
(b) Two-component, temperature–composition, solid–liquid diagram,
one compound AB formed that melts congruently, negligible solid–solid
solubility
...
4 Draw phase diagrams for the following types of systems
...
(a) Two-component, temperature–composition, solid–liquid diagram,
one compound of formula AB2 that melts incongruently, negligible
solid–solid solubility; (b) two-component, constant temperature–
composition, liquid–vapour diagram, formation of an azeotrope at xB = 0
...

6
...
6
...
State what substances

(if compounds give their formulas) exist in each region
...

6
...
6
...
State what substances

(if compounds give their formulas) exist in each region
...


195

0
...
33

Temperature, T

Temperature, T

EXERCISES

0

0
...
4

xB

0
...
8

0

1

Fig
...
37

0
...
4

xB

0
...
8

1

Fig
...
38

Exercises
6
...
3 kPa and that of
1,2-dimethylbenzene is 20
...
What is the composition of a liquid mixture
that boils at 90°C when the pressure is 0
...

θ /°C 110
...
0
114
...
8
117
...
0
121
...
0

xM

0
...
795

0
...
527

0
...
300

0
...
097

6
...
923

0
...
698

0
...
527

0
...
297

0
...
What is the composition of a liquid mixture
that boils at 90°C when the pressure is 19 kPa? What is the composition of the
vapour produced?
6
...
7 kPa and that of
pure liquid B is 52
...
These two compounds form ideal liquid and gaseous
mixtures
...
350
...

6
...
8 kPa and that of

pure liquid B is 82
...
These two compounds form ideal liquid and gaseous
mixtures
...
612
...

6
...
6589 is 88°C
...
6 kPa and 50
...
(a) Is this solution ideal? (b) What is
the initial composition of the vapour above the solution?
6
...
6°C and 125
...
Plot the
temperature/composition diagram for the mixture
...
250
and (b) x O = 0
...
5(b) The following temperature/composition data were obtained for a

mixture of two liquids A and B at 1
...

θ /°C
125
130
135
140
145
150
xA

0
...
65

0
...
30

0
...
098

yA

0
...
91

0
...
61

0
...
25

The boiling points are 124°C for A and 155°C for B
...
What is the composition of the vapour
in equilibrium with the liquid of composition (a) xA = 0
...
33?
6
...

(a) NaH2PO4 in water at equilibrium with water vapour but disregarding
the fact that the salt is ionized
...


xA = 0
...
At this temperature the vapour pressures of pure A and B
are 110
...
5 kPa, respectively
...
6(b) State the number of components for a system in which AlCl3 is

6
...
9 kPa at 358 K) and dibromopropene
DE
(DP, p* = 17
...
If zDE = 0
...
How many phases and components are present in an otherwise empty
heated container?

6
...
Consider an

equimolar solution of benzene and toluene
...
9 kPa and 2
...
The solution is
boiled by reducing the external pressure below the vapour pressure
...
Assume that the rate of vaporization is low enough for the
temperature to remain constant at 20°C
...
5(a) The following temperature/composition data were obtained for a

mixture of octane (O) and methylbenzene (M) at 1
...

6
...
7(b) Ammonium chloride, NH4Cl, decomposes when it is heated
...
How many components and phases are present?
6
...
(a) How many phases and
components are present
...

6
...
8a is not saturated
...
(b) What is the variance
(the number of degrees of freedom) of the system? Identify the independent
variables
...
9(a) Methylethyl ether (A) and diborane, B2H6 (B), form a compound
that melts congruently at 133 K
...

The melting points of pure A and B are 131 K and 110 K, respectively
...
Assume negligible solid–solid
solubility
...
9(b) Sketch the phase diagram of the system NH3 /N2H4 given that the

two substances do not form a compound with each other, that NH3 freezes
at −78°C and N2H4 freezes at +2°C, and that a eutectic is formed when the
mole fraction of N2H4 is 0
...


T1
300°C

T2

0

0
...
4

Temperature, T

Temperature, T

T1

0
...
4

xB

0
...
8

xB

0
...
8

1

Fig
...
41

b

0

b

400°C

6
...
39 shows the phase diagram for two partially miscible liquids,

which can be taken to be that for water (A) and 2-methyl-1-propanol (B)
...
8 is
heated, at each stage giving the number, composition, and relative amounts of
the phases present
...
2

0
...
6

0
...
6
...
6
...
10(b) Figure 6
...
Label the regions,

and describe what will be observed when liquids of compositions a and b are
cooled to 200 K
...
14(a) Figure 6
...

6
...
6
...


1000

b

6
...
6
...
6
...
11(a) Indicate on the phase diagram in Fig
...
41 the feature that denotes

incongruent melting
...
11(b) Indicate on the phase diagram in Fig
...
42 the feature that denotes
incongruent melting
...
12(a) Sketch the cooling curves for the isopleths a and b in Fig
...
41
...
12(b) Sketch the cooling curves for the isopleths a and b in Fig
...
42
...
(a) Label the regions of the
diagrams as to which phases are present
...
(c) What is the
vapour pressure of the solution at 70°C when just one drop of liquid remains
...
(e) What are the mole fractions for
the conditions of part c? (f) At 85°C and 760 Torr, what are the amounts of
substance in the liquid and vapour phases when zheptane = 0
...
14(b) Uranium tetrafluoride and zirconium tetrafluoride melt at 1035°C
and 912°C, respectively
...
77
...
28 is in equilibrium
with a solid solution of composition x(ZrF4) = 0
...
At 850°C the two
compositions are 0
...
90, respectively
...
40 is cooled slowly from 900°C to 500°C
...
15(a) Methane (melting point 91 K) and tetrafluoromethane (melting

point 89 K) do not form solid solutions with each other, and as liquids they
are only partially miscible
...
43 and the eutectic temperature is 84 K at x(CF4)
= 0
...
At 86 K, the phase in equilibrium with the tetrafluoromethane-rich

PROBLEMS

solution changes from solid methane to a methane-rich liquid
...
10 and x(CF4) = 0
...
Sketch the phase diagram
...
15(b) Describe the phase changes that take place when a liquid mixture of
4
...
0 mol CH3OCH3 (melting point
135 K) is cooled from 140 K to 90 K
...
The system exhibits one
eutectic at x(B2H6) = 0
...
90 and 104 K
...
16(a) Refer to the information in Exercise 6
...
2

0
...
6

0
...
10, (b) 0
...
50,
(d) 0
...
95
...
16(b) Refer to the information in Exercise 6
...
10, (b) 0
...
50,
(d) 0
...
95
...
17(a) Hexane and perfluorohexane show partial miscibility below 22
...


90

The critical concentration at the upper critical temperature is x = 0
...
At 22
...
24 and x = 0
...
5°C the mole fractions are 0
...
51
...
Describe the phase changes that occur
when perfluorohexane is added to a fixed amount of hexane at (a) 23°C, (b) 22°C
...
17(b) Two liquids, A and B, show partial miscibility below 52
...
The
critical concentration at the upper critical temperature is x = 0
...
At 40
...
22 and x = 0
...
5°C the mole fractions are 0
...
48
...
Describe the phase changes that occur
when B is added to a fixed amount of A at (a) 48°C, (b) 52
...


60
0

0
...
4

0
...
8

1

Fig
...
43

Problems*
Numerical problems
6
...
The mole fraction of 1-butanol in the liquid (x) and vapour (y) phases
at 1
...
Artigas,
C
...
Cea, F
...
Royo, and J
...
Urieta, J
...
Eng
...


T/K

396
...
94

391
...
15

389
...
66

6
...


388
...
1065

0
...
2646

0
...
5017

0
...
7171

y

0
...
3691

0
...
5138

0
...
6409

0
...
86 K
...
(b) Estimate the temperature at
which a solution whose mole fraction of 1-butanol is 0
...

(c) State the compositions and relative proportions of the two phases present
after a solution initially 0
...
94 K
...
2‡ An et al
...
An, H
...
Fuguo, and W
...
Chem
...
Mole fractions of N,Ndimethylacetamide in the upper (x1) and lower (x2) phases of a two-phase
region are given below as a function of temperature:
T/K

(a) Plot the phase diagram
...
750 mol of N,N-dimethylacetamide with
0
...
0 K
...
3

78

80

82

84

86

88

90
...

6
...
The best

characterized are P4S3, P4S7, and P4S10, all of which melt congruently
...
Label each region of the
diagram with the substance that exists in that region and indicate its phase
...
The melting point of pure phosphorus is 44°C
and that of pure sulfur is 119°C
...
28
...
2
and negligible solid–solid solubility
...
820

309
...
031

308
...
686

x1

0
...
400

0
...
326

0
...
529

0
...
625

0
...
690

304
...
803

299
...
000

294
...
255

0
...
193

0
...
157

6
...
724

0
...
783

0
...
814

cooling curves of two metals A and B
...


6 PHASE DIAGRAMS
0

0

-

D mixG /(kJ mol 1)

with the data of these curves
...
Give the probable formulas of any
compounds that form
...
0

1060

700

20
...
0

940

700

400

40
...
0

750

700

400

60
...
0

550

-3

1500 K

-4
0

0
...
6

0
...
6
...
6 Consider the phase diagram in Fig
...
44, which represents a solid–liquid

Liquid

Temperature, q/°C

equilibrium
...
Indicate the number of species and
phases present at the points labelled b, d, e, f, g, and k
...
16, 0
...
57, 0
...
84
...
402
1324

1314

1300

0
...
0

Ca2Si

450

1030

0
...
2

0
...
6

0
...
84

0
...
6
...
57

0
...
16

Temperature, T

0
...
0
90
...
6
...
7 Sketch the phase diagram for the Mg/Cu system using the following

information: θf (Mg) = 648°C, θf (Cu) = 1085°C; two intermetallic compounds
are formed with θf (MgCu2) = 800°C and θf (Mg2Cu) = 580°C; eutectics of
mass percentage Mg composition and melting points 10 per cent (690°C),
33 per cent (560°C), and 65 per cent (380°C)
...
Describe what will be observed if the melt is cooled
slowly to room temperature
...

6
...
45 shows ∆mixG(xPb, T) for a mixture of copper and lead
...
1, (ii) xPb = 0
...
What is the equilibrium
composition of the final mixture? Include an estimate of the relative amounts
of each phase
...
9‡ The temperature–composition diagram for the Ca/Si binary system is
shown in Fig
...
46
...
(b) If a 20 per cent by atom composition
melt of silicon at 1500°C is cooled to 1000°C, what phases (and phase
composition) would be at equilibrium? Estimate the relative amounts of each
phase
...
What phases, and relative
amounts, would be at equilibrium at a temperature (i) slightly higher than
1030°C, (ii) slightly lower than 1030°C? Draw a graph of the mole percentages
of both Si(s) and CaSi2(s) as a function of mole percentage of melt that is
freezing at 1030°C
...
10 Iron(II) chloride (melting point 677°C) and potassium chloride
(melting point 776°C) form the compounds KFeCl3 and K2FeCl4 at elevated
temperatures
...
Eutectics are formed with compositions x = 0
...
54 (melting point 393°C), where x is the mole
fraction of FeCl2
...
34
...
State the phases that are in equilibrium
when a mixture of composition x = 0
...


Theoretical problems
6
...

6
...


Applications: to biology, materials science, and chemical
engineering
6
...
For example, urea, CO(NH2)2, competes for
NH and CO groups and interferes with hydrogen bonding in a polypeptide
...


Unfolded

Temperature

0
...
8

6
...
For example, liquid crystalline Kevlar (3) is strong enough
to be the material of choice for bulletproof vests and is stable at temperatures
up to 600 K
...
7
Native

0
...
5
0

0
...
2
Denaturant

0
...
6
...
6
...
It shows three structural regions: the native
form, the unfolded form, and a ‘molten globule’ form, a partially unfolded
but still compact form of the protein
...
1? (ii) Describe what
happens to the polymer as the native form is heated in the presence of
denaturant at concentration 0
...


Temperature, q/°C

6
...

3
The hydrophobic chains stack together to form an extensive bilayer about
5 nm across, leaving the polar groups exposed to the aqueous environment on
either side of the membrane (see Chapter 19 for details)
...
Biological cell membranes exist as liquid crystals at
physiological temperatures
...
6
...

The two components are dielaidoylphosphatidylcholine (DEL) and
dipalmitoylphosphatidylcholine (DPL)
...
5 is cooled from 45°C
...
18 The technique of float zoning, which is similar to zone refining
(Impact I6
...
Consult a textbook of materials
science or metallurgy and prepare a discussion of the principles, advantages,
and disadvantages of float zoning
...
19 Magnesium oxide and nickel oxide withstand high temperatures
...
Draw the temperature–composition diagram for the system using
the data below, where x is the mole fraction of MgO in the solid and y its
mole fraction in the liquid
...
35

0
...
83

1
...
18

0
...
65

1
...
30, (b) the composition
and proportion of the phases present when a solid of composition x = 0
...
70 will begin to solidify
...
20 The bismuth–cadmium phase diagram is of interest in metallurgy,
and its general form can be estimated from expressions for the depression
of freezing point
...
5 K, Tf (Cd) = 594 K, ∆fus H(Bi) = 10
...
07 kJ mol−1
...

Use the phase diagram to state what would be observed when a liquid of
composition x(Bi) = 0
...
What are the relative
abundances of the liquid and solid at (a) 460 K and (b) 350 K? Sketch the
cooling curve for the mixture
...
17 Use a phase diagram like that shown in Fig
...
36 to indicate how zone
levelling may be described
...
21‡ Carbon dioxide at high pressure is used to separate various compounds
in citrus oil
...
2 K is given below for a variety of pressures (Y
...
Morotomi, K
...
Koga, and Y
...
Chem
...
Data 41, 951 (1996))
...
15 The compound p-azoxyanisole forms a liquid crystal
...
0 g of the solid
was placed in a tube, which was then evacuated and sealed
...
94

6
...
97

8
...
27

x

0
...
4541

0
...
7744

0
...
6
...
9982

0
...
9973

0
...
9922

(a) Plot the portion of the phase diagram represented by these data
...
02 MPa at 323
...


7
Spontaneous chemical
reactions
7
...
2 The description of equilibrium

The response of equilibria to
the conditions
7
...
4 The response of equilibria to

temperature
I7
...
The equilibrium composition
corresponds to a minimum in the Gibbs energy plotted against the extent of reaction, and
by locating this minimum we establish the relation between the equilibrium constant
and the standard Gibbs energy of reaction
...
The principles
of thermodynamics established in the preceding chapters can be applied to the description of the thermodynamic properties of reactions that take place in electrochemical
cells, in which, as the reaction proceeds, it drives electrons through an external circuit
...
There are two major topics
developed in this connection
...


extraction of metals from their
oxides
Equilibrium electrochemistry
7
...
6 Varieties of cells
7
...
8 Standard potentials
7
...
2 Impact on biochemistry:

Chemical reactions tend to move towards a dynamic equilibrium in which both reactants and products are present but have no further tendency to undergo net change
...
However, in many important cases the equilibrium mixture has
significant concentrations of both reactants and products
...
Because many reactions of ions involve the transfer of electrons, they can
be studied (and utilized) by allowing them to take place in an electrochemical cell
...


Energy conversion in biological
cells

Spontaneous chemical reactions
Checklist of key ideas
Further reading
Discussion questions
Exercises

We have seen that the direction of spontaneous change at constant temperature and
pressure is towards lower values of the Gibbs energy, G
...


Problems

7
...


7
...
Even though this reaction looks trivial, there are
many examples of it, such as the isomerization of pentane to 2-methylbutane and the
conversion of l-alanine to d-alanine
...
The quantity ξ (xi) is called the extent of reaction;
it has the dimensions of amount of substance and is reported in moles
...
So, if
initially 2
...
5 mol, then the amount of A
remaining will be 0
...

The reaction Gibbs energy, ∆ rG, is defined as the slope of the graph of the Gibbs
energy plotted against the extent of reaction:
∆rG =

A ∂G D
C ∂ξ F p,T

[7
...
However, to see that there is a close relationship with the
normal usage, suppose the reaction advances by dξ
...
2)

We see that ∆rG can also be interpreted as the difference between the chemical
potentials (the partial molar Gibbs energies) of the reactants and products at the composition of the reaction mixture
...
Moreover, because
the reaction runs in the direction of decreasing G (that is, down the slope of G plotted
against ξ), we see from eqn 7
...
The slope is zero, and the
reaction is spontaneous in neither direction, when
∆rG = 0

(7
...
7
...
It follows that, if we can find the
composition of the reaction mixture that ensures µB = µA, then we can identify the
composition of the reaction mixture at equilibrium
...

If ∆rG > 0, the reverse reaction is spontaneous
...


As the reaction advances
(represented by motion from left to right
along the horizontal axis) the slope of the
Gibbs energy changes
...


Fig
...
1

202

7 CHEMICAL EQUILIBRIUM

If two weights are coupled as shown
here, then the heavier weight will move the
lighter weight in its non-spontaneous
direction: overall, the process is still
spontaneous
...


Fig
...
2

A reaction for which ∆rG < 0 is called exergonic (from the Greek words for work producing)
...
A
simple mechanical analogy is a pair of weights joined by a string (Fig
...
2): the lighter
of the pair of weights will be pulled up as the heavier weight falls down
...
In biological cells, the oxidation of carbohydrates act
as the heavy weight that drives other reactions forward and results in the formation of
proteins from amino acids, muscle contraction, and brain activity
...
The reaction can
be made to occur only by doing work on it, such as electrolysing water to reverse its
spontaneous formation reaction
...

7
...

(a) Perfect gas equilibria

When A and B are perfect gases we can use eqn 5
...
4)°

pA

If we denote the ratio of partial pressures by Q, we obtain
∆rG = ∆rG 7 + RT ln Q
Comment 7
...


Q=

pB
pA

(7
...
It ranges from 0 when pB = 0 (corresponding to pure A) to infinity when pA = 0 (corresponding to pure B)
...
For our
reaction
7
7
7
7
∆rG 7 = G B,m − G A,m = µ B − µA

(7
...
6 we saw that the difference in standard molar Gibbs energies of the products and reactants is equal to the difference in their standard Gibbs energies of formation, so in practice we calculate ∆rG 7 from
∆rG 7 = ∆f G 7(B) − ∆f G 7(A)

(7
...
The ratio of partial pressures at equilibrium is denoted K,
and eqn 7
...
8)°

7
...


In molecular terms, the minimum in the Gibbs energy, which corresponds to ∆rG
= 0, stems from the Gibbs energy of mixing of the two gases
...

Consider a hypothetical reaction in which A molecules change into B molecules
without mingling together
...
7
...
There is no intermediate minimum in the graph
...
We have seen that the contribution of a mixing process to the change in
Gibbs energy is given by eqn 5
...
This expression makes a U-shaped contribution to the total change in Gibbs energy
...
7
...

We see from eqn 7
...

Therefore, at equilibrium the partial pressure of A exceeds that of B, which means that
the reactant A is favoured in the equilibrium
...
Now the product B is favoured in the equilibrium
...
8 to a general reaction
...

Consider the reaction 2 A + B → 3 C + D
...
This equation has the form
0=

∑ νJJ

(7
...
In our example, these numbers have the values
νA = −2, νB = −1, νC = +3, and νD = +1
...
Then we define ξ so that, if it changes by ∆ξ, then the
change in the amount of any species J is νJ∆ξ
...
1 Identifying stoichiometric numbers

To express the equation
N2(g) + 3 H2(g) → 2 NH3(g)

(7
...
1 The approach to equilibrium

203

Including
mixing
0

Extent of reaction, x

Mixing

If the mixing of reactants and
products is ignored, then the Gibbs energy
changes linearly from its initial value (pure
reactants) to its final value (pure products)
and the slope of the line is ∆rG 7
...
The sum
of the two contributions has a minimum
...

Fig
...
3

204

7 CHEMICAL EQUILIBRIUM
in the notation of eqn 7
...

Therefore, if initially there is 10 mol N2 present, then when the extent of reaction
changes from ξ = 0 to ξ = 1 mol, implying that ∆ξ = +1 mol, the amount of N2
changes from 10 mol to 9 mol
...

When ∆ξ = +1 mol, the amount of H2 changes by −3 × (1 mol) = −3 mol and the
amount of NH3 changes by +2 × (1 mol) = +2 mol
...
Few, however, make the distinction
between the two types of quantity
...
1
...
11)

with the standard reaction Gibbs energy calculated from

∑ ν∆ f G 7 − ∑ ν∆ f G 7

∆rG 7 =

Products

(7
...
12b)

J

The reaction quotient, Q, has the form
Q=

activities of products
activities of reactants

(7
...
More formally, to write the general expression for Q we introduce the symbol Π to denote the
product of what follows it (just as ∑ denotes the sum), and define Q as
Q=

Π aνJ
J

[7
...
Recall from Table 5
...

Illustration 7
...
The reaction quotient is then
−2 −3
2
Q = a A a B aCaD =

2
aCaD
3
a2 aB
A

7
...
1 The dependence of the reaction Gibbs energy on the reaction
quotient

Consider the reaction with stoichiometric numbers νJ
...
The resulting
infinitesimal change in the Gibbs energy at constant temperature and pressure is
A
D
dG = ∑ µJdnJ = ∑ µJνJ dξ = B ∑ νJ µJE dξ
C J
F
J
J

(7
...
15)

To make further progress, we note that the chemical potential of a species J is related
7
to its activity by eqn 5
...
When this expression is substituted
into eqn 7
...
2

Recall that a ln x = ln x a and ln x + ln y

= ∆rG 7 + RT ln Q

A
+ · · · = ln xy · · · , so ∑ ln xi = ln B
C
i

with Q given by eqn 7
...


Now we conclude the argument based on eqn 7
...
At equilibrium, the slope
of G is zero: ∆rG = 0
...
16]

J

equilibrium

This expression has the same form as Q, eqn 7
...
From now on, we shall not write the ‘equilibrium’ subscript explicitly, and
will rely on the context to make it clear that for K we use equilibrium values and for Q
we use the values at the specified stage of the reaction
...
Note that, because activities are dimensionless numbers, the thermodynamic equilibrium constant is also dimensionless
...
16 are often replaced by the
numerical values of molalities (that is, by replacing aJ by bJ/b7, where b7 = 1 mol kg−1),
molar concentrations (that is, as [J]/c 7, where c 7 = 1 mol dm−3), or the numerical
values of partial pressures (that is, by pJ/p7, where p7 = 1 bar)
...
The approximation is particularly severe for
electrolyte solutions, for in them activity coefficients differ from 1 even in very dilute
solutions (Section 5
...


D

ΠxiE
...
3 Writing an equilibrium constant
...
3)
...

Comment 7
...
17 may be expressed
in terms of spectroscopic data for gasphase species; so this expression also
provides a link between spectroscopy
and equilibrium composition
...
11 and replace Q by K
...
17)

This is an exact and highly important thermodynamic relation, for it enables us to
predict the equilibrium constant of any reaction from tables of thermodynamic data,
and hence to predict the equilibrium composition of the reaction mixture
...
1 Calculating an equilibrium constant

Calculate the equilibrium constant for the ammonia synthesis reaction, eqn 7
...

Method Calculate the standard reaction Gibbs energy from eqn 7
...
17
...
16, and because the gases are taken to
be perfect, we replace each activity by the ratio p/p7, where p is a partial pressure
...
5 kJ mol−1)
Then,
ln K = −

2 × (−16
...
3145 J K−1 mol−1) × (298 K)

=

2 × 16
...
3145 × 298

Hence, K = 6
...
This result is thermodynamically exact
...
2 THE DESCRIPTION OF EQUILIBRIUM
and this ratio has exactly the value we have just calculated
...
1 Evaluate the equilibrium constant for N2O4(g) 5 2 NO2(g) at 298 K
...
15]

Example 7
...
The standard Gibbs energy of reaction for the decomposition
1
H2O(g) → H2(g) + – O2(g) is +118
...
What is the degree of dis2
sociation of H2O at 2300 K and 1
...
17, so the task is to relate the degree of dissociation, α, to K
and then to find its numerical value
...
Because the standard Gibbs
energy of reaction is large and positive, we can anticipate that K will be small, and
hence that α < 1, which opens the way to making approximations to obtain its
<
numerical value
...
17 in the form

∆rG 7

ln K = −
=−

(+118
...
3145 J K−1 mol−1) × (2300 K)
RT
118
...
3145 × 2300

It follows that K = 2
...
The equilibrium composition can be expressed in
terms of α by drawing up the following table:
H2O

H2

O2

Initial amount

n

0

0

Change to reach equilibrium

−αn

+αn

1
+ –αn
2

Amount at equilibrium

(1 − α)n

αn

1
–αn
2

1−α

α

1
1 + –α
2

Mole fraction, xJ

1
1 + –α
2

(1 − α)p

Partial pressure, pJ

1+

1
–α
2

αp
1
1 + –α
2

1
Total: (1 + –α)n
2

1
–α
2
1
1 + –α
2
1
–αp
2
1
1 + –α
2

where, for the entries in the last row, we have used pJ = xJ p (eqn 1
...
The equilibrium constant is therefore
K=

pH2 p1/2
O2
pH2O

=

α 3/2p1/2
(1 − α)(2 + α)1/2

207

208

7 CHEMICAL EQUILIBRIUM
In this expression, we have written p in place of p/p7, to keep the notation simple
...
00 bar (that is, p/p7 = 1
...
0205
...

A note on good practice Always check that the approximation is consistent with
the final answer
...

<
Comment 7
...


Self-test 7
...
2 kJ mol−1 for the same reaction, suppose that steam at 200 kPa is passed
through a furnace tube at that temperature
...

[0
...
To do so, we need to
know the activity coefficients, and then to use aJ = γJ xJ or aJ = γJbJ /b7 (recalling that the
activity coefficients depend on the choice)
...
18)
γAγB bAbB
The activity coefficients must be evaluated at the equilibrium composition of the mixture (for instance, by using one of the Debye–Hückel expressions, Section 5
...
In elementary applications, and to
begin the iterative calculation of the concentrations in a real example, the assumption
is often made that the activity coefficients are all so close to unity that Kγ = 1
...

aAaB

=

Molecular interpretation 7
...
The bulk of
the population is associated with the
species A, so that species is dominant at
equilibrium
...
7
...
1)
...
The atoms distribute themselves over
both sets of energy levels in accord with the Boltzmann distribution (Fig
...
4)
...

It can be appreciated from the illustration that, if the reactants and products
both have similar arrays of molecular energy levels, then the dominant species in a
reaction mixture at equilibrium will be the species with the lower set of energy

7
...
However, the fact that the Gibbs energy occurs in the expression is a signal
that entropy plays a role as well as energy
...
7
...
We see that, although the B energy levels lie higher than the A energy
levels, in this instance they are much more closely spaced
...
Closely spaced energy levels correlate with a high entropy (see
Molecular interpretation 3
...
This competition is mirrored in eqn 7
...
19)

Note that a positive reaction enthalpy results in a lowering of the equilibrium
constant (that is, an endothermic reaction can be expected to have an equilibrium
composition that favours the reactants)
...

Population
(d) Equilibria in biological systems

Even though the reaction A → B is
endothermic, the density of energy levels in
B is so much greater than that in A that the
population associated with B is greater than
that associated with A, so B is dominant at
equilibrium
...
7
...
7 that for biological systems it is appropriate to adopt the biological standard state, in which aH+ = 10−7 and pH = −log aH+ = 7
...
56
that the relation between the thermodynamic and biological standard Gibbs energies
of reaction for a reaction of the form
A + ν H+(aq) → P

(7
...
20b)

is

Note that there is no difference between the two standard values if hydrogen ions are
not involved in the reaction (ν = 0)
...
4 Using the biological standard state

Consider the reaction
NADH(aq) + H+(aq) → NAD+(aq) + H2(g)
at 37°C, for which ∆rG 7 = −21
...
NADH is the reduced form of nicotinamide adenine dinucleotide and NAD+ is its oxidized form; the molecules play an
important role in the later stages of the respiratory process
...
1,
∆rG ⊕ = −21
...
1 × (8
...
7 kJ mol−1
Note that the biological standard value is opposite in sign (in this example) to the
thermodynamic standard value: the much lower concentration of hydronium ions
(by seven orders of magnitude) at pH = 7 in place of pH = 0, has resulted in the
reverse reaction becoming spontaneous
...
3 For a particular reaction of the form A → B + 2 H+ in aqueous

solution, it was found that ∆rG 7 = +20 kJ mol−1 at 28°C
...

[−61 kJ mol−1]

210

7 CHEMICAL EQUILIBRIUM

The response of equilibria to the conditions
Equilibria respond to changes in pressure, temperature, and concentrations of reactants
and products
...
As we shall see in detail in Sections
22
...
*, catalysts increase the rate at which equilibrium is attained but do not
affect its position
...

7
...
The value of ∆rG 7, and hence of K, is therefore independent of the
pressure at which the equilibrium is actually established
...
21)

The conclusion that K is independent of pressure does not necessarily mean that
the equilibrium composition is independent of the pressure, and its effect depends on
how the pressure is applied
...
However, so long as the gases are perfect, this addition of
gas leaves all the partial pressures of the reacting gases unchanged: the partial pressures of a perfect gas is the pressure it would exert if it were alone in the container, so
the presence of another gas has no effect
...
Alternatively, the pressure of the system may be increased by
confining the gases to a smaller volume (that is, by compression)
...
Consider, for instance, the perfect gas equilibrium A 5 2 B, for
which the equilibrium constant is
K=

p2
B
pA p7

The right-hand side of this expression remains constant only if an increase in pA cancels an increase in the square of pB
...
Then the number of A molecules will increase as the volume of the container is
decreased and its partial pressure will rise more rapidly than can be ascribed to a
simple change in volume alone (Fig
...
6)
...
1 Le Chatelier’s principle states that:
(a)

(b)

When a reaction at equilibrium is
compressed (from a to b), the reaction
responds by reducing the number of
molecules in the gas phase (in this case by
producing the dimers represented by the
linked spheres)
...
7
...
This it can do by reducing the
number of particles in the gas phase, which implies a shift A ← 2 B
...


7
...
0

Extent of dissociation, a

To treat the effect of compression quantitatively, we suppose that there is an
amount n of A present initially (and no B)
...
It follows that the mole fractions present at equilibrium are

211

100

0
...
6

10

0
...
2

1
...
1

which rearranges to

0
1/2

0

4

8
p /p

12

16

Æ

A
D
1
α=
7F
C 1 + 4p/Kp

(7
...
The
value α = 0 corresponds to pure A; α = 1
corresponds to pure B
...
7
...
7
...
It also shows that as p is increased, α
decreases, in accord with Le Chatelier’s principle
...
5 Predicting the effect of compression

To predict the effect of an increase in pressure on the composition of the ammonia
synthesis at equilibrium, eqn 7
...
So, Le Chatelier’s principle predicts that an increase in pressure will favour the product
...
Therefore, doubling the pressure must increase Kx
by a factor of 4 to preserve the value of K
...
4 Predict the effect of a tenfold pressure increase on the equilibrium
composition of the reaction 3 N2(g) + H2(g) → 2 N3H(g)
...
4 The response of equilibria to temperature
Le Chatelier’s principle predicts that a system at equilibrium will tend to shift in the
endothermic direction if the temperature is raised, for then energy is absorbed as heat
and the rise in temperature is opposed
...
These conclusions can be summarized as follows:
Exothermic reactions: increased temperature favours the reactants
...

We shall now justify these remarks and see how to express the changes quantitatively
...


212

7 CHEMICAL EQUILIBRIUM
(a) The van ’t Hoff equation

The van ’t Hoff equation, which is derived in the Justification below, is an expression
for the slope of a plot of the equilibrium constant (specifically, ln K) as a function of
temperature
...
23)

Justification 7
...
17, we know that
ln K = −

∆rG 7
RT

Differentiation of ln K with respect to temperature then gives
d ln K
dT

=−

1 d(∆rG 7/T)
R

dT

The differentials are complete because K and ∆rG 7 depend only on temperature, not
on pressure
...
53) in the form
d(∆rG 7/T)
dT

=−

∆r H 7
T2

where ∆r H 7 is the standard reaction enthalpy at the temperature T
...
23a
...
23a can be rewritten as

−

d ln K
2

T d(1/T)

=

∆r H 7
RT 2

which simplifies into eqn 7
...


Equation 7
...
A negative slope means
that ln K, and therefore K itself, decreases as the temperature rises
...
The opposite occurs in the case of endothermic reactions
...

When the reaction is exothermic, −∆r H 7/T corresponds to a positive change of
entropy of the surroundings and favours the formation of products
...
As a result, the equilibrium lies less to the right
...
The importance of the unfavourable change of entropy of the surroundings is
reduced if the temperature is raised (because then ∆r H 7/T is smaller), and the reaction is able to shift towards products
...
4 THE RESPONSE OF EQUILIBRIA TO TEMPERATURE
A

B

A

Energy

Energy

B

Low temperature

Low temperature

High temperature

Population

(a)

213

High temperature

Population

(b)

Fig
...
8 The effect of temperature on a
chemical equilibrium can be interpreted in
terms of the change in the Boltzmann
distribution with temperature and the
effect of that change in the population of
the species
...
(b) In an
exothermic reaction, the opposite happens
...
3 The temperature dependence of the equilibrium constant

8

6

- ln K

The typical arrangement of energy levels for an endothermic reaction is shown in
Fig
...
8a
...
The change corresponds to an increased
population of the higher energy states at the expense of the population of the lower
energy states
...
Therefore, the total population of B
states increases, and B becomes more abundant in the equilibrium mixture
...
7
...


4

2

Example 7
...
Calculate the standard reaction
enthalpy of the decomposition
...
0

2
...
4

2
...
8

2
...
98 × 10−4

1
...
86 × 10−1

1
...
23b that, provided the reaction enthalpy can be

assumed to be independent of temperature, a plot of −ln K against 1/T should be a
straight line of slope ∆r H 7/R
...
86

2
...
22

2
...
83

4
...
68

−0
...
7
...
The slope of the graph is +9
...
6 × 103 K) × R = +80 kJ mol−1

When −ln K is plotted against 1/T, a
straight line is expected with slope equal to
∆r H 7/R if the standard reaction enthalpy
does not vary appreciably with
temperature
...

Fig
...
9

Exploration The equilibrium
constant of a reaction is found to
fit the expression ln K = a + b/(T/K ) +
c/(T/K )3 over a range of temperatures
...

(b) Plot ln K against 1/T between 400 K and
600 K for a = −2
...
0 × 103, and
c = 2
...


214

7 CHEMICAL EQUILIBRIUM
Self-test 7
...
0 × 1024 at 300 K, 2
...
0 × 104 at 700 K
...

[−200 kJ mol−1]

The temperature dependence of the equilibrium constant provides a noncalorimetric method of determining ∆ r H 7
...

However, the temperature dependence is weak in many cases, so the plot is reasonably
straight
...

(b) The value of K at different temperatures

To find the value of the equilibrium constant at a temperature T2 in terms of its value
K1 at another temperature T1, we integrate eqn 7
...
24)

1/T1

If we suppose that ∆ r H 7 varies only slightly with temperature over the temperature
range of interest, then we may take it outside the integral
...
25)

T1 F

Illustration 7
...
1 × 105 for the reaction as written in eqn 7
...
7 in the Data section by
using ∆r H 7 = 2∆f H 7(NH3, g), and assume that its value is constant over the range
of temperatures
...
2 kJ mol−1, from eqn 7
...
1 × 105) −

(−92
...
3145 J K mol

1

C 500 K

−

1 D
298 K F

= −1
...
18, a lower value than at 298 K, as expected for this exothermic reaction
...
6 The equilibrium constant for N2O4(g) 5 2 NO2(g) was calculated in

Self-test 7
...
Estimate its value at 100°C
...
For example,
synthetic chemists can improve the yield of a reaction by changing the temperature
of the reaction mixture
...


7
...
As we shall see, these equilibria can be discussed in
terms of the thermodynamic functions for the reactions

–300

1
3 Al2O3
1
2 TiO2

Reaction (iii)

1
2 SiO2

Reaction (iv)
ZnO

Reaction (ii)

–200

Æ

–1

MO(s) + C(s) 5 M(s) + CO(g)

DrG /(kJ mol )

I7
...
This is the case when the line for reaction
(i) lies below (is more positive than) the line for one of the reactions (ii) to (iv)
...
For example, CuO can be reduced to
Cu at any temperature above room temperature
...
On the
other hand, Al2O3 is not reduced by carbon until the temperature has been raised to
above 2000°C
...
Because in reaction (iii)
there is a net increase in the amount of gas, the standard reaction entropy is large and
positive; therefore, its ∆rG 7 decreases sharply with increasing temperature
...
In reaction (ii), the amount of gas is constant, so the
entropy change is small and ∆rG 7 changes only slightly with temperature
...
7
...
Note that
∆rG 7 decreases upwards!
At room temperature, ∆ rG 7 is dominated by the contribution of the reaction
enthalpy (T∆ rS 7 being relatively small), so the order of increasing ∆ rG 7 is the same as
the order of increasing ∆ r H 7 (Al2O3 is most exothermic; Ag2O is least)
...
As a result, the temperature dependence
of the standard Gibbs energy of oxidation should be similar for all metals, as is shown
by the similar slopes of the lines in the diagram
...

Successful reduction of the oxide depends on the outcome of the competition of the
carbon for the oxygen bound to the metal
...
7
...


216

7 CHEMICAL EQUILIBRIUM

Equilibrium electrochemistry
We shall now see how the foregoing ideas, with certain changes of technical detail, can
be used to describe the equilibrium properties of reactions taking place in electrochemical cells
...

An electrochemical cell consists of two electrodes, or metallic conductors, in contact with an electrolyte, an ionic conductor (which may be a solution, a liquid, or a
solid)
...
The two
electrodes may share the same compartment
...
1
...
If the electrolytes are different, the two compartments may
be joined by a salt bridge, which is a tube containing a concentrated electrolyte solution (almost always potassium chloride in agar jelly) that completes the electrical
circuit and enables the cell to function
...
An
electrolytic cell is an electrochemical cell in which a non-spontaneous reaction is
driven by an external source of current
...
5 Half-reactions and electrodes
It will be familiar from introductory chemistry courses that oxidation is the removal
of electrons from a species, a reduction is the addition of electrons to a species, and a
redox reaction is a reaction in which there is a transfer of electrons from one species
to another
...
The reducing agent (or ‘reductant’) is the electron donor;
the oxidizing agent (or ‘oxidant’) is the electron acceptor
...
Even reactions that are not redox reactions may often be expressed as the difference of two
reduction half-reactions
...
In general we write a couple as Ox/Red and the corresponding reduction half-reaction as
Ox + ν e− → Red

(7
...
1 Varieties of electrode
Electrode type

Designation

Redox couple

Half-reaction

Metal/metal ion

M(s) | M +(aq)

M+/M

M+(aq) + e− → M(s)

Gas

Pt(s) | X2(g) | X +(aq)

X+/X2

1
X+(aq) + e− → – X2(g)
2

Pt(s) | X2(g) | X −(aq)

X2/X −

Metal/insoluble salt
Redox

−

M(s) | MX(s) | X (aq)
+

Pt(s) | M (aq),M (aq)
2+

1
– X2(g) + e− → X−(aq)
2

MX/M,X
2+

+

M /M

−

MX(s) + e− → M(s) + X−(aq)
M2+(aq) + e− → M+(aq)

7
...
7 Expressing a reaction in terms of half-reactions

The dissolution of silver chloride in water AgCl(s) → Ag+(aq) + Cl−(aq), which is
not a redox reaction, can be expressed as the difference of the following two reduction half-reactions:
AgCl(s) + e− → Ag(s) + Cl−(aq)
Ag+(aq) + e− → Ag(s)
The redox couples are AgCl/Ag,Cl− and Ag+/Ag, respectively
...
7 Express the formation of H2O from H2 and O2 in acidic solution (a
redox reaction) as the difference of two reduction half-reactions
...
This quotient is
defined like the reaction quotient for the overall reaction, but the electrons are ignored
...
8 Writing the reaction quotient of a half-reaction

The reaction quotient for the reduction of O2 to H2O in acid solution, O2(g) +
4 H+(aq) + 4 e− → 2 H2O(l), is
Q=

2
aH2O
4
a H+aO2

≈

p7

Electrons

4
a H+pO2

The approximations used in the second step are that the activity of water is 1 (because the solution is dilute) and the oxygen behaves as a perfect gas, so aO2 ≈ pO2/p7
...
8 Write the half-reaction and the reaction quotient for a chlorine gas

electrode
...
As the reaction proceeds, the electrons released in the oxidation Red1 →
Ox1 + ν e− at one electrode travel through the external circuit and re-enter the cell
through the other electrode
...

The electrode at which oxidation occurs is called the anode; the electrode at which
reduction occurs is called the cathode
...
7
...
At the anode, oxidation results in the transfer of
electrons to the electrode, so giving it a relative negative charge (corresponding to a
low potential)
...
6 Varieties of cells
The simplest type of cell has a single electrolyte common to both electrodes (as in
Fig
...
11)
...
7
...

Note that the + sign of the cathode can be
interpreted as indicating the electrode at
which electrons enter the cell, and the −
sign of the anode is where the electrons
leave the cell
...
7
...
The
copper electrode is the cathode and the zinc
electrode is the anode
...


as in the ‘Daniell cell’ in which the redox couple at one electrode is Cu2+/Cu and at the
other is Zn2+/Zn (Fig
...
12)
...
In an electrode concentration cell the electrodes themselves have different concentrations,
either because they are gas electrodes operating at different pressures or because they
are amalgams (solutions in mercury) with different concentrations
...
This potential is called the liquid junction potential, E lj
...

At the junction, the mobile H+ ions diffuse into the more dilute solution
...
The potential then settles down to a value such that, after that brief
initial period, the ions diffuse at the same rates
...

The contribution of the liquid junction to the potential can be reduced (to about 1
to 2 mV) by joining the electrolyte compartments through a salt bridge (Fig
...
13)
...

(b) Notation

In the notation for cells, phase boundaries are denoted by a vertical bar
...
7
...
7
...
Thus the cell in Fig
...
13 is denoted

Salt bridge

Zn

Zn(s) | ZnSO4(aq)ӇCuSO4(aq) | Cu(s)

Cu

Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu(s)
An example of an electrolyte concentration cell in which the liquid junction potential
is assumed to be eliminated is
Pt(s) | H2(g) | HCl(aq, b1) || HCl(aq, b2) | H2(g) | Pt(s)
...
7 The electromotive force

ZnSO4(aq)

CuSO4(aq)

Electrode
compartments
Fig
...
13 The salt bridge, essentially an
inverted U-tube full of concentrated salt
solution in a jelly, has two opposing liquid
junction potentials that almost cancel
...
The cell reaction is the reaction in the cell written on the
assumption that the right-hand electrode is the cathode, and hence that the spontaneous reaction is one in which reduction is taking place in the right-hand compartment
...
If the left-hand electrode turns out to
be the cathode, then the reverse of the corresponding cell reaction is spontaneous
...

Then we subtract from it the left-hand reduction half-reaction (for, by implication, that

7
...
Thus, in the cell Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu(s)
the two electrodes and their reduction half-reactions are
Right-hand electrode: Cu2+(aq) + 2 e− → Cu(s)
Left-hand electrode: Zn2+(aq) + 2 e− → Zn(s)
Hence, the overall cell reaction is the difference:
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
(a) The Nernst equation

A cell in which the overall cell reaction has not reached chemical equilibrium can do
electrical work as the reaction drives electrons through an external circuit
...
This potential difference is called the cell potential and is
measured in volts, V (1 V = 1 J C −1 s)
...

When the cell potential is small, the same number of electrons can do only a small
amount of work
...

According to the discussion in Section 3
...
38 (we,max = ∆G), with ∆G identified (as we shall show) with
the Gibbs energy of the cell reaction, ∆rG
...
Moreover, we
saw in Section 7
...
Therefore, to make use of ∆rG we must
ensure that the cell is operating reversibly at a specific, constant composition
...
The resulting potential difference is called the electromotive force (emf), E, of the cell
...
27)

where F is Faraday’s constant, F = eNA, and ν is the stoichiometric coefficient of
the electrons in the half-reactions into which the cell reaction can be divided
...
It will be the basis of all that follows
...
3 The relation between the electromotive force and the reaction
Gibbs energy

We consider the change in G when the cell reaction advances by an infinitesimal
amount dξ at some composition
...
15 we can write (at constant temperature and pressure)
dG = ∆rGdξ
The maximum non-expansion (electrical) work that the reaction can do as it
advances by dξ at constant temperature and pressure is therefore
dwe = ∆rGdξ

219

Gibbs energy, G

220

7 CHEMICAL EQUILIBRIUM

DrG < 0,
E>0

DrG > 0,
E<0

This work is infinitesimal, and the composition of the system is virtually constant
when it occurs
...
The total charge transported between the electrodes when
this change occurs is −νeNAdξ (because νdξ is the amount of electrons and the
charge per mole of electrons is −eNA)
...
The work done when an infinitesimal charge −νFdξ travels from
the anode to the cathode is equal to the product of the charge and the potential
difference E (see Table 2
...
7
...

When expressed in terms of a cell potential,
the spontaneous direction of change can be
expressed in terms of the cell emf, E
...

The reverse reaction is spontaneous when
E < 0
...


When we equate this relation to the one above (dwe = ∆rGdξ ), the advancement dξ
cancels, and we obtain eqn 7
...


It follows from eqn 7
...
Note that a negative reaction Gibbs energy, corresponding to a spontaneous cell reaction, corresponds to a
positive cell emf
...
27 is that it shows that
the driving power of a cell (that is, its emf) is proportional to the slope of the Gibbs
energy with respect to the extent of reaction
...
7
...
When the slope is close to zero (when the cell
reaction is close to equilibrium), the emf is small
...
9 Converting between the cell emf and the reaction Gibbs energy

Equation 7
...
Conversely, if we know the value of ∆ rG at a particular composition, then we can predict the emf
...
6485 × 104 C mol−1)

=1V

where we have used 1 J = 1 C V
...
We know that the reaction Gibbs energy is related to the composition of the
reaction mixture by eqn 7
...
28]

and called the standard emf of the cell
...
It follows that

7
...
29)

6

This equation for the emf in terms of the composition is called the Nernst equation;
the dependence of cell potential on composition that it predicts is summarized in
Fig
...
15
...
9c)
...
29 that the standard emf (which will shortly move to centre stage
of the exposition) can be interpreted as the emf when all the reactants and products in
the cell reaction are in their standard states, for then all activities are 1, so Q = 1 and
ln Q = 0
...
28) should always be kept in mind and underlies all its applications
...
10 Using the Nernst equation

Because RT/F = 25
...
7 mV

ν

ln Q

It then follows that, for a reaction in which ν = 1, if Q is increased by a factor of 10,
then the emf decreases by 59
...


(b) Cells at equilibrium

A special case of the Nernst equation has great importance in electrochemistry and
provides a link to the earlier part of the chapter
...

However, a chemical reaction at equilibrium cannot do work, and hence it generates
zero potential difference between the electrodes of a galvanic cell
...
30)

This very important equation (which could also have been obtained more directly
by substituting eqn 7
...
17) lets us predict equilibrium constants from
measured standard cell potentials
...

Illustration 7
...
10 V, the equilibrium constant
for the cell reaction Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq), for which ν = 2, is
K = 1
...
We conclude that the displacement of copper by zinc goes
virtually to completion
...


-2 -1

1
0
log Q

2

3

Fig
...
15 The variation of cell emf with the
value of the reaction quotient for the cell
reaction for different values of ν (the
number of electrons transferred)
...
69 mV, so the vertical scale refers
to multiples of this value
...
Does the cell
emf become more or less sensitive to
composition as the temperature increases?

222

7 CHEMICAL EQUILIBRIUM
7
...
Although it is not
possible to measure the contribution of a single electrode, we can define the potential
of one of the electrodes as zero and then assign values to others on that basis
...
31]

at all temperatures
...
The standard potential, E 7, of another couple is then
assigned by constructing a cell in which it is the right-hand electrode and the standard
hydrogen electrode is the left-hand electrode
...
The measurement is made on the ‘Harned
cell’:
Pt(s) | H2(g) | HCl(aq) | AgCl(s) | Ag(s)

1
– H2(g) + AgCl(s) → HCl(aq) + Ag(s)
2

for which the Nernst equation is
E = E 7(AgCl/Ag, Cl −) −

RT
F

ln

aH+a Cl −
a1/2
H2

We shall set aH2 = 1 from now on, and for simplicity write the standard potential as E 7;
then
E = E7 −

RT
F

ln aH+a Cl −

The activities can be expressed in terms of the molality b of HCl(aq) through aH+ =
γ±b/b 7 and aCl− = γ±b/b 7 as we saw in Section 5
...
This expression rearranges to
E+

2RT
F

ln b = E 7 −

2RT
F

ln γ ±

{7
...
9; a 1,1electrolyte is a solution of singly charged M+ and X− ions), we know that ln γ ± ∝ −b1/2
...
69 (because ln x = ln 10 log x = 2
...
Therefore, with the constant of proportionality in this relation written as (F/2RT)C, eqn 7
...
33}

The expression on the left is evaluated at a range of molalities, plotted against b1/2,
and extrapolated to b = 0
...
In precise work, the b1/2 term is brought to the left, and a
higher-order correction term from the extended Debye–Hückel law is used on the
right
...
8 STANDARD POTENTIALS
0
...
12 Determining the standard emf of a cell

The emf of the cell Pt(s) | H2(g, p7) | HCl(aq, b) | AgCl(s) | Ag(s) at 25°C has the following values:
3
...
52053

5
...
49257

9
...
46860

25
...
41824

To determine the standard emf of the cell we draw up the following table, using
2RT/F = 0
...
051 39 ln b

3
...
793
0
...
2256

5
...
370
0
...
2263

9
...
023
0
...
2273

25
...
063
0
...
2299

0
...
2280

E /V + 0
...
2270
0
...
2250
0
...
7
...
2232 V
...
Determine the standard emf of the cell
...
042
0
...
444
0
...
19
0
...
071 V]

Table 7
...
An important feature of standard emf
of cells and standard potentials of electrodes is that they are unchanged if the chemical equation for the cell reaction or a half-reaction is multiplied by a numerical factor
...

However, it also increases the number of electrons transferred by the same factor, and
by eqn 7
...
A practical consequence is that a cell
emf is independent of the physical size of the cell
...

The standard potentials in Table 7
...
However, to do so, we must take into account the fact that
different couples may correspond to the transfer of different numbers of electrons
...


3
...
0 5
...
9 The data below are for the cell Pt(s) | H2(g, p ) | HBr(aq, b) |

0
...
0 2
...
7
...
The intercept at b1/2 = 0
is E 7
...
12 results
in a plot that deviates from linearity
...
4 Evaluating a standard potential from two others

Given that the standard potentials of the Cu2+/Cu and Cu+/Cu couples are
+0
...
522 V, respectively, evaluate E 7(Cu2+,Cu+)
...
Therefore, we should convert the E 7 values to
∆G 7 values by using eqn 7
...
27 again
...

−

(a) Cu (aq) + 2 e → Cu(s)
(b) Cu+(aq) + e− → Cu(s)

7

E = +0
...
522 V,

Couple

E 7/V

Ce4+(aq) + e− → Ce3+(aq)

+1
...
34

1
H+(aq) + e− → – H2(g)
2

0

7

so ∆rG = −2(0
...
522 V)F

AgCl(s) + e− → Ag(s) + Cl −(aq)

+0
...
2* Standard
potentials at 298 K

−0
...
71

* More values are given in the Data section
...
158 V) × F
Therefore, E 7 = +0
...
Note that the generalization of the calculation we just
performed is

νc E 7(c) = νaE 7(a) + νbE 7(b)

(7
...

Self-test 7
...


[−0
...
9 Applications of standard potentials
Cell emfs are a convenient source of data on equilibrium constants and the Gibbs
energies, enthalpies, and entropies of reactions
...

(a) The electrochemical series
Table 7
...
35a)

that the cell reaction
Red1 + Ox2 → Ox1 + Red2

(7
...
Because in the cell reaction Red1 reduces Ox2, we can conclude that
7
7
Red1 has a thermodynamic tendency to reduce Ox2 if E 1 < E 2

More briefly: low reduces high
...
13 Using the electrochemical series

Zinc

Because E 7(Zn2+,Zn) = −0
...
34 V, zinc has a thermodynamic tendency to reduce Cu2+ ions in aqueous solution
...
2
...
3 shows a part of the electrochemical series, the metallic elements (and
hydrogen) arranged in the order of their reducing power as measured by their standard potentials in aqueous solution
...
This conclusion is qualitative
...
For example, to determine whether zinc can displace
magnesium from aqueous solutions at 298 K, we note that zinc lies above magnesium

7
...
Zinc can reduce hydrogen ions, because hydrogen lies higher in the series
...

The reactions of the electron transport chains of respiration are good applications
of this principle
...
2 Energy conversion in biological cells

The whole of life’s activities depends on the coupling of exergonic and endergonic reactions, for the oxidation of food drives other reactions forward
...
The essence of the action of ATP is its ability to lose its terminal phosphate
group by hydrolysis and to form adenosine diphosphate (ADP):
ATP(aq) + H2O(l) → ADP(aq) + P i−(aq) + H3O+(aq)
−
where Pi− denotes an inorganic phosphate group, such as H2PO 4
...
The hydrolysis is therefore
exergonic (∆rG ⊕ < 0) under these conditions and 31 kJ mol−1 is available for driving
other reactions
...
In view of its exergonicity the ADP-phosphate
bond has been called a ‘high-energy phosphate bond’
...

In fact, even in the biological sense it is not of very ‘high energy’
...
Thus ATP acts as a phosphate donor to
a number of acceptors (for example, glucose), but is recharged by more powerful
phosphate donors in a number of biochemical processes
...
The process
begins with glycolysis, a partial oxidation of glucose by nicotinamide adenine dinu−
cleotide (NAD+, 2) to pyruvate ion, CH3COCO2 , continues with the citric acid cycle,
which oxidizes pyruvate to CO2, and ends with oxidative phosphorylation, which reduces
O2 to H2O
...
The citric acid cycle and
oxidative phosphorylation are the main mechanisms for the extraction of energy from
carbohydrates during aerobic metabolism, a form of metabolism in which inhaled O2
does play a role
...
At blood temperature, ∆rG ⊕ =
−147 kJ mol−1 for the oxidation of glucose by NAD+ to pyruvate ions
...
The
reaction is exergonic, and therefore spontaneous: the oxidation of glucose is used to
‘recharge’ the ATP
...
2 Very strenuous exercise, such as bicycle racing, can
decrease sharply the concentration of O2 in muscle cells and the condition known as
muscle fatigue results from increased concentrations of lactate ion
...
In the presence of O2, pyruvate
is oxidized further during the citric acid cycle and oxidative phosphorylation, which
occur in a special compartment of the cell called the mitochondrion
...
The citric acid cycle and oxidative phosphorylation generate
as many as 38 ATP molecules for each glucose molecule consumed
...


7
...
Therefore, aerobic oxidation of glucose is
much more efficient than glycolysis
...
For example, the biosynthesis of sucrose
from glucose and fructose can be driven by plant enzymes because the reaction is
endergonic to the extent ∆rG ⊕ = +23 kJ mol−1
...
For instance, the formation of a peptide link is
endergonic, with ∆rG ⊕ = +17 kJ mol−1, but the biosynthesis occurs indirectly and is
equivalent to the consumption of three ATP molecules for each link
...

The respiratory chain

In the exergonic oxidation of glucose 24 electrons are transferred from each C6H12O6
molecule to six O2 molecules
...
We have already seen that, in
biological cells, glucose is oxidized to CO2 by NAD+ and FAD during glycolysis and
the citric acid cycle:
C6H12O6(s) + 10 NAD+ + 2 FAD + 4 ADP + 4 P − + 2 H2O
i
→ 6 CO2 + 10 NADH + 2 FADH2 + 4 ATP + 6 H+

227

228

7 CHEMICAL EQUILIBRIUM

In the respiratory chain, electrons from the powerful reducing agents NADH and
FADH2 pass through four membrane-bound protein complexes and two mobile electron carriers before reducing O2 to H2O
...

The respiratory chain begins in complex I (NADH-Q oxidoreductase), where
NADH is oxidized by coenzyme Q (Q, 4) in a two-electron reaction:
–– ––→ NAD+ + QH2
H+ + NADH + Q – ––
complex I

E ⊕ = +0
...
015 V,

∆rG ⊕ = −2
...
Cytochrome c contains the
haem c group (5), the central iron ion of which can exist in oxidation states +3 and +2
...
15 V,
∆rG ⊕ = −30 kJ mol−1
complex III

Reduced cytochrome c carries electrons from complex III to complex IV (cytochrome
c oxidase), where O2 is reduced to H2O:
1
complex IV
2 Fe2+(Cyt c) + 2 H+ + – O2
→ 2 Fe3+(Cyt c) + H2O
2
⊕
⊕
E = +0
...
7
...


Oxidative phosphorylation

The reactions that occur in complexes I, III, and IV are sufficiently exergonic to drive
the synthesis of ATP in the process called oxidative phosphorylation:
ADP + P − + H + → ATP
i

∆rG ⊕ = +31 kJ mol−1

We saw above that the phosphorylation of ADP to ATP can be coupled to the exergonic dephosphorylation of other molecules
...
However, oxidative
phosphorylation operates by a different mechanism
...
7
...
The protein complexes
associated with the electron transport chain span the inner membrane and phosphorylation takes place in the intermembrane space
...
9 APPLICATIONS OF STANDARD POTENTIALS
in complexes I, III, and IV is first used to do the work of moving protons across the
mitochondrial membrane
...
For example, the oxidation of NADH by Q in complex
I is coupled to the transfer of four protons across the membrane
...
Then the enzyme H+-ATPase
uses the energy stored in the proton gradient to phosphorylate ADP to ATP
...
The ATP is
then hydrolysed on demand to perform useful biochemical work throughout the cell
...
The energy stored in a transmembrane proton gradient come
from two contributions
...
The charge
difference across a membrane per mole of H+ ions is NAe, or F, where F = eNA
...
3, that the molar Gibbs energy difference is then ∆Gm,2 = F∆φ
...
This equation also provides an estimate of the Gibbs energy available for phosphorylation of ADP
...
4 and ∆φ ≈ 0
...
5 kJ mol−1
...

(b) The determination of activity coefficients

Once the standard potential of an electrode in a cell is known, we can use it to
determine mean activity coefficients by measuring the cell emf with the ions at the
concentration of interest
...
32 in the form
ln γ ± =

E7 − E
2RT/F

− ln b

{7
...

(c) The determination of equilibrium constants

The principal use for standard potentials is to calculate the standard emf of a cell
formed from any two electrodes
...
37)

Because ∆G 7 = −νFE 7, it then follows that, if the result gives E 7 > 0, then the corresponding cell reaction has K > 1
...
14 Calculating an equilibrium constant from standard potentials

Silver/
silver chloride
electrode

A disproportionation is a reaction in which a species is both oxidized and reduced
...
52 V

Left-hand electrode:
Pt(s) | Cu2+(aq),Cu+(aq)

Phosphate
buffer
solution

Cu2+(aq) + e− → Cu+(s)

E 7 = +0
...
The standard emf of the cell
is therefore

The glass electrode
...

Fig
...
18

E 7 = +0
...
16 V = +0
...
Because ν = 1,
from eqn 7
...
36 V
0
...
36
0
...
2 × 106
...
11 Calculate the solubility constant (the equilibrium constant for the

reaction Hg2Cl2(s) 5 Hg2+(aq) + 2 Cl−(aq)) and the solubility of mercury(I) chlor2
ide at 298
...
Hint
...

2
[2
...
7 × 10−7 mol kg−1]

Hydrated
silica

(d) Species-selective electrodes

Outside

Inside

50 mm

Glass permeable
+
+
to Li and Na ions
Fig
...
19 A section through the wall of a
glass electrode
...
An example is the glass electrode (Fig
...
18),
which is sensitive to hydrogen ion activity, and has a potential proportional to pH
...
It is necessary to calibrate the glass electrode before
use with solutions of known pH
...
The membrane itself is permeable to Na+ and Li+ ions but not to H+
ions
...
2)
...
7
...
The hydrogen ions in the test solution modify this layer to an extent that
depends on their activity in the solution, and the charge modification of the outside
layer is transmitted to the inner layer by the Na+ and Li+ ions in the glass
...


7
...
The glass can also be made re+
sponsive to Na+, K+, and NH 4 ions by being doped with Al2O3 and B2O3
...
A simple form of a gas-sensing electrode consists of a glass electrode contained
in an outer sleeve filled with an aqueous solution and separated from the test solution
by a membrane that is permeable to gas
...
The presence of an enzyme that converts a compound,
such as urea or an amino acid, into ammonia, which then affects the pH, can be used
to detect these organic compounds
...
In one
arrangement, a porous lipophilic (hydrocarbon-attracting) membrane is attached to
a small reservoir of a hydrophobic (water-repelling) liquid, such as dioctylphenylphosphonate, that saturates it (Fig
...
20)
...
The complex’s ions are able to migrate through the
lipophilic membrane, and hence give rise to a transmembrane potential, which is
detected by a silver/silver chloride electrode in the interior of the assembly
...

In theory, the transmembrane potential should be determined entirely by differences in the activity of the species that the electrode was designed to detect
...
The
asymmetry potential is due to the fact that it is not possible to manufacture a membrane material that has the same structure and the same chemical properties throughout
...
For example, a Na+ selective electrode also responds, albeit less effectively, to
the activity of K+ ions in the test solution
...
38)

where Eap is the asymmetry potential, β is an experimental parameter that captures
deviations from the Nernst equation, and kX,Y is the selectivity coefficient of the electrode and is related to the response of the electrode to the interfering species Y+
...
The selectivity coefficient, and hence
interference effects, can be minimized when designing and manufacturing a speciesselective electrode
...

(e) The determination of thermodynamic functions

The standard emf of a cell is related to the standard reaction Gibbs energy through
eqn 7
...
Therefore, by measuring E 7 we can obtain this important
thermodynamic quantity
...
6
...
7
...
Chelated ions are able to migrate
through the lipophilic membrane
...
15 Determining the Gibbs energy of formation of an ion electrochemically

The cell reaction taking place in
Pt(s) | H2 | H+(aq) || Ag+(aq) | Ag(s)

E 7 = +0
...
15 kJ mol−1
which is in close agreement with the value in Table 2
...


The temperature coefficient of the standard cell emf, dE 7/dT, gives the standard entropy of the cell reaction
...
27, which combine to give
dE 7
dT

=

∆rS 7

(7
...
Hence
we have an electrochemical technique for obtaining standard reaction entropies and
through them the entropies of ions in solution
...
40)

This expression provides a non-calorimetric method for measuring ∆ r H 7 and,
through the convention ∆ f H 7(H+, aq) = 0, the standard enthalpies of formation of
ions in solution (Section 2
...
Thus, electrical measurements can be used to calculate
all the thermodynamic properties with which this chapter began
...
5 Using the temperature coefficient of the cell potential

The standard emf of the cell Pt(s) | H2(g) | HBr(aq) | AgBr(s) | Ag(s) was measured
over a range of temperatures, and the data were fitted to the following polynomial:
E 7/V = 0
...
99 × 10−4(T/K − 298) − 3
...

Method The standard Gibbs energy of reaction is obtained by using eqn 7
...
The standard entropy of reaction
is obtained by using eqn 7
...
The reaction enthalpy is obtained by combining the values of the standard Gibbs energy and entropy
...
07131 V, so

∆rG 7 = −νFE 7 = −(1) × (9
...
07131 V)
= −6
...
880 kJ mol−1

CHECKLIST OF KEY IDEAS

233

The temperature coefficient of the cell potential is
dE 7
dT

= −4
...
45 × 10−6)(T/K − 298) V K−1

At T = 298 K this expression evaluates to
dE
dT

= −4
...
39, the reaction entropy is
∆rS 7 = 1 × (9
...
99 × 10−4 V K−1)
= −48
...
880 kJ mol−1 + (298 K) × (−0
...
2 kJ mol−1
One difficulty with this procedure lies in the accurate measurement of small temperature coefficients of cell potential
...

Self-test 7
...

[+0
...
The extent of reaction (ξ) is defined such that, when the
extent of reaction changes by a finite amount ∆ξ, the amount
of A present changes from nA,0 to nA,0 − ∆ξ
...
The reaction Gibbs energy is the slope of the graph of the
Gibbs energy plotted against the extent of reaction: ∆rG =
(∂G/∂ξ)p,T ; at equilibrium, ∆rG = 0
...
An exergonic reaction is a reaction for which ∆rG < 0; such a
reaction can be used to drive another process
...

4
...

5
...

6
...

7
...

J
C J
F equilibrium

Π

8
...

9
...
However, partial pressures and concentrations
can change in response to a change in pressure
...
Le Chatelier’s principle states that a system at equilibrium,
when subjected to a disturbance, responds in a way that tends
to minimize the effect of the disturbance
...
Increased temperature favours the reactants in exothermic
reactions and the products in endothermic reactions
...
The temperature dependence of the equilibrium constant is
given by the van ‘t Hoff equation: d ln K/dT = ∆ r H 7/RT 2
...

13
...
An electrolytic cell is an electrochemical cell in which
a non-spontaneous reaction is driven by an external source of
current
...
Oxidation is the removal of electrons from a species;
reduction is the addition of electrons to a species; a redox

234

7 CHEMICAL EQUILIBRIUM
reaction is a reaction in which there is a transfer of electrons
from one species to another
...
The anode is the electrode at which oxidation occurs
...


20
...

21
...


16
...


22
...


17
...


23
...


18
...


24
...


19
...


Further reading
Articles and texts

P
...
Atkins and J
...
de Paula, Physical chemistry for the life sciences
...
H
...

A
...
Bard and L
...
Faulkner, Electrochemical methods
...

M
...
Blandamer, Chemical equilibria in solution
...

W
...
Cramer and D
...
Knaff, Energy transduction in biological
membranes, a textbook of bioenergetics
...

D
...
Crow, Principles and applications of electrochemistry
...


C
...
Hamann, A
...
Vielstich, Electrochemistry
...

Sources of data and information

M
...
Antelman, The encyclopedia of chemical electrode potentials,
Plenum, New York (1982)
...
J
...
Parsons, and J
...
), Standard potentials in
aqueous solution
...

R
...
Goldberg and Y
...
Tewari, Thermodynamics of enzymecatalyzed reactions
...
Phys
...
Ref
...
Part 1: 22, 515 (1993)
...
Part 3: 23, 1035 (1994)
...
Part 5: 24, 1765 (1995)
...
Denbigh, The principles of chemical equilibrium, with applications
in chemistry and chemical engineering
...


Discussion questions
7
...


diagram in Fig
...
10 to identify the lowest temperature at which zinc oxide can
be reduced to zinc metal by carbon
...
2 Suggest how the thermodynamic equilibrium constant may respond

7
...


why the latter is related to thermodynamic quantities
...
3 Account for Le Chatelier’s principle in terms of thermodynamic

quantities
...
4 Explain the molecular basis of the van ’t Hoff equation for the

temperature dependence of K
...
5 (a) How may an Ellingham diagram be used to decide whether one metal

may be used to reduce the oxide of another metal? (b) Use the Ellingham

7
...
1
...
8 Describe a method for the determination of a standard potential of a

redox couple
...
9 Devise a method for the determination of the pH of an aqueous solution
...
8(a) Calculate the percentage change in Kx for the reaction H2CO(g) 5

7
...
00 atm total pressure, water is 1
...
Calculate
(a) K, (b) ∆rG 7, and (c) ∆rG at this temperature
...
0 bar to 2
...


7
...
8(b) Calculate the percentage change in Kx for the reaction CH3OH(g) +

7
...
46 per cent dissociated at 25°C and 1
...
Calculate (a) K at 25°C, (b) ∆rG 7,
(c) K at 100°C given that ∆rH 7 = +57
...


7
...
106
...
50 g of
borneol and 14
...
0 dm3 is heated to
503 K and allowed to come to equilibrium
...


αe, at 298 K is 0
...
00 bar total pressure
...


7
...
00 bar in

the equilibrium Br2(g) 5 2 Br(g)
...


7
...
Assume that the reaction
enthalpy is independent of temperature
...
3(b) From information in the Data section, calculate the standard Gibbs

energy and the equilibrium constant at (a) 25°C and (b) 50°C for the reaction
CH4(g) + 3 Cl2(g) 5 CHCl3(l) + 3 HCl(g)
...

7
...
00 mol A, 2
...
00 mol D were mixed and allowed to come to
equilibrium at 25°C, the resulting mixture contained 0
...
00 bar
...

7
...
00 mol A, 1
...
00 mol D were mixed and allowed to come to
equilibrium at 25°C, the resulting mixture contained 0
...
00 bar
...

7
...
The
standard reaction Gibbs energy is +33 kJ mol−1 at 1280 K
...


7
...
The standard reaction Gibbs
energy is +22 kJ mol−1 at 1120 K
...


NOCl(g) 5 HCl(g) + CH3NO2(g) when the total pressure is increased from
1
...
0 bar at constant temperature
...
9(b) The equilibrium constant for the reaction N2(g) + O2(g) 5 2 NO(g) is
1
...
A mixture consisting of 5
...
0 g of
oxygen in a container of volume 1
...
Calculate the mole fraction of NO at equilibrium
...
10(a) What is the standard enthalpy of a reaction for which the equilibrium

constant is (a) doubled, (b) halved when the temperature is increased by 10 K
at 298 K?
7
...
11(a) The standard Gibbs energy of formation of NH3(g) is −16
...
What is the reaction Gibbs energy when the partial pressures of the
N2, H2, and NH3 (treated as perfect gases) are 3
...
0 bar, and 4
...
11(b) The dissociation vapour pressure of NH4Cl at 427°C is 608 kPa but
at 459°C it has risen to 1115 kPa
...
Assume that the vapour behaves
as a perfect gas and that ∆H 7 and ∆S 7 are independent of temperature in the
range given
...
12(a) Estimate the temperature at which CaCO3(calcite) decomposes
...
12(b) Estimate the temperature at which CuSO4·5H2O undergoes
dehydration
...
13(a) For CaF2(s) 5 Ca2+(aq) + 2 F−(aq), K = 3
...
Calculate
the standard Gibbs energy of formation of CaF2(aq)
...
6(a) The equilibrium constant of the reaction 2 C3H6(g) 5 C2H4(g) +
C4H8(g) is found to fit the expression ln K = A + B/T + C/T 2 between 300 K
and 600 K, with A = −1
...
51 × 105 K2
...


7
...
4 × 10−8 at 25°C and the
standard Gibbs energy of formation of PbI2(s) is −173
...
Calculate
the standard Gibbs energy of formation of PbI2(aq)
...
6(b) The equilibrium constant of a reaction is found to fit the expression

7
...
04, B = −1176 K,
and C = 2
...
Calculate the standard reaction enthalpy and standard
reaction entropy at 450 K
...
7(a) The standard reaction Gibbs energy of the isomerization of borneol

(C10H17OH) to isoborneol in the gas phase at 503 K is +9
...
Calculate
the reaction Gibbs energy in a mixture consisting of 0
...
30 mol of isoborneol when the total pressure is 600 Torr
...
7(b) The equilibrium pressure of H2 over solid uranium and uranium

hydride, UH3, at 500 K is 139 Pa
...


(a) Zn | ZnSO4(aq) || AgNO3(aq)|Ag
(b) Cd | CdCl2(aq) || HNO3(aq)|H2(g) | Pt
(c) Pt | K3[Fe(CN)6](aq),K4[Fe(CN)6](aq) || CrCl3(aq) | Cr
7
...
15(a) Devise cells in which the following are the reactions and calculate the

standard emf in each case:
(a) Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

equation for the cell reaction
...

(c) Assuming that the Debye–Hückel limiting law holds at this concentration,
calculate E 7(AgCl, Ag)
...
17(a) Calculate the equilibrium constants of the following reactions at 25°C

(c) 2 H2(g) + O2(g) → 2 H2O(l)

from standard potential data:

7
...
16(a) Use the Debye–Hückel limiting law and the Nernst equation to

(a) Sn(s) + Sn4+(aq) 5 2 Sn2+(aq)
(b) Sn(s) + 2 AgCl(s) 5 SnCl2(aq) + 2 Ag(s)
7
...
050 mol kg−1)
|| Cd(NO3)2(aq, 0
...


7
...
9509 V at 25°C
...


7
...
18(b) The emf of the cell Bi|Bi2S3(s)|Bi2S3(aq)|Bi is −0
...

Calculate (a) the solubility product of Bi2S3 and (b) its solubility
...
At 25°C and a
molality of HCl of 0
...
4658 V
...
1 The equilibrium constant for the reaction, I2(s) + Br2(g) 5 2 IBr(g)

is 0
...
(a) Calculate ∆rG 7 for this reaction
...
The pressure and
temperature are held at 0
...
Find the partial
pressure of IBr(g) at equilibrium
...
(c) In fact, solid
iodine has a measurable vapour pressure at 25°C
...
2 Consider the dissociation of methane, CH4(g), into the elements H2(g)

and C(s, graphite)
...
85 kJ mol−1 and that
∆ f S 7(CH4, g) = −80
...
(b) Assuming that ∆ f H 7 is independent of
temperature, calculate K at 50°C
...
010 bar
...

7
...
32, B =
−1
...
65
...

p
7
...
44

2
...
71

Assuming ∆ r H 7 to be constant over this temperature range, calculate K, ∆rG 7,
∆r H 7, and ∆rS 7
...

7
...
The equilibrium pressure of NH3 in the presence of CaCl2·NH3 is

1
...
Find an expression for the temperature dependence of ∆rG 7
in the same range
...
6 Calculate the equilibrium constant of the reaction CO(g) + H2(g) 5

H2CO(g) given that, for the production of liquid formaldehyde, ∆rG 7 =
+28
...

7
...
45 cm3 at 437 K and
at an external pressure of 101
...
The
mass of acid present in the sealed container was 0
...
The experiment was
repeated with the same container but at 471 K, and it was found that 0
...
Calculate the equilibrium constant for the
dimerization of the acid in the vapour and the enthalpy of vaporization
...
8 A sealed container was filled with 0
...
400 mol I2(g), and

0
...
00 bar
...

7
...
244

7
...
181

104nI

2
...
4555

2
...
68 cm3
...

7
...
carried out a study of Cl2O(g) by photoelectron ionization

(R
...
Thorn, L
...
Stief, S
...
Kuo, and R
...
Klemm, J
...
Chem
...
From their measurements, they report ∆ f H 7(Cl2O) = +77
...

They combined this measurement with literature data on the reaction
Cl2O (g) + H2O(g)→ 2 HOCl(g), for which K = 8
...


237

PROBLEMS
+16
...
Calculate that value
...


Determine the standard emf of the cell and the mean activity coefficient of
HCl at these molalities
...
)

7
...

For example, the lower value was cited in the review article by R
...

Chem
...
14, 246 (1981)); Walsh later leant towards the upper end of the
range (H
...
Frey, R
...
M
...
Chem
...
, Chem
...

1189 (1986))
...
-K
...
L
...
Phys
...
90, 1507 (1986)
...
18 Careful measurements of the emf of the cell Pt | H2(g, p7) | NaOH(aq,

7
...

Hydrogen and carbon monoxide have been investigated for use in fuel cells,
so their solubilities in molten salts are of interest
...
Desimoni and
P
...
Zambonin, J
...
Soc
...
1, 2014 (1973)) with the
following results:

log sH2 = −5
...
98 −
−3

980

0
...
01125 mol kg−1) | AgCl(s) | Ag have been
reported (C
...
Bezboruah, M
...
G
...
C
...
K
...
V
...
Chem
...
Faraday Trans
...
Among the data is the
following information:

θ/°C

20
...
0

30
...
04774

1
...
04942

Calculate pKw at these temperatures and the standard enthalpy and entropy of
the autoprotolysis of water at 25
...

7
...
Sen, J
...
Soc
...
I 69, 2006 (1973)) and some
values for LiCl are given below
...
Base
your answer on the following version of the extended Debye–Hückel law:

T/K

−1

where s is the solubility in mol cm bar
...

7
...
7 kJ mol−1 for the reaction in the Daniell cell at

25°C, and b(CuSO4) = 1
...
0 × 10−3 mol kg−1,
calculate (a) the ionic strengths of the solutions, (b) the mean ionic activity
coefficients in the compartments, (c) the reaction quotient, (d) the standard
cell potential, and (e) the cell potential
...
)
7
...
What is the emf of a cell
fuelled by (a) hydrogen and oxygen, (b) the combustion of butane at 1
...
15 Although the hydrogen electrode may be conceptually the simplest
electrode and is the basis for our reference state of electrical potential in
electrochemical systems, it is cumbersome to use
...
One of these alternatives is the
quinhydrone electrode (quinhydrone, Q · QH2, is a complex of quinone,
C6H4O2 = Q, and hydroquinone, C6H4O2H2 = QH2)
...
6994 V
...
190 V, what is the pH of the HCl solution? Assume that the
Debye–Hückel limiting law is applicable
...
16 Consider the cell, Zn(s)| ZnCl2 (0
...

Given that E 7 (Zn2+,Zn) = −0
...
2676 V, and that
the emf is +1
...
Determine
(b) the standard emf, (c) ∆rG, ∆rG 7, and K for the cell reaction, (d) the mean
ionic activity and activity coefficient of ZnCl2 from the measured cell
potential, and (e) the mean ionic activity coefficient of ZnCl2 from the
Debye–Hückel limiting law
...
52 × 10−4 V K−1
...

7
...
J
...
J
...
Ives, J
...
Soc
...
6077

3
...
0403

7
...
9474

E/V

0
...
56825

0
...
52267

0
...
461, B = 1
...
20, and I = b/b 7
...
09141 mol kg−1:
b1/(mol kg−1)

0
...
09141*

0
...
2171

1
...
350

E/V

−0
...
0000

0
...
0379

0
...
1336

7
...
G
...
E
...
Res
...
Bur
...
53, 283 (1954)) and the results were found to fit the expression
E 7/V = 0
...
8564 × 10−4(θ/°C) − 3
...
869 × 10−9(θ/°C)3
Calculate the standard Gibbs energy and enthalpy of formation of Cl−(aq) and
its entropy at 298 K
...
21‡ (a) Derive a general relation for (∂E/∂p)T,n for electrochemical cells

employing reactants in any state of matter
...
Cohen and K
...
Physik
...
167A, 365 (1933)) calculated the change in volume for the
reaction TlCl(s) + CNS−(aq) → TlCNS(s) + Cl−(aq) at 30°C from density data
and obtained ∆rV = −2
...
080 cm3 mol−1
...
Their results are given in the following table:
p/atm

1
...
56

9
...
98

10
...
39

12
...
82

From this information, obtain (∂E/∂p)T,n at 30°C and compare to the value
obtained from ∆ rV
...
How
constant is (∂E/∂p)T,n? (d) From the polynomial, estimate an effective
isothermal compressibility for the cell as a whole
...
22‡ The table below summarizes the emf observed for the cell
Pd | H2(g, 1 bar) | BH(aq, b), B(aq, b) | AgCl(s) | Ag
...
The data are for 25°C and it is found that E 7 = 0
...

Use the data to determine pKa for the acid at 25°C and the mean activity
coefficient (γ±) of BH as a function of molality (b) and ionic strength (I)
...
5091 and B and k are parameters that depend upon the ions
...
04 mol kg−1 and
0 ≤ I ≤ 0
...

b/(mol kg−1)

0
...
02

0
...
04

0
...
74452

0
...
71928

0
...
70809

b/(mol kg )

0
...
07

0
...
09

0
...
70380

0
...
69790

0
...
69338

−1

Hint
...

7
...
Shortly before it was (falsely) believed that the first
superheavy element had been discovered, an attempt was made to predict the
chemical properties of ununpentium (Uup, element 115, O
...
Keller, C
...

Nestor, and B
...
Phys
...
78, 1945 (1974))
...
5 eV, I(Uup) = 5
...
22 eV, S (Uup+, aq) =
+1
...
69 meV K−1
...

7
...
In a fluoride-selective
electrode used in the analysis of water samples a crystal of LaF3 doped with
Eu2+, denoted as Eu2+:LaF3, provides a semipermeable barrier between the test
solution and the solution inside the electrode (the filling solution), which
contains 0
...
1 mol kg−1 NaCl(aq)
...
It follows that the half-cell for a fluoride-selective
electrode is represented by
Ag(s) | AgCl(s) | NaCl(aq, b1), NaF (aq, b1) | Eu2+:LaF3 (s) | F−(aq, b2)
where b1 and b2 are the molalities of fluoride ion in the filling and test
solutions, respectively
...

(b) The fluoride-selective electrode just described is not sensitive to HF(aq)
...
1
...
5 × 10−4 at 298 K to specify a range
of pH values in which the electrode responds accurately to the activity of F− in
the test solution at 298 K
...
25 Express the equilibrium constant of a gas-phase reaction A + 3 B 5 2 C

in terms of the equilibrium value of the extent of reaction, ξ, given that
initially A and B were present in stoichiometric proportions
...


7
...
2
...

7
...
606AC

1/2

C

when Ks is small (in a sense to be specified)
...
28 Here we investigate the molecular basis for the observation that the
hydrolysis of ATP is exergonic at pH = 7
...
(a) It is thought that the
exergonicity of ATP hydrolysis is due in part to the fact that the standard
entropies of hydrolysis of polyphosphates are positive
...
This observation has been used to support the hypothesis that
electrostatic repulsion between adjacent phosphate groups is a factor that
controls the exergonicity of ATP hydrolysis
...
Do these
electrostatic effects contribute to the ∆r H or ∆rS terms that determine the
exergonicity of the reaction? Hint
...

7
...
0 and the ATP, ADP, and P − concentrations are all
i
1
...

7
...
(a) What is the percentage efficiency of aerobic
respiration under biochemical standard conditions? (b) The following
conditions are more likely to be observed in a living cell: pCO2 = 5
...
132 atm, [glucose] = 5
...
0 × 10−4 mol dm−3, pH = 7
...
Assuming that activities can be
replaced by the numerical values of molar concentrations, calculate the
efficiency of aerobic respiration under these physiological conditions
...
2)
...
Why is biological energy conversion more or less efficient than energy
conversion in a diesel engine?
7
...
Could
a bacterium evolve to use the ethanol/nitrate pair instead of the glucose/O2
pair as a source of metabolic energy?
7
...
33 The standard potentials of proteins are not commonly measured by
the methods described in this chapter because proteins often lose their native
structure and function when they react on the surfaces of electrodes
...
The standard potential of the
protein is then determined from the Nernst equation, the equilibrium
concentrations of all species in solution, and the known standard potential
of the electron donor
...
The one-electron reaction between cytochrome c, cyt, and
2,6-dichloroindophenol, D, can be followed spectrophotometrically because
each of the four species in solution has a distinct colour, or absorption
spectrum
...

7
7
(a) Consider E cyt and E D to be the standard potentials of cytochrome c and D,
respectively
...
(b) The following data were obtained for
the reaction between oxidized cytochrome c and reduced D in a pH 6
...
The ratios [Dox]eq/[Dred]eq and [cytox]eq/[cytred]eq were adjusted by
titrating a solution containing oxidized cytochrome c and reduced D with a
solution of sodium ascorbate, which is a strong reductant
...
237 V, determine the standard potential
cytochrome c at pH 6
...


7
...

Worsnop et al
...
R
...
E
...
S
...
C
...
Standard
reaction Gibbs energies can be computed for the following reactions at 190 K
from their data:

[Dox]eq/[Dred]eq

0
...
00843

0
...
0497

0
...
238 0
...
0106

0
...
0894

0
...
335

0
...
39

Which solid is thermodynamically most stable at 190 K if pH2O = 1
...
1×10−10 bar? Hint
...


7
...

The following equilibrium constants are based on measurements by R
...
Cox
and C
...
Hayman (Nature 332, 796 (1988)) on the reaction 2ClO (g) →
(ClO)2 (g)
...
13 × 10

8

258

5
...
45 × 10

7

T/K

288

295

4
...
67 × 105

5
...
20 × 10

6

9
...
02 × 104

(a) Derive the values of ∆ r H 7 and ∆rS 7 for this reaction
...
8 kJ mol−1 and S m(ClO) = 226
...


(i)

H2O (g)→ H2O (s)

∆rG 7 = −23
...
2 kJ mol−1

(iii) 2 H2O (g) + HNO3 (g)→ HNO3·2H2O (s)

∆rG 7 = −85
...
8 kJ mol−1

7
...
(a) What pressure is needed? (b) Now suppose that a new
catalyst is developed that is most cost-effective at 400°C when the pressure
gives the same value of ∆rG
...
Isotherms of
∆rG(T, p) in the pressure range 100 atm ≤ p ≤ 400 atm are needed to derive
the answer
...
In Part 2 we examine the structures and properties of
individual atoms and molecules from the viewpoint of quantum mechanics
...


8

Quantum theory: introduction and principles

9

Quantum theory: techniques and applications

10 Atomic structure and atomic spectra
11 Molecular structure
12 Molecular symmetry
13 Molecular spectroscopy 1: rotational and vibrational spectra
14 Molecular spectroscopy 2: electronic transitions
15 Molecular spectroscopy 3: magnetic resonance
16 Statistical thermodynamics 1: the concepts
17 Statistical thermodynamics 2: applications
18 Molecular interactions
19 Materials 1: macromolecules and aggregates
20 Materials 2: the solid state

This page intentionally left blank

Quantum theory:
introduction and
principles
This chapter introduces some of the basic principles of quantum mechanics
...
These experiments led to the conclusion that particles may not have an arbitrary energy and that
the classical concepts of ‘particle’ and ‘wave’ blend together
...
In quantum mechanics, all the properties of a system are expressed
in terms of a wavefunction that is obtained by solving the Schrödinger equation
...
Finally, we introduce some of the techniques of quantum
mechanics in terms of operators, and see that they lead to the uncertainty principle, one of
the most profound departures from classical mechanics
...
1 The failures of classical physics
8
...
1 Impact on biology: Electron

microscopy
The dynamics of microscopic
systems
8
...
However, towards the end of the nineteenth
century, experimental evidence accumulated showing that classical mechanics failed
when it was applied to particles as small as electrons, and it took until the 1920s to
discover the appropriate concepts and equations for describing them
...


8
...
5 The information in a

wavefunction
8
...
7 The postulates of quantum

mechanics

The origins of quantum mechanics

Checklist of key ideas
Further reading

The basic principles of classical mechanics are reviewed in Appendix 2
...
These conclusions agree with everyday experience
...

We shall also investigate the properties of light
...
Such waves are generated by the acceleration
of electric charge, as in the oscillating motion of electrons in the antenna of a radio
transmitter
...
As its name suggests, an electromagnetic field has two components, an electric field that acts on charged particles (whether stationary or
moving) and a magnetic field that acts only on moving charged particles
...
8
...
The frequency is measured in hertz, where 1 Hz = 1 s−1
...
1)

Therefore, the shorter the wavelength, the higher the frequency
...
(b) The wave is
shown travelling to the right at a speed c
...
The frequency, ν, is the number of
cycles per second that occur at a given
point
...
8
...
2]

λ

Wavenumbers are normally reported in reciprocal centimetres (cm−1)
...
2 summarizes the electromagnetic spectrum, the description and classification of the electromagnetic field according to its frequency and wavelength
...
Our eyes perceive different wavelengths
of radiation in this range as different colours, so it can be said that white light is a
mixture of light of all different colours
...
So, just as
our view of particles (and in particular small particles) needs to be adjusted, a new
view of light also has to be developed
...
1

Harmonic waves are waves with
displacements that can be expressed as
sine or cosine functions
...


8
...
In particular, we shall see that observations of the radiation emitted by hot bodies, heat capacities, and the spectra of atoms
and molecules indicate that systems can take up energy only in discrete amounts
...
At high temperatures, an appreciable
proportion of the radiation is in the visible region of the spectrum, and a higher
Wavelength/m

Molecular
rotation
Fig
...
2

Far
infrared

Molecular
vibration

–7

10

–8

10–9

10

–10

10

–11

1 nm

700 nm

10

Vacuum
ultraviolet

Electronic
excitation

The electromagnetic spectrum and the classification of the spectral regions
...
1 THE FAILURES OF CLASSICAL PHYSICS

dE = ρdλ

ρ=

8πkT

(8
...
381 × 10−23 J K−1)
...
A high density of states at the wavelength λ simply means that there is a lot of energy associated with wavelengths lying
between λ and λ + dλ
...
3 over all wavelengths between zero and infinity, and
the total energy (in joules) within the region is obtained by multiplying that total
energy density by the volume of the region
...
When a high
frequency, short wavelength oscillator
(a) is excited, that frequency of radiation
is present
...


Fig
...
5

An experimental representation of a
black-body is a pinhole in an otherwise
closed container
...
Radiation leaking
out through the pinhole is characteristic
of the radiation within the container
...
8
...

This behaviour is seen when a heated iron bar glowing red hot becomes white hot
when heated further
...
8
...
The curves are those
of an ideal emitter called a black body, which is an object capable of emitting and
absorbing all frequencies of radiation uniformly
...
8
...

The explanation of black-body radiation was a major challenge for nineteenthcentury scientists, and in due course it was found to be beyond the capabilities of
classical physics
...
He regarded the presence of radiation of frequency ν (and
therefore of wavelength λ = c/ν) as signifying that the electromagnetic oscillator of
that frequency had been excited (Fig
...
5)
...
2) to calculate the average energy of each oscillator as kT
...
Note
how the energy density increases in the
region of shorter wavelengths as the
temperature is raised, and how the peak
shifts to shorter wavelengths
...


Fig
...
3

246

8 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
Unfortunately (for Rayleigh, Jeans, and classical physics), although the Rayleigh–
Jeans law is quite successful at long wavelengths (low frequencies), it fails badly at
short wavelengths (high frequencies)
...
8
...
The equation therefore predicts that oscillators of very
short wavelength (corresponding to ultraviolet radiation, X-rays, and even γ-rays) are
strongly excited even at room temperature
...
According to classical physics,
even cool objects should radiate in the visible and ultraviolet regions, so objects
should glow in the dark; there should in fact be no darkness
...
3)
predicts an infinite energy density at short
wavelengths
...


Fig
...
6

25

The German physicist Max Planck studied black-body radiation from the viewpoint
of thermodynamics
...
This proposal is quite contrary
to the viewpoint of classical physics (on which the equipartition principle used by
Rayleigh is based), in which all possible energies are allowed
...
In particular, Planck
found that he could account for the observed distribution of energy if he supposed
that the permitted energies of an electromagnetic oscillator of frequency ν are integer
multiples of hν :
E = nhν

r/{8p(kT )5/(hc)4}

20

n = 0, 1, 2,
...
4)

where h is a fundamental constant now known as Planck’s constant
...
5

1
...
5

2
...
5)
accounts very well for the experimentally
determined distribution of black-body
radiation
...
The distribution coincides
with the Rayleigh–Jeans distribution at
long wavelengths
...
8
...
5 predicts the
behaviour summarized by Fig
...
2
...
5)

− 1)

(For references to the derivation of this expression, see Further reading
...
8
...
The currently accepted value for h is
6
...

The Planck distribution resembles the Rayleigh–Jeans law (eqn 8
...
For short wavelengths, hc/λkT
> 1 and ehc/λkT → ∞ faster than λ5 → 0; therefore ρ → 0 as λ → 0 or ν → ∞
...

For long wavelengths, hc/λkT < 1, and the denominator in the Planck distribution
<
can be replaced by

A
C

ehc/λkT − 1 = 1 +

hc

λkT

D
F

+··· −1≈

hc

λkT

When this approximation is substituted into eqn 8
...

It is quite easy to see why Planck’s approach was successful while Rayleigh’s was not
...
According to classical mechanics, all the oscillators of the
field share equally in the energy supplied by the walls, so even the highest frequencies
are excited
...
According to Planck’s hypothesis, however, oscillators are excited only if

8
...
This energy is too large for the walls to supply
in the case of the very high frequency oscillators, so the latter remain unexcited
...

(c) Heat capacities

In the early nineteenth century, the French scientists Pierre-Louis Dulong and AlexisThérèse Petit determined the heat capacities of a number of monatomic solids
...

Dulong and Petit’s law is easy to justify in terms of classical physics
...
According to this principle, the mean energy of an atom as it
oscillates about its mean position in a solid is kT for each direction of displacement
...
The contribution of this motion to the
molar internal energy is therefore
Um = 3NAkT = 3RT
because NAk = R, the gas constant
...
3) is then predicted to be

A ∂Um D
= 3R
C ∂T F V

CV,m =

(8
...
9 J K−1 mol−1, is in striking accord with Dulong and Petit’s
value
...
It was found that the molar heat capacities of all
monatomic solids are lower than 3R at low temperatures, and that the values approach
zero as T → 0
...
He then
invoked Planck’s hypothesis to assert that the energy of oscillation is confined to
discrete values, and specifically to nhν, where n is an integer
...
4) and obtained
Um =

3NAhν
hν/kT

e

−1

in place of the classical expression 3RT
...
The resulting expression is now known as the
Einstein formula:
2

CV,m = 3Rf

A θ E D A eθ E/2T D
f=
C T F C eθ E /T − 1F

2

(8
...


247

Comment 8
...
If
2
x < 1, a good approximation is e x ≈
<
1 + x
...
01 = 1
...

≈ 1 + 0
...


Comment 8
...

The constant-volume heat capacity is
defined as CV = (∂U/∂T)V
...


248

8 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
At high temperatures (when T > θE) the exponentials in f can be expanded as
>
1 + θ E/T + · · · and higher terms ignored (see Comment 8
...
The result is

3

f=

CV,m /R

2

0

(8
...
At low
temperatures, when T < θE,
<

1

0

2
2
A θ E D 1 1 + θ E/2T + · · · 5
2
6 ≈1
C T F 3 (1 + θ E/T + · · · ) − 1 7

2

2

2

A θE D A eθE /2T D A θE D −θ /T
f≈
=
e E
C T F C eθE /T F C T F
0
...
5

2

Experimental low-temperature
molar heat capacities and the temperature
dependence predicted on the basis of
Einstein’s theory
...
7)
accounts for the dependence fairly well,
but is everywhere too low
...
8
...
7, plot
CV,m against T for several values
of the Einstein temperature θE
...
6
...
8b)

The strongly decaying exponential function goes to zero more rapidly than 1/T
goes to infinity; so f → 0 as T → 0, and the heat capacity therefore approaches zero
too
...
The physical reason for this success is that at low temperatures only a
few oscillators possess enough energy to oscillate significantly
...

Figure 8
...
The general shape of the curve is satisfactory, but the numerical
agreement is in fact quite poor
...
This complication is taken into
account by averaging over all the frequencies present, the final result being the Debye
formula:
CV,m = 3Rf

f=3

A TD
C θD F

3 θ /T
D

Ύ

0

x4ex
(ex − 1)2

dx

(8
...

The integral in eqn 8
...
The details of this modification, which, as Fig
...
9 shows, gives
improved agreement with experiment, need not distract us at this stage from the main
conclusion, which is that quantization must be introduced in order to explain the
thermal properties of solids
...
1 Assessing the heat capacity

The Debye temperature for lead is 105 K, corresponding to a vibrational frequency
of 2
...
6 × 1013 Hz
...
8
...
For lead at 25°C, corresponding to T/θD = 2
...
99 and the heat capacity has almost its classical value
...
13, corresponding to f = 0
...


(d) Atomic and molecular spectra

The most compelling evidence for the quantization of energy comes from spectroscopy, the detection and analysis of the electromagnetic radiation absorbed, emitted,
or scattered by a substance
...
2 WAVE–PARTICLE DUALITY

249

3
Absorption intensity

Debye
Einstein

CV,m /R

Emission intensity

2

1

200
0

0
...
5
1
T/qE or T/q D

2

Debye’s modification of Einstein’s
calculation (eqn 8
...
For copper,
T/θD = 2 corresponds to about 170 K, so
the detection of deviations from Dulong
and Petit’s law had to await advances in
low-temperature physics
...
8
...
9), plot dCV,m/dT,
the temperature coefficient of CV,m, against
T for θD = 400 K
...
8
...


or scattered by a molecule as a function of frequency (ν), wavelength (λ), or wavenumber
(# = ν/c) is called its spectrum (from the Latin word for appearance)
...
8
...
8
...
The obvious feature of both is that radiation is emitted or absorbed
at a series of discrete frequencies
...
8
...
Then, if the energy of an atom
decreases by ∆E, the energy is carried away as radiation of frequency ν, and an emission
‘line’, a sharply defined peak, appears in the spectrum
...
8
...
This spectrum is part of that
due to the electronic, vibrational, and
rotational excitation of sulfur dioxide
(SO2) molecules
...


E3
hn = E3 - E2

Energy

0

E2
hn = E2 - E1
hn = E3 - E1

(8
...
We develop the principles and applications of atomic spectroscopy in
Chapter 10 and of molecular spectroscopy in Chapters 13–15
...
2 Wave–particle duality
At this stage we have established that the energies of the electromagnetic field and of
oscillating atoms are quantized
...

One experiment shows that electromagnetic radiation—which classical physics treats
as wave-like—actually also displays the characteristics of particles
...


E1

Fig
...
12 Spectroscopic transitions, such as
those shown above, can be accounted for if
we assume that a molecule emits a photon
as it changes between discrete energy levels
...


250

8 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
(a) The particle character of electromagnetic radiation

The observation that electromagnetic radiation of frequency ν can possess only the
energies 0, hν, 2hν,
...

particles, each particle having an energy hν
...
These particles of
electromagnetic radiation are now called photons
...

Example 8
...
0 s
...

Method Each photon has an energy hν, so the total number of photons needed to
produce an energy E is E/hν
...
The latter
is given by the product of the power (P, in watts) and the time interval for which
the lamp is turned on (E = P∆t)
...
60 × 10−7 m) × (100 J s−1) × (1
...
626 × 10−34 J s) × (2
...
8 × 1020

Note that it would take nearly 40 min to produce 1 mol of these photons
...
Moreover, an analytical result may be used for other data
without having to repeat the entire calculation
...
1 How many photons does a monochromatic (single frequency)
infrared rangefinder of power 1 mW and wavelength 1000 nm emit in 0
...
This effect
is the ejection of electrons from metals when they are exposed to ultraviolet radiation
...

2 The kinetic energy of the ejected electrons increases linearly with the frequency
of the incident radiation but is independent of the intensity of the radiation
...


Kinetic energy
of ejected
electron

Ru
bid
ium
Po
tas
So
siu
diu
m
m

2
...
81 ´ 104 cm-1, 551 nm)
2
...
86 ´ 104 cm-1, 539 nm)

2
...
69 ´ 104 cm-1, 593 nm)

Kinetic energy of photoelectron, EK

8
...

Fig
...
13

Exploration Calculate the value of
Planck’s constant given that the
following kinetic energies were observed
for photoejected electrons irradiated by
radiation of the wavelengths noted
...
8
...
(a) The energy of the photon is
insufficient to drive an electron out of the
metal
...


λi/nm 320 330 345
360
385
EK/eV 1
...
05 0
...
735 0
...
13 illustrates the first and second characteristics
...
If we suppose that the
projectile is a photon of energy hν, where ν is the frequency of the radiation, then the
conservation of energy requires that the kinetic energy of the ejected electron should
obey
1
– mev 2 = hν − Φ
2

(8
...
8
...
Photoejection cannot occur
if hν < Φ because the photon brings insufficient energy: this conclusion accounts for
observation (1)
...
11 predicts that the kinetic energy of an ejected electron
should increase linearly with frequency, in agreement with observation (2)
...
A practical application of eqn 8
...
8
...


Electron
beam

(b) The wave character of particles
Diffracted
electrons
Nickel crystal
Fig
...
15 The Davisson–Germer experiment
...


Comment 8
...


Although contrary to the long-established wave theory of light, the view that light consists of particles had been held before, but discarded
...
Nevertheless, experiments carried out in
1925 forced people to consider that possibility
...
8
...
Diffraction is the interference caused by
an object in the path of waves
...

Davisson and Germer’s success was a lucky accident, because a chance rise of temperature caused their polycrystalline sample to anneal, and the ordered planes of atoms
then acted as a diffraction grating
...
P
...
Electron diffraction is the basis for special techniques in microscopy
used by biologists and materials scientists (Impact I8
...
4)
...

We have also seen that waves of electromagnetic radiation have particle-like properties
...
When examined on an atomic
scale, the classical concepts of particle and wave melt together, particles taking on the
characteristics of waves, and waves the characteristics of particles
...
12)

That is, a particle with a high linear momentum has a short wavelength (Fig
...
16)
...


(b)

Example 8
...

Method To use the de Broglie relation, we need to know the linear momentum,

p, of the electrons
...
At the end of the period of acceleration, all the
acquired energy is in the form of kinetic energy, EK = p2/2me, so we can determine p by setting p2/2me equal to eV
...


8
...
626 × 10−34 J s
{2 × (9
...
609 × 10−19 C) × (4
...
1 × 10−12 m
where we have used 1 V C = 1 J and 1 J = 1 kg m2 s−2
...
1 pm is
shorter than typical bond lengths in molecules (about 100 pm)
...
4)
...
2 Calculate (a) the wavelength of a neutron with a translational kinetic

energy equal to kT at 300 K, (b) a tennis ball of mass 57 g travelling at 80 km/h
...
2 × 10−34 m]

We now have to conclude that, not only has electromagnetic radiation the
character classically ascribed to particles, but electrons (and all other particles) have
the characteristics classically ascribed to waves
...
Duality strikes at the heart of
classical physics, where particles and waves are treated as entirely distinct entities
...
In classical mechanics, in contrast, energies could be varied continuously
...
A new mechanics had to be devised to take its place
...
1 Electron microscopy

The basic approach of illuminating a small area of a sample and collecting light with a
microscope has been used for many years to image small specimens
...
1)
...

There is great interest in the development of new experimental probes of very small
specimens that cannot be studied by traditional light microscopy
...
One technique that is often used to image nanometre-sized
objects is electron microscopy, in which a beam of electrons with a well defined de
Broglie wavelength replaces the lamp found in traditional light microscopes
...
In transmission
electron microscopy (TEM), the electron beam passes through the specimen and the

Fig
...
16 An illustration of the de Broglie
relation between momentum and
wavelength
...
A particle
with high momentum has a wavefunction
with a short wavelength, and vice versa
...
8
...

Chloroplasts are typically 5 µm long
...
)

image is collected on a screen
...
An image of the surface is then obtained by scanning the
electron beam across the sample
...
Electron
wavelengths in typical electron microscopes can be as short as 10 pm, but it is not
possible to focus electrons well with magnetic lenses so, in the end, typical resolutions
of TEM and SEM instruments are about 2 nm and 50 nm, respectively
...
2 nm)
...
The measurements must be conducted under high vacuum
...
A consequence of these requirements
is that neither technique can be used to study living cells
...
8
...


The dynamics of microscopic systems
Quantum mechanics acknowledges the wave–particle duality of matter by supposing
that, rather than travelling along a definite path, a particle is distributed through space
like a wave
...
The mathematical representation of the wave that in quantum mechanics
replaces the classical concept of trajectory is called a wavefunction, ψ (psi)
...
3 The Schrödinger equation
In 1926, the Austrian physicist Erwin Schrödinger proposed an equation for finding
the wavefunction of any system
...
13)

The factor V(x) is the potential energy of the particle at the point x; because the total
energy E is the sum of potential and kinetic energies, the first term must be related
(in a manner we explore later) to the kinetic energy of the particle; $ (which is read
h-cross or h-bar) is a convenient modification of Planck’s constant:
$=

h
2π

= 1
...
14)

For a partial justification of the form of the Schrödinger equation, see the Justification
below
...
For the present, treat the equation as a
quantum-mechanical postulate
...
1
...


8
...
1 The Schrödinger equation
For one-dimensional systems:
−

$2 d2ψ
+ V(x)ψ = Eψ
2m dx2

Where V(x) is the potential energy of the particle and E is its total energy
...
1 Using the Schrödinger equation to develop the de Broglie relation

Although the Schrödinger equation should be regarded as a postulate, like Newton’s
equations of motion, it can be seen to be plausible by noting that it implies the
de Broglie relation for a freely moving particle in a region with constant potential
energy V
...
13 into
d2ψ
dx 2

=−

2m
$2

(E − V)ψ

General strategies for solving differential equations of this and other types that
occur frequently in physical chemistry are treated in Appendix 2
...
5

Complex numbers and functions are
discussed in Appendix 2
...

Similarly, a complex function of the
form f = g + ih, where g and h are
functions of real arguments, has a real
part Re(f ) = g and an imaginary part
Im(f ) = h
...
In the present case, we can use the relation eiθ = cos θ + i sin θ to write

ψ = cos kx + i sin kx
The real and imaginary parts of ψ are drawn in Fig
...
18, and we see that the
imaginary component Im(ψ) = sin kx is shifted in the direction of the particle’s
motion
...

Now we recognize that cos kx (or sin kx) is a wave of wavelength λ = 2π/k, as can
be seen by comparing cos kx with the standard form of a harmonic wave, cos(2πx/λ)
...
Because EK = p2/2m, it follows that
p = k$
Therefore, the linear momentum is related to the wavelength of the wavefunction by
p=

2π

λ

×

h
2π

=

h

λ

which is the de Broglie relation
...

Fig
...
18

dx
|y |2

Probability
= |y |2dx

x x + dx

8
...
Here we concentrate on the
information it carries about the location of the particle
...
He made use of an analogy with the wave
theory of light, in which the square of the amplitude of an electromagnetic wave in a
region is interpreted as its intensity and therefore (in quantum terms) as a measure of
the probability of finding a photon present in the region
...
It states that the value of | ψ |2 at a point is proportional
to the probability of finding the particle in a region around that point
...
8
...


The wavefunction ψ is a
probability amplitude in the sense that its
square modulus (ψ *ψ or |ψ |2) is a
probability density
...
We represent
the probability density by the density of
shading in the superimposed band
...
8
...
6

To form the complex conjugate, ψ *, of a
complex function, replace i wherever it
occurs by −i
...
If the
wavefunction is real, |ψ |2 = ψ 2
...
The wavefunction ψ itself is called
the probability amplitude
...
8
...

The Born interpretation does away with any worry about the significance of a
negative (and, in general, complex) value of ψ because |ψ |2 is real and never negative
...
4 THE BORN INTERPRETATION OF THE WAVEFUNCTION
probability of finding a particle in a region (Fig
...
21)
...


z
dt

dz

r
Example 8
...
(Notice that this wavefunction depends only on
this distance, not the angular position relative to the nucleus
...
0 pm3,
which is small even on the scale of the atom, located at (a) the nucleus, (b) a
distance a0 from the nucleus
...
8
...


Method The region of interest is so small on the scale of the atom that we can

ignore the variation of ψ within it and write the probability, P, as proportional to
the probability density (ψ 2; note that ψ is real) evaluated at the point of interest
multiplied by the volume of interest, δV
...

Answer In each case δV = 1
...
(a) At the nucleus, r = 0, so

Wavefunction
Probability
density

P ∝ e0 × (1
...
0) × (1
...
0 pm3) = (0
...
0 pm3)
Therefore, the ratio of probabilities is 1
...
14 = 7
...
Note that it is more probable
(by a factor of 7) that the electron will be found at the nucleus than in a volume
element of the same size located at a distance a0 from the nucleus
...

A note on good practice The square of a wavefunction is not a probability: it is a

probability density, and (in three dimensions) has the dimensions of 1/length3
...
In general, we have
to take into account the variation of the amplitude of the wavefunction over the
volume of interest, but here we are supposing that the volume is so small that the
variation of ψ in the region can be ignored
...
3 The wavefunction for the electron in its lowest energy state in the ion

He+ is proportional to e−2r/a0
...
Any comment?
[55; more compact wavefunction]

(a) Normalization

A mathematical feature of the Schrödinger equation is that, if ψ is a solution, then so
is Nψ, where N is any constant
...
13, so any constant factor can be cancelled
...


Fig
...
21 The sign of a wavefunction has no
direct physical significance: the positive
and negative regions of this wavefunction
both correspond to the same probability
distribution (as given by the square
modulus of ψ and depicted by the density
of shading)
...
Furthermore, the sum over all space of these individual
probabilities must be 1 (the probability of the particle being somewhere is 1)
...
15)

−∞

Almost all wavefunctions go to zero at sufficiently great distances so there is rarely
any difficulty with the evaluation of this integral, and wavefunctions for which the
integral in eqn 8
...
It follows that
1

N=

r sin q d q
dr

r df
q

z

A
C

∞

Ύ

ψ *ψ dx

−∞

D
F

1/2

(8
...
From now on, unless we state otherwise, we always use wavefunctions that have been normalized to 1; that is, from now on we assume that ψ
already includes a factor that ensures that (in one dimension)
∞

Ύ

f

ψ *ψ dx = 1

(8
...
17b)

or, more succinctly, if

The spherical polar coordinates
used for discussing systems with spherical
symmetry
...
8
...
For
systems with spherical symmetry it is best to work in spherical polar coordinates r, θ,
and φ (Fig
...
22): x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ
...
To cover all space, the radius r
ranges from 0 to ∞, the colatitude, θ, ranges from 0 to π, and the azimuth, φ, ranges
from 0 to 2π (Fig
...
23), so the explicit form of eqn 6
...
17c)

π

2π

ΎΎΎ
0

0

ψ *ψ r 2 sin θdrdθdφ = 1

(8
...
4 Normalizing a wavefunction

p

The surface of a sphere is covered
by allowing θ to range from 0 to π, and
then sweeping that arc around a complete
circle by allowing φ to range from 0 to 2π
...
8
...
3
...
17c

is equal to 1
...
17d
...
4 THE BORN INTERPRETATION OF THE WAVEFUNCTION

259

limits on the first integral sign refer to r, those on the second to θ, and those on the
third to φ
...
1
...

If Example 8
...

Given (from Section 10
...
9 pm, the results are (a) 2
...
9 × 10−7, corresponding to 1 chance in 3
...

Self-test 8
...
3
...
The principal constraint is that ψ must not be infinite anywhere
...
17 would be infinite (in other words, ψ would not be
square-integrable) and the normalization constant would be zero
...
The requirement that ψ is finite everywhere rules out many possible
solutions of the Schrödinger equation, because many mathematically acceptable
solutions rise to infinity and are therefore physically unacceptable
...

The requirement that ψ is finite everywhere is not the only restriction implied
by the Born interpretation
...
6a will meet) a
solution of the Schrödinger equation that gives rise to more than one value of |ψ |2 at
a single point
...
This restriction is expressed by saying that the wavefunction must be
single-valued; that is, have only one value at each point of space
...
7

Infinitely sharp spikes are acceptable
provided they have zero width, so it is
more appropriate to state that the
wavefunction must not be infinite over
any finite region
...


260

8 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES

y

(a)

x (b)
¥

(c)

¥

(d)

Fig
...
24 The wavefunction must satisfy
stringent conditions for it to be acceptable
...


Comment 8
...
These cases arise when the
potential energy has peculiar properties,
such as rising abruptly to infinity
...
There are only two
cases of this behaviour in elementary
quantum mechanics, and the peculiarity
will be mentioned when we meet them
...
Because it is a second-order differential equation,
the second derivative of ψ must be well-defined if the equation is to be applicable
everywhere
...
8
...

At this stage we see that ψ must be continuous, have a continuous slope, be
single-valued, and be square-integrable
...
These are such
severe restrictions that acceptable solutions of the Schrödinger equation do not in
general exist for arbitrary values of the energy E
...

That is, the energy of a particle is quantized
...
That is the task of the next chapter
...
We have seen that the Born interpretation tells us as much as we can
know about location, but how do we find any additional information?
8
...
13 by setting V = 0, and is
−

$2 d2ψ
2m dx 2

= Eψ

(8
...
19)

where A and B are constants
...
18, we simply
substitute it into the left-hand side of the equation and confirm that we obtain Eψ:
−

$2 d2ψ
2m dx 2

=−
=−
=

$2 d2
2m dx 2
$2
2m

$2k2
2m

(Aeikx + Be−kx)

{A(ik)2eikx + B(−ik)2e−ikx }

(Aeikx + Be−ikx) = Eψ

(a) The probability density

We shall see later what determines the values of A and B; for the time being we can
treat them as arbitrary constants
...
19, then the wavefunction is simply

ψ = Aeikx

(8
...
5 THE INFORMATION IN A WAVEFUNCTION
Where is the particle? To find out, we calculate the probability density:
|ψ |2 = (Aeikx)*(Aeikx) = (A*e−ikx)(Aeikx) = | A |2

|y | = 1
2

ikx

Im e = sin kx

(8
...
8
...
In other words, if the
wavefunction of the particle is given by eqn 8
...
The same would be true if the wavefunction in eqn 8
...

Now suppose that in the wavefunction A = B
...
19 becomes

ψ = A(eikx + e−ikx) = 2A cos kx

261

(a)

ikx

Re e = cos kx
cos kx

cos2 kx

(8
...
23)

This function is illustrated in Fig
...
25b
...
The locations where the probability density is zero
correspond to nodes in the wavefunction: particles will never be found at the nodes
...
The location
where a wavefunction approaches zero without actually passing through zero is not a
node
...
The probability density, of course, never passes through zero because it cannot
be negative
...
8
...
(b) The
probability distribution corresponding
to the superposition of states of equal
magnitude of linear momentum but
opposite direction of travel
...
13 and 8
...
24a)

with (in one dimension)
@=−

$2 d2
2m dx 2

+ V(x)

(8
...
In this case, the operation is to take the second derivative of ψ and
(after multiplication by −$2/2m) to add the result to the outcome of multiplying ψ
by V
...
The hamiltonian operator is the
operator corresponding to the total energy of the system, the sum of the kinetic and
potential energies
...
3—
that the first term in eqn 8
...
When the Schrödinger equation is written as
in eqn 8
...
25a)

If we denote a general operator by ) (where Ω is uppercase omega) and a constant
factor by ω (lowercase omega), then an eigenvalue equation has the form
)ψ = ωψ

(8
...
9

If the probability density of a particle
is a constant, then it follows that,
with x ranging from −∞ to +∞, the
normalization constants, A or B, are 0
...
We ignore this
complication here
...
The eigenvalue in eqn 8
...
The function ψ in an equation of this kind is called an eigenfunction
of the operator ) and is different for each eigenvalue
...
24a
is the wavefunction corresponding to the energy E
...
The wavefunctions are the
eigenfunctions of the hamiltonian operator, and the corresponding eigenvalues are
the allowed energies
...
5 Identifying an eigenfunction

Show that eax is an eigenfunction of the operator d/dx, and find the corresponding
2
eigenvalue
...

Method We need to operate on the function with the operator and check whether

the result is a constant factor times the original function
...
For
2
ψ = eax ,
)ψ =

d
dx

eax = 2axeax = 2ax × ψ
2

2

which is not an eigenvalue equation even though the same function ψ occurs
on the right, because ψ is now multiplied by a variable factor (2ax), not a constant
2
factor
...

Self-test 8
...

Thus, it is often the case that we can write
(Operator corresponding to an observable)ψ = (value of observable) × ψ
The symbol ) in eqn 8
...

Therefore, if we know both the wavefunction ψ and the operator ) corresponding
to the observable Ω of interest, and the wavefunction is an eigenfunction of the
operator ), then we can predict the outcome of an observation of the property Ω
(for example, an atom’s energy) by picking out the factor ω in the eigenvalue
equation, eqn 8
...


8
...
26]

i dx

That is, the operator for location along the x-axis is multiplication (of the wavefunction)
by x and the operator for linear momentum parallel to the x-axis is proportional to
taking the derivative (of the wavefunction) with respect to x
...
6 Determining the value of an observable

What is the linear momentum of a particle described by the wavefunction in
eqn 8
...
26), and inspect the result
...

Answer (a) With the wavefunction given in eqn 8
...
25b we find that
px = +k$
...

Self-test 8
...
What is the
angular momentum of a particle described by the wavefunction e−2iφ?
[lz = −2$]

We use the definitions in eqn 8
...
For example, suppose we wanted the operator for a potential energy of the form
1
V = –kx 2, with k a constant (later, we shall see that this potential energy describes the
2
vibrations of atoms in molecules)
...
26 that the operator
corresponding to V is multiplication by x 2:
1
W = –kx 2 ×
2

(8
...
To construct the operator for
kinetic energy, we make use of the classical relation between kinetic energy and linear

263

Comment 8
...
26 apply
to observables that depend on spatial
variables; intrinsic properties, such as
spin (see Section 9
...


264

8 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
2
momentum, which in one dimension is EK = px /2m
...
26 we find:

ÊK =
High curvature,
high kinetic
energy

2m C i dx F C i dx F

=−

$2 d2
2m dx 2

(8
...
8
...
This illustration shows two
wavefunctions: the sharply curved function
corresponds to a higher kinetic energy than
the less sharply curved function
...
11

We are using the term ‘curvature’
informally: the precise technical
definition of the curvature of a function
f is (d2f /dx 2)/{1 + (df /dx)2}3/2
...
8
...
Sharply curved regions
contribute a high kinetic energy to the
average; slightly curved regions contribute
only a small kinetic energy
...
29)

with W the multiplicative operator in eqn 8
...

The expression for the kinetic energy operator, eqn 8
...
In
mathematics, the second derivative of a function is a measure of its curvature: a large
second derivative indicates a sharply curved function (Fig
...
26)
...
This interpretation is consistent
with the de Broglie relation, which predicts a short wavelength (a sharply curved
wavefunction) when the linear momentum (and hence the kinetic energy) is high
...
8
...
The curvature of a wavefunction in
general varies from place to place
...
8
...
Wherever the wavefunction is not sharply curved, its contribution to the overall kinetic energy is low
...
Hence, we can expect a particle
to have a high kinetic energy if the average curvature of its wavefunction is high
...
22)
...
For example, suppose we need to know the wavefunction of a particle with a
given total energy and a potential energy that decreases with increasing x (Fig
...
28)
...
We can therefore guess that the wavefunction
will look like the function sketched in the illustration, and more detailed calculation
confirms this to be so
...
An hermitian operator is one for
which the following relation is true:
1
5*
Hermiticity: ψ i*)ψj dx = 2 ψ j*)ψi dx 6
[8
...


Justification 8
...
26
...
8
...
Only the real part of the
wavefunction is shown, the imaginary part
is similar, but displaced to the right
...
5 THE INFORMATION IN A WAVEFUNCTION

ψ i*Yxψj dx =

−∞

∞

$

iΎ

ψ i*

−∞

=

$
i

ψ i*ψ j

dψj
dx
∞

dx

−
−∞

$

∞

iΎ

ψj

dψ *
i

−∞

dx

dx

The first term on the right is zero, because all wavefunctions are zero at infinity in either direction, so we are left with
∞

∞

1$
ψ i*Yxψj dx = −
ψj
dx = 2
i −∞
dx
3i
−∞
$

Ύ

1
=2
3

Ύ

dψ i*

∞

Ύ

ψ*
j

−∞

5*
dx 6
dx 7

dψi

∞

5*
ψ j*Yxψi dx 6
7
−∞

Ύ

as we set out to prove
...
7 Confirm that the operator d2/dx 2 is hermitian
...
All observables have real values (in the mathematical sense, such as
x = 2 m and E = 10 J), so all observables are represented by hermitian operators
...
31]

For example, the hamiltonian operator is hermitian (it corresponds to an observable,
the energy)
...


266

8 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
Justification 8
...
The conclusion that
ω * = ω confirms that ω is real
...
To verify that the two wavefunctions are mutually orthogonal, we integrate the product (sin x)(sin 2x) over all
space, which we may take to span from x = 0 to x = 2π, because both functions
repeat themselves outside that range
...
8
...
A useful integral for this calculation is

0
...
5
-1
0

Illustration 8
...
8
...

The function—and the value of the
integral—repeats itself for all replications
of the section between 0 and 2π, so the
integral from −∞ to ∞ is zero
...

Self-test 8
...


G
H
I

∞

Ύ

−∞

J

sin x sin 3xdx = 0 K

L

(d) Superpositions and expectation values

Suppose now that the wavefunction is the one given in eqn 8
...
What
is the linear momentum of the particle it describes? We quickly run into trouble if
we use the operator technique
...
32)

This expression is not an eigenvalue equation, because the function on the right
(sin kx) is different from that on the left (cos kx)
...
5 THE INFORMATION IN A WAVEFUNCTION
When the wavefunction of a particle is not an eigenfunction of an operator, the
property to which the operator corresponds does not have a definite value
...
We say that the
total wavefunction is a superposition of more than one wavefunction
...
However, because the two component wavefunctions
occur equally in the superposition, half the measurements will show that the particle
is moving to the right (px = +k$), and half the measurements will show that it is
moving to the left (px = −k$)
...

The same interpretation applies to any wavefunction written as a linear combination of eigenfunctions of an operator
...
33)

k

where the c k are numerical (possibly complex) coefficients and the ψk correspond to
different momentum states
...

Then according to quantum mechanics:
1 When the momentum is measured, in a single observation one of the eigenvalues
corresponding to the ψk that contribute to the superposition will be found
...

3 The average value of a large number of observations is given by the expectation
value, ͗Ω ͘, of the operator ) corresponding to the observable of interest
...
34]

This formula is valid only for normalized wavefunctions
...


267

Comment 8
...
In a
sum, c1 = c2 = 1
...
567f + 1
...


8 QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
Justification 8
...
The interpretation of this
expression is that, because every observation of the property Ω results in the value
ω (because the wavefunction is an eigenfunction of )), the mean value of all the
observations is also ω
...
For simplicity, suppose the wavefunction is the sum of two eigenfunctions (the general case, eqn 8
...
Then

Ύ
= (c ψ + c ψ )*(c )ψ + c )ψ )dτ
Ύ
= (c ψ + c ψ )*(c ω ψ + c ω ψ )dτ
Ύ

͗Ω ͘ = (c1ψ1 + c2ψ2)*)(c1ψ1 + c2ψ2)dτ

1 1

2 2

1

1

1 1

2 2

1

1 1

2

2

2

2 2

5
6
7

1

5
6
7

1

Ύ

Ύ

0

0

5
6
7

= c1 1ω 1 ψ 1 ψ1dτ + c 2 2ω 2 ψ 2 ψ2dτ
*c
*
*c
*
5
6
7

268

Ύ

Ύ

+ c2 1ω1 ψ2 ψ1dτ + c1 2ω 2 ψ 1 ψ2dτ
*c
*
*c
*
The first two integrals on the right are both equal to 1 because the wavefunctions are
individually normalized
...
We can conclude that
͗Ω ͘ = | c1 |2ω 1 + | c2 |2ω 2
This expression shows that the expectation value is the sum of the two eigenvalues
weighted by the probabilities that each one will be found in a series of measurements
...


Example 8
...

Method The average radius is the expectation value of the operator corresponding

to the distance from the nucleus, which is multiplication by r
...
4) and then evaluate
the integral in eqn 8
...


8
...
Using the normalized
function in Example 8
...
9 pm (see Section 10
...
4 pm
...
4 pm
...

Self-test 8
...
6 pm]

the nucleus in the hydrogen atom
...
28
...
35)

This conclusion confirms the previous assertion that the kinetic energy is a kind of
average over the curvature of the wavefunction: we get a large contribution to the
observed value from regions where the wavefunction is sharply curved (so d2ψ /dx 2
is large) and the wavefunction itself is large (so that ψ * is large too)
...
6 The uncertainty principle
We have seen that, if the wavefunction is Aeikx, then the particle it describes has a
definite state of linear momentum, namely travelling to the right with momentum
px = +k$
...
In other words, if the momentum
is specified precisely, it is impossible to predict the location of the particle
...

Before discussing the principle further, we must establish its other half: that, if we
know the position of a particle exactly, then we can say nothing about its momentum
...

If we know that the particle is at a definite location, its wavefunction must be large
there and zero everywhere else (Fig
...
30)
...
In other words, we can create a sharply localized

x
Location
of particle
Fig
...
30 The wavefunction for a particle at a
well-defined location is a sharply spiked
function that has zero amplitude
everywhere except at the particle’s position
...
The superposition of a
few harmonic functions gives a wavefunction that spreads over a range of locations
(Fig
...
31)
...
When an infinite
number of components is used, the wave packet is a sharp, infinitely narrow spike,
which corresponds to perfect localization of the particle
...
However, we have lost all information about its momentum because, as we
saw above, a measurement of the momentum will give a result corresponding to any
one of the infinite number of waves in the superposition, and which one it will give
is unpredictable
...

A quantitative version of this result is
1
∆p∆q ≥ –$
2

Fig
...
31 The wavefunction for a particle
with an ill-defined location can be
regarded as the superposition of several
wavefunctions of definite wavelength that
interfere constructively in one place but
destructively elsewhere
...
An infinite number of waves
is needed to construct the wavefunction of
a perfectly localized particle
...
Explore how the probability
density ψ 2(x) changes with the value of N
...
36a)

In this expression ∆p is the ‘uncertainty’ in the linear momentum parallel to the
axis q, and ∆q is the uncertainty in position along that axis
...
36b)

If there is complete certainty about the position of the particle (∆q = 0), then the only
way that eqn 8
...
Conversely, if the momentum parallel to an axis is known
exactly (∆p = 0), then the position along that axis must be completely uncertain
(∆q = ∞)
...
36 refer to the same direction in space
...
The restrictions that
the uncertainty principle implies are summarized in Table 8
...

Example 8
...
0 g is known to within 1 µm s−1
...

Method Estimate ∆p from m∆v, where ∆v is the uncertainty in the speed; then use
eqn 8
...

Answer The minimum uncertainty in position is

∆q =
=

$
2m∆v
1
...
0 × 10 −3 kg) × (1 × 10 −6 m s−1)

= 5 × 10 −26 m

where we have used 1 J = 1 kg m2 s−2
...
However, if the mass is
that of an electron, then the same uncertainty in speed implies an uncertainty in

8
...


Table 8
...
10 Estimate the minimum uncertainty in the speed of an electron in a

Variable 2

one-dimensional region of length 2a0
...
36 suggests
...
Specifically, two observables Ω1
and Ω2 are complementary if
)1()2ψ) ≠ )2()1ψ)

(8
...
The different outcomes of the effect of applying )1
and )2 in a different order are expressed by introducing the commutator of the two
operators, which is defined as
[)1,)2] = )1 )2 − )2 )1

z
px
py
pz
* Pairs of observables that cannot be determined
simultaneously with arbitrary precision are
marked with a white rectangle; all others are
unrestricted
...
38]

We can conclude from Illustration 8
...
39)

Illustration 8
...
The second expression is clearly different from the first, so the two
operators do not commute
...
39
...
39 is of such vital significance in quantum mechanics
that it is taken as a fundamental distinction between classical mechanics and quantum mechanics
...
26
...
13

For two functions f and g,
d(fg) = fdg + gdf
...
14

The ‘modulus’ notation |
...

For example, |−2| = 2, |3i| = 3, and
|−2 + 3i| = {(−2 − 3i)(−2 + 3i)}1/2 = 131/2
...
40 ensures that the product of
uncertainties has a real, non-negative
value
...
For any two pairs of observables, Ω1 and Ω2, the
uncertainties (to be precise, the root mean square deviations of their values from the
mean) in simultaneous determinations are related by
1
∆Ω1∆Ω2 ≥ – | ͗[)1,)2]͘ |
2

(8
...
36a when we identify the observables with x and px
and use eqn 8
...

Complementary observables are observables with non-commuting operators
...
Classical mechanics supposed, falsely as we now know,
that the position and momentum of a particle could be specified simultaneously
with arbitrary precision
...

The realization that some observables are complementary allows us to make
considerable progress with the calculation of atomic and molecular properties, but
it does away with some of the most cherished concepts of classical physics
...
7 The postulates of quantum mechanics
For convenience, we collect here the postulates on which quantum mechanics is based
and which have been introduced in the course of this chapter
...
All dynamical information is contained in the wavefunction ψ
for the system, which is a mathematical function found by solving the Schrödinger
equation for the system
...
If the wavefunction of a particle has the value ψ at
some point r, then the probability of finding the particle in an infinitesimal volume
dτ = dxdydz at that point is proportional to |ψ |2dτ
...
An acceptable wavefunction must be continuous, have
a continuous first derivative, be single-valued, and be square-integrable
...
Observables, Ω, are represented by operators, ), built from position
and momentum operators of the form
X=x×

Yx =

$ d
i dx

or, more generally, from operators that satisfy the commutation relation [X, Yx] = i$
...
It is impossible to specify simultaneously,
with arbitrary precision, both the momentum and the position of a particle and, more
generally, any pair of observable with operators that do not commute
...
In classical physics, radiation is described in terms of an
oscillating electromagnetic disturbance that travels through
vacuum at a constant speed c = λν
...
The Born interpretation of the wavefunction states that the
value of |ψ |2, the probability density, at a point is proportional
to the probability of finding the particle at that point
...
A black body is an object that emits and absorbs all
frequencies of radiation uniformly
...
Quantization is the confinement of a dynamical observable
to discrete values
...
The variation of the energy output of a black body with
wavelength is explained by invoking quantization of energy,
the limitation of energies to discrete values, which in turn
leads to the Planck distribution, eqn 8
...


13
...


4
...
7
and 8
...

5
...

6
...

7
...

8
...

9
...

10
...


14
...
The position and momentum
operators are X = x × and Yx = ($/i)d/dx, respectively
...
The hamiltonian operator is the operator for the total energy
of a system, @ψ = Eψ and is the sum of the operators for
kinetic energy and potential energy
...
An eigenvalue equation is an equation of the form )ψ = ωψ
...

17
...

18
...
The eigenvalues of hermitian operators are real
and correspond to observables, measurable properties of a
system
...

19
...

2
20
...

21
...

22
...

2

Further reading
Articles and texts

P
...
Atkins, Quanta: A handbook of concepts
...

P
...
Atkins and R
...
Friedman, Molecular quantum mechanics
...

D
...
Dover, New York (1989)
...
P
...
B
...
Sands, The Feynman lectures on
physics
...
Addison–Wesley, Reading (1965)
...
S
...
and L
...
Pedersen, Problems and solutions in
quantum chemistry and physics
...

L
...
B
...
Dover, New York (1985)
...
1 Summarize the evidence that led to the introduction of quantum

8
...


system and how those properties may be predicted
...
2 Explain why Planck’s introduction of quantization accounted for the

8
...


momentum in terms of the shape of the wavefunction
...
3 Explain why Einstein’s introduction of quantization accounted for the

8
...


solving the Schrödinger equation explicitly
...
1a To what speed must an electron be accelerated for it to have a
wavelength of 3
...
1b To what speed must a proton be accelerated for it to have a wavelength

8
...
14 eV
...


of 3
...
8b The work function for metallic rubidium is 2
...
Calculate the

8
...


matter; its approximate value is 1/137
...
)
8
...


What speed does a hydrogen molecule need to travel to have the same linear
momentum?
−1

8
...
45 Mm s
...
0100 per cent, what uncertainty in its
location must be tolerated?
8
...
If the uncertainty in its

momentum is to be reduced to 0
...
4a Calculate the energy per photon and the energy per mole of photons

for radiation of wavelength (a) 600 nm (red), (b) 550 nm (yellow),
(c) 400 nm (blue)
...
4b Calculate the energy per photon and the energy per mole of photons

for radiation of wavelength (a) 200 nm (ultraviolet), (b) 150 pm (X-ray),
(c) 1
...

8
...
4a
...
5b Calculate the speed to which a stationary 4He atom (mass 4
...
4b
...
6a A glow-worm of mass 5
...
10 W entirely in the backward direction
...
6b A photon-powered spacecraft of mass 10
...
50 kW entirely in the backward
direction
...
0 y if released into
free space?
8
...
How many photons does
it emit each second if its power is (a) 1
...
7b A laser used to read CDs emits red light of wavelength 700 nm
...
10 W,
(b) 1
...
9a Calculate the size of the quantum involved in the excitation of (a) an
electronic oscillation of period 1
...
0 s
...

8
...
50 fs, (b) a molecular vibration of period
2
...
0 ms
...

8
...
0 g travelling at
1
...

8
...
0 kV, (c) 100 kV
...
11a Confirm that the operator Zz = ($/i)d/dφ, where φ is an angle, is
hermitian
...
11b Show that the linear combinations  + iU and  − iU are not hermitian

if  and U are hermitian operators
...
12a Calculate the minimum uncertainty in the speed of a ball of mass

500 g that is known to be within 1
...
What is the
minimum uncertainty in the position of a bullet of mass 5
...
000 01 m s−1 and 350
...
12b An electron is confined to a linear region with a length of the same

order as the diameter of an atom (about 100 pm)
...

8
...
4 Mm s−1
...

8
...
9 Mm s−1
...

8
...

8
...


PROBLEMS

275

Problems*
8
...


Numerical problems
8
...
Calculate the energy density in the range 650 nm to
655 nm inside a cavity of volume 100 cm3 when its temperature is (a) 25°C,
(b) 3000°C
...
11 Use the Planck distribution to deduce the Stefan–Boltzmann law that
the total energy density of black-body radiation is proportional to T 4, and
find the constant of proportionality
...
2 For a black body, the temperature and the wavelength of emission
1
maximum, λmax, are related by Wien’s law, λmaxT = – c2, where c2 = hc/k
5
(see Problem 8
...
Values of λmax from a small pinhole in an electrically
heated container were determined at a series of temperatures, and the
results are given below
...


θ/°C

1000

1500

2000

2500

3000

3500

λmax/nm

2181

1600

1240

1035

878

763

8
...
12‡ Prior to Planck’s derivation of the distribution law for black-body
radiation, Wien found empirically a closely related distribution function
which is very nearly but not exactly in agreement with the experimental
results, namely, ρ = (a/λ5)e−b/λkT
...
(a) By fitting Wien’s empirical formula
to Planck’s at short wavelengths determine the constants a and b
...
10) and with the Stefan–Boltzmann law (Problem 8
...


temperature θE, where θE = hν/k
...
Evaluate θE for (a) diamond,
for which ν = 46
...
15 THz
...
13 Normalize the following wavefunctions: (a) sin(nπx/L) in the range
0 ≤ x ≤ L, where n = 1, 2, 3,
...

Hint: The volume element in three dimensions is dτ = r 2dr sin θ dθ dφ,
with 0 ≤ r < ∞, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π
...
4
...
4 The ground-state wavefunction for a particle confined to a

8
...
(b) Confirm that these two functions are
mutually orthogonal
...
0 nm long
...
95 nm and 5
...
95 nm and 2
...
90 nm and 10
...


8
...
Give the
corresponding eigenvalue where appropriate
...
5 The ground-state wavefunction of a hydrogen atom is

8
...
State the eigenvalue of î when relevant
...
17 Which of the functions in Problem 8
...


where a0 = 53 pm (the Bohr radius)
...
0 pm
centred on the nucleus
...
What is the probability that the electron is inside it?
8
...
and 0 ≤ φ ≤ 2π
...

8
...
18 A particle is in a state described by the wavefunction ψ = (cos χ)eikx +
(sin χ)e−ikx, where χ (chi) is a parameter
...
Verify that the value of the product
∆p∆x is consistent with the predictions from the uncertainty principle
...
19 Evaluate the kinetic energy of the particle with wavefunction given in
Problem 8
...


8
...
20 Calculate the average linear momentum of a particle described by the
2
following wavefunctions: (a) eikx, (b) cos kx, (c) e−αx , where in each one x
ranges from −∞ to +∞
...
Determine the expectation value of the
commutator of the position and momentum operators
...
21 Evaluate the expectation values of r and r 2 for a hydrogen atom with

Theoretical problems

wavefunctions given in Problem 8
...


8
...
22 Calculate (a) the mean potential energy and (b) the mean kinetic energy
of an electron in the ground state of a hydrogenic atom
...


* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady
...
23 Use mathematical software to construct superpositions of cosine
functions and determine the probability that a given momentum will be
observed
...
Evaluate the root mean
square location of the packet, ͗x 2͘1/2
...
24 Show that the expectation value of an operator that can be written as the

square of an hermitian operator is positive
...
25 (a) Given that any operators used to represent observables must satisfy
the commutation relation in eqn 8
...
(b) With the
identification of X in this representation, what would be the operator for 1/x?
Hint
...


Applications: to biology, environmental science, and
astrophysics
8
...

On the assumption that the human eye evolved to be most sensitive at the
wavelength of light corresponding to the maximum in the Sun’s radiant
energy distribution, determine the colour of light to which the eye is most
sensitive
...
10
...
27 We saw in Impact I8
...
For electrons
moving at speeds close to c, the speed of light, the expression for the de
Broglie wavelength (eqn 8
...
(a) Use the expression above to
calculate the de Broglie wavelength of electrons accelerated through 50 kV
...
28‡ Solar energy strikes the top of the Earth’s atmosphere at a rate of
343 W m−2
...
The Earth–atmosphere system absorbs
the remaining energy and re-radiates it into space as black-body radiation
...
Use Wien’s law, Problem 8
...

8
...
Kulkarni, K
...
R
...
Nakajima (Science 270, 1478 (1995))
...
The mass
of the star, as determined from its gravitational effect on a companion star,
is roughly 20 times the mass of Jupiter
...
(a) From available thermodynamic data, test
the stability of methane at temperatures above 1000 K
...


9

Quantum theory:
techniques and
applications
To find the properties of systems according to quantum mechanics we need to solve the
appropriate Schrödinger equation
...
We shall see that only certain
wavefunctions and their corresponding energies are acceptable
...
The
solutions bring to light a number of highly nonclassical, and therefore surprising, features of
particles, especially their ability to tunnel into and through regions where classical physics
would forbid them to be found
...
The chapter concludes with an introduction to the experimental techniques used to probe the quantization of energy in molecules
...
Gas-phase molecules, for instance, undergo translational
motion and their kinetic energy is a contribution to the total internal energy of a
sample
...
Energy is
also stored as molecular vibration and transitions between vibrational states are
responsible for the appearance of infrared and Raman spectra
...
4

The energy levels

9
...
1

A particle in a box

9
...
3

Tunnelling

I9
...
6

Rotation in two dimensions: a
particle on a ring

9
...
2 Impact on nanoscience:

Translational motion

Quantum dots
9
...
5 introduced the quantum mechanical description of free motion in one
dimension
...
1a)

@=−

$2

theory

d2

2m dx 2

(9
...
1 are

ψk = Aeikx + Be−ikx

Time-independent
perturbation theory

9
...
9

Further information 9
...
2)

Note that we are now labelling both the wavefunctions and the energies (that is,
the eigenfunctions and eigenvalues of @) with the index k
...
2: Perturbation
theory
Discussion questions
Exercises
Problems

278

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
¥

Potential energy, V

¥

0
0
Wall

L

x
Wall

A particle in a one-dimensional
region with impenetrable walls
...


Fig
...
1

functions are solutions by substituting ψk into the left-hand side of eqn 9
...
In this case, all values of k, and therefore all
values of the energy, are permitted
...

We saw in Section 8
...
That
is, eikx is an eigenfunction of the operator Yx with eigenvalue +k$, and e−ikx is an eigenfunction with eigenvalue −k$
...
This conclusion is consistent with the uncertainty principle because, if the momentum is certain, then the
position cannot be specified (the operators X and Yx do not commute, Section 8
...

9
...
9
...
This model is an idealization of
the potential energy of a gas-phase molecule that is free to move in a one-dimensional
container
...
The
particle in a box is also used in statistical thermodynamics in assessing the contribution of the translational motion of molecules to their thermodynamic properties
(Chapter 16)
...
1), so the general solutions given in eqn 9
...
However, we can us e±ix = cos x ± i sin x to write

ψk = Aeikx + Be−ikx = A(cos kx + i sin kx) + B(cos kx − i sin kx)
= (A + B) cos kx + (A − B)i sin kx
If we absorb all numerical factors into two new coefficients C and D, then the general
solutions take the form

ψk(x) = C sin kx + D cos kx

Ek =

k 2$2
2m

(9
...
However, when the particle is confined within a region, the acceptable wavefunctions must
satisfy certain boundary conditions, or constraints on the function at certain locations
...
This behaviour is consistent with
the fact that it is physically impossible for the particle to be found with an infinite
potential energy
...
The continuity of the wavefunction then requires it to vanish just
inside the well at x = 0 and x = L
...
These boundary conditions imply quantization, as we show in the following Justification
...
1 A PARTICLE IN A BOX
Justification 9
...
The permitted wavelengths satisfy
1
L = n × –λ
2

n = 1, 2,
...

n
According to the de Broglie relation, these wavelengths correspond to the momenta
h nh
p= =
λ 2L
The particle has only kinetic energy inside the box (where V = 0), so the permitted
energies are
p2
n2h2
E=
=
with n = 1, 2,
...
Consider the wall at
x = 0
...
3, ψ (0) = D (because sin 0 = 0 and cos 0 = 1)
...
It follows that the wavefunction must be of the form
ψk(x) = C sin kx
...
Taking C = 0 would give ψk(x) = 0 for all x, which would conflict
with the Born interpretation (the particle must be somewhere)
...


The value n = 0 is ruled out, because it implies k = 0 and ψk(x) = 0 everywhere (because
sin 0 = 0), which is unacceptable
...
The wavefunctions are therefore

ψn(x) = C sin(nπx/L)

n = 1, 2,
...
)
Because k and Ek are related by eqn 9
...


We conclude that the energy of the particle in a one-dimensional box is quantized
and that this quantization arises from the boundary conditions that ψ must satisfy if
it is to be an acceptable wavefunction
...
So far, only energy has been quantized; shortly
we shall see that other physical observables may also be quantized
...
To do so, we look for the value of C that ensures
that the integral of ψ 2 over all the space available to the particle (that is, from x = 0 to
x = L) is equal to 1:

Ύ

L

0

Ύ

L

ψ 2 dx = C 2

0

sin2

nπx
L

= C2 ×

L
2

= 1,

so C =

A 2 D 1/2
C LF

279

280

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
n
10
Classically
allowed
energies
9

100

81

for all n
...
4b)

2

2

5

for 0 ≤ x ≤ L

6

25

E /(h /8mL )

7

A 2D
A nπx D
sin
C LF
C L F

8

49

(9
...


8mL2

0

The allowed energy levels for a
particle in a box
...


Fig
...
2

54 3 2

1

y
x
The first five normalized
wavefunctions of a particle in a box
...


Fig
...
3

Exploration Plot the probability
density for a particle in a box with
n = 1, 2,
...
How do your
plots illustrate the correspondence
principle?

Self-test 9
...
Hint
...


The energies and wavefunctions are labelled with the ‘quantum number’ n
...
For a particle in a box there is an infinite number of acceptable solutions, and the quantum number n specifies the one of interest (Fig
...
2)
...
4)
...
3 shows some of the wavefunctions of a particle in a box: they are all sine
functions with the same amplitude but different wavelengths
...
Note that the number of nodes (points
where the wavefunction passes through zero) also increases as n increases, and that the
wavefunction ψn has n − 1 nodes
...

The linear momentum of a particle in a box is not well defined because the wavefunction sin kx is a standing wave and, like the example of cos kx treated in Section
8
...
However, each wavefunction is a superposition of momentum eigenfunctions:

A 2 D 1/2 nπx 1 A 2 D 1/2 ikx −ikx
ψn =
sin
=
(e − e )
C LF
L
2i C L F

k=

nπ
L

(9
...
This detection of opposite directions of travel with equal probability is the quantum mechanical version of
the classical picture that a particle in a box rattles from wall to wall, and in any given
period spends half its time travelling to the left and half travelling to the right
...
2 What is (a) the average value of the linear momentum of a particle in

a box with quantum number n, (b) the average value of p2?
[(a) ͗p͘ = 0, (b) ͗p2͘ = n2h2/4L2]
Comment 9
...
1 A PARTICLE IN A BOX
E1 =

h2

(9
...
The physical origin
of the zero-point energy can be explained in two ways
...
Hence it has nonzero kinetic energy
...

The separation between adjacent energy levels with quantum numbers n and n + 1 is
(n + 1)2h2 n2h2
h2
En+1 − En =
−
= (2n + 1)
(9
...
The separation of adjacent levels
becomes zero when the walls are infinitely far apart
...
The translational energy of completely free
particles (those not confined by walls) is not quantized
...
1 Accounting for the electronic absorption spectra of polyenes

β-Carotene (1) is a linear polyene in which 10 single and 11 double bonds alternate
along a chain of 22 carbon atoms
...
294 nm
...
From eqn 9
...
60 × 10

−19

J

(6
...
110 × 10−31 kg) × (2
...
10, ∆E = hν) that the frequency of radiation required to cause this transition is
∆E

1
...
41 × 1014 s−1
h 6
...
03 × 1014 s−1 (λ = 497 nm), corresponding to radiation in the visible range of the electromagnetic spectrum
...
3 Estimate a typical nuclear excitation energy by calculating the first

excitation energy of a proton confined to a square well with a length equal to the
diameter of a nucleus (approximately 1 fm)
...
6 GeV]

281

282

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
The probability density for a particle in a box is

n=1
Wall

Wall

n=2

(a)

n=1
n=2

(b)

n=2

ψ 2(x) =

2
L

sin2

nπx

(9
...
The nonuniformity is pronounced when n is small (Fig
...
4),
but—provided we ignore the increasingly rapid oscillations—ψ 2(x) becomes more
uniform as n increases
...
That the quantum result corresponds to the classical prediction at high
quantum numbers is an illustration of the correspondence principle, which states
that classical mechanics emerges from quantum mechanics as high quantum numbers are reached
...
1 Using the particle in a box solutions

(c)

n=1

(a) The first two wavefunctions,
(b) the corresponding probability
distributions, and (c) a representation of
the probability distribution in terms of the
darkness of shading
...
9
...
What is the probability, P, of locating the
electron between x = 0 (the left-hand end of a molecule) and x = 0
...
0 nm?

ψ 2dx is the probability of finding the particle in the small
region dx located at x ; therefore, the total probability of finding the electron in the
specified region is the integral of ψ 2dx over that region
...
4b with n = 1
...
2 nm, which gives P = 0
...
The result corresponds to a
chance of 1 in 20 of finding the electron in the region
...

Self-test 9
...
25L and x = 0
...

[0
...
5
...
Specifically,
the functions ψn and ψn′ are orthogonal if

Ύψ *ψ dτ = 0
n

n′

(9
...
A general feature of quantum mechanics,
which we prove in the Justification below, is that wavefunctions corresponding to different
energies are orthogonal; therefore, we can be confident that all the wavefunctions of a
particle in a box are mutually orthogonal
...
1
...
2 MOTION IN TWO AND MORE DIMENSIONS

283

Justification 9
...
Then we can write
@ψn = Enψn

@ψm = Emψm

Now multiply the first of these two Schrödinger equations by ψ m and the second by
*
ψ n and integrate over all space:
*

Ύψ * @ψ dτ = E Ύψ * ψ dτ Ύψ *@ψ dτ = E Ύψ *ψ dτ
m

n

n

m n

n

m

m

n m

Next, noting that the energies themselves are real, form the complex conjugate of
the second expression (for the state m) and subtract it from the first expression (for
the state n):
A

D*

Ύψ * @ψ dτ − BC Ύψ *@ψ dτEF
m

n

n

m

Ύ

Ύ

= En ψ m ψndτ − Em ψnψ m dτ
*
*

By the hermiticity of the hamiltonian (Section 8
...

Two functions are orthogonal if the
integral of their product is zero
...
The integral is
equal to the total area beneath the graph of
the product, and is zero
...
9
...
2 Verifying the orthogonality of the wavefunctions for a particle in a box

We can verify the orthogonality of wavefunctions of a particle in a box with n = 1
and n = 3 (Fig
...
5):

Ύ

L

ψ 1 ψ 3dx =
*

0

sin
sin
LΎ
L
2

L

πx

0

3πx
L

dx = 0

The property of orthogonality is of great importance in quantum mechanics
because it enables us to eliminate a large number of integrals from calculations
...
Sets of functions that are normalized and mutually
orthogonal are called orthonormal
...
4b are orthonormal
...
2
...
2 Motion in two and more dimensions
Next, we consider a two-dimensional version of the particle in a box
...
9
...

The wavefunction is now a function of both x and y and the Schrödinger equation is
−

$2 A ∂2ψ
2m C ∂x 2

+

∂2ψ D
∂y 2 F

= Eψ

(9
...
The
particle is confined to the plane bounded
by impenetrable walls
...


Fig
...
6

284

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
We need to see how to solve this partial differential equation, an equation in more
than one variable
...
An important application of this procedure, as we
shall see, is the separation of the Schrödinger equation for the hydrogen atom into
equations that describe the radial and angular variation of the wavefunction
...
10 can be found by writing the wavefunction as a
product of functions, one depending only on x and the other only on y:

ψ (x,y) = X(x)Y(y)
With this substitution, we show in the Justification below that eqn 9
...
11)

The quantity EX is the energy associated with the motion of the particle parallel to the
x-axis, and likewise for EY and motion parallel to the y-axis
...
3 The separation of variables technique applied to the particle in a
two-dimensional box

The first step in the justification of the separability of the wavefunction into the
product of two functions X and Y is to note that, because X is independent of y and
Y is independent of x, we can write
∂2ψ
∂x 2

=

∂2XY
∂x 2

=Y

d2X

∂2ψ

dx 2

∂y 2

=

∂2XY
∂y 2

=X

d2Y
dy 2

Then eqn 9
...
But the sum of these two terms is a constant given by the right-hand side
of the equation; therefore, even the second term cannot change when y is changed
...
By a similar
argument, the first term is a constant when x changes, and we write it −2mEX /$2,
and E = EX + EY
...
11
...
2 MOTION IN TWO AND MORE DIMENSIONS

285

The wavefunctions for a particle
confined to a rectangular surface depicted
as contours of equal amplitude
...

Fig
...
7

+
+

(a)

-

-

+

+

-

+

-

(b)

(c)

(d)

Each of the two ordinary differential equations in eqn 9
...
We can therefore adapt the results in
eqn 9
...
12a)

2
A n2 n2 D h2
1
En1n2 = 2 + 2
C L1 L2 F 8m

with the quantum numbers taking the values n1 = 1, 2,
...
independently
...
9
...
They are the two-dimensional
versions of the wavefunctions shown in Fig
...
3
...

We treat a particle in a three-dimensional box in the same way
...
Solution of the Schrödinger equation by the separation of variables technique
then gives

ψn1,n2,n3(x,y,z) =

A 8 D 1/2 n1πx
n2πy
n3πz
sin
sin
sin
C L1L 2L 3 F
L1
L2
L3
0 ≤ x ≤ L1, 0 ≤ y ≤ L2, 0 ≤ z ≤ L3 (9
...
Then eqn 9
...
13)

Exploration Use mathematical
software to generate threedimensional plots of the functions in this
illustration
...


286

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
Consider the cases n1 = 1, n2 = 2 and n1 = 2, n2 = 1:

ψ1,2 =

-

+

ψ2,1 =

(a)
M

+

(b)
M

The wavefunctions for a particle
confined to a square surface
...

The two functions correspond to the same
energy
...


Wavefunction

Fig
...
8

E

V
x

A particle incident on a barrier from
the left has an oscillating wave function,
but inside the barrier there are no
oscillations (for E < V)
...
(Only the real component of
the wavefunction is shown
...
9
...
In this case, in which there are two degenerate wavefunctions, we say that the energy level 5(h2/8mL2) is ‘doubly degenerate’
...
Figure 9
...
Because the
box is square, we can convert one wavefunction into the other simply by rotating the
plane by 90°
...
Similar arguments account
for the degeneracy of states in a cubic box
...
4b)
...
3 Tunnelling
If the potential energy of a particle does not rise to infinity when it is in the walls of the
container, and E < V, the wavefunction does not decay abruptly to zero
...
9
...

Hence the particle might be found on the outside of a container even though according
to classical mechanics it has insufficient energy to escape
...

The Schrödinger equation can be used to calculate the probability of tunnelling of
a particle of mass m incident on a finite barrier from the left
...
2 we can
write

ψ = Aeikx + Be−ikx

k$ = (2mE)1/2

(9
...
15)

We shall consider particles that have E < V (so, according to classical physics, the
particle has insufficient energy to pass over the barrier), and therefore V − E is
positive
...
16)

as we can readily verify by differentiating ψ twice with respect to x
...
To the right of the barrier (x > L), where V = 0 again, the wavefunctions are

9
...
17)

Incident wave

The complete wavefunction for a particle incident from the left consists of
an incident wave, a wave reflected from the barrier, the exponentially changing
amplitudes inside the barrier, and an oscillating wave representing the propagation of
the particle to the right after tunnelling through the barrier successfully (Fig
...
10)
...
4b
...
19)

At this stage, we have four equations for the six unknown coefficients
...
Therefore, we can set B′ = 0, which removes one more
unknown
...

The probability that a particle is travelling towards positive x (to the right) on the
left of the barrier is proportional to | A |2, and the probability that it is travelling to the
right on the right of the barrier is | A′ |2
...
After some algebra (see Problem 9
...
9
...


V

(9
...
This function is plotted in Fig
...
12; the transmission coefficient for
E > V is shown there too
...
20a
>
simplifies to
T ≈ 16ε (1 − ε)e−2κ L

(9
...
It follows that particles of low mass are more able to tunnel through
barriers than heavy ones (Fig
...
13)
...


0
...
18)

Their slopes (their first derivatives) must also be continuous there (Fig
...
11):

287

0
...
9
...
The conditions for continuity
enable us to connect the wavefunctions in
the three zones and hence to obtain
relations between the coefficients that
appear in the solutions of the Schrödinger
equation
...
4

x

0
...
6
0
...
2
0
...
2 0
...
6 0
...
0
Incident energy, E /V

10
0
...
9
...
The horizontal
axis is the energy of the incident particle
expressed as a multiple of the barrier
height
...
The graph on the left
is for E < V and that on the right for E > V
...
However, T < 1 for E > V,
whereas classically T would be 1
...


288

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
Heavy Light
particle particle

Wavefunction

n=2

V
n=1
V
0

x
Fig
...
13 The wavefunction of a heavy
particle decays more rapidly inside a
barrier than that of a light particle
...


Fig
...
14

0

L

x

A potential well with a finite depth
...
9
...


Fig
...
15

A number of effects in chemistry (for example, the isotope-dependence of some
reaction rates) depend on the ability of the proton to tunnel more readily than the
deuteron
...
Tunnelling of protons between acidic and basic groups is also an
important feature of the mechanism of some enzyme-catalysed reactions
...

A problem related to the one just considered is that of a particle in a square-well
potential of finite depth (Fig
...
14)
...
The wavefunctions are found by ensuring, as in the discussion of tunnelling,
that they and their slopes are continuous at the edges of the potential
...
9
...
A further difference from the solutions
for an infinitely deep well is that there is only a finite number of bound states
...

Detailed consideration of the Schrödinger equation for the problem shows that in
general the number of levels is equal to N, with
N−1<

(8mVL)1/2
h


(9
...
We see that the deeper and wider the well, the greater the number of bound states
...

IMPACT ON NANOSCIENCE

I9
...
The future economic impact of nanotechnology could
be very significant
...
1 IMPACT ON NANOSCIENCE: SCANNING PROBE MICROSCOPY
devices has driven the design of ever smaller and more powerful microprocessors
...
As the
ability to process data increases with the number of circuits in a chip, it follows that
soon chips and the devices that use them will have to become bigger if processing
power is to increase indefinitely
...

We will explore several concepts of nanoscience throughout the text
...

Consequently, SPM has far better resolution than electron microscopy (Impact I8
...

One modality of SPM is scanning tunnelling microscopy (STM), in which a platinum–
rhodium or tungsten needle is scanned across the surface of a conducting solid
...
9
...
In the constant-current mode of operation, the stylus
moves up and down corresponding to the form of the surface, and the topography of
the surface, including any adsorbates, can be mapped on an atomic scale
...
In the constant-z
mode, the vertical position of the stylus is held constant and the current is monitored
...

Figure 9
...
Each ‘bump’ on the surface corresponds to an atom
...

In atomic force microscopy (AFM) a sharpened stylus attached to a cantilever is
scanned across the surface
...
9
...
The deflection is
monitored either by interferometry or by using a laser beam
...
A spectacular demonstration of the power of AFM is given in
Fig
...
19, which shows individual DNA molecules on a solid surface
...
9
...

That current is very sensitive to the
distance of the tip above the surface
...

Fig
...
17

Fig
...
18 In atomic force microscopy, a laser
beam is used to monitor the tiny changes in
the position of a probe as it is attracted to
or repelled from atoms on a surface
...
9
...
(Courtesy of
Veeco Instruments
...
2 Exploring the origin of the current in scanning tunnelling microscopy

To get an idea of the distance dependence of the tunnelling current in STM,
suppose that the wavefunction of the electron in the gap between sample and
needle is given by ψ = Be−κx, where κ = {2me(V − E)/$2}1/2; take V − E = 2
...

By what factor would the current drop if the needle is moved from L1 = 0
...
60 nm from the surface?
Method We regard the tunnelling current to be proportional to the transmission

probability T, so the ratio of the currents is equal to the ratio of the transmission
probabilities
...
20a or 9
...
20b
...
50 nm and V − E = 2
...
20 × 10−19 J the value of κ L is

1 2me(V − E) 51/2
6 L1
κ L1 = 2
$2
3
7
1 2 × (9
...
20 × 10−19 J) 51/2
6 × (5
...
054 × 10−34 J s)2
3
7
= (7
...
0 × 10−10 m) = 3
...
20b to calculate the transmission probabilities at the
two distances
...
0×10−10 m)

= e−2×(7
...
23

We conclude that, at a distance of 0
...
50 nm
...
5 The ability of a proton to tunnel through a barrier contributes to the

rapidity of proton transfer reactions in solution and therefore to the properties of
acids and bases
...
0 eV (1
...
9 eV
...
7 × 102; we expect proton transfer reactions to be much
faster than deuteron transfer reactions
...
22)

where k is the force constant: the stiffer the ‘spring’, the greater the value of k
...
22 corresponds to a potential energy
1
V = – kx 2
2

(9
...
9
...
The Schrödinger equation for the particle is therefore
$2 d2ψ
2m dx 2

1
+ – kx 2ψ = Eψ
2

(9
...
4 The energy levels
Equation 9
...

Quantization of energy levels arises from the boundary conditions: the oscillator will
not be found with infinitely large compressions or extensions, so the only allowed
solutions are those for which ψ = 0 at x = ±∞
...


(9
...
It
follows that the separation between adjacent levels is
Ev+1 − Ev = $ω

Fig
...
20 The parabolic potential energy
1
V = –kx 2 of a harmonic oscillator, where x
2
is the displacement from equilibrium
...


(9
...
Therefore, the energy levels form a uniform ladder of
spacing $ω (Fig
...
21)
...

Because the smallest permitted value of v is 0, it follows from eqn 9
...
27)

The mathematical reason for the zero-point energy is that v cannot take negative
values, for if it did the wavefunction would be ill-behaved
...
We can picture this zero-point state as one in which the
particle fluctuates incessantly around its equilibrium position; classical mechanics
would allow the particle to be perfectly still
...
3 Calculating a molecular vibrational absorption frequency

Atoms vibrate relative to one another in molecules with the bond acting like a
spring
...
That is, only the H atom moves, vibrating as a
simple harmonic oscillator
...
25 describes the allowed vibrational
energy levels of a X-H bond
...
For example k = 516
...
Because
the mass of a proton is about 1
...
25 gives
ω ≈ 5
...
4 × 102 THz)
...
26 that the separation of
adjacent levels is $ω ≈ 5
...
36 eV)
...
From eqn 9
...
2 eV, or 15 kJ mol−1
...
4 THE ENERGY LEVELS

7
6
5
4
3

hw

2
1

0
0
Displacement, x
Fig
...
21 The energy levels of a harmonic
oscillator are evenly spaced with separation
$ω, with ω = (k/m)1/2
...


292

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
The excitation of the vibration of the bond from one level to the level immediately
above requires 57 zJ
...
5 µm
...
We shall see in Chapter 13 that the
concepts just described represent the starting point for the interpretation of vibrational spectroscopy, an important technique for the characterization of small and
large molecules in the gas phase or in condensed phases
...
5 The wavefunctions

1

It is helpful at the outset to identify the similarities between the harmonic oscillator
and the particle in a box, for then we shall be able to anticipate the form of the oscillator
wavefunctions without detailed calculation
...
9
...
20)
...

First, because the potential energy climbs towards infinity only as x 2 and not abruptly,
the wavefunction approaches zero more slowly at large displacements than for the
particle in a box
...


exp(-x 2)

0
...
6

0
...
2
0
-2

-1

0
x

1

2

Fig
...
22 The graph of the Gaussian
2
function, f(x) = e−x
...
24 shows that the wavefunction for a harmonic oscillator has the form

ψ (x) = N × (polynomial in x) × (bell-shaped Gaussian function)
where N is a normalization constant
...
9
...
The precise form of the wavefunctions are

Comment 9
...

They satisfy the recursion relation
HV+1 − 2yHV + 2VHV−1 = 0
An important integral is
∞

10
if V′ ≠ V
2
HV′ HVe−y dy = 2
3 π1/22VV! if V′ = V
−∞

Ύ

Hermite polynomials are members of a
class of functions called orthogonal
polynomials
...
See Further reading for a reference
to their properties
...
28)

The factor Hv(y) is a Hermite polynomial (Table 9
...
For instance, because H0(y) =
1, the wavefunction for the ground state (the lowest energy state, with v = 0) of the
harmonic oscillator is

ψ0(x) = N0e−y /2 = N0e−x /2α
2

2

2

(9
...
29b)

The wavefunction and the probability distribution are shown in Fig
...
23
...

The wavefunction for the first excited state of the oscillator, the state with v = 1, is
obtained by noting that H1(y) = 2y (note that some of the Hermite polynomials are
very simple functions!):

ψ1(x) = N1 × 2ye−y /2
2

(9
...
5 THE WAVEFUNCTIONS

293

Table 9
...
0
The normalized wavefunction and
probability distribution (shown also by
shading) for the lowest energy state of a
harmonic oscillator
...


0 1
23

Fig
...
24

This function has a node at zero displacement (x = 0), and the probability density has
maxima at x = ±α, corresponding to y = ±1 (Fig
...
24)
...

In the case of the harmonic oscillator wavefunctions in eqn 9
...
The Gaussian function goes very strongly to zero as the displacement increases
(in either direction), so all the wavefunctions approach zero at large displacements
...
The exponent y 2 is proportional to x 2 × (mk)1/2, so the wavefunctions decay
more rapidly for large masses and stiff springs
...
As v increases, the Hermite polynomials become larger at large displacements
(as x v), so the wavefunctions grow large before the Gaussian function damps them
down to zero: as a result, the wavefunctions spread over a wider range as v increases
...
9
...
The shading in Fig
...
26
that represents the probability density is based on the squares of these functions
...
5
Wavefunction, y

Fig
...
23

0

-0
...
0

-4

-2

0
y

2

4

Fig
...
25 The normalized wavefunctions for
the first five states of a harmonic oscillator
...


20
4
3
2
1
0

Fig
...
26 The probability distributions for
the first five states of a harmonic oscillator
and the state with v = 20
...


Exploration To gain some insight into
the origins of the nodes in the
harmonic oscillator wavefunctions, plot the
Hermite polynomials Hv(y) for v = 0
through 5
...
We see classical properties emerging in the correspondence
limit of high quantum numbers, for a classical particle is most likely to be found at the
turning points (where it travels most slowly) and is least likely to be found at zero displacement (where it travels most rapidly)
...
3 Normalizing a harmonic oscillator wavefunction

Find the normalization constant for the harmonic oscillator wavefunctions
...
16
...
In this one-dimensional problem, the volume
element is dx and the integration is from −∞ to +∞
...
The integrals required are given in
Comment 9
...

Answer The unnormalized wavefunction is

ψv(x) = Hv(y)e−y /2
2

It follows from the integrals given in Comment 9
...
1
...


ψ0 and ψ1 are
orthogonal
...
2
...
6 Confirm, by explicit evaluation of the integral, that

(b) The properties of oscillators

With the wavefunctions that are available, we can start calculating the properties of
a harmonic oscillator
...
31)

−∞

(Here and henceforth, the wavefunctions are all taken to be normalized to 1
...
For instance, we show in the
following example that the mean displacement, ͗x͘, and the mean square displacement, ͗x 2͘, of the oscillator when it is in the state with quantum number v are
͗x͘ = 0

1
͗x 2͘ = (v + –)
2

$
(mk)1/2

(9
...
5 THE WAVEFUNCTIONS

295

The result for ͗x͘ shows that the oscillator is equally likely to be found on either side
of x = 0 (like a classical oscillator)
...
This increase is apparent from the probability densities in
Fig
...
26, and corresponds to the classical amplitude of swing increasing as the oscillator becomes more highly excited
...
4 Calculating properties of a harmonic oscillator

We can imagine the bending motion of a CO2 molecule as a harmonic oscillation
relative to the linear conformation of the molecule
...
Calculate the mean displacement of the
oscillator when it is in a quantum state v
...


The operator for position along x is multiplication by the value of x (Section 8
...

The resulting integral can be evaluated either by inspection (the integrand is the
product of an odd and an even function), or by explicit evaluation using the
formulas in Comment 9
...
To give practice in this type of calculation, we illustrate
the latter procedure
...

∞

͗x͘ =

Ύ

∞

Ύ

ψ *xψvdx = N 2
v
v

−∞

(Hve−y /2)x(Hve−y /2)dx
2

2

−∞

∞

Ύ
=α N Ύ
= α 2N 2
v
2

2
v

(Hve−y /2)y(Hve−y /2)dy
2

2

−∞
∞

Hv yHve−y dy
2

−∞

Now use the recursion relation (see Comment 9
...
As remarked in the text, the mean displacement
is zero because the displacement occurs equally on either side of the equilibrium
position
...

Self-test 9
...
(Use the recursion relation twice
...
32]

1
The mean potential energy of an oscillator, the expectation value of V = –kx 2, can
2
now be calculated very easily:
1
1
1
͗V͘ = ͗ –kx 2͘ = –(v + –)$
2
2
2

A k D 1/2 1
1
= –(v + –)$ω
2
2
C mF

Comment 9
...
The product of an
odd and even function is itself odd, and
the integral of an odd function over a
symmetrical range about x = 0 is zero
...
33)

296

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
1
Because the total energy in the state with quantum number v is (v + –)$ω, it follows
2
that
1
͗V͘ = –Ev
2

(9
...
34b)

The result that the mean potential and kinetic energies of a harmonic oscillator are
equal (and therefore that both are equal to half the total energy) is a special case of the
virial theorem:
If the potential energy of a particle has the form V = ax b, then its mean potential
and kinetic energies are related by
2͗EK͘ = b͗V͘

(9
...
The virial theorem is
a short cut to the establishment of a number of useful results, and we shall use it again
...
For example, it follows from
the shape of the wavefunction (see the Justification below) that in its lowest energy
state there is about an 8 per cent chance of finding an oscillator stretched beyond its
classical limit and an 8 per cent chance of finding it with a classically forbidden compression
...
The probability of being found in classically forbidden regions
decreases quickly with increasing v, and vanishes entirely as v approaches infinity, as
we would expect from the correspondence principle
...
Molecules,
however, are normally in their vibrational ground states, and for them the probability
is very significant
...
4 Tunnelling in the quantum mechanical harmonic oscillator

According to classical mechanics, the turning point, xtp, of an oscillator occurs when
1
its kinetic energy is zero, which is when its potential energy – kx 2 is equal to its total
2
energy E
...
25
...
6 ROTATION IN TWO DIMENSIONS: A PARTICLE ON A RING
∞

∞

P=

Ύ

297

2
2
ψ 0 dx = α N 0

Ύe

−y 2

Table 9
...
01

0
...
05

0
...
10

0
...
50

0
...
00

0
...
50

0
...
00

0
...
2
...
843) = 0
...
9 per cent of a large number of observations, any oscillator in the
state v = 0 will be found stretched to a classically forbidden extent
...

The total probability of finding the oscillator tunnelled into a classically forbidden
region (stretched or compressed) is about 16 per cent
...


Rotational motion
The treatment of rotational motion can be broken down into two parts
...

It may be helpful to review the classical description of rotational motion given in
Appendix 3, particularly the concepts of moment of inertia and angular momentum
...
6 Rotation in two dimensions: a particle on a ring

Jz

We consider a particle of mass m constrained to move in a circular path of radius r in
the xy-plane (Fig
...
27)
...
We can therefore write E = p2/2m
...
Because mr 2 is the moment of
inertia, I, of the mass on its path, it follows that
E=

2
Jz

2I

(9
...

(a) The qualitative origin of quantized rotation

Because Jz = ±pr, and, from the de Broglie relation, p = h/λ , the angular momentum
about the z-axis is
Jz = ±

hr

λ

Opposite signs correspond to opposite directions of travel
...
It follows that, if we can see why the

x

p

r
m

y

Fig
...
27 The angular momentum of a
particle of mass m on a circular path of
radius r in the xy-plane is represented by
a vector J with the single non-zero
component Jz of magnitude pr
perpendicular to the plane
...

Suppose for the moment that λ can take an arbitrary value
...
9
...
When φ increases
beyond 2π, the wavefunction continues to change, but for an arbitrary wavelength it
gives rise to a different value at each point, which is unacceptable (Section 8
...
An
acceptable solution is obtained only if the wavefunction reproduces itself on successive circuits, as in Fig
...
28b
...
Hence, the energy of the particle is quantized
...
The value ml = 0 corresponds to λ = ∞; a ‘wave’ of infinite wavelength has
a constant height at all values of φ
...
The
circumference has been opened out into
a straight line; the points at φ = 0 and 2π
are identical
...
Moreover, on successive circuits it
interferes destructively with itself, and does
not survive
...

Fig
...
28

Jz = ±

hr

λ

=

ml hr
2πr

=

ml h
2π

where we have allowed ml to have positive or negative values
...


(9
...
9
...
It then follows from eqn 9
...
38a)

We shall see shortly that the corresponding normalized wavefunctions are

ψml(φ) =
ml > 0

(a)

(b)

ml < 0

Fig
...
29 The angular momentum of a
particle confined to a plane can be
represented by a vector of length |ml | units
along the z-axis and with an orientation
that indicates the direction of motion of
the particle
...


eimlφ
(2π)1/2

(9
...

We have arrived at a number of conclusions about rotational motion by combining
some classical notions with the de Broglie relation
...
However, to be sure that the correct solutions have been
obtained, and to obtain practice for more complex problems where this less formal
approach is inadequate, we need to solve the Schrödinger equation explicitly
...

Justification 9
...
10:
@=−

$2 A ∂2
∂2 D
B 2 + 2E
2m C ∂x
∂y F

9
...
It is always a good idea to use coordinates that reflect the full symmetry of
the system, so we introduce the coordinates r and φ (Fig
...
30), where x = r cos φ and
y = r sin φ
...
39)

r 2 ∂φ 2

r
x

f

y

d2

2mr 2 dφ 2

The moment of inertia I = mr 2 has appeared automatically, so H may be written
@=−

z

2

However, because the radius of the path is fixed, the derivatives with respect to r can
be discarded
...
40)

2I dφ 2

Fig
...
30 The cylindrical coordinates z, r,
and φ for discussing systems with axial
(cylindrical) symmetry
...


and the Schrödinger equation is
d2ψ
dφ

2

=−

2IE
$2

ψ

(9
...
42)

|ml| = 2

The quantity ml is just a dimensionless number at this stage
...
That is, the
wavefunction ψ must satisfy a cyclic boundary condition, and match at points
separated by a complete revolution: ψ (φ + 2π) = ψ (φ)
...
43)

Because we require (−1) = 1, 2ml must be a positive or a negative even integer
(including 0), and therefore ml must be an integer: ml = 0, ±1, ±2,
...
The energy is quantized and
restricted to the values given in eqn 9
...
The occurrence of ml as its
square means that the energy of rotation is independent of the sense of rotation (the
sign of ml), as we expect physically
...
Although the result
has been derived for the rotation of a single mass point, it also applies to any body of
moment of inertia I constrained to rotate about one axis
...
37 (Jz = ml$)
...
9
...
As shown in the following

ml = 0
Fig
...
31 The real parts of the wavefunctions
of a particle on a ring
...


Comment 9
...


300

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS

Comment 9
...
It follows that the
z-component of the angular momentum
has a magnitude given by eqn 9
...
For
more information on vectors, see
Appendix 2
...
5
...
6 The quantization of angular momentum

In the discussion of translational motion in one dimension, we saw that the opposite signs in the wavefunctions eikx and e−ikx correspond to opposite directions of
travel, and that the linear momentum is given by the eigenvalue of the linear
momentum operator
...
In classical mechanics the
orbital angular momentum lz about the z-axis is defined as
lz = xpy − ypx

[9
...
The operators for the two linear momentum components are given in eqn 8
...
45)

When expressed in terms of the coordinates r and φ, by standard manipulations this
equation becomes
Zz =

Angular
momentum

$ ∂

(9
...
38b
...
47)

That is, ψml is an eigenfunction of Zz, and corresponds to an angular momentum ml$
...
These features are the origin of the vector representation of
angular momentum, in which the magnitude is represented by the length of a vector and the direction of motion by its orientation (Fig
...
32)
...
9
...


To locate the particle given its wavefunction in eqn 9
...
9
...
Hence the
location of the particle is completely indefinite, and knowing the angular momentum
precisely eliminates the possibility of specifying the particle’s location
...
6), and the inability to specify them simultaneously with arbitrary precision is another example of the uncertainty principle
...
7 ROTATION IN THREE DIMENSIONS: THE PARTICLE ON A SPHERE
9
...
We shall need the results of this calculation when we come to
describe rotating molecules and the states of electrons in atoms and in small clusters
of atoms
...
9
...


Im y

f
2p

0

(a) The Schrödinger equation
0 2p

The hamiltonian for motion in three dimensions (Table 8
...
48)

The symbol ∇2 is a convenient abbreviation for the sum of the three second derivatives; it is called the laplacian, and read either ‘del squared’ or ‘nabla squared’
...
The wavefunction is therefore a function of the colatitude, θ,
and the azimuth, φ (Fig
...
35), and we write it ψ (θ,φ)
...
9
...


q

(9
...
50)

where Θ is a function only of θ and Φ is a function only of φ
...
7 The separation of variables technique applied to the particle on
a sphere

The laplacian in spherical polar coordinates is (see Further reading)
∇2 =

∂2
∂r

2

+

2 ∂

+

r ∂r

1
r2

Λ2

Fig
...
34 The wavefunction of a particle on
the surface of a sphere must satisfy two
cyclic boundary conditions; this
requirement leads to two quantum
numbers for its state of angular
momentum
...
51a)

z

q

where the legendrian, Λ2, is
Λ =
2

1

∂2

+

sin2θ ∂φ 2

1

∂

sin θ ∂θ

sin θ

∂
∂θ

r

(9
...
9
...
For a
particle confined to the surface of a sphere,
only the colatitude, θ, and the azimuth, φ,
can change
...
3 The spherical harmonics

1

ml

Yl,ml(θ,ϕ)

0

0

A 1 D 1/2
B E
C 4π F

1

0

A 3 D 1/2
B E cos θ
C 4π F

∂φ

2

2

+

∂

1

sin θ ∂θ

sin θ

∂(ΦΘ)
∂θ

= −εΘΦ

We now use the fact that Θ and Φ are each functions of one variable, so the partial
derivatives become complete derivatives:

1/2

±1

A 3D
,B E
C 8π F

0

A 5 D 1/2
B
E (3 cos2θ − 1)
C 16π F

±1

A 15 D 1/2
, B E cos θ sin θ e±iφ
C 8π F

±2

A 15 D 1/2 2 ±2iφ
B
E sin θ e
C 32π F

0

3

∂2(ΘΦ)

sin θ

l

2

To verify that this expression is separable, we substitute ψ = ΘΦ :

A 7 D 1/2
B
E (5 cos3θ − 3 cos θ)
C 16π F

sin θ e±iφ

1/2

±1

A 21 D
E
,B
C 64π F

±2

A 105 D 1/2 2
B
E sin θ cos θ e±2iφ
C 32π F

±3

A 35 D 1/2 3 ±3iφ
E sin θ e
,B
C 64π F

(5 cos2θ − 1)sin θ e ±iφ

Θ

d2Φ

sin θ dφ
2

2

+

Φ

d

sin θ dθ

sin θ

dΘ
dθ

= −εΘΦ

Division through by ΘΦ, multiplication by sin2θ, and minor rearrangement gives

Φ

d2Φ
dφ

2

+

sin θ d

Θ dθ

sin θ

dΘ
dθ

+ ε sin2θ = 0

The first term on the left depends only on φ and the remaining two terms depend
only on θ
...
3), and by the same argument, the complete equation can be
separated
...
5, so it has the
same solutions (eqn 9
...
The second is much more complicated to solve, but the
solutions are tabulated as the associated Legendre functions
...
The presence of the quantum number ml in the
second equation implies, as we see below, that the range of acceptable values of ml is
restricted by the value of l
...
6

The spherical harmonics are orthogonal
and normalized in the following sense:
π 2π

ΎΎ
0

Yl′,ml′(θ,φ)*Yl,ml(θ,φ) sin θ dθ dφ

0

= δl′lδml′ml
An important ‘triple integral’ is
π 2π

ΎΎ

0 0

Yl′,ml″(θ,φ)*Yl′,ml′(θ,φ)Yl,ml (θ,φ)

sin θ dθ dφ = 0
unless ml″ = m′ + ml and we can form a
l
triangle with sides of lengths l″, l′, and l
(such as 1, 2, and 3 or 1, 1, and 1, but not
1, 2, and 4)
...
7

The real and imaginary components of
the Φ component of the wavefunctions,
eimlφ = cos mlφ + i sin mlφ, each have | ml |
angular nodes, but these nodes are not
seen when we plot the probability
density, because |eimlφ |2 = 1
...
7, solution of the Schrödinger equation shows that the
acceptable wavefunctions are specified by two quantum numbers l and ml that are restricted to the values
l = 0, 1, 2,
...
, −l

(9
...
The normalized wavefunctions are usually denoted Yl,ml(θ,φ) and are
called the spherical harmonics (Table 9
...

Figure 9
...
There are no angular nodes around the z-axis for functions with ml = 0, which corresponds to there
being no component of orbital angular momentum about that axis
...
37 shows
the distribution of the particle of a given angular momentum in more detail
...
Note how, for a given value of l, the most probable location of the particle migrates towards the xy-plane as the value of | ml | increases
...


(9
...
7 ROTATION IN THREE DIMENSIONS: THE PARTICLE ON A SPHERE

303

l=0

l = 0, ml = 0

l = 1, ml = 0

l=1

l = 2, ml = 0

l=2
l = 3, ml = 0

l = 4, ml = 0
Fig
...
36 A representation of the
wavefunctions of a particle on the surface
of a sphere which emphasizes the location
of angular nodes: dark and light shading
correspond to different signs of the
wavefunction
...

All these wavefunctions correspond to
ml = 0; a path round the vertical z-axis
of the sphere does not cut through any
nodes
...
9
...
The
distance of a point on the surface from the origin is proportional to the square modulus of the
amplitude of the wavefunction at that point
...
Which of the following statements are true: (a) for a given value of r, the
energy separation between adjacent levels decreases with increasing l, (b) increasing r leads to
an decrease in the value of the energy for each level, (c) the energy difference between adjacent
levels increases as r increases
...
Because there are
2l + 1 different wavefunctions (one for each value of ml) that correspond to the same
energy, it follows that a level with quantum number l is (2l + 1)-fold degenerate
...
Therefore, by comparing this equation with eqn 9
...


(9
...
, −l

(9
...
We can also see that the states corresponding to high
angular momentum around the z-axis are those in which most nodal lines cut the
equator: a high kinetic energy now arises from motion parallel to the equator because
the curvature is greatest in that direction
...
4 Calculating the frequency of a molecular rotational transition

Under certain circumstances, the particle on a sphere is a reasonable model for
the description of the rotation of diatomic molecules
...
The moment of inertia of 1H127I is then
I = mHr 2 = 4
...
It follows that
$2
2I

=

(1
...
288 × 10−47 kg m2)

= 1
...
1297 zJ
...
09 J mol−1
...
53, the first few
rotational energy levels are therefore 0 (l = 0), 0
...
7782 zJ (l = 2), and
1
...
The degeneracies of these levels are 1, 3, 5, and 7, respectively (from
2l + 1), and the magnitudes of the angular momentum of the molecule are 0, 21/2$,
61/2$, and (12)1/2$ (from eqn 9
...
It follows from our calculations that the l = 0
and l = 1 levels are separated by ∆E = 0
...
A transition between these two
rotational levels of the molecule can be brought about by the emission or absorption
of a photon with a frequency given by the Bohr frequency condition (eqn 8
...
594 × 10−22 J
6
...
915 × 1011 Hz = 391
...
Because the transition energies depend on the moment
of inertia, microwave spectroscopy is a very accurate technique for the determination of bond lengths
...


9
...
8 Repeat the calculation for a 2H127I molecule (same bond length as
1

H127I)
...
9
...
We shall
see soon that this representation is too
specific because the azimuthal orientation
of the vector (its angle around z) is
indeterminate
...
, −l for a given value of l means
that the component of angular momentum about the z-axis may take only 2l + 1
values
...
In classical terms, this restriction means
that the plane of rotation of the particle can take only a discrete range of orientations
(Fig
...
38)
...

The quantum mechanical result that a rotating body may not take up an arbitrary
orientation with respect to some specified axis (for example, an axis defined by the
direction of an externally applied electric or magnetic field) is called space quantization
...
9
...
The idea behind the experiment was that a rotating, charged
body behaves like a magnet and interacts with the applied field
...
Because the direction in which the magnet is driven by the inhomogeneous field depends on the magnet’s orientation, it follows that a broad band of atoms is expected to emerge from the region where the
magnetic field acts
...

In their first experiment, Stern and Gerlach appeared to confirm the classical prediction
...
When the experiment was repeated with a beam of very low
intensity (so that collisions were less frequent) they observed discrete bands, and so
confirmed the quantum prediction
...
9
...

(b) The classically expected result
...


306

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
z

about the two axes perpendicular to z)
...
45:

+2
ml
+1

Zx =

0

∂

i C ∂z

[Zx,Zy] = i$Zz

-2

z

y

−z

∂D
∂y F

Zy =

$A

z

∂

i C ∂x

−x

∂D
∂z F

Zz =

$A

x

∂

i C ∂y

−y

∂D
∂x F

(9
...
27, these three operators do not commute
with one another:

-1

(a)

$A

[Zy,Zz] = i$Zx

[Zz,Zx] = i$Zy

(9
...
In other words,
lx, ly, and lz are complementary observables
...
9
...
9
...

However, because the azimuthal angle
of the vector around the z-axis is
indeterminate, a better representation is
as in (b), where each vector lies at an
unspecified azimuthal angle on its cone
...
56b)

where Λ2 is the legendrian in eqn 9
...
This operator does commute with all three
components:
[Z 2,Zq] = 0

q = x, y, and z

(9
...
29
...
It follows that the illustration in Fig
...
38, which
is summarized in Fig
...
40a, gives a false impression of the state of the system, because
it suggests definite values for the x- and y-components
...

The vector model of angular momentum uses pictures like that in Fig
...
40b
...
Each cone has a definite projection (of ml units) on the z-axis,
representing the system’s precise value of lz
...
The vector representing the state of angular momentum can be thought of
as lying with its tip on any point on the mouth of the cone
...

IMPACT ON NANOSCIENCE

I9
...
1 we outlined some advantages of working in the nanometre regime
...
Here we focus on the origins and consequences of these quantum mechanical effects
...
It carries an electrical current
because the electrons are delocalized over all the atomic nuclei
...
Immediately,
we predict from eqn 9
...
However, consider a nanocrystal, a small cluster of atoms with dimensions
in the nanometre scale
...
6, we predict that the electronic energies
are quantized and that the separation between energy levels decreases with increasing
size of the cluster
...
2 IMPACT ON NANOSCIENCE: QUANTUM DOTS
shape
...
39 that the energy levels of
an electron in a sphere of radius R are given by
En =

n2h2

(9
...
Transfer of energy to a semiconductor increases the
mobility of electrons in the material
...
The
holes are also mobile, so to describe electrical conductivity in semiconductors we need
to consider the movement of electron–hole pairs, also called excitons, in the material
...
6
can give us qualitative insight into the origins of conductivity in semiconductors
...
Now we explore the impact of energy quantization on the optical and electronic properties of semiconducting nanocrystals
...
They can be made in solution or by depositing atoms
on a surface, with the size of the nanocrystal being determined by the details of the
synthesis (see, for example, Impact I20
...
A quantitative but approximate treatment
that leads to the energy of the exciton begins with the following hamiltonian for a
spherical quantum dot of radius R:
@=−

$2
2me

∇2 −
e

$2
2mh

∇2 + V(re,rh)
h

(9
...
Taking into account only the Coulomb attraction between
the hole, with charge +e, and the electron, with charge −e, we write (see Chapter 9 and
Appendix 3 for details):
V(re,rh) = −

e2
4πε | re − rh |

(9
...
Solving the Schrödinger equation in this case is not a trivial task,
but the final expression for the energy of the exciton, Eex, is relatively simple (see
Further reading for details):
Eex =

h2 A 1
8R2 C me

+

1 D
mh F

−

1
...
60)

As expected, we see that the energy of the exciton decreases with increasing radius
of the quantum dot
...

The expression for Eex has important consequences for the optical properties of
quantum dots
...
The

307

308

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
electrical properties of large, macroscopic samples of semiconductors cannot be tuned
in this way
...
Therefore, we predict that, as the radius of the quantum dot decreases, the excitation wavelength increases
...
This phenomenon is indeed observed in suspensions of CdSe
quantum dots of different sizes
...
But the
special optical properties of quantum dots can also be exploited
...
This
property forms the basis for the use of quantum dots in the visualization of biological
cells at work
...
When the other end of the spacer reacts specifically with a cellular component, such as a protein, nucleic acid, or membrane, the
cell becomes labelled with a light-emitting quantum dot
...
Though this technique has been used extensively with organic
molecules as labels, quantum dots are more stable and are stronger light emitters
...
8 Spin
Stern and Gerlach observed two bands of Ag atoms in their experiment
...
The conflict was resolved by the
suggestion that the angular momentum they were observing was not due to orbital
angular momentum (the motion of an electron around the atomic nucleus) but arose
instead from the motion of the electron about its own axis
...
The explanation of the existence of spin
emerged when Dirac combined quantum mechanics with special relativity and established the theory of relativistic quantum mechanics
...
To
distinguish this spin angular momentum from orbital angular momentum we use the
spin quantum number s (in place of l; like l, s is a non-negative number) and ms, the
spin magnetic quantum number, for the projection on the z-axis
...
An α electron (top) is an
1
electron with ms = + – ; a β electron
2
1
(bottom) is an electron with ms = − –
...


Fig
...
41

ms = s, s − 1,
...
61)

The detailed analysis of the spin of a particle is sophisticated and shows that the
property should not be taken to be an actual spinning motion
...
However, the picture of an actual
spinning motion can be very useful when used with care
...
866$
...
The spin may lie in 2s + 1 = 2 different orientations (Fig
...
41)
...
8 SPIN
1
One orientation corresponds to ms = + – (this state is often denoted α or ↑); the other
2
1
orientation corresponds to ms = − – (this state is denoted β or ↓)
...

Why the atoms behave like this is explained in Chapter 10 (but it is already probably
familiar from introductory chemistry that the ground-state configuration of a silver
atom is [Kr]4d105s1, a single unpaired electron outside a closed shell)
...
For example,
1
1
protons and neutrons are spin- – particles (that is, s = – ) and invariably spin with
2
2
3 1/2
angular momentum (–) $ = 0
...
Because the masses of a proton and a neutron
4
are so much greater than the mass of an electron, yet they all have the same spin
angular momentum, the classical picture would be of these two particles spinning
much more slowly than an electron
...
Some mesons are spin-1 particles
(as are some atomic nuclei), but for our purposes the most important spin-1 particle
is the photon
...
We shall see the importance of photon
spin in the next chapter
...
Thus, electrons and protons are fermions and photons are bosons
...
Photons, for example, transmit the electromagnetic force that binds together electrically charged particles
...

The properties of angular momentum that we have developed are set out in
Table 9
...
As mentioned there, when we use the quantum numbers l and ml we shall
mean orbital angular momentum (circulation in space)
...
When we use j
and mj we shall mean either (or, in some contexts to be described in Chapter 10, a
combination of orbital and spin momenta)
...
4 Properties of angular momentum
Quantum number

Symbol

Values*

Specifies

Orbital angular momentum

l

0, 1, 2,
...
, −l

Component on z-axis, ml $

Spin

s

1
–
2

Magnitude, {s(s + 1)}1/2$

Spin magnetic

ms

1
±–
2

Component on z-axis, ms $

Total

j

l + s, l + s − 1,
...
, −j

Component on z-axis, mj $

To combine two angular momenta, use the Clebsch–Gordan series:
j = j1 + j2, j1 + j2 − 1,
...

*Note that the quantum numbers for magnitude (l, s, j, etc
...


309

310

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS

Techniques of approximation
All the applications treated so far have had exact solutions
...

To make progress with these problems we need to develop techniques of approximation
...

Variation theory is most commonly encountered in the context of molecular orbital
theory, and we consider it there (Chapter 11)
...

9
...
62)

In time-independent perturbation theory, the perturbation is always present and
unvarying
...

In time-independent perturbation theory, we suppose that the true energy of the
system differs from the energy of the simple system, and that we can write
E = E (0) + E (1) + E (2) +
...
63)

where E (1) is the ‘first-order correction’ to the energy, a contribution proportional to
@ (1), and E (2) is the ‘second-order correction’ to the energy, a contribution proportional to @ (1)2, and so on
...


(9
...
As we show in Further information 9
...
65a)

and

E (2) =
0

∑

n≠0

2

Ύ

ψ (0)*@ (1)ψ (0)dτ
0
0
(0)
E (0) − E n
0

=

| H (1) |2

n0
∑ E(0) − E (0)

n≠0

0

(9
...
65c]

in a convenient compact notation for integrals that we shall use frequently
...
We can
interpret E (1) as the average value of the perturbation, calculated by using the unperturbed wavefunction
...
10 TIME-DEPENDENT PERTURBATION THEORY
when small weights are hung along its length
...
Those hanging at the antinodes, however,
have a pronounced effect (Fig
...
42a)
...
In terms of the violin analogy, the average is now taken over the distorted
waveform of the vibrating string, in which the nodes and antinodes are slightly shifted
(Fig
...
42b)
...
65b:
1
...

2
...

3
...
The
opposite is true when the energy levels lie close together
...
9
...
(b) The second-order energy
is a similar average, but over the distortion
induced by the perturbation
...
5 Using perturbation theory

Find the first-order correction to the ground-state energy for a particle in a well
with a variation in the potential of the form V = −ε sin(πx /L), as in Fig
...
43
...
65a
...


V

V

Answer The perturbation hamiltonian is

@ (1) = −ε sin(πx/L)
–e sin(p x /L)

Therefore, the first-order correction to the energy is
4L/3π

0

5
4
6
4
7

Ύ ψ@
L

E (1) =
0

1

0

(1)

ψ1dx = −

2ε
L

Ύ sin
L

3

0

πx
L

dx = −

L

x

8ε
3π

Note that the energy is lowered by the perturbation, as would be expected for the
shape shown in Fig
...
43
...
9 Suppose that only ψ3 contributes to the distortion of the wavefunction: calculate the coefficient c3 and the second-order correction to the energy by
using eqn 9
...
76 in Further information 9
...

[c3 = −8εmL2/15πh2, E (2) = −64ε 2mL2/225π2h2]
0

9
...
Many of the perturbations
encountered in chemistry are time-dependent
...


Fig
...
43 The potential energy for a particle
in a box with a potential that varies as
−ε sin(πx/L) across the floor of the box
...


312

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
Classically, for a molecule to be able to interact with the electromagnetic field and
absorb or emit a photon of frequency ν, it must possess, at least transiently, a dipole
oscillating at that frequency
...
66)

where @ (t) is the time-dependent perturbation
...
8

An electric dipole consists of two electric
charges +q and −q separated by a
distance R
...


(9
...
We suppose that the perturbation is absent until t = 0, and then it is turned on
...
2 that the rate of change of population of the state
ψf due to transitions from state ψi, wf←i, is proportional to the square modulus of the
matrix element of the perturbation between the two states:
wf←i ∝ | H (1) |2
fi

(9
...
67), we conclude that
wf←i ∝ | µz,fi |2E2

(9
...
70]

The size of the transition dipole can be regarded as a measure of the charge redistribution that accompanies a transition
...
This
result will be the basis of most of our subsequent discussion of spectroscopy in
Chapters 10 and 13–15 and of the kinetics of electron transfer in Chapter 24
...
The wavefunction of a free particle is ψk = Aeikx + Be−ikx,
Ek = k 2$2/2m
...
The wavefunctions and energies of a particle in a onedimensional box of length L are, respectively, ψn(x) = (2/L)1/2
sin(nπx /L) and En = n2h2/8mL2, n = 1,2,
...

3
...

4
...
Orthonormal functions are sets of functions
that are normalized and mutually orthogonal
...
The wavefunctions and energies of a particle in a twodimensional box are given by eqn 9
...


6
...

7
...
The transmission probability is given by
eqn 9
...

8
...
As a consequence, V = – kx 2
...
The wavefunctions and energy levels of a quantum
mechanical harmonic oscillator are given by eqns 9
...
25, respectively
...
The virial theorem states that, if the potential energy of a
particle has the form V = ax b, then its mean potential and
kinetic energies are related by 2͗EK͘ = b͗V ͘
...
Angular momentum is the moment of linear momentum
around a point
...
The wavefunctions and energies of a particle on a ring are,
respectively, ψml(φ) = (1/2π)1/2eimlφ and E = ml2$2/2I, with
I = mr 2 and ml = 0, ±1, ±2,
...
The wavefunctions of a particle on a sphere are the spherical
harmonics, the functions Yl,ml(θ,φ) (Table 9
...
The energies
are E = l(l + 1)$2/2I, l = 0, 1, 2,
...
For a particle on a sphere, the magnitude of the angular
momentum is {l(l + 1)}1/2$ and the z-component of the
angular momentum is ml $, ml = l, l − 1,
...

15
...

16
...
A fermion is a particle with a half-integral spin
quantum number; a boson is a particle with an integral spin
quantum number
...
For an electron, the spin quantum number is s = –
...
The spin magnetic quantum number is ms = s, s − 1,
...

2
2
19
...

20
...
The first- and second-order
corrections to the energy are given by eqns 9
...
65b,
respectively
...

21
...


Further reading
D
...
McQuarrie, Mathematical methods for scientists and engineers
...


Articles and texts

P
...
Atkins and R
...
Friedman, Molecular quantum mechanics
...


J
...
C
...
J
...
Educ
...


C
...
Johnson, Jr
...
G
...
Dover, New York (1986)
...
N
...
Prentice–Hall, Upper Saddle River
(2000)
...
Pauling and E
...
Wilson, Introduction to quantum mechanics with
applications to chemistry
...


Further information
commonly called ‘matrix elements’, are incorporated into the bracket
notation by writing

Further information 9
...
9 is often written
͗n | n′͘ = 0

(n′ ≠ n)

This Dirac bracket notation is much more succinct than writing out
the integral in full
...
Thus, the bra ͗n | corresponds to
ψ n and the ket | n′͘ corresponds to the wavefunction ψn′
...
Similarly, the normalization condition in eqn
8
...
These two expressions can be combined into
one:
͗n | n′͘ = δnn′

Ύ

͗n | ) | m ͘ = ψ n ψmdτ
*)

[9
...
An
integration is implied whenever a complete bracket is written
...
73)

with the bra and the ket corresponding to the same state (with
quantum number n and wavefunction ψn)
...
30) if
͗n | ) | m ͘ = ͗m | ) | n͘*

(9
...
71)

Here δnn′, which is called the Kronecker delta, is 1 when n′ = n and 0
when n′ ≠ n
...
9) and which are

Further information 9
...
Our first task is to
develop the results of time-independent perturbation theory, in
which a system is subjected to a perturbation that does not vary with

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS

1 Time-independent perturbation theory

To develop expressions for the corrections to the wavefunction and
energy of a system subjected to a time-independent perturbation, we
write

ψ = ψ (0) + λψ (1) + λ2ψ (2) +
...

= E(0)ψ (0) + λ(E(0)ψ (1) + E(1)ψ (0)) + λ2(E(2)ψ (0) + E(1)ψ (1)
+ E(0)ψ (2)) +
...

The equations we have derived are applicable to any state of the
system
...
The first equation, which we now write
@ (0)ψ (0) = E(0)ψ (0)
0
0
0
is the Schrödinger equation for the ground state of the unperturbed
system, which we assume we can solve (for instance, it might be the
equation for the ground state of the particle in a box, with the
solutions given in eqn 9
...
To solve the next equation, which is now
written
@ (1)ψ (0) + @ (0)ψ (1) = E (0)ψ (1) + E (1)ψ (0)
0
0
0
0
0
we suppose that the first-order correction to the wavefunction can be
expressed as a linear combination of the wavefunctions of the
unperturbed system, and write

∑ cnψ (0)
n
n

(9
...
65a
...
)
= (E(0) + λE(1) + λ2E(2) +
...
)

ψ (1) =
0

E(0) if n=0, 0 otherwise
0

∑ cnE (0) Ύψ (0)*ψ (0)dτ + E (1) Ύψ (0)*ψ (0)dτ
0
k
n
0
k
0
n

That is,

Ύψ

(0)
(1) (0)
k *@ ψ 0 dτ

+ ck E(0) = ck E (0)
k
0

which we can rearrange into

ck = −

Ύψ

(0)
(1) (0)
k *@ ψ 0 dτ

(9
...


Terms in λ0:

but 1 if n = 0

That is,

and

(1)

if n ≠ 0,

5
4
6
4
7

@ = @ (0) + λ@ (1)

(0)

(0)
(0)
0 *ψ n dτ = 0

1 if n=0, 0 otherwise

where the power of λ indicates the order of the correction
...
Then, we go on to discuss time-dependent perturbation theory,
in which a perturbation is turned on at a specific time and the system
is allowed to evolve
...
However, we can go one step
further by substituting eqn 9
...
76 now produces the
final result, eqn 9
...

2 Time-dependent perturbation theory

To cope with a perturbed wavefunction that evolves with time, we
need to solve the time-dependent Schrödinger equation,
@Ψ = i$

∂Ψ

(9
...
78a)

1
i$

ΎH
t

(1)
iωn0t
dt
n0 (t)e

(9
...
78 is quite lengthy (see Further
reading)
...
78b, a perturbation
that is switched on very slowly to a constant value gives the same
expression for the coefficients as we obtained for time-independent
perturbation theory
...
9
...
Substitution
of this expression into eqn 9
...
We also suppose that we are interested in the
coefficients long after the perturbation has settled down into its final
value, when t >> τ (so that the exponential in the second numerator is
close to zero and can be ignored)
...
9
...

A large value of τ corresponds to very slow switching
...
78a, we obtain the
time-independent expression, eqn 9
...

In accord with the general rules for the interpretation of
wavefunctions, the probability that the system will be found in the
state n is proportional to the square modulus of the coefficient of the
state, | cn(t) |2
...
68
...
1 Discuss the physical origin of quantization energy for a particle confined

to moving inside a one-dimensional box or on a ring
...
2 Discuss the correspondence principle and provide two examples
...
5 Distinguish between a fermion and a boson
...
3 Define, justify, and provide examples of zero-point energy
...


9
...
Why is

9
...


Exercises
9
...
0 nm
...
8b Calculate the zero-point energy of a harmonic oscillator consisting of a

9
...
82 zJ
...


electronvolts, and reciprocal centimetres between the levels (a) n = 3 and
n = 1, (b) n = 7 and n = 6 of an electron in a box of length 1
...

9
...
49L

and 0
...
Take the
wavefunction to be a constant in this range
...
2b Calculate the probability that a particle will be found between 0
...
67L in a box of length L when it has (a) n = 1, (b) n = 2
...


particle of mass 5
...


9
...
33 × 10−25 kg, the difference

9
...
88 × 10−25 kg, the difference

in adjacent energy levels is 3
...
Calculate the force constant of the
oscillator
...
10a Calculate the wavelength of a photon needed to excite a transition
between neighbouring energy levels of a harmonic oscillator of effective mass
equal to that of a proton (1
...

9
...
3a Calculate the expectation values of p and p2 for a particle in the state

n = 1 in a square-well potential
...
9949 u) and force constant 544 N m−1
...
3b Calculate the expectation values of p and p2 for a particle in the state
n = 2 in a square-well potential
...
11a Refer to Exercise 9
...


9
...
What would be the
length of the box such that the zero-point energy of the electron is equal to its
rest mass energy, mec 2? Express your answer in terms of the parameter
λC = h/mec, the ‘Compton wavelength’ of the electron
...
11b Refer to Exercise 9
...

9
...
4b Repeat Exercise 9
...


1
...


9
...
12b Calculate the minimum excitation energies of (a) the 33 kHz quartz

state n = 3?

crystal of a watch, (b) the bond between two O atoms in O2, for which
k = 1177 N m−1
...
5b What are the most likely locations of a particle in a box of length L in the

state n = 5?
9
...
What is the degeneracy of the level
that has an energy three times that of the lowest level?
9
...
What is the degeneracy of the level

9
...
1 is a solution of the
1
Schrödinger equation for the oscillator and that its energy is – $ω
...
13b Confirm that the wavefunction for the first excited state of a one-

–
–
that has an energy 14 times that of the lowest level?
3

dimensional linear harmonic oscillator given in Table 9
...

2

9
...


9
...


3

9
...
00 m
...
8a Calculate the zero-point energy of a harmonic oscillator consisting of a
particle of mass 2
...


9
...

9
...
9688 u
...
15b Assuming that the vibrations of a 14N2 molecule are equivalent to those

of a harmonic oscillator with a force constant k = 2293
...
0031 u
...
16a The wavefunction, ψ (φ), for the motion of a particle in a ring is of the
form ψ = Neimlφ
...

9
...

9
...


317

9
...

1
1
9
...


9
...


Problems*
Numerical problems
9
...
0 cm
...
2 The mass to use in the expression for the vibrational frequency of a

diatomic molecule is the effective mass µ = mAmB /(mA + mB), where mA and
mB are the masses of the individual atoms
...
Herzberg, van Nostrand (1950):
H35Cl

H81Br

HI

CO

NO

2990

2650

2310

2170

1904

expression for the first-order correction to the ground-state energy, E (1)
...

9
...


Suppose it is vertical; then the potential energy of the particle depends on x
because of the presence of the gravitational field
...
Account for the result
...
The energy of the
particle depends on its height as mgh, where g = 9
...
Because g is so
small, the energy correction is small; but it would be significant if the box were
near a very massive star
...
7 Calculate the second-order correction to the energy for the system

Calculate the force constants of the bonds and arrange them in order of
increasing stiffness
...
3 The rotation of an 1H127I molecule can be pictured as the orbital motion

of an H atom at a distance 160 pm from a stationary I atom
...
) Suppose that the molecule rotates
only in a plane
...
What, apart from 0, is the minimum angular momentum of the
molecule?
9
...

2

9
...
9
...
(a) Write a general

described in Problem 9
...

Account for the shape of the distortion caused by the perturbation
...

The following integrals are useful

Ύ
Ύ

x sin ax sin bx dx = −
cos ax sin bx dx =

d
da

Ύ

cos ax sin bx dx

cos(a − b)x
2(a − b)

−

cos(a + b)x
2(a + b)

+ constant

Theoretical problems
9
...
0 mol perfect gas molecules all occupy the lowest energy

level of a cubic box
...
9 Derive eqn 9
...


a

e
0

Fig
...
45

L /2

x

L

9
...
They are: V = 0 for −∞ < x
≤ 0, 0, V = V2 for 0 ≤ x ≤ L, and V = V3 for L ≤ x < ∞
...
In region 3 the
wavefunction has only a forward component, eik3x, which represents a particle
that has traversed the barrier
...
The transmission probability, T, is the ratio of the
square modulus of the region 3 amplitude to the square modulus of the
incident amplitude
...
(b) Show that the general
equation for T reduces to eqn 9
...


318

9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS

V1 = V3 = 0
...

9
...


Calculate (a) the probability that the particle is inside the barrier and (b) the
average penetration depth of the particle into the barrier
...
12 Confirm that a function of the form e

9
...
Show that any two of the
components do not mutually commute, and find their commutator
...
28 Starting from the operator lz = xpy − ypx , prove that in spherical polar

coordinates lz = −i$∂/∂φ
...


9
...
13 Calculate the mean kinetic energy of a harmonic oscillator by using the
relations in Table 9
...


9
...


9
...
1
...
15 Determine the values of δx = (͗x 2͘ − ͗x͘2)1/2 and δp = (͗p2͘ − ͗p͘2)1/2 for

(a) a particle in a box of length L and (b) a harmonic oscillator
...

9
...
Use the relations between
Hermite polynomials given in Table 9
...

9
...
Show that for small
displacements the motion of the group is harmonic and calculate the energy of
excitation from v = 0 to v = 1
...
18 Show that, whatever superposition of harmonic oscillator states is used
to construct a wavepacket, it is localized at the same place at the times 0, T,
2T,
...

9
...

9
...

9
...
Although r varies with angle ϕ, the
two are related by r 2 = a2 sin2φ + b2 cos2φ
...
22 Use mathematical software to construct a wavepacket of the form

Ψ(φ,t) =

ml,max

∑c

ml = 0

i(ml φ −Emlt/$)
mle

Em = m2$2/2I
l
l

with coefficients c of your choice (for example, all equal)
...

9
...


calculation, justify the remark that [l2,lq] = 0 for all q = x, y, and z
...
(b) Show that for large values of n, a
quantum particle approaches the classical values
...


Applications: to biology and nanotechnology
9
...
1)
...
In the ground state of retinal, each
level up to n = 6 is occupied by two electrons
...
(c) Using your results and
Illustration 9
...

9
...
This tunnelling occurs over distances that are often
greater than 1
...
For a specific combination of donor and acceptor, the rate of
electron tunnelling is proportional to the transmission probability, with
κ ≈ 7 nm−1 (eqn 9
...
By what factor does the rate of electron tunnelling
between two co-factors increase as the distance between them changes from
2
...
0 nm?
9
...
Estimate the vibrational frequency of CO bound to
myoglobin by using the data in Problem 9
...
O bond
...
25 Derive an expression in terms of l and ml for the half-angle of the apex of
the cone used to represent an angular momentum according to the vector
model
...
Show that the minimum
possible angle approaches 0 as l → ∞
...
34 Of the four assumptions made in Problem 9
...
Suppose that the first two assumptions are still reasonable
and that you have at your disposal a supply of myoglobin, a suitable buffer in
which to suspend the protein, 12C16O, 13C16O, 12C18O, 13C18O, and an infrared
spectrometer
...
O bond, describe a set of experiments that: (a) proves
which atom, C or O, binds to the haem group of myoglobin, and (b) allows
for the determination of the force constant of the C
...


9
...


9
...
24 Confirm that Y3,+3 is normalized to 1
...
)

PROBLEMS

structural basis of the haem group and the chlorophylls
...
As in Illustration 9
...
(a) Calculate the energy and angular momentum
of an electron in the highest occupied level
...

9
...
For a one-dimensional random coil of
N units, the restoring force at small displacements and at a temperature T is
F=−

kT
2l

AN + nD
ln B
E
CN − nF

where l is the length of each monomer unit and nl is the distance between the
ends of the chain (see Section 19
...
Show that for small extensions (n < N)
<
the restoring force is proportional to n and therefore the coil undergoes
harmonic oscillation with force constant kT/Nl2
...

9
...
To get an idea of the magnitudes of
these forces, calculate the force acting between two electrons separated by
2
...
Hints
...
854 × 10−12 C2 J−1 m−1 is the vacuum permittivity
...


319

9
...
2 that quantum
mechanical effects need to be invoked in the description of the electronic
properties of metallic nanocrystals, here modelled as three-dimensional boxes
...
Show that the
Schrödinger equation is separable
...
(c) Specialize the result from
part (b) to an electron moving in a cubic box of side L = 5 nm and draw an
energy diagram resembling Fig
...
2 and showing the first 15 energy levels
...
(d)
Compare the energy level diagram from part (c) with the energy level diagram
for an electron in a one-dimensional box of length L = 5 nm
...
39 We remarked in Impact I9
...
Here, we justify eqn 9
...
(a) The Hamiltonian for a particle free to
move inside a sphere of radius R is

@=−

$
2m

∇2

Show that the Schrödinger equation is separable into radial and angular
components
...
Then show that the Schrödinger
equation can be separated into two equations, one for X, the radial equation,
and the other for Y, the angular equation:

−

$2 A d2X(r) 2 dX(r) D l(l + 1)$2
B
E +
+
X(r) = EX(r)
2m C dr 2
r dr F
2mr 2

Λ2Y = −l(l + 1)Y
You may wish to consult Further information 10
...

(c) Consider the case l = 0
...
54 after substituting me for m) also
applies when l ≠ 0
...
1 The structure of hydrogenic

atoms
10
...
3 Spectroscopic transitions and

selection rules
The structures of many-electron
atoms
10
...
5 Self-consistent field orbitals

The spectra of complex atoms
I10
...
6 Quantum defects and

ionization limits
10
...
We see what experimental information is
available from a study of the spectrum of atomic hydrogen
...
The wavefunctions obtained are the ‘atomic orbitals’ of hydrogenic atoms
...
In conjunction with the Pauli exclusion principle, we account for the periodicity of atomic properties
...
We see in the closing sections of the chapter how such spectra are
described by using term symbols, and the origin of the finer details of their appearance
...
The concepts we
meet are of central importance for understanding the structures and reactions of
atoms and molecules, and hence have extensive chemical applications
...
A hydrogenic atom is a one-electron atom or
ion of general atomic number Z; examples of hydrogenic atoms are H, He+, Li2+, O7+,
and even U91+
...
So even He,
with only two electrons, is a many-electron atom
...
They also provide a set of
concepts that are used to describe the structures of many-electron atoms and, as we
shall see in the next chapter, the structures of molecules too
...
8 Spin–orbit coupling
10
...
1:
The separation of motion
Discussion questions
Exercises
Problems

The structure and spectra of hydrogenic atoms
When an electric discharge is passed through gaseous hydrogen, the H2 molecules are
dissociated and the energetically excited H atoms that are produced emit light of discrete frequencies, producting a spectrum of a series of ‘lines’ (Fig
...
1)
...
1)

with n1 = 1 (the Lyman series), 2 (the Balmer series), and 3 (the Paschen series), and that
in each case n2 = n1 + 1, n1 + 2,
...


100

120

150

200

300

400

500

321

l /nm

Visible

1000
800
600

2000

10
...
10
...
Both the observed
spectrum and its resolution into
overlapping series are shown
...


Paschen
Brackett

Self-test 10
...


[821 nm]

The form of eqn 10
...
2)

The Ritz combination principle states that the wavenumber of any spectral line is the
difference between two terms
...
3)

Thus, if each spectroscopic term represents an energy hcT, the difference in energy
when the atom undergoes a transition between two terms is ∆E = hcT1 − hcT2 and,
according to the Bohr frequency conditions (Section 8
...
This expression rearranges into the Ritz
formula when expressed in terms of wavenumbers (on division by c)
...

Because spectroscopic observations show that electromagnetic radiation is absorbed
and emitted by atoms only at certain wavenumbers, it follows that only certain energy
states of atoms are permitted
...

10
...
4)

322

10 ATOMIC STRUCTURE AND ATOMIC SPECTRA
where r is the distance of the electron from the nucleus and ε 0 is the vacuum permittivity
...
5)

4πε0r

The subscripts on ∇2 indicate differentiation with respect to the electron or nuclear
coordinates
...
We show in Further information
10
...
6)

where differentiation is now with respect to the coordinates of the electron relative to
the nucleus
...
The reduced mass is very similar to the electron mass because mN, the mass of the nucleus, is much larger than the
mass of an electron, so 1/µ ≈ 1/me
...

Because the potential energy is centrosymmetric (independent of angle), we can
suspect that the equation is separable into radial and angular components
...
7)

and examine whether the Schrödinger equation can be separated into two equations,
one for R and the other for Y
...
1, the equation does
separate, and the equations we have to solve are
Λ2Y = −l(l + 1)Y
−

$2 d2u
2µ dr 2

(10
...
9)

where
Veff = −

Ze2
4πε0r

+

l(l + 1)$2
2µr 2

(10
...
8 is the same as the Schrödinger equation for a particle free to move
round a central point, and we considered it in Section 9
...
The solutions are the
spherical harmonics (Table 9
...

We consider them in more detail shortly
...
9 is called the radial wave
equation
...

(b) The radial solutions

We can anticipate some features of the shapes of the radial wavefunctions by analysing
the form of Veff
...
10 is the Coulomb potential energy of the

10
...
1 The shape of the radial wavefunction

When r is very small (close to the nucleus), u ≈ 0, so the right-hand side of eqn 10
...
9 and write
−

d2u
dr

2

+

l(l + 1)
r2

u≈0

The solution of this equation (for r ≈ 0) is
u ≈ Ar l+1 +

B
rl

0

Effective potential energy, Veff

electron in the field of the nucleus
...
When l = 0, the electron has no angular momentum, and the effective potential energy is purely Coulombic and attractive at all radii
(Fig
...
2)
...
When the electron is close to the nucleus (r ≈ 0), this
repulsive term, which is proportional to 1/r 2, dominates the attractive Coulombic
component, which is proportional to 1/r, and the net effect is an effective repulsion of
the electron from the nucleus
...
However,
they are similar at large distances because the centrifugal contribution tends to zero
more rapidly (as 1/r 2) than the Coulombic contribution (as 1/r)
...
We show in the Justification below that close to the nucleus the
radial wavefunction is proportional to r l, and the higher the orbital angular momentum, the less likely the electron is to be found (Fig
...
3)
...


323

l¹0

l=0

Radius, r
Fig
...
2 The effective potential energy of
an electron in the hydrogen atom
...
When
the electron has nonzero orbital angular
momentum, the centrifugal effect gives rise
to a positive contribution that is very large
close to the nucleus
...


Exploration Plot the effective potential

energy against r for several nonzero
values of the orbital angular momentum l
...
Because
2

du
dr

2

2

=

d (rR)
dr

2

2

=r

dR
dr

2

+2

dR
dr

2

tr

dR
dr

2

Wavefunction, y

Because R = u/r, and R cannot be infinite at r = 0, we must set B = 0, and hence
obtain R ≈ Ar l
...
9 becomes
l=0
1
2
3

this equation has the form
−

$2 d2R
2µ dr 2

t ER

The acceptable (finite) solution of this equation (for r large) is
−(2µ | E |/$2)r

Rte

and the wavefunction decays exponentially towards zero as r increases
...
10
...
Electrons
are progressively excluded from the
neighbourhood of the nucleus as l
increases
...


324

10 ATOMIC STRUCTURE AND ATOMIC SPECTRA
decaying form at great distances (see Further reading)
...
11)

32π2ε 2 $2n2
0

with n = 1, 2,
...
12)

These functions are most simply written in terms of the dimensionless quantity ρ
(rho), where

ρ=

2Zr
na0

a0 =

4πε 0$2

(10
...
9 pm; it is so called because the same quantity
appeared in Bohr’s early model of the hydrogen atom as the radius of the electron
orbit of lowest energy
...
14)

where L is a polynomial in ρ called an associated Laguerre polynomial: it links the r ≈ 0
solutions on its left (corresponding to R ∝ ρ l ) to the exponentially decaying function
on its right
...
1)
...

Table 10
...
For an infinitely heavy nucleus (or one that may be assumed to be so), µ = me
and a = a0, the Bohr radius
...
3
...
1 THE STRUCTURE OF HYDROGENIC ATOMS
2

0
...
4
0
...
5

n = 1, l = 0
1

R /(Z /a0)3/2

0
...
2

n = 3, l = 0
0
...
5

0
0

0
(a)

0

1

2

3

–0
...
5

15
Zr /a0

22
...
05

0

7

...
8

Zr /a0

15

22
...
15

0
...
04

R /(Z /a0)3/2

0
...
2

0
...
6

0
...
03
0
...
01
–0
...
5

Zr /a0

15

22
...
10
...
Note that the orbitals with l = 0 have a
nonzero and finite value at the nucleus
...


Exploration Use mathematical software to find the locations of the radial nodes in hydrogenic wavefunctions with n up to 3
...

3 The associated Laguerre polynomial is a function that oscillates from positive to
negative values and accounts for the presence of radial nodes
...
1 and illustrated in
Fig
...
4
...
1 Calculating a probability density

To calculate the probability density at the nucleus for an electron with n = 1, l = 0,
and ml = 0, we evaluate ψ at r = 0:

A ZD
ψ1,0,0(0,θ,φ) = R1,0(0)Y0,0(θ,φ) = 2
C a0 F

3/2

A 1D
C 4π F

1/2

Comment 10
...
Nodes
at the nucleus are all angular nodes
...
15 × 10−6 pm−3 when Z = 1
...
2 Evaluate the probability density at the nucleus of the electron for an
electron with n = 2, l = 0, ml = 0
...
2 Atomic orbitals and their energies
An atomic orbital is a one-electron wavefunction for an electron in an atom
...
When an electron is described by one of these wavefunctions, we say that it
‘occupies’ that orbital
...

For instance, an electron described by the wavefunction ψ1,0,0 and in the state |1,0,0͘
is said to occupy the orbital with n = 1, l = 0, and ml = 0
...
and determines the energy of the electron:

Energy of widely
separated stationary
electron and nucleus
Continuum

n
+

0

-

H +e

¥

3

-hcRH
4

2

Energy

-hcRH
9

An electron in an orbital with quantum number n has an energy given by eqn 10
...

The two other quantum numbers, l and ml , come from the angular solutions, and
specify the angular momentum of the electron around the nucleus:
An electron in an orbital with quantum number l has an angular momentum of
magnitude {l(l + 1)}1/2$, with l = 0, 1, 2,
...

An electron in an orbital with quantum number ml has a z-component of angular
momentum ml $, with ml = 0, ±1, ±2,
...


Classically
allowed
energies

Note how the value of the principal quantum number, n, controls the maximum value
of l and l controls the range of values of ml
...
We saw in Section 9
...
The value of s is fixed at – for an electron, so we do
2
1
1
not need to consider it further at this stage
...
It follows that, to specify the state of an electron in a hydrogenic
atom, we need to give the values of four quantum numbers, namely n, l, ml, and ms
...
10
...
The values are relative to an
infinitely separated, stationary electron and
a proton
...
11 are depicted in Fig
...
5
...
All the energies given by eqn 10
...
They refer to the
bound states of the atom, in which the energy of the atom is lower than that of the
infinitely separated, stationary electron and nucleus (which corresponds to the zero of
energy)
...

These solutions correspond to unbound states of the electron, the states to which an
electron is raised when it is ejected from the atom by a high-energy collision or photon
...


10
...
11 is consistent with the spectroscopic result summarized by eqn 10
...
15]

32π2ε 2 $2
0

where µ H is the reduced mass for hydrogen
...
16]

8ε 2 h3c
0

Insertion of the values of the fundamental constants into the expression for RH gives
almost exact agreement with the experimental value
...


327

Comment 10
...
3, is a primitive but useful
model that gives insight into the bound
and unbound states of the electron in a
hydrogenic atom
...
15 shows that
the energies of a particle (for example,
an electron in a hydrogenic atom) are
quantized when its total energy, E, is
lower than its potential energy, V (the
Coulomb interaction energy between
the electron and the nucleus)
...


(b) Ionization energies

The ionization energy, I, of an element is the minimum energy required to remove an
electron from the ground state, the state of lowest energy, of one of its atoms
...
10
...
17)
−18

), which corres-

Example 10
...
Determine (a) the ionization energy of the lower state, (b) the value of
the Rydberg constant
...
If the upper state lies at an energy −hcRH/n2, then, when
the atom makes a transition to Elower, a photon of wavenumber
#=−

RH
2

n

−

Elower

I
hc

−

105
100
95
90

85

hc

is emitted
...
179 aJ (a, for atto, is the prefix that denotes 10
ponds to 13
...


RH
n2

A plot of the wavenumbers against 1/n should give a straight line of slope −RH and
intercept I/hc
...

2

Answer The wavenumbers are plotted against 1/n2 in Fig
...
6
...
1788 aJ (1312
...


80

0

0
...
2

Fig
...
6 The plot of the data in Example
10
...


Exploration The initial value of n was

not specified in Example 10
...
Show
that the correct value can be determined by
making several choices and selecting the
one that leads to a straight line
...
A similar
extrapolation procedure can be used for many-electron atoms (see Section 10
...

Self-test 10
...
Determine (a) the ionization energy of the lower state, (b) the ionization
energy of the ground state, (c) the mass of the deuteron (by expressing the Rydberg
constant in terms of the reduced mass of the electron and the deuteron, and solving for the mass of the deuteron)
...
1 kJ mol−1, (b) 1312
...
8 × 10 −27 kg,
a result very sensitive to RD]

(c) Shells and subshells

All the orbitals of a given value of n are said to form a single shell of the atom
...
It is common to refer to successive shells by letters:
n=

2

3

4
...


Thus, all the orbitals of the shell with n = 2 form the L shell of the atom, and so on
...
These subshells are generally referred to by letters:
s

p

d

f

l=

2 2s
[1]

3d
[5]

2p
[3]

Energy

3p
[3]

1

2

3

4

5

6
...


The letters then run alphabetically (j is not used)
...
7 is a version of Fig
...
5
which shows the subshells explicitly
...
Thus, when n = 1, there is only one subshell, the one with l = 0
...

When n = 1 there is only one subshell, that with l = 0, and that subshell contains
only one orbital, with ml = 0 (the only value of ml permitted)
...
When n = 3 there are nine orbitals (one with l = 0, three with l = 1,
and five with l = 2)
...
10
...
In general, the number of orbitals in a shell of principal quantum number n
is n2, so in a hydrogenic atom each energy level is n2-fold degenerate
...
From
Table 10
...
10
...
In hydrogenic atoms, all orbitals
of a given shell have the same energy
...
18)

This wavefunction is independent of angle and has the same value at all points of constant radius; that is, the 1s orbital is spherically symmetrical
...
It follows
0
that the most probable point at which the electron will be found is at the nucleus itself
...
2 ATOMIC ORBITALS AND THEIR ENERGIES

329

Subshellls

p

Low potential energy
but
high kinetic energy

d

M shell, n = 3

Energy

s

Lowest total energy
Low kinetic energy
but
high potential energy

L shell, n = 2

c

(a) 1s

a b

K shell, n = 1
Shells

Orbitals

Fig
...
8 The organization of orbitals (white
squares) into subshells (characterized by l)
and shells (characterized by n)
...
10
...
(a) The sharply curved
but localized orbital has high mean kinetic
energy, but low mean potential energy;
(b) the mean kinetic energy is low, but the
potential energy is not very favourable;
(c) the compromise of moderate kinetic
energy and moderately favourable potential
energy
...
The closer the electron is to the nucleus on average, the lower its average potential energy
...
10
...
However, this shape implies a high kinetic
energy, because such a wavefunction has a very high average curvature
...
However, such a wavefunction spreads to great distances from the nucleus
and the average potential energy of the electron will be correspondingly high
...

The energies of ns orbitals increase (become less negative; the electron becomes less
tightly bound) as n increases because the average distance of the electron from the
1
nucleus increases
...
35), ͗E K͘ = − – ͗V ͘ so, even
2
though the average kinetic energy decreases as n increases, the total energy is equal to
1
– ͗V͘, which becomes less negative as n increases
...
10
...
A simpler procedure is to show only the boundary
surface, the surface that captures a high proportion (typically about 90 per cent) of
the electron probability
...
10
...


(b) 2s
Fig
...
10 Representations of the 1s and 2s
hydrogenic atomic orbitals in terms of their
electron densities (as represented by the
density of shading)

Fig
...
11 The boundary surface of an s
orbital, within which there is a 90 per cent
probability of finding the electron
...
19)
2
C
n2 F 7 Z
3

60

40

The variation with n and l is shown in Fig
...
12
...


Zár ñ/a 0

s
p
d

20
Example 10
...

0
1

2

3

n

4

5

6

The variation of the mean radius
of a hydrogenic atom with the principal
and orbital angular momentum quantum
numbers
...

Fig
...
12

Method The mean radius is the expectation value

Ύ

Ύ

͗r͘ = ψ *rψ dτ = r| ψ |2 dτ
We therefore need to evaluate the integral using the wavefunctions given in
Table 10
...
The angular parts of the wavefunction are
normalized in the sense that
π

2π

ΎΎ
0

| Yl,ml | 2 sin θ dθ dφ = 1
0

The integral over r required is given in Example 8
...

Answer With the wavefunction written in the form ψ = RY, the integration is
∞ π

͗r͘ =

2π

ΎΎΎ
0

0

∞

2
rR n,l |Yl,ml | 2r 2 dr sin θ dθ dφ =
0

ΎrR

3 2
n,l dr

0

For a 1s orbital,
1/2

A Z D −Zr/a
0
R1,0 = 2 3
e
C a0 F
Hence
͗r͘ =

4Z
a3
0

∞

Ύ

0

r 3e−2Zr/a0 dr =

3a0
2Z

Self-test 10
...
19
...

For example, the 1s, 2s, and 3s orbitals have 0, 1, and 2 radial nodes, respectively
...

Self-test 10
...
1) is equal to zero, and locate the radial node at 2a0 /Z (see
Fig
...
4)
...

[(a) 2a0 /Z; (b)1
...
10a0 /Z]

10
...
We can imagine a probe with a volume dτ and sensitive to electrons, and which we can move around near the nucleus of a hydrogen atom
...
10
...

Now consider the probability of finding the electron anywhere between the two
walls of a spherical shell of thickness dr at a radius r
...
10
...
The probability that the electron will be found
between the inner and outer surfaces of this shell is the probability density at the
radius r multiplied by the volume of the probe, or |ψ |2 × 4πr 2dr
...
20)

The more general expression, which also applies to orbitals that are not spherically
symmetrical, is
P(r) = r 2R(r)2

y *ydt

(e) Radial distribution functions

r
Radius
Fig
...
13 A constant-volume electronsensitive detector (the small cube) gives its
greatest reading at the nucleus, and a
smaller reading elsewhere
...


(10
...

0
...
2 The general form of the radial distribution function

π

P(r)dr =

2π

ΎΎ
0

0
...
4

P /(Z /a0)3

The probability of finding an electron in a volume element dτ when its wavefunction is ψ = RY is |RY |2dτ with dτ = r 2dr sin θ dθ dφ
...
For a 1s orbital,
4Z 3
a3
0

4

r 2e−2Zr/a0

The radial distribution function P
gives the probability that the electron will
be found anywhere in a shell of radius r
...
The value of P is equivalent to
the reading that a detector shaped like a
spherical shell would give as its radius is
varied
...
10
...
2)
...


P(r) =

2
r /a0

2π

ΎΎ
0

0

(10
...

2 As r → ∞, P(r) → 0 on account of the exponential term
...
10
...


332

10 ATOMIC STRUCTURE AND ATOMIC SPECTRA
The maximum of P(r), which can be found by differentiation, marks the most probable radius at which the electron will be found, and for a 1s orbital in hydrogen occurs
at r = a0, the Bohr radius
...
2a0 = 275 pm
...

Example 10
...

Method We find the radius at which the radial distribution function of the hydro-

genic 1s orbital has a maximum value by solving dP/dr = 0
...

Answer The radial distribution function is given in eqn 10
...
It follows that

dP
dr

=

4Z 3 A
a3 C
0

2r −

2Zr 2D
a0 F

e−2Zr/a0

This function is zero where the term in parentheses is zero, which is at
r* =

a0
Z

Then, with a0 = 52
...
9

26
...
6

13
...
82

7
...
61

5
...
29

10
...
At uranium the most probable radius is only 0
...
(On a scale where r* = 10 cm for H, r* = 1 mm for U
...

Self-test 10
...

[(3 + 51/2)a0 /Z]

(f) p Orbitals

The three 2p orbitals are distinguished by the three different values that ml can take
when l = 1
...
The orbital with ml = 0, for
instance, has zero angular momentum around the z-axis
...
The
wavefunction of a 2p-orbital with ml = 0 is

10
...
10
...
A nodal plane passes through the nucleus and
separates the two lobes of each orbital
...


Exploration Use mathematical software to plot the boundary surfaces of the real parts of
the spherical harmonics Y1,m (θ,φ)
...

5/2

A ZD
ψp0 = R2,1(r)Y1,0(θ,φ) =
r cos θ e−Zr/2a0
1/2 C
4(2π)
a0 F
1

= r cos θ f(r)
where f(r) is a function only of r
...
23)

All p orbitals with ml = 0 have wavefunctions of this form regardless of the value of n
...
10
...
The wavefunction is zero everywhere in the xy-plane,
where z = 0, so the xy-plane is a nodal plane of the orbital: the wavefunction changes
sign on going from one side of the plane to the other
...
In the present case, the functions correspond to non-zero angular momentum
about the z-axis: e+iφ corresponds to clockwise rotation when viewed from below, and
e−iφ corresponds to counter-clockwise rotation (from the same viewpoint)
...
To draw the functions it is usual to represent them as
standing waves
...
24)

These linear combinations are indeed standing waves with no net orbital angular
momentum around the z-axis, as they are superpositions of states with equal and
opposite values of ml
...
10
...

The wavefunction of any p orbital of a given shell can be written as a product of x, y,
or z and the same radial function (which depends on the value of n)
...
3 The linear combination of degenerate wavefunctions

We justify here the step of taking linear combinations of degenerate orbitals when
we want to indicate a particular point
...

Suppose ψ1 and ψ2 are both solutions of the Schrödinger equation with energy E;
then we know that
Hψ1 = Eψ1

Hψ2 = Eψ2

Now consider the linear combination

ψ = c1ψ1 + c2ψ2
where c1 and c2 are arbitrary coefficients
...


(g) d Orbitals

When n = 3, l can be 0, 1, or 2
...
The five d orbitals have ml = +2, +1, 0, −1, −2 and correspond to five different angular momenta around the z-axis (but the same magnitude
of angular momentum, because l = 2 in each case)
...
10
...
The real combinations have the following
forms:
dxy = xyf(r)
dyz = yzf(r)
dzx = zxf(r)
1
1
–(x 2 − y 2)f(r)
dz 2 = (–√3)(3z 2 − r 2)f(r)
dx 2−y 2 = 2
2

(10
...
10
...
Two nodal planes in each orbital
intersect at the nucleus and separate the
lobes of each orbital
...


Exploration To gain insight into the
shapes of the f orbitals, use
mathematical software to plot the
boundary surfaces of the spherical
harmonics Y3,m (θ,φ)
...
3 SPECTROSCOPIC TRANSITIONS AND SELECTION RULES
10
...
11
...
10)
...
However, this is not so, because a photon has an intrinsic spin angular
momentum corresponding to s = 1 (Section 9
...
The change in angular momentum
of the electron must compensate for the angular momentum carried away by the
photon
...

Similarly, an s electron cannot make a transition to another s orbital, because there
would then be no change in the electron’s angular momentum to make up for the
angular momentum carried away by the photon
...

A selection rule is a statement about which transitions are allowed
...
The selection rules for hydrogenic atoms are
∆ml = 0, ±1

∆l = ±1

(10
...

Justification 10
...
10 that the rate of transition between two states is proportional
to the square of the transition dipole moment, µ fi, between the initial and final
states, where (using the notation introduced in Further information 9
...
27]

and µ is the electric dipole moment operator
...
If the transition
dipole moment is zero, the transition is forbidden; the transition is allowed if the
transition moment is non-zero
...

To evaluate a transition dipole moment, we consider each component in turn
...
28)

To evaluate the integral, we note from Table 9
...
6) that the
integral
π

s

p

d

Paschen
Balmer

15 328 cm-1 (Ha)
20 571 cm-1 (Hb)
23 039 cm-1 (Hg)
-1
24 380 cm (Hd)

102 824 cm-1
97 492 cm-1
82 259 cm-1
Lyman

2π

0

0

ΎΎ

Y*,ml Y1,mYli,ml sin θ dθ dφ
lf
f

i

is zero unless lf = li ± 1 and ml,f = ml,i + m
...
The same
procedure, but considering the x- and y-components, results in the complete set of
rules
...
2 Applying selection rules

To identify the orbitals to which a 4d electron may make radiative transitions, we
first identify the value of l and then apply the selection rule for this quantum number
...
Thus, an electron may make
a transition from a 4d orbital to any np orbital (subject to ∆ml = 0, ±1) and to any
nf orbital (subject to the same rule)
...

Self-test 10
...
10
...
The
thicker the line, the more intense the
transition
...
10
...
The thicknesses of the transition lines in the diagram
denote their relative intensities in the spectrum; we see how to determine transition
intensities in Section 13
...


The structures of many-electron atoms
The Schrödinger equation for a many-electron atom is highly complicated because all
the electrons interact with one another
...
We shall adopt a simple approach based on what we
already know about the structure of hydrogenic atoms
...

10
...
), where ri is the
vector from the nucleus to electron i
...
) = ψ (r1)ψ (r2)
...
29)

10
...
This description is only approximate, but it is a useful model for discussing
the chemical properties of atoms, and is the starting point for more sophisticated
descriptions of atomic structure
...
5 The orbital approximation

The orbital approximation would be exact if there were no interactions between
electrons
...
We shall now
show that if ψ (r1) is an eigenfunction of @1 with energy E1, and ψ (r2) is an eigenfunction of @2 with energy E2, then the product ψ (r1,r2) = ψ (r1)ψ (r2) is an eigenfunction of the combined hamiltonian @
...
This is the result we need to prove
...


(a) The helium atom

The orbital approximation allows us to express the electronic structure of an atom by
reporting its configuration, the list of occupied orbitals (usually, but not necessarily,
in its ground state)
...

The He atom has two electrons
...
The first electron occupies a 1s hydrogenic orbital, but because Z = 2 that orbital is more compact
than in H itself
...


1
ms = +2

(b) The Pauli principle

Lithium, with Z = 3, has three electrons
...
The third electron,
however, does not join the first two in the 1s orbital because that configuration is forbidden by the Pauli exclusion principle:

1
ms = -2

No more than two electrons may occupy any given orbital, and if two do occupy
one orbital, then their spins must be paired
...
Specifically, one
1
1
electron has ms = + –, the other has ms = − – and they are orientated on their respective
2
2
cones so that the resultant spin is zero (Fig
...
18)
...
It was proposed by Wolfgang Pauli in 1924 when he was trying to account for the

Fig
...
18 Electrons with paired spins have
zero resultant spin angular momentum
...


338

10 ATOMIC STRUCTURE AND ATOMIC SPECTRA
absence of some lines in the spectrum of helium
...

The Pauli exclusion principle in fact applies to any pair of identical fermions (particles with half integral spin)
...
It does not apply to iden2
2
tical bosons (particles with integral spin), which include photons (spin 1), 12C nuclei
(spin 0)
...

The Pauli exclusion principle is a special case of a general statement called the Pauli
principle:
When the labels of any two identical fermions are exchanged, the total wavefunction changes sign; when the labels of any two identical bosons are exchanged, the
total wavefunction retains the same sign
...
To see that the Pauli principle implies the Pauli exclusion principle, we
consider the wavefunction for two electrons ψ (1,2)
...
30)

Suppose the two electrons in an atom occupy an orbital ψ, then in the orbital approximation the overall wavefunction is ψ (1)ψ (2)
...
There are several
possibilities for two spins: both α, denoted α(1)α(2), both β, denoted β(1)β(2), and
one α the other β, denoted either α(1)β(2) or α(2)β(1)
...
3

A stronger justification for taking linear
combinations in eqn 10
...

See Section 10
...


(10
...
The total wavefunction of the system is therefore the product of the orbital
part and one of the four spin states:

ψ (1)ψ (2)α(1)α(2) ψ (1)ψ (2)β(1)β(2) ψ (1)ψ (2)σ+(1,2) ψ (1)ψ (2)σ−(1,2)
The Pauli principle says that for a wavefunction to be acceptable (for electrons), it
must change sign when the electrons are exchanged
...
The
same is true of α(1)α(2) and β(1)β(2)
...
The combination σ+(1,2) changes to

σ+(2,1) = (1/21/2){α(2)β(1) + β(2)α(1)} = σ+(1,2)
because it is simply the original function written in a different order
...
Finally, consider σ−(1,2):

σ−(2,1) = (1/21/2){α(2)β(1) − β(2)α(1)}
= −(1/21/2){α(1)β(2) − β(1)α(2)} = −σ−(1,2)
This combination does change sign (it is ‘antisymmetric’)
...


10
...
This is the content of the
Pauli exclusion principle
...
Nevertheless, even then the overall wavefunction must
still be antisymmetric overall, and must still satisfy the Pauli principle itself
...
In general, for N electrons in orbitals
ψa, ψb,
...
, N) =

1
(N!)1/2

ψa(1)α (1)
ψa(1)β (1)
ψb(1)α (1)
Ӈ
ψz(1)β (1)

ψa(2)α (2)
ψa(2)β (2)
ψb(2)α (2)
Ӈ
ψz(2)β (2)

ψa(3)α (3)
ψa(3)β (3)
ψb(3)α (3)
Ӈ
ψz(3)β (3)


...


...


ψa(N)α (N)
ψa(N)β (N)
ψb(N)α (N)
Ӈ
ψz(N)β (N)
[10
...
23
...
In Li (Z = 3), the third electron cannot enter the 1s
orbital because that orbital is already full: we say the K shell is complete and that the
two electrons form a closed shell
...
The third electron is excluded from the K shell and must
occupy the next available orbital, which is one with n = 2 and hence belonging to the
L shell
...


No net effect
of these
electrons

(c) Penetration and shielding

Unlike in hydrogenic atoms, the 2s and 2p orbitals (and, in general, all subshells of
a given shell) are not degenerate in many-electron atoms
...
If it is at a distance r from the nucleus,
it experiences an average repulsion that can be represented by a point negative charge
located at the nucleus and equal in magnitude to the total charge of the electrons
within a sphere of radius r (Fig
...
19)
...
In everyday parlance, Zeff itself is commonly referred to as the ‘effective nuclear charge’
...
33]

The electrons do not actually ‘block’ the full Coulombic attraction of the nucleus:
the shielding constant is simply a way of expressing the net outcome of the nuclear

r

Net effect
equivalent to
a point charge
at the centre
Fig
...
19 An electron at a distance r from
the nucleus experiences a Coulombic
repulsion from all the electrons within a
sphere of radius r and which is equivalent
to a point negative charge located on the
nucleus
...


10 ATOMIC STRUCTURE AND ATOMIC SPECTRA

Radial distribution function, P

340

3p
3s

0

4

8

Zr/a0

12

16

Fig
...
20 An electron in an s orbital (here a
3s orbital) is more likely to be found close
to the nucleus than an electron in a p
orbital of the same shell (note the closeness
of the innermost peak of the 3s orbital to
the nucleus at r = 0)
...


Exploration Calculate and plot the
graphs given above for n = 4
...
2* Effective nuclear
charge, Zeff = Z – σ
Element

Z

Orbital

Zeff

He

2

1s

1
...
6727

2s

3
...
1358

* More values are given in the Data section
...

The shielding constant is different for s and p electrons because they have different
radial distributions (Fig
...
20)
...
Because only electrons inside the sphere defined by the location of the electron (in effect, the core electrons) contribute to shielding, an s electron
experiences less shielding than a p electron
...
Similarly, a d electron penetrates less than a p electron of the same shell
(recall that the wavefunction of a d orbital varies as r 2 close to the nucleus, whereas a
p orbital varies as r), and therefore experiences more shielding
...
2)
...

We return to this point shortly
...

We can now complete the Li story
...
This occupation results in the ground-state
configuration 1s22s1, with the central nucleus surrounded by a complete helium-like
shell of two 1s electrons, and around that a more diffuse 2s electron
...
Thus,
the valence electron in Li is a 2s electron and its other two electrons belong to its core
...
In brief, we imagine the bare nucleus of atomic number Z, and then
feed into the orbitals Z electrons in succession
...
As an example, consider the
carbon atom, for which Z = 6 and there are six electrons to accommodate
...
Hence the ground-state configuration of
C is 1s22s22p2, or more succinctly [He]2s22p2, with [He] the helium-like 1s2 core
...
Thus, one electron can be thought of
as occupying the 2px orbital and the other the 2py orbital (the x, y, z designation is
arbitrary, and it would be equally valid to use the complex forms of these orbitals),
and the lowest energy configuration of the atom is [He]2s22p1 2p1
...
4 THE ORBITAL APPROXIMATION
applies whenever degenerate orbitals of a subshell are available for occupation
...

For instance, nitrogen (Z = 7) has the configuration [He]2s22p1 2p1 2p1 , and only when
x
y
z
we get to oxygen (Z = 8) is a 2p orbital doubly occupied, giving [He]2s22p2 2p1 2p1
...

The explanation of Hund’s rule is subtle, but it reflects the quantum mechanical property of spin correlation, that electrons with parallel spins behave as if they have a
tendency to stay well apart, and hence repel each other less
...
We can now conclude that, in the ground state
of the carbon atom, the two 2p electrons have the same spin, that all three 2p electrons
in the N atoms have the same spin, and that the two 2p electrons in different orbitals
in the O atom have the same spin (the two in the 2px orbital are necessarily paired)
...
6 Spin correlation

Suppose electron 1 is described by a wavefunction ψa(r1) and electron 2 is described
by a wavefunction ψb(r2); then, in the orbital approximation, the joint wavefunction of the electrons is the product ψ = ψa(r1)ψb(r2)
...
According to quantum mechanics, the
correct description is either of the two following wavefunctions:

ψ± = (1/21/2){ψa(r1)ψb(r2) ± ψb(r1)ψa(r2)}
According to the Pauli principle, because ψ+ is symmetrical under particle interchange, it must be multiplied by an antisymmetric spin function (the one denoted
σ−)
...
Conversely, ψ− is antisymmetric, so it must be multiplied by one of the three symmetric spin states
...
7
for an explanation)
...
We see that ψ− vanishes, which means that there is zero probability of finding the two electrons at the same point in space when they have parallel
spins
...
Because the two electrons have different relative spatial distributions
depending on whether their spins are parallel or not, it follows that their Coulombic
interaction is different, and hence that the two states have different energies
...

This closed-shell configuration is denoted [Ne], and acts as a core for subsequent elements
...
Like lithium with the configuration
[He]2s1, sodium has a single s electron outside a complete core
...
The L shell is completed by eight electrons, so the element with Z = 3 (Li) should have similar properties to the element
with Z = 11 (Na)
...


341

10 ATOMIC STRUCTURE AND ATOMIC SPECTRA

Energy

342

Fig
...
21 Strong electron–electron
repulsions in the 3d orbitals are minimized
in the ground state of Sc if the atom has the
configuration [Ar]3d14s 2 (shown on the
left) instead of [Ar]3d24s1 (shown on the
right)
...


Comment 10
...


Ten electrons can be accommodated in the five 3d orbitals, which accounts for the
electron configurations of scandium to zinc
...
5 show that for these atoms the energies of the 3d orbitals are always lower
than the energy of the 4s orbital
...
To understand this observation, we have to consider the nature of electron–electron repulsions in 3d and 4s
orbitals
...
As a result, Sc has the configuration [Ar]3d14s2 rather than the two alternatives,
for then the strong electron–electron repulsions in the 3d orbitals are minimized
...
10
...
The effect just described is generally true for
scandium through zinc, so their electron configurations are of the form [Ar]3dn4s2,
where n = 1 for scandium and n = 10 for zinc
...

At gallium, the building-up principle is used in the same way as in preceding
periods
...
Because 18 electrons have intervened since argon, this period is
the first ‘long period’ of the periodic table
...
A similar intrusion of
the f orbitals in Periods 6 and 7 accounts for the existence of the f block of the periodic
table (the lanthanoids and actinoids)
...
First, we remove valence p electrons, then valence s electrons, and then as many d electrons as are necessary to achieve the specified charge
...
It is reasonable that we remove the more energetic 4s electrons
in order to form the cation, but it is not obvious why the [Ar]3d3 configuration is preferred in V2+ over the [Ar]3d14s2 configuration, which is found in the isoelectronic Sc
atom
...
As Zeff increases, transfer of a 4s electron to a 3d orbital becomes more
favourable because the electron–electron repulsions are compensated by attractive
interactions between the nucleus and the electrons in the spatially compact 3d orbital
...
This conclusion explains why V2+ has a [Ar]3d3 configuration and
also accounts for the observed [Ar]4s03dn configurations of the M2+ cations of Sc
through Zn
...
Thus, the configuration of the O2− ion is
achieved by adding two electrons to [He]2s22p4, giving [He]2s22p6, the same as the
configuration of neon
...
The second ionization
energy, I2, is the minimum energy needed to remove a second electron (from the

10
...
10
...


singly charged cation)
...
10
...
3
...
As shown in the Justification below, the two are related by
5
∆ionH 7(T) = I + –RT
2

Synoptic table 10
...
34)

It follows from Kirchhoff’s law (Section 2
...
36) that the reaction enthalpy
for

I1/(kJ mol−1)

H

1312
2372

Mg

Justification 10
...
20 kJ mol−1
...


M(g) → M+(g) + e−(g)
at a temperature T is related to the value at T = 0 by
∆r H 7(T) = ∆rH 7(0) +

Τ

Ύ ∆ C dT
r

7
p

0

5
The molar constant-pressure heat capacity of each species in the reaction is – R, so
2
5
5
7
∆rC p = + – R
...
The reac2
2
tion enthalpy at T = 0 is the same as the (molar) ionization energy, I
...
33
then follows
...
4* Electron
affinities, Ea /(kJ mol−1)
Cl

349

F

322

H

73

O

141

O−

7

∆r H (T) = I1 + I2 + 5RT

* More values are given in the Data section
...
4)
...

It follows from a similar argument to that given in the Justification above that the
standard enthalpy of electron gain, ∆eg H 7, at a temperature T is related to the electron affinity by
5
∆eg H 7(T) = −Eea − –RT
2

(10
...
In typical thermodynamic cycles the
that appears in
eqn 10
...
34, so ionization energies and electron affinities can be
used directly
...
36)

–844

344

10 ATOMIC STRUCTURE AND ATOMIC SPECTRA
As ionization energy is often easier to measure than electron affinity, this relation can
be used to determine numerical values of the latter
...
The former is more regular and we concentrate on it
...
3, compared with Z = 3)
...
The ionization energy increases from boron to nitrogen on account
of the increasing nuclear charge
...
The explanation is that at oxygen a 2p
orbital must become doubly occupied, and the electron–electron repulsions are
increased above what would be expected by simple extrapolation along the row
...

(The kink is less pronounced in the next row, between phosphorus and sulfur because
their orbitals are more diffuse
...

The outermost electron in sodium is 3s
...
As a result, the ionization
energy of sodium is substantially lower than that of neon
...

Electron affinities are greatest close to fluorine, for the incoming electron enters a
vacancy in a compact valence shell and can interact strongly with the nucleus
...
The incoming electron is repelled by the charge
already present
...

10
...
The potential energy of the electrons is
V=−

Ze 2

∑ 4πε r
i

0 i

1
+–
2

∑
i,j

2
′ e
4πε0rij

(10
...
The first
term is the total attractive interaction between the electrons and the nucleus
...
It is hopeless to expect to find analytical solutions of a
Schrödinger equation with such a complicated potential energy term, but computational techniques are available that give very detailed and reliable numerical solutions
for the wavefunctions and energies
...
R
...
Fock to take
into account the Pauli principle correctly
...


10
...
In the Ne atom,
for instance, the orbital approximation suggests the configuration 1s22s22p6 with the
orbitals approximated by hydrogenic atomic orbitals
...
A Schrödinger equation can be written for this electron by ascribing to it
a potential energy due to the nuclear attraction and the repulsion from the other electrons
...
38)

A similar equation can be written for the 1s and 2s orbitals in the atom
...

2 The second takes into account the potential energy of the electron of interest due
to the electrons in the other occupied orbitals
...

There is no hope of solving eqn 10
...
However, it can be solved
numerically if we guess an approximate form of the wavefunctions of all the orbitals
except 2p
...
This sequence of calculations gives the form of the 2p, 2s, and 1s
orbitals, and in general they will differ from the set used initially to start the calculation
...
The recycling continues until the orbitals
and energies obtained are insignificantly different from those used at the start of the
current cycle
...

Figure 10
...
They show the grouping of electron density into shells, as was anticipated
by the early chemists, and the differences of penetration as discussed above
...
They also considerably extend that discussion by providing
detailed wavefunctions and precise energies
...


2p
2s

3s

0

K L

1 r/a
0

2

M

Fig
...
23 The radial distribution functions
for the orbitals of Na based on SCF
calculations
...


The spectra of complex atoms
The spectra of atoms rapidly become very complicated as the number of electrons
increases, but there are some important and moderately simple features that make
atomic spectroscopy useful in the study of the composition of samples as large and as
complex as stars (Impact I10
...
The general idea is straightforward: lines in the spectrum (in either emission or absorption) occur when the atom undergoes a transition
with a change of energy |∆E|, and emits or absorbs a photon of frequency ν = |∆E |/h
and # = |∆E |/hc
...
However, the actual energy levels are not given solely
by the energies of the orbitals, because the electrons interact with one another in
various ways, and there are contributions to the energy in addition to those we have
already considered
...
5

The web site for this text contains links
to databases of atomic spectra
...
1 Spectroscopy of stars

The bulk of stellar material consists of neutral and ionized forms of hydrogen and
helium atoms, with helium being the product of ‘hydrogen burning’ by nuclear fusion
...
It is generally accepted that the
outer layers of stars are composed of lighter elements, such as H, He, C, N, O, and Ne
in both neutral and ionized forms
...
The core itself contains the heaviest elements and 56Fe is particularly abundant because it is a very stable
nuclide
...
For example, the temperature is estimated to be 3
...

Astronomers use spectroscopic techniques to determine the chemical composition
of stars because each element, and indeed each isotope of an element, has a characteristic spectral signature that is transmitted through space by the star’s light
...
Nuclear reactions
in the dense stellar interior generate radiation that travels to less dense outer layers
...
To a good approximation, the distribution of energy emitted
from a star’s photosphere resembles the Planck distribution for a very hot black
body (Section 8
...
For example, the energy distribution of our Sun’s photosphere
may be modelled by a Planck distribution with an effective temperature of 5
...

Superimposed on the black-body radiation continuum are sharp absorption and emission lines from neutral atoms and ions present in the photosphere
...
The
data can also reveal the presence of small molecules, such as CN, C2, TiO, and ZrO, in
certain ‘cold’ stars, which are stars with relatively low effective temperatures
...
The photosphere, chromosphere, and corona comprise a
star’s ‘atmosphere’
...
The reasons for this increase
in temperature are not fully understood
...
5 MK, so black-body emission is strong from the X-ray to the radiofrequency region of the spectrum
...
The most intense emission lines in the visible range are
from the Fe13+ ion at 530
...
4 nm, and the Ca4+ ion at 569
...

Because only light from the photosphere reaches our telescopes, the overall chemical composition of a star must be inferred from theoretical work on its interior and
from spectral analysis of its atmosphere
...
8 per cent helium
...
2 per cent is due to heavier elements, among which C, N, O, Ne, and Fe are the most abundant
...
27) and their effective temperatures (Problem 13
...

10
...

However, we cannot use the procedure illustrated in Example 10
...
7 SINGLET AND TRIPLET STATES

347

because the energy levels of a many-electron atom do not in general vary as 1/n2
...
Typical values of Zeff are a little more than 1, so we expect binding
energies to be given by a term of the form −hcR/n2, but lying slightly lower in energy
than this formula predicts
...
The quantum defect is best regarded as a purely empirical
quantity
...
In such cases we can write
#=

I
hc

−

R

(10
...
If the lower state is
not the ground state (a possibility if we wish to generalize the concept of ionization
energy), the ionization energy of the ground state can be determined by adding the
appropriate energy difference to the ionization energy obtained as described here
...
7 Singlet and triplet states
Suppose we were interested in the energy levels of a He atom, with its two electrons
...
The two electrons need not be paired because
they occupy different orbitals
...
Both states are permissible, and can contribute to the spectrum of the atom
...
In the paired case, the two spin momenta cancel each other, and there is zero net
spin (as was depicted in Fig
...
18)
...

Its spin state is the one we denoted σ− in the discussion of the Pauli principle:

σ−(1,2) = (1/21/2){α(1)β(2) − β(1)α(2)}

(10
...
As illustrated in Fig
...
24, there are three
ways of achieving a nonzero total spin, but only one way to achieve zero spin
...
40b)

The fact that the parallel arrangement of spins in the 1s 2s configuration of the He
atom lies lower in energy than the antiparallel arrangement can now be expressed
by saying that the triplet state of the 1s12s1 configuration of He lies lower in energy
than the singlet state
...
The origin of the energy difference lies in the effect of
spin correlation on the Coulombic interactions between electrons, as we saw in the
case of Hund’s rule for ground-state configurations
...
The two states of 1s12s1
He, for instance, differ by 6421 cm−1 (corresponding to 0
...


1
ms = + 2

MS = 0
1
ms = - 2

1
ms = - 2

1
ms = - 2

MS = -1
Fig
...
24 When two electrons have parallel
spins, they have a nonzero total spin
angular momentum
...

Note that, although we cannot know the
orientation of the spin vectors on the
cones, the angle between the vectors is the
same in all three cases, for all three
arrangements have the same total spin
angular momentum (that is, the resultant
of the two vectors has the same length in
each case, but points in different
directions)
...
10
...
Note that, whereas two paired spins
are precisely antiparallel, two ‘parallel’
spins are not strictly parallel
...
6

667

...
56
52
...
71
58
...
Note that there are no
transitions between the singlet and
triplet levels
...
10
...
5

The spectrum of atomic helium is more complicated than that of atomic hydrogen,
but there are two simplifying features
...

Excitation of two electrons requires an energy greater than the ionization energy of
the atom, so the He+ ion is formed instead of the doubly excited atom
...
Thus, there is
a spectrum arising from transitions between singlet states (including the ground
state) and between triplet states, but not between the two
...
The Grotrian diagram for
helium in Fig
...
25 shows the two sets of transitions
...
6

We have already remarked that the
electron’s spin is a purely quantum
mechanical phenomenon that has no
classical counterpart
...
Namely, the
magnetic field generated by a spinning
electron, regarded classically as a
moving charge, induces a magnetic
moment
...


10
...
10
...
Similarly, an
electron with orbital angular momentum (that is, an electron in an orbital with l > 0)
is in effect a circulating current, and possesses a magnetic moment that arises from its
orbital momentum
...
The
strength of the coupling, and its effect on the energy levels of the atom, depend on
the relative orientations of the spin and orbital magnetic moments, and therefore
on the relative orientations of the two angular momenta (Fig
...
27)
...

Thus, when the spin and orbital angular momenta are nearly parallel, the total angular

10
...
10
...
When the angular
momenta are parallel, as in (a), the
magnetic moments are aligned
unfavourably; when they are opposed, as
in (b), the interaction is favourable
...


momentum is high; when the two angular momenta are opposed, the total angular
momentum is low
...
10
...
The different values of j that can arise
2
1
for a given value of l label levels of a term
...
When l = 1, j may be
3
1
either – (the spin and orbital angular momenta are in the same sense) or – (the spin
2
2
and angular momenta are in opposite senses)
...
For

=

2

l = 2

Identify the levels that may arise from the configurations (a) d1, (b) s1
...
4 Identifying the levels of a configuration

1
–
2
s =

j =–
5
2

Fig
...
26 Angular momentum gives rise to
a magnetic moment (µ)
...
For spin angular
momentum, there is a factor 2, which
increases the magnetic moment to twice
its expected value (see Section 10
...


Low
energy

these one-electron systems, the total angular momentum is the sum and difference
of the orbital and spin momenta
...
(b) For an s electron l = 0, so only
2
2
2
2
1
one level is possible, and j = –
...
8 Identify the levels of the configurations (a) p1 and (b) f 1
...
10
...


350

10 ATOMIC STRUCTURE AND ATOMIC SPECTRA
The dependence of the spin–orbit interaction on the value of j is expressed in terms
of the spin–orbit coupling constant, A (which is typically expressed as a wavenumber)
...
41)

Justification 10
...
7

The scalar product (or dot product) u·1
of two vectors u and 1 with magnitudes
u and 1 is u·1 = u1 cos θ, where θ is the
angle between the two vectors
...
If the magnetic field arises from the orbital angular momentum of
the electron, it is proportional to l; if the magnetic moment µ is that of the electron
spin, then it is proportional to s
...
Next, we note that the total angular momentum is the vector sum of the spin and
orbital momenta: j = l + s
...

The preceding equation is a classical result
...
42)

At this point, we calculate the first-order correction to the energy by evaluating the
expectation value:
1
1
͗ j,l,s|£· ™| j,l,s͘ = –͗ j,l,s|¡ 2 − ™ 2 − s 2 | j,l,s͘ = –{j(j + 1) − l(l + 1) − s(s + 1)}$2 (10
...
42
...


Illustration 10
...
Note that the
low-j level lies below the high-j level in
energy
...
10
...

2
Because the orbital angular momentum is zero in this state, the spin–orbit coupling energy is zero (as is confirmed by setting j = s and l = 0 in eqn 10
...
When the
electron is excited to an orbital with l = 1, it has orbital angular momentum and can
give rise to a magnetic field that interacts with its spin
...
10
...
Note that the baricentre (the
‘centre of gravity’) of the levels is unchanged, because there are four states of energy
1
– hcA and two of energy −hcA
...
8 SPIN–ORBIT COUPLING

~
n /cm-1
16 973

Example 10
...
10
...

Calculate the spin–orbit coupling constant for the upper configuration of the Na
atom
...
10
...
This separation can
2
2
be expressed in terms of A by using eqn 10
...
Therefore, set the observed splitting
equal to the energy separation calculated from eqn 10
...


Answer The two levels are split by
1 3 3
1 1
3
∆# = A –{–(– + 1) − – (– + 1)} = –A
2 2 2
2 2
2

The experimental value is 17
...
2 cm−1) = 11
...
23 cm−1,
K: 38
...
Note the increase of A with atomic number (but more slowly than Z 4 for these many-electron atoms)
...
9 The configuration
...
56

cm−1 and 25 703
...
What is the spin–orbit coupling
constant in this excited state?
[1
...
16 nm

D1

(b) Fine structure

Two spectral lines are observed when the p electron of an electronically excited alkali
metal atom undergoes a transition and falls into a lower s orbital
...
The two lines are an example of the fine struc2
ture of a spectrum, the structure in a spectrum due to spin–orbit coupling
...
The yellow line at
589 nm (close to 17 000 cm−1) is actually a doublet composed of one line at 589
...
2 cm−1) and another at 589
...
4 cm−1); the components of this
doublet are the ‘D lines’ of the spectrum (Fig
...
30)
...


2

589
...
To understand why this is so, imagine riding on the orbiting electron and seeing a charged
nucleus apparently orbiting around us (like the Sun rising and setting)
...
The greater the nuclear charge, the
greater this current, and therefore the stronger the magnetic field we detect
...

The coupling increases sharply with atomic number (as Z 4)
...
4 cm−1), in heavy
atoms like Pb it is very large (giving shifts of the order of thousands of reciprocal
centimetres)
...
10
...
The
splitting of the spectral lines (by 17 cm−1)
reflects the splitting of the levels of the
2
P term
...
9 Term symbols and selection rules

Configuration

3
We have used expressions such as ‘the j = – level of a configuration’
...
The convention of using lowercase letters to label orbitals and uppercase letters to label overall states applies throughout spectroscopy, not just to atoms
...


Spin
correlation +
electrostatic

5

3

P

2 The left superscript in the term symbol (the 2 in 2P3/2) gives the multiplicity of
the term
...


1

P

P

Magnetic
(spin- orbit)

3

3

3

P2

We shall now say what each of these statements means; the contributions to the energies which we are about to discuss are summarized in Fig
...
31
...
For light atoms, magnetic
interactions are small, but in heavy atoms
they may dominate the electrostatic
(charge–charge) interactions
...
10
...
, |l1 − l2 |
l =
1

1

l = 2

=

L

L

=

2

1

L =
3
l = 2

l =

l = 2

When several electrons are present, it is necessary to judge how their individual orbital
angular momenta add together or oppose each other
...
It has 2L + 1 orientations distinguished by the quantum number ML, which can take the values L, L − 1,
...
Similar remarks apply to the total
spin quantum number, S, and the quantum number MS, and the total angular
momentum quantum number, J, and the quantum number MJ
...
10
...


(10
...
The maximum
value, L = l1 + l2, is obtained when the two orbital angular momenta are in the same
direction; the lowest value, |l1 − l2 |, is obtained when they are in opposite directions
...
10
...
For two p electrons (for which l1 = l2 = 1), L = 2, 1, 0
...

designation of orbitals, but uses uppercase Roman letters:
L:

0

1

2

3

4

5

6
...


Thus, a p2 configuration can give rise to D, P, and S terms
...

A closed shell has zero orbital angular momentum because all the individual orbital
angular momenta sum to zero
...
In the case of a single electron outside
a closed shell, the value of L is the same as the value of l; so the configuration [Ne]3s1
has only an S term
...
9 TERM SYMBOLS AND SELECTION RULES

353

Example 10
...

Method Use the Clebsch–Gordan series and begin by finding the minimum value

of L (so that we know where the series terminates)
...

Answer (a) The minimum value is | l1 − l2 | = |2 − 2| = 0
...
, 0 = 4, 3, 2, 1, 0
corresponding to G, F, D, P, S terms, respectively
...
Therefore,
L′ = 1 + 1, 1 + 1 − 1,
...
The overall result is
L = 3, 2, 2, 1, 1, 1, 0
giving one F, two D, three P, and one S term
...
10 Repeat the question for the configurations (a) f 1d1 and (b) d3
...

Once again, we use the Clebsch–Gordan series in the form

Throughout our discussion of atomic
spectroscopy, distinguish italic S, the
total spin quantum number, from
Roman S, the term label
...
45)

(c) The total angular momentum

As we have seen, the quantum number j tells us the relative orientation of the spin
and orbital angular momenta of a single electron
...
If there is a single electron outside a closed shell, J = j, with j either l + – or
2
1
1
1
1
|l − – |
...
The [Ne]3p1 configuration has l = 1; therefore

1
–
2
s =

(a)

S

1
s = 2

=

1

S = 0
1
s = 2

1
to decide on the value of S, noting that each electron has s = –, which gives S = 1, 0 for
2
two electrons (Fig
...
33)
...

2
2
The multiplicity of a term is the value of 2S + 1
...
A single electron has S = s = –, so a configuration such as [Ne]3s1 can
2
2
give rise to a doublet term, S
...

When there are two unpaired electrons S = 1, so 2S + 1 = 3, giving a triplet term, such
as 3D
...
7 and saw
that their energies differ on account of the different effects of spin correlation
...
, |s1 − s2 |

Comment 10
...
10
...
The state with S = 0
can have only one value of MS (MS = 0) and
is a singlet; the state with S = 1 can have any
of three values of MS (+1, 0, −1) and is a
triplet
...
10
...
24, respectively
...
These levels lie at
2
2
different energies on account of the magnetic spin–orbit interaction
...
This complicated problem can be
simplified when the spin–orbit coupling is weak (for atoms of low atomic number),
for then we can use the Russell–Saunders coupling scheme
...
We therefore imagine that all the
orbital angular momenta of the electrons couple to give a total L, and that all the spins
are similarly coupled to give a total S
...
The permitted values of J are given by the Clebsch–Gordan series

J = L + S, L + S − 1,
...
46)

For example, in the case of the 3D term of the configuration [Ne]2p13p1, the permitted values of J are 3, 2, 1 (because 3D has L = 2 and S = 1), so the term has three levels,
3
D3, 3D2, and 3D1
...
For example, a 2P
term has the two levels 2P3/2 and 2P1/2, and 3D has the three levels 3D3, 3D2, and 3D1
...

Example 10
...

Method Begin by writing the configurations, but ignore inner closed shells
...
Next, couple L and S
to find J
...
For
F, for which the valence configuration is 2p5, treat the single gap in the closed-shell
2p6 configuration as a single particle
...
Because L = l = 0 and S = s = –, it is possible for J = j = s = – only
...
(b) For F, the configuration is [He]2s 2p , which we can treat
as [Ne]2p−1 (where the notation 2p−1 signifies the absence of a 2p electron)
...
Two values of J = j are allowed: J = –, –
...
(c) We are treating an excited configuration of carbon because, in the ground configuration, 2p2, the Pauli principle
forbids some terms, and deciding which survive (1D, 3P, 1S, in fact) is quite complicated
...
For information about how to deal with
equivalent electrons, see Further reading
...
This is a two-electron problem, and l1 = l2 = 1,
1
s1 = s2 = –
...
The terms are therefore 3D and 1D,
2
3
1
P and P, and 3S and 1S
...
For 1D, L = 2 and S = 0, so the single level is 1D2
...
For the 3S term there is only
one level, 3S1 (because J = 1 only), and the singlet term is 1S0
...
9 TERM SYMBOLS AND SELECTION RULES

355

Self-test 10
...

[(a) 3P2, 3P1, 3P0, 1P1;
(b) F4, F3, F2, F3, D3, D2, D1, 1D2, 3P1, 3P0, 1P1]
3

3

3

1

3

3

3

Russell–Saunders coupling fails when the spin–orbit coupling is large (in heavy
atoms)
...
This scheme is called jj-coupling
...
If the spin and the orbital angular momentum
2
2
of each electron are coupled together strongly, it is best to consider each electron as a
3
1
particle with angular momentum j = – or –
...

Although jj-coupling should be used for assessing the energies of heavy atoms, the
term symbols derived from Russell–Saunders coupling can still be used as labels
...
Such a correlation
diagram is shown in Fig
...
34
...

(d) Selection rules

Any state of the atom, and any spectral transition, can be specified by using term symbols
...
10
...
The corresponding absorptions
are therefore denoted
2

Pure
Russell–Saunders
coupling

1

D2

P1/2 ← 2S1/2
3

(The configurations have been omitted
...
They can
therefore be expressed in terms of the term symbols, because the latter carry information about angular momentum
...
47)

where the symbol ←|→ denotes a forbidden transition
...

The rules about ∆L and ∆l express the fact that the orbital angular momentum of an

P2
P1
P0

3
3

C Si Ge

Sn

Pb

Fig
...
34 The correlation diagram for some
of the states of a two-electron system
...


356

10 ATOMIC STRUCTURE AND ATOMIC SPECTRA
individual electron must change (so ∆l = ±1), but whether or not this results in an
overall change of orbital momentum depends on the coupling
...
If we insist on labelling the terms of heavy atoms with symbols like 3D,
then we shall find that the selection rules progressively fail as the atomic number
increases because the quantum numbers S and L become ill defined as jj-coupling
becomes more appropriate
...
For
this reason, transitions between singlet and triplet states (for which ∆S = ±1), while
forbidden in light atoms, are allowed in heavy atoms
...
A hydrogenic atom is a one-electron atom or ion of general
atomic number Z
...

2
...

3
...


13
...

14
...

15
...


4
...


16
...


5
...


17
...


6
...


18
...


7
...

8
...

9
...

10
...

11
...
) = ψ (r1)ψ (r2)
...
The Pauli exclusion principle states that no more than two
electrons may occupy any given orbital and, if two do occupy
one orbital, then their spins must be paired
...
The first ionization energy I1 is the minimum energy
necessary to remove an electron from a many-electron atom
in the gas phase; the second ionization energy I2 is the
minimum energy necessary to remove an electron from an
ionized many-electron atom in the gas phase
...
The electron affinity Eea is the energy released when an
electron attaches to a gas-phase atom
...
A singlet term has S = 0; a triplet term has S = 1
...
Spin–orbit coupling is the interaction of the spin magnetic
moment with the magnetic field arising from the orbital
angular momentum
...
Fine structure is the structure in a spectrum due to spin–orbit
coupling
...
A term symbol is a symbolic specification of the state of an
atom, 2S+1{L}J
...
The allowed values of a combined angular momenta are
obtained by using the Clebsch–Gordan series: J = j1 + j2,
j1 + j2 − 1,
...


357

FURTHER INFORMATION
26
...

28
...

29
...


30
...

31
...


Further reading
Articles and texts

P
...
Atkins, Quanta: a handbook of concepts
...

P
...
Bernath, Spectra of atoms and molecules
...

K
...
Happer, Atomic spectroscopy
...
G
...
Trigg), 2, 245
...

E
...
Condon and H
...
Cambridge
University Press (1980)
...
W
...
J
...

Educ
...

C
...
Johnson, Jr
...
G
...
Dover, New York (1986)
...
C
...
W
...
Kirby, and D
...
In Encyclopedia of applied physics
(ed
...
L
...
VCH, New York (1993)
...
Shenkuan, The physical basis of Hund’s rule: orbital contraction
effects
...
Chem
...
69, 800 (1992)
...
G
...
Pierloot, and D
...
J
...
Educ
...

Sources of data and information

S
...
O
...
, Atomic energy levels and Grotrian
diagrams
...

D
...
Lide (ed
...


Further information
x1

Further information 10
...
The total
energy is
E=

2
p1

2m1

+

2
p2

2m2

+V

where p1 = m1R1 and p2 = m2 R2, the dot signifying differentiation with
respect to time
...
10
...
It follows that
x1 = X +

m2
m

x

x2 = X −

m1
m

x

The linear momenta of the particles can be expressed in terms of the
rates of change of x and X:

m1

m2
x

Fig
...
35 The coordinates used for discussing the separation of the
relative motion of two particles from the motion of the centre
of mass
...
6
...
m
...

Now we write the overall wavefunction as the product ψtotal =
ψc
...
ψ, where the first factor is a function of only the centre of mass
coordinates and the second is a function of only the relative
coordinates
...
2a
and 9
...
m
...


$2 A ∂ 2 2 ∂ 1 2 D
B
+
+ Λ E RY + VRY = ERY
2µ C ∂r 2 r ∂r r 2 F

Because R depends only on r and Y depends only on the angular
coordinates, this equation becomes
−

$2 A d2R 2Y dR R 2 D
BY
+
+ Λ Y E + VRY = ERY
F
2µ C dr 2 r dr r 2

If we multiply through by r 2/RY, we obtain
−

$2 A 2 d2R
dR D
$2 2
Br
E + Vr 2 −
+ 2r
Λ Y = Er 2
2
dr F
2µR C dr
2µY

At this point we employ the usual argument
...

When we write this constant as $2l(l + 1)/2µ, eqn 10
...


The separation of angular and radial motion

The laplacian in three dimensions is given in eqn 9
...
It follows that
the Schrödinger equation in eqn 10
...
1 Describe the separation of variables procedure as it is applied to simplify
the description of a hydrogenic atom free to move through space
...
5 Outline the electron configurations of many-electron atoms in terms of
their location in the periodic table
...
2 List and describe the significance of the quantum numbers needed to
specify the internal state of a hydrogenic atom
...
6 Describe and account for the variation of first ionization energies along
Period 2 of the periodic table
...
3 Specify and account for the selection rules for transitions in hydrogenic
atoms
...
7 Describe the orbital approximation for the wavefunction of a manyelectron atom
...
4 Explain the significance of (a) a boundary surface and (b) the radial
distribution function for hydrogenic orbitals
...
8 Explain the origin of spin–orbit coupling and how it affects the

appearance of a spectrum
...
1a When ultraviolet radiation of wavelength 58
...
59 Mm s−1
...

10
...
4 nm from a helium lamp

is directed on to a sample of xenon, electrons are ejected with a speed of
1
...
Calculate the ionization energy of xenon
...
4b The wavefunction for the 2s orbital of a hydrogen atom is
N(2 – r/a0)e−r/2a0
...

10
...

10
...


10
...


10
...
2b By differentiation of the 3s radial wavefunction, show that it has three

electron in a hydrogenic atom and determine the radius at which the electron
is most likely to be found
...

10
...

10
...
1, the radial wavefunction is proportional to 20 − 10ρ + ρ 2
...
4a The wavefunction for the ground state of a hydrogen atom is Ne− r/a0
...


10
...

10
...


PROBLEMS
10
...


10
...
8a What is the orbital angular momentum of an electron in the orbitals
(a) 1s, (b) 3s, (c) 3d? Give the numbers of angular and radial nodes in each case
...
14b Which of the following transitions are allowed in the normal
electronic emission spectrum of an atom: (a) 5d → 2s, (b) 5p → 3s,
(c) 6p → 4f?

10
...

10
...

10
...

3
10
...
What is its orbital angular momentum quantum number in each case?
2

10
...
15a (a) Write the electronic configuration of the Ni2+ ion
...
15b (a) Write the electronic configuration of the V2+ ion
...
16a Suppose that an atom has (a) 2, (b) 3 electrons in different orbitals
...
16b Suppose that an atom has (a) 4, (b) 5 electrons in different orbitals
...
11a State the orbital degeneracy of the levels in a hydrogen atom that have

What are the possible values of the total spin quantum number S? What is the
multiplicity in each case?

1
1
–
energy (a) –hcRH; (b) – –hcRH; (c) – 25 hcRH
...
17a What atomic terms are possible for the electron configuration ns1nd1 ?

10
...

10
...
12b What information does the term symbol 3F4 provide about the

angular momentum of an atom?
10
...
13b At what radius in the H atom does the radial distribution function of

the ground state have (a) 50 per cent, (b) 75 per cent of its maximum value?

Which term is likely to lie lowest in energy?
10
...
18a What values of J may occur in the terms (a) 1S, (b) 2P, (c) 3P? How

many states (distinguished by the quantum number MJ) belong to each level?
10
...
19a Give the possible term symbols for (a) Li [He]2s1, (b) Na [Ne]3p1
...
19b Give the possible term symbols for (a) Sc [Ar]3d 14s 2,

(b) Br [Ar]3d 104s 24p 5
...
1 The Humphreys series is a group of lines in the spectrum of atomic

hydrogen
...
4 nm
...
2 A series of lines in the spectrum of atomic hydrogen lies at 656
...
27 nm, 434
...
29 nm
...
3 The Li2+ ion is hydrogenic and has a Lyman series at 740 747 cm−1,
877 924 cm−1, 925 933 cm−1, and beyond
...
Go on to predict the
wavenumbers of the two longest-wavelength transitions of the Balmer series
of the ion and find the ionization energy of the ion
...
4 A series of lines in the spectrum of neutral Li atoms rise from

combinations of 1s 22p1 2P with 1s 2nd 1 2D and occur at 610
...
29 nm,
and 413
...
The d orbitals are hydrogenic
...
78 nm above the ground state, which is 1s 22s 1 2S
...


10
...
P
...
H
...
A
...
Rev
...
The two contending configurations are
[Rn]5f 147s27p1 and [Rn]5f 146d7s2
...
Which
level would be lowest according to a simple estimate of spin–orbit coupling?
10
...
On close inspection, the line is found to have two closely spaced
components, one at 766
...
11 nm
...

10
...
098 cm−1 whereas that of D lies at 82 281
...

Calculate the ratio of the ionization energies of H and D
...
8 Positronium consists of an electron and a positron (same mass, opposite
charge) orbiting round their common centre of mass
...
Predict the wavenumbers of the first
three lines of the Balmer series of positronium
...


360

10 ATOMIC STRUCTURE AND ATOMIC SPECTRA

10
...
It arises from the interaction between
applied magnetic fields and the magnetic moments due to orbital and spin
angular momenta (recall the evidence provided for electron spin by the
Stern–Gerlach experiment, Section 9
...
To gain some appreciation for the socalled normal Zeeman effect, which is observed in transitions involving singlet
states, consider a p electron, with l = 1 and ml = 0, ±1
...
When a field of magnitude B
is present, the degeneracy is removed and it is observed that the state with
ml = +1 moves up in energy by µ BB, the state with ml = 0 is unchanged, and
the state with ml = –1 moves down in energy by µ BB, where µB = e$/2me =
9
...
1)
...
(a) Calculate the splitting in reciprocal centimetres between
the three spectral lines of a transition between a 1S0 term and a 1P1 term in the
presence of a magnetic field of 2 T (where 1 T = 1 kg s−2 A−1)
...
Is the line splitting caused by the
normal Zeeman effect relatively small or relatively large?
10
...
Its atomic number was
believed to be 126
...
)

Theoretical problems
10
...
12 Show by explicit integration that (a) hydrogenic 1s and 2s orbitals,

(b) 2px and 2py orbitals are mutually orthogonal
...
13‡ Explicit expressions for hydrogenic orbitals are given in Tables 10
...
3
...
(b) Determine the positions of both the
radial nodes and nodal planes of the 3s, 3px, and 3dxy orbitals
...
(d) Draw a graph of the radial distribution
function for the three orbitals (of part (b)) and discuss the significance of the
graphs for interpreting the properties of many-electron atoms
...
Construct
the boundary plots so that the distance from the origin to the surface is the
absolute value of the angular part of the wavefunction
...
g
...

10
...
If not,

does a linear combination exist that is an eigenfunction of lz?
10
...
What is the significance of this result?
10
...
Calculate the ‘size’ of a hydrogen
atom in its ground state according to this definition
...

10
...
Evaluate the expectation value of 1/r for (a) a
hydrogen 1s orbital, (b) a hydrogenic 2s orbital, (c) a hydrogenic 2p orbital
...
18 One of the most famous of the obsolete theories of the hydrogen atom
was proposed by Bohr
...
In the Bohr atom, an electron travels in a circle around
the nucleus
...
Bohr proposed that the angular
momentum is limited to integral values of $
...
Calculate the energies of a hydrogenic atom using the Bohr model
...
19 The Bohr model of the atom is specified in Problem 10
...
What
features of it are untenable according to quantum mechanics? How does the
Bohr ground state differ from the actual ground state? Is there an
experimental distinction between the Bohr and quantum mechanical models
of the ground state?
10
...
The usual choice is that of a hydrogen atom, with the unit of
length being the Bohr radius, a0, and the unit of energy being the (negative of
the) energy of the 1s orbital
...
21 Some of the selection rules for hydrogenic atoms were derived in
Justification 10
...
Complete the derivation by considering the x- and ycomponents of the electric dipole moment operator
...
22‡ Stern–Gerlach splittings of atomic beams are small and require either

large magnetic field gradients or long magnets for their observation
...
9), L is the length of the magnet, EK is the average kinetic energy
of the atoms in the beam, and dB/dz is the magnetic field gradient across the
beam
...
(b) Calculate the magnetic field
gradient required to produce a splitting of 1
...

10
...
4b)
...

Use this property to show that (a) the wavefunction is antisymmetric under
particle exchange, (b) no two electrons can occupy the same orbital with the
same spin
...
24 Hydrogen is the most abundant element in all stars
...
Account for
this observation
...
25 The distribution of isotopes of an element may yield clues about the
nuclear reactions that occur in the interior of a star
...

10
...
Such Rydberg atoms have unique properties and are of interest to
astrophysicists
...
Calculate this separation for n = 100; also calculate
the average radius, the geometric cross-section, and the ionization energy
...
27 The spectrum of a star is used to measure its radial velocity with respect
to the Sun, the component of the star’s velocity vector that is parallel to a
vector connecting the star’s centre to the centre of the Sun
...
When a star emitting
electromagnetic radiation of frequency ν moves with a speed s relative to an
observer, the observer detects radiation of frequency νreceding = ν f or νapproaching
= ν/f, where f = {(1 – s/c)/(1 + s/c)}1/2 and c is the speed of light
...

Furthermore, νapproaching > ν and an approaching star is characterized by a blue
shift of its spectrum with respect to the spectrum of an identical, but stationary
source
...
Measurement of the same spectral line in a star
gives νstar and the speed of recession or approach may be calculated from the
value of ν and the equations above
...
882 nm,
441
...
020 nm
...
392 nm, 440
...
510 nm in the spectrum of an Earth-bound iron arc
...
(b) What additional
information would you need to calculate the radial velocity of HDE 271 182
with respect to the Sun?
10
...
For this reason, they are found in many oxidoreductases and
in several proteins of oxidative phosphorylation and photosynthesis (Impact
I7
...
2)
...

10
...
Aluminium,
which causes anaemia and dementia, is also a member of the group but its
chemical properties are dominated by the +3 oxidation state
...
Explain the trends you observe
...

The third ionization energy, I3, is the minimum energy needed to remove an
electron from the doubly charged cation: E2+(g) → E3+(g) + e−(g), I3 = E(E3+ )
− E(E2+)
...


11
The Born–Oppenheimer
approximation
Valence-bond theory
11
...
2 Polyatomic molecules

Molecular structure
The concepts developed in Chapter 10, particularly those of orbitals, can be extended to
a description of the electronic structures of molecules
...
In valence-bond theory, the starting
point is the concept of the shared electron pair
...
The theory introduces the concepts of σ and π bonds, promotion, and
hybridization that are used widely in chemistry
...


Molecular orbital theory
11
...
4 Homonuclear diatomic

molecules
11
...
1 Impact on biochemistry:

The biochemical reactivity
of O2, N2, and NO
Molecular orbitals for
polyatomic systems
11
...
7 Computational chemistry
11
...
The quantum mechanical description of chemical bonding has become highly developed through the use of computers,
and it is now possible to consider the structures of molecules of almost any complexity
...
N
...
We
shall see, however, that the other principal type of bond, an ionic bond, in which the
cohesion arises from the Coulombic attraction between ions of opposite charge, is
also captured as a limiting case of a covalent bond between dissimilar atoms
...

There are two major approaches to the calculation of molecular structure, valencebond theory (VB theory) and molecular orbital theory (MO theory)
...
Valence-bond theory, though, has left its imprint on the language of chemistry, and it is important to know the significance of terms that chemists
use every day
...
First, we set out the
concepts common to all levels of description
...
Next, we present the basic
ideas of MO theory
...


The Born–Oppenheimer approximation
All theories of molecular structure make the same simplification at the outset
...
We therefore adopt the Born–Oppenheimer approximation in which it is supposed that the nuclei, being so much heavier than an electron, move relatively slowly and may be treated as stationary while the electrons move
in their field
...

The approximation is quite good for ground-state molecules, for calculations suggest that the nuclei in H2 move through only about 1 pm while the electron speeds
through 1000 pm, so the error of assuming that the nuclei are stationary is small
...
4) and mass spectrometry
...
Then we choose a different separation and repeat the
calculation, and so on
...
11
...
When more than one molecular parameter is changed in a polyatomic molecule, we obtain a potential energy surface
...
Once the curve has been calculated or determined experimentally (by using
the spectroscopic techniques described in Chapters 13 and 14), we can identify the
equilibrium bond length, Re, the internuclear separation at the minimum of the
curve, and the bond dissociation energy, D0, which is closely related to the depth, De,
of the minimum below the energy of the infinitely widely separated and stationary
atoms
...
The language it introduced, which includes concepts such as spin pairing,
orbital overlap, σ and π bonds, and hybridization, is widely used throughout chemistry,
especially in the description of the properties and reactions of organic compounds
...

11
...

To understand why this pairing leads to bonding, we have to examine the wavefunction for the two electrons that form the bond
...

The spatial wavefunction for an electron on each of two widely separated H atoms
is

ψ = χH1sA(r1)χH1sB(r2)
if electron 1 is on atom A and electron 2 is on atom B; in this chapter we use χ (chi) to
denote atomic orbitals
...

When the atoms are close, it is not possible to know whether it is electron 1 that is
on A or electron 2
...
When two outcomes are equally probable,

363

Energy

11
...
11
...
The equilibrium bond length
corresponds to the energy minimum
...
1

The dissociation energy differs from the
depth of the well by an energy equal to
the zero-point vibrational energy of the
1
bonded atoms: D0 = De − – $ω , where ω
2
is the vibrational frequency of the bond
(Section 13
...


364

11 MOLECULAR STRUCTURE
quantum mechanics instructs us to describe the true state of the system as a superposition of the wavefunctions for each possibility (Section 8
...

However, this illustration is an attempt
...
The top illustration represents
A(1)B(2), and the middle illustration
represents the contribution A(2)B(1)
...


Fig
...
2

(11
...
2)

The formation of the bond in H2 can be pictured as due to the high probability that
the two electrons will be found between the two nuclei and hence will bind them
together
...
11
...

The electron distribution described by the wavefunction in eqn 11
...
A σ bond has cylindrical symmetry around the internuclear axis, and is
so called because, when viewed along the internuclear axis, it resembles a pair of
electrons in an s orbital (and σ is the Greek equivalent of s)
...
The origin of the role of spin is that the wavefunction given in eqn 11
...

Spin pairing is not an end in itself: it is a means of achieving a wavefunction (and the
probability distribution it implies) that corresponds to a low energy
...
1 Electron pairing in VB theory

The Pauli principle requires the wavefunction of two electrons to change sign when
the labels of the electrons are interchanged (see Section 10
...
The total VB wavefunction for two electrons is

ψ (1,2) = {A(1)B(2) + A(2)B(1)}σ (1,2)
where σ represents the spin component of the wavefunction
...
The combination of two spins that has this property is

σ−(1,2) = (1/21/2){α (1)β(2) − α (2)β(1)}
which corresponds to paired electron spins (Section 10
...
Therefore, we conclude
that the state of lower energy (and hence the formation of a chemical bond) is
achieved if the electron spins are paired
...
11
...


The VB description of H2 can be applied to other homonuclear diatomic molecules,
such as nitrogen, N2
...
It is conventional
z
to take the z-axis to be the internuclear axis, so we can imagine each atom as having a
2pz orbital pointing towards a 2pz orbital on the other atom (Fig
...
3), with the 2px
and 2py orbitals perpendicular to the axis
...
Its spatial wavefunction is given by
eqn 11
...


11
...
Instead, they merge to form two π
bonds
...
11
...
It is so called because, viewed along the internuclear axis, a π bond resembles a pair of electrons in a p orbital (and π is the Greek
equivalent of p)
...
The overall
bonding pattern in N2 is therefore a σ bond plus two π bonds (Fig
...
5), which is consistent with the Lewis structure :N
...

11
...

Likewise, π bonds are formed by pairing electrons that occupy atomic orbitals of the
appropriate symmetry
...
The valence electron configuration
2 1
of an O atom is 2s 22px 2py 2p1
...
Because the 2py and 2pz orbitals lie at 90° to each other, the two σ bonds
also lie at 90° to each other (Fig
...
6)
...
However, the theory predicts a bond angle of 90°,
whereas the actual bond angle is 104
...


365

Nodal
plane

Internuclear
axis
Fig
...
4 A π bond results from orbital
overlap and spin pairing between electrons
in p orbitals with their axes perpendicular
to the internuclear axis
...


Self-test 11
...

[A trigonal pyramidal molecule with each N-H bond 90°; experimental: 107°]

Another deficiency of VB theory is its inability to account for carbon’s tetravalence
1
(its ability to form four bonds)
...
This deficiency is overcome by allowing for promotion, the excitation of an electron to an orbital of higher energy
...
These electrons may pair with four
electrons in orbitals provided by four other atoms (such as four H1s orbitals if the
molecule is CH4), and hence form four σ bonds
...
Promotion, and
the formation of four bonds, is a characteristic feature of carbon because the promotion energy is quite small: the promoted electron leaves a doubly occupied 2s orbital
and enters a vacant 2p orbital, hence significantly relieving the electron–electron
repulsion it experiences in the former
...

The description of the bonding in CH4 (and other alkanes) is still incomplete
because it implies the presence of three σ bonds of one type (formed from H1s and C2p

Fig
...
5 The structure of bonds in a
nitrogen molecule: there is one σ bond and
two π bonds
...


H
O

H

Fig
...
6 A first approximation to the
valence-bond description of bonding in an
H2O molecule
...
This model suggests that the
bond angle should be 90°, which is
significantly different from the
experimental value
...
2

A characteristic property of waves is
that they interfere with one another,
resulting in a greater displacement
where peaks or troughs coincide, giving
rise to constructive interference, and a
smaller displacement where peaks
coincide with troughs, giving rise to
destructive interference
...


orbitals) and a fourth σ bond of a distinctly different character (formed from H1s and
C2s)
...
The
origin of the hybridization can be appreciated by thinking of the four atomic orbitals
centred on a nucleus as waves that interfere destructively and constructively in different regions, and give rise to four new shapes
...
3)

As a result of the interference between the component orbitals, each hybrid orbital
consists of a large lobe pointing in the direction of one corner of a regular tetrahedron
(Fig
...
7)
...
47°
...

It is now easy to see how the valence-bond description of the CH4 molecule leads to
a tetrahedral molecule containing four equivalent C-H bonds
...
For
example, the (un-normalized) wavefunction for the bond formed by the hybrid
orbital h1 and the 1sA orbital (with wavefunction that we shall denote A) is

ψ = h1(1)A(2) + h1(2)A(1)
Because each sp3 hybrid orbital has the same composition, all four σ bonds are identical apart from their orientation in space (Fig
...
8)
...
11
...
As a result, the bond strength is greater than for a bond formed

Resultant
+

2p
H

Constructive
interference

C
+

109
...
There are four such hybrids:
each one points towards the corner of a
regular tetrahedron
...


Fig
...
7

Each sp3 hybrid orbital forms a σ
bond by overlap with an H1s orbital
located at the corner of the tetrahedron
...

Fig
...
8

Fig
...
9 A more detailed representation of
the formation of an sp3 hybrid by
interference between wavefunctions
centred on the same atomic nucleus
...
)

11
...
This increased bond strength is another factor that helps
to repay the promotion energy
...
An ethene molecule is planar,
with HCH and HCC bond angles close to 120°
...
However, instead of using all
four orbitals to form hybrids, we form sp2 hybrid orbitals:
h1 = s + 21/2py

3
1
h2 = s + (– )1/2px − (– )1/2py
2
2

3
1
h3 = s − (– )1/2px − (– )1/2py (11
...
11
...

The third 2p orbital (2pz) is not included in the hybridization; its axis is perpendicular to the plane in which the hybrids lie
...

Thus, in the first of these hybrids the ratio of s to p contributions is 1:2
...
The different signs of the coefficients ensure that constructive interference
takes place in different regions of space, so giving the patterns in the illustration
...
The sp2-hybridized C
atoms each form three σ bonds by spin pairing with either the h1 hybrid of the other
C atom or with H1s orbitals
...
When the two CH2 groups lie in the same plane, the two
electrons in the unhybridized p orbitals can pair and form a π bond (Fig
...
11)
...

A similar description applies to ethyne, HC
...
Now the C
atoms are sp hybridized, and the σ bonds are formed using hybrid atomic orbitals of
the form
h1 = s + pz

h2 = s − pz

(a)

(b)

(a) An s orbital and two p orbitals
can be hybridized to form three equivalent
orbitals that point towards the corners of
an equilateral triangle
...


Fig
...
10

(11
...
The electrons in them pair either
with an electron in the corresponding hybrid orbital on the other C atom or with an
electron in one of the H1s orbitals
...
11
...


Fig
...
11 A representation of the structure
of a double bond in ethene; only the π
bond is shown explicitly
...
2 Hybrid orbitals do not always form bonds
...
Use valence-bond theory to suggest possible shapes for the
hydrogen peroxide molecule, H2O2
...
8°); rotation around the O-O bond is possible, so the molecule
interconverts between planar and non-planar geometries at high temperatures
...
1)
...
For example, sp3d 2 hybridization results in six equivalent hybrid orbitals pointing towards the corners of a regular octahedron and is sometimes invoked to account
for the structure of octahedral molecules, such as SF6
...
11
...
The overall electron
density has cylindrical symmetry around
the axis of the molecule
...
1* Some hybridization schemes
Coordination number

Arrangement

Composition

2

Linear

sp, pd, sd

Angular

sd

Trigonal planar

sp2, p2d

Unsymmetrical planar

spd

Trigonal pyramidal

pd 2

Tetrahedral

sp3, sd 3

Irregular tetrahedral

spd 2, p3d, dp3

Square planar

p2d 2, sp2d

Trigonal bipyramidal

sp3d, spd 2

Tetragonal pyramidal

sp2d 2, sd 4, pd 4, p3d 2

Pentagonal planar

p2d 3

Octahedral

sp3d2

Trigonal prismatic

spd 4, pd 5

Trigonal antiprismatic

p3d 2

3

4

5

6

* Source: H
...
Walter, and G
...
Kimball, Quantum chemistry, Wiley (1944)
...
This
theory has been more fully developed than VB theory and provides the language that
is widely used in modern discussions of bonding
...
In this chapter we use the simplest molecular species of all, the hydrogen
+
molecule-ion, H 2 , to introduce the essential features of bonding, and then use it as a
guide to the structures of more complex systems
...

11
...
6)

where rA1 and rB1 are the distances of the electron from the two nuclei (1) and R is the
distance between the two nuclei
...

The one-electron wavefunctions obtained by solving the Schrödinger equation
Hψ = Eψ are called molecular orbitals (MO)
...
3 THE HYDROGEN MOLECULE-ION

369

the value of |ψ |2, the distribution of the electron in the molecule
...

+
The Schrödinger equation can be solved analytically for H 2 (within the Born–
Oppenheimer approximation), but the wavefunctions are very complicated functions;
moreover, the solution cannot be extended to polyatomic systems
...

(a) Linear combinations of atomic orbitals

If an electron can be found in an atomic orbital belonging to atom A and also in an
atomic orbital belonging to atom B, then the overall wavefunction is a superposition
of the two atomic orbitals:

ψ± = N(A ± B)

(11
...

The technical term for the superposition in eqn 11
...
An approximate molecular orbital formed from a linear
combination of atomic orbitals is called an LCAO-MO
...


(a)

Example 11
...
7
...

Answer When we substitute the wavefunction, we find

5
ψ *ψ dτ = N
A dτ + B dτ + 2 AB dτ6 = N 2(1 + 1 + 2S)
7
3
1

Ύ

22

Ύ

2

Ύ

2

Ύ

where S = ∫AB dτ
...
11
...


Exploration Plot the 1σ orbital for
different values of the internuclear
distance
...


1
Boundary
surface

{2(1 + S)}1/2

+
In H 2 , S ≈ 0
...
56
...
3 Normalize the orbital ψ− in eqn 11
...


[N = 1/{2(1 − S)}1/2, so N = 1
...
13 shows the contours of constant amplitude for the two molecular
orbitals in eqn 11
...
11
...
Plots like these
are readily obtained using commercially available software
...
In this case, we use

Nuclei
Fig
...
14 A general indication of the shape
of the boundary surface of a σ orbital
...
3

The law of cosines states that for a
triangle such as that shown in (2) with
sides rA, rB, and R, and angle θ facing
2
2
side rB we may write: r B = r A + R2 −
2rAR cos θ
...
8)

(πa3 )1/2
0

and note that rA and rB are not independent (2), but related by the law of cosines (see
Comment 11
...
9)

To make this plot, we have taken N = 0
...
1)
...
The probability density corresponding to the (real) wavefunction ψ+ in eqn 11
...
10)

This probability density is plotted in Fig
...
15
...
According to eqn 11
...

2 B2, the probability density if the electron were confined to the atomic orbital B
...


Fig
...
15 The electron density calculated by
forming the square of the wavefunction
used to construct Fig
...
13
...


This last contribution, the overlap density, is crucial, because it represents an
enhancement of the probability of finding the electron in the internuclear region
...

We shall frequently make use of the result that electrons accumulate in regions where
atomic orbitals overlap and interfere constructively
...
Hence, the energy
of the molecule is lower than that of the separate atoms, where each electron can
interact strongly with only one nucleus
...
The modern (and still controversial)
explanation does not emerge from the simple LCAO treatment given here
...
This orbital shrinkage improves the electron–nucleus attraction more
than it is decreased by the migration to the internuclear region, so there is a net lowering of potential energy
...
Throughout the following discussion
we ascribe the strength of chemical bonds to the accumulation of electron density in
the internuclear region
...

The σ orbital we have described is an example of a bonding orbital, an orbital
which, if occupied, helps to bind two atoms together
...
An electron that occupies a σ orbital is called a σ electron, and if that is the only electron present in the molecule (as in the ground state of
+
H 2 ), then we report the configuration of the molecule as 1σ 1
...
3 THE HYDROGEN MOLECULE-ION

371

The energy E1σ of the 1σ orbital is (see Problem 11
...
11)

where
2
1
R 1 A R D 5 −R/a
6e 0
S = AB dτ = 2 1 + + –
3
C a0 F 7
a0
3
2
2
A
e
A
e2 1
R D −2R/a 5
06
21 − 1 +
j =
dτ =
e
C
4πε0 rB
4πε0R 3
a0 F
7

Ύ

Ύ

k=

4πε Ύ r
e2

AB

0

dτ =

B

A
R D −R/a
1+
e 0
4πε0a0 C
a0 F
e2

(11
...
12b)

(11
...

2 The integral j is a measure of the interaction between a nucleus and electron density centred on the other nucleus
...

Figure 11
...

The energy of the 1σ orbital decreases as the internuclear separation decreases from
large values because electron density accumulates in the internuclear region as the
constructive interference between the atomic orbitals increases (Fig
...
17)
...
In addition, the nucleus–nucleus repulsion
(which is proportional to 1/R) becomes large
...
Calcula+
tions on H2 give Re = 130 pm and De = 1
...
6 eV, so this simple LCAO-MO description of the molecule, while
inaccurate, is not absurdly wrong
...
11
...
The alternative
g,u notation is introduced in Section 11
...


Region of
constructive
interference

(c) Antibonding orbitals

The linear combination ψ− in eqn 11
...

Because it is also a σ orbital we label it 2σ
...
11
...
19)
...
13)

There is a reduction in probability density between the nuclei due to the −2AB term
(Fig
...
20); in physical terms, there is destructive interference where the two atomic
orbitals overlap
...

The energy E2σ of the 2σ antibonding orbital is given by (see Problem 11
...
14)

where the integrals S, j, and k are given by eqn 11
...
The variation of E2σ with R is
shown in Fig
...
16, where we see the destabilizing effect of an antibonding electron
...
11
...


372

11 MOLECULAR STRUCTURE
Region of
destructive
interference

(a)

Fig
...
18 A representation of the
destructive interference that occurs when
two H1s orbitals overlap and form an
antibonding 2σ orbital
...
11
...

Note the internuclear node
...
Point to the features of the 2σ
orbital that lead to antibonding
...
11
...
(a) In a
bonding orbital, the nuclei are attracted to
the accumulation of electron density in the
internuclear region
...


Fig
...
22 The parity of an orbital is even (g)
if its wavefunction is unchanged under
inversion through the centre of symmetry
of the molecule, but odd (u) if the
wavefunction changes sign
...


Fig
...
20 The electron density calculated by
forming the square of the wavefunction
used to construct Fig
...
19
...


The effect is partly due to the fact that an antibonding electron is excluded from the
internuclear region, and hence is distributed largely outside the bonding region
...
11
...
Figure 11
...
This important conclusion
stems in part from the presence of the nucleus–nucleus repulsion (e 2/4πε0 R): this
contribution raises the energy of both molecular orbitals
...

For homonuclear diatomic molecules, it is helpful to describe a molecular orbital
by identifying its inversion symmetry, the behaviour of the wavefunction when it is
inverted through the centre (more formally, the centre of inversion) of the molecule
...
11
...
This so-called gerade symmetry
(from the German word for ‘even’) is denoted by a subscript g, as in σg
...
This ungerade symmetry (‘odd symmetry’)
is denoted by a subscript u, as in σu
...
When using the g,u
notation, each set of orbitals of the same inversion symmetry are labelled separately

11
...
The general rule is that
each set of orbitals of the same symmetry designation is labelled separately
...
4 Homonuclear diatomic molecules
In Chapter 10 we used the hydrogenic atomic orbitals and the building-up principle
to deduce the ground electronic configurations of many-electron atoms
...

The general procedure is to construct molecular orbitals by combining the available
atomic orbitals
...
As in the case of atoms, if several degenerate molecular orbitals
are available, we add the electrons singly to each individual orbital before doubly
occupying any one orbital (because that minimizes electron–electron repulsions)
...
4) that, if electrons
do occupy different degenerate orbitals, then a lower energy is obtained if they do so
with parallel spins
...
Each H atom con+
tributes a 1s orbital (as in H 2 ), so we can form the 1σg and 1σu orbitals from them, as
we have seen already
...
11
...
Note that from two atomic orbitals we can build two molecular orbitals
...

There are two electrons to accommodate, and both can enter 1σg by pairing their
spins, as required by the Pauli principle (see the following Justification)
...
This approach shows that an electron pair,
which was the focus of Lewis’s account of chemical bonding, represents the maximum
number of electrons that can enter a bonding molecular orbital
...
4

When treating homonuclear diatomic
molecules, we shall favour the more
modern notation that focuses attention
on the symmetry properties of the
orbital
...


Fig
...
23 A molecular orbital energy level
diagram for orbitals constructed from the
overlap of H1s orbitals; the separation of
the levels corresponds to that found at the
equilibrium bond length
...


2s (1 u)
s

He1s

He1s

Justification 11
...
7, is ψ (1)ψ (2)
...
To satisfy the Pauli
principle, it must be multiplied by the antisymmetric spin state α(1)β(2) − β(1)α(2)
to give the overall antisymmetric state

ψ (1,2) = ψ (1)ψ (2){α(1)β(2) − β(1)α(2)}
Because α(1)β (2) − β(1)α(2) corresponds to paired electron spins, we see that two
electrons can occupy the same molecular orbital (in this case, the bonding orbital)
only if their spins are paired
...
Each He
atom contributes a 1s orbital, so 1σg and 1σu molecular orbitals can be constructed
...
There are
four electrons to accommodate
...
11
...
The ground electronic configuration

1s (1sg)

Fig
...
24 The ground electronic
configuration of the hypothetical fourelectron molecule He2 has two bonding
electrons and two antibonding electrons
...


Comment 11
...


374

11 MOLECULAR STRUCTURE
2s

A

2s

2pz

B

2pz

Fig
...
25 According to molecular orbital
theory, σ orbitals are built from all orbitals
that have the appropriate symmetry
...
From these four
orbitals, four molecular orbitals can be built
...
We see that there is one bond and one antibond
...

We shall now see how the concepts we have introduced apply to homonuclear
diatomic molecules in general
...

A general principle of molecular orbital theory is that all orbitals of the appropriate
symmetry contribute to a molecular orbital
...
These orbitals include the 2s orbitals on each atom and the 2pz orbitals
on the two atoms (Fig
...
25)
...

The procedure for calculating the coefficients will be described in Section 11
...
At
this stage we adopt a simpler route, and suppose that, because the 2s and 2pz orbitals
have distinctly different energies, they may be treated separately
...

These illustrations are schematic
...
11
...
6

Note that we number only the molecular
orbitals formed from atomic orbitals
in the valence shell
...

Centre of
inversion
-

+

pu

(11
...
11
...
The figure also shows that
the bonding π orbital has odd parity, whereas
the antiboding π orbital has even parity
...
16a)

and another consisting of two orbitals of the form

ψ = cA2pz χA2pz + cB2pz χB2pz

(11
...
Therefore, the two sets of orbitals have the form χA2s ± χB2s
and χA2pz ± χB2pz
...
The two 2pz orbitals directed along the internuclear axis overlap strongly
...
11
...
These two σ orbitals are labelled 2σg and
2σu, respectively
...

(b) π orbitals

Now consider the 2px and 2py orbitals of each atom
...
This overlap may be constructive or destructive, and results in a bonding or an antibonding π orbital (Fig
...
27)
...
The two 2px orbitals overlap to give a bonding and antibonding πx orbital, and the two 2py orbitals overlap to give two πy orbitals
...

We also see from Fig
...
27 that a bonding π orbital has odd parity and is denoted πu
and an antibonding π orbital has even parity, denoted πg
...
4 HOMONUCLEAR DIATOMIC MOLECULES

375

1

0
...
6

S
0
...
2
-

(a) When two orbitals are on atoms
that are far apart, the wavefunctions are small
where they overlap, so S is small
...
Note that S will decrease again as
the two atoms approach more closely than
shown here, because the region of negative
amplitude of the p orbital starts to overlap the
positive overlap of the s orbital
...

Fig
...
28

0
0

2

R /a0

4

6

The overlap integral, S, between
two H1s orbitals as a function of their
separation R
...
11
...

Fig
...
30

(c) The overlap integral

The extent to which two atomic orbitals on different atoms overlap is measured by the
overlap integral, S:

Ύ

S = χ A χB dτ
*

Atom

[11
...
11
...
If χA and χB are simultaneously large in some
region of space, then S may be large
...
In some cases, simple
formulas can be given for overlap integrals and the variation of S with bond length
plotted (Fig
...
29)
...
59 for two H1s orbitals at the equilibrium
+
bond length in H 2 , which is an unusually large value
...
2 to 0
...

Now consider the arrangement in which an s orbital is superimposed on a px orbital
of a different atom (Fig
...
30)
...
Therefore, there is no net overlap between
the s and p orbitals in this arrangement
...
In some cases, π orbitals are less strongly bonding
than σ orbitals because their maximum overlap occurs off-axis
...
11
...
However, we must remember that we have assumed that 2s and 2pz orbitals

Molecule
2su

Atom

1pg
2p

2p
1pu

2sg
1su
2s

2s
1sg

Fig
...
31 The molecular orbital energy level
diagram for homonuclear diatomic
molecules
...
As remarked in the text,
this diagram should be used for O2 (the
configuration shown) and F2
...
11
...


Atom

Molecule
2su

Atom

1pg
2p

2sg

2p

1pu

contribute to different sets of molecular orbitals whereas in fact all four atomic
orbitals contribute jointly to the four σ orbitals
...
11
...
The order
shown in Fig
...
33 is appropriate as far as N2, and Fig
...
31 applies for O2 and F2
...
The consequent switch in order occurs at about N2
...
Anionic species (such as the peroxide ion, O 2 ) need more
electrons than the parent neutral molecules; cationic species (such as O +) need fewer
...
Two electrons pair, occupy, and fill the
1σg orbital; the next two occupy and fill the 1σu orbital
...
There
are two 1πu orbitals, so four electrons can be accommodated in them
...
Therefore, the ground-state configuration of N2 is 1σ g1σ u1π u2σ g
...
11
...
As remarked in the
text, this diagram should be used for
diatomics up to and including N2 (the
configuration shown)
...
18]

where n is the number of electrons in bonding orbitals and n* is the number of electrons in antibonding orbitals
...
For H2,
b = 1, corresponding to a single bond, H-H, between the two atoms
...
In N2, b = – (8 − 2) = 3
...
N:)
...
11
...
Its bond order is 2
...
Because the electrons are in different
orbitals, they will have parallel spins
...
7, be in a triplet state
...
This prediction, which VB theory does not make, is confirmed by experiment
...
4 HOMONUCLEAR DIATOMIC MOLECULES
Synoptic table 11
...
7

Synoptic table 11
...
14

NN

3

109
...
1

3

941
...
45

NN

CH

1

114

HCl

1

427
...
Paramagnetism, the rarer
property, arises when the molecules
have unpaired electron spins
...


368

CC

377

962

CC

3

* More values will be found in the Data section
...


* More values will be found in the Data section
...


An F2 molecule has two more electrons than an O2 molecule
...
We conclude that F2 is a singly-bonded
g
molecule, in agreement with its Lewis structure
...

The zero bond order is consistent with the monatomic nature of Ne
...
For bonds between atoms
of a given pair of elements:

2 The greater the bond order, the greater the bond strength
...
2 lists some typical bond lengths in diatomic and polyatomic molecules
...
Table 11
...

Example 11
...


Method Because the molecule with the larger bond order is likely to have the larger

dissociation energy, compare their electronic configurations and assess their bond
orders
...
11
...
The experimental dissociation energies are 945 kJ mol−1 for N2
+
and 842 kJ mol−1 for N 2
...
4 Which can be expected to have the higher dissociation energy, F2
+
or F 2 ?

Bond dissociation energies are
commonly used in thermodynamic
cycles, where bond enthalpies, ∆bondH 7,
should be used instead
...
7 concerning ionization
enthalpies, that
3
X2(g) → 2 X(g) ∆bondH 7(T) = De + –RT
2

1 The greater the bond order, the shorter the bond
...
8

+
[F 2 ]

To derive this relation, we have
supposed that the molar constant7
pressure heat capacity of X2 is – R
2
(Molecular interpretation 2
...


378

11 MOLECULAR STRUCTURE
-

(e) Photoelectron spectroscopy

EK(e )

hn – Ii
X

hn

+

Ii

Orbital i

X

An incoming photon carries an
energy hν ; an energy Ii is needed to remove
an electron from an orbital i, and the
difference appears as the kinetic energy of
the electron
...
11
...
The technique is also used to study solids, and
in Chapter 25 we shall see the important information that it gives about species at or
on surfaces
...
11
...
19)

This equation (which is like the one used for the photoelectric effect, Section 8
...
First, photoelectrons may originate from one of a number of
different orbitals, and each one has a different ionization energy
...
20)

where Ii is the ionization energy for ejection of an electron from an orbital i
...
Photoelectron spectra are interpreted in
terms of an approximation called Koopmans’ theorem, which states that the ionization energy Ii is equal to the orbital energy of the ejected electron (formally: Ii = −ε i)
...
Similarly, the energy of unfilled (‘virtual orbitals’) is related to the electron affinity
...

The ionization energies of molecules are several electronvolts even for valence electrons, so it is essential to work in at least the ultraviolet region of the spectrum and
with wavelengths of less than about 200 nm
...
43
nm, corresponding to a photon energy of 21
...
Its use gives rise to the technique
of ultraviolet photoelectron spectroscopy (UPS)
...

The kinetic energies of the photoelectrons are measured using an electrostatic
deflector that produces different deflections in the paths of the photoelectrons as they
pass between charged plates (Fig
...
35)
...
The electron flux
can be recorded and plotted against kinetic energy to obtain the photoelectron spectrum
...
1 Interpreting a photoelectron spectrum

A photoelectron spectrometer
consists of a source of ionizing radiation
(such as a helium discharge lamp for UPS
and an X-ray source for XPS), an
electrostatic analyser, and an electron
detector
...

Fig
...
35

Photoelectrons ejected from N2 with He(I) radiation had kinetic energies of
5
...
5 cm−1)
...
43 nm has
wavenumber 1
...
22 eV
...
20, 21
...
63 eV + Ii , so Ii = 15
...
This ionization
energy is the energy needed to remove an electron from the occupied molecular
orbital with the highest energy of the N2 molecule, the 2σg bonding orbital (see
Fig
...
33)
...
5 HETERONUCLEAR DIATOMIC MOLECULES

379

Self-test 11
...
53 eV
...

[16
...
5 Heteronuclear diatomic molecules
The electron distribution in the covalent bond between the atoms in a heteronuclear
diatomic molecule is not shared evenly because it is energetically favourable for the
electron pair to be found closer to one atom than the other
...
The bond in HF, for instance, is polar, with the electron pair closer to the
F atom
...
There is a matching partial positive charge, δ +, on the H atom
...
98y(H)
- 0
...
19y (H)
+ 0
...
22)

where χH is an H1s orbital and χF is an F2p orbital
...
6 eV below
the zero of energy (the separated proton and electron) and the F2p orbital lies at
18
...
11
...
Hence, the bonding σ orbital in HF is mainly F2p and the antibonding σ orbital is mainly H1s orbital in character
...


Fig
...
36 The atomic orbital energy levels of
H and F atoms and the molecular orbitals
they form
...
4* Pauling
electronegativities

(b) Electronegativity

The charge distribution in bonds is commonly discussed in terms of the electronegativity, χ, of the elements involved (there should be little danger of confusing this use
of χ with its use to denote an atomic orbital, which is another common convention)
...
Pauling
used valence-bond arguments to suggest that an appropriate numerical scale of electronegativities could be defined in terms of bond dissociation energies, D, in electronvolts and proposed that the difference in electronegativities could be expressed as
1
| χA − χB | = 0
...
6 eV

with unequal coefficients
...
A nonpolar bond has |cA |2 = |cB |2 and a pure ionic bond has one
coefficient zero (so the species A+ B − would have cA = 0 and cB = 1)
...

The opposite is true of the antibonding orbital, for which the dominant component
comes from the atomic orbital with higher energy
...
The general form of the
molecular orbitals is

18
...
21)

13
...
4 eV

A polar bond consists of two electrons in an orbital of the form

[11
...

A list of Pauling electronegativities is given in Table 11
...
The most electronegative

Element

χP

H

2
...
6

N

3
...
4

F

4
...
2

Cs

0
...


380

11 MOLECULAR STRUCTURE
elements are those close to fluorine; the least are those close to caesium
...
The difference for HF, for instance, is 1
...
51
...
He argued that an element is likely to be highly electronegative if it has
a high ionization energy (so it will not release electrons readily) and a high electron
affinity (so it is energetically favorable to acquire electrons)
...
24]

where I is the ionization energy of the element and Eea is its electron affinity (both
in electronvolts, Section 10
...
The Mulliken and Pauling scales are approximately in line with each other
...
35χ1/2 − 1
...

M
(c) The variation principle

A more systematic way of discussing bond polarity and finding the coefficients in the
linear combinations used to build molecular orbitals is provided by the variation
principle:
If an arbitrary wavefunction is used to calculate the energy, the value calculated is
never less than the true energy
...
9

The name ‘secular’ is derived from the
Latin word for age or generation
...


This principle is the basis of all modern molecular structure calculations (Section 11
...

The arbitrary wavefunction is called the trial wavefunction
...
We might get a lower energy if we use a
more complicated wavefunction (for example, by taking a linear combination of several atomic orbitals on each atom), but we shall have the optimum (minimum energy)
molecular orbital that can be built from the chosen basis set, the given set of atomic
orbitals
...
21
...
25a)

(β − ES)cA + (αB − E)cB = 0

(11
...
It is negative and can be interpreted
as the energy of the electron when it occupies A (for αA) or B (for α B)
...
The parameter β is called a resonance integral
(for classical reasons)
...

Justification 11
...
21 is real but not normalized because at this stage
the coefficients can take arbitrary values
...
The energy of the trial wavefunction is the expectation value
of the energy operator (the hamiltonian, @, Section 8
...
5 HETERONUCLEAR DIATOMIC MOLECULES

Ύψ *@ψ dτ
E=
Ύψ *ψ dτ

(11
...
This is a standard problem in calculus, and is solved by finding the
coefficients for which
∂E
∂cA

∂E

=0

∂cB

=0

The first step is to express the two integrals in terms of the coefficients
...
17)
...
27]

Ύ

β = A@B dτ = B@A dτ (by the hermiticity of @)
Then

Ύψ

2
2
@ψ dτ = c AαA + c B α B + 2cAcB β

The complete expression for E is
E=

2
2
c AαA + c Bα B + 2cAcB β

(11
...
After a bit of work, we obtain
∂E
∂cA
∂E
∂cB

=
=

2 × (cAαA − cAE + cB β − cBSE )
2
2
c A + c B + 2cAcBS

2 × (cBα B − cBE + cA β − cASE )
2
2
c A + c B + 2cAcBS

=0
=0

381

382

11 MOLECULAR STRUCTURE
For the derivatives to vanish, the numerators of the expressions above must vanish
...
25)
...
As for any set of simultaneous equations, the secular equations have a
solution if the secular determinant, the determinant of the coefficients, is zero; that
is, if

αA − E
β − ES
Comment 11
...
29)

This determinant expands to a quadratic equation in E (see Illustration 11
...
Its two
roots give the energies of the bonding and antibonding molecular orbitals formed
from the atomic orbitals and, according to the variation principle, the lower root is
the best energy achievable with the given basis set
...
2 Using the variation principle (1)

To find the energies E of the bonding and antibonding orbitals of a homonuclear
diatomic molecule set with αA = α B = α in eqn 11
...
The
lower energy (E+ in the Illustration) gives the coefficients for the bonding molecular
orbital, the upper energy (E−) the coefficients for the antibonding molecular orbital
...
This equation is
obtained by demanding that the best wavefunction should also be normalized
...
30)

Illustration 11
...
2, we use eqn 11
...
5 HETERONUCLEAR DIATOMIC MOLECULES
2
2
Now we use the normalization condition, eqn 11
...

We saw in Illustrations 11
...
3 that, when the two atoms are the same, and we
can write α A = α B = α, the solutions are
E+ =
E− =

α+β

cA =

1+S

α−β

cA =

1−S

1
{2(1 + S)}1/2
1
{2(1 − S)}1/2

c B = cA

(11
...
31b)

In this case, the bonding orbital has the form

ψ+ =

A+B
{2(1 + S)}1/2

(11
...
32b)

in agreement with the discussion of homonuclear diatomics we have already given,
but now with the normalization constant in place
...
The secular determinant is then

αA − E
β

β
= (αA − E)(α B − E) − β 2 = 0
αB − E

The solutions can be expressed in terms of the parameter ζ (zeta), with
1
ζ = – arctan
2

2| β |

α B − αA

(11
...
34a)

E+ = αA + β tan ζ

ψ+ = A cos ζ + B sin ζ

(11
...

We show in the following Justification that, when the energy difference is very large, in
the sense that | α B − αA | > 2| β |, the energies of the resulting molecular orbitals differ
>
only slightly from those of the atomic orbitals, which implies in turn that the bonding
and antibonding effects are small
...
The difference in energy between core and valence orbitals is the justification for neglecting
the contribution of core orbitals to bonding
...

Justification 11
...
33, ζ ≈ |β |/(αB − αA)
...
Noting that β is normally a negative number, so that β /| β | = −1, we
can use eqn 11
...
11

For x << 1, we can write: sin x ≈ x, cos x ≈
1, tan x ≈ x, and arctan x = tan−1 x ≈ x
...
25 you are invited to derive these expressions via a different route
...

Now we consider the behaviour of the wavefunctions in the limit of large |α B − αA |,
when ζ << 1
...
34, we write ψ− ≈ B
and ψ+ ≈ A
...


Example 11
...
0 eV and the following ionization energies: H1s: 13
...
2 eV,
F2p: 17
...

Method Because the F2p and H1s orbitals are much closer in energy than the F2s

and H1s orbitals, to a first approximation neglect the contribution of the F2s
orbital
...
34, we need to know the values of the Coulomb integrals
α H and α F
...
Calculate ζ from eqn 11
...
34
...
6 eV and α F = −17
...
58; so ζ = 13
...


Then

I11
...
4 eV
E+ = −17
...
97χH − 0
...
24χH + 0
...
6 eV) has a composition that is more F2p orbital than H1s, and that the opposite is true of the higher
energy, antibonding orbital
...
6 The ionization energy of Cl is 13
...
0 eV
...
8 eV, ψ− = −0
...
79χCl; E+ = −13
...
79χH + 0
...
1 The biochemical reactivity of O2 , N2 , and NO

We can now see how some of these concepts are applied to diatomic molecules that
play a vital biochemical role
...
1 per cent O2
and 75
...
Molecular orbital theory predicts—correctly—that O2
has unpaired electron spins and, consequently, is a reactive component of the Earth’s
atmosphere; its most important biological role is as an oxidizing agent
...
So taxing is the process that only certain bacteria and archaea are capable of carrying it out, making nitrogen available first to plants and other microorganisms in the form of ammonia
...

The reactivity of O2, while important for biological energy conversion, also poses
serious physiological problems
...
The ground-state electronic configuration of O 2 is 1σ g 1σ u2σ g 1π u1π g ,
3
so the ion is a radical with a bond order b = –
...

The enzyme superoxide dismutase protects cells by catalysing the disproportionation
−
(or dismutation) of O 2 into O2 and H2O2:
−
2 O 2 + 2 H+ → H2O2 + O2

However, H2O2 (hydrogen peroxide), formed by the reaction above and by leakage of
electrons out of the respiratory chain, is a powerful oxidizing agent and also harmful
to cells
...
A catalase catalyses the
reaction
2 H2O2 → 2 H2O + O2
and a peroxidase reduces hydrogen peroxide to water by oxidizing an organic molecule
...
11
...


stroke, inflammatory disease, and other conditions
...

Important examples of antioxidants are vitamin C (ascorbic acid), vitamin E (αtocopherol), and uric acid
...
The molecule is synthesized from
the amino acid arginine in a series of reactions catalysed by nitric oxide synthase and
requiring O2 and NADPH
...
37 shows the bonding scheme in NO and illustrates a number of points
we have made about heteronuclear diatomic molecules
...
The 3σ and 1π orbitals are predominantly of O character as that is
the more electronegative element
...
It follows that NO is a radical with an unpaired electron that can be regarded as localized more on the N atom
than on the O atom
...

Because NO is a radical, we expect it to be reactive
...
As we saw above,
there is a biochemical price to be paid for the reactivity of biological radicals
...
Indeed, the
−
radicals O 2 and NO combine to form the peroxynitrite ion:
−
NO · + O 2 · → ONOO−

where we have shown the unpaired electrons explicitly
...
Note that the structure of the ion is consistent with the bonding
scheme in Fig
...
37: because the unpaired electron in NO is slightly more localized
−
on the N atom, we expect that atom to form a bond with an O atom from the O 2 ion
...
As for diatomic molecules, polyatomic molecular orbitals spread
over the entire molecule
...
35)

i

where χi is an atomic orbital and the sum extends over all the valence orbitals of all the
atoms in the molecule
...

The principal difference between diatomic and polyatomic molecules lies in the
greater range of shapes that are possible: a diatomic molecule is necessarily linear, but

11
...
The shape of a polyatomic molecule—the specification of its bond
lengths and its bond angles—can be predicted by calculating the total energy of the
molecule for a variety of nuclear positions, and then identifying the conformation
that corresponds to the lowest energy
...
6 The Hückel approximation
Molecular orbital theory takes large molecules and extended aggregates of atoms,
such as solid materials, in its stride
...
Although the classification of an orbital as σ or π is strictly valid only in linear
molecules, as will be familiar from introductory chemistry courses, it is also used to
denote the local symmetry with respect to a given A-B bond axis
...
In his
approach, the π orbitals are treated separately from the σ orbitals, and the latter form
a rigid framework that determines the general shape of the molecule
...
For example, in ethene, we take the σ bonds as
fixed, and concentrate on finding the energies of the single π bond and its companion
antibond
...
In ethene, for instance, we would write

ψ = cAA + cBB

(11
...
Next, the optimum coefficients and
energies are found by the variation principle as explained in Section 11
...
That is,
we have to solve the secular determinant, which in the case of ethene is eqn 11
...
37)

The roots of this determinant can be found very easily (they are the same as those
in Illustration 11
...
In a modern computation all the resonance integrals and overlap
integrals would be included, but an indication of the molecular orbital energy level
diagram can be obtained very readily if we make the following additional Hückel
approximations:
1 All overlap integrals are set equal to zero
...

3 All remaining resonance integrals are set equal (to β )
...
The assumptions result in the following structure of the secular determinant:
1 All diagonal elements: α − E
...

3 All other elements: 0
...
38)

The roots of the equation are
E± = α ± β

Fig
...
38 The Hückel molecular orbital
energy levels of ethene
...


(11
...
11
...
We see the effect of
neglecting overlap by comparing this result with eqn 11
...

The building-up principle leads to the configuration 1π 2, because each carbon atom
supplies one electron to the π system
...
These two orbitals
jointly form the frontier orbitals of the molecule
...
For example, we can estimate that 2| β | is the π * ← π excitation
energy of ethene, the energy required to excite an electron from the 1π to the 2π orbital
...
4 eV (−230 kJ mol−1)
...
We have seen that the secular equations that we have to solve for a twoatom system have the form
(HAA − Ei SAA)ci,A + (HAB − Ei SAB)ci,B = 0

(11
...
40b)

where the eigenvalue Ei corresponds to a wavefunction of the form ψi = ci,AA + ci,BB
...
25)
...
41a)

(HBA − E1SBA)c1,A + (HBB − E1SBB)c1,B = 0

(11
...
41c)

(HBA − E2SBA)c2,A + (HBB − E2SBB)c2,B = 0

(11
...
42)

then each pair of equations may be written more succinctly as
(H − Ei S)ci = 0

or

Hci = Sci Ei

(11
...
To proceed with the
calculation of the eigenvalues and coefficients, we introduce the matrices
C = (c1 c2) =

A c1,A
C c1,B

c2,A D
c2,B F

E=

A E1 0 D
C 0 E2F

[11
...
6 THE HÜCKEL APPROXIMATION
for then the entire set of equations we have to solve can be expressed as
HC = SCE

(11
...
7 Show by carrying out the necessary matrix operations that eqn 11
...
41(a)–(d)
...
Then
HC = CE
At this point, we multiply from the left by the inverse matrix C −1, and find
C −1HC = E

(11
...
In other words, to find the eigenvalues Ei, we have to
find a transformation of H that makes it diagonal
...
The diagonal elements then correspond to the eigenvalues Ei and the
columns of the matrix C that brings about this diagonalization are the coefficients of
the members of the basis set, the set of atomic orbitals used in the calculation, and
hence give us the composition of the molecular orbitals
...
As a result, we have to solve N equations of the form
Hci = Sci E i by diagonalization of the N × N matrix H, as directed by eqn 11
...

Example 11
...

Method The matrices will be four-dimensional for this four-atom system
...
Find the
matrix C that diagonalizes H: for this step, use mathematical software
...

Solution

⎛ H11
⎜H
H = ⎜ 21
⎜ H31
⎜H
⎝ 41

H12
H 22
H32
H 42

H13
H 23
H33
H 43

H14 ⎞ ⎛ α β 0 0 ⎞
H 24 ⎟ ⎜ β α β 0 ⎟
⎟=
H34 ⎟ ⎜ 0 β α β ⎟
⎜
⎟
H 44 ⎟ ⎝ 0 0 β α ⎠
⎠

Mathematical software then diagonalizes this matrix to
⎛ α + 1
...
62β
0
0
E=⎜
⎟
0
0
α − 0
...
62β ⎠
⎝
and the matrix that achieves the diagonalization is
⎛ 0
...
602 0
...
372⎞
⎜ 0
...
372 −0
...
602⎟
C=⎜
0
...
372 −0
...
602⎟
⎜
⎟
⎝ 0
...
602 0
...
372⎠

389

390

11 MOLECULAR STRUCTURE
We can conclude that the energies and molecular orbitals are
E1 = α + 1
...
62β
E3 = α − 0
...
62β

ψ1 = 0
...
602χB + 0
...
372χD
ψ2 = 0
...
372χB − 0
...
602χD
ψ3 = 0
...
372χB − 0
...
602χD
ψ4 = −0
...
602χB − 0
...
372χD

where the C2p atomic orbitals are denoted by χA,
...
Note that the orbitals are
mutually orthogonal and, with overlap neglected, normalized
...
8 Repeat the exercise for the allyl radical, · CH2-CH=CH2
...
62β,
a - 1
...
62 b

C2p
2p

a + 0
...
62β

(11
...
11
...
Note that the greater the
number of internuclear nodes, the higher the energy of the orbital
...
The frontier
orbitals of butadiene are the 2π orbital (the HOMO, which is largely bonding) and the
3π orbital (the LUMO, which is largely antibonding)
...
‘Largely antibonding’ indicates that the antibonding effects dominate
...
In ethene the total energy is
Eπ = 2(α + β ) = 2α + 2β
In butadiene it is

1p

a + 1
...
11
...
The four p
electrons (one supplied by each C) occupy
the two lower π orbitals
...


Eπ = 2(α + 1
...
62β ) = 4α + 4
...
48β (about 110 kJ
mol−1) than the sum of two individual π bonds
...
A closely related quantity is the
π -bond formation energy, the energy released when a π bond is formed
...
48)

where N is the number of carbon atoms in the molecule
...
48β
...
5 Estimating the delocalization energy

Use the Hückel approximation to find the energies of the π orbitals of cyclobutadiene, and estimate the delocalization energy
...
6 THE HÜCKEL APPROXIMATION

391

Method Set up the secular determinant using the same basis as for butadiene, but

note that atoms A and D are also now neighbours
...
For the delocalization energy,
subtract from the total π-bond energy the energy of two π-bonds
...
Two occupy the lowest orbital (of energy
α + 2β ), and two occupy the doubly degenerate orbitals (of energy α)
...
Two isolated π bonds would have an energy 4α + 4β ;
therefore, in this case, the delocalization energy is zero
...
9 Repeat the calculation for benzene
...
Benzene is often expressed in a mixture
of valence-bond and molecular orbital terms, with typically valence-bond language
used for its σ framework and molecular orbital language used to describe its π
electrons
...
The six C atoms are regarded as sp2 hybridized,
with a single unhydridized perpendicular 2p orbital
...
11
...
The internal angle of a regular hexagon is 120°,
so sp2 hybridization is ideally suited for forming σ bonds
...

Now consider the molecular orbital component of the description
...
Their energies are
calculated within the Hückel approximation by diagonalizing the hamiltonian matrix
⎛α β 0 0 0 β ⎞
⎜ β α β 0 0 0⎟
⎜ 0 β α β 0 0⎟
⎟
H =⎜
⎜ 0 0 β α β 0⎟
⎜ 0 0 0 β α β⎟
⎜
⎟
⎝ β 0 0 0 β α⎠

b2g

e2u

e1g

a2u

The MO energies, the eigenvalues of this matrix, are simply
E = α ± 2β, α ± β, α ± β

Fig
...
40 The σ framework of benzene is
formed by the overlap of Csp2 hybrids,
which fit without strain into a hexagonal
arrangement
...
49)

as shown in Fig
...
41
...
Note that the lowest energy orbital is bonding between all
neighbouring atoms, the highest energy orbital is antibonding between each pair of
neighbours, and the intermediate orbitals are a mixture of bonding, nonbonding, and
antibonding character between adjacent atoms
...
11
...
The
symmetry labels are explained in Chapter
12
...
In
the ground state, only the bonding orbitals
are occupied
...
There are six electrons to
accommodate (one from each C atom), so the three lowest orbitals (a2u and the doubly2 4
degenerate pair e1g) are fully occupied, giving the ground-state configuration a 2ue1g
...

The π-electron energy of benzene is
Eπ = 2(α + 2β ) + 4(α + β ) = 6α + 8β
If we ignored delocalization and thought of the molecule as having three isolated
π bonds, it would be ascribed a π-electron energy of only 3(2α + 2β ) = 6α + 6β
...
The π-bond formation energy in benzene is 8β
...
First, the shape of the regular hexagon is ideal for the formation of strong σ
bonds: the σ framework is relaxed and without strain
...

11
...
12

The web site contains links to sites
where you may perform semi-empirical
and ab initio calculations on simple
molecules directly from your web
browser
...
The full treatment of molecular electronic structure is quite
easy to formulate but difficult to implement
...
John Pople and Walter Kohn were awarded the Nobel Prize in Chemistry for
1998 for their contributions to the development of computational techniques for the
elucidation of molecular structure and reactivity
...
ψz,β(N)
This is the wavefunction for an N-electron closed-shell molecule in which electron 1
occupies molecular orbital ψa with spin α , electron 2 occupies molecular orbital ψa
with spin β, and so on
...
To achieve this
behaviour, we write the wavefunction as a sum of all possible permutations with the
appropriate sign:

Ψ = ψa,α(1)ψa,β (2)
...
ψz,β (N) + · · ·
There are N! terms in this sum, and the entire sum can be written as a determinant:

Ψ =

ψa,α (1) ψa,β (1) L L ψ z,β (1)
ψa,α (2) ψa,β (2) L L ψ z,β (2)
1
M
M
M
N!
M
M
M
ψa,α (N ) ψa,β (N ) L L ψ z,β (N )

(11
...
7 COMPUTATIONAL CHEMISTRY
The initial factor ensures that the wavefunction is normalized if the component
molecular orbitals are normalized
...
ψz,β (N)|

(11
...
5c), the optimum wavefunctions, in the sense of corresponding to the
lowest total energy, must satisfy the Hartree–Fock equations:
f1ψa,σ(1) = εψa,σ(1)

(11
...
The Fock operator f1 is
f1 = h1 + ∑j{2Jj(1) − Kj(1)}

(11
...
53a]

0 ni

the Coulomb operator J, where

Ύ

Jj(1)ψa(1) = ψ j*(2)ψj(2)

A e2 D
ψ (1)dτ2
C 4πε0r12 F a

[11
...
53c]

Although the Hartree–Fock equations look deceptively simple, with the Fock operator looking like a hamiltonian, we see from these definitions that f actually depends on
the wavefunctions of all the electrons
...
That process is then continued using the
newly found wavefunctions until each cycle of calculation leaves the energies and
wavefunctions unchanged to within a chosen criterion
...

The difficulty in this procedure is in the solution of the Hartree–Fock equations
...
54)

where F is the matrix formed from the Fock operator:

Ύ

Fij = χ *(1)f1χj(1)dτ
i

(11
...
55b)

393

11 MOLECULAR STRUCTURE
Justification 11
...
51, which gives
M

M

i=1

f1

i=1

∑ ciα χi(1) = εα ∑ ciα χi(1)

Now multiply from the left by χ j*(1) and integrate over the coordinates of electron 1:
Sji

5
4
6
4
7

Fji

5
4
4
6
4
4
7

394

∑ ciα Ύχj(1)* f(1)χi(1)dr1 = εα ∑ ciα Ύχj(1)*χi(1)dr1
M

i=1

M

i=1

That is,
M

M

i=1

i=1

∑ Fji ciα = εα ∑ Sji ciα
This expression has the form of the matrix equation in eqn 11
...


(b) Semi-empirical and ab initio methods

There are two main strategies for continuing the calculation from this point
...
In the ab initio methods, an attempt is
made to calculate all the integrals that appear in the Fock and overlap matrices
...

The Fock matrix has elements that consist of integrals of the form

Ύ

(AB|CD) = A(1)B(1)

A e2 D
C(2)D(2)dτ1dτ2
C 4πε 0r12 F

(11
...
It can be appreciated that, if there are several dozen atomic orbitals used to
build the molecular orbitals, then there will be tens of thousands of integrals of this
form to evaluate (the number of integrals increases as the fourth power of the number
of atomic orbitals in the basis)
...
The surviving integrals are then adjusted until the energy levels are in good agreement with
experiment
...
These procedures are now readily available in commercial software
packages and can be used with very little detailed knowledge of their mode of calculation
...
The latter is important when assessing, for instance, the likelihood that a
given molecule will bind to an active site in an enzyme
...
Here the problem
is to evaluate as efficiently as possible thousands of integrals
...
7 COMPUTATIONAL CHEMISTRY
Gaussian orbitals
...
The
advantage of GTOs over the correct orbitals (which for hydrogenic systems are proportional to e−ζr) is that the product of two Gaussian functions is itself a Gaussian
function that lies between the centres of the two contributing functions (Fig
...
42)
...
56 become two-centre integrals of the form

395

2

A e2 D
(AB|CD) = X(1)
Y(2)dτ1dτ2
C 4πε 0r12 F

Ύ

G1G 2
3

G1
y (x)

(11
...
Integrals of this form are much easier and faster to evaluate numerically than the original four-centre integrals
...

(c) Density functional theory

A technique that has gained considerable ground in recent years to become one of the
most widely used techniques for the calculation of molecular structure is density functional theory (DFT)
...

The central focus of DFT is the electron density, ρ, rather than the wavefunction ψ
...
The exact ground-state energy of an n-electron molecule is
E[ρ] = EK + EP;e,N + EP;e,e + EXC[ρ]

x
Fig
...
42 The product of two Gaussian
functions (the purple curves) is itself a
Gaussian function located between the two
contributing Gaussians
...
58)

where EK is the total electron kinetic energy, EP;e,N the electron–nucleus potential
energy, EP;e,e the electron–electron potential energy, and EXC[ρ] the exchange–
correlation energy, which takes into account all the effects due to spin
...
59)

i=1

are calculated from the Kohn–Sham equations, which are found by applying the variation principle to the electron energy, and are like the Hartree–Fock equations except
for a term VXC, which is called the exchange–correlation potential:
Electron–electron
⎧ Kinetic
⎫
Electron–nucleus
repulsion
4 4
7
⎪6energy 8 6 attraction 8 64748 Exchange– ⎪
4 4
7
correlation
N
678 ⎪
⎪ 2
Z je 2
ρ(r2)e 2
⎪ $
⎪
2
∇1 −
+
dr2 + VXC(r1)⎬ψi(r1) = ε iψi(r1)
⎨−
4πε 0rj1
4πε 0r12
j=1
⎪ 2me
⎪
⎪
⎪
⎪
⎪
⎩
⎭

∑

Ύ

Comment 11
...
60)

The exchange–correlation potential is the ‘functional derivative’ of the exchange–
correlation energy:
VXC[ρ] =

δEXC[ρ]
δρ

Consider the functional G[f ] where f is a
function of x
...
By
analogy with the derivative of a
function, the functional derivative is
then defined as
δG
δf

(11
...
First, we
guess the electron density
...
See
Appendix 2 for more details and
examples
...
Then the exchange–correlation potential is calculated by assuming
an approximate form of the dependence of the exchange–correlation energy on the
electron density and evaluating the functional derivative in eqn 11
...
For this step,
the simplest approximation is the local-density approximation and to write

Ύ

EXC[ρ] = ρ(r)ε XC[ρ(r)]dr

(11
...
Next, the Kohn–Sham equations are solved to obtain an initial set of
orbitals
...
59) and the process is repeated until the density and the
exchange–correlation energy are constant to within some tolerance
...
8 The prediction of molecular properties

(a)

The results of molecular orbital calculations are only approximate, with deviations
from experimental values increasing with the size of the molecule
...
In the next sections we give a brief
summary of strategies used by computational chemists for the prediction of molecular properties
...
11
...


Fig
...
44

An elpot diagram of ethanol
...

The raw output of a molecular structure calculation is a list of the coefficients of
the atomic orbitals in each molecular orbital and the energies of these orbitals
...

Different signs of the wavefunctions are represented by different colours
...
The total electron density at any point is then the sum of the
squares of the wavefunctions evaluated at that point
...
11
...

As shown in the illustration, there are several styles of representing an isodensity surface, as a solid form, as a transparent form with a ball-and-stick representation of the
molecule within, or as a mesh
...

One of the most important aspects of a molecule other than its geometrical shape is
the distribution of charge over its surface
...
11
...

Representations such as those we have illustrated are of critical importance in a
number of fields
...
Such considerations are important for assessing the
pharmacological activity of potential drugs
...
8 THE PREDICTION OF MOLECULAR PROPERTIES
(b) Thermodynamic and spectroscopic properties

We already saw in Section 2
...
The computational approach also makes
it possible to gain insight into the effect of solvation on the enthalpy of formation
without conducting experiments
...

Computational methods are available that allow for the inclusion of several solvent
molecules around a solute molecule, thereby taking into account the effect of molecular interactions with the solvent on the enthalpy of formation of the solute
...
As an example, consider the amino acid glycine, which can exist in a
neutral (4) or zwitterionic (5) form, in which the amino group is protonated and the
carboxyl group is deprotonated
...

However, in water the opposite is true because of strong interactions between the
polar solvent and the charges in the zwitterion
...
Several experimental and computational studies of aromatic hydrocarbons indicate that decreasing the energy of the
LUMO enhances the ability of a molecule to accept an electron into the LUMO, with an
attendant increase in the value of the molecule’s standard potential
...
For example, stepwise substitution of the hydrogen atoms in p-benzoquinone
by methyl groups (-CH3) results in a systematic increase in the energy of the LUMO
and a decrease in the standard potential for formation of the semiquinone radical (6):

The standard potentials of naturally occurring quinones are also modified by the presence of different substituents, a strategy that imparts specific functions to specific
quinones
...
2)
...
The
transition of lowest energy (and longest wavelength) occurs between the HOMO and
LUMO
...
For example, consider the linear polyenes shown in Table 11
...
The table also shows that, as expected, the
wavelength of the lowest-energy electronic transition decreases as the energy separation between the HOMO and LUMO increases
...
5 Ab initio calculations and spectroscopic data
{E(HOMO) − E(LUMO)}/eV
18
...
7

252

11
...
5

Polyene

304

polyene in the group
...
Extrapolation of
the trend suggests that a sufficiently long linear polyene should absorb light in the
visible region of the electromagnetic spectrum
...
The ability of β-carotene to absorb visible
light is part of the strategy employed by plants to harvest solar energy for use in photosynthesis (Chapter 23)
...
In the Born–Oppenheimer approximation, nuclei are treated
as stationary while electrons move around them
...
In valence-bond theory (VB theory), a bond is regarded as
forming when an electron in an atomic orbital on one atoms
pairs its spin with that of an electron in an atomic orbital on
another atom
...
A valence bond wavefunction with cylindrical symmetry
around the internuclear axis is a σ bond
...

4
...

5
...

6
...
An
antibonding orbital is a molecular orbital that, if occupied,
decreases the strength of a bond between two atoms
...
A σ molecular orbital has zero orbital angular momentum
about the internuclear axis
...

8
...
11
...
33
...
When constructing molecular orbitals, we need to consider
only combinations of atomic orbitals of similar energies and
of the same symmetry around the internuclear axis
...
The bond order of a diatomic molecule is b = –(n − n*),
2
where n and n* are the numbers of electrons in bonding
and antibonding orbitals, respectively
...
The electronegativity, χ, of an element is the power of its
atoms to draw electrons to itself when it is part of a
compound
...
In a bond between dissimilar atoms, the atomic orbital
belonging to the more electronegative atom makes the larger
contribution to the molecular orbital with the lowest energy
...


EXERCISES
13
...
The overlap matrix, S, is formed of all
i
Sij = ∫ψ *ψj dτ
...
The variation principle states that if an arbitrary wavefunction
is used to calculate the energy, the value calculated is never
less than the true energy
...
In the Hückel method, all Coulomb integrals Hii are set equal
(to α), all overlap integrals are set equal to zero, all resonance
integrals Hij between non-neighbours are set equal to zero,
and all remaining resonance integrals are set equal (to β)
...
The π-electron binding energy is the sum of the energies of
each π electron
...
The delocalization energy is
the extra stabilization of a conjugated system
...
In the self-consistent field procedure, an initial guess about
the composition of the molecular orbitals is successively
refined until the solution remains unchanged in a cycle of
calculations
...
In semi-empirical methods for the determination of electronic
structure, the Schrödinger equation is written in terms of
parameters chosen to agree with selected experimental
quantities
...


Further reading
Articles and texts

T
...
Albright and J
...
Burdett, Problems in molecular orbital theory
...

P
...
Atkins and R
...
Friedman, Molecular quantum mechanics
...

I
...
Levine, Quantum chemistry
...

D
...
McQuarrie, Mathematical methods for scientists and engineers
...

R
...
Mebane, S
...
Schanley, T
...
Rybolt, and C
...
Bruce, The
correlation of physical properties of organic molecules with
computed molecular surface areas
...
Chem
...
76, 688 (1999)
...
Pauling, The nature of the chemical bond
...

C
...
Quinn, Computational quantum chemistry: an interactive guide
to basis set theory
...

Sources of data and information

D
...
Lide (ed
...

P
...
Scott and W
...
Richards, Energy levels in atoms and molecules
...


Discussion questions
11
...

11
...


11
...

11
...


11
...


11
...


11
...


11
...


Exercises
11
...


11
...


−
11
...
2b Give the ground-state electron configurations of (a) ClF, (b) CS, and
−
(c) O 2
...


400

11 MOLECULAR STRUCTURE

which molecule should have the greater bond dissociation energy
...


11
...
9b Suppose that the function ψ = Ae−ar , with A being the normalization

to be stabilized by (a) the addition of an electron to form AB , (b) the removal
of an electron to form AB+?

constant and a being an adjustable parameter, is used as a trial wavefunction
for the 1s orbital of the hydrogen atom
...
3a From the ground-state electron configurations of B2 and C2, predict

−

11
...
Is XeF likely to have a shorter bond
length than XeF+?

2

E=

3a$2
2µ

−

1/2
e2 A a D
B 3E
ε 0 C2π F

11
...
Is BrCl likely to have a shorter bond
length than BrCl−?

where e is the electron charge, and µ is the effective mass of the H atom
...
5a Use the electron configurations of NO and N2 to predict which is likely

of ionization energy 11
...


11
...
5b Arrange the species O 2 , O2, O2 , O2 in order of increasing bond length
...
10b What is the energy of an electron that has been ejected from an orbital
of ionization energy 4
...
6a Show that the sp2 hybrid orbital (s + 21/2p)/31/2 is normalized to 1 if the s

11
...


the basis that the molecule is formed from the appropriately hybridized CH2
or CH fragments
...
6b Normalize the molecular orbital ψA + λψB in terms of the parameter λ

and the overlap integral S
...
7a Confirm that the bonding and antibonding combinations ψA ± ψB are

mutually orthogonal in the sense that their mutual overlap is zero
...
7b Suppose that a molecular orbital has the form N(0
...
844B)
...

11
...
4)
...


11
...

11
...

11
...
Estimate the π-electron binding energy in each case
...
13a Write down the secular determinants (a) anthracene (8),

(b) phenanthrene (9) within the Hückel approximation and using the C2p
orbitals as the basis set
...
8b Can the function ψ = x 2(L − 2x) be used as a trial wavefunction for the

n = 1 state of a particle with mass m in a one-dimensional box of length L? If
the answer is yes, then express the energy of this trial wavefunction in terms of
h, m, and L and compare it with the exact result (eqn 9
...
If the answer is no,
explain why this is not a suitable trial wavefunction
...
9a Suppose that the function ψ = Ae−ar , with A being the normalization
2

constant and a being an adjustable parameter, is used as a trial wavefunction
for the 1s orbital of the hydrogen atom
...
13b Use mathematical software to estimate the π-electron binding energy

of (a) anthracene (7), (b) phenanthrene (8) within the Hückel approximation
...
1 Show that, if a wave cos kx centred on A (so that x is measured from A)

interferes with a similar wave cos k′x centred on B (with x measured from B)
a distance R away, then constructive interference occurs in the intermediate
1
3
region when k = k′ = π/2R and destructive interference if kR = – π and k′R = – π
...
2 The overlap integral between two H1s orbitals on nuclei separated by a
1
distance R is S = {1 + (R/a0) + – (R/a0)2}e−R/a0
...

3
11
...
The
overlap integral between an H1s orbital and an H2p orbital on nuclei

separated by a distance R and forming a σ orbital is S = (R/a0){1 + (R/a0) +
1
– (R/a0)2}e−R/a0
...

11
...
Plot the two amplitudes for positions along the
molecular axis both inside and outside the internuclear region
...
5 Repeat the calculation in Problem 11
...
Then form the difference density, the difference between
1
2
2
ψ 2 and – {ψ A + ψ B}
...


PROBLEMS
11
...
(a) Make a probability
density plot, and both surface and contour plots of the xz-plane amplitudes of
the 2pzσ and 2pzσ* molecular orbitals
...
Include
plots for both internuclear distances, R, of 10a0 and 3a0, where a0 = 52
...

Interpret the graphs, and describe why this graphical information is useful
...
13‡ Set up and solve the Hückel secular equations for the π electrons of
−
NO3
...
Determine the delocalization energy of the ion
...
7 Imagine a small electron-sensitive probe of volume 1
...
14 In the ‘free electron molecular orbital’ (FEMO) theory, the electrons in
a conjugated molecule are treated as independent particles in a box of length
L
...
The
tetraene CH2=CHCH=CHCH=CHCH=CH2 can be treated as a box of
length 8R, where R ≈ 140 pm (as in this case, an extra half bond-length is often
added at each end of the box)
...
Estimate the colour a sample of
the compound is likely to appear in white light
...
8 The energy of H2 with internuclear separation R is given by the

11
...
14) of conjugated molecules is rather
crude and better results are obtained with simple Hückel theory
...
9):

+
into an H2 molecule-ion in its ground state
...
Do the same for the
molecule-ion the instant after the electron has been excited into the
antibonding LCAO-MO
...
The values are given below
...

R/a0

0

1

2

3

4

V1/Eh

11
...
729

10
...
330

10
...
000

10
...
406

10
...
092

1
...
858

0
...
349

0
...
3 eV and a 0 = 52
...

2

11
...
8 may be used to calculate the molecular

potential energy curve for the antibonding orbital, which is given by
E = EH +

e

2

4πε0 R

−

V1 − V2
1−S

Plot the curve
...
10‡ J
...
Dojahn, E
...
M
...
E
...
Phys
...
100,

9649 (1996)) characterized the potential energy curves of homonuclear
diatomic halogen molecules and molecular anions
...
411

916
...
60

F2−

1
...
0

1
...

11
...
Speculate about the existence of ‘hyper Rydberg’ H2
formed from two H atoms with 100s electrons
...
Is such a molecule likely to exist
under any circumstances?
11
...
21 eV photons,
electrons were ejected with kinetic energies of 11
...
23 eV, and 5
...

Sketch the molecular orbital energy level diagram for the species, showing the
ionization energies of the three identifiable orbitals
...
, N

Use this expression to determine a reasonable empirical estimate of the
resonance integral β for the homologous series consisting of ethene,
butadiene, hexatriene, and octatetraene given that π *←π ultraviolet
absorptions from the HOMO to the LUMO occur at 61 500, 46 080, 39 750,
and 32 900 cm−1, respectively
...
(c) In the context of
this Hückel model, the π molecular orbitals are written as linear combinations
of the carbon 2p orbitals
...
, N

Determine the values of the coefficients of each of the six 2p orbitals in each
of the six π molecular orbitals of hexatriene
...
Comment on trends that
relate the energy of a molecular orbital with its ‘shape’, which can be inferred
from the magnitudes and signs of the coefficients in the linear combination
that describes the molecular orbital
...
16 For monocyclic conjugated polyenes (such as cyclobutadiene and
benzene) with each of N carbon atoms contributing an electron in a 2p orbital,
simple Hückel theory gives the following expression for the energies Ek of the
resulting π molecular orbitals:

Ek = α + 2β cos

2kπ
N

k = 0, ±1, ±2,
...
, ±(N − 1)/2 (odd N)

(a) Calculate the energies of the π molecular orbitals of benzene and
cyclooctatetraene
...
(b) Calculate and compare the delocalization energies of benzene
(using the expression above) and hexatriene (see Problem 11
...
What do
you conclude from your results? (c) Calculate and compare the delocalization
energies of cyclooctaene and octatetraene
...
17 If you have access to mathematical software that can perform matrix
diagonalization, use it to solve Problems 11
...
16, disregarding the
expressions for the energies and coefficients given there
...
18 Molecular orbital calculations based on semi-empirical, ab initio, and

DFT methods describe the spectroscopic properties of conjugated molecules a

402

11 MOLECULAR STRUCTURE

bit better than simple Hückel theory
...

(b) Plot the HOMO–LUMO energy separations against the experimental
frequencies for π *←π ultraviolet absorptions for these molecules (Problem
11
...
Use mathematical software to find the polynomial equation that best
fits the data
...
(d) Discuss why the calibration
procedure of part (b) is necessary
...
19 Electronic excitation of a molecule may weaken or strengthen some
bonds because bonding and antibonding characteristics differ between the
HOMO and the LUMO
...
Therefore, promotion of an electron from the HOMO to the
LUMO weakens this carbon–carbon bond in the excited electronic state,
relative to the ground electronic state
...
15 and discuss in detail any changes in bond order
that accompany the π *←π ultraviolet absorptions in these molecules
...
20 As mentioned in Section 2
...
(a) Using molecular modelling software and a semi-empirical
method of your choice, calculate the standard enthalpy of formation of
ethene, butadiene, hexatriene, and octatetraene in the gas phase
...
(c) A good thermochemical database will also report the
uncertainty in the experimental value of the standard enthalpy of formation
...


Theoretical problems
11
...

Show that it has its maximum amplitude in the direction specified
...
22 Use the expressions in Problems 11
...
9 to show that the

antibonding orbital is more antibonding than the bonding orbital is bonding
at most internuclear separations
...
23 Derive eqns 11
...
14 by working with the normalized
+
LCAO-MOs for the H2 molecule-ion (Section 11
...
Proceed by evaluating
the expectation value of the hamiltonian for the ion
...


αA − E

B

β

αB − E

=0

where αA ≠ αB and we have taken S = 0
...
34a and 11
...
Here, we shall develop the result for the case
(α B − αA)2 >> β 2
...
(b) Now use the expansion
(1 + x)1/2 = 1 +

x
2

−

x3
8

+···

to show that
E− = α B +

β2
α B − αA

E+ = αA −

β2
α B − αA

which is the limiting result used in Justification 11
...


Applications: to astrophysics and biology
11
...
12a you were invited to set up the Hückel secular

determinant for linear and cyclic H3
...
The molecular ion H3 was discovered as long
ago as 1912 by J
...
Thomson, but only more recently has the equivalent
equilateral triangular structure been confirmed by M
...
Gaillard et al
...

+
Rev
...
The molecular ion H3 is the simplest polyatomic
species with a confirmed existence and plays an important role in chemical
reactions occurring in interstellar clouds that may lead to the formation of
+
water, carbon monoxide, and ethyl alcohol
...
(a) Solve the Hückel secular
equations for the energies of the H3 system in terms of the parameters α and
β, draw an energy level diagram for the orbitals, and determine the binding
+
−
energies of H3 , H3, and H3
...
D
...
N
...
Chem
...
65, 3547 (1976)) give the
+
dissociation energy for the process H3 → H + H + H+ as 849 kJ mol−1
...
3, calculate the enthalpy of the reaction
+
H+(g) + H2(g) → H3 (g)
...
Then go on to calculate
the bind energies of the other H3 species in (a)
...
27‡ There is some indication that other hydrogen ring compounds and

ions in addition to H3 and D3 species may play a role in interstellar chemistry
...
S
...
A
...
Phys
...
96, 10793 (1992)),
−
+
+
H5 , H6, and H7 are particularly stable whereas H4 and H5 are not
...

11
...
Specifically, we shall describe
the factors that stabilize the planar conformation of the peptide group
...
24 Take as a trial function for the ground state of the hydrogen atom
2
(a) e−kr, (b) e−kr and use the variation principle to find the optimum value of k
in each case
...
The only part of the laplacian
that need be considered is the part that involves radial derivatives (eqn 9
...

11
...
5 that, to find the energies of the bonding and
antibonding orbitals of a heteronuclear diatomic molecule, we need to solve
the secular determinant

2
The web site contains links to molecular modelling freeware and to other sites where you may perform molecular orbital calculations directly from your web
browser
...
Furthermore, a
number of studies indicate that there is a linear correlation between the
LUMO energy and the reduction potential of aromatic hydrocarbons (see, for
example, J
...
Lowe, Quantum chemistry, Chapter 8, Academic Press (1993))
...
023

CH3

H

CH3

H

−0
...
165

CH3

CH3

CH3

CH3

−0
...
078

ψ3 = fψO − gψC + hψN

where the coefficients a through h are all positive
...
In a non-bonding molecular orbital, a pair of electrons
resides in an orbital confined largely to one atom and not appreciably
involved in bond formation
...
(c) Draw a diagram showing
the relative energies of these molecular orbitals and determine the occupancy
of the orbitals
...
Convince yourself that there are four electrons to be
distributed among the molecular orbitals
...
The LCAO-MOs are given by

ψ4 = aψO + bψC

R5

CH3

ψ1 = aψO + bψC + cψN

R3

H

It follows that we can model the peptide group with molecular orbital theory
by making LCAO-MOs from 2p orbitals perpendicular to the plane defined by
the O, C, and N atoms
...
Also, draw an energy level diagram
and determine the occupancy of the orbitals
...
(g) Use
your results from parts (a)–(f) to construct arguments that support the planar
model for the peptide link
...
29 Molecular orbital calculations may be used to predict trends in the
standard potentials of conjugated molecules, such as the quinones and flavins,
that are involved in biological electron transfer reactions (Impact I17
...
It is

Using molecular modelling software and the computational method
of your choice (semi-empirical, ab initio, or density functional theory
methods), calculate ELUMO, the energy of the LUMO of each substituted
1,4-benzoquinone, and plot ELUMO against E 7
...
2)
...
(c) The 1,4-benzoquinone
for which R2 = R3 = R5 = CH3 and R6 = H is a suitable model of plastoquinone,
a component of the photosynthetic electron transport chain (Impact I7
...

Determine ELUMO of this quinone and then use your results from part (a) to
estimate its standard potential
...
2 and I23
...


12
The symmetry elements of
objects
12
...
2 The symmetry classification of

molecules
12
...
We see how to classify any
molecule according to its symmetry and how to use this classification to discuss molecular
properties
...
These symmetry labels
are used to identify integrals that necessarily vanish
...
By knowing which atomic orbitals may have nonzero overlap,
we can decide which ones can contribute to molecular orbitals
...

Finally, by considering the symmetry properties of integrals, we see that it is possible to
derive the selection rules that govern spectroscopic transitions
...
4 Character tables and symmetry

labels
12
...
6 Vanishing integrals and

selection rules

The systematic discussion of symmetry is called group theory
...
However, because
group theory is systematic, its rules can be applied in a straightforward, mechanical
way
...
In
some cases, though, it leads to unexpected results
...
A sphere is more symmetrical than
a cube because it looks the same after it has been rotated through any angle about any
diameter
...
12
...
Similarly, an NH3 molecule is ‘more symmetrical’ than an
H2O molecule because NH3 looks the same after rotations of 120° or 240° about the
axis shown in Fig
...
2, whereas H2O looks the same only after a rotation of 180°
...
Typical symmetry operations include rotations, reflections,
and inversions
...
For instance, a rotation (a symmetry operation) is carried out around an
axis (the corresponding symmetry element)
...
1 OPERATIONS AND SYMMETRY ELEMENTS

405

C3
N
H

C2
C3
(a)
C2

C4

O
(b)
H

Fig
...
1 Some of the symmetry elements of
a cube
...


(a) An NH3 molecule has a
threefold (C3) axis and (b) an H2O
molecule has a twofold (C2) axis
...

Fig
...
2

possess the same set of symmetry elements
...


F

12
...

There are five kinds of symmetry operation (and five kinds of symmetry element) of
this kind
...
These more extensive groups are called space groups
...
Because every molecule is indistinguishable from itself if nothing is
done to it, every object possesses at least the identity element
...

An n-fold rotation (the operation) about an n-fold axis of symmetry, Cn (the corresponding element) is a rotation through 360°/n
...
An H2O molecule has one
twofold axis, C2
...
A pentagon has a C5 axis, with two (clockwise and counterclockwise) rotations through 72° associated with it
...
A cube has three C4 axes, four C3 axes, and six C2 axes
...
If a molecule
possesses several rotation axes, then the one (or more) with the greatest value of n is
called the principal axis
...


C

I

Cl

Br

1 CBrClFI

C6

2 Benzene, C6H6

Comment 12
...


406

12 MOLECULAR SYMMETRY
Centre of
inversion,i

s´
v

sv

Fig
...
3 An H2O molecule has two mirror
planes
...
e
...


S4
C4

sh

(a)
S6

C6

sh

sd

sd

sd

Dihedral mirror planes (σd) bisect
the C2 axes perpendicular to the principal
axis
...
12
...
12
...


A reflection (the operation) in a mirror plane, σ (the element), may contain the
principal axis of a molecule or be perpendicular to it
...
An H2O molecule has two vertical
planes of symmetry (Fig
...
3) and an NH3 molecule has three
...
12
...
When the plane of symmetry is perpendicular to the principal axis it
is called ‘horizontal’ and denoted σh
...

In an inversion (the operation) through a centre of symmetry, i (the element), we
imagine taking each point in a molecule, moving it to the centre of the molecule, and
then moving it out the same distance on the other side; that is, the point (x, y, z) is
taken into the point (−x, −y, −z)
...
A C6H6 molecule does have
a centre of inversion, as does a regular octahedron (Fig
...
5); a regular tetrahedron
and a CH4 molecule do not
...
The first component is a rotation through
360°/n, and the second is a reflection through a plane perpendicular to the axis of that
rotation; neither operation alone needs to be a symmetry operation
...
12
...

12
...
(b) The
staggered form of ethane has an S6 axis
composed of a 60° rotation followed by a
reflection
...
12
...
This procedure puts CH4
and CCl4, which both possess the same symmetry elements as a regular tetrahedron,
into the same group, and H2O into another group
...
There are two systems of notation (Table 12
...
The Schoenflies
system (in which a name looks like C4v) is more common for the discussion of individual molecules, and the Hermann–Mauguin system, or International system (in
which a name looks like 4mm), is used almost exclusively in the discussion of crystal
symmetry
...
12
...
12
...


12
...
12
...
Start at the top and
answer the question posed in each diamond (Y = yes, N = no)
...
It belongs to Ci if it has the identity and the inversion alone (3), and to Cs if it has
the identity and a mirror plane alone (4)
...
1 The notation for point groups*
Ci

⁄

Cs

m

C1

1

C2

2

C3

3

C4

4

C6

6

C2v

2mm

C3v

3m

C4v

4mm

C6v

6mm

C2h

2m

C3h

fl

C4h

4 /m

C6h

6/m

D2

222

D3

32

D4

422

D6

622

D2h

mmm

D3h

fl2m

D4h

4 /mmm

D6h

D2d

›2m

D3d

‹m

S4

› /m

S6

6/mmm
–
3

Th

m3

T

23

Td

›3m

O

432

Oh

m3m

* In the International system (or Hermann–Mauguin system) for point groups, a number n denotes the
presence of an n-fold axis and m denotes a mirror plane
...
It is important to distinguish symmetry elements of the same type but of
different classes, as in 4/mmm, in which there are three classes of mirror plane
...
The only groups listed here are the so-called ‘crystallographic
point groups’ (Section 20
...


n= 2

3

4

5

6

¥

Cn

Dn
Cone
Cnv
(pyramid)

Cnh

Dnh
(plane or bipyramid)
Dnd

S2n

Fig
...
8 A summary of the shapes corresponding to different point groups
...
12
...


12
...
Note that symbol Cn
is now playing a triple role: as the label of a symmetry element, a symmetry operation,
and a group
...

If in addition to the identity and a Cn axis a molecule has n vertical mirror planes σv,
then it belongs to the group Cnv
...
An NH3 molecule has the elements E, C3, and 3σv, so it belongs to the group C3v
...
Other members of the group
C∞v include the linear OCS molecule and a cone
...
An example is trans-CHCl=CHCl
(6), which has the elements E, C2, and σh, so belongs to the group C2h; the molecule
B(OH)3 in the conformation shown in (7) belongs to the group C3h
...
12
...

(c) The groups Dn, Dnh, and Dnd

We see from Fig
...
7 that a molecule that has an n-fold principal axis and n twofold
axes perpendicular to Cn belongs to the group Dn
...
The planar trigonal BF3 molecule has the elements E, C3, 3C2, and σh (with one C2 axis along each B-F bond), so belongs to D3h
(8)
...
All homonuclear diatomic
molecules, such as N2, belong to the group D∞h because all rotations around the axis
are symmetry operations, as are end-to-end rotation and end-to-end reflection; D∞h
is also the group of the linear OCO and HCCH molecules and of a uniform cylinder
...

A molecule belongs to the group Dnd if in addition to the elements of Dn it possesses
n dihedral mirror planes σd
...


C2

sh

i

Fig
...
9 The presence of a twofold axis and
a horizontal mirror plane jointly imply the
presence of a centre of inversion in the
molecule
...
2

The prime on 3C2 indicates that the
′
three C2 axes are different from the other
three C2 axes
...


F

(d) The groups Sn

Molecules that have not been classified into one of the groups mentioned so far, but
that possess one Sn axis, belong to the group Sn
...
Molecules belonging to Sn with n > 4 are
rare
...


B

8 Boron trifluoride, BF3

410

12 MOLECULAR SYMMETRY
C2

C3

sh

C2

C2
–

Cl

Cl

C2
C2

P

C2

Au

C2
C4

sh
C2
9 Ethene, CH2=CH2 (D2h)

C2

sh

10 Phosphorus pentachloride,
PCl5 (D3h)

11 Tetrachloroaurate(III) ion,
[AuCl4] (D4h)

sd
C2
Ph
C2

C3,S3

S4

C2, S4

12 Allene, C3H4 (D2d)

13 Ethane, C2H6 (D3d)

14 Tetraphenylmethane,
C(C6H5)4 (S4)

(e) The cubic groups

15 Buckminsterfullerene, C60 (I )

Fig
...
10 (a) Tetrahedral, (b) octahedral,
and (c) icosahedral molecules are drawn in
a way that shows their relation to a cube:
they belong to the cubicgroups Td, Oh, and
Ih, respectively
...
g
...
Most belong to the cubic groups, and in particular to the tetrahedral
groups T, Td, and Th (Fig
...
10a) or to the octahedral groups O and Oh (Fig
...
10b)
...
12
...
The groups Td and Oh are the groups of the regular tetrahedron
(for instance, CH4) and the regular octahedron (for instance, SF6), respectively
...
12
...
The group Th is based on T but also contains a centre of inversion (Fig
...
12)
...
3 SOME IMMEDIATE CONSEQUENCES OF SYMMETRY

(a) T

411

(b) O

Fig
...
11 Shapes corresponding to the point groups (a) T and (b) O
...


Fig
...
12 The shape of an object belonging
to the group Th
...
A sphere and an
atom belong to R3, but no molecule does
...


Ru

Example 12
...

Method Use the flow diagram in Fig
...
7
...
12
...
Because the molecule has a fivefold axis, it belongs to the group
D5h
...


Fe

Self-test 12
...


[D5d]

17 Ferrocene, Fe(cp)2

12
...
3

Some statements about the properties of a molecule can be made as soon as its point
group has been identified
...


(a) Polarity

A polar molecule is one with a permanent electric dipole moment (HCl, O3, and NH3
are examples)
...
12
...
For example,
the perpendicular component of the dipole associated with one O-H bond in H2O is
cancelled by an equal but opposite component of the dipole of the second O-H bond,
so any dipole that the molecule has must be parallel to the twofold symmetry axis
...
The arrows represent local
contributions to the overall electric dipole,
such as may arise from bonds between
pairs of neighbouring atoms with different
electronegativities
...
12
...
12
...
Any molecule containing an
inversion also possesses at least an S2
element because i and S2 are equivalent
...
12
...

The same remarks apply generally to the group Cnv, so molecules belonging to any
of the Cnv groups may be polar
...
, there are
symmetry operations that take one end of the molecule into the other
...
We can conclude that only molecules belonging to the groups Cn , Cnv , and
Cs may have a permanent electric dipole moment
...
Thus ozone,
O3, which is angular and belongs to the group C2v, may be polar (and is), but carbon
dioxide, CO2, which is linear and belongs to the group D∞h, is not
...
An achiral molecule is a molecule that can be
superimposed on its mirror image
...
A chiral molecule and its mirror-image partner constitute an enantiomeric pair of isomers and rotate the plane of polarization in equal but opposite
directions
...
However, we need to be aware that such an axis may
be present under a different name, and be implied by other symmetry elements that
are present
...
Any molecule containing a centre of inversion, i, also possesses an S2 axis, because i is equivalent to C2 in conjunction with σh, and that combination of elements is S2 (Fig
...
14)
...
Similarly, because S1 = σ, it follows
that any molecule with a mirror plane is achiral
...
However, a molecule may be achiral even though it does not have a centre of inversion
...


COOH

COOH

H3 C
H

S4

H
CH3

+

N
H

C

H
NH2

CH3
18 L-Alanine, NH2CH(CH3)COOH

C

H
NH2

19 Glycine, NH2CH2COOH

H3 C

H

H

CH3

20 N(CH2CH(CH3)CH(CH3)CH2)2 +

12
...
This material will enable us to discuss the formulation and
labelling of molecular orbitals and selection rules in spectroscopy
...
4 Character tables and symmetry labels
We saw in Chapter 11 that molecular orbitals of diatomic and linear polyatomic
molecules are labelled σ, π, etc
...
Thus, a
σ orbital does not change sign under a rotation through any angle, a π orbital changes
sign when rotated by 180°, and so on (Fig
...
15)
...
For example,
we can speak of an individual pz orbital as having σ symmetry if the z-axis lies along
the bond, because pz is cylindrically symmetrical about the bond
...


s

p

Fig
...
15 A rotation through 180° about the
internuclear axis (perpendicular to the
page) leaves the sign of a σ orbital
unchanged but the sign of a π orbital is
changed
...


(a) Representations and characters

Labels analogous to σ and π are used to denote the symmetries of orbitals in polyatomic molecules
...
11
...
As we shall see, these
labels indicate the behaviour of the orbitals under the symmetry operations of the
relevant point group of the molecule
...

Thus, to assign the labels σ and π, we use the table shown in the margin
...
The entry +1 shows that the
orbital remains the same and the entry −1 shows that the orbital changes sign under
the operation C2 at the head of the column (as illustrated in Fig
...
15)
...

The entries in a complete character table are derived by using the formal techniques
of group theory and are called characters, χ (chi)
...
12
...

Under σv, the change (pS, pB, pA) ← (pS, pA, pB) takes place
...
e
...
e
...
e
...

Fig
...
16

(12
...
4

The matrix D(σv) is called a representative of the operation σv
...


See Appendix 2 for a summary of the
rules of matrix algebra
...
For instance, C2 has the effect (−pS, −pB, −pA) ← (pS, pA, pB), and its representative is
⎛ −1 0 0 ⎞
D(C2) = ⎜ 0 0 −1⎟
⎜ 0 −1 0 ⎟
⎝
⎠

(12
...
3)

The identity operation has no effect on the basis, so its representative is the 3 × 3 unit
matrix:
⎛ 1 0 0⎞
D(E) = ⎜ 0 1 0⎟
⎜ 0 0 1⎟
⎝
⎠

(12
...
We denote this three-dimensional representation by Γ (3)
...

The character of an operation in a particular matrix representation is the sum of
the diagonal elements of the representative of that operation
...

Inspection of the representatives shows that they are all of block-diagonal form:
⎛[n] 0 0 ⎞
D = ⎜ 0 [n] [n]⎟
⎜ 0 [n] [n]⎟
⎝
⎠
The block-diagonal form of the representatives show us that the symmetry operations
of C2v never mix pS with the other two functions
...
It is readily verified
that the pS orbital itself is a basis for the one-dimensional representation
D(E ) = 1

D(C2) = −1

D(σv) = 1

D(σ v) = −1
′

which we shall call Γ (1)
...
We say that the original threedimensional representation has been reduced to the ‘direct sum’ of a one-dimensional
representation ‘spanned’ by pS, and a two-dimensional representation spanned by
(pA, pB)
...
4 CHARACTER TABLES AND SYMMETRY LABELS
orbital plays a role different from the other two
...
5)

The one-dimensional representation Γ (1) cannot be reduced any further, and is
called an irreducible representation of the group (an ‘irrep’)
...
These
combinations are sketched in Fig
...
17
...

In this way we find the following representation in the new basis:
⎛ 1 0⎞
D(E ) = ⎜
⎟
⎝ 0 1⎠

⎛ −1 0⎞
D(C2) = ⎜
⎟
⎝ 0 1⎠

⎛1 0 ⎞
D(σv) = ⎜
⎟
⎝ 0 −1⎠

D(σv) = 1

D(C2) = −1

D(σ v = −1
′)

which is the same one-dimensional representation as that spanned by pS, and p2 spans
D(E ) = 1

D(σv) = −1

D(C2) = 1

D(σ v = −1
′)

which is a different one-dimensional representation; we shall denote it Γ (1)′
...
2)
...
An A or a B is used to denote a one-dimensional representation; A is
used if the character under the principal rotation is +1, and B is used if the character
is −1
...
When higher dimensional irreducible representations are permitted,
E denotes a two-dimensional irreducible representation and T a three-dimensional
irreducible representation; all the irreducible representations of C2v are one-dimensional
...
6)

Symmetry operations fall into the same class if they are of the same type (for example,
rotations) and can be transformed into one another by a symmetry operation of the

Table 12
...


-

+

-

+

A

-

+
B

A
+

-

B

⎛ −1 0 ⎞
D(σ v = ⎜
′)
⎟
⎝ 0 −1⎠

The new representatives are all in block-diagonal form, and the two combinations are
not mixed with each other by any operation of the group
...
Thus, p1
spans
D(E ) = 1

415

h=4
z

z 2, y 2, x 2
xy

Fig
...
17 Two symmetry-adapted linear
combinations of the basis orbitals shown in
Fig
...
16
...


416

12 MOLECULAR SYMMETRY
Table 12
...


group
...
The character table in
Table 12
...


sv

+

C3

-

C3

(b) The structure of character tables

sv
²

sv
¢

Fig
...
18 Symmetry operations in the same
class are related to one another by the
symmetry operations of the group
...


Comment 12
...


In general, the columns in a character table are labelled with the symmetry operations
of the group
...
3)
...
In the C3v character table we see that the two threefold rotations (clockwise and counter-clockwise rotations by 120°) belong to the same class: they are
related by a reflection (Fig
...
18)
...
The two reflections of the group C2v fall into different classes: although
they are both reflections, one cannot be transformed into the other by any symmetry
operation of the group
...
The
order of the group C3v, for instance, is 6
...
They are labelled with the symmetry species (the analogues of the
labels σ and π)
...
By convention, irreducible representations are labelled with upper case
Roman letters (such as A1 and E) but the orbitals to which they apply are labelled with
the lower case italic equivalents (so an orbital of symmetry species A1 is called an a1
orbital)
...
12
...

(c) Character tables and orbital degeneracy

The character of the identity operation E tells us the degeneracy of the orbitals
...
Any
doubly degenerate pair of orbitals in C3v must be labelled e because, in this group, only
E symmetry species have characters greater than 1
...
This last
point is a powerful result of group theory, for it means that, with a glance at the
character table of a molecule, we can state the maximum possible degeneracy of its
orbitals
...
4 CHARACTER TABLES AND SYMMETRY LABELS

417

Example 12
...
The maximum number in the column headed
by the identity E is the maximum orbital degeneracy possible in a molecule of that
point group
...
atoms, and decide which number can be used to form a molecule that
can have orbitals of symmetry species T
...
Reference to

the character table for this group shows that the maximum degeneracy is 2, as no
character exceeds 2 in the column headed E
...
A tetrahedral molecule (symmetry group T) has an irreducible
representation with a T symmetry species
...

Self-test 12
...
What is the maximum possible degree of degeneracy of its orbitals?
[5]

(d) Characters and operations

The characters in the rows labelled A and B and in the columns headed by symmetry
operations other than the identity E indicate the behaviour of an orbital under the
corresponding operations: a +1 indicates that an orbital is unchanged, and a −1 indicates that it changes sign
...

For the rows labelled E or T (which refer to the behaviour of sets of doubly and
triply degenerate orbitals, respectively), the characters in a row of the table are the
sums of the characters summarizing the behaviour of the individual orbitals in the
basis
...
12
...
Care must be exercised with these characters because the transformations of orbitals can be quite complicated; nevertheless, the sums of the individual
characters are integers
...
Because H2O belongs to the point
group C2v, we know by referring to the C2v character table (Table 12
...
We can decide the appropriate label for
O2px by noting that under a 180° rotation (C2) the orbital changes sign (Fig
...
21), so
it must be either B1 or B2, as only these two symmetry types have character −1 under
C2
...
As we shall see, any molecular orbital built from this atomic orbital will also be a
b1 orbital
...

The behaviour of s, p, and d orbitals on a central atom under the symmetry operations of the molecule is so important that the symmetry species of these orbitals

e

e

Fig
...
19 Typical symmetry-adapted linear
combinations of orbitals in a C3v molecule
...
12
...


418

12 MOLECULAR SYMMETRY
sv

C2

sv
´
+
-

Fig
...
21 A px orbital on the central atom of
a C2v molecule and the symmetry elements
of the group
...
12
...


+

χ(C3) = 1

χ(σv) = 1

Comparison with the C3v character table shows that ψ1 is of symmetry species A1, and
therefore that it contributes to a1 molecular orbitals in NH3
...
3 Identifying the symmetry species of orbitals

Identify the symmetry species of the orbital ψ = ψA − ψB in a C2v NO2 molecule,
where ψA is an O2px orbital on one O atom and ψB that on the other O atom
...
The same
technique may be applied to linear combinations of orbitals on atoms that are related
by symmetry transformations of the molecule, such as the combination ψ1 = ψA + ψB
+ ψC of the three H1s orbitals in the C3v molecule NH3 (Fig
...
22)
...
To make these allocations, we look at the
symmetry species of x, y, and z, which appear on the right-hand side of the character
table
...
3 shows that pz (which is proportional to
zf (r)), has symmetry species A1 in C3v, whereas px and py (which are proportional to
xf (r) and yf (r), respectively) are jointly of E symmetry
...
An s orbital
on the central atom always spans the fully symmetrical irreducible representation
(typically labelled A1 but sometimes A 1 of a group as it is unchanged under all sym′)
metry operations
...
We can see at a glance that in C3v, dxy and dx 2 − y 2 on a
central atom jointly belong to E and hence form a doubly degenerate pair
...
We need to consider how the combination changes under each operation of
the group, and then write the character as +1, −1, or 0 as specified above
...


+

Fig
...
23 One symmetry-adapted linear
combination of O2px orbitals in the C2v
NO− ion
...
12
...
Under C2, ψ changes into itself,
implying a character of +1
...
Under σ v ψ → −ψ, so the character for this
′,
operation is also −1
...


D

C

−
Self-test 12
...
Identify the symmetry type of the combination ψA − ψB
+ ψC − ψD
...
5 VANISHING INTEGRALS AND ORBITAL OVERLAP

419

y

12
...
7)

where f1 and f2 are functions
...
If we knew that the integral is zero, we could say at once that a molecular
orbital does not result from (A,B) overlap in that molecule
...


(a)
y

(a) The criteria for vanishing integrals

The key point in dealing with the integral I is that the value of any integral, and of an
overlap integral in particular, is independent of the orientation of the molecule (Fig
...
24)
...
Because the volume element dτ is invariant under any symmetry operation, it follows that the integral is nonzero only if the integrand itself, the
product f1 f2, is unchanged by any symmetry operation of the molecular point group
...
It follows that the
only contribution to a nonzero integral comes from functions for which under any
symmetry operation of the molecular point group f1 f2 → f1 f2, and hence for which the
characters of the operations are all equal to +1
...

We use the following procedure to deduce the symmetry species spanned by the
product f1 f2 and hence to see whether it does indeed span A1
...

That is, I is a basis of a representation of
symmetry species A1 (or its equivalent)
...
12
...

2 Multiply the numbers in each column, writing the results in the same order
...
The integral must be zero if this sum does not contain A1
...
12
...
It follows
that the integral must be zero
...
12
...
Had we taken f1 = sN and f2 = s1
instead, where s1 = sA + sB + sC, then because each spans A1 with characters 1,1,1:
f1:
f2:
f1 f2:

1
1
1

1
1
1

1
1
1

sC

-

sB
+

Fig
...
25 A symmetry-adapted linear
combination that belongs to the symmetry
species E in a C3v molecule such as NH3
...
12
...


420

12 MOLECULAR SYMMETRY
The characters of the product are those of A1 itself
...
A shortcut that works when f1 and f2 are bases for irreducible representations of a group is to note their symmetry species: if they are different, then the
integral of their product must vanish; if they are the same, then the integral may be
nonzero
...
For example, the N-H distance in ammonia may be so great that the
(s1, sN) overlap integral is zero simply because the orbitals are so far apart
...
4 Deciding if an integral must be zero (1)

May the integral of the function f = xy be nonzero when evaluated over a region the
shape of an equilateral triangle centred on the origin (Fig
...
26)?

-

+

Method First, note that an integral over a single function f is included in the previ-

The integral of the function f = xy
over the tinted region is zero
...
The
insert shows the shape of the function in
three dimensions
...
12
...
7
...
To decide that, we identify the point group and then examine
the character table to see whether f belongs to A1 (or its equivalent)
...
If we refer to
the character table of the group, we see that xy is a member of a basis that spans the
irreducible representation E′
...

Self-test 12
...
12
...
For instance, in C2v we may find the characters 2, 0, 0, −2 when we multiply
the characters of f1 and f2 together
...
12
...
The insert shows the
shape of the function in three dimensions
...
This expression is symbolic
...
Because the sum on the right does not include a component that is a basis for an
irreducible representation of symmetry species A1, we can conclude that the integral
of f1 f2 over all space is zero in a C2v molecule
...
For example, if we found the characters 8, −2, −6, 4, it would
not be obvious that the sum contains A1
...
The recipe is as follows:

12
...

2 In the first row write down the characters of the symmetry species we want to
analyse
...

4 Multiply the two rows together, add the products together, and divide by the
order of the group
...

Illustration 12
...
When the
procedure is repeated for all four symmetry species, we find that f1 f2 spans A1 + 2A2
+ 5B2
...
5 Does A2 occur among the symmetry species of the irreducible repre-

sentations spanned by a product with characters 7, −3, −1, 5 in the group C2v?
[No]

(a)

(b) Orbitals with nonzero overlap

The rules just given let us decide which atomic orbitals may have nonzero overlap in
a molecule
...
12
...
The general rule is that only orbitals of the same symmetry species
may have nonzero overlap, so only orbitals of the same symmetry species form bonding and antibonding combinations
...
We are therefore at the point of contact between group theory and the material introduced in that
chapter
...
Thus, the (sN, s1)-overlap orbitals are called a1 orbitals (or a 1 if we wish to
*,
emphasize that they are antibonding)
...

Does the N atom have orbitals that have nonzero overlap with them (and give rise to
e molecular orbitals)? Intuition (as supported by Figs
...
28b and c) suggests that
N2px and N2py should be suitable
...
Therefore, N2px and N2py also belong to E, so may have nonzero
overlap with s2 and s3
...
12
...

This diagram illustrates the three bonding
orbitals that may be constructed from
(N2s, H1s) and (N2p, H1s) overlap in a C3v
molecule
...
(There are also three antibonding
orbitals of the same species
...
The two e orbitals that result are shown in Fig
...
28 (there are
also two antibonding e orbitals)
...
As explained earlier, reference to the C3v character
table shows that dz 2 has A1 symmetry and that the pairs (dx 2−y 2, dxy) and (dyz,dzx) each
transform as E
...
Whether or not the d
orbitals are in fact important is a question group theory cannot answer because the
extent of their involvement depends on energy considerations, not symmetry
...
5 Determining which orbitals can contribute to bonding

The four H1s orbitals of methane span A1 + T2
...

Answer An s orbital spans A1, so it may have nonzero overlap with the A1 combina-

tion of H1s orbitals
...
The dxy, dyz, and dzx orbitals span T2, so they may overlap
the same combination
...
It follows that in methane there are (C2s,H1s)overlap a1 orbitals and (C2p,H1s)-overlap t2 orbitals
...
The lowest energy configuration is probably a 1 t 2 , with all
bonding orbitals occupied
...
6 Consider the octahedral SF6 molecule, with the bonding arising
from overlap of S orbitals and a 2p orbital on each F directed towards the central S
atom
...
What s orbitals have nonzero overlap? Suggest
what the ground-state configuration is likely to be
...
) that
have a particular symmetry
...
), as input and generates combinations
of the specified symmetry
...

Symmetry-adapted linear combinations are the building blocks of LCAO molecular
orbitals, for they include combinations such as those used to construct molecular
orbitals in benzene
...

The technique for building SALCs is derived by using the full power of group
theory
...


12
...

(ii) Add together all the orbitals in each column with the factors as determined
in (i)
...

For example, from the (sN,sA,s B,sC) basis in NH3 we form the table shown in the
margin
...
Applying the same technique to the
column under sA gives
1
1
ψ = –(sA + s B + sC + sA + s B + sC ) = –(sA + s B + sC)
6
3

The same combination is built from the other two columns, so they give no further
information
...

We now form the overall molecular orbital by forming a linear combination of all
the SALCs of the specified symmetry species
...
The coefficients are found by solving the
Schrödinger equation; they do not come directly from the symmetry of the system
...
This problem can be illustrated as follows
...
The difference of the second and third gives
1
1
–(sB − sC), and this combination and the first, –(2sA − s B − sC) are the two (now linearly
2
6
independent) SALCs we have used in the discussion of e orbitals
...
6 Vanishing integrals and selection rules
Integrals of the form

Ύ

I = f1 f2 f3dτ

(12
...
5d), and it is important to know when they are necessarily zero
...
To test whether this is so, the characters of all three functions are multiplied together in the same way as in the rules set out above
...
6 Deciding if an integral must be zero (2)

Does the integral ∫(3dz2)x(3dxy)dτ vanish in a C2v molecule?
Method We must refer to the C2v character table (Table 12
...

Answer We draw up the following table:

E
1
1
1
1

f3 = dxy
f2 = x
f1 = dz 2
f1 f2 f3

σv
−1
1
1
−1

C2
1
−1
1
−1

σv
′
−1
−1
1
1

A2
B1
A1

The characters are those of B2
...

Self-test 12
...
The z-component of this vector is defined
through

Ύ

µz,fi = −e ψ f*zψi dτ

A1

A2

B1

y-polarized

x-polarized

x-polarized

Forbidden

[12
...
The transition moment has the form of the
integral in eqn 12
...

As an example, we investigate whether an electron in an a1 orbital in H2O (which
belongs to the group C2v) can make an electric dipole transition to a b1 orbital
(Fig
...
29)
...
8 as x, y, and z in turn
...
The three calculations run as follows:
x-component

Forbidden
B2

y-component

C2

σv

σ′v

E

C2

f3

The polarizations of the allowed
transitions in a C2v molecule
...
The
perspective view of the molecule makes it
look rather like a door stop; however, from
the side, each ‘door stop’ is in fact an
isosceles triangle
...
12
...
Therefore, we conclude that the electric dipole

CHECKLIST OF KEY IDEAS

425

transitions between a1 and b1 are allowed
...

Example 12
...

Answer The procedure works out as follows:

f3(py)
f2(q)
f1(px)
f1 f2 f3

E
3
3
3
27

8C3
0
0
0
0

3C2
−1
−1
−1
−1

6σd
−1
−1
−1
−1

6S4
1
1
1
1

T2
T2
T2

We can use the decomposition procedure described in Section 12
...

A more detailed analysis (using the matrix representatives rather than the characters) shows that only q = z gives an integral that may be nonzero, so the transition
is z-polarized
...

Self-test 12
...
We shall see that the techniques of group theory greatly simplify the analysis of molecular structure and spectra
...
A symmetry operation is an action that leaves an object
looking the same after it has been carried out
...
A symmetry element is a point, line, or plane with respect to
which a symmetry operation is performed
...
A point group is a group of symmetry operations that leaves at
least one common point unchanged
...


7
...
The basis is
the set of functions on which the representative acts
...
A character, χ, is the sum of the diagonal elements of a matrix
representative
...
A character table characterizes the different symmetry types
possible in the point group
...
The notation for point groups commonly used for molecules
and solids is summarized in Table 12
...


10
...
An irreducible representation cannot be
reduced further
...
To be polar, a molecule must belong to Cn, Cnv, or Cs (and
have no higher symmetry)
...
Symmetry species are the labels for the irreducible
representations of a group
...
A molecule may be chiral only if it does not possess an axis of
improper rotation, Sn
...
Decomposition of the direct product is the reduction of a
product of symmetry species to a sum of symmetry species,
Γ × Γ ′ = Γ (1) + Γ (2) + · · ·

426

12 MOLECULAR SYMMETRY

13
...

14
...


15
...


Further reading
Articles and texts

P
...
Atkins and R
...
Friedman, Molecular quantum mechanics
...

F
...
Cotton, Chemical applications of group theory
...

R
...
Saunders, Philadelphia
(1992)
...
C
...
D
...
Dover,
New York (1989)
...
F
...
Kettle, Symmetry and structure: readable group theory for
chemists
...

Sources of data and information

G
...
Breneman, Crystallographic symmetry point group notation
flow chart
...
Chem
...
64, 216 (1987)
...
W
...
S
...
S
...
Phillips, Tables for group theory
...


Discussion questions
12
...


12
...


12
...


12
...


12
...


12
...


12
...

12
...


Exercises
12
...
List the symmetry
elements of the group and locate them in the molecule
...
4a Show that the transition A1 → A2 is forbidden for electric dipole

12
...
List the symmetry
elements of the group and locate them in the molecule
...
4b Is the transition A1g → E2u forbidden for electric dipole transitions in a
D6h molecule?

12
...


12
...


12
...


12
...


(b) HW2(CO)10 (D4h), (c) SnCl4 (Td)
...
6a Molecules belonging to the point groups D2h or C3h cannot be chiral
...
3a Use symmetry properties to determine whether or not the integral
∫px zpzdτ is necessarily zero in a molecule with symmetry C4v
...
6b Molecules belonging to the point groups Th or Td cannot be chiral
...
3b Use symmetry properties to determine whether or not the integral

∫px zpzdτ is necessarily zero in a molecule with symmetry D6h
...
7a The group D2 consists of the elements E, C2, C2 and C2 where the
′,
″,

three twofold rotations are around mutually perpendicular axes
...


PROBLEMS
12
...

Construct the group multiplication table
...
12a Consider the C2v molecule NO2
...
Is there any orbital
of the central N atom that can have a nonzero overlap with that combination
of O orbitals? What would be the case in SO2, where 3d orbitals might be
available?

12
...


−
12
...
Is there any orbital of the central N atom

12
...
What would be the
case in SO3, where 3d orbitals might be available?

(a) a sharpened cylindrical pencil, (b) a three-bladed propellor, (c) a
four-legged table, (d) yourself (approximately)
...
9a List the symmetry elements of the following molecules and name
the point groups to which they belong: (a) NO2, (b) N2O, (c) CHCl3,
(d) CH2=CH2, (e) cis-CHBr=CHBr, (f) trans-CHCl=CHCl
...
9b List the symmetry elements of the following molecules and name the
point groups to which they belong: (a) naphthalene, (b) anthracene, (c) the
three dichlorobenzenes
...
10a Assign (a) cis-dichloroethene and (b) trans-dichloroethene to point

groups
...
13a The ground state of NO2 is A1 in the group C2v
...
13b The ClO2 molecule (which belongs to the group C2v) was trapped in a
solid
...
Light polarized parallel to the y-axis
(parallel to the OO separation) excited the molecule to an upper state
...
14a What states of (a) benzene, (b) naphthalene may be reached by

12
...


12
...
11a Which of the molecules in Exercises 12
...
10a can be (a) polar,

12
...


12
...
9b and 12
...
15b Determine whether the integral over f1 and f2 in Exercise 12
...


Problems*
12
...
Which of these molecules can be
(i) polar, (ii) chiral?

up the 6 × 6 matrices that represent the group in this basis
...
Confirm, by calculating the traces of the matrices: (a) that
′
symmetry elements in the same class have the same character, (b) that the
representation is reducible, and (c) that the basis spans 3A1 + B1 + 2B2
...
2 The group C2h consists of the elements E, C2, σ h, i
...
5 Confirm that the z-component of orbital angular momentum is a basis

multiplication table and find an example of a molecule that belongs to the
group
...

12
...
Show that the group must therefore have a centre of
inversion
...
Use the character table to
confirm these remarks
...
4 Consider the H2O molecule, which belongs to the group C2v
...
7 Construct the multiplication table of the Pauli spin matrices, σ, and the

12
...


428

12 MOLECULAR SYMMETRY

⎛ 0 1⎞
σx = ⎜
⎟
⎝ 1 0⎠

⎛ 0 −i⎞
σy = ⎜
⎟
⎝i 0⎠

⎛1 0 ⎞
σz = ⎜
⎟
⎝ 0 −1⎠

⎛ 1 0⎞
σ0 = ⎜
⎟
⎝ 0 1⎠

Do the four matrices form a group under multiplication?
12
...
9 Suppose that a methane molecule became distorted to (a) C3v symmetry
by the lengthening of one bond, (b) C2v symmetry, by a kind of scissors action
in which one bond angle opened and another closed slightly
...
10‡ B
...
Bovenzi and G
...
Pearse, Jr
...
Chem
...
Dalton Trans
...
Reaction with NiSO4 produced a
complex in which two of the essentially planar ligands are bonded at right
angles to a single Ni atom
...


27

28

derived from a tetrahedron by a distortion shown in (27)
...
(c) What is the point
group of the distorted octahedron? (d) What is the symmetry species of the
distortion considered as a vibration in the new, less symmetric group?
12
...
Identify the irreducible
representations spanned by these orbitals in (a) C2v, (b) C3v, (c) Td, (d) Oh
...
What sets of orbitals do the seven f orbitals split into?
12
...
11‡ R
...
Hoge, and D
...
Brauer (Inorg
...
36, 1464 (1997))

12
...
In
the complex anion [trans-Ag(CF3)2(CN)2]−, the Ag-CN groups are collinear
...
(b) Now
suppose the CF3 groups cannot rotate freely (because the ion was in a solid, for
example)
...
Name the point group of the complex if each CF3 group
has a CF bond in that plane (so the CF3 groups do not point to either CN
group preferentially) and the CF3 groups are (i) staggered, (ii) eclipsed
...
Taking as a basis the N2s, N2p, and O2p orbitals, identify the
irreducible representations they span, and construct the symmetry-adapted
linear combinations
...
16 Construct the symmetry-adapted linear combinations of C2pz orbitals

for benzene, and use them to calculate the Hückel secular determinant
...
6d
...
17 The phenanthrene molecule (29) belongs to the group C2v with

the C2 axis perpendicular to the molecular plane
...
(b) Use your results from part (a) to calculate
the Hückel secular determinant
...
12‡ A computational study by C
...
Marsden (Chem
...
Lett
...
For example, most of the AM4 structures were not
tetrahedral but had two distinct values for MAM bond angles
...
18‡ In a spectroscopic study of C60, F
...
Orlandi, and F
...
Phys
...
100, 10849 (1996)) assigned peaks in the fluorescence
spectrum
...
The ground electronic
state is A1g, and the lowest-lying excited states are T1g and Gg
...
(b) What if the transition is accompanied by a
vibration that breaks the parity?

PROBLEMS
Applications: to astrophysics and biology
12
...
(a) Identify the symmetry elements and determine the point group
of this molecule
...
(c) Obtain
the group multiplication table by explicit multiplication of the matrices
...

12
...
The H4
analogues have not yet been found, and the square planar structure is thought
to be unstable with respect to vibration
...

12
...
2) and protection against
harmful biological oxidations
...
(a) To what point group does this model
of β-carotene belong? (b) Classify the irreducible representations spanned by

429

the carbon 2pz orbitals and find their symmetry-adapted linear combinations
...

(d) What states of this model of β-carotene may be reached by electric dipole
transitions from its (totally symmetrical) ground state?
12
...
2) and
the haem groups of cytochromes (Impact I7
...
The ground
electronic state is A1g and the lowest-lying excited state is Eu
...


13
General features of spectroscopy

Experimental techniques
13
...
3 Linewidths
I13
...
1

Pure rotation spectra

Moments of inertia
The rotational energy levels
13
...
7 Rotational Raman spectra
13
...
4
13
...
11 Anharmonicity
13
...
13 Vibrational Raman spectra of
diatomic molecules
13
...
Rotational energy levels are considered first, and we see how to derive
expressions for their values and how to interpret rotational spectra in terms of molecular
dimensions
...
Next, we consider the vibrational energy levels of diatomic molecules, and see that we
can use the properties of harmonic oscillators developed in Chapter 9
...
We also see that the symmetry properties of the
vibrations of polyatomic molecules are helpful for deciding which modes of vibration can
be studied spectroscopically
...
10

The vibrations of polyatomic
molecules
13
...
15
I13
...
16
I13
...
17

Normal modes
Infrared absorption spectra of
polyatomic molecules
Impact on environmental
science: Global warming
Vibrational Raman spectra of
polyatomic molecules
Impact on biochemistry:
Vibrational microscopy
Symmetry aspects of molecular
vibrations

Checklist of key ideas
Further reading
Further information 13
...
2: Selection rules
for rotational and vibrational
spectroscopy
Discussion questions
Exercises
Problems

The origin of spectral lines in molecular spectroscopy is the absorption, emission, or
scattering of a photon when the energy of a molecule changes
...
Molecular spectra are therefore more complex than atomic spectra
...
They also provide a way of
determining a variety of molecular properties, particularly molecular dimensions,
shapes, and dipole moments
...

Pure rotational spectra, in which only the rotational state of a molecule changes,
can be observed in the gas phase
...

Electronic spectra, which are described in Chapter 14, show features arising from
simultaneous vibrational and rotational transitions
...


13
...
In
emission spectroscopy, a molecule undergoes a transition from a state of high energy
E1 to a state of lower energy E2 and emits the excess energy as a photon
...
We say
net absorption, because it will become clear that, when a sample is irradiated, both
absorption and emission at a given frequency are stimulated, and the detector measures the difference, the net absorption
...
10)
...
We shall discuss emission spectroscopy in
Chapter 14; here we focus on absorption spectroscopy, which is widely employed in
studies of electronic transitions, molecular rotations, and molecular vibrations
...
Vibrational and rotational
transitions, the focus of the discussion in this chapter, can be induced in two ways
...
Second, vibrational and rotational
energy levels can be explored by examining the frequencies present in the radiation
scattered by molecules in Raman spectroscopy
...
These scattered photons constitute the lower-frequency Stokes radiation from
the sample (Fig
...
1)
...

The component of radiation scattered without change of frequency is called Rayleigh
radiation
...
13
...
The process can be regarded as
taking place by an excitation of the
molecule to a wide range of states
(represented by the shaded band), and the
subsequent return of the molecule to a
lower state; the net energy change is then
carried away by the photon
...
1 Experimental techniques
A spectrometer is an instrument that detects the characteristics of light scattered,
emitted, or absorbed by atoms and molecules
...
2 shows the general layouts
of absorption and emission spectrometers operating in the ultraviolet and visible
ranges
...
In most

Detector
Detector

Beam
combiner
Sample

Scattered
radiation
Source

Source
Grating

Sample

Reference
(a)

(b)

Fig
...
2 Two examples of spectrometers:
(a) the layout of an absorption
spectrometer, used primarily for studies in
the ultraviolet and visible ranges, in which
the exciting beams of radiation pass
alternately through a sample and a
reference cell, and the detector is
synchronized with them so that the relative
absorption can be determined, and (b) a
simple emission spectrometer, where light
emitted or scattered by the sample is
detected at right angles to the direction of
propagation of an incident beam of
radiation
...
1

The principles of operation of radiation
sources, dispersing elements, Fourier
transform spectrometers, and detectors
are described in Further information
13
...


Sample
cell
Laser

spectrometers, light transmitted, emitted, or scattered by the sample is collected by
mirrors or lenses and strikes a dispersing element that separates radiation into different frequencies
...
In a typical Raman spectroscopy experiment, a monochromatic incident
laser beam is passed through the sample and the radiation scattered from the front
face of the sample is monitored (Fig
...
3)
...

Modern spectrometers, particularly those operating in the infrared and nearinfrared, now almost always use Fourier transform techniques of spectral detection
and analysis
...
The
total signal from a sample is like a chord played on a piano, and the Fourier transform
of the signal is equivalent to the separation of the chord into its individual notes, its
spectrum
...
2 The intensities of spectral lines
Monochromator
or interferometer
Fig
...
3 A common arrangement adopted
in Raman spectroscopy
...
The focused beam strikes
the sample and scattered light is both
deflected and focused by the mirror
...


The ratio of the transmitted intensity, I, to the incident intensity, I0, at a given
frequency is called the transmittance, T, of the sample at that frequency:
T=

I

[13
...
2)

The quantity ε is called the molar absorption coefficient (formerly, and still widely,
the ‘extinction coefficient’)
...
Its
dimensions are 1/(concentration × length), and it is normally convenient to express it
in cubic decimetres per mole per centimetre (dm3 mol−1 cm−1)
...
This change of units demonstrates that
ε may be regarded as a molar cross-section for absorption and, the greater the crosssectional area of the molecule for absorption, the greater its ability to block the passage of the incident radiation
...
2, we introduce the absorbance, A, of the sample at a given
wavenumber as
A = log

I0
I

or

A = −log T

[13
...
4)

The product ε[J]l was known formerly as the optical density of the sample
...
4 suggests that, to achieve sufficient absorption, path lengths through gaseous samples must be very long, of the order of metres, because concentrations are low
...
Conversely, path lengths through liquid samples can be
significantly shorter, of the order of millimetres or centimetres
...
2 THE INTENSITIES OF SPECTRAL LINES

433

Justification 13
...
However, it is simple to account for its
form
...
We can therefore write
dI = −κ [J]Idl
where κ (kappa) is the proportionality coefficient, or equivalently
dI

= −κ [J]dl

I

This expression applies to each successive layer into which the sample can be
regarded as being divided
...


Illustration 13
...
However, as absorption bands generally spread over a range
of wavenumbers, quoting the absorption coefficient at a single wavenumber might
not give a true indication of the intensity of a transition
...
13
...
5]

band

For lines of similar widths, the integrated absorption coefficients are proportional to
the heights of the lines
...
If the transmittance is 0
...
1)2 = 0
...


Area =
integrated
absorption
coefficient

~
Wavenumber, n
Fig
...
4 The integrated absorption
coefficient of a transition is the area under a
plot of the molar absorption coefficient
against the wavenumber of the incident
radiation
...
Stimulated
absorption is the transition from a low energy state to one of higher energy that is
driven by the electromagnetic field oscillating at the transition frequency
...
10 that the transition rate, w, is the rate of change of probability of the
molecule being found in the upper state
...
13
...
Einstein wrote the transition rate as
w = Bρ

The processes that account for
absorption and emission of radiation and
the attainment of thermal equilibrium
...

Fig
...
5

Comment 13
...
5 and 13
...


(13
...
When the molecule is exposed to black-body radiation from a source
of temperature T, ρ is given by the Planck distribution (eqn 8
...
7)

For the time being, we can treat B as an empirical parameter that characterizes the
transition: if B is large, then a given intensity of incident radiation will induce transitions strongly and the sample will be strongly absorbing
...

Einstein considered that the radiation was also able to induce the molecule in the
upper state to undergo a transition to the lower state, and hence to generate a photon
of frequency ν
...
8)

where B′ is the Einstein coefficient of stimulated emission
...
However, he realized that stimulated emission was not the only means by which
the excited state could generate radiation and return to the lower state, and suggested
that an excited state could undergo spontaneous emission at a rate that was independent of the intensity of the radiation (of any frequency) that is already present
...
9)

The constant A is the Einstein coefficient of spontaneous emission
...
10)

where N′ is the population of the upper state
...
11)

13
...
2 The relation between the Einstein coefficients

At thermal equilibrium, the total rates of emission and absorption are equal, so
NBρ = N′(A + B′ρ)
This expression rearranges into

ρ=

N′A
NB − N′B′

=

A/B
N/N′ − B′/B

=

A/B
e

hν/kT

− B′/B

We have used the Boltzmann expression (Molecular interpretation 3
...
7), which describes
the radiation density at thermal equilibrium
...
11
...
5)
...

Spontaneous emission can be largely ignored at the relatively low frequencies of
rotational and vibrational transitions, and the intensities of these transitions can
be discussed in terms of stimulated emission and absorption
...
12)

(a)

and is proportional to the population difference of the two states involved in the
transition
...
3 and 12
...
Selection rules also apply to molecular
spectra, and the form they take depends on the type of transition
...
We saw in Section 9
...
13]

where ¢ is the electric dipole moment operator
...
13
...

We know from time-dependent perturbation theory (Section 9
...
It follows that the coefficient of stimulated absorption

(b)
Fig
...
6 (a) When a 1s electron becomes a
2s electron, there is a spherical migration of
charge; there is no dipole moment
associated with this migration of charge;
this transition is electric-dipole forbidden
...
(There are subtle effects arising
from the sign of the wavefunction that give
the charge migration a dipolar character,
which this diagram does not attempt to
convey
...
A detailed analysis gives
B=

| µfi |2

(13
...
It follows that, to identify the selection rules, we must establish the conditions
for which µ fi ≠ 0
...
For instance, we shall see that a molecule gives a
rotational spectrum only if it has a permanent electric dipole moment
...
A detailed study of the transition moment leads to the specific selection rules that express the allowed transitions in terms of the changes in quantum
numbers
...
3), such as the rule ∆l = ±1 for the angular
momentum quantum number
...
3 Linewidths
A number of effects contribute to the widths of spectroscopic lines
...
Other contributions cannot be changed, and represent an inherent limitation on resolution
...
13
...
Notice that the line broadens
as the temperature is increased
...
Plot the resulting line shape for
various temperatures
...
In some cases, meaningful spectroscopic data can be obtained
only from gaseous samples
...

One important broadening process in gaseous samples is the Doppler effect, in
which radiation is shifted in frequency when the source is moving towards or away
from the observer
...
15)

where c is the speed of light (see Further reading for derivations)
...
16)

Molecules reach high speeds in all directions in a gas, and a stationary observer detects
the corresponding Doppler-shifted range of frequencies
...
The detected spectral ‘line’ is the absorption or emission profile arising from all the resulting Doppler
shifts
...

The Doppler line shape is therefore also a Gaussian (Fig
...
7), and we show in the

13
...
17)

For a molecule like N2 at room temperature (T ≈ 300 K), δν/ν ≈ 2
...
For a
typical rotational transition wavenumber of 1 cm−1 (corresponding to a frequency of
30 GHz), the linewidth is about 70 kHz
...
Therefore, to obtain
spectra of maximum sharpness, it is best to work with cold samples
...
This usage is doubly wrong
...

Second, ‘wavenumber’ is not a unit, it is an observable with the dimensions of
1/length and commonly reported in reciprocal centimetres (cm−1)
...
3 Doppler broadening

We know from the Boltzmann distribution (Molecular interpretation 3
...
The observed frequen2
cies, νobs, emitted or absorbed by the molecule are related to its speed by eqn 13
...
When s << c, the Doppler shift in the frequency is

νobs − ν ≈ ±νs/c
which implies a symmetrical distribution of observed frequencies with respect to
molecular speeds
...
The width at half-height can be calculated directly from the exponent (see Comment 13
...
17
...
The same is true of the spectra of samples in condensed phases and solution
...
Specifically,
when the Schrödinger equation is solved for a system that is changing with time, it is
found that it is impossible to specify the energy levels exactly
...
3

A Gaussian function of the general form
2
2
y(x) = ae−(x−b) /2σ , where a, b, and σ are
constants, has a maximum y(b) = a and a
width at half-height δx = 2σ (2 ln 2)1/2
...
18)

τ

This expression is reminiscent of the Heisenberg uncertainty principle (eqn 8
...
When
the energy spread is expressed as a wavenumber through δE = hcδ#, and the values of
the fundamental constants introduced, this relation becomes
δ# ≈

5
...
19)

No excited state has an infinite lifetime; therefore, all states are subject to some lifetime broadening and, the shorter the lifetimes of the states involved in a transition, the
broader the corresponding spectral lines
...
The dominant one for low frequency transitions is collisional deactivation, which arises from
collisions between molecules or with the walls of the container
...
Because τcol = 1/z, where z is the collision frequency, and from the kinetic
model of gases (Section 1
...
The collisional linewidth can
therefore be minimized by working at low pressures
...
Hence it is a natural limit to
the lifetime of an excited state, and the resulting lifetime broadening is the natural
linewidth of the transition
...
Natural linewidths
depend strongly on the transition frequency (they increase with the coefficient of
spontaneous emission A and therefore as ν 3), so low frequency transitions (such as the
microwave transitions of rotational spectroscopy) have very small natural linewidths,
and collisional and Doppler line-broadening processes are dominant
...
For example, a typical electronic
excited state natural lifetime is about 10−8 s (10 ns), corresponding to a natural width
of about 5 × 10−4 cm−1 (15 MHz)
...

IMPACT ON ASTROPHYSICS

I13
...
8
...
726 ± 0
...
This cosmic microwave
background radiation is the residue of energy released during the Big Bang, the event
that brought the Universe into existence
...

The interstellar space in our galaxy is a little warmer than the cosmic background
and consists largely of dust grains and gas clouds
...
Interstellar clouds are significant
because it is from them that new stars, and consequently new planets, are formed
...
Colder clouds range from 0
...
1 IMPACT ON ASTROPHYSICS: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY OF INTERSTELLAR SPACE

Fig
...
8 Rotational spectrum of the Orion nebula, showing spectral fingerprints of diatomic and polyatomic molecules present in the
interstellar cloud
...
A
...
, Astrophys
...
315, 621 (1987)
...
There are also colder and
denser clouds, some with masses greater than 500 000 solar masses, densities greater
than 109 particles m−3, and temperatures that can be lower than 10 K
...
There are also trace amounts of larger molecules
...

It follows from the the Boltzmann distribution and the low temperature of a molecular cloud that the vast majority of a cloud’s molecules are in their vibrational and
electronic ground states
...
As a result, the spectrum of the cloud in the
radiofrequency and microwave regions consists of sharp lines corresponding to rotational transitions (Fig
...
8)
...
Earth-bound radiotelescopes are often located at the tops of high mountains, as
atmospheric water vapour can reabsorb microwave radiation from space and hence
interfere with the measurement
...
The experiments have revealed the presence
of trace amounts (with abundances of less than 10−8 relative to hydrogen) of neutral
molecules, ions, and radicals
...
The largest molecule detected by rotational
spectroscopy is the nitrile HC11N
...
The experiments are conducted

439

440

13 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Table 13
...
Diatomic molecules

R
mA

I = µR2

mB

µ=

mAmB
m

2
...
Symmetric rotors

mC
R¢

I|| = 2mA(1 − cos θ)R2

mB

mA

R

mA

q

+

mA
mB

q

mC
1
{(3mA + mB)R′ + 6mAR[– (1 + 2cos θ)] 1/2}R′
3
m

I|| = 2mA(1 − cos θ)R2
I⊥ = mA(1 − cos θ)R2 +

mA

mC
R¢
mA
mA

mB R

mA

I|| = 4mAR2
I⊥ = 2mAR2 + 2mCR′2

mA
mC

4
...


mAmB
(1 + 2 cos θ)R2
m

13
...
2)
...
The data can detect the presence of gas and solid water, CO, and CO2 in
molecular clouds
...
However,
infrared emissions can be observed if molecules are occasionally excited by highenergy photons emitted by hot stars in the vicinity of the cloud
...


2
2
I = 3mAr A + 3mDr D

mA

rA
mA

mA
mB
mC

The general strategy we adopt for discussing molecular spectra and the information
they contain is to find expressions for the energy levels of molecules and then to
calculate the transition frequencies by applying the selection rules
...
In this section we illustrate the strategy by considering
the rotational states of molecules
...
4 Moments of inertia
The key molecular parameter we shall need is the moment of inertia, I, of the
molecule (Section 9
...
The moment of inertia of a molecule is defined as the mass of
each atom multiplied by the square of its distance from the rotational axis through the
centre of mass of the molecule (Fig
...
9):

∑ mi r 2
i

mD

rD

Pure rotation spectra

I=

441

mD

mD

Fig
...
9 The definition of moment of
inertia
...
In this
example, the centre of mass lies on an axis
passing through the B and C atom, and the
perpendicular distances are measured
from this axis
...
20]

i

where ri is the perpendicular distance of the atom i from the axis of rotation
...

In general, the rotational properties of any molecule can be expressed in terms of
the moments of inertia about three perpendicular axes set in the molecule (Fig
...
10)
...
For linear molecules, the moment of inertia around the internuclear
axis is zero
...
1
...
1 Calculating the moment of inertia of a molecule

Ia

Ic
Ib
Fig
...
10 An asymmetric rotor has three
different moments of inertia; all three
rotation axes coincide at the centre
of mass of the molecule
...
The HOH bond angle is 104
...
7 pm
...
20, the moment of inertia is the sum of the masses

f

multiplied by the squares of their distances from the axis of rotation
...


R
rH

1

442

13 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Answer From eqn 13
...
It follows that
I = 2mHR2 sin2 φ
Substitution of the data gives
I = 2 × (1
...
57 × 10−11 m)2 × sin2 52
...
91 × 10−47 kg m2
Note that the mass of the O atom makes no contribution to the moment of inertia
for this mode of rotation as the atom is immobile while the H atoms circulate
around it
...

Self-test 13
...
The C-Cl bond length is 177 pm and
the HCCl angle is 107°; m(35Cl) = 34
...

[4
...
Rigid rotors can be classified into four types (Fig
...
11):
Spherical rotors have three equal moments of inertia (examples: CH4, SiH4, and
SF6)
...

Linear rotors have one moment of inertia (the one about the molecular axis) equal
to zero (examples: CO2, HCl, OCS, and HC
...

Asymmetric rotors have three different moments of inertia (examples: H2O,
H2CO, and CH3OH)
...

Fig
...
11

0

I
I

I

Asymmetric Ic
rotor
Ia

I

Ib

13
...
5 The rotational energy levels
The rotational energy levels of a rigid rotor may be obtained by solving the appropriate Schrödinger equation
...

The classical expression for the energy of a body rotating about an axis a is
1
Ea = –Iaω 2
a
2

(13
...
A body free to rotate about three axes has an
energy

J
16

1
1
1
2
E = –Iaω 2 + –Ibω b + –Icω 2
a
c
2
2
2

15

Because the classical angular momentum about the axis a is Ja = Iaωa, with similar
expressions for the other axes, it follows that
2Ia

+

J2
b
2Ib

+

J2
c

(13
...
We described the quantum mechanical properties of angular
momentum in Section 9
...


14

Energy

E=

J2
a

13
12
11

(a) Spherical rotors

10

When all three moments of inertia are equal to some value I, as in CH4 and SF6, the
classical expression for the energy is
E=

2
2
Ja + Jb + J2
c

2I

=

J

9

2

8B

2I
6B

where J = + + is the square of the magnitude of the angular momentum
...


(13
...


1
0

7
6
5
4
3
2
1
0

Fig
...
12 The rotational energy levels of a
linear or spherical rotor
...


Comment 13
...
25)

The rotational constant as defined by eqn 13
...
The energy of a
rotational state is normally reported as the rotational term, F( J), a wavenumber, by
division by hc:
F(J) = BJ(J + 1)

2B

8

[13
...
13
...
The energy is normally
expressed in terms of the rotational constant, B, of the molecule, where
hcB =

2

J = 0, 1, 2,
...
26)

The definition of B as a wavenumber
is convenient when we come to
vibration–rotation spectra
...
Then B = $/4πcI and the
energy is E = hBJ(J + 1)
...
27)

Because the rotational constant decreases as I increases, we see that large molecules
have closely spaced rotational energy levels
...
85 × 10−45 kg m2, and hence B = 0
...

(b) Symmetric rotors

In symmetric rotors, two moments of inertia are equal but different from the third (as
in CH3Cl, NH3, and C6H6); the unique axis of the molecule is its principal axis (or
figure axis)
...
If I|| > I⊥, the rotor is classified as oblate (like a
pancake, and C6H6); if I|| < I⊥ it is classified as prolate (like a cigar, and CH3Cl)
...
22, becomes
E=

2
Jb + J2
c

2I|

+

J2
a
2I||

2
2
Again, this expression can be written in terms of J 2 = J a + J b + J 2:
c

E=

J2 − J2
a
2I⊥

+

J2
a
2I||

=

J2
2I

+

A 1
1 D 2
−
J
C 2I|| 2I⊥ F a

(13
...
We also know from the quantum theory of
angular momentum (Section 9
...
, ± J
...
) Therefore, we also replace J 2 by K2$2
...


K = 0, ±1,
...
29)

with
A=
J
K»J
(a)

J
K=0

(b)
Fig
...
13 The significance of the quantum
number K
...
(b) When
K = 0 the molecule has no angular
momentum about its principal axis: it is
undergoing end-over-end rotation
...
30]

Equation 13
...
When K = 0, there is
no component of angular momentum about the principal axis, and the energy levels
depend only on I⊥ (Fig
...
13)
...
The sign of K does not affect the energy because opposite values of K correspond to
opposite senses of rotation, and the energy does not depend on the sense of rotation
...
2 Calculating the rotational energy levels of a molecule

A 14NH3 molecule is a symmetric rotor with bond length 101
...
7°
...

A note on good practice To calculate moments of inertia precisely, we need to

specify the nuclide
...
5 THE ROTATIONAL ENERGY LEVELS

445

Method Begin by calculating the rotational constants A and B by using the expres-

sions for moments of inertia given in Table 13
...
Then use eqn 13
...

Answer Substitution of mA = 1
...
0031 u, R = 101
...
7° into the second of the symmetric rotor expressions in Table 13
...
4128 × 10−47 kg m2 and I⊥ = 2
...
Hence, A = 6
...
977 cm−1
...
29 that
F(J,K)/cm−1 = 9
...
633K2
Upon multiplication by c, F(J,K) acquires units of frequency:
F(J,K)/GHz = 299
...
9K 2
For J = 1, the energy needed for the molecule to rotate mainly about its figure axis
(K = ±J) is equivalent to 16
...
3 GHz), but end-over-end rotation (K = 0)
corresponds to 19
...
1 GHz)
...
2 A CH335Cl molecule has a C-Cl bond length of 178 pm, a C-H

bond length of 111 pm, and an HCH angle of 110
...
Calculate its rotational energy
terms
...
444J(J + 1) + 4
...
3J(J + 1) + 137K2]
(a)

(c) Linear rotors

For a linear rotor (such as CO2, HCl, and C2H2), in which the nuclei are regarded as
mass points, the rotation occurs only about an axis perpendicular to the line of atoms
and there is zero angular momentum around the line
...
0 in eqn 13
...
The rotational terms of a linear molecule are therefore
F(J) = BJ(J + 1)

J = 0, 1, 2,
...
31)

This expression is the same as eqn 13
...
0 but for a spherical rotor A = B
...
However,
we must not forget that the angular momentum of the molecule has a component
on an external, laboratory-fixed axis
...
, ±J, giving 2J + 1 values in all (Fig
...
14)
...
Consequently, all 2J + 1
orientations of the rotating molecule have the same energy
...
A linear rotor has K fixed at 0, but the angular momentum may still have
2J + 1 components on the laboratory axis, so its degeneracy is 2J + 1
...
Therefore, as well as having a (2J + 1)fold degeneracy arising from its orientation in space, the rotor also has a (2J + 1)-fold
degeneracy arising from its orientation with respect to an arbitrary axis in the

MJ = 0

(c)
Fig
...
14 The significance of the quantum
number MJ
...

(b) An intermediate value of MJ
...


446

13 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Field MJ
on 0
±1

±2
±3

±4

Field
off

molecule
...
This degeneracy increases very rapidly: when J = 10, for instance,
there are 441 states of the same energy
...
g
...
13
...
The splitting of states by an electric field
is called the Stark effect
...
32a)

where (see Further reading for a derivation)
±5

±6

±7

The effect of an electric field on
the energy levels of a polar linear rotor
...

Fig
...
15

a(J,MJ) =

J(J + 1) − 3M 2
J
2hcBJ(J + 1)(2J − 1)(2J + 3)

Note that the energy of a state with quantum number MJ depends on the square of the
permanent electric dipole moment, µ
...
However, as spectra
can be recorded for samples at pressures of only about 1 Pa and special techniques
(such as using an intense laser beam or an electrical discharge) can be used to vaporize even some quite nonvolatile substances, a wide variety of samples may be studied
...

(e) Centrifugal distortion

We have treated molecules as rigid rotors
...
13
...
The effect of centrifugal distortion on a
diatomic molecule is to stretch the bond and hence to increase the moment of inertia
...
The effect
is usually taken into account largely empirically by subtracting a term from the energy
and writing
F(J) = BJ(J + 1) − DJ J2(J + 1)2

Centrifugal
force

(13
...
It is large when the bond is
easily stretched
...
22)
DJ =

Fig
...
16 The effect of rotation on a
molecule
...
The effect is to increase the
moment of inertia of the molecule and
hence to decrease its rotational constant
...
32b)

4B3
#2

(13
...

13
...
1 to10 cm−1 (for example, 0
...
59 cm−1 for HCl), so rotational transitions lie in the
microwave region of the spectrum
...
Modulation of the transmitted intensity can be
achieved by varying the energy levels with an oscillating electric field
...
6 ROTATIONAL TRANSITIONS
modulation, an electric field of about 105 V m−1 and a frequency of between 10 and
100 kHz is applied to the sample
...
2) that the gross selection rule for the observation of a pure rotational spectrum is that a molecule must have a permanent electric
dipole moment
...
The classical basis of this rule is that a polar molecule appears to possess a
fluctuating dipole when rotating, but a nonpolar molecule does not (Fig
...
17)
...
Homonuclear
diatomic molecules and symmetrical linear molecules such as CO2 are rotationally
inactive
...
An example of
a spherical rotor that does become sufficiently distorted for it to acquire a dipole
moment is SiH4, which has a dipole moment of about 8
...
1 D; molecular dipole moments and their units are discussed in Section 18
...
The pure rotational
spectrum of SiH4 has been detected by using long path lengths (10 m) through highpressure (4 atm) samples
...
13
...
This picture is
the classical origin of the gross selection
rule for rotational transitions
...
2 Identifying rotationally active molecules

Of the molecules N2, CO2, OCS, H2O, CH2 =CH2, C6H6, only OCS and H2O are
polar, so only these two molecules have microwave spectra
...
3 Which of the molecules H2, NO, N2O, CH4 can have a pure rotational spectrum?
[NO, N2O]

The specific rotational selection rules are found by evaluating the transition dipole
moment between rotational states
...
2 that, for a
linear molecule, the transition moment vanishes unless the following conditions are
fulfilled:
∆J = ±1

∆MJ = 0, ±1

(13
...
The allowed change in J in each case arises from the conservation of angular momentum when a photon, a spin-1 particle, is emitted or absorbed
(Fig
...
18)
...
36)

where µ0 is the permanent electric dipole moment of the molecule
...

For symmetric rotors, an additional selection rule states that ∆K = 0
...
13
...
If the
molecule is rotating in the same sense as
the spin of the incoming photon, then J
increases by 1
...
Such a molecule cannot be accelerated into different states
of rotation around the figure axis by the absorption of radiation, so ∆K = 0
...


(13
...
38)

However, because the second term is typically very small compared with the first, the
appearance of the spectrum closely resembles that predicted from eqn 13
...

Example 13
...

Method We calculated the energy levels in Example 13
...
The 14NH3 molecule is a

polar symmetric rotor, so the selection rules ∆J = ±1 and ∆K = 0 apply
...
37
...
977 cm−1, we can draw up the
following table for the J + 1 ← J transitions
...
95
598
...
91
1197

2
59
...
82
2393


...


...
95 cm−1 (598
...

Self-test 13
...
2 for details)
...
888 cm−1 (26
...
13
...
The intensities reflect
the populations of the initial level in each
case and the strengths of the transition
dipole moments
...
37 is shown in Fig
...
19
...
and of separation 2B
...
Because the
masses of the atoms are known, it is a simple matter to deduce the bond length of a
diatomic molecule
...
This difficulty
can be overcome by using isotopically substituted molecules, such as ABC and A′BC;
then, by assuming that R(A-B) = R(A′-B), both A-B and B-C bond lengths can be
extracted from the two moments of inertia
...
10
...

The intensities of spectral lines increase with increasing J and pass through a maximum before tailing off as J becomes large
...

The Boltzmann distribution (Molecular interpretation 3
...
7 ROTATIONAL RAMAN SPECTRA

449

increases
...
The
value of J corresponding to a maximum of this expression is found by treating J as a
continuous variable, differentiating with respect to J, and then setting the result equal
to zero
...
24)
Jmax ≈

A kT D
C 2hcB F

1/2
1
−–
2

(13
...
2 cm−1) at room temperature,
kT ≈ 1000hcB, so Jmax ≈ 30
...
36) and on the population difference
between the two states involved in the transition (see Section 13
...
Hence the
value of J corresponding to the most intense line is not quite the same as the value of
J for the most highly populated level
...
7 Rotational Raman spectra
The gross selection rule for rotational Raman transitions is that the molecule must be
anisotropically polarizable
...
A formal derivation of this rule is given in Further information 13
...

The distortion of a molecule in an electric field is determined by its polarizability,
α (Section 18
...
More precisely, if the strength of the field is E, then the molecule
acquires an induced dipole moment of magnitude

µ = αE

(13
...
An atom is isotropically
polarizable
...
The polarizability of a spherical rotor is also isotropic
...
13
...

The electron distribution in H2, for example, is more distorted when the field is applied
parallel to the bond than when it is applied perpendicular to it, and we write α || > α⊥
...
This activity is one
reason for the importance of rotational Raman spectroscopy, for the technique can be
used to study many of the molecules that are inaccessible to microwave spectroscopy
...
This inactivity does not mean that such molecules are never
found in rotationally excited states
...

We show in Further information 13
...
41)
∆K = 0

The ∆J = 0 transitions do not lead to a shift of the scattered photon’s frequency in pure
rotational Raman spectroscopy, and contribute to the unshifted Rayleigh radiation
...
13
...
The polarizability may be
different when the field is applied
(a) parallel or (b) perpendicular to the
molecular axis (or, in general, in different
directions relative to the molecule); if that
is so, then the molecule has an anisotropic
polarizability
...
13
...
When the molecule
makes a transition with ∆J = +2, the scattered radiation leaves the molecule in a higher
rotational state, so the wavenumber of the incident radiation, initially #i, is decreased
...
42a)

The Stokes lines appear to low frequency of the incident radiation and at displacements 6B, 10B, 14B,
...
When the molecule makes a
transition with ∆J = −2, the scattered photon emerges with increased energy
...
42b)

The anti-Stokes lines occur at displacements of 6B, 10B, 14B,
...
;
J = 2 is the lowest state that can contribute under the selection rule ∆J = −2) to high
frequency of the incident radiation
...


Frequency
Example 13
...
The
form of a typical rotational Raman
spectrum is also shown
...

Fig
...
21

Predict the form of the rotational Raman spectrum of 14N2, for which B =
1
...
732 nm laser radiation
...
The Stokes and
anti-Stokes lines are given by eqn 13
...

Answer Because λ i = 336
...
2 cm−1, eqns 13
...
42b give the following line positions:
J
Stokes lines
#/cm−1
λ /nm
Anti-Stokes lines
#/cm−1
λ /nm

0

1

2

3

29 685
...
868

29 677
...
958

29 669
...
048

29 661
...
139

29 709
...
597

29 717
...
507

There will be a strong central line at 336
...
The spread of the entire spectrum is very small, so the
incident light must be highly monochromatic
...
5 Repeat the calculation for the rotational Raman spectrum of NH3

(B = 9
...


[Stokes lines at 29 637
...
4, 29 557
...
6 cm−1,
anti-Stokes lines at 29 757
...
0 cm−1
...
8 Nuclear statistics and rotational states
If eqn 13
...


13
...
and not 2 ← 0, 3 ← 1,
...
The form of the Pauli principle
given in Justification Section 10
...
In
particular, when a CO2 molecule rotates through 180°, two identical O nuclei are
interchanged, so the overall wavefunction of the molecule must remain unchanged
...
orbitals of atoms) shows that they change sign by (−1)J
under such a rotation (Fig
...
22)
...

The selective occupation of rotational states that stems from the Pauli principle is
termed nuclear statistics
...
However, the consequences are not always as
simple as for CO2 because there are complicating features when the nuclei have
nonzero spin: there may be several different relative nuclear spin orientations consistent with even values of J and a different number of spin orientations consistent with
odd values of J
...
13
...
In
general, for a homonuclear diatomic molecule with nuclei of spin I, the numbers of
ways of achieving states of odd and even J are in the ratio
Number of ways of achieving odd J 1 (I + 1)/I for half-integral spin nuclei
=2
Number of ways of achieving even J 3 I/(I + 1) for integral spin nuclei
(13
...
13
...
Wavefunctions with
J even do not change sign; those with J odd
do change sign
...
For N2, with I = 1, the ratio is 1:2
...
4 The effect of nuclear statistics on rotational spectra

Hydrogen nuclei are fermions, so the Pauli principle requires the overall wavefunction to change sign under particle interchange
...

For the overall wavefunction of the molecule to change sign when the spins are
parallel, the rotational wavefunction must change sign
...
In contrast, if the nuclear spins are paired, their wavefunction is
α(A)β(B) − α(B)β(A), which changes sign when α and β are exchanged in order to
bring about a simple A ↔ B interchange overall (Fig
...
24)
...
Hence, only even values of J are allowed if the nuclear spins are paired
...
10
...
10
...


Different relative nuclear spin orientations change into one another only very
slowly, so an H2 molecule with parallel nuclear spins remains distinct from one with

Frequency
Fig
...
23 The rotational Raman spectrum
of a diatomic molecule with two identical
1
spin-– nuclei shows an alternation in
2
intensity as a result of nuclear statistics
...


452

13 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA

B

A
(-1)J

B

Change
sign

A
Change
sign if
antiparallel

paired nuclear spins for long periods
...
The form with parallel nuclear spins is called
ortho-hydrogen and the form with paired nuclear spins is called para-hydrogen
...
13
...
This
energy is of some concern to manufacturers of liquid hydrogen, for the slow conversion of ortho-hydrogen into para-hydrogen (which can exist with J = 0) as nuclear
spins slowly realign releases rotational energy, which vaporizes the liquid
...
One such technique is to pass hydrogen over a metal surface: the
molecules adsorb on the surface as atoms, which then recombine in the lower energy
para-hydrogen form
...
13
...
The relabelling
can be thought of as occurring in two steps:
the first is a rotation of the molecule; the
second is the interchange of unlike spins
(represented by the different colours of the
nuclei)
...


Lowest
rotational
state of
ortho -hydrogen

J=1

In this section, we adopt the same strategy of finding expressions for the energy levels,
establishing the selection rules, and then discussing the form of the spectrum
...

13
...
13
...
11
...
In regions close to Re (at the minimum of the
curve) the potential energy can be approximated by a parabola, so we can write
1
V = – kx 2
2

J=0

When hydrogen is cooled, the
molecules with parallel nuclear spins
accumulate in their lowest available
rotational state, the one with J = 0
...
This is
a slow process under normal
circumstances, so energy is slowly released
...
13
...
44)

where k is the force constant of the bond
...

To see the connection between the shape of the molecular potential energy curve
and the value of k, note that we can expand the potential energy around its minimum
by using a Taylor series:
V(x) = V(0) +

Lowest
rotational
state of
para -hydrogen

x = R − Re

A dV D
A d2V D 2
1
x+ –
x +···
2
C dx F 0
C dx 2 F 0

(13
...
The first derivative of V is 0 at the minimum
...
For small displacements we can ignore all the higher terms, and so write
1
V(x) ≈ –
2

A d2V D 2
x
C dx 2 F 0

(13
...
47]

We see that if the potential energy curve is sharply curved close to its minimum, then
k will be large
...
13
...


13
...
5

Parabola

It is often useful to express a function
f(x) in the vicinity of x = a as an infinite
Taylor series of the form:

Potential energy, V

Molecular potential energy

k large

A df D
f(x) = f(a) + B E (x − a)
C dx F a
+
k small

Re

Fig
...
26 A molecular potential energy
curve can be approximated by a parabola
near the bottom of the well
...

At high excitation energies the parabolic
approximation is poor (the true potential is
less confining), and it is totally wrong near
the dissociation limit
...
13
...

A strongly confining well (one with steep
sides, a stiff bond) corresponds to high
values of k
...
48)

where meff is the effective mass:
meff =

m1m2

[13
...
1, but here
the separation of variables procedure is used to separate the relative motion of the
atoms from the motion of the molecule as a whole
...

The Schrödinger equation in eqn 13
...
24 for a particle of mass
m undergoing harmonic motion
...
4 to
write down the permitted vibrational energy levels:
1
Ev = (v + – )$ω
2

ω=

A k D
C meff F

1/2

v = 0, 1, 2,
...
50)

The vibrational terms of a molecule, the energies of its vibrational states expressed in
wavenumbers, are denoted G(v), with Ev = hcG(v), so
1
G(v) = (v + – )#
2

#=

1 A k D
2πc C meff F

+

1 A dnf D
B
E (x − a)n + · · ·
n! C dx n F a

where n = 0, 1, 2,
...
51)

The vibrational wavefunctions are the same as those discussed in Section 9
...

It is important to note that the vibrational terms depend on the effective mass of the
molecule, not directly on its total mass
...
The vibration would then be that of a light atom relative to that of a
stationary wall (this is approximately the case in HI, for example, where the I atom
barely moves and meff ≈ mH)
...

2
Illustration 13
...
The effective mass of 1H35Cl is 1
...
67 × 10−27 kg, so the Cl atom is acting like a brick wall)
...
63 × 1014 s−1, ν = 89
...
35 µm
...


13
...
13
...


The gross selection rule for a change in vibrational state brought about by absorption
or emission of radiation is that the electric dipole moment of the molecule must change
when the atoms are displaced relative to one another
...
The classical basis of this rule is that the molecule can shake the electromagnetic field into oscillation if its dipole changes as it vibrates, and vice versa
(Fig
...
28); its formal basis is given in Further information 13
...
Note that the
molecule need not have a permanent dipole: the rule requires only a change in dipole
moment, possibly from zero
...
g
...

Homonuclear diatomic molecules are infrared inactive because their dipole moments
remain zero however long the bond; heteronuclear diatomic molecules are infrared
active
...
4 Identifying infrared active molecules

Of the molecules N2, CO2, OCS, H2O, CH2=CH2, and C6H6, all except N2 possess
at least one vibrational mode that results in a change of dipole moment, so all
except N2 can show a vibrational absorption spectrum
...
For example, the symmetric stretch of
CO2, in which the O-C-O bonds stretch and contract symmetrically is inactive
because it leaves the dipole moment unchanged (at zero)
...
6 Which of the molecules H2, NO, N2O, and CH4 have infrared active

vibrations?

[NO, N2O, CH4]

The specific selection rule, which is obtained from an analysis of the expression for
the transition moment and the properties of integrals over harmonic oscillator wavefunctions (as shown in Further information 13
...
52)

Transitions for which ∆v = +1 correspond to absorption and those with ∆v = −1
correspond to emission
...
11 ANHARMONICITY

455

It follows from the specific selection rules that the wavenumbers of allowed vibrational transitions, which are denoted ∆Gv+ –– for the transition v + 1 ← v, are
1
2
∆Gv+ –– = G(v + 1) − G(v) = #
1
2

As we have seen, # lies in the infrared region of the electromagnetic spectrum, so
vibrational transitions absorb and generate infrared radiation
...
It follows from the Boltzmann distribution that
almost all the molecules will be in their vibrational ground states initially
...
As a result,
the spectrum is expected to consist of a single absorption line
...
may also
appear (in emission)
...
However, as we shall now show, the
breakdown of the harmonic approximation causes the transitions to lie at slightly
different frequencies, so several lines are observed
...
53)

D0 De

13
...
53 are only approximate because they are based on a
parabolic approximation to the actual potential energy curve
...
At high vibrational excitations the swing of the atoms (more precisely, the spread of the vibrational
wavefunction) allows the molecule to explore regions of the potential energy curve
where the parabolic approximation is poor and additional terms in the Taylor expansion of V (eqn 13
...
The motion then becomes anharmonic, in the
sense that the restoring force is no longer proportional to the displacement
...


0 Displacement
Fig
...
29 The dissociation energy of a
molecule, D0, differs from the depth of the
potential well, De, on account of the zeropoint energy of the vibrations of the bond
...
54)

a$
2

xe =

2meff ω

3

4

2
1

where De is the depth of the potential minimum (Fig
...
29)
...
54
allows for dissociation at large displacements
...
The Morse
potential energy is

=

#
4De

v=0
0
–1

0

1 2 3
a(R – Re)

4

5

(13
...
The number of vibrational
levels of a Morse oscillator is finite, and v = 0, 1, 2,
...
13
...
26)
...


Fig
...
30 The Morse potential energy curve
reproduces the general shape of a
molecular potential energy curve
...
The number of bound levels is
finite
...
13
...


1
1
1
G(v) = (v + –)# − (v + –)2 xe# + (v + –)3ye# + · · ·
2
2
2

where xe, ye,
...

When anharmonicities are present, the wavenumbers of transitions with ∆v = +1 are
∆Gv+ –– = # − 2(v + 1)xe# + · · ·
1
2

~
DG v+1/2 = n (v+1 ¬ v)

Linear
extrapolation

(13
...
57 shows that, when xe > 0, the transitions move to lower wavenumbers
as v increases
...
, even though these first,
second,
...
The first overtone,
for example, gives rise to an absorption at
G(v + 2) − G(v) = 2# − 2(2v + 3)xe# + · · ·

(13
...
Therefore, the selection rule is also only an
approximation
...

(b) The Birge–Sponer plot

When several vibrational transitions are detectable, a graphical technique called a
Birge–Sponer plot may be used to determine the dissociation energy, D0, of the bond
...
56)

∑ ∆Gv +
v

1
––
2

(13
...
13
...

The construction in Fig
...
32 shows that the area under the plot of ∆Gv + –– against
1
2
1
v + – is equal to the sum, and therefore to D0
...
Most actual plots differ
from the linear plot as shown in Fig
...
32, so the value of D0 obtained in this way is
usually an overestimate of the true value
...
5 Using a Birge–Sponer plot

True curve

--1 3 5
2 2 2

v+1
2

The area under a plot of transition
wavenumber against vibrational quantum
number is equal to the dissociation energy
of the molecule
...

Fig
...
32

The observed vibrational intervals of H+ lie at the following values for 1 ← 0, 2 ← 1,
2

...
Determine the dissociation energy of the molecule
...

Answer The points are plotted in Fig
...
33, and a linear extrapolation is shown as

a dotted line
...
Each square corresponds to 100 cm−1 (refer to the scale
of the vertical axis); hence the dissociation energy is 21 400 cm−1 (corresponding to
256 kJ mol−1)
...
12 VIBRATION–ROTATION SPECTRA
2500
P-branch

R-branch
Q-branch

2000

1

37

H Cl
35
H Cl

Absorbance

1

~
n /cm-1

1500
1000
500

2800

0
1 3
––
2 2

1
v+ –
2

39
2

–

Fig
...
33 The Birge–Sponer plot used in
Example 13
...
The area is obtained simply
by counting the squares beneath the line or
using the formula for the area of a right
triangle
...
13
...
The lines appear
in pairs because H35Cl and H37Cl both
contribute (their abundance ratio is 3:1)
...


Self-test 13
...
7 (which corresponds to the 1 ← 0 transition), 965
...
4, and
172 cm−1
...

[35
...
12 Vibration–rotation spectra
Each line of the high resolution vibrational spectrum of a gas-phase heteronuclear
diatomic molecule is found to consist of a large number of closely spaced components
(Fig
...
34)
...
The separation
between the components is less than 10 cm−1, which suggests that the structure is due
to rotational transitions accompanying the vibrational transition
...
Just as ice-skaters rotate
more rapidly when they bring their arms in, and more slowly when they throw them out,
so the molecular rotation is either accelerated or retarded by a vibrational transition
...
If the molecule also possesses angular
momentum about its axis, as in the case of the electronic orbital angular momentum
of the paramagnetic molecule NO, then the selection rules also allow ∆J = 0
...
60)

If we ignore anharmonicity and centrifugal distortion,
1
S(v,J) = (v + –)# + BJ(J + 1)
2

(13
...
with an intensity distribution reflecting both the populations of the rotational levels and the magnitude of the J −1 ← J
transition moment (Fig
...
35)
...

The intensities reflect the populations of
the initial rotational levels
...
This branch, when it is allowed (as in NO), appears at the vibrational
transition wavenumber
...
13
...
The R branch consists of lines with ∆J = +1:
#R(J) = S(v + 1, J + 1) − S(v, J) = # + 2B(J + 1)

J

P

~
n R(J )
~
n (J )

J-1

~
nP(J + 1)

~
n R(J - 1)

(13
...
As a result,
the Q branch (if it exists) consists of a series of closely spaced lines
...
62b)

(b) Combination differences

Fig
...
35

B1

(13
...

The separation between the lines in the P and R branches of a vibrational transition
gives the value of B
...
However, the latter is more precise
...
We
shall continue with the simple expression initially
...
The absorptions then fall into three groups called
branches of the spectrum
...
13
...


(13
...
This procedure is used widely in spectroscopy to extract information about a particular state
...

As can be seen from Fig
...
36, the transitions #R(J − 1) and #P(J + 1) have a common upper state, and hence can be anticipated to depend on B0
...
63 that
1
#R(J − 1) − #P(J + 1) = 4B0(J + –)
2

(13
...
(Any deviation from a straight line is a consequence of centrifugal distortion, so that effect can be investigated too
...
13 VIBRATIONAL RAMAN SPECTRA OF DIATOMIC MOLECULES
lower state, and hence their combination difference gives information about the upper
state:
1
#R(J) − #P(J) = 4B1(J + –)
2

9
8
7
6
5
4
3
2
1
0

v=1

(13
...
440 cm−1 and
B1 = 10
...


459

13
...
65)

#S(J) = #i − # − 6B − 4BJ
Note that, unlike in infrared spectroscopy, a Q branch is obtained for all linear
molecules
...
13
...

The information available from vibrational Raman spectra adds to that from
infrared spectroscopy because homonuclear diatomics can also be studied
...
2
...
2* Properties of diatomic molecules
#/cm−1
1

H2

1

4401

B/cm−1

k/(N m−1)

De /(kJ mol−1)

74

60
...
59

516

428

H127I

2308

161

6
...
244

323

239

H Cl

1

35

Cl2

* More values are given in the Data section
...
As homonuclear and heteronuclear diatomic
molecules swell and contract during a vibration, the control of the nuclei over the
electrons varies, and hence the molecular polarizability changes
...
The specific selection rule for
vibrational Raman transitions in the harmonic approximation is ∆v = ±1
...
2
...
The lines to low frequency, the Stokes lines, correspond to ∆v = +1
...
It follows
that anti-Stokes lines are usually weak because very few molecules are in an excited
vibrational state initially
...
13
...
The selection rules are ∆J = 0, ±2 (as in pure rotational Raman
spectroscopy), and give rise to the O branch (∆J = −2), the Q branch (∆J = 0), and the
S branch (∆J = +2):

O

Q

S

Frequency of
scattered radiation

Fig
...
37 The formation of O, Q, and S
branches in a vibration–rotation Raman
spectrum of a linear rotor
...
13
...


460

13 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA

The vibrations of polyatomic molecules

O

Q

S

There is only one mode of vibration for a diatomic molecule, the bond stretch
...
Nonetheless, we
shall see that infrared and Raman spectroscopy can be used to obtain information
about the structure of systems as large as animal and plant tissues (see Impact I13
...

13
...
We then see that we can choose combinations of these atomic displacements that give the simplest description of the vibrations
...
If the molecule is linear,
there are 3N − 5 independent vibrational modes
...
5 The number of vibrational modes
Fig
...
38 The structure of a vibrational line
in the vibrational Raman spectrum of
carbon monoxide, showing the O, Q, and S
branches
...

Each atom may change its location by varying one of its three coordinates (x, y,
and z), so the total number of displacements available is 3N
...
For example, three coordinates
are needed to specify the location of the centre of mass of the molecule, so three
of the 3N displacements correspond to the translational motion of the molecule
as a whole
...

Two angles are needed to specify the orientation of a linear molecule in space:
in effect, we need to give only the latitude and longitude of the direction in which
the molecular axis is pointing (Fig
...
39a)
...
13
...

Therefore, two (linear) or three (nonlinear) of the 3N − 3 internal displacements
are rotational
...
It follows that
the number of modes of vibration Nvib is 3N − 5 for linear molecules and 3N − 6 for
nonlinear molecules
...
5 Determining the number of vibrational modes

(b)
Fig
...
39 (a) The orientation of a linear
molecule requires the specification of two
angles
...


Water, H2O, is a nonlinear triatomic molecule, and has three modes of vibration
(and three modes of rotation); CO2 is a linear triatomic molecule, and has four
modes of vibration (and only two modes of rotation)
...


The next step is to find the best description of the modes
...
13
...
This illustration shows
the stretching of one bond (the mode ν L), the stretching of the other (ν R), and the
two perpendicular bending modes (ν2)
...
15 INFRARED ABSORPTION SPECTRA OF POLYATOMIC MOLECULES
disadvantage: when one CO bond vibration is excited, the motion of the C atom sets
the other CO bond in motion, so energy flows backwards and forwards between ν L
and ν R
...

The description of the vibrational motion is much simpler if linear combinations of
ν L and ν R are taken
...
13
...
In this mode, the C atom is buffeted simultaneously from each
side and the motion continues indefinitely
...
Both modes are independent in the sense that, if one is excited,
then it does not excite the other
...
The two other normal modes
are the bending modes ν2
...

The four normal modes of CO2, and the Nvib normal modes of polyatomics in
general, are the key to the description of molecular vibrations
...
66)

where #q is the wavenumber of mode q and depends on the force constant kq for the
mode and on the effective mass mq of the mode
...
For example, in the symmetric stretch of
CO2, the C atom is stationary, and the effective mass depends on the masses of only
the O atoms
...
The three normal modes of H2O are shown in
Fig
...
41: note that the predominantly bending mode (ν2) has a lower frequency than
the others, which are predominantly stretching modes
...
One point
that must be appreciated is that only in special cases (such as the CO2 molecule) are the
normal modes purely stretches or purely bends
...
Another point in this
connection is that heavy atoms generally move less than light atoms in normal modes
...
13
...
(a) The stretching
modes are not independent, and if one
C-O group is excited the other begins
to vibrate
...
(b) The
symmetric and antisymmetric stretches
are independent, and one can be excited
without affecting the other: they are
normal modes
...


n1 (3652 cm-1)

13
...
Deciding whether
this is so can sometimes be done by inspection
...
13
...
The antisymmetric stretch, however, changes the dipole moment
because the molecule becomes unsymmetrical as it vibrates, so this mode is infrared
active
...
Both
bending modes are infrared active: they are accompanied by a changing dipole perpendicular to the principal axis, so transitions involving them lead to a perpendicular
band in the spectrum
...


-

1
n2 (1595 cm )

-1
n3 (3756 cm )

The three normal modes of H2O
...


Fig
...
41

462

13 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA

Comment 13
...


Comment 13
...


Synoptic table 13
...


The active modes are subject to the specific selection rule ∆vq = ±1 in the harmonic
approximation, so the wavenumber of the fundamental transition (the ‘first harmonic’)
of each active mode is #q
...
The force field may also be estimated by using the semi-empirical, ab
initio, and DFT computational techniques described in Section 11
...
Superimposed
on the simple force field scheme are the complications arising from anharmonicities
and the effects of molecular rotation
...
In a liquid, for example, a molecule
may be able to rotate through only a few degrees before it is struck by another, so it
changes its rotational state frequently
...

The lifetimes of rotational states in liquids are very short, so in most cases the
rotational energies are ill-defined
...
19) of more than 1 cm−1 can easily
result
...

One very important application of infrared spectroscopy to condensed phase samples, and for which the blurring of the rotational structure by random collisions is a
welcome simplification, is to chemical analysis
...
The intensities of the vibrational bands that can be identified with the
motions of small groups are also transferable between molecules
...
3)
...
2 Global warming1

Solar energy strikes the top of the Earth’s atmosphere at a rate of 343 W m−2
...

The Earth–atmosphere system absorbs the remaining energy and re-emits it into space
as black-body radiation, with most of the intensity being carried by infrared radiation
in the range 200–2500 cm−1 (4–50 µm)
...

The trapping of infrared radiation by certain gases in the atmosphere is known as
the greenhouse effect, so called because it warms the Earth as if the planet were enclosed
in a huge greenhouse
...
The major constituents to the Earth’s atmosphere, O2
and N2, do not contribute to the greenhouse effect because homonuclear diatomic
molecules cannot absorb infrared radiation
...
H
...
, New York (2005)
...
2 IMPACT ON ENVIRONMENTAL SCIENCE: GLOBAL WARMING

H2O

Light intensity

greenhouse effect (Fig
...
42)
...
3–7
...
6–2
...
2–4
...

Increases in the levels of greenhouse gases, which also include methane, dinitrogen
oxide, ozone, and certain chlorofluorocarbons, as a result of human activity have the
potential to enhance the natural greenhouse effect, leading to significant warming of
the planet
...

The concentration of water vapour in the atmosphere has remained steady over
time, but concentrations of some other greenhouse gases are rising
...
The concentration of methane, CH4, has more
than doubled during this time and is now at its highest level for 160 000 years (160 ka;
a is the SI unit denoting 1 year)
...

Human activities are primarily responsible for the rising concentrations of atmospheric
CO2 and CH4
...
The additional methane comes mainly from the petroleum industry and from agriculture
...
5 K since the
late nineteenth century (Fig
...
43)
...
The Intergovernmental Panel on
Climate Change (IPCC) estimated in 1995 that, by the year 2100, the Earth will undergo
an increase in temperature of 3 K
...
To place a temperature rise of
3 K in perspective, it is useful to consider that the average temperature of the Earth
during the last ice age was only 6 K colder than at present
...
One example of a significant change in the

463

O3

CH4
H2O

7

...
0

16
...
13
...
The purple line is the
intensity of the radiation actually emitted
...


0
...
4
0
...
2
-0
...
6
-0
...
6

Fig
...
43 The average change in surface
temperature of the Earth from 1855
to 2002
...
5 m, which is sufficient to alter weather patterns and submerge currently coastal
ecosystems
...
Clearly, in order to reverse global warming trends, we need to develop alternatives to fossil fuels, such as hydrogen (which can be used in fuel cells, Impact I25
...

13
...
It is sometimes quite difficult to judge by inspection when this is so
...
The other modes of CO2 leave the polarizability unchanged, so
they are Raman inactive
...

I||
I^
Scattered
radiation

The definition of the planes used
for the specification of the depolarization
ratio, ρ, in Raman scattering
...
13
...
) Because it is often possible to judge intuitively if a
mode changes the molecular dipole moment, we can use this rule to identify modes
that are not Raman active
...
In general, it is necessary to use group theory to predict whether a mode is infrared or Raman active (Section 13
...

(a) Depolarization

The assignment of Raman lines to particular vibrational modes is aided by noting the
state of polarization of the scattered light
...
13
...
A photon is
emitted when the excited state returns to a
state close to the ground state
...
67]

To measure ρ, the intensity of a Raman line is measured with a polarizing filter (a
‘half-wave plate’) first parallel and then perpendicular to the polarization of the incident beam
...
13
...
A line is classified as depolarized if it has ρ close to or greater than 0
...
75
...
Vibrations that are not
totally symmetrical give rise to depolarized lines because the incident radiation can
give rise to radiation in the perpendicular direction too
...
13
...
The

technique is then called resonance Raman spectroscopy
...
Furthermore, because it is often the
case that only a few vibrational modes contribute to the more intense scattering, the
spectrum is greatly simplified
...
Examples include the
pigments β-carotene and chlorophyll, which capture solar energy during plant photosynthesis (see Impact I23
...
The resonance Raman spectra of Fig
...
46 show vibrational transitions from only the few pigment molecules that are bound to very large
proteins dissolved in an aqueous buffer solution
...

Comparison of the spectra in Figs
...
46a and 13
...

(c) Coherent anti-Stokes Raman spectroscopy

The intensity of Raman transitions may be enhanced by coherent anti-Stokes Raman
spectroscopy (CARS, Fig
...
47)
...
68)

Suppose that ν2 is varied until it matches any Stokes line from the sample, such as the
one with frequency ν1 − ∆ν ; then the coherent emission will have frequency

ν ′ = 2ν1 − (ν1 − ∆ν) = ν1 + ∆ν

465

Light intensity

13
...
13
...
(a) Laser
excitation of the sample at 407 nm shows
Raman bands due to both chlorophyll a
and β-carotene bound to the protein
because both pigments absorb light at this
wavelength
...
(Adapted
from D
...
Ghanotakis et al
...

Biophys
...
)

which is the frequency of the corresponding anti-Stokes line
...

An advantage of CARS is that it can be used to study Raman transitions in the presence of competing incoherent background radiation, and so can be used to observe
the Raman spectra of species in flames
...
48
...
69)

Dye
laser

n2
n1

Fig
...
47

n3
Sample

The experimental arrangement for the CARS experiment
...
13
...
The peaks correspond to
the Q branch of the vibration–rotation
spectrum of N2 gas
...
F
...
, J
...
Educ
...
)

466

13 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
IMPACT ON BIOCHEMISTRY

I13
...
13
...
Adapted from N
...
, Proc
...
Acad
...
USA 95, 4837
(1998)
...

The techniques of vibrational microscopy provide details of cellular events that cannot
be observed with traditional light or electron microscopy
...

The sample is then moved by very small increments along a plane perpendicular to
the direction of illumination and the process is repeated until vibrational spectra for
all sections the sample are obtained
...
Up to a point, the smaller
the area that is illuminated, the smaller the area from which a spectrum can be obtained
...
For this reason, lasers and synchrotron radiation (see
Further information 13
...

In a conventional light microscope, an image is constructed from a pattern of
diffracted light waves that emanate from the illuminated object
...
Ultimately, this diffraction limit prevents the study of samples that are much
smaller than the wavelength of light used as a probe
...
61λ /a, where λ is the wavelength of the incident beam of radiation and a is the numerical aperture of the objective lens, the lens
that collects light scattered by the object
...
Use of
the best equipment makes it possible to probe areas as small as 9 µm2 by vibrational
microscopy
...
49 shows the infrared spectra of a single mouse cell, living and dying
...
The dying cell shows an additional absorption at 1730 cm−1,
which is due to the ester carbonyl group from an unidentified compound
...

Vibrational microscopy has also been used in biomedical and pharmaceutical
laboratories
...

13
...
Each normal mode must

13
...


467

C 2,S4
z

sd

1

C3

Example 13
...

Method The first step in the procedure is to identify the symmetry species of the

irreducible representations spanned by all the 3N displacements of the atoms,
using the characters of the molecular point group
...
Next, subtract the
symmetry species of the translations
...
Finally, subtract the symmetry species of the rotations,
which are also given in the character table (and denoted there by R x, R y, or R z)
...

Fig
...
50

Answer There are 3 × 5 = 15 degrees of freedom, of which (3 × 5) − 6 = 9 are vibra-

tions
...
13
...
Under E, no displacement coordinates are changed, so
the character is 15
...
Under the C2 indicated, the z-displacement of the central atom is left
unchanged, whereas its x- and y-components both change sign
...
Under the S4 indicated, the z-displacement of the central atom is reversed, so χ(S4) = −1
...
The characters are therefore 15, 0, −1, −1, 3
...
5a), we find that this representation spans A1 + E + T1 + 3T2
...
Hence, the nine vibrations span
A1 + E + 2T2
...
13
...
We shall see that symmetry analysis gives a quick way of deciding which modes are active
...
8 Establish the symmetry species of the normal modes of H2O
...
This is easily done by checking the character table of the molecular point group
for the symmetry species of the irreducible representations spanned by x, y, and z, for
their species are also the symmetry species of the components of the electric dipole
moment
...


Justification 13
...
The
x-component is

Fig
...
51 Typical normal modes of
vibration of a tetrahedral molecule
...


468

13 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA

Ύ

µx,10 = −e ψ *xψ0 dτ
1

(13
...

2
The ground-state vibrational wavefunction is a Gaussian function of the form e−x ,
so it is symmetrical in x
...
Consequently, the excited state wavefunction
must have the same symmetry as the displacement x
...
7 Identifying infrared active modes

Which modes of CH4 are infrared active?
Method Refer to the Td character table to establish the symmetry species of x, y,
and z for this molecule, and then use the rule given above
...
We found in Example13
...
Therefore, only the T2 modes
are infrared active
...
The A1 mode, which is inactive, is the symmetrical ‘breathing’
mode of the molecule
...
9 Which of the normal modes of H2O are infrared active? [All three]

(b) Raman activity of normal modes

Group theory provides an explicit recipe for judging the Raman activity of a normal
mode
...
) listed in
the character table are noted (they transform in the same way as the polarizability),
and then we use the following rule:
If the symmetry species of a normal mode is the same as the symmetry species of a
quadratic form, then the mode is Raman active
...
6 Identifying Raman active modes

To decide which of the vibrations of CH4 are Raman active, refer to the Td character table
...
6 that the symmetry species of the
normal modes are A1 + E + 2T2
...
By combining this information with that in
Example 13
...
The
assignment of spectral features to the T2 modes is straightforward because these are
the only modes that are both infrared and Raman active
...
Measurement of the depolarization
ratio distinguishes between these modes because the A1 mode, being totally symmetric, is polarized and the E mode is depolarized
...
10 Which of the vibrational modes of H2O are Raman active?

[All three]

CHECKLIST OF KEY IDEAS

469

Checklist of key ideas
1
...

2
...

Stokes and anti-Stokes radiation are scattered radiation at a
lower and higher frequency, respectively, than the incident
radiation
...

3
...

4
...

5
...

6
...
Stimulated
emission is the radiation-driven transition from a high energy
state to one of lower energy
...

7
...
Spectral lines are affected by Doppler broadening,
lifetime broadening, and collisional deactivation of excited
states
...
A rigid rotor is a body that does not distort under the stress
of rotation
...
A symmetric rotor is a rigid rotor
with two equal moments of inertia
...
An
asymmetric rotor is a rigid rotor with three different
moments of inertia
...
The rotational terms of a spherical rotor are F(J) = BJ(J + 1)
with B = $/4πcI, J = 0, 1, 2,
...

10
...
In an oblate top, I|| > I⊥
...

11
...
, K = 0, ±1,
...

12
...
and are (2J + 1)-fold degenerate
...
The centrifugal distortion constant, DJ , is the empirical
constant in the expression F(J) = BJ(J + 1) − DJ J2(J + 1)2 that
takes into account centrifugal distortion, DJ ≈ 4B3/# 2
...
The gross rotational selection rule for microwave spectra is:
for a molecule to give a pure rotational spectrum, it must be

polar
...
The rotational wavenumbers in the
absence and presence of centrifugal distortion are given by
eqns 13
...
38, respectively
...
The gross selection rule for rotational Raman spectra is: the
molecule must be anisotropically polarizable
...

16
...

17
...

2

18
...
The specific selection
rule is: ∆v = ±1
...
Morse potential energy, eqn 13
...
55
...
A Birge–Sponer plot is a graphical procedure for determining
the dissociation energy of a bond
...
The P branch consists of vibration–rotation infrared
transitions with ∆J = −1; the Q branch has transitions with
∆J = 0; the R branch has transitions with ∆J = +1
...
The gross selection rule for vibrational Raman spectra is: the
polarizability must change as the molecule vibrates
...

23
...
The number of
normal modes is 3N − 6 (for nonlinear molecules) or 3N − 5
(linear molecules)
...
A symmetric stretch is a symmetrically stretching vibrational
mode
...

25
...

26
...
A
depolarized line is a line with ρ close to or greater than 0
...
A
polarized line is a line with ρ < 0
...

27
...

Coherent anti-Stokes Raman spectroscopy (CARS) is a

470

13 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Raman technique that relies on the use of two incident beams
of radiation
...
A normal mode is infrared active if its symmetry species is the
same as any of the symmetry species of x, y, or z, then the

mode is infrared active
...


Further reading
Articles and texts

L
...
Part I
...
J
...
Educ
...
Part II
...
J
...
Educ
...

H
...
Gremlich and B
...
Marcel Dekker, New York (2001)
...
B
...
C
...
C
...
Dover,
New York (1980)
...
M
...
Carrington, Rotational spectroscopy of diatomic
molecules
...

Sources of data and information

G
...
Spectra of
diatomic molecules
...


M
...
Jacox, Vibrational and electronic energy levels of polyatomic
transient molecules
...
3 (1994)
...
Herzberg, Molecular spectra and molecular structure II
...
Van Nostrand–Reinhold,
New York (1945)
...
P
...
Herzberg, Molecular spectra and molecular
structure IV
...
Van NostrandReinhold, New York (1979)
...
C
...
E
...
L
...
), Encyclopedia of
spectroscopy and spectrometry
...


B
...
VCH, New York
(1989)
...
P
...
J
...
Educ
...


G
...
Wiley, New York (2000)
...
1 Spectrometers

Here we provide additional detail on the principles of operation of
spectrometers, describing radiation sources, dispersing elements,
detectors, and Fourier transform techniques
...
Monochromatic sources
that can be tuned over a range of frequencies include the klystron and
the Gunn diode, which operate in the microwave range, and lasers,
which are discussed in Chapter 14
...
For far infrared radiation with 35 cm−1
< # < 200 cm−1, a typical source is a mercury arc inside a quartz
envelope, most of the radiation being generated by the hot quartz
...
The Nernst filament
consists of a ceramic filament of lanthanoid oxides that is heated to
temperatures ranging from 1200 to 2000 K
...


A quartz–tungsten–halogen lamp consists of a tungsten filament
that, when heated to about 3000 K, emits light in the range 320 nm <
λ < 2500 nm
...

A gas discharge lamp is a common source of ultraviolet and visible
radiation
...

At pressures exceeding 1 kPa, the output consists of sharp lines on
a broad, intense background due to emission from a mixture of ions
formed by the electrical discharge
...
In a deuterium lamp, excited D2 molecules dissociate into
electronically excited D atoms, which emit intense radiation between
200–400 nm
...
As electrons travelling in a circle are constantly accelerated by
the forces that constrain them to their path, they generate radiation
(Fig
...
52)
...
13
...
The electrons injected into the
ring from the linear accelerator and booster synchrotron are
accelerated to high speed in the main ring
...


including the infrared and X-rays
...

The dispersing element

The dispersing element in most absorption spectrometers operating
in the ultraviolet to near-infrared region of the spectrum is a
diffraction grating, which consists of a glass or ceramic plate into
which fine grooves have been cut and covered with a reflective
aluminium coating
...
71)

where n = 1, 2,
...
13
...
For given values of n and θ, larger differences in φ are
Scattered
beam
Incident
beam

q

Fig
...
54 A polychromatic beam is dispersed by a diffraction
grating into three component wavelengths λ1, λ 2, and λ 3
...
Rotating the diffraction grating as
shown by the double arrows allows λ 1 or λ 3 to reach the detector
...
Wide angular separation
results in wide spatial separation between wavelengths some distance
away from the grating, where a detector is placed
...
13
...
Turning the grating
around an axis perpendicular to the incident and diffracted beams
allows different wavelengths to be analysed; in this way, the
absorption spectrum is built up one narrow wavelength range at a
time
...
In a polychromator, there is
no slit and a broad range of wavelengths can be analysed
simultaneously by array detectors, such as those discussed below
...
13
...
When the two components
Movable
mirror, M1
Beam
splitter

Mirror, M2

f

Diffraction grating
Fig
...
53 One common dispersing element is a diffraction grating,
which separates wavelengths spatially as a result of the scattering of light
by fine grooves cut into a coated piece of glass
...
71)
...
13
...
The beam-splitting element
divides the incident beam into two beams with a path difference that
depends on the location of the mirror M1
...


13 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA

Intensity, I

Intensity, I

472

0

1

2
~
np

3

4

0

Fig
...
56 An interferogram produced as the path length p is changed
in the interferometer shown in Fig
...
55
...


Exploration Referring to Fig
...
55, the mirror M1 moves in
finite distance increments, so the path difference p is also
incremented in finite steps
...
That is, draw plots of I(p)/I0 against
#p, each with a different number of data points spanning the same
total distance path taken by the movable mirror M1
...
13
...


Exploration For a signal consisting of only a few
monochromatic beams, the integral in eqn 13
...
Use this
information to draw your own version of Fig
...
57
...

∞

Ύ {I(p) − –I(0)} cos 2π#p dp

I(#) = 4

1
2

(13
...
72)

Hence, the interferometer converts the presence of a particular
wavenumber component in the signal into a variation in intensity
of the radiation reaching the detector
...
13
...
The detected signal oscillates as the two
components alternately come into and out of phase as the path
difference is changed (Fig
...
56)
...
73 with p = 0
...
13
...


(13
...
This step is a standard technique of mathematics, and
is the ‘Fourier transformation’ step from which this form of
spectroscopy takes its name
...
13
...
13
...
This
spectrum is the Fourier transform of the interferogram, and is a
depiction of the contributing frequencies
...


FURTHER INFORMATION

473

A major advantage of the Fourier transform procedure is that all
the radiation emitted by the source is monitored continuously
...
As a result, Fourier transform
spectrometers have a higher sensitivity than conventional
spectrometers
...
In Raman microscopy, a CCD detector can be used in a
variation of the technique known as Raman imaging: a special optical
filter allows only one Stokes line to reach the two-dimensional
detector, which then contains a map of the distribution of the
intensity of that line in the illuminated area
...
Detectors
may consist of a single radiation sensing element or of several small
elements arranged in one or two-dimensional arrays
...
The most common
detectors found in commercial infrared spectrometers are sensitive
in the mid-infrared region
...
In a
pyroelectric device the capacitance is sensitive to temperature and
hence the presence of infrared radiation
...
2a) is used to generate an electrical signal proportional to
the intensity of light that strikes the detector
...
Each electron ejected from the
photocathode is accelerated by a potential difference to another
metallic surface, called the dynode, from which more electrons are
ejected
...
Depending on how the detector is constructed, a
PMT can produce up to 108 electrons per photon that strikes the
photocathode
...
At room
temperature, a small number of electrons on the surfaces of the
photocathode and dynodes have sufficient energy to be ejected even
in the dark
...
To minimize the dark
current, it is common to lower the temperature of the PMT
detector
...
In an avalanche photodiode,
the photo-generated electrons are accelerated through a very large
electrical potential difference
...

The charge-coupled device (CCD) is a two-dimensional array of
several million small photodiode detectors
...
CCD detectors are the
imaging devices in digital cameras, but are also used widely in

A number of factors determine a spectrometer’s resolution, the
smallest observable separation between two closely spaced spectral
bands
...
Furthermore, the
distance between the grating and the slit placed in front of a detector
must be long enough and the slit’s width must be narrow enough
so full advantage can be taken of the grating’s dispersing ability
(Fig
...
54)
...

The resolution of Fourier transform spectrometers is determined
by the maximum path length difference, pmax, of the interferometer:
∆# =

1
2pmax

(13
...
1 cm−1 requires a maximum path length
difference of 5 cm
...
3b)
...

Further information 13
...
The
starting point for our discussion is the total wavefunction for a
molecule, which can be written as

ψtotal = ψc
...
ψ
where ψc
...
describes the motion of the centre of mass and ψ
describes the internal motion of the molecule
...
7)
...
1
...
76)

and our task is to explore conditions for which this integral vanishes
or has a non-zero value
...
Equation 13
...

The transition dipole moment has three components, given by:

µfi,x = µ0͗YJ,f,MJ,f |sin θ cos φ |YJ,i,MJ,i͘
µfi,y = µ0͗YJ,f,MJ,f |sin θ sin φ |YJ,i,MJ,i͘
µfi,z = µ0͗YJ,f,MJ,f |cos θ |YJ,i,MJ,i͘
We see immediately that the molecule must have a permanent dipole
moment in order to have a microwave spectrum
...

For the specific selection rules we need to examine the conditions
for which the integrals do not vanish, and we must consider each
component
...
3)
...
8), this integral vanishes unless Jf − Ji = ±1 and MJ,f − MJ,i = 0
...
35
...
8

An important ‘triple integral’ involving the spherical harmonics is
π

ΎΎ
0

2π

Yl ″,m l″(θ,φ)*Yl ′,m′l(θ,φ)Yl,ml (θ,φ) sin θ dθ dφ = 0

0

unless ml″ = ml′ + ml and lines of length l″, l′, and l can form a triangle
...
It follows that

µfi,x ∝ ͗YJ,f,MJ,f |(Y1,1 + Y1,−1)|YJ,i,MJ,i͘
µfi,y ∝ ͗YJ,f,MJ,f |(Y1,1 − Y1,−1)|YJ,i,MJ,i͘
According to the properties of the spherical harmonics, these
integrals vanish unless Jf − Ji = ±1 and MJ,f − MJ,i = ±1
...
35
...
The incident electric field of a wave of electromagnetic
radiation of frequency ω i induces a molecular dipole moment that is
given by

µind = αE(t) = αE cos ω it

If the molecule is rotating at a circular frequency ω R, to an external
observer its polarizability is also time dependent (if it is anisotropic),
and we can write

α = α0 + ∆α cos 2ωRt
where ∆α = α|| − α⊥ and α ranges from α0 + ∆α to α0 − ∆α as the
molecule rotates
...
13
...
Substituting this
expression into the expression for the induced dipole moment gives

µind = (α0 + ∆α cos 2ω Rt) × (E cos ω it)
= α0E cos ωit + E∆α cos 2ω Rt cos ω it
1
= α0E cos ωit + –E∆α{cos(ω i + 2ω R)t + cos(ω i − 2ω R)t}
2
This calculation shows that the induced dipole has a component
oscillating at the incident frequency (which generates Rayleigh
radiation), and that it also has two components at ω i ± 2ω R, which
give rise to the shifted Raman lines
...
This is the gross selection rule for rotational Raman
spectroscopy
...
This is the origin of the specific
selection rule ∆J = ±2
...
First, we write the x-, y-, and z-components of the
induced dipole moment as

µind,x = µx sin θ cos φ

µind,y = µy sin θ sin φ

µind,z = µz cos θ

where µx, µy , and µz are the components of the electric dipole
moment of the molecule and the z-axis is coincident with the
molecular figure axis
...
40 and the preceding equations, it follows that

µind = α⊥Ex sin θ cos φ + α⊥Ey sin θ sin φ + α||E cos θ
= α⊥E sin2θ + α||E cos2θ
By using the spherical harmonic Y2,0(θ,φ) = (5/16π)1/2(3 cos2θ − 1)
and the relation sin2θ = 1 − cos2θ, it follows that:
1/2
5
1
A πD
1
2
4
µind = 2 –α || + –α⊥ + – B E ∆αY2,0(θ,φ) 6 E
3
3
3
C 5F
7
3
For a transition between two rotational states, we calculate the
integral ͗YJ,f,MJ,f | µind |YJ,i,MJ,i͘, which has two components:
1
2
(–α || + –α⊥)͗YJf , MJ,f |(YJi,MJ,i͘ and E∆α ͗YJf , MJ,f |Y2,0YJi,MJ,i͘
3
3

According to the properties of the spherical harmonics (Table 9
...
These are the gross and specific
selection rules for linear rotors
...
76 when the molecule does not change electronic or rotational

475

FURTHER INFORMATION

E
Electric dipole moment, m

a^

0

Linear
approximation

E

Actual

0
Extension, x

a ||

90°

Fig
...
60 The electric dipole moment of a heteronuclear diatomic
molecule varies as shown by the purple curve
...


E
(Fig
...
60)
...
It then follows that, with f ≠ i,
͗vf | N|vi͘ = µ0͗vf |vi͘ + δq͗vf |x|vi͘
The term proportional to µ0 is zero because the states with different
values of v are orthogonal
...
13
...
This is the origin of the ∆J = ±2 selection rule in
rotational Raman spectroscopy
...
For simplicity, we shall consider a one-dimensional oscillator
(like a diatomic molecule)
...
This is the gross selection rule for infrared
spectroscopy
...
We need to write out the wavefunctions in terms of the
Hermite polynomials given in Section 9
...
4 should be reviewed, for it gives further
details of the calculation)
...
28; note that in this context α is not the polarizability)
...
It follows that the transition dipole moment is zero
unless ∆v = ±1
...
9

An important integral involving Hermite polynomials is
∞

10
Hv′Hve dy = 2 1/2 v
3 π 2 v!
−∞

Ύ

−y 2

if v′ ≠ v
if v′ = v

Vibrational Raman spectra

The gross selection rule for vibrational Raman spectroscopy is based
on an analysis of the transition dipole moment ͗εvf | ¢|εvi͘, which is
written from eqn 13
...

For simplicity, we consider a one-dimensional harmonic oscillator
(like a diatomic molecule)
...
40 to write the transition dipole moment as

µ fi = ͗εvf | N|εvi͘ = ͗εvf | α |εvi͘E = ͗vf |α(x)|vi͘E
where α(x) = ͗ε | α |ε͘ is the polarizability of the molecule, which we
expect to be a function of small displacements x from the equilibrium
bond length of the molecule (Section 13
...
Next, we expand α(x) as
a Taylor series, so the transition dipole moment becomes
A dα D
µfi = vf α(0) + B E x + · · · vi E
C dx F 0
A dα D
= ͗vf |vi͘α(0)E + B E ͗vf |x|vi͘E + · · ·
C dx F 0
The term containing ͗vf |vi͘ vanishes for f ≠ i because the harmonic
oscillator wavefunctions are orthogonal
...
Therefore, the
polarizability of the molecule must change during the vibration; this
is the gross selection rule of Raman spectroscopy
...


Discussion questions
13
...

13
...

13
...


13
...
Do you expect excitation to high
rotational energy levels to change the equilibrium bond length of this
molecule? Justify your answer
...
5 Suppose that you wish to characterize the normal modes of benzene in
the gas phase
...
1a Calculate the ratio of the Einstein coefficients of spontaneous and
stimulated emission, A and B, for transitions with the following characteristics:
(a) 70
...


molecule vibrationally and (b) that one collision in 100 is effective
...


13
...
0 cm microwave radiation
...
Calculate the width
(in hertz) of rotational transitions in the molecule
...
2a What is the Doppler-shifted wavelength of a red (660 nm) traffic light
approached at 80 km h−1?

13
...
The equilibrium bond length is 115 pm
...
2b At what speed of approach would a red (660 nm) traffic light appear

green (520 nm)?

13
...
The equilibrium bond length is 112
...


13
...
10 cm−1, (b) 1
...


13
...
3b Estimate the lifetime of a state that gives rise to a line of width

(a) 100 MHZ, (b) 2
...


13
...
0 × 10 collisions in each
second
...
4b A molecule in a gas undergoes about 1
...


considered as a rigid rotator is 63
...
6b If the wavenumber of the J = 1 ← 0 rotational transition of 1H81Br

considered as a rigid rotator is 16
...
7a Given that the spacing of lines in the microwave spectrum of 27Al1H is
constant at 12
...
9815 u)
...
7b Given that the spacing of lines in the microwave spectrum of Cl F is
constant at 1
...
9688 u, m(19F) = 18
...

−1

127 35

13
...
1142 cm
...
9688 u, m(127I) = 126
...


477

13
...
7 cm−1) in
the ground and first excited vibrational states at (a) 298 K, (b) 500 K
...
17b Calculate the relative numbers of Br2 molecules (# = 321 cm−1) in the

second and first excited vibrational states at (a) 298 K, (b) 800 K
...
18a The hydrogen halides have the following fundamental vibrational

wavenumbers: 4141
...
9 cm−1 (H35Cl); 2649
...
5 cm−1 (H127I)
...


13
...
39021 cm−1
...
9949 u)
...
18b From the data in Exercise 13
...


13
...
19a For 16O2, ∆G values for the transitions v = 1 ← 0, 2 ← 0, and 3 ← 0

constants B( H C N) = 44
...
208 GHz
...
9b Determine the CO and CS bond lengths in OCS from the rotational
constants B(16O12C32S) = 6081
...
8 MHz
...
10a The wavenumber of the incident radiation in a Raman spectrometer is

20 487 cm−1
...
10b The wavenumber of the incident radiation in a Raman spectrometer is

20 623 cm−1
...
11a The rotational Raman spectrum of Cl2 (m( Cl) = 34
...
9752 cm−1 and a similar series of antiStokes lines
...

35

35

13
...
9984 u) shows a
−1

series of Stokes lines separated by 3
...
Calculate the bond length of the molecule
...
12a Which of the following molecules may show a pure rotational

microwave absorption spectrum: (a) H2, (b) HCl, (c) CH4, (d) CH3Cl,
(e) CH2Cl2?
13
...
13a Which of the following molecules may show a pure rotational Raman

spectrum: (a) H2, (b) HCl, (c) CH4, (d) CH3Cl?
13
...
14a An object of mass 1
...
22, 3088
...
21 cm−1
...

Assume ye to be zero
...
19b For 14N2, ∆G values for the transitions v = 1 ← 0, 2 ← 0, and 3 ← 0

are, respectively, 2345
...
40, and 6983
...
Calculate # and xe
...


13
...
86, 4367
...
04, 9826
...
8 cm−1
...

13
...
83, 3374
...
51, 7596
...
35 cm−1
...

13
...


What is the wavenumber of the line originating from the rotational state with
J = 2? Use the information in Table 13
...

13
...

What is the wavenumber of the line originating from the rotational state with
J = 2? Use the information in Table 13
...

13
...
22b Which of the following molecules may show infrared absorption
spectra: (a) CH3CH3, (b) CH4, (c) CH3Cl, (d) N2?
13
...
23b How many normal modes of vibration are there for the following
molecules: (a) C6H6, (b) C6H6CH3, (c) HC
...
CH
...
0 Hz
...


13
...
14b An object of mass 2
...
0 Hz
...


13
...
15a Calculate the percentage difference in the fundamental vibration
wavenumber of 23Na35Cl and 23Na37Cl on the assumption that their force
constants are the same
...
25a Consider the vibrational mode that corresponds to the uniform

Raman active when it is (a) angular, (b) linear?
active when it is (a) trigonal planar, (b) trigonal pyramidal?
expansion of the benzene ring
...
15b Calculate the percentage difference in the fundamental vibration

13
...
Is it (a) Raman, (b) infrared active?

wavenumber of 1H35Cl and 2H37Cl on the assumption that their force
constants are the same
...
26a The molecule CH2Cl2 belongs to the point group C2v
...
16a The wavenumber of the fundamental vibrational transition of 35Cl2 is

564
...
Calculate the force constant of the bond (m(35Cl) = 34
...

13
...
2 cm−1
...
9183 u, m(81Br) = 80
...

79

displacements of the atoms span 5A1 + 2A2 + 4B1 + 4B2
...
26b A carbon disulfide molecule belongs to the point group D∞h
...
What are
the symmetries of the normal modes of vibration?

478

13 MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA

Problems*
Numerical problems
13
...
Calculate the
total energy density in the visible region (700 nm to 400 nm) for a black body
at (a) 1500 K, a typical operating temperature for globars, (b) 2500 K, a typical
operating temperature for tungsten filament lamps, (c) 5800 K, the surface
temperature of the Sun
...
2 Calculate the Doppler width (as a fraction of the transition wavelength)
for any kind of transition in (a) HCl, (b) ICl at 25°C
...
1142 cm−1 and #(ICl)
= 384 cm−1 and additional information in Table 13
...


occurred at 84 421
...
25, and 96 476
...
Calculate
the rotational constant and bond length of CuBr
...
10 The microwave spectrum of 16O12CS (C
...
Townes, A
...
Holden, and

F
...
Merritt, Phys
...
74, 1113 (1948)) gave absorption lines (in GHz) as
follows:
1

2

32

J

24
...
488 82

34

23
...
651 64

60
...
462 40

Use the expressions for moments of inertia in Table 13
...

13
...


13
...
Find an
expression for the collision-limited lifetime of an excited state assuming that
every collision is effective
...
30 nm2) at 25°C and 1
...
To what value must the pressure of the
gas be reduced in order to ensure that collision broadening is less important
than Doppler broadening?

Tanaka, M
...
Tanaka (J
...
Phys
...
4 The rotational constant of NH3 is equivalent to 298 GHz
...
4 pm
and a bond angle of 106
...


13
...
81,
1
427
...
31, and 991
...
Show that they fit the expression (v + –)# −
2
1
(v + –)2x#, and deduce the force constant, zero-point energy, and dissociation
2
energy of the molecule
...
5 The rotational constant for CO is 1
...
6116 cm−1 in the

13
...
By how much does the
internuclear distance change as a result of this transition?
13
...
189 60 cm−1, B(C6D6) =
0
...
The moments of inertia of the molecules about any axis
perpendicular to the C6 axis were calculated from these data as I(C6H6) =
1
...
7845 × 10−45 kg m2
...

13
...
L
...
A
...
Chem
...
21, 1340 (1953)):
83
...
13, 124
...
37, 165
...
23, 206
...
86 cm−1
...
Predict the
positions of the corresponding lines in 2H35Cl
...
8 Is the bond length in HCl the same as that in DCl? The wavenumbers of
the J = 1 ← 0 rotational transitions for H35Cl and 2H35Cl are 20
...
7840 cm−1, respectively
...
007825 u and
2
...
The mass of 35Cl is 34
...
Based
on this information alone, can you conclude that the bond lengths are the
same or different in the two molecules?
13
...
This problem was overcome by flowing the halogen gas over
copper heated to 1100 K (E
...
Manson, F
...
de Lucia, and W
...
Chem
...
63, 2724 (1975))
...
7 223 379
...
2 240 584
...
5 257 793
...
Also, estimate the value of J
for the most highly populated rotational energy level at 298 K and at 100 K
...
It has one Raman active vibrational mode at 1400 cm−1,
two strong IR active modes at 2360 and 540 cm−1, and one weak IR mode at
3735 cm−1
...

13
...
Upon closer
examination at higher resolution, this band is observed to be split into two sets
of closely spaced peaks, one on each side of the centre of the spectrum at
2143
...
The separation between the peaks immediately to the right and
left of the centre is 7
...
Make the harmonic oscillator and rigid rotor
approximations and calculate from these data: (a) the vibrational
wavenumber of a CO molecule, (b) its molar zero-point vibrational energy,
(c) the force constant of the CO bond, (d) the rotational constant B, and (e)
the bond length of CO
...
15 The HCl molecule is quite well described by the Morse potential with
De = 5
...
7 cm−1, and x# = 52
...
Assuming that the
potential is unchanged on deuteration, predict the dissociation energies (D0)
of (a) HCl, (b) DCl
...
16 The Morse potential (eqn 13
...
When RbH was
studied, it was found that # = 936
...
15 cm−1
...
7 pm
...
Plot these curves on the same diagram for
J = 40, 80, and 100, and observe how the dissociation energy is affected by the

* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady
...
(Taking B = 3
...
)
13
...
Luo, G
...
McBane, G
...
F
...
R
...
Chem
...
98, 3564 (1993)) reported experimental observation of the He2 complex,
a species that had escaped detection for a long time
...
51 × 10−23
J, hcD0 about 2 × 10−26 J, and Re about 297 pm
...

(b) Such a weakly bound complex is hardly likely to be rigid
...

13
...
15, the semi-empirical, ab initio, and DFT

methods discussed in Chapter 11 can be used to estimate the force field of a
molecule
...
(a) Using
molecular modelling software3 and the computational method of your choice
(semi-empirical, ab initio, or DFT methods), calculate the fundamental
vibrational wavenumbers and visualize the vibrational normal modes of SO2
in the gas phase
...

Compare the calculated and experimental values
...
19 Consider the molecule CH3Cl
...
20 Suppose that three conformations are proposed for the nonlinear
molecule H2O2 (2, 3, and 4)
...
The Raman spectrum of
the same sample has bands at 877, 1408, 1435, and 3407 cm−1
...
(a) If
H2O2 were linear, how many normal modes of vibration would it have? (b)
Give the symmetry point group of each of the three proposed conformations
of nonlinear H2O2
...
Explain your reasoning
...
Begin the derivation by letting the
particles experience a restoring force of magnitude k(rc − re) that is countered
perfectly by a centrifugal force meffω 2rc, where ω is the angular velocity of the
rotating molecule
...
Finally, write an expression for the energy
of the rotating molecule, compare it with eqn 13
...
For help with the classical aspects of this derivation, see Appendix 3
...
23 In the group theoretical language developed in Chapter 12, a spherical
rotor is a molecule that belongs to a cubic or icosahedral point group, a
symmetric rotor is a molecule with at least a threefold axis of symmetry, and
an asymmetric rotor is a molecule without a threefold (or higher) axis
...
Classify each of the following molecules as a
spherical, symmetric, linear, or asymmetric rotor and justify your answers
with group theoretical arguments: (a) CH4, (b) CH3CN, (c) CO2, (d) CH3OH,
(e) benzene, (f) pyridine
...
24 Derive an expression for the value of J corresponding to the most
highly populated rotational energy level of a diatomic rotor at a temperature
T remembering that the degeneracy of each level is 2J + 1
...
1142 cm−1) at 25°C
...
Evaluate the expression for CH4
(for which B = 5
...

13
...
Nevertheless, the peaks of both branches can be identified
and have been used to measure the rotational constants of the molecules
(R
...
H
...
M
...
Chem
...
Faraday Soc
...
Show, from a knowledge of the value of J corresponding to the
intensity maximum, that the separation of the peaks of the O and S branches is
given by the Placzek–Teller relation δ# = (32BkT/hc)1/2
...
8

15
...
4

Calculate the bond lengths in the three molecules
...
26 Confirm that a Morse oscillator has a finite number of bound states, the
states with V < hcDe
...


Applications: to biology, environmental science, and
astrophysics
13
...
21 Show that the moment of inertia of a diatomic molecule composed
of atoms of masses mA and mB and bond length R is equal to meff R2, where
meff = mAmB/(mA + mB)
...
22 Derive eqn 13
...
Treat the bond as an elastic spring

some invertebrates
...
The Fe2O2
group of oxygenated haemerythrin is coloured and has an electronic
absorption band at 500 nm
...
(a) Why
is resonance Raman spectroscopy and not infrared spectroscopy the method
of choice for the study of the binding of O2 to haemerythrin? (b) Proof that
the 844 cm−1 band arises from a bound O2 species may be obtained by
conducting experiments on samples of haemerythrin that have been mixed
with 18O2, instead of 16O2
...
(c) The fundamental vibrational wavenumbers for the

3
The web site contains links to molecular modelling freeware and to other sites where you may perform molecular orbital calculations directly from your web
browser
...
Explain this trend in terms
−
2−
of the electronic structures of O2, O2 , and O 2
...
4
...
(e)
2
2
The resonance Raman spectrum of haemerythrin mixed with 16O18O has two
bands that can be attributed to the O-O stretching mode of bound oxygen
...


radiation
...
95 u) in a distant star was found
to be shifted from 654
...
5 nm and to be broadened to 61
...

What is the speed of recession and the surface temperature of the star?
13
...
Dalgarno, in Chemistry in the interstellar medium, Frontiers of

Astrophysics, E
...
Avrett (ed
...
Demonstrate through a calculation why
CH would not be as useful for this purpose as CN
...
190 cm−1
...
31‡ There is a gaseous interstellar cloud in the constellation Ophiuchus

5

6

8

+
13
...
28‡ A mixture of carbon dioxide (2
...
00 bar and

298 K in a gas cell of length 10 cm has an infrared absorption band centred at
2349 cm−1 with absorbances, A(#), described by:
A(#) =

a1
1 + a2(# − a3)2

that is illuminated from behind by the star ζ-Ophiuci
...
S
...
A
...
J
...
A strong absorption line in the
ultraviolet region at λ = 387
...
Unexpectedly, a second strong absorption line with 25 per
cent of the intensity of the first was found at a slightly longer wavelength (∆λ =
0
...
Calculate
the temperature of the CN molecules
...
3 K
...
If he had, his prize might have been for the discovery of the
cosmic microwave background radiation
...
932, a2 = 0
...
504, a5 = 0
...
(a) Draw graphs of A(#) and
ε(#)
...
The
CO bond length is 116
...
(c) Within what height, h, is basically all the
infrared emission from the Earth in this band absorbed by atmospheric
carbon dioxide? The mole fraction of CO2 in the atmosphere is 3
...
0065(h/m) below 10 km
...

13
...
27, we saw that Doppler shifts of atomic spectral lines
are used to estimate the speed of recession or approach of a star
...
3a, it is easy to see that Doppler broadening of an
atomic spectral line depends on the temperature of the star that emits the

the atmospheres of Jupiter, Saturn, and Uranus
...
29, with C replacing A,
when centrifugal distortion and other complications are ignored
...
6 cm−1,
B = 43
...
71 cm−1
...
The rather large discrepancy with the
experimental values is due to factors ignored in eqn 13
...
(b) Calculate an
+
approximate value of the H-H bond length in H3
...
B
...
Chem
...
96, 3702 (1991)) is 87
...
Use this result to calculate the
values of the rotational constants B and C
...
The molecular ion D3 was first produced by J
...
Shy, J
...

Farley, W
...
Lamb Jr, and W
...
Wing (Phys
...
Lett 45, 535 (1980)) who
observed the ν2(E′) band in the infrared
...
33 The space immediately surrounding stars, also called the circumstellar
space, is significantly warmer because stars are very intense black-body emitters
with temperatures of several thousand kelvin
...
What new features in the spectrum of
CO can be observed in gas ejected from and still near a star with temperatures
of about 1000 K, relative to gas in a cloud with temperature of about 10 K?
Explain how these features may be used to distinguish between circumstellar
and interstellar material on the basis of the rotational spectrum of CO
...
A common
theme throughout the chapter is that electronic transitions occur within a stationary nuclear
framework
...
A specially important example of stimulated
radiative decay is that responsible for the action of lasers, and we see how this stimulated
emission may be achieved and employed
...
1 The electronic spectra of

diatomic molecules
14
...
1 Impact on biochemistry:

The energies needed to change the electron distributions of molecules are of the order
of several electronvolts (1 eV is equivalent to about 8000 cm−1 or 100 kJ mol−1)
...
1)
...
Lasers have brought unprecedented
precision to spectroscopy, made Raman spectroscopy a widely useful technique, and
have made it possible to study chemical reactions on a femtosecond time scale
...


Vision
The fates of electronically
excited states
14
...
2 Impact on biochemistry:

Fluorescence microscopy
14
...

Immediately after an electronic transition they are subjected to different forces and

14
...
6 Applications of lasers in

chemistry
Checklist of key ideas
Further reading

Synoptic table 14
...
1: Examples of
practical lasers

λ /nm

Infrared

>1000

<3
...
3

170

Exercises

Yellow

580

5
...


ν /(1014 Hz)

E/(kJ mol−1)

Colour

6
...
The resulting vibrational structure of
electronic transitions can be resolved for gaseous samples, but in a liquid or solid the
lines usually merge together and result in a broad, almost featureless band (Fig
...
1)
...
The electronic spectra of gaseous samples are therefore very complicated but rich in information
...
1 The electronic spectra of diatomic molecules
We examine some general features of electronic transitions by using diatomic
molecules as examples
...
Then we use the symmetry designations to formulate selection rules
...

400

500

600
l /nm

700

The absorption spectrum of
chlorophyll in the visible region
...

Fig
...
1

Comment 14
...


(a) Term symbols

The term symbols of linear molecules (the analogues of the symbols 2P, etc
...

The value of | Λ | is denoted by the symbols Σ, Π, ∆,
...
, respectively
...
for atoms
...
A single electron
in a σ orbital has λ = 0: the orbital is cylindrically symmetrical and has no angular
nodes when viewed along the internuclear axis
...
The term symbol for H 2 is therefore Σ
...
The component of total spin angular momentum about the internuclear
+
axis is denoted Σ, where Σ = S, S − 1, S − 2,
...
For H2 , because there is only one
1
1
2
electron, S = s = – (Σ = ± –) and the term symbol is Σ, a doublet term
...
For H 2 , the parity of the only occupied
orbital is g (Section 11
...
If there
are several electrons, the overall parity is calculated by using
g×g=g

u×u=g

u×g=u

(14
...
The term symbol for the
ground state of any closed-shell homonuclear diatomic molecule is 1Σg because the
spin is zero (a singlet term in which all electrons paired), there is no orbital angular
momentum from a closed shell, and the overall parity is g
...
If there are two π electrons (as in the ground state of O2, with
4
2
configuration 1π u1π g ) then the term symbol may be either Σ (if the electrons are
travelling in opposite directions, which is the case if they occupy different π orbitals,
one with λ = +1 and the other with λ = −1) or ∆ (if they are travelling in the same
direction, which is the case if they occupy the same π orbital, both λ = +1, for
instance)
...
The overall parity of the molecule is
(closed shell) × g × g = g
The term symbol is therefore 3Σg
...
1 THE ELECTRONIC SPECTRA OF DIATOMIC MOLECULES

483

Table 14
...
74

7 882
...
55

1433

122
...
9
35 713

819

142

49 363

700

-

160

+

* Adapted from G
...
C
...
D
...


(closed shell) × (+) × (−) = (−)
−
and the full term symbol of the ground electronic state of O2 is 3Σg
...
For
example, the term symbol for the excited state of O2 formed by placing two electrons
in a 1πg,x (or in a 1πg,y) orbital is 1∆g because | Λ | = 2 (two electrons in the same π
orbital), the spin is zero (all electrons are paired), and the overall parity is (closed
shell) × g × g = g
...
2 and Fig
...
3 summarize the configurations, term
symbols, and energies of the ground and some excited states of O2
...
The selection rules concerned with changes in angular
momentum are
∆S = 0

O(3P) + O(3P)

+
Su

3

(b) Selection rules

∆Λ = 0, ±1

O(3P) + O(1D)

60
-

For Σ terms, a ± superscript denotes the behaviour of the molecular wavefunction
under reflection in a plane containing the nuclei (Fig
...
2)
...
14
...


~
Wavenumber, n /(103 cm 1)

2
2 1
† The configuration π u π 1π g π g should also give rise to a 3∆u term, but electronic transitions to or from
u
this state have not been observed
...
14
...


∆Ω = 0, ±1

where Ω = Λ + Σ is the quantum number for the component of total angular
momentum (orbital and spin) around the internuclear axis (Fig
...
4)
...
9), the origins of these rules are conservation of angular momentum during a transition and the fact that a photon has a spin of 1
...
First, for Σ
terms, only Σ+ ↔ Σ+ and Σ− ↔ Σ− transitions are allowed
...

That is, u → g and g → u transitions are allowed, but g → g and u → u transitions are
forbidden
...
14
...


484

14 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
Justification 14
...
The three components of the dipole moment operator transform like
x, y, and z, and are all u
...
Likewise, for a u → u
transition, the overall parity is u × u × u = u, so the transition dipole moment must
also vanish
...
The zcomponent of the dipole moment operator, the only component of µ responsible
for Σ ↔ Σ transitions, has (+) symmetry
...
Therefore, for Σ terms, Σ+ ↔ Σ− transitions are not allowed
...
14
...
However, a vibration of the
molecule can destroy the inversion
symmetry of the molecule and the g,u
classification no longer applies
...


A forbidden g → g transition can become allowed if the centre of symmetry is eliminated by an asymmetrical vibration, such as the one shown in Fig
...
5
...
A transition that derives its intensity from an asymmetrical vibration of a molecule is called a vibronic transition
...
1 Which of the following electronic transitions are allowed in O2:
−
+
−
−
+
−
3 −
Σg ↔ 1∆g , 3Σg ↔ 1Σg , 3Σg ↔ 3∆u, 3Σg ↔ 3Σu , 3Σg ↔ 3Σ−?
u

−
[3Σ− ↔ 3Σu ]
g

(c) Vibrational structure

To account for the vibrational structure in electronic spectra of molecules (Fig
...
6),
we apply the Franck–Condon principle:

e /(dm3 mol–1 cm–1)

400

Because the nuclei are so much more massive than the electrons, an electronic
transition takes place very much faster than the nuclei can respond
...
14
...
Shown here is the ultraviolet
spectrum of gaseous SO2 at 298 K
...


As a result of the transition, electron density is rapidly built up in new regions of the
molecule and removed from others
...
The stationary equilibrium separation
of the nuclei in the initial electronic state therefore becomes a stationary turning point
in the final electronic state (Fig
...
7)
...
Before the absorption, the molecule is in the lowest vibrational state of its lowest
electronic state (Fig
...
8); the most probable location of the nuclei is at their equilibrium separation, Re
...
When the transition occurs, the molecule is excited to the
state represented by the upper curve
...
14
...
The vertical line is the origin of the
expression vertical transition, which is used to denote an electronic transition that
occurs without change of nuclear geometry
...
The level marked * is the one in which the nuclei are most probably at

14
...
The dipole moment operator is a sum
over all nuclei and electrons in the molecule:

∑ ri + e∑ ZI RI

¢ = −e

Electronic
excited state

Electronic
ground state

Nuclei stationary
Internuclear separation
Fig
...
7 According to the Franck–Condon
principle, the most intense vibronic
transition is from the ground vibrational
state to the vibrational state lying vertically
above it
...


(14
...
The
intensity of the transition is proportional to the square modulus, | µ fi |2, of the magnitude
of the transition dipole moment (eqn 9
...
This overlap
integral is a measure of the match between the vibrational wavefunctions in the upper
and lower electronic states: S = 1 for a perfect match and S = 0 when there is no similarity
...
2 The Franck–Condon approximation

The overall state of the molecule consists of an electronic part, |ε͘, and a vibrational
part, |v͘
...
Therefore,

∑ ͗εf |ri | εi͗͘vf |vi͘ = µε ,ε S(vf ,vi)
f

i

(14
...
However, it is not the only accessible vibrational state because
several nearby states have an appreciable probability of the nuclei being at the separation Re
...

The vibrational structure of the spectrum depends on the relative horizontal position of the two potential energy curves, and a long vibrational progression, a lot of
vibrational structure, is stimulated if the upper potential energy curve is appreciably
displaced horizontally from the lower
...

The separation of the vibrational lines of an electronic absorption spectrum depends
on the vibrational energies of the upper electronic state
...
2)
...
4)

The matrix element µε f,εi is the electric-dipole transition moment arising from the
redistribution of electrons (and a measure of the ‘kick’ this redistribution gives to
the electromagnetic field, and vice versa for absorption)
...


Fig
...
8 In the quantum mechanical
version of the Franck–Condon principle,
the molecule undergoes a transition to the
upper vibrational state that most closely
resembles the vibrational wavefunction of
the vibrational ground state of the lower
electronic state
...


486

14 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
Because the transition intensity is proportional to the square of the magnitude of
the transition dipole moment, the intensity of an absorption is proportional to |S(vf ,vi)| 2,
which is known as the Franck–Condon factor for the transition
...

This conclusion is the basis of the illustration in Fig
...
8, where we see that the
vibrational wavefunction of the ground state has the greatest overlap with the vibrational states that have peaks at similar bond lengths in the upper electronic state
...
1 Calculating a Franck–Condon factor

Consider the transition from one electronic state to another, their bond lengths
being Re and R e and their force constants equal
...


1

Method We need to calculate S(0,0), the overlap integral of the two ground-state

S (0,0)2

vibrational wavefunctions, and then take its square
...
1)
...
5

ψ0 =

A 1 D 1/2 −x 2/2α 2
e
C α π1/2 F

ψ0 =
′

A 1 D 1/2 −x′2/2α 2
e
C α π1/2 F

where x = R − Re and x ′ = R − R′ , with α = ($2/mk)1/4 (Section 9
...
The overlap
e
integral is
∞

S(0,0) = ͗0|0͘ =
0

0

2
(Re – Re )/21/2 a
¢

4

The Franck–Condon factor for the
arrangement discussed in Example 14
...

Fig
...
9

Ύ

ψ 0ψ0dR =
′

−∞

1

απ

1/2

∞

Ύ

e−(x

+x′2)/2α 2

2

dx

−∞

1
We now write α z = R − – (Re + R′ ), and manipulate this expression into
e
2

S(0,0) =

1
π

1/2

∞

Ύ

−(Re−R′)2/4α 2
e

e

e−z dz
2

−∞
1/2

The value of the integral is π
...
14
...

For Br2, Re = 228 pm and there is an upper state with R′ = 266 pm
...
1 × 10−10, so the intensity of
the 0–0 transition is only 5
...

Self-test 14
...
14
...
2
...
14
...
Find the corresponding Franck–Condon factors when the centres are
coincident and W′ < W
...
2 THE ELECTRONIC SPECTRA OF POLYATOMIC MOLECULES
(e) Rotational structure

P

R

P

487
R

Just as in vibrational spectroscopy, where a vibrational transition is accompanied by
rotational excitation, so rotational transitions accompany the excitation of the vibrational excitation that accompanies electronic excitation
...
However, the principal difference is that electronic excitation can result in
much larger changes in bond length than vibrational excitation causes alone, and the
rotational branches have a more complex structure than in vibration–rotation spectra
...
The rotational energy levels of the initial and final states are
E(J) = hcBJ(J + 1)

E(J′) = hcB′J′(J′ + 1)

and the rotational transitions occur at the following positions relative to the vibrational transition of wavenumber # that they accompany:
P branch (∆J = −1):

#P(J) = # − (B′ + B)J + (B′ − B)J 2

Q branch (∆J = 0):

#Q(J) = # + (B′ − B)J(J + 1)

R branch (∆J = +1):

#R(J) = # + (B′ + B)(J + 1) + (B′ − B)(J + 1)

(14
...
5b)
2

(14
...
63
...
In this case the lines of the R branch converge with increasing J and when
J is such that |B′ − B|(J + 1) > B′ + B the lines start to appear at successively decreasing
wavenumbers
...
14
...
When the bond is
shorter in the excited state than in the ground state, B′ > B and B′ − B is positive
...
14
...


(a) B' < B

(b) B' > B

Fig
...
11 When the rotational constants of
a diatomic molecule differ significantly in
the initial and final states of an electronic
transition, the P and R branches show a
head
...


14
...
For example, when a carbonyl group (> C=O) is present, an absorption at
about 290 nm is normally observed, although its precise location depends on the
nature of the rest of the molecule
...
3)
...
In a d-metal complex,
where the immediate environment of the atom is no longer spherical, the d orbitals
Synoptic table 14
...


ε /(dm3 mol−1 cm−1)

7 000

Comment 14
...


488

14 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
z

y

x

dz

eg

H2O

2
2
-y

3+

t2g

Ti

Fig
...
12

1 [Ti(OH2)6]

3+

eg
DO

d

dx

2

3
- DO
5

dzx

dyz

dxy

The classification of d-orbitals in an octahedral environment
...
We show in the following Justification that in an octahedral complex, such as
[Ti(OH2)6]3+ (1), the five d orbitals of the central atom are split into two sets (2), a
triply degenerate set labelled t2g and a doubly degenerate set labelled eg
...


2
- DO
5

t2g

Justification 14
...
14
...


In an octahedral d-metal complex, six identical ions or molecules, the ligands, are at
the vertices of a regular octahedron, with the metal ion at its centre
...
Figure 14
...
An electron occupying an orbital
of the former group has a less favourable potential energy than when it occupies any
of the three orbitals of the other group, and so the d-orbitals split into the two sets
shown in (2) with an energy difference ∆O: a triply degenerate set comprising the
dxy, dyz, and dzx orbitals and labelled t2g, and a doubly degenerate set comprising the
dx 2−y 2 and dz 2 orbitals and labelled eg
...
Neither ∆O nor ∆T
is large, so transitions between the two sets of orbitals typically occur in the visible
region of the spectrum
...
As an example, the spectrum of [Ti(OH2)6]3+
near 20 000 cm−1 (500 nm) is shown in Fig
...
13, and can be ascribed to the promotion of its single d electron from a t2g orbital to an eg orbital
...
5 eV
...
2 THE ELECTRONIC SPECTRA OF POLYATOMIC MOLECULES

489

According to the Laporte rule (Section 14
...
However, d–d transitions become weakly allowed as vibronic transitions
as a result of coupling to asymmetrical vibrations such as that shown in Fig
...
5
...
In such charge-transfer
transitions the electron moves through a considerable distance, which means that the
transition dipole moment may be large and the absorption is correspondingly intense
...
In this oxoanion, the electron migrates from an orbital that
is largely confined to the O atom ligands to an orbital that is largely confined to the
Mn atom
...
The reverse migration, a metal-to-ligand charge-transfer transition
(MLCT), can also occur
...
The resulting excited state may have a very long
lifetime if the electron is extensively delocalized over several aromatic rings, and such
species can participate in photochemically induced redox reactions (see Section 23
...

The intensities of charge-transfer transitions are proportional to the square of the
transition dipole moment, in the usual way
...
However, because the
integrand in the transition dipole is proportional to the product of the initial and
final wavefunctions, it is zero unless the two wavefunctions have nonzero values in
the same region of space
...
17)
...


p*

p
A C=C double bond acts as a
chromophore
...

Fig
...
14

(c) π * ← π and π * ← n transitions

Absorption by a C=C double bond results in the excitation of a π electron into an
antibonding π * orbital (Fig
...
14)
...
Its energy is about
7 eV for an unconjugated double bond, which corresponds to an absorption at 180 nm
(in the ultraviolet)
...

An important example of an π * ← π transition is provided by the photochemical
mechanism of vision (Impact I14
...

The transition responsible for absorption in carbonyl compounds can be traced
to the lone pairs of electrons on the O atom
...
One
of these electrons may be excited into an empty π * orbital of the carbonyl group
(Fig
...
15), which gives rise to a π * ← n transition (an ‘n to π-star transition’)
...
Because π * ← n transitions in
carbonyls are symmetry forbidden, the absorptions are weak
...


Fig
...
15

490

14 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
z

(d) Circular dichroism spectroscopy

Magnetic field

x

y

Electric field

Propagation
Fig
...
16 Electromagnetic radiation
consists of a wave of electric and magnetic
fields perpendicular to the direction of
propagation (in this case the x-direction),
and mutually perpendicular to each other
...


Electronic spectra can reveal additional details of molecular structure when experiments are conducted with polarized light, electromagnetic radiation with electric and
magnetic fields that oscillate only in certain directions
...
14
...
The plane
of polarization may be oriented in any direction around the direction of propagation
(the x-direction in Fig
...
16), with the electric and magnetic fields perpendicular to
that direction (and perpendicular to each other)
...

When plane-polarized radiation passes through samples of certain kinds of matter,
the plane of polarization is rotated around the direction of propagation
...
3b)
...
In a circularly
polarized ray of light, the electric field describes a helical path as the wave travels
through space (Fig
...
17), and the rotation may be either clockwise or counterclockwise
...
In terms of the absorbances for the two components, AL and AR,
the circular dichroism of a sample of molar concentration [J] is reported as
∆ε = εL − εR =

(a)

R

AL − AR
[J]l

(14
...

Circular dichroism is a useful adjunct to visible and UV spectroscopy
...
14
...

Moreover, CD spectra can be used to assign the absolute configuration of complexes
by comparing the observed spectrum with the CD spectrum of a similar complex of
known handedness
...
In these
cases the spectrum of the polymer chain arises from the chirality of individual monomer units and, in addition, a contribution from the three-dimensional structure of
the polymer itself
...
14
...
The arrays
of arrows in these illustrations show the
view of the electric field when looking
toward the oncoming ray: (a) rightcircularly polarized, (b) left-circularly
polarized light
...
1 Vision

The eye is an exquisite photochemical organ that acts as a transducer, converting
radiant energy into electrical signals that travel along neurons
...

Indeed, a single type of protein, rhodopsin, is the primary receptor for light throughout the animal kingdom, which indicates that vision emerged very early in evolutionary history, no doubt because of its enormous value for survival
...
The ocular fluid is principally water, and passage of light
through this medium is largely responsible for the chromatic aberration of the eye,
the blurring of the image as a result of different frequencies being brought to slightly
different focuses
...
The pigments in this
region are the carotene-like xanthophylls (3), which absorb some of the blue light and

I14
...
They also protect the photoreceptor molecules from
too great a flux of potentially dangerous high energy photons
...

About 57 per cent of the photons that enter the eye reach the retina; the rest are
scattered or absorbed by the ocular fluid
...
A rhodopsin molecule consists of an opsin protein molecule to
which is attached a 11-cis-retinal molecule (4)
...
The attachment
is by the formation of a protonated Schiff’s base, utilizing the -CHO group of the
chromophore and the terminal NH2 group of the sidechain, a lysine residue from
opsin
...
The rhodopsin
molecules are situated in the membranes of special cells (the ‘rods’ and the ‘cones’)
that cover the retina
...
14
...

Immediately after the absorption of a photon, the 11-cis-retinal molecule undergoes photoisomerization into all-trans-retinal (5)
...
14
...
The left- and right-handed forms
of these isomers give identical absorption
spectra
...


492

14 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS

Fig
...
19 The structure of the rhodopsin
molecule, consisting of an opsin protein to
which is attached an 11-cis-retinal molecule
embedded in the space surrounded by the
helical regions
...
The process occurs because the π * ← π excitation of an electron loosens
one of the π-bonds (the one indicated by the arrow in 5), its torsional rigidity is lost,
and one part of the molecule swings round into its new position
...
The
straightened tail of all-trans-retinal results in the molecule taking up more space than
11-cis-retinal did, so the molecule presses against the coils of the opsin molecule that
surrounds it
...
25–0
...

In a sequence of biochemical events known as the biochemical cascade, metarhodopsin
II activates the protein transducin, which in turn activates a phosphodiesterase enzyme
that hydrolyses cyclic guanine monophosphate (cGMP) to GMP
...
2 for a discussion of transmembrane potentials)
...

The resting state of the rhodopsin molecule is restored by a series of nonradiative
chemical events powered by ATP
...
The free all-trans-retinol molecule now undergoes enzyme-catalysed isomerization into 11-cis-retinol followed by dehydrogenation to form 11-cis-retinal, which is then delivered back into an opsin molecule
...


Phosphorescence

Illumination on

Emission intensity

The fates of electronically excited states
A radiative decay process is a process in which a molecule discards its excitation
energy as a photon
...
This thermal degradation converts the excitation energy completely into
thermal motion of the environment (that is, to ‘heat’)
...

14
...
14
...


In fluorescence, spontaneous emission of radiation occurs within a few nanoseconds
after the exciting radiation is extinguished (Fig
...
20)
...
The difference suggests that fluorescence is a fast
conversion of absorbed radiation into re-emitted energy, and that phosphorescence
involves the storage of energy in a reservoir from which it slowly leaks
...
21 shows the sequence of steps involved in fluorescence
...
14
...
The excited molecule

14
...
14
...
After the initial absorption,
the upper vibrational states undergo
radiationless decay by giving up energy to
the surroundings
...


Wavelength
Fig
...
22 An absorption spectrum (a)
shows a vibrational structure characteristic
of the upper state
...


is subjected to collisions with the surrounding molecules, and as it gives up energy
nonradiatively it steps down the ladder of vibrational levels to the lowest vibrational
level of the electronically excited molecular state
...
It might therefore survive long enough to
undergo spontaneous emission, and emit the remaining excess energy as radiation
...
14
...

Provided they can be seen, the 0–0 absorption and fluorescence transitions can be
expected to be coincident
...
transitions that occur at progressively higher wavenumber and with intensities governed by
the Franck–Condon principle
...

downward transitions that hence occur with decreasing wavenumbers
...
Because the solvent molecules
do not have time to rearrange during the transition, the absorption occurs in an environment characteristic of the solvated ground state; however, the fluorescence occurs
in an environment characteristic of the solvated excited state (Fig
...
23)
...
The vivid oranges and greens of fluorescent
dyes are an everyday manifestation of this effect: they absorb in the ultraviolet and
blue, and fluoresce in the visible
...
14
...
On the left we see
that the absorption occurs with the solvent
(the ellipses) in the arrangement
characteristic of the ground electronic state
of the molecule (the sphere)
...


494

14 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS

Molecular potential energy

Intersystem
crossing

Singlet

electronic and vibrational quanta
...
We
examine the mechanisms of fluorescence quenching in Chapter 23
...
14
...
The important step is the
intersystem crossing, the switch from a
singlet state to a triplet state brought about
by spin–orbit coupling
...


S1

T1
IC
ISC

20

(p, p *)

15
10

nce
sce
e
ore
enc
sph oresc
Pho
Flu

25

471 nm

30 (p, p*)

~
n /(103 cm-1)

S0

ISC

5
0
Fig
...
25 A Jablonski diagram (here, for
naphthalene) is a simplified portrayal of
the relative positions of the electronic
energy levels of a molecule
...
The ground vibrational states of
each electronic state are correctly located
vertically but the other vibrational states
are shown only schematically
...
)

Figure 14
...
The first steps are the same as in fluorescence, but the presence of a triplet excited state plays a decisive role
...

Hence, if there is a mechanism for unpairing two electron spins (and achieving the conversion of ↑↓ to ↑↑), the molecule may undergo intersystem crossing, a nonradiative
transition between states of different multiplicity, and become a triplet state
...
9d) that singlet–triplet transitions may
occur in the presence of spin–orbit coupling, and the same is true in molecules
...

If an excited molecule crosses into a triplet state, it continues to deposit energy into
the surroundings
...
7)
...
1)
...
The molecules are therefore able to emit weakly, and the emission may continue
long after the original excited state was formed
...
It also suggests (as is confirmed experimentally) that phosphorescence should be most intense from solid samples: energy transfer is then less efficient and intersystem crossing has time to occur as the singlet excited
state steps slowly past the intersection point
...
The confirmation of
the mechanism is the experimental observation (using the sensitive magnetic resonance techniques described in Chapter 15) that the sample is paramagnetic while the
reservoir state, with its unpaired electron spins, is populated
...
14
...

IMPACT ON BIOCHEMISTRY

I14
...
Four notable exceptions are the amino acids tryptophan (λabs ≈ 280 nm and
λfluor ≈ 348 nm in water), tyrosine (λabs ≈ 274 nm and λfluor ≈ 303 nm in water), and
phenylalanine (λabs ≈ 257 nm and λfluor ≈ 282 nm in water), and the oxidized form of
the sequence serine–tyrosine–glycine (6) found in the green fluorescent protein (GFP)
of certain jellyfish
...


14
...
A common fluorescent label is GFP
...
However, great care is
required to eliminate fluorescent impurities from the sample
...
4 Dissociation and predissociation
Another fate for an electronically excited molecule is dissociation, the breaking of
bonds (Fig
...
26)
...

Absorption occurs in a continuous band above this dissociation limit because the
final state is an unquantized translational motion of the fragments
...

In some cases, the vibrational structure disappears but resumes at higher photon
energies
...
14
...
When a molecule is excited to a vibrational level,
its electrons may undergo a redistribution that results in it undergoing an internal
conversion, a radiationless conversion to another state of the same multiplicity
...
The state into which the molecule converts may be dissociative, so the
states near the intersection have a finite lifetime, and hence their energies are imprecisely defined
...
When the incoming photon brings enough energy to excite the molecule

Internuclear separation
Fig
...
26 When absorption occurs to
unbound states of the upper electronic
state, the molecule dissociates and the
absorption is a continuum
...


Dissociation
limit

Molecular potential energy

Molecular potential energy

Continuum

Continuum
Dissociation
limit

Continuum

Internuclear separation
Fig
...
27 When a dissociative state crosses a
bound state, as in the upper part of the
illustration, molecules excited to levels near
the crossing may dissociate
...


495

496

14 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
to a vibrational level high above the intersection, the internal conversion does not occur
(the nuclei are unlikely to have the same geometry)
...


Lasers
Lasers have transformed chemistry as much as they have transformed the everyday
world
...
In this section, we discuss the mechanisms of laser action, and then explore their applications in
chemistry
...
1, we discuss the modes of operation of a number of commonly available laser systems
...
5 General principles of laser action
I

Pump

Laser action

A

(a) Population inversion

X
Fig
...
28 The transitions involved in one
kind of three-level laser
...

The laser transition is the stimulated
emission A → X
...
In stimulated emission, an excited state is stimulated to emit a photon by
radiation of the same frequency; the more photons that are present, the greater the
probability of the emission
...


A¢
X
Fig
...
29 The transitions involved in a fourlevel laser
...


One requirement of laser action is the existence of a metastable excited state, an
excited state with a long enough lifetime for it to participate in stimulated emission
...
Because at thermal equilibrium the opposite is true, it is necessary to achieve a population inversion in which there are more molecules in the
upper state than in the lower
...
14
...
The molecule
is excited to an intermediate state I, which then gives up some of its energy nonradiatively and changes into a lower state A; the laser transition is the return of A to the
ground state X
...
In practice, I consists of many states, all of which can convert to the upper of the two laser states A
...
The pumping is often achieved
with an electric discharge through xenon or with the light of another laser
...

The disadvantage of this three-level arrangement is that it is difficult to achieve
population inversion, because so many ground-state molecules must be converted to
the excited state by the pumping action
...
14
...
Because A′ is unpopulated initially, any population
in A corresponds to a population inversion, and we can expect laser action if A is
sufficiently metastable
...


14
...
The cavity is essentially a region between two mirrors, which reflect the
light back and forth
...
As in the treatment of a particle in a box
(Section 9
...
7)

where n is an integer and L is the length of the cavity
...
In addition, not all wavelengths that can be sustained by the
cavity are amplified by the laser medium (many fall outside the range of frequencies
of the laser transitions), so only a few contribute to the laser radiation
...

Photons with the correct wavelength for the resonant modes of the cavity and the
correct frequency to stimulate the laser transition are highly amplified
...
It stimulates the
emission of another photon, which in turn stimulates more (Fig
...
30)
...
Some of this radiation can be withdrawn if one
of the mirrors is partially transmitting
...
Only photons that are travelling strictly parallel to the axis of
the cavity undergo more than a couple of reflections, so only they are amplified, all
others simply vanishing into the surroundings
...
It may also be polarized, with its electric vector in a
particular plane (or in some other state of polarization), by including a polarizing
filter into the cavity or by making use of polarized transitions in a solid medium
...
In spatial coherence the waves are in step across the cross-section of the beam
emerging from the cavity
...
The latter is normally expressed in terms of a coherence length, lC, the distance
over which the waves remain coherent, and is related to the range of wavelengths, ∆ λ
present in the beam:
lC =

(a)Thermal equilibrium

(14
...
When
many wavelengths are present, the waves get out of step in a short distance and the
coherence length is small
...

(c) Q-switching

A laser can generate radiation for as long as the population inversion is maintained
...
When overheating is a problem, the
laser can be operated only in pulses, perhaps of microsecond or millisecond duration,
so that the medium has a chance to cool or the lower state discard its population
...
14
...
(a) The
Boltzmann population of states (see
Molecular interpretation 3
...
(b) When the
initial state absorbs, the populations are
inverted (the atoms are pumped to the
excited state)
...

The radiation is coherent (phases in step)
...
One way of achieving
pulses is by Q-switching, the modification of the resonance characteristics of the
laser cavity
...

Example 14
...
10 J can generate radiation in 3
...
Assuming that the pulses are rectangular, calculate the peak power
output and the average power output of this laser
...
The
excited state is populated while the cavity
is nonresonant
...


Fig
...
31

duration of the interval, and is expressed in watts (1 W = 1 J s−1)
...
The average power output, Paverage, is the total energy
released by a large number of pulses divided by the duration of the time interval
over which the total energy was measured
...


Answer From the data,

Ppeak =
Circularly
polarized
radiation +
-

(a)

Pockels
cell
+
-

Laser
medium

0
...
0 × 10−9 s

= 3
...
The pulse repetition rate is 10 Hz, so ten
pulses are emitted by the laser for every second of operation
...
10 J × 10 s−1 = 1
...
0 W

Plane
polarized
radiation

The peak power is much higher than the average power because this laser emits
light for only 30 ns during each second of operation
...
3 Calculate the peak power and average power output of a laser with

a pulse energy of 2
...

[Ppeak = 67 MW, Paverage = 76 kW]
(b)

-

+

(c)
Fig
...
32 The principle of a Pockels cell
...
In this sequence, (a) the
plane polarized ray becomes circularly
polarized, (b) is reflected, and (c) emerges
from the Pockels cell with perpendicular
plane polarization
...


The aim of Q-switching is to achieve a healthy population inversion in the absence
of the resonant cavity, then to plunge the population-inverted medium into a cavity,
and hence to obtain a sudden pulse of radiation
...
14
...
One technique is to use
a Pockels cell, which is an electro-optical device based on the ability of some crystals,
such as those of potassium dihydrogenphosphate (KH2PO4), to convert plane-polarized
light to circularly polarized light when an electrical potential difference is applied
...
14
...
As
a result, the reflected light does not stimulate more emission
...
An alternative technique is to use a saturable absorber, typically a solution of a dye that loses its ability

14
...
The dye
then suddenly becomes transparent and the cavity becomes resonant
...


1ns

499

1ps

(d) Mode locking

The technique of mode locking can produce pulses of picosecond duration and less
...

The resonant modes differ in frequency by multiples of c/2L (as can be inferred from
eqn 14
...
Normally, these modes have random phases relative to each
other
...
Then interference occurs
to give a series of sharp peaks, and the energy of the laser is obtained in short bursts
(Fig
...
33)
...
In a laser with a cavity of
length 30 cm, the peaks are separated by 2 ns
...


Time
Fig
...
33 The output of a mode-locked
laser consists of a stream of very narrow
pulses separated by an interval equal to the
time it takes for light to make a round trip
inside the cavity
...
4 The origin of mode locking

The general expression for a (complex) wave of amplitude E 0 and frequency ω is
E 0eiω t
...
A wave formed by superimposing N modes with
n = 0, 1,
...
14
...
We see that it is a series of peaks with maxima
separated by t = 2L/c, the round-trip transit time of the light in the cavity, and that
the peaks become sharper as N is increased
...
The modulation can be pictured as the opening of a shutter in synchrony
with the round-trip travel time of the photons in the cavity, so only photons making
the journey in that time are amplified
...
The transducer sets up standing-wave vibrations in the prism and modulates the

0

1

3
2
ct /2L

4

5

Fig
...
34 The function derived in
Justification 14
...


500

14 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
Table 14
...
We also see in Section 20
...

14
...
4)
...
What follows is only an initial listing of applications of lasers to chemistry
...

(a) Multiphoton spectroscopy

The large number of photons in an incident beam generated by a laser gives rise to a
qualitatively different branch of spectroscopy, for the photon density is so high that
more than one photon may be absorbed by a single molecule and give rise to multiphoton processes
...
For example, in one-photon spectroscopy,
only g ↔ u transitions are observable; in two-photon spectroscopy, however, the overall outcome of absorbing two photons is a g → g or a u → u transition
...
An intense excitation beam increases the intensity of scattered radiation, so the use of laser sources
increases the sensitivity of Raman spectroscopy
...
The monochromaticity of laser radiation is also a great
advantage, for it makes possible the observation of scattered light that differs by only
fractions of reciprocal centimetres from the incident radiation
...
Monochromaticity

14
...
1) and resonance Raman spectroscopy (Section 13
...

The availability of nondivergent beams makes possible a qualitatively different kind
of spectroscopy
...
This configuration is employed in the technique called stimulated Raman spectroscopy
...
This multiple scattering results
in lines of frequency νi ± 2νM, νi ± 3νM, and so on, where νi is the frequency of the
incident radiation and νM the frequency of a molecular excitation
...
One consequence
of state-specificity for photochemistry is that the illumination of a sample may be
photochemically precise and hence efficient in stimulating a reaction, because its
frequency can be tuned exactly to an absorption
...
The rate of a reaction is generally increased by raising the temperature
because the energies of the various modes of motion of the molecule are enhanced
...
With a laser we can excite the kinetically significant mode, so rate enhancement is achieved most efficiently
...
Isotope separation is possible because two isotopomers, or species
that differ only in their isotopic composition, have slightly different energy levels and
hence slightly different absorption frequencies
...
Direct photoionization by the absorption of a single
photon does not distinguish between isotopomers because the upper level belongs to
a continuum; to distinguish isotopomers it is necessary to deal with discrete states
...
In the first step, a photon excites an atom
to a higher state; in the second step, a photon achieves photoionization from that state
(Fig
...
35)
...
6 µm CO2 laser radiation results in the formation of products at room temperature
without a catalyst
...

A related application is the study of state-to-state reaction dynamics, in which a
specific state of a reactant molecule is excited and we monitor not only the rate at
which it forms products but also the states in which they are produced
...


Isotopomer 1
Isotopomer 2

Fig
...
35 In one method of isotope
separation, one photon excites an
isotopomer to an excited state, and then a
second photon achieves photoionization
...


14 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
U atoms

235

+

U ions

Dye laser
+

-

Copper
vapour
laser
Fig
...
36 An experimental arrangement for
isotope separation
...


UV photon

Isotopomer 1
Isotopomer 2
IR photon
IR photon
Internuclear separation

Fig
...
37 Isotopomers may be separated by
making use of their selective absorption of
infrared photons followed by
photodissociation with an ultraviolet
photon
...
14
...


depends on the nuclear mass
...
An example of this procedure is the
photoionization of uranium vapour, in which the incident laser is tuned to excite 235U
but not 238U
...
14
...

This procedure is being used in the latest generation of uranium separation plants
...
The
key problem is to achieve both mass selectivity (which requires excitation to take place
between discrete states) and dissociation (which requires excitation to continuum
states)
...
14
...
An alternative procedure is to make use of
multiphoton absorption within the ground electronic state (Fig
...
38); the efficiency
of absorption of the first few photons depends on the match of their frequency to the
energy level separations, so it is sensitive to nuclear mass
...
The isotopomers 32SF6 and 34SF6 have been separated in this way
...
This procedure has
been employed successfully to separate isotopes of B, N, O, and, most efficiently, H
...
Once again, the initial absorption, which is isotope selective, opens the way to subsequent further absorption and the formation of a geometrical isomer that can be separated chemically
...

A different, more physical approach, that of photodeflection, is based on the recoil
that occurs when a photon is absorbed by an atom and the linear momentum of the
photon (which is equal to h/λ) is transferred to the atom
...
6 APPLICATIONS OF LASERS IN CHEMISTRY

503

its original path only if the absorption actually occurs, and the incident radiation can
be tuned to a particular isotope
...
For
instance, if a Ba atom absorbs about 50 photons of 550 nm light, it will be deflected
by only about 1 mm after a flight of 1 m
...
Q-switched lasers produce
nanosecond pulses, which are generally fast enough to study reactions with rates controlled by the speed with which reactants can move through a fluid medium
...
These
timescales are available from mode-locked lasers
...
It is also possible to study energy transfer, molecular rotations,
vibrations, and conversion from one mode of motion to another
...

Here, we describe some of the experimental techniques that employ pulsed lasers
...
14
...
1)
...
The rates of appearance and disappearance of the various species are determined by observing timedependent changes in the absorption spectrum of the sample during the course of the
reaction
...
Pulsed ‘white’ light can be
generated directly from the laser pulse by the phenomenon of continuum generation, in which focusing an ultrafast laser pulse on a vessel containing a liquid such as
water, carbon tetrachloride, CaF, or sapphire results in an outgoing beam with a wide
distribution of frequencies
...
For example, a difference in travel distance of ∆d = 3 mm
corresponds to a time delay ∆t = ∆d/c ≈10 ps between two beams, where c is the speed
of light
...
14
...

Variations of the arrangement in Fig
...
39 allow for the observation of fluorescence
decay kinetics of A* and time-resolved Raman spectra during the course of the reaction
...
In this case, continuum generation is not necessary
...
14
...
The time delay
between the pump and probe pulses may
be varied by moving the motorized stage in
the direction shown by the double arrow
...
Also in this case continuum generation is not necessary
...
1) or by stimulated Raman scattering of the laser pulse in a medium
such as H2(g) or CH4(g)
...
14
...
5 µm × 4
...
Each peak corresponds to a
single dye molecule
...
S
...
Acc
...
Res
...


There is great interest in the development of new experimental probes of very small
specimens
...
On
the other hand, techniques that can probe the structure, dynamics, and reactivity of
single molecules will be needed to advance research on nanometre-sized materials
(Impact I20
...

We saw in Impact I13
...
Fluorescence microscopy (Impact I14
...
3)
...

The bulk of the work done in the field of single-molecule spectroscopy is based on
fluorescence microscopy with laser excitation
...
Two techniques are commonly used to circumvent the diffraction limit
...
Second, special strategies are used to illuminate very small volumes
...
It is possible to construct fibres with tip diameters in the
range of 50 to 100 nm, which are indeed smaller than visible wavelengths
...
Figure 14
...
5 µm × 4
...
Each peak corresponds to a single dye molecule
...

This illumination scheme is limited by diffraction and, as a result, data from far-field
microscopy have less structural detail than data from NSOM
...

In the wide-field epifluorescence method, a two-dimensional array detector (Further information 13
...
14
...
If the fluorescing molecules are well separated in
the specimen, then it is possible to obtain a map of the distribution of fluorescent
molecules in the illuminated area
...
14
...

Though still a relatively new technique, single-molecule spectroscopy has already
been used to address important problems in chemistry and biology
...
Single-molecule methods allow a chemist to study the nature

CHECKLIST OF KEY IDEAS

CCD
From
laser
Optical
filter
Lens

(a)

(b)

of distributions of physical and chemical properties in an ensemble of molecules
...
Such studies
have shown that not every molecule in a sample has the same fluorescence lifetime,
probably because each molecule interacts with its immediate environment in a slightly
different way
...


505

Fig
...
41 (a) Layout of an epifluorescence
microscope
...
A CCD detector (see Further
information 13
...
(b) Observation of
fluorescence from single MHC proteins
that have been labelled with a fluorescent
marker and are bound to the surface of a
cell (the area shown has dimensions of
12 µm × 12 µm)
...
E
...


Checklist of key ideas
1
...


8
...


2
...


9
...


3
...

4
...

5
...

6
...

7
...


10
...

11
...

12
...
Qswitching is the modification of the resonance characteristics
of the laser cavity and, consequently, of the laser output
...
Mode locking is a technique for producing pulses of
picosecond duration and less by matching the phases of many
resonant cavity modes
...
Applications of lasers in chemistry include multiphoton
spectroscopy, Raman spectroscopy, precision-specified
transitions, isotope separation, time-resolved spectroscopy,
and single-molecule spectroscopy
...
Herzberg, Molecular spectra and molecular structure I
...
Krieger, Malabar (1989)
...
Herzberg, Molecular spectra and molecular structure III
...
Van
Nostrand–Reinhold, New York (1966)
...
R
...
Kluwer/Plenum,
New York (1999)
...
C
...
E
...
L
...
), Encyclopedia of
spectroscopy and spectrometry
...


G
...
H
...
Gallmann, N
...

Keller, Frontiers in ultrashort pulse generation: pushing the limits
in linear and nonlinear optics
...

Sources of data and information

M
...
Jacox, Vibrational and electronic energy levels of polyatomic
transient molecules
...
3 (1994)
...
R
...
), CRC Handbook of Chemistry and Physics, Sections 9
and 10, CRC Press, Boca Raton (2000)
...
R
...
K
...

Jones and Bartlett, Boston (1998)
...
1 Examples of practical lasers

Gas lasers

Figure 14
...

In practice, the requirements can be satisfied by using a variety of
different systems, and this section reviews some that are commonly
available
...
Noticeably absent from this discussion are
solid state lasers (including the ubiquitous diode lasers), which we
discuss in Chapter 20
...
The pumping is
normally achieved using a gas that is different from the gas
responsible for the laser emission itself
...
14
...
The initial
step is the excitation of an He atom to the metastable 1s12s1
Helium

Comment 14
...


1s12s1 1S
1
1 3
1s 2s S

Neon
3
...
2 mm

Population inversion

Slow relaxation

Efficient pumping

Metastable
state

632
...
The
pumping (of the neon) depends on a coincidental matching of the
helium and neon energy separations, so excited He atoms can
transfer their excess energy to Ne atoms during a collision
...
14
...
14
...


Ar

2

+

n3

CO2
n1

507

n2

Bend

+

Radiative,
72 nm

Ar

N2
3

Symmetric stretch

Ar +

454 nm
to
514 nm

Antisymmetric stretch

FURTHER INFORMATION

1
10
...
14
...


configuration by using an electric discharge (the collisions of
electrons and ions cause transitions that are not restricted by electricdipole selection rules)
...
Laser action generating 633 nm radiation (among
about 100 other lines) then occurs
...
14
...
The discharge results in the formation of Ar+ and Ar2+ ions
in excited states, which undergo a laser transition to a lower state
...
One of the design problems is to find
materials that can withstand this damaging residual radiation
...

The krypton-ion laser works similarly
...
Both lasers are
widely used in laser light shows (for this application argon and
krypton are often used simultaneously in the same cavity) as well as
laboratory sources of high-power radiation
...
14
...
2 µm and 10
...
6 µm, in the infrared) arises from vibrational
transitions
...
The vibrational levels happen to coincide with the ladder
of antisymmetric stretch (ν3, see Fig
...
40) energy levels of CO2,
which pick up the energy during a collision
...
This transition is allowed by anharmonicities in the
molecular potential energy
...
14
...
The
pumping also depends on the coincidental matching of energy
separations; in this case the vibrationally excited N2 molecules have
excess energies that correspond to a vibrational excitation of the
antisymmetric stretch of CO2
...


help remove energy from this state and maintain the population
inversion
...

Chemical and exciplex lasers

Chemical reactions may also be used to generate molecules with
nonequilibrium, inverted populations
...
The latter then attacks Cl2 to
produce vibrationally excited (‘hot’) HCl molecules
...

Such processes are remarkable examples of the direct conversion of
chemical energy into coherent electromagnetic radiation
...
This odd situation is
achieved by forming an exciplex, a combination of two atoms that
survives only in an excited state and which dissociates as soon as the
excitation energy has been discarded
...
An
electric discharge through the mixture produces excited Cl atoms,
which attach to the Xe atoms to give the exciplex XeCl*
...
As soon as XeCl* has discarded a

508

14 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS

Absorption

Exciplex, AB*

Laser
transition
Dissociative
state, AB

e /(105 dm3 mol -1 cm-1)

Molecular potential energy

1

Fluorescence

Laser
region

A-B distance

300 400 500
l /nm

600

The molecular potential energy curves for an exciplex
...
Because only the upper
state can exist, there is never any population in the lower state
...
14
...


photon, the atoms separate because the molecular potential energy
curve of the ground state is dissociative, and the ground state of the
exciplex cannot become populated (Fig
...
46)
...


the optical path need be through the dye
...
14
...


Fig
...
46

Comment 14
...
An exciplex has the form
AB* whereas an excimer, an excited dimer, is AA*
...
The tuning problem is
overcome by using a titanium sapphire laser (see above) or a dye
laser, which has broad spectral characteristics because the solvent
broadens the vibrational structure of the transitions into bands
...
A commonly used dye is rhodamine 6G in
methanol (Fig
...
47)
...
14
...
The dye is flowed
through the cell inside the laser cavity
...


Discussion questions
−
14
...


14
...

14
...
4 Explain how colour can arise from molecules
...
5 Describe the mechanism of fluorescence
...
6 What is the evidence for the correctness of the mechanism of
fluorescence?
14
...

14
...


EXERCISES

509

Exercises
Molar absorption coefficient, e

+
+
14
...
What is the total

spin and total orbital angular momentum of the molecule? Show that the term
symbol agrees with the electron configuration that would be predicted using
the building-up principle
...
1b One of the excited states of the C2 molecule has the valence electron
2
2
3
configuration 1σ g 1σ u1π u1π 1
...

g
14
...
Calculate the percentage
reduction in intensity when light of that wavelength passes through 2
...
25 mmol dm−3
...
2b The molar absorption coefficient of a substance dissolved in hexane is
known to be 327 dm3 mol−1 cm−1 at 300 nm
...
50 mm
of a solution of concentration 2
...

14
...
00 cm transmits 20
...
If the concentration of the component is
0
...
3b When light of wavelength 400 nm passes through 3
...
667 mmol dm−3, the
transmission is 65
...
Calculate the molar absorption coefficient of the
solute at this wavelength and express the answer in cm2 mol−1
...
4a The molar absorption coefficient of a solute at 540 nm is 286 dm3 mol−1

cm−1
...
5 mm cell containing a
solution of the solute, 46
...
What is the
concentration of the solution?
14
...
When light of that wavelength passes through a
7
...
3 per cent of the light is
absorbed
...
5a The absorption associated with a particular transition begins at
230 nm, peaks sharply at 260 nm, and ends at 290 nm
...
21 × 104 dm3 mol−1 cm−1
...
5)
...
5b The absorption associated with a certain transition begins at 199 nm,
peaks sharply at 220 nm, and ends at 275 nm
...
25 × 104 dm3 mol−1 cm−1
...
14
...
5)
...
6a The two compounds, 2,3-dimethyl-2-butene and 2,5-dimethyl-2,4hexadiene, are to be distinguished by their ultraviolet absorption spectra
...
Match the maxima to the compounds and justify the assignment
...
6b 1,3,5-hexatriene (a kind of ‘linear’ benzene) was converted into
benzene itself
...
7a The following data were obtained for the absorption by Br2 in carbon

tetrachloride using a 2
...
Calculate the molar absorption coefficient of
bromine at the wavelength employed:
[Br2]/(mol dm−3)

0
...
0050

0
...
0500

T/(per cent)

81
...
6

12
...
0 × 10 −3

emax

~
~ ~
e (n ) = emax{1 - k (n - nmax)2}

~
nmax
~
Wavenumber, n

Fig
...
49
14
...
50 mm cell
...
0010

0
...
0100

0
...
2

1
...
8a A 2
...
The concentration of the benzene was 0
...
Calculate the molar absorption coefficient of benzene at this
wavelength given that the transmission was 48 per cent
...
0-mm cell at the same wavelength?
14
...
50-mm cell was filled with a solution of a dye
...
5 mmol dm−3
...
What
will the transmittance be in a 4
...
9a A swimmer enters a gloomier world (in one sense) on diving to greater
depths
...
2 × 10−3 dm3 mol−1 cm−1, calculate the depth at which a
diver will experience (a) half the surface intensity of light, (b) one tenth the
surface intensity
...
9b Given that the maximum molar absorption coefficient of a molecule

containing a carbonyl group is 30 dm3 mol−1 cm−1 near 280 nm, calculate the
thickness of a sample that will result in (a) half the initial intensity of
radiation, (b) one-tenth the initial intensity
...
10a The electronic absorption bands of many molecules in solution have

half-widths at half-height of about 5000 cm−1
...

14
...
54 × 104 dm3 mol−1 cm−1
...

+
14
...
Explain
why the intensity of the v = 2 ← 0 transition is stronger than that of the 0 ← 0
transition
...
11b The photoionization of F2 by 21 eV photons produces F 2
...


510

14 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS

Problems*
Numerical problems
14
...
Given that the
separation in energy between the minima in their respective potential energy
curves of these two electronic states is 6
...

14
...
1
...

14
...
Energy levels for the
iodine and bromine atoms occur at 0, 7598; and 0, 3685 cm−1, respectively
...
What possibilities exist for
the numerical value of the dissociation energy of IBr? Decide which is the
correct possibility by calculating this quantity from ∆ f H 7(IBr, g) =
+40
...


14
...

Assume such a line shape, and show that A ≈ 1
...
The absorption spectrum of azoethane (CH3CH2N2)
between 24 000 cm−1 and 34 000 cm−1 is shown in Fig
...
50
...
Then integrate the absorption
band graphically
...
A more
sophisticated procedure would be to use mathematical software to fit a

10

polynomial to the absorption band (or a Gaussian), and then to integrate the
result analytically
...
5 A lot of information about the energy levels and wavefunctions of small
inorganic molecules can be obtained from their ultraviolet spectra
...
14
...
Estimate the integrated absorption
coefficient for the transition
...
What electronic states are accessible
from the A1 ground state of this Cv molecule by electric dipole transitions?
14
...
G
...
C
...
Chen, and W
...
Wentworth (J
...
Chem
...
These anions have
+
a 2 Σ u ground state and 2Πg, 2Πu, and 2 Σ + excited states
...

14
...
The wavenumbers (in
cm−1) of transitions from the ground state to the vibrational levels of the first
−
excited state (3 Σ u ) are 50 062
...
4, 51 369
...
6, 52 579
...
4, 53 679
...
0, 54 641
...
2, 55 460
...
1, 56 107
...
3, 56 570
...
What is the dissociation energy of the upper electronic
state? (Use a Birge–Sponer plot
...
(This excited atom is responsible for a
great deal of photochemical mischief in the atmosphere
...
Use this information to calculate the
dissociation energy of ground-state O2 from the Schumann–Runge data
...
8 The compound CH3CH=CHCHO has a strong absorption in the

ultraviolet at 46 950 cm−1 and a weak absorption at 30 000 cm−1
...

14
...
The hydrocarbon acts as an
electron donor and I2 as an electron acceptor
...
14
...
654

3
...
288

2
...
890

14
...
0 ns
...
If the ratio of the transition probabilities
for stimulated emission for the S* → S to the T → S transitions is 1
...
184

Investigate the hypothesis that there is a correlation between the energy of the
HOMO of the hydrocarbon (from which the electron comes in the
charge–transfer transition) and hνmax
...
1

-

e /(dm3 mol 1 cm 1)

hνmax /eV

8

14
...
(a) What is the molar
concentration of a solution in which there is, on average, one solute molecule
in 1
...
0 fL) of solution? (b) It is important to use pure solvents in
single-molecule spectroscopy because optical signals from fluorescent

* Problems denoted with the symbol ‡ were supplied by Charles Trapp and Carmen Giunta
...

1

PROBLEMS

B

~
n /cm

-1

25 000

20 000

A

15 000

14
...
A molecule of a fluorescent
dye commonly used to label biopolymers can withstand about 106 excitations
by photons before light-induced reactions destroy its π system and the
molecule no longer fluoresces
...
0 mW of 488 nm radiation from a
continuous-wave argon ion laser? You may assume that the dye has an
absorption spectrum that peaks at 488 nm and that every photon delivered by
the laser is absorbed by the molecule
...
Suppose
that water containing a fluorescent impurity of molar mass 100 g mol−1 is used
as solvent and that analysis indicates the presence of 0
...
0 kg of solvent
...
0 µm3 of solution? You may take the density of water as 1
...

Comment on the suitability of this solvent for single-molecule spectroscopy
experiments
...
14
...
13 Assume that the electronic states of the π electrons of a conjugated
molecule can be approximated by the wavefunctions of a particle in a onedimensional box, and that the dipole moment can be related to the
displacement along this length by µ = −ex
...
Hint
...
14 Use a group theoretical argument to decide which of the following
transitions are electric-dipole allowed: (a) the π * ← π transition in ethene,
(b) the π * ← n transition in a carbonyl group in a C2v environment
...
15 Suppose that you are a colour chemist and had been asked to intensify
the colour of a dye without changing the type of compound, and that the dye
in question was a polyene
...
16 One measure of the intensity of a transition of frequency ν is the

oscillator strength, f, which is defined as
f=

8π 2meν | µ fi |2
3he 2

Consider an electron in an atom to be oscillating harmonically in one
dimension (the three-dimensional version of this model was used in early
attempts to describe atomic structure)
...
1
...

3
14
...
16) of a chargetransfer transition modelled as the migration of an electron from an H1s
orbital on one atom to another H1s orbital on an atom a distance R away
...
Sketch the oscillator strength as a function of R using the
curve for S given in Fig
...
29
...
18 The line marked A in Fig
...
51 is the fluorescence spectrum of

benzophenone in solid solution in ethanol at low temperatures observed
when the sample is illuminated with 360 nm light
...
Now a component
of fluorescence from naphthalene can be detected
...

14
...
The
absorption spectrum rises sharply from zero to a maximum at 360 nm with a
trail of peaks of lessening intensity at 345 nm, 330 nm, and 305 nm
...

14
...
4)
...
Consider a sample of J that is illuminated with a
beam of intensity I0(#) at the wavenumber #
...

However, not all of the absorbed intensity is emitted and the intensity of
fluorescence depends on the fluorescence quantum yield, φf , the efficiency of
photon emission
...
Because of a Stokes shift of magnitude
∆#Stokes, fluorescence occurs at a wavenumber #f , with #f + ∆#Stokes = #
...

(a) Use the Beer–Lambert law to express Iabs(#) in terms of I0(#), [J], l, and
ε (#), the molar absorption coefficient of J at #
...

14
...
What atom multiplicities are permitted when (a) an O2 molecule, (b)
an N2 molecule dissociates into atoms?

Applications: to biochemistry, environmental science,
and astrophysics
14
...
Each protein molecule has two Fe2+ ions that are in
very close proximity and work together to bind one molecule of O2
...
Figure 14
...
What may be inferred from the spectrum?

512

14 MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
14
...
Compute
the absorbance of UV radiation at 300 nm expected for an ozone abundance
of 300 DU (a typical value) and 100 DU (a value reached during seasonal
Antarctic ozone depletions) given a molar absorption coefficient of
476 dm3 mol−1 cm−1
...
27‡ G
...
G
...
Horansky, and V
...
Phys
...
100,

Her(OH)8
~
Wavenumber, n
Fig
...
52

14
...
Of these photons, 30 per cent are absorbed or scattered by
the atmosphere and 25 per cent of the surviving photons are scattered by the
surface of the cornea of the eye
...
The area of the pupil at night is about 40 mm2 and the response time
of the eye is about 0
...
Of the photons passing through the pupil, about 43
per cent are absorbed in the ocular medium
...
1 s? For a continuation of this
story, see R
...
Rodieck, The first steps in seeing, Sinauer, Sunderland (1998)
...
24 Use molecule (7) as a model of the trans conformation of the
chromophore found in rhodopsin
...
(a)
Using molecular modelling software and the computational method of your
instructor’s choice, calculate the energy separation between the HOMO and
LUMO of (7)
...
(c) Based
on your results from parts (a) and (b), do you expect the experimental
frequency for the π * ← π visible absorption of the trans form of (7) to be
higher or lower than that for the 11-cis form of (7)?

11559 (1996)) examined the UV absorption spectrum of CH3I, a species of
interest in connection with stratospheric ozone chemistry
...
(a) Compute the integrated absorption
coefficient over a triangular lineshape in the range 31 250 to 34 483 cm−1 and a
maximal molar absorption coefficient of 150 dm3 mol−1 cm−1 at 31 250 cm−1
...
4 Torr and 373 K
exists as dimers
...
0 cm
...
Compute the absorbance expected at
31 250 cm−1 in a sample cell of length 12
...

14
...
M
...
-W
...
Wassermann, U
...
Flesch, and E
...
Phys
...
100, 10070 (1996)) measured
the ionization energies of Cl2O2 by photoelectron spectroscopy in which the
ionized fragments were detected using a mass spectrometer
...
05 eV and the enthalpy
of the dissociative ionization Cl2O2 → Cl + OClO+ + e− is 10
...
They used
this information to make some inferences about the structure of Cl2O2
...
The
Cl2O2 in the photoionization step is the lowest energy isomer, whatever its
structure may be, and its enthalpy of formation had previously been reported
as +133 kJ mol−1
...
Given ∆ f H 7(OClO+) = +1096 kJ mol−1 and ∆ f H 7(e−) = 0,
determine whether the Cl2O2 in the dissociative ionization is the same as that
in the photoionization
...
29‡ One of the principal methods for obtaining the electronic spectra of

14
...
This spectral range,
denoted UV-B, spans the wavelengths of about 290 nm to 320 nm
...
B
...
P
...
M
...
F
...
J
...
J
...
R
...
E
...
J
...

λ /nm

292
...
3 300
...
4 310
...
0 320
...
9 69
...
5

Compute the integrated absorption coefficient of ozone over the wavelength
range 290–320 nm
...
ε (#) can be fitted to an exponential function
quite well
...
Many radical spectra have been found in comets, including
that due to CN
...
Subsequently, their
fluorescence is excited by sunlight of longer wavelength
...
One
such study is that of the fluorescence spectrum of CN in the comet at large
heliocentric distances by R
...
Wagner and D
...
Schleicher (Science 275, 1918
(1997)), in which the authors determine the spatial distribution and rate of
production of CN in the coma
...
6 nm and the weaker (1–1) band with relative intensity 0
...
4 nm
...
3 and
421
...
From these data, calculate the energy of the excited
S1 state relative to the ground S0 state, the vibrational wavenumbers and the
difference in the vibrational wavenumbers of the two states, and the relative
populations of the v = 0 and v = 1 vibrational levels of the S1 state
...
Only
eight rotational levels of the S1 state are thought to be populated
...
The chapter begins with an account of conventional nuclear
magnetic resonance, which shows how the resonance frequency of a magnetic nucleus is
affected by its electronic environment and the presence of magnetic nuclei in its vicinity
...
The experimental techniques for electron paramagnetic resonance resemble those
used in the early days of NMR
...


15
The effect of magnetic fields on
electrons and nuclei

The energies of electrons in
magnetic fields
15
...
3 Magnetic resonance
spectroscopy
15
...
6 The fine structure
15
...
4

When two pendulums share a slightly flexible support and one is set in motion, the
other is forced into oscillation by the motion of the common axle
...
The energy transfer occurs most efficiently when
the frequencies of the two pendulums are identical
...

Resonance is the basis of a number of everyday phenomena, including the response
of radios to the weak oscillations of the electromagnetic field generated by a distant
transmitter
...


15
...
8
15
...
1
15
...
11
15
...
13

The Stern–Gerlach experiment (Section 9
...

It turns out that many nuclei also possess spin angular momentum
...
First, we establish how the energies of electrons and nuclei depend on the
strength of an external field
...


15
...
1 The energies of electrons in magnetic fields
Classically, the energy of a magnetic moment m in a magnetic field ; is equal to the
scalar product

Electron paramagnetic
resonance
15
...
16 Hyperfine structure
I15
...
1: Fourier
transformation of the FID curve
Discussion questions
Exercises
Problems

514

15 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE

Comment 15
...
The (non-SI) unit
gauss, G, is also occasionally used:
1 T = 104 G
...
2

The scalar product (or ‘dot product’)
of two vectors a and b is given by
a·b = ab cos q, where a and b are the
magnitudes of a and b, respectively,
and q is the angle between them
...
1)

Quantum mechanically, we write the hamiltonian as
@ = −¢·;

(15
...
8 that the magnetic moment is
proportional to the angular momentum
...
3)

where ™ is the orbital angular momentum operator and

γe = −

e

(15
...
26)
...
3 becomes
Nz = γe Zz

@ = −γeB0Zz = −NzB0

and

(15
...
5b)

where the Bohr magneton, µ B, is

µ B = −γe $ =

e$
2me

= 9
...
6]

The Bohr magneton is often regarded as the fundamental quantum of magnetic
moment
...
8), is also proportional to its spin angular momentum
...
3, the spin magnetic moment and hamiltonian operators are,
respectively,
¢ = g eγ e £

and

@ = −g eγ e;·£

(15
...
002 319
...
For a magnetic field B 0 in the z-direction,
Nz = g eγ e Sz

and

@ = −g eγ eB 0 Sz

(15
...
8b)

1
with ms = ± –
...
When a field is present, the degeneracy is removed: the state with ms = + –
2
1
1
1
moves up in energy by – g e µ BB 0 and the state with ms = − – moves down by – g e µ BB 0
...
2 THE ENERGIES OF NUCLEI IN MAGNETIC FIELDS

515

z

Table 15
...
)

even

odd

1 3 5
half-integer (–, – , – ,
...
)
2 2 2

* The spin of a nucleus may be different if it is in an excited state; throughout this chapter we deal only with the
ground state of nuclei
...
15
...
9)

2π

Equation 15
...
For a field of 1 T, the Larmor frequency is 30 GHz
...
2 The energies of nuclei in magnetic fields
The spin quantum number, I, of a nucleus is a fixed characteristic property of a
nucleus and is either an integer or a half-integer (Table 15
...
A nucleus with spin
quantum number I has the following properties:
1
...

2
...
, −I
...
If I > 0, a magnetic moment with a constant magnitude and an orientation that
is determined by the value of mI
...
A proton has I = –
2
14
and its spin may adopt either of two orientations; a N nucleus has I = 1 and its spin
may adopt any of three orientations; both 12C and 16O have I = 0 and hence zero magnetic moment
...
3:
¢ = γÎ

and

H = −γ ;·Î

(15
...
2)
...
10b)

As for electrons, the nuclear spin may be pictured as precessing around the direction
of the applied field at a rate proportional to the applied field
...
9, with γe replaced by γ ) of about
40 MHz
...

Fig
...
1

516

15 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
Synoptic table 15
...
826

−18
...
165

1

99
...
586

26
...
577

2

0
...
857

4
...
536

13

1
...
405

6
...
705

14

99
...
404

1
...
076

1

n
H
H
C
N

* More values are given in the Data section
...
051 × 10−27 J T −1

[15
...
Nuclear g-factors vary between −6 and +6 (see
Table 15
...

For the remainder of this chapter we shall assume that γ is positive, as is the case for
the majority of nuclei
...
The nuclear magneton is about 2000 times smaller
than the Bohr magneton, so nuclear magnetic moments—and consequently the
energies of interaction with magnetic fields—are about 2000 times weaker than the
electron spin magnetic moment
...
15
...
Note that the β state is lower in
energy than the α state (because the
magnetogyric ratio of an electron is
negative)
...


15
...
From eqn 15
...
12a)

If the sample is exposed to radiation of frequency ν, the energy separations come into
resonance with the radiation when the frequency satisfies the resonance condition
(Fig
...
2):
hν = g e µ BB 0

(15
...
Electron
paramagnetic resonance (EPR), or electron spin resonance (ESR), is the study of
molecules and ions containing unpaired electrons by observing the magnetic fields at
which they come into resonance with monochromatic radiation
...
3 T (the value used in most commercial EPR spectrometers) correspond
to resonance with an electromagnetic field of frequency 10 GHz (1010 Hz) and wavelength 3 cm
...


15
...
13a)

and resonant absorption occurs when the resonance condition (Fig
...
3)
hν = γ $B 0

(15
...
Because γ $B 0 /h is the Larmor frequency of the nucleus, this resonance
occurs when the frequency of the electromagnetic field matches the Larmor frequency
(ν = νL )
...
Larmor frequencies of
nuclei at the fields normally employed (about 12 T) typically lie in the radiofrequency
region of the electromagnetic spectrum (close to 500 MHz), so NMR is a radiofrequency technique
...
As well as protons, which are the most common nuclei
1
studied by NMR, spin- – nuclei include 13C, 19F, and 31P
...
15
...
Resonance occurs when the energy
separation of the levels matches the energy
of the photons in the electromagnetic field
...
However, they have proved invaluable in chemistry, for they reveal so much
structural information
...

15
...
In simple
instruments, the magnetic field is provided by a permanent magnet
...
15
...
The sample is placed in the cylindrically wound magnet
...
However, sample spinning can lead to irreproducible results, and is often avoided
...

The intensity of an NMR transition depends on a number of factors
...
14a)

where
Nα − Nβ ≈

Superconducting
magnet

Transmitter

Nγ $B0
2kT

(15
...
It follows that decreasing the temperature increases the intensity by increasing the population difference
...

We shall also see (Section 15
...
15
...
The link from the
transmitter to the detector indicates that
the high frequency of the transmitter is
subtracted from the high frequency
received signal to give a low frequency
signal for processing
...
We also conclude that absorptions of nuclei with large magnetogyric ratios (1H, for instance) are
more intense than those with small magnetogyric ratios (13C, for instance)
Justification 15
...
2, we know
that the rate of absorption of electromagnetic radiation is proportional to the
population of the lower energy state (Nα in the case of a proton NMR transition)
and the rate of stimulated emission is proportional to the population of the upper
state (Nβ)
...
Therefore, the net rate of absorption is proportional to the difference in populations, and we can write
Rate of absorption ∝ Nα − Nβ
The intensity of absorption, the rate at which energy is absorbed, is proportional to
the product of the rate of absorption (the rate at which photons are absorbed) and
the energy of each photon, and the latter is proportional to the frequency ν of the
incident radiation (through E = hν)
...
14a
...
1) to write the ratio of populations as
Nβ
Nα
Comment 15
...
If x << 1,
2
−x
then e ≈ 1 − x
...
The expansion of the exponential term is appropriate for
∆E < kT, a condition usually met for nuclear spins
...
14b
...
5 The chemical shift
Nuclear magnetic moments interact with the local magnetic field
...
This additional field is proportional to the
applied field, and it is conventional to write
δB = −σ B 0

[15
...
The ability of the applied field to induce an
electronic current in the molecule, and hence affect the strength of the resulting local
magnetic field experienced by the nucleus, depends on the details of the electronic structure near the magnetic nucleus of interest, so nuclei in different chemical groups have

15
...
The calculation of reliable values of the shielding constant
is very difficult, but trends in it are quite well understood and we concentrate on them
...
16)

the nuclear Larmor frequency is

νL =

γ B loc
2π

= (1 − σ )

γ B0

(15
...
Hence, different
nuclei, even of the same element, come into resonance at different frequencies
...
18]

The standard for protons is the proton resonance in tetramethylsilane (Si(CH3)4,
commonly referred to as TMS), which bristles with protons and dissolves without
reaction in many liquids
...
For 13C, the
reference frequency is the 13C resonance in TMS; for 31P it is the 31P resonance in
85 per cent H3PO4(aq)
...

Illustration 15
...
18,

ν − ν ° = ν °δ × 10−6
A nucleus with δ = 1
...
00 × 10−6 = 500 Hz
In a spectrometer operating at 100 MHz, the shift relative to the reference would be
only 100 Hz
...
This
practice is unnecessary
...
17 into eqn 15
...
19)

As the shielding, σ, gets smaller, δ increases
...
Some typical chemical shifts are given in
Fig
...
5
...
15
...


CH3CH2OH

CH3CH2OH

CH3CH2OH

300

-

C-X in ArX
R-C ºN

R-COOH
R=C=R
R3C+

200

100

0

d

very different ranges of chemical shifts
...

By convention, NMR spectra are plotted with δ increasing from right to left
...
In the original continuous wave (CW) spectrometers, in which the
radiofrequency was held constant and the magnetic field varied (a ‘field sweep experiment’), the spectrum was displayed with the applied magnetic field increasing from
left to right: a nucleus with a small chemical shift experiences a relatively low local
magnetic field, so it needs a higher applied magnetic field to bring it into resonance
with the radiofrequency field
...

(b) Resonance of different groups of nuclei

4
...
6

1
...

The bold letters denote the protons giving
rise to the resonance peak, and the step-like
curve is the integrated signal
...
15
...
15
...
The CH3 protons form one group of nuclei with δ ≈ 1
...
Finally, the OH proton is in another environment, and has a chemical shift of δ ≈ 4
...
It reduces the electron density of the distant methyl protons least, and
those nuclei are least deshielded
...
5 THE CHEMICAL SHIFT
The relative intensities of the signals (the areas under the absorption lines) can be
used to help distinguish which group of lines corresponds to which chemical group
...
Spectrometers can integrate the absorption automatically (as indicated
in Fig
...
6)
...
Counting
the number of magnetic nuclei as well as noting their chemical shifts helps to identify
a compound present in a sample
...
Nevertheless, considerable success has been achieved with the calculation for diatomic molecules and small
molecules such as H2O and CH4 and even large molecules, such as proteins, are within
the scope of some types of calculation
...

The empirical approach supposes that the observed shielding constant is the sum of
three contributions:

σ = σ (local) + σ (neighbour) + σ (solvent)

(15
...
The neighbouring group contribution,
σ (neighbour), is the contribution from the groups of atoms that form the rest of the
molecule
...

(d) The local contribution

It is convenient to regard the local contribution to the shielding constant as the sum
of a diamagnetic contribution, σd, and a paramagnetic contribution, σp:

σ (local) = σd + σp

(15
...
A paramagnetic contribution to σ (local) reinforces the applied
magnetic field and deshields the nucleus in question
...
The
total local contribution is positive if the diamagnetic contribution dominates, and is
negative if the paramagnetic contribution dominates
...

The circulation generates a magnetic field that opposes the applied field and hence
shields the nucleus
...
22)

where µ 0 is the vacuum permeability (a fundamental constant, see inside the front
cover) and r is the electron–nucleus distance
...
2 Calculating the diamagnetic contribution to the chemical shift of a
proton

To calculate σd for the proton in a free H atom, we need to calculate the expectation value of 1/r for a hydrogen 1s orbital
...
1,
and a useful integral is given in Example 8
...
Because dτ = r 2dr sin θ dθdφ, we
can write
1
r

=

Ύ

ψ *ψ
r

dτ =

1
πa 3
0

2π

Ύ Ύ
0

∞

π

dφ

sinθ dθ

0

Ύ

0

re−2r/a0dr =

4
a3
0

∞

Ύ re

−2r/a0

0

dr =

1
a0

Therefore,

σd =

e2µ 0
12πmea 0

With the values of the fundamental constants inside the front cover, this expression evaluates to 1
...

The diamagnetic contribution is the only contribution in closed-shell free atoms
...
Thus, it is the only contribution to the local shielding from inner cores of atoms, for cores remain spherical even though the atom may
be a component of a molecule and its valence electron distribution highly distorted
...
It follows that the shielding is decreased if the
electron density on the atom is reduced by the influence of an electronegative atom
nearby
...
15
...
That is, as the electronegativity increases, δ decreases
...
It is zero in free atoms and around the axes of linear
molecules (such as ethyne, HC
...
We
can expect large paramagnetic contributions from small atoms in molecules with lowlying excited states
...


Chemical shift relative to CH4, d

1
CH3CH2X
2

3
CH3CH2X
4

I Br

Cl

F

2
...
5
2
4
Electronegativity of halogen
Fig
...
7 The variation of chemical shielding
with electronegativity
...

However, to emphasize that chemical shifts
are subtle phenomena, notice that the
trend for the methyl protons is opposite to
that expected
...


(e) Neighbouring group contributions

The neighbouring group contribution arises from the currents induced in nearby
groups of atoms
...
The applied field generates currents in the electron distribution of X and gives rise to an induced magnetic moment proportional to the
applied field; the constant of proportionality is the magnetic susceptibility, χ (chi), of
the group X
...

First, the strength of the additional magnetic field the proton experiences is inversely
proportional to the cube of the distance r between H and X
...
We assume that the magnetic susceptibility of X has two components, χ|| and χ⊥, which are parallel and perpendicular to the
axis of symmetry of X, respectively
...
5 THE CHEMICAL SHIFT

q

523

r

c||

B
B

c|

3

2

1

vector connecting X to H (1, where X is represented by the ellipse and H is represented
by the circle)
...
Some of the time the H-X axis will be perpendicular to the applied field
and then only χ⊥ will contribute to the induced magnetic moment that shields X from
the applied field
...

When the applied field is parallel to the H-X axis, only χ|| contributes to the induced
magnetic moment at X
...
We conclude that,
as the molecule tumbles and the H-X axis takes all possible angles with respect to the
applied field, the effects of anisotropic magnetic susceptibility do not average to zero
because χ|| ≠ χ⊥
...
1 For a tumbling H-X molecule, show that when θ = 90°: (a) contributions from the χ⊥ component lead to shielding of H, or σ (neighbour) > 0,
and (b) contributions from the χ|| component lead to deshielding of H, or
σ (neighbour) < 0
...

[Draw diagrams similar to 2 and 3 where the χ⊥ component is parallel to the
H-X axis and then analyse the problem as above
...
23)

where χ|| and χ⊥ are both negative for a diamagnetic group X
...
23 shows
that the neighbouring group contribution may be positive or negative according to
the relative magnitudes of the two magnetic susceptibilities and the relative orientation of the nucleus with respect to X
...
7° < θ
< 125
...
15
...

A special case of a neighbouring group effect is found in aromatic compounds
...
15
...


p

524

15 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
B

ability of the field to induce a ring current, a circulation of electrons around the ring,
when it is applied perpendicular to the molecular plane
...
15
...


Ring
current

(f) The solvent contribution
Magnetic
field

Fig
...
9 The shielding and deshielding
effects of the ring current induced in the
benzene ring by the applied field
...


A solvent can influence the local magnetic field experienced by a nucleus in a variety
of ways
...
The anisotropy of the magnetic susceptibility of the solvent
molecules, especially if they are aromatic, can also be the source of a local magnetic
field
...
15
...
We shall see that the NMR spectra of
species that contain protons with widely different chemical shifts are easier to interpret than those in which the shifts are similar, so the appropriate choice of solvent
may help to simplify the appearance and interpretation of a spectrum
...
6 The fine structure

H

Fig
...
10 An aromatic solvent (benzene
here) can give rise to local currents that
shield or deshield a proton in a solvent
molecule
...


The splitting of resonances into individual lines in Fig
...
6 is called the fine structure
of the spectrum
...
The
strength of the interaction is expressed in terms of the scalar coupling constant, J, and
reported in hertz (Hz)
...
The constant of proportionality in this expression is hJ/$2, because
each angular momentum is proportional to $
...
If the resonance line of a particular nucleus is split by a certain amount by a second nucleus, then
the resonance line of the second nucleus is split by the first to the same extent
...
In NMR, letters far apart
in the alphabet (typically A and X) are used to indicate nuclei with very different
chemical shifts; letters close together (such as A and B) are used for nuclei with
similar chemical shifts
...

1
The energy level diagram for a single spin-– nucleus and its single transition were
2
1
shown in Fig
...
3, and nothing more needs to be said
...
24)

15
...
15
...
The four levels on the left are those of the two
spins in the absence of spin–spin coupling
...
The transitions shown are for β ← α of A
or X, the other nucleus (X or A, respectively) remaining unchanged
...


where νA and νX are the Larmor frequencies of A and X and mA and mX are their
quantum numbers
...
15
...
The
spin–spin coupling depends on the relative orientation of the two nuclear spins, so
it is proportional to the product mAmX
...
15
...

15
...
Once again, we have exaggerated the
effect of spin–spin coupling
...
26b)

It follows that the X resonance also consists of two lines of separation J, but they are
centred on the chemical shift of X (as shown in Fig
...
13)
...
26a)

Therefore, the A resonance consists of a doublet of separation J centred on the chemical shift of A (Fig
...
13)
...
15
...

The transition energies are
1
∆E = hνX ± –hJ
2

aAaX

(15
...
A higher energy is obtained if both spins are α or both spins are β
...
The resulting energy level diagram (for J > 0) is shown on
1
the right of Fig
...
11
...

4
When a transition of nucleus A occurs, nucleus X remains unchanged
...
There are two such
transitions, one in which βA ← αA occurs when the X nucleus is α X, and the other in
which βA ← αA occurs when the X nucleus is β X
...
15
...
15
...
The energies of the transitions are
1
∆E = hνA ± –hJ
2

aA b X

A resonance

E = −hνAmA − hνXmX + hJmAmX

bAaX

J

J

dA

dX

Fig
...
13 The effect of spin–spin coupling
on an AX spectrum
...
The pairs of
resonances are centred on the chemical
shifts of the protons in the absence of
spin–spin coupling
...
Instead of a single line from A, we get a doublet of lines separated by
J and centred on the chemical shift characteristic of A
...

These features are summarized in Fig
...
13
...
As we shall explain below, a group of equivalent nuclei resonates like a single nucleus
...
15
...

The A resonance in an AXn species, though, is quite different from the A resonance in
an AX species
...

The resonance of A is split into a doublet of separation J by one X, and each line of that
doublet is split again by the same amount by the second X (Fig
...
15)
...
The A resonance of an A n X 2 species would also be a 1:2:1
triplet of splitting J, the only difference being that the intensity of the A resonance
would be n times as great as that of AX2
...
15
...
The X resonance, though, is still
1
a doublet of separation J
...
The easiest way of constructing the pattern of fine structure is to
draw a diagram in which successive rows show the splitting of a subsequent proton
...
15
...
15
...
16
...
15
...

2

X resonance
in AX2

X resonance
in AX

dA
J

dX
The X resonance of an AX2
species is also a doublet, because the two
equivalent X nuclei behave like a single
nucleus; however, the overall absorption is
twice as intense as that of an AX species
...
15
...
15
...
The
resonance of A is split into two by coupling
with one X nucleus (as shown in the inset),
and then each of those two lines is split into
two by coupling to the second X nucleus
...


dA
Fig
...
16 The origin of the 1:3:3:1 quartet in
the A resonance of an AX3 species
...
15
...


15
...
15
...
n protons, as in Figs
...
15 and 15
...
The resulting intensity
distribution has a binomial distribution
and is given by the integers in the
corresponding row of Pascal’s triangle
...
Four protons,
in AX4, split the A resonance into a
1:4:6:4:1 quintet
...
15
...
Two
equivalent spin-1 nuclei give rise to a
1:2:3:2:1 quintet
...
1 Accounting for the fine structure in a spectrum

Account for the fine structure in the NMR spectrum of the C-H protons of ethanol
...
There is no
splitting within groups of equivalent protons
...

Answer The three protons of the CH3 group split the resonance of the CH2 protons into a 1:3:3:1 quartet with a splitting J
...
The CH2 and CH3 protons all interact with the OH proton, but these
couplings do not cause any splitting because the OH protons migrate rapidly from
molecule to molecule and their effect averages to zero
...
2 What fine-structure can be expected for the protons in 14NH +? The
4

spin quantum number of nitrogen is 1
...
Thus, 1JCH is the coupling constant for a
proton joined directly to a 13C atom, and 2JCH is the coupling constant when the same
two nuclei are separated by two bonds (as in 13C-C-H)
...
Both 3J and 4J can give detectable effects in a spectrum, but couplings over larger numbers of bonds can generally

527

528

15 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
be ignored
...
4 Hz
between the CH3 and CH2 protons in CH3C
...
CC
...

The sign of JXY indicates whether the energy of two spins is lower when they are parallel (J < 0) or when they are antiparallel (J > 0)
...
An additional point is that J
varies with the angle between the bonds (Fig
...
19)
...
It follows that the measurement of 3JHH in a
series of related compounds can be used to determine their conformations
...
15
...


Exploration Draw a family of curves
showing the variation of 3JHH with φ
for which A = +7
...
0 Hz, and C
varies slightly from a typical value of
+5
...
What is the effect of changing the
value of the parameter C on the shape of
the curve? In a similar fashion, explore the
effect of the values of A and B on the shape
of the curve
...
4

The average (or mean value) of a
function f(x) over the range x = a to x = b
b
is Úa f(x)dx/(b − a)
...

Therefore the average value of (1 –
p
3 cos2q) is Ú 0 (1 – 3 cos2q) sin q dq/p = 0
...
27)

160

125

(d) The origin of spin–spin coupling

Spin–spin coupling is a very subtle phenomenon, and it is better to treat J as an
empirical parameter than to use calculated values
...

A nucleus with spin projection mI gives rise to a magnetic field with z-component
Bnuc at a distance R, where, to a good approximation,
Bnuc = −

γ $µ 0
4πR3

(1 − 3 cos2θ)mI

(15
...
The magnitude of this field is about 0
...
3 nm, corresponding to a splitting of resonance signal of about 104 Hz, and is of
the order of magnitude of the splitting observed in solid samples (see Section 15
...

In a liquid, the angle θ sweeps over all values as the molecule tumbles, and 1–3 cos2θ
averages to zero
...
The direct interaction does make an important contribution to the spectra of solid samples and is a
very useful indirect source of structure information through its involvement in spin
relaxation (Section 15
...

Spin–spin coupling in molecules in solution can be explained in terms of the
polarization mechanism, in which the interaction is transmitted through the bonds
...
15
...
The coupling mechanism depends on the fact
that in some atoms it is favourable for the nucleus and a nearby electron spin to be
parallel (both α or both β), but in others it is favourable for them to be antiparallel
(one α and the other β)
...
A pictorial description of the Fermi
contact interaction is as follows
...
15
...
Far from the nucleus the field generated by this loop is
indistinguishable from the field generated by a point magnetic dipole
...
The magnetic interaction
between this non-dipolar field and the electron’s magnetic moment is the contact

15
...
15
...
The two
arrangements have slightly different
energies
...


Fig
...
21 The origin of the Fermi contact
interaction
...
However, if an electron
can sample the field close to the region
indicated by the sphere, the field
distribution differs significantly from that
of a point dipole
...


interaction
...
15
...
The lower energy state of an electron spin in such a field is the β state
...
We shall suppose that it is
energetically favourable for an electron spin and a nuclear spin to be antiparallel (as is
the case for a proton and an electron in a hydrogen atom)
...
The second electron in the bond,
which must have α spin if the other is β, will be found mainly at the far end of the bond
(because electrons tend to stay apart to reduce their mutual repulsion)
...
The opposite is true when X is β, for now the α spin of Y has the
lower energy
...
That is, 1JHH is positive
...
In this case (Fig
...
22), an X nucleus with α spin polarizes
the electrons in its bond, and the α electron is likely to be found closer to the C nucleus
...
4d), so the more favourable arrangement is for
the α electron of the neighbouring bond to be close to the C nucleus
...
Hence, according to this mechanism, the lower Larmor frequency of Y will be obtained if its spin is parallel to that of X
...


Fermi

Pauli

Y

Hund

C

Pauli
Fermi

X

The polarization mechanism for
JHH spin–spin coupling
...
In this case, J < 0,
corresponding to a lower energy when the
nuclear spins are parallel
...
15
...
These nuclei may also interact by a dipolar mechanism with the
electron magnetic moments and with their orbital motion, and there is no simple way
of specifying whether J will be positive or negative
...
Chemically equivalent nuclei
are nuclei that would be regarded as ‘equivalent’ according to ordinary chemical
criteria
...

The difference between chemical and magnetic equivalence is illustrated by CH2F2
and H2C=CF2
...
However, although
the protons in CH2F2 are magnetically equivalent, those in CH2=CF2 are not
...
In constrast, in CH2F2 both protons
are connected to a given F nucleus by identical bonds, so there is no distinction
between them
...
However, they are in practice made magnetically equivalent
by the rapid rotation of the CH3 group, which averages out any differences
...

An important feature of chemically equivalent magnetic nuclei is that, although
they do couple together, the coupling has no effect on the appearance of the spectrum
...
15
...
Then, because the relative orientations of
nuclear spins are not changed in any transition, the magnitude of the coupling between
them is undetectable
...


a

(a) A group of two equivalent
nuclei realigns as a group, without change
of angle between the spins, when a
resonant absorption occurs
...
(b)
Three equivalent nuclei also realign as a
group without change of their relative
orientations
...
15
...
6 THE FINE STRUCTURE

531

Comment 15
...
2 The energy levels of an A2 system
1
Consider an A2 system of two chemically equivalent spin-– nuclei
...
There are four spin states that
(just as for two electrons) can be classified according to their total spin I (the analogue of S for two electrons) and their total projection MI on the z-axis
...
15
...

As remarked in Section 15
...
The scalar
product can be expressed in terms of the total nuclear spin by noting that
2
2
I2 = (I1 + I2)·(I1 + I2) = I 1 + I 2 + 2I1·I2

As in Section 10
...
2 are those
with a definite resultant, and hence a
well defined value of I
...
See Fig
...
24
...
15
...
We see that three of the states move in energy in one direction and the
fourth (the one with antiparallel spins) moves three times as much in the opposite
direction
...
15
...


The NMR spectrum of the A2 species arises from transitions between the levels
...
Put another way, the allowed transitions are subject to the selection
rule ∆I = 0
...
The allowed
transitions are shown in Fig
...
24: we see that there are only two transitions, and that
they occur at the same resonance frequency that the nuclei would have in the absence
of spin–spin coupling
...

(f) Strongly coupled nuclei

NMR spectra are usually much more complex than the foregoing simple analysis suggests
...
15
...
When spin–spin
coupling is taken into account, the energy
levels on the right are obtained
...
The only allowed
transitions are those that preserve the angle
between the spins, and so take place
between the three states with I = 1
...


532

15 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE

n °Dd << J

n°D d » J

n°Dd » J

n°Dd >> J
The NMR spectra of an A2 system
(top) and an AX system (bottom) are
simple ‘first-order’ spectra
...
Note how the inner two lines of
the bottom spectrum move together, grow
in intensity, and form the single central line
of the top spectrum
...


Fig
...
25

H

O
C
N

are much greater than the spin–spin coupling constants
...
The spectra that result are called firstorder spectra
...
The complicated
spectra that are then obtained are called strongly coupled spectra (or ‘second-order
spectra’) and are much more difficult to analyse (Fig
...
25)
...

A clue to the type of analysis that is appropriate is given by the notation for the
types of spins involved
...
An AB system, on the other
hand (with two nuclei of similar chemical shifts), gives a spectrum typical of a strongly
coupled system
...
AX may also denote a homonuclear spin system in
which the nuclei are of the same element but in markedly different environments
...
7 Conformational conversion and exchange processes
The appearance of an NMR spectrum is changed if magnetic nuclei can jump rapidly
between different environments
...
15
...
When the jumping
rate is low, the spectrum shows two sets of lines, one each from molecules in each conformation
...
At intermediate inversion rates, the line is very broad
...
Coalescence of the two
lines occurs when

τ=

Fig
...
26 When a molecule changes from
one conformation to another, the positions
of its protons are interchanged and jump
between magnetically distinct
environments
...
29)

Example 15
...
The two CH3 resonances are separated by 390 Hz in a 600 MHz
spectrometer
...
29 for the average lifetimes of the conformations
...

Answer With δν = 390 Hz,

τ=

2
= 1
...
8 THE MAGNETIZATION VECTOR
It follows that the signal will collapse to a single line when the interconversion rate
exceeds about 830 s−1
...

Self-test 15
...
3 ms]

A similar explanation accounts for the loss of fine structure in solvents able to exchange protons with the sample
...
When this chemical exchange occurs, a molecule ROH with
an α-spin proton (we write this ROHα) rapidly converts to ROHβ and then perhaps
to ROHα again because the protons provided by the solvent molecules in successive
exchanges have random spin orientations
...
15
...
The effect is
observed when the lifetime of a molecule due to this chemical exchange is so short that
the lifetime broadening is greater than the doublet splitting
...
1 s for the splitting to be observable
...
In dry
dimethylsulfoxide (DMSO), the exchange rate may be slow enough for the splitting to
be detected
...

One of the best analogies that has been suggested to illustrate the difference between
the old and new ways of observing an NMR spectrum is that of detecting the spectrum
of vibrations of a bell
...
A lot of time would be spent getting zero response when the stimulating frequency was between the bell’s vibrational modes
...
The equivalent in NMR is to monitor the
radiation nuclear spins emit as they return to equilibrium after the appropriate stimulation
...
Moreover, multiple-pulse FTNMR gives chemists unparalleled control over the information content and display of
spectra
...
These features are generally expressed
in terms of the vector model of angular momentum introduced in Section 9
...

15
...
As we saw in Section
2
9
...
15
...

2
The angle around the z-axis is
indeterminate
...
15
...
(a) In the absence of
an externally applied field, there are equal
numbers of α and β spins at random angles
around the z-axis (the field direction) and
the magnetization is zero
...

As a result, there is a net magnetization
along the z-axis
...
As the uncertainty principle
does not allow us to specify the x- and y-components of the angular momentum, all
1
we know is that the vector lies somewhere on a cone around the z-axis
...
15
...

2
In the absence of a magnetic field, the sample consists of equal numbers of α and β
nuclear spins with their vectors lying at random angles on the cones
...
The magnetization, M, of the sample, its net nuclear magnetic moment, is zero (Fig
...
28a)
...
First, the
energies of the two orientations change, the α spins moving to low energy and the
β spins to high energy (provided γ > 0)
...
This motion is a pictorial representation of the difference in energy of the
spin states (it is not an actual representation of reality)
...
Secondly, the populations of the two spin states (the numbers of α and β spins) at thermal equilibrium
change, and there will be more α spins than β spins
...
That is, there is only a tiny imbalance of
populations, and it is even smaller for other nuclei with their smaller magnetogyric
ratios
...
15
...

(b) The effect of the radiofrequency field

We now consider the effect of a radiofrequency field circularly polarized in the xyplane, so that the magnetic component of the electromagnetic field (the only component we need to consider) is rotating around the z-direction, the direction of the
applied field B0, in the same sense as the Larmor precession
...
Suppose we choose the frequency of this field to be equal to
the Larmor frequency of the spins, νL = (γ /2π)B0; this choice is equivalent to selecting
the resonance condition in the conventional experiment
...
15
...
Just as the spins precess about the strong static field B0 at a frequency
γ B0/2π, so under the influence of the field B1 they precess about B1 at a frequency
γ B1/2π
...
If we were to imagine
stepping on to a platform, a so-called rotating frame, that rotates around the direction of the applied field at the radiofrequency, then the nuclear magnetization appears
stationary if the radiofrequency is the same as the Larmor frequency (Fig
...
29b)
...
15
...
The duration of the pulse depends on the strength of the B1 field, but is
typically of the order of microseconds
...
To an external observer (the role
played by a radiofrequency coil) in this stationary frame, the magnetization vector is
now rotating at the Larmor frequency in the xy-plane (Fig
...
30b)
...
8 THE MAGNETIZATION VECTOR

535

M

M

Signal

90° pulse

nL
n = nL

(a)

Time

B1

B1

M
(a)
B0

Detecting
coil

M

B1
(b)
Fig
...
29 (a) In a resonance experiment, a
circularly polarized radiofrequency
magnetic field B1 is applied in the xy-plane
(the magnetization vector lies along the
z-axis)
...
When the two frequencies
coincide, the magnetization vector of the
sample rotates around the direction of the
B1 field
...
15
...
(b) To an external stationary
observer (the coil), the magnetization
vector is rotating at the Larmor frequency,
and can induce a signal in the coil
...
In practice, the processing takes place after subtraction of a constant high frequency component (the radiofrequency used for B1), so that
all the signal manipulation takes place at frequencies of a few kilohertz
...
The form of the signal that we can expect is therefore the oscillatingdecaying free-induction decay (FID) shown in Fig
...
31
...
30)

We have considered the effect of a pulse applied at exactly the Larmor frequency
...
If the difference in frequency is small compared to the inverse of
the duration of the 90° pulse, the magnetization will end up in the xy-plane
...
The detected
signal shows that a particular resonant frequency is present
...
15
...


536

15 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE

Signal

(c) Time- and frequency-domain signals

Time

(a)

(b)

Frequency

Fig
...
32 (a) A free induction decay signal
of a sample of AX species and (b) its
analysis into its frequency components
...
Explore the effect
on the shape of the FID curve of changing
the chemical shifts (and therefore the
Larmor frequencies) of the A and X nuclei
...
6

Signal

The web site for this text contains links
to databases of NMR spectra and to sites
that allow for interactive simulation of
NMR spectra
...
15
...
Its Fourier transform is
the frequency-domain spectrum shown in
Fig
...
6
...


We can think of the magnetization vector of a homonuclear AX spin system with
J = 0 as consisting of two parts, one formed by the A spins and the other by the X spins
...

However, because the A and X nuclei precess at different frequencies, they induce two
signals in the detector coils, and the overall FID curve may resemble that in Fig
...
32a
...

The problem we must address is how to recover the resonance frequencies present
in a free-induction decay
...
2 and 15
...
When the signal in
Fig
...
32a is transformed in this way, we get the frequency-domain spectrum shown
in Fig
...
32b
...

The FID curve in Fig
...
33 is obtained from a sample of ethanol
...
15
...
We can now see why the FID curve in Fig
...
33 is so
complex: it arises from the precession of a magnetization vector that is composed of
eight components, each with a characteristic frequency
...
9 Spin relaxation
There are two reasons why the component of the magnetization vector in the xy-plane
shrinks
...
The return to equilibrium is the process called spin relaxation
...

Now consider the effect of a 180° pulse, which may be visualized in the rotating
frame as a flip of the net magnetization vector from one direction along the z-axis to
the opposite direction
...
After the pulse, the populations revert to their thermal equilibrium values exponentially
...
15
...
31)

Because this relaxation process involves giving up energy to the surroundings (the
‘lattice’) as β spins revert to α spins, the time constant T1 is also called the spin–lattice
relaxation time
...
Such fields
can arise from the tumbling motion of molecules in a fluid sample
...
Only if the molecule tumbles at
about the resonance frequency will the fluctuating magnetic field be able to induce
spin changes effectively, and only then will T1 be short
...
9 SPIN RELAXATION

537

t + T1

t

b

a

a

Fig
...
34 In longitudinal relaxation the
spins relax back towards their thermal
equilibrium populations
...
On
the right, which represents the sample a
long time after a time T1 has elapsed, the
populations are those characteristic of a
Boltzmann distribution (see Molecular
interpretation 3
...
In actuality, T1 is the
time constant for relaxation to the
arrangement on the right and T1 ln 2 is the
half-life of the arrangement on the left
...
15
...

Note that the populations of the states
remain the same; only the relative phase of
the spins relaxes
...


increases with temperature and with reducing viscosity of the solvent, so we can
expect a dependence like that shown in Fig
...
35
...
15
...
The magnetization vector is large when
all the spins are bunched together immediately after a 90° pulse
...
At that stage, the component
of magnetization vector in the plane would be zero
...
32)

Relaxation time

b

T1

T2
Low
temperature,
high viscosity

High
temperature,
low viscosity

Rate of motion
Fig
...
35 The variation of the two
relaxation times with the rate at which the
molecules move (either by tumbling or
migrating through the solution)
...
Note
that, at rapid rates of motion, the two
relaxation times coincide
...
8
0
...
4

0
...
15
...

The width at half-height is inversely
proportional to the parameter T2 and the
longer the transverse relaxation time, the
narrower the line
...
Explore the
effect of the parameter T2 on the width and
the maximal intensity of a Lorentzian line
...


180° pulse

M

∆ν1/2 =

T* =
2

Signal

Interval t

Relaxation
from -M0
to M0

The result of applying a 180° pulse
to the magnetization in the rotating frame
and the effect of a subsequent 90° pulse
...


1

(15
...
1 Hz can be anticipated, in broad agreement with observation
...
In practice, the magnet is not perfect, and the field is different at
different locations in the sample
...
It is common to express the extent of inhomogeneous broadening
in terms of an effective transverse relaxation time, T 2 by using a relation like
*,
eqn 15
...
15
...
Any relaxation process that changes the balance
between α and β spins will also contribute to this randomization, so the time constant
T2 is almost always less than or equal to T1
...
When the fluctuations are
slow, each molecule lingers in its local magnetic environment and the spin orientations randomize quickly around the applied field direction
...
In other words, slow molecular motion corresponds to short T2 and fast
motion corresponds to long T2 (as shown in Fig
...
35)
...

If the y-component of magnetization decays with a time constant T2, the spectral
line is broadened (Fig
...
37), and its width at half-height becomes

1

[15
...
As an example, consider a line in a spectrum with a width of
10 Hz
...
34 that the effective transverse relaxation time is
T* =
2

1
π × (10 s−1)

= 32 ms

(b) The measurement of T1

The longitudinal relaxation time T1 can be measured by the inversion recovery technique
...
A 180° pulse is achieved by
applying the B1 field for twice as long as for a 90° pulse, so the magnetization vector
precesses through 180° and points in the −z-direction (Fig
...
38)
...
The β spins begin to relax back into α spins, and the
magnetization vector first shrinks exponentially, falling through zero to its thermal
equilibrium value, Mz
...
The frequency-domain
spectrum is then obtained by Fourier transformation
...
9 SPIN RELAXATION
The intensity of the spectrum obtained in this way depends on the length of the
magnetization vector that is rotated into the xy-plane
...
We can therefore measure T1
by fitting an exponential curve to the series of spectra obtained with different values
of τ
...
15
...


- t/T2

e
Echo

The measurement of T2 (as distinct from T 2 depends on being able to eliminate the
*)
effects of inhomogeneous broadening
...

A spin echo is the magnetic analogue of an audible echo: transverse magnetization
is created by a radiofrequency pulse, decays away, is reflected by a second pulse, and
grows back to form an echo
...
15
...
We can
consider the overall magnetization as being made up of a number of different magnetizations, each of which arises from a spin packet of nuclei with very similar precession frequencies
...
The precession frequencies also differ if there is more than one chemical shift present
...

First, a 90° pulse is applied to the sample
...
The spin packets now begin to fan out because they have different Larmor frequencies, with some above the radiofrequency and some below
...

After an evolution period τ, a 180° pulse is applied to the sample; this time, about
the y-axis of the rotating frame (the axis of the pulse is changed from x to y by a 90°
phase shift of the radiofrequency radiation)
...
Thus, as the vectors continue to precess, the fast vectors
are now behind the slow; the fan begins to close up again, and the resultant signal
begins to grow back into an echo
...
Because the effects of field
inhomogeneities have been suppressed by the refocusing, the echo signal will have
been attenuated by the factor e−2τ /T2 caused by spin–spin relaxation alone
...

2
The important feature of the technique is that the size of the echo is independent of
any local fields that remain constant during the two τ intervals
...
Hence, the size of the echo
is independent of inhomogeneities in the magnetic field, for these remain constant
...
Hence, the effects of the true relaxation are not refocused, and the size of the echo
decays with the time constant T2 (Fig
...
40)
...
15
...


540

15 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
IMPACT ON MEDICINE

Sample

I15
...

Magnetic resonance imaging (MRI) is a portrayal of the concentrations of protons in a
solid object
...

If an object containing hydrogen nuclei (a tube of water or a human body) is placed
in an NMR spectrometer and exposed to a homogeneous magnetic field, then a single
resonance signal will be detected
...
15
...
Then the water protons will be resonant at the
frequencies

νL(z) =

γ
2π

(B0 + Gz z)

(15
...
15
...
The
resulting intensity pattern is a map of the
numbers in all the slices, and portrays the
shape of the sample
...


(Similar equations may be written for gradients along the x- and y-directions
...
This is an
example of slice selection, the application of a selective 90° pulse that excites nuclei in
a specific region, or slice, of the sample
...

The image of a three-dimensional object such as a flask of water can be obtained if the
slice selection technique is applied at different orientations (see Fig
...
41)
...

In practice, the NMR signal is not obtained by direct analysis of the FID curve after
application of a single 90° pulse
...
9c)
...
The first step consists of a 90° pulse that results in slice selection along the z-direction
...
At each position
along the gradient, a spin packet will precess at a different Larmor frequency due to
chemical shift effects and the field inhomogeneity, so each packet will dephase to
a different extent by the end of the evolution period
...
1 For
each value of τ, the next steps are application of the 180° pulse and then of a read
gradient, a field gradient along the x-direction, during detection of the echo
...
Therefore Fourier transformation of the FID gives different signals for
protons at different positions along x
...

One strategy for solving this problem takes advantage of the fact that the relaxation
times of water protons are shorter for water in biological tissues than for the pure
liquid
...
A T1-weighted image is obtained by repeating the spin echo sequence
1
For technical reasons, it is more common to vary the magnitude of the phase gradient
...


15
...
Under
these conditions, differences in signal intensities are directly related to differences in
T1
...
Each point on the image is an echo signal that behaves in the manner shown in
Fig
...
40, so signal intensities are strongly dependent on variations in T2
...
Another strategy involves the use of contrast
agents, paramagnetic compounds that shorten the relaxation times of nearby protons
...

The MRI technique is used widely to detect physiological abnormalities and to
observe metabolic processes
...
The technique
is based on differences in the magnetic properties of deoxygenated and oxygenated
haemoglobin, the iron-containing protein that transports O2 in red blood cells
...
Because there is greater blood flow in
active regions of the brain than in inactive regions, changes in the intensities of proton resonances due to changes in levels of oxygenated haemoglobin can be related to
brain activity
...
15
...
In fact, the invisibility of hard structures in MRI is an advantage, as
it allows the imaging of structures encased by bone, such as the brain and the spinal
cord
...

15
...
1 per
cent)
...
Hence, it
is not normally necessary to take into account 13C-13C spin–spin coupling within a
molecule
...
If we were observing a 13C-NMR spectrum, we would obtain a very
complex spectrum on account of the coupling of the one 13C nucleus with many of
the protons that are present
...
Thus, if the CH3 protons of
ethanol are irradiated with a second, strong, resonant radiofrequency pulse, they
undergo rapid spin reorientations and the 13C nucleus senses an average orientation
...
Proton decoupling
has the additional advantage of enhancing sensitivity, because the intensity is concentrated into a single transition frequency instead of being spread over several transition
frequencies (see Section 15
...
If care is taken to ensure that the other parameters
on which the strength of the signal depends are kept constant, the intensities of
proton-decoupled spectra are proportional to the number of 13C nuclei present
...


541

Fig
...
42 The great advantage of MRI is
that it can display soft tissue, such as in this
cross-section through a patient’s head
...
)

15 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
15
...

Dipole–dipole interactions are discussed
in Chapter 18
...
15
...
Each grey square above the
line represents an excess population and
each white square below the line represents
a population deficit
...


We have seen already that one advantage of protons in NMR is their high magnetogyric
ratio, which results in relatively large Boltzmann population differences and hence
greater resonance intensities than for most other nuclei
...

To understand the effect, we consider the populations of the four levels of a homonuclear (for instance, proton) AX system; these were shown in Fig
...
12
...
The thermal equilibrium absorption intensities reflect these populations as
shown in Fig
...
43
...
When we saturate the X transition, the populations of the X levels are equalized (Nα X = Nβ X) and all transitions involving α X ↔ β X spin flips are no
longer observed
...
If
that were all there were to happen, all we would see would be the loss of the X resonance and no effect on the A resonance
...
Relaxation can occur in a variety of ways
if there is a dipolar interaction between the A and X spins
...
However,
the populations of the αA β X and βAα X levels remain unchanged at the values characteristic of saturation
...
15
...
Another possibility is for the dipolar interaction between the
two spins to cause α to flip to β and β to flop to α
...

Now we see from the illustration that the population differences in the states involved
in the A transitions are decreased, so the resonance absorption is diminished
...
9, the efficiency of the intensityenhancing βA β X ↔ αAαX relaxation is high if the dipole field oscillates at a frequency
close to the transition frequency, which in this case is about 2ν ; likewise, the efficiency

aA bX
te
ra
tu
Sa

X

(a)

X

A

bAaX

aA bX

A

aAaX

X
(b)

Relax

A

bA b X

te
ra
tu
Sa

Energy

bA b X

X

bA bX
A

aA bX

bAaX
A

aAaX

X

bAaX

X
(c)

En
ha
nc
e

Comment 15
...
15
...
15
...
(b) Dipole–dipole relaxation relaxes the populations of the highest and lowest states, and they
regain their original populations
...
15
...


15
...
15
...
15
...
(b) Dipole–dipole relaxation relaxes the populations of the two intermediate states, and they regain their original
populations
...
15
...


of the intensity-diminishing αA β X ↔ βAα X relaxation is high if the dipole field is
stationary (as there is no frequency difference between the initial and final states)
...
15
...
A small molecule rotating rapidly can be expected to
have substantial motion at 2ν, and a consequent enhancement of the signal
...
36]

Here I A and IA are the intensities of the NMR signals due to nucleus A before and after
°
application of the long (> T1) radiofrequency pulse that saturates transitions due to
the X nucleus
...
However, η also depends on the
2
values of the magnetogyric ratios of A and X
...
37)

where γA and γ X are the magnetogyric ratios of nuclei A and X, respectively
...
99, which shows that an enhancement of about a factor of 2 can be achieved
...
The Overhauser enhancement of a proton A generated by saturating a spin X depends on the fraction of A’s
spin–lattice relaxation that is caused by its dipolar interaction with X
...
The determination
of the structure of a small protein in solution involves the use of several hundred
NOE measurements, effectively casting a net over the protons present
...


544

15 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
15
...
Even a first-order spectrum is complex, for the fine structure of
different groups of lines can overlap
...
This separation is essentially what is achieved
in two-dimensional NMR
...


Time

Fig
...
46 The pulse sequence used in
correlation spectroscopy (COSY)
...

Acquisitions of free-induction decays are
taken during t2 for a set of different
evolution times t1
...
52
...
8

A vector, V, of length 1, in the xy-plane
and its two components, 1x and 1y , can
be thought of as forming a right-angled
triangle, with 1 the length of the
hypotenuse (see the illustration)
...


Now we shall see how the PEMD pulse structure can be used to devise experiments
that reveal spin–spin couplings and internuclear distances in small and large molecules
...
The basic COSY
experiment uses the simplest of all two-dimensional pulse sequences, consisting of
two consecutive 90° pulses (Fig
...
46)
...
Figure 15
...
A 90° pulse applied in the x-direction (in
the stationary frame) tilts the magnetization vector toward the y-axis
...
At a time t1 the vector will have swept through an angle 2πν t1 and the magnitude
of the magnetization will have decayed by spin–spin relaxation to M = M0e−t1/T2
...
38a)

Application of the second 90° pulse parallel to the x-axis tilts the magnetization
again and the resulting vector has components with magnitudes (once again, in the
stationary frame)
Mx = M sin 2πν t1

My = 0

Mz = M cos 2πν t1

(15
...
The signal
intensity is related to Mx, the magnitude of the magnetization that is rotating around
the xy-plane at the time of application of the detection pulse, so it follows that the signal
strength varies sinusoidally with the duration of the evolution period
...
15
...


15
...

(b) A plot of the maximum intensity of
each absorption line against t1
...

Fig
...
48

z

n

(b)

n2

n

(a) The two-dimensional NMR
spectrum of the sample discussed in
Figs
...
47 and 15
...
See the text for
an explanation of how the spectrum is
obtained from a series of Fourier
transformations of the data
...

Fig
...
49

A plot of the maximum intensity of each absorption band in Fig
...
48a against t1
has the form shown in Fig
...
48b
...
If we continue the process by first plotting signal
intensity against t1 for several frequencies along the ν2 axis and then carrying out
Fourier transformations, we generate a family of curves that can be pooled together
into a three-dimensional plot of I(ν1,ν2), the signal intensity as a function of the
frequencies ν1 and ν2 (Fig
...
49a)
...
The most
common representation of the data is as a contour plot, such as the one shown in
Fig
...
49b
...

However, when the one-dimensional spectrum is complex, the COSY experiment shows
which spins are related by spin–spin coupling
...

From our discussion so far, we know that the key to the COSY technique is the
effect of the second 90° pulse
...
15
...
At thermal equilibrium, the population of the αAα X level is the greatest, and that of the βA β X level is the
least; the other two levels have the same energy and an intermediate population
...
If a second 90° pulse

M cos 2pnt1
y

M sin 2pnt1
x

(c)

2pnt1

M
90o pulse
z
y

M sin 2pnt1
x

(d)

M cos 2pnt1

Fig
...
47 (a) The effect of the pulse
sequence shown in Fig
...
46 on the
magnetization M0 of a sample of a
compound with only one proton
...
(c)
After a time t1 has elapsed, the vector will
have swept through an angle 2πν t1 and the
magnitude of the magnetization will have
decayed to M
...
(d) Application
of the second 90° pulse parallel to the x-axis
tilts the magnetization again and the
resulting vector has components with
magnitude Mx = M sin 2πν t1, My = 0, and
Mz = M cos 2πν t1
...


546

15 MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE

Fig
...
50 An example of the change in the
population of energy levels of an AX spin
system that results from the second 90°
pulse of a COSY experiment
...

In this example, we imagine that the pulse
affects the X spins first, and then the A
spins
...
As a result,
excitation of the A spins by the pulse
generates an FID in which one of the two
A transitions has increased in intensity
and the other has decreased
...
Similar schemes
can be written to show that magnetization
can be transferred from the A spins to the X
spins
...
15
...


X transition

bAa X
X transition

aAa X

bA b X
A transition

bA b X

aA b X

bAa X

aA b X
A transition

FID

aAa X

is applied at a time t1 that is short compared to the spin–lattice relaxation time T1, the
extra input of energy causes further changes in the populations of the four states
...
It is
difficult to visualize these changes because the A spins are affecting the X spins and
vice-versa
...
Depending on the evolution time t1, the 90° pulse may leave the population differences across each of the two X transitions unchanged, inverted, or somewhere in
between
...
15
...
Excitation of the A transitions will now generate an FID in which one of the two A transitions has increased in intensity (because
the population difference is now greater), and the other has decreased (because the
population difference is now smaller)
...
As the evolution time t1 is increased,
the intensities of the signals from A spins oscillate with frequencies determined by the
frequencies of the two X transitions
...

This transfer of information between spins is at the heart of two-dimensional NMR
spectroscopy: it leads to the correlation between different signals in a spectrum
...

So, just as before, if we conduct a series of experiments in which t1 is incremented,
Fourier transformation of the FIDs on t2 yields a set of spectra I(t1,F2) in which the
signal amplitudes oscillate as a function of t1
...
The signals
are spread out in F1 according to their precession frequencies during the detection
period
...
15
...
15
...
Each group consists of a block of
four signals separated by J
...
That is, the spectrum along the diagonal is
equivalent to the one-dimensional spectrum obtained with the conventional NMR
technique (Fig
...
13)
...

Although information from two-dimensional NMR spectroscopy is trivial in an AX
system, it can be of enormous help in the interpretation of more complex spectra,
leading to a map of the couplings between spins and to the determination of the bonding network in complex molecules
...
Below we illustrate
the procedure by assigning the resonances in the COSY spectrum of an amino acid
...
12 TWO-DIMENSIONAL NMR

547

d

1
...
0
3
...
0

3
...
0
d

1
...
3 The COSY spectrum of isoleucine

Figure 15
...

We begin the assignment process by considering which protons should be interacting by spin–spin coupling
...
The Ca-H proton is coupled only to the Cb-H proton
...
The Cb-H protons are coupled to the Ca-H, Cc-H, and Cd-H protons
...
The inequivalent Cd-H protons are coupled to the Cb-H and Ce-H protons
...
6 shares a cross-peak with only one other resonance
at δ = 1
...
4, 1
...
9
...
6 and 1
...

• The proton with resonance at δ = 0
...
8 to the Ce-H protons
...
4 and 1
...
9
...
9 to the
Cc-H protons and the resonances at δ = 1
...
2 to the inequivalent Cd-H
protons
...
For example, the second 90° pulse actually mixes the spin state transitions
caused by the first 90° pulse (hence the term ‘mixing period’)
...
The latter transitions cannot generate any signal in the receiver coil
of the spectrometer, but their existence can be demonstrated by applying a third pulse
to mix them back into the four observable single quantum transitions
...


Fig
...
52 Proton COSY spectrum of
isoleucine
...
E
...
C
...
S
...
508, Prentice Hall,
Upper Saddle River (1998)
...
We have seen that the steady-state nuclear Overhauser effect can provide
information about internuclear distances through analysis of enhancement patterns
in the NMR spectrum before and after saturation of selected resonances
...
15
...
The results of double Fourier transformation is a
spectrum in which the cross-peaks form a map of all the NOE interactions in a
molecule
...
11), NOESY data reveal internuclear
distances up to about 0
...

15
...
Nevertheless, there are good reasons for seeking to
overcome these difficulties
...
Moreover, many species are intrinsically interesting as solids, and it is
important to determine their structures and dynamics
...
This kind of information is crucial to an interpretation of the bulk properties of
the polymer in terms of its molecular characteristics
...

Problems of resolution and linewidth are not the only features that plague NMR
studies of solids, but the rewards are so great that considerable efforts have been made
to overcome them and have achieved notable success
...
Hence, in a pulse experiment, there is a need for lengthy
delays—of several seconds—between successive pulses so that the spin system has
time to revert to equilibrium
...
Moreover, because lines are so broad, very high powers of radiofrequency radiation may be required to achieve saturation
...

(a) The origins of linewidths in solids

There are two principal contributions to the linewidths of solids
...
As we saw in the discussion of
spin–spin coupling, a nuclear magnetic moment will give rise to a local magnetic field,
which points in different directions at different locations around the nucleus
...
14 THE EPR SPECTROMETER
B loc = −

γ $µ 0mI
4πR3

(1 – 3 cos2θ)

549

(15
...
Many nuclei may contribute to the total local field experienced by a nucleus of interest, and different nuclei
in a sample may experience a wide range of fields
...

A second source of linewidth is the anisotropy of the chemical shift
...
In general, this ability depends on the orientation of the molecule
relative to the applied field
...
However, the anisotropy is not averaged
to zero for stationary molecules in a solid, and molecules in different orientations
have resonances at different frequencies
...

(b) The reduction of linewidths

Fortunately, there are techniques available for reducing the linewidths of solid samples
...
The ‘magic
angle’ is the angle at which 1 – 3 cos2θ = 0, and corresponds to 54
...
In the technique,
the sample is spun at high speed at the magic angle to the applied field (Fig
...
53)
...
The difficulty with MAS is that the
spinning frequency must not be less than the width of the spectrum, which is of the
order of kilohertz
...

Pulsed techniques similar to those described in the previous section may also be
used to reduce linewidths
...
However, because the range of coupling strengths is so large,
radiofrequency power of the order of 1 kW is required
...


Electron paramagnetic resonance
Electron paramagnetic resonance (EPR) is less widely applicable than NMR because
it cannot be detected in normal, spin-paired molecules and the sample must possess
unpaired electron spins
...
3b)
...

15
...
The FT-EPR instrument is based on the concepts developed in Section 15
...
The
layout of the more common CW-EPR spectrometer is shown in Fig
...
54
...
74°

54
...
15
...
74° (that is, arccos (–)1/2)
3
to the applied magnetic field
...


Microwave
source

Detector

Sample
cavity

Electromagnet
Modulation
input

Phasesensitive
detector

Fig
...
54 The layout of a continuous-wave
EPR spectrometer
...
3 T, which requires 9 GHz (3 cm)
microwaves for resonance
...
3 T
...
15
...
The peculiar appearance
of the spectrum, which is in fact the first-derivative of the absorption, arises from the
detection technique, which is sensitive to the slope of the absorption curve (Fig
...
56)
...
15 The g-value
Field
strength

Fig
...
55 The EPR spectrum of the benzene
−
radical anion, C6H6 , in fluid solution
...


1
Equation 15
...
0023
...
Consequently, the resonance condition is normally written as

hν = gµBB0

(15
...

Absorption, A

Illustration 15
...
40 mT in a
spectrometer operating at 9
...
Its g-value is therefore
g=

Field, B
First derivative of
absorption, dA /dB
Fig
...
56 When phase-sensitive detection is
used, the signal is the first derivative of the
absorption intensity
...


Fig
...
57 An applied magnetic field can
induce circulation of electrons that makes
use of excited state orbitals (shown as a
green outline)
...
626 08 × 10−34 J s) × (9
...
2740 × 10−24 J T −1) × (0
...
0027

Self-test 15
...
000 GHz (radiation belonging to the Q band
of the microwave region)?
[1
...
Therefore, the
g-value gives some information about electronic structure and plays a similar role in
EPR to that played by shielding constants in NMR
...
15
...
This additional path for circulation of electrons gives rise to a local
magnetic field that adds to the applied field
...
As we saw in Section 10
...
That is, the local field strength
is proportional to the molecular spin–orbit coupling constant, ξ
...
This proportionality is widely observed
...
0027 and inorganic radicals have g-values typically in the range 1
...
1
...
16 HYPERFINE STRUCTURE

551

varying from 0 to 6, because in them ∆E is small (on account of the splitting of dorbitals brought about by interactions with ligands, as we saw in Section 14
...

Just as in the case of the chemical shift in NMR spectroscopy, the g-value is
anisotropic: that is, its magnitude depends on the orientation of the radical with
respect to the applied field
...
Therefore, anisotropy of the g-value is
observed only for radicals trapped in solids
...
16 Hyperfine structure
The most important feature of EPR spectra is their hyperfine structure, the splitting
of individual resonance lines into components
...
The source of the hyperfine structure in EPR is the magnetic interaction
between the electron spin and the magnetic dipole moments of the nuclei present in
the radical
...
The proton spin is a source of magnetic field, and depending on the orientation of the nuclear spin, the field it generates adds to or subtracts from the applied
field
...
41)

where a is the hyperfine coupling constant
...
42a)

No hyperfine
splitting
Hyperfine splitting
due to one proton

1
The other half (which have mI = − –) resonate when
2
1
hν = gµB(B − –a),
2

or

B=

hν
g µB

1
+ –a
2

aN

(15
...
15
...

If the radical contains an 14N atom (I = 1), its EPR spectrum consists of three lines
of equal intensity, because the 14N nucleus has three possible spin orientations, and
each spin orientation is possessed by one-third of all the radicals in the sample
...

When there are several magnetic nuclei present in the radical, each one contributes
to the hyperfine structure
...
It is not
hard to show that, if the radical contains N equivalent protons, then there are N + 1
hyperfine lines with a binomial intensity distribution (the intensity distribution given
by Pascal’s triangle)
...
15
...
More generally, if the radical contains N equivalent
nuclei with spin quantum number I, then there are 2NI + 1 hyperfine lines with an
intensity distribution based on a modified version of Pascal’s triangle as shown in the
following Example
...
15
...
As a result, the spectrum
consists of two lines (of equal intensity)
instead of one
...
The diagonal lines show the
energies of the states as the applied field is
increased, and resonance occurs when the
separation of states matches the fixed
energy of the microwave photon
...
3 Predicting the hyperfine structure of an EPR spectrum
1
...
61 mT and two
1
equivalent protons (I = – ) with hyperfine constant 0
...
Predict the form of the
2
EPR spectrum
...
35 mT

Method We should consider the hyperfine structure that arises from each type

1:2:1

1:2:1

1:2:1

Fig
...
59 The analysis of the hyperfine
structure of radicals containing one 14N
nucleus (I = 1) and two equivalent protons
...
So, split a line with one
nucleus, then each of those lines is split by a second nucleus (or group of nuclei),
and so on
...

Answer The 14N nucleus gives three hyperfine lines of equal intensity separated by

1
...
Each line is split into doublets of spacing 0
...
35 mT splitting
(Fig
...
59)
...
35 mT
...

Self-test 15
...
15
...


1

3

6

7

6

3

Fig
...
60 The analysis of the hyperfine
structure of radicals containing three
equivalent 14N nuclei
...
Moreover, because the magnitude of
the splitting depends on the distribution of the unpaired electron near the magnetic
nuclei present, the spectrum can be used to map the molecular orbital occupied by the
−
unpaired electron
...
375 mT,
and one proton is close to a C atom with one-sixth the unpaired electron spin density
(because the electron is spread uniformly around the ring), the hyperfine splitting
caused by a proton in the electron spin entirely confined to a single adjacent C atom
should be 6 × 0
...
25 mT
...
43)

with Q = 2
...
In this equation, ρ is the spin density on a C atom and a is the
hyperfine splitting observed for the H atom to which it is attached
...
5 Using the McConnell equation

The hyperfine structure of the EPR spectrum of the radical anion (naphthalene)−
can be interpreted as arising from two groups of four equivalent protons
...
490 mT and for those in the 2, 3, 6,
and 7 positions have a = 0
...
The densities obtained by using the McConnell
equation are 0
...
08, respectively (8)
...
6 The spin density in (anthracene)− is shown in (9)
...

[A 1:2:1 triplet of splitting 0
...
22 mT, split into a 1:4:6:4:1 quintet of
plitting 0
...
2 IMPACT ON BIOCHEMISTRY: SPIN PROBES
(b) The origin of the hyperfine interaction

The hyperfine interaction is an interaction between the magnetic moments of the
unpaired electron and the nuclei
...

An electron in a p orbital does not approach the nucleus very closely, so it experiences a field that appears to arise from a point magnetic dipole
...
The contribution of a magnetic nucleus
to the local field experienced by the unpaired electron is given by an expression like
that in eqn 15
...
A characteristic of this type of interaction is that it is anisotropic
...
Therefore, hyperfine structure due to the
dipole–dipole interaction is observed only for radicals trapped in solids
...
However, because
an s electron has a nonzero probability of being at the nucleus, it is incorrect to treat
the interaction as one between two point dipoles
...
6d is a magnetic interaction that occurs when the point dipole approximation fails
...

The dipole–dipole interactions of p electrons and the Fermi contact interaction of s
electrons can be quite large
...
4 mT from the 14N nucleus
...
More values are listed in Table 15
...
The magnitudes of the
contact interactions in radicals can be interpreted in terms of the s orbital character of
the molecular orbital occupied by the unpaired electron, and the dipole–dipole interaction can be interpreted in terms of the p character
...
11)
...
The sample is fluid, and as the radicals are tumbling
the hyperfine structure cannot be due to the dipole–dipole interaction
...
The explanation lies in a
polarization mechanism similar to the one responsible for spin–spin coupling in NMR
...
15
...
The electron with opposite spin is therefore more likely to be close to the
C atom at the other end of the bond
...
Calculation using this model leads to a hyperfine interaction in agreement with the observed
value of 2
...


553

Synoptic table 15
...
8 (1s)

2

H

7
...
2 (2s)

19

F

1720 (2s)

3
...
4 (2p)

* More values are given in the Data section
...
15
...
The arrangement in (a) is lower in
energy than that in (b), so there is an
effective coupling between the unpaired
electron and the proton
...
2 Spin probes

We saw in Sections 15
...
15 that anisotropy of the g-value and of the nuclear
hyperfine interactions can be observed when a radical is immobilized in a solid
...
62 shows the variation of the lineshape of the EPR spectrum of the di-tertbutyl nitroxide radical (10) with temperature
...
At 77 K,

ESR spectra of the di-tert-butyl
nitroxide radical at 292 K (top) and 77 K
(bottom)
...
R
...
M
...
R
...
C
...
), Wiley, New York (1972)
...
15
...
Both isotropic and anisotropic hyperfine couplings
determine the appearance of the spectrum, which now consists of three broad peaks
...
The ideal
spin probe is one with a spectrum that broadens significantly as its motion is restricted
to a relatively small extent
...

Just as chemical exchange can broaden proton NMR spectra (Section 15
...
Therefore, the distance
between two spin probe molecules may be measured from the linewidths of their
EPR spectra
...
For example,
the kinetics of association of two polypeptides labelled with the synthetic amino
acid 2,2,6,6,-tetramethylpiperidine-1-oxyl-4-amino-4-carboxylic acid (11) may be
studied by measuring the change in linewidth of the label with time
...


Checklist of key ideas
1
...
The energy of a nucleus in a magnetic field B0 is
EmI = −γ $B 0mI, where γ is the nuclear magnetogyric ratio
...
The resonance condition for an electron in a magnetic field is
hν = ge µ BB 0
...

3
...

4
...

5
...

0
6
...

7
...

8
...


1
9
...


10
...

11
...

12
...

13
...

14
...

15
...

16
...
As a result, its resonance is a
single line and not a group of lines
...
The nuclear Overhauser effect (NOE) is the modification of
one resonance by the saturation of another
...
In two-dimensional NMR, spectra are displayed in two axes,
with resonances belonging to different groups lying at
different locations on the second axis
...
Another example is nuclear Overhauser effect
spectroscopy (NOESY), in which internuclear distances up to
about 0
...

19
...
74° to the applied magnetic field
...
The EPR resonance condition is written hν = g µBB, where g is
the g-value of the radical; the deviation of g from ge = 2
...

21
...

22
...

23
...
25 mT
...
Chem
...
67, A93 (1990); Part 4
...
Ibid
...
Ibid
...


Articles and texts

N
...
Atherton, Principles of electron spin resonance
...

E
...
Becker, High resolution NMR: theory and chemical applications
...


M
...
Vlaardingerbroek and J
...
de Boer, Magnetic resonance imaging:
theory and practice
...

M
...
Levitt, Spin dynamics
...


R
...

Oxford University Press (1998)
...
C
...
E
...
L
...
), Encyclopedia of
spectroscopy and spectrometry
...


D
...
Lide (ed
...


R
...
King and K
...
Williams
...

Part 1
...
J
...
Educ
...
Nuclear magnetic resonance: The single
pulse experiment
...
A243; Part 3
...


C
...
Poole, Jr
...
A
...
), Handbook of electron spin
resonance: data sources, computer technology, relaxation, and
ENDOR
...
1–2
...


Further information
Further information 15
...
2)
...
44)

Because e
= cos(2πν t) + i sin(2πν t), the expression above is a sum
over harmonically oscillating functions, with each one weighted by
the intensity I(ν)
...
45)

0

where Re means take the real part of the following expression
...
The integration is carried out at a series of frequencies ν on a
computer that is built into the spectrometer
...
1 Discuss in detail the origins of the local, neighbouring group, and
solvent contributions to the shielding constant
...
4 Discuss the origins of diagonal and cross peaks in the COSY spectrum of
an AX system
...
2 Discuss in detail the effects of a 90° pulse and of a 180° pulse on a system
1
of spin- – nuclei in a static magnetic field
...
5 Discuss how the Fermi contact interaction and the polarization
mechanism contribute to spin–spin couplings in NMR and hyperfine
interactions in EPR
...
3 Suggest a reason why the relaxation times of 13C nuclei are typically

much longer than those of 1H nuclei
...
6 Suggest how spin probes could be used to estimate the depth of a crevice
in a biopolymer, such as the active site of an enzyme
...
1a What is the resonance frequency of a proton in a magnetic field of
14
...
1b What is the resonance frequency of a 19F nucleus in a magnetic field
of 16
...
2a

33

3
S has a nuclear spin of – and a nuclear g-factor of 0
...
Calculate the
2

energies of the nuclear spin states in a magnetic field of 7
...

15
...
404
...
50 T
...
3a Calculate the frequency separation of the nuclear spin levels of a 13C

nucleus in a magnetic field of 14
...
73 × 107 T −1 s−1
...
9a The chemical shift of the CH3 protons in acetaldehyde (ethanal) is

δ = 2
...
80
...
5 T, (b) 15 T?
15
...
16 and
that of the CH2 protons is 3
...
What is the difference in local magnetic field
between the two regions of the molecule when the applied field is (a) 1
...
5 T?
15
...
3b Calculate the frequency separation of the nuclear spin levels of a 14N

nucleus in a magnetic field of 15
...
93 × 107 T −1 s−1
...
90 Hz and the data in Exercise 15
...

15
...
4a In which of the following systems is the energy level separation the

largest? (a) A proton in a 600 MHz NMR spectrometer, (b) a deuteron in the
same spectrometer
...
4b In which of the following systems is the energy level separation the

largest? (a) A 14N nucleus in (for protons) a 600 MHz NMR spectrometer,
(b) an electron in a radical in a field of 0
...

15
...
00 T magnetic field
...
5b Calculate the magnetic field needed to satisfy the resonance condition
for unshielded protons in a 150
...

15
...
2 to predict the magnetic fields at which (a) 1H, (b) 2H,

(c) 13C come into resonance at (i) 250 MHz, (ii) 500 MHz
...
8b What are the relative values of the chemical shifts observed for nuclei
in the spectrometers mentioned in Exercise 15
...
6b Use Table 15
...


J = 6
...
9b in a spectrometer operating at
(a) 350 MHz, (b) 650 MHz
...
11a Two groups of protons are made equivalent by the isomerization

of a fluxional molecule
...
0 and the other has δ = 5
...
At what rate of
interconversion will the two signals merge in a spectrometer operating at
250 MHz?
15
...
At low temperatures, where the interconversion is slow,
one group has δ = 5
...
8
...
12a Sketch the form of the 19F-NMR spectra of a natural sample of

−
−
tetrafluoroborate ions, BF 4, allowing for the relative abundances of 10BF 4
−
and 11BF 4
...
7a Calculate the relative population differences (δN/N) for protons in
fields of (a) 0
...
5 T, and (c) 10 T at 25°C
...
12b From the data in Table 15
...
Sketch the proton and 31P resonances in the NMR spectrum of PH 4
...
7b Calculate the relative population differences (δN/N) for 13C nuclei in

15
...
50 T, (b) 2
...
5 T at 25°C
...
8a The first generally available NMR spectrometers operated at a

frequency of 60 MHz; today it is not uncommon to use a spectrometer that
operates at 800 MHz
...


15
...

15
...


PROBLEMS

557

15
...


for each proton? What is the g-value of the radical given that the spectrometer
is operating at 9
...
15a The duration of a 90° or 180° pulse depends on the strength of the B1

constants 2
...
6 mT gives a spectrum centred on 332
...
At what
fields do the hyperfine lines occur and what are their relative intensities?

field
...
19a A radical containing two inequivalent protons with hyperfine

field
...
5 µs, what is the strength of the B1 field? How
long would the corresponding 90° pulse require?

15
...
11 mT, 2
...
89 mT gives a spectrum centred on
332
...
At what fields do the hyperfine lines occur and what are their
relative intensities?

15
...
20a Predict the intensity distribution in the hyperfine lines of the EPR

X–band spectrometer (9 GHz) to observe 1H-NMR and a 300 MHz
spectrometer to observe EPR?

spectra of (a) ·CH3, (b) ·CD3
...
15b The duration of a 90° or 180° pulse depends on the strength of the B1

15
...
What magnetic field is needed to satisfy the resonance
condition?
15
...
12 mT

in a spectrometer operating at 9
...
What is the g-value of the electron
in the atom?
15
...
02 mT in a spectrometer operating at 9
...
What is the g-value
of the electron in the atom?
15
...
The lines occur at 330
...
5 mT, and 334
...
What is the hyperfine coupling constant for each
proton? What is the g-value of the radical given that the spectrometer is
operating at 9
...
18b A radical containing three equivalent protons shows a four–line

spectrum with an intensity distribution 1:3:3:1
...
4 mT,
333
...
8 mT, and 338
...
What is the hyperfine coupling constant

15
...

15
...
0025
...
302 GHz,
(b) 33
...
21b The naphthalene radical anion has g = 2
...
At what field should
you search for resonance in a spectrometer operating at (a) 9
...
88 GHz?
15
...
What is the nuclear spin of the nucleus?
15
...
What is the
spin of the nuclei?
15
...

2

15
...

2

Problems*
Numerical problems
15
...
What field
is required for resonance? What is the relative population difference at room
temperature? Which is the lower energy spin state of the neutron?

spectrum shows five resonance peaks
...
The ratio of the integrated intensities of peak II to peaks I, III,
IV, and V is approximately 10 to 1
...
2 Two groups of protons have δ = 4
...
2 and are interconverted
by a conformational change of a fluxional molecule
...
What is the activation energy of the interconversion?

0°C
-30°C
-60°C

15
...
15
...
10 s
...
4‡ In a classic study of the application of NMR to the measurement of

rotational barriers in molecules, P
...
Nair and J
...
Roberts (J
...
Chem
...
79, 4565 (1957)) obtained the 40 MHz 19F-NMR spectrum of
F2BrCCBrCl2
...
15
...
At 193 K the

III

IV

V
160 Hz

Fig
...
63

* Problems denoted with the symbol ‡ were supplied by Charles Trapp and Carmen Giunta
...
Explain the spectrum and its change with temperature
...

15
...
27) have been used to
correlate data on vicinal proton coupling constants in systems of the type
R1R2CHCHR3R4
...
Karplus, J
...
Chem
...
85,
2870 (1963)), is 3JHH = A cos2 φHH + B
...
3 Hz;
when R3 = CH3 and R4 = H, 3JHH = 8
...
2
Hz
...

15
...
T
...
Mitchell and B
...

Reson
...
33, 325 (1995)) have studied the relation between 3JHH and
3
JSnSn in compounds of the type Me3SnCH2CHRSnMe3 and find that 3JSnSn =
78
...
84 Hz
...
(b) Obtain the Karplus equation for 3JSnSn and
plot it as a function of the dihedral angle
...


four equivalent protons
...
148 mT and a(H) = 0
...

15
...
2 mT with the nucleus
...
7 mT
...
4 mT
...
3 mT
...
12 The hyperfine coupling constants observed in the radical anions (12),
(13), and (14) are shown (in millitesla, mT)
...


15
...
64 shows the proton COSY spectrum of 1-nitropropane
...


NO2CH2CH2CH3

1
2

d
3
4
5
5

4

3 d

2

1

The COSY spectrum of 1-nitropropane (NO2CH2CH2CH3)
...
(Spectrum
provided by Prof
...
Morris
...
15
...
8 The angular NO2 molecule has a single unpaired electron and can be
trapped in a solid matrix or prepared inside a nitrite crystal by radiation
−
damage of NO2 ions
...
64 mT in a spectrometer operating at
9
...
When the field lies along the bisector of the ONO angle, the
resonance lies at 331
...
What are the g-values in the two orientations?
15
...
3 mT
...
3 to predict the splitting between the hyperfine lines of the
spectrum of ·CD3
...
10 The p-dinitrobenzene radical anion can be prepared by reduction of
p-dinitrobenzene
...
13 Calculate σd for a hydrogenic atom with atomic number Z
...
14 In this problem you will use the molecular electronic structure methods
described in Chapter 11 to investigate the hypothesis that the magnitude of
the 13C chemical shift correlates with the net charge on a 13C atom
...
(b) The 13C chemical shifts of the para C atoms in each of the
molecules that you examined in part (a) are given below:

Substituent
δ

OH
130
...
4

H
128
...
9

CN
129
...
4

Is there a linear correlation between net charge and 13C chemical shift of the
para C atom in this series of molecules? (c) If you did find a correlation in part
(b), use the concepts developed in this chapter to explain the physical origins
of the correlation
...


PROBLEMS
15
...
28
...
In gypsum,
for instance, the splitting in the H2O resonance can be interpreted in terms of
a magnetic field of 0
...
What is the separation of the protons in the H2O molecule?
15
...
28
...
The volume
element is sin θ dθ dφ in polar coordinates
...
17 The shape of a spectral line, I(ω), is related to the free induction decay

signal G(t) by
∞

Ύ

I(ω) = a Re

G(t)eiω tdt

0

where a is a constant and ‘Re’ means take the real part of what follows
...

15
...
17, show that, if G(t) = (a cos ω1t
+ b cos ω 2t)e−t/τ, then the spectrum consists of two lines with intensities
proportional to a and b and located at ω = ω1 and ω 2, respectively
...
When chemical exchange is fast, the NMR spectrum of the same
proton in I consists of a single peak with a resonance frequency ν given by
ν = fIνI + fEIνEI, where fI = [I]/([I] + [EI]) and fEI = [EI]/([I] + [EI]) are,
respectively, the fractions of free I and bound I
...
Show that, when the initial concentration of I, [I]0, is much
greater than the initial concentration of E, [E]0, a plot of [I]0 against δν −1 is a
straight line with slope [E]0∆ν and y-intercept −KI
...
24 The molecular electronic structure methods described in Chapter 11
may be used to predict the spin density distribution in a radical
...
2)
...
(a) The phenoxy radical shown in (15) is a
suitable model of the tyrosine radical
...
(b) Predict the form of the EPR spectrum of (15)
...
19 EPR spectra are commonly discussed in terms of the parameters that
occur in the spin-hamiltonian, a hamiltonian operator that incorporates
various effects involving spatial operators (like the orbital angular
momentum) into operators that depend on the spin alone
...
65), the eigenvalues of the spin are
the same as those of the spin-hamiltonian Hspin = −g γeB0 sz (note the g in place
of ge) and find an expression for g
...
20 Interpret the following features of the NMR spectra of hen lysozyme:
(a) saturation of a proton resonance assigned to the side chain of methionine105 changes the intensities of proton resonances assigned to the side chains of
tryptophan-28 and tyrosine-23; (b) saturation of proton resonances assigned
to tryptophan-28 did not affect the spectrum of tyrosine-23
...
21 When interacting with a large biopolymer or even larger organelle,
a small molecule might not rotate freely in all directions and the dipolar
interaction might not average to zero
...
Average the dipolar field
over this restricted range of orientations and confirm that the average vanishes
when θ ′ = π (corresponding to rotation over an entire sphere)
...
15
if it is bound to a biopolymer that enables it to rotate up to θ ′ = 30°?
15
...

15
...
25 Sketch the EPR spectra of the di-tert-butyl nitroxide radical (10) at
292 K in the limits of very low concentration (at which electron exchange is
negligible), moderate concentration (at which electron exchange effects begin
to be observed), and high concentration (at which electron exchange effects
predominate)
...

15
...
What field gradient (in
microtesla per metre, µT m−1) is required to produce a separation of 100 Hz
between two protons separated by the long diameter of a human kidney
(taken as 8 cm) given that they are in environments with δ = 3
...
4 T
...
27 Suppose a uniform disk-shaped organ is in a linear field gradient, and
that the MRI signal is proportional to the number of protons in a slice of
width δx at each horizontal distance x from the centre of the disk
...


16
The distribution of molecular
states
16
...
2 The molecular partition

function
I16
...
3 The internal energy
16
...
5 The canonical ensemble
16
...
7 Independent molecules

Checklist of key ideas
Further reading
Further information 16
...
2:
The Boltzmann formula
Further information 16
...
Two key ideas are introduced in this chapter
...
In this chapter we see its derivation in terms of the distribution of particles over
available states
...
We see how to
interpret the partition function and how to calculate it in a number of simple cases
...
In the final part
of the chapter, we generalize the discussion to include systems that are composed of
assemblies of interacting particles
...


The preceding chapters of this part of the text have shown how the energy levels
of molecules can be calculated, determined spectroscopically, and related to their
structures
...
To do so, we now introduce
the concepts of statistical thermodynamics, the link between individual molecular
properties and bulk thermodynamic properties
...
For example, the pressure of a gas
depends on the average force exerted by its molecules, and there is no need to specify
which molecules happen to be striking the wall at any instant
...
The fluctuations in pressure are very small compared with the steady pressure: it is highly improbable that there will be a sudden lull
in the number of collisions, or a sudden surge
...

This chapter introduces statistical thermodynamics in two stages
...
We can use statistical thermodynamics
once we have deduced the Boltzmann distribution
...
5) we extend
the arguments to systems composed of interacting particles
...
1 CONFIGURATIONS AND WEIGHTS

561

The distribution of molecular states
We consider a closed system composed of N molecules
...
Collisions result in the ceaseless redistribution of energy not only
between the molecules but also among their different modes of motion
...
The populations of the states remain
almost constant, but the precise identities of the molecules in each state may change
at every collision
...
The only restriction is that the molecules should be independent, in the sense that the total energy
of the system is a sum of their individual energies
...
We also adopt the principle of equal a priori probabilities, the assumption that all possibilities for the distribution of energy are equally
probable
...
We have no reason
to presume otherwise than that, for a collection of molecules at thermal equilibrium,
vibrational states of a certain energy, for instance, are as likely to be populated as
rotational states of the same energy
...
That is, statistical thermodynamics provides a molecular justification for the concept of temperature and some insight into this crucially important quantity
...
1 Configurations and weights
Any individual molecule may exist in states with energies ε0, ε1,
...
To obtain the actual internal energy, U, we may have to add a
constant to the calculated energy of the system
...

(a) Instantaneous configurations

At any instant there will be n0 molecules in the state with energy ε0, n1 with ε1, and so
on
...
in the form {n0, n1,
...
The instantaneous
configuration fluctuates with time because the populations change
...
One, for example, might be
{N,0,0,
...
Another
might be {N − 2,2,0,0,
...

The latter configuration is intrinsically more likely to be found than the former
because it can be achieved in more ways: {N,0,0,
...
} can be achieved in –N(N − 1) different ways (Fig
...
1; see
2
Justification 16
...
At this stage in the argument, we are ignoring the requirement
that the total energy of the system should be constant (the second configuration has
a higher energy than the first)
...


Fig
...
1 Whereas a configuration
{5,0,0,
...
} can be
achieved in the ten different ways shown
here, where the tinted blocks represent
different molecules
...
However, 3! of the
selections that put three molecules in the
first receptacle are equivalent, 6! that put
six molecules into the second receptacle are
equivalent, and so on
...

Fig
...
2

Comment 16
...

In eqn 16
...
1, and by definition
0! = 1
...
} and {N − 2,2,0,
...
In other words, a system free to switch
between the two configurations would show properties characteristic almost exclusively of the second configuration
...
} can be achieved
in W different ways, where W is called the weight of the configuration
...
} is given by the expression
W=

N!

(16
...


1
Equation 16
...


Justification 16
...
One candidate for
promotion to an upper state can be selected in N ways
...
However, we
should not distinguish the choice (Jack, Jill) from the choice (Jill, Jack) because they
lead to the same configurations
...

2
Now we generalize this remark
...
The first ball can be selected in N different ways, the next ball
in N − 1 different ways for the balls remaining, and so on
...
1 = N! ways of selecting the balls for distribution over the bins
...
16
...
Similarly, there are
n1! ways in which the n1 balls in the bin labelled ε1 can be chosen, and so on
...
regardless of the order in which the balls were
chosen is N!/n0!n1!
...
1
...
1 Calculating the weight of a distribution

To calculate the number of ways of distributing 20 identical objects with the
arrangement 1, 0, 3, 5, 10, 1, we note that the configuration is {1,0,3,5,10,1} with
N = 20; therefore the weight is
W=

20!
1!0!3!5!10!1!

= 9
...
1 Calculate the weight of the configuration in which 20 objects are
distributed in the arrangement 0, 1, 5, 0, 8, 0, 3, 2, 0, 1
...
19 × 1010]

16
...
We shall therefore need the expression
ln W = ln

N!
n0!n1!n2!
...
One reason for introducing ln W is that it is easier to make approximations
...
2)

1

∑ (ni ln ni − ni) = N ln N − ∑ ni ln ni
i

(16
...
3 is derived by noting that the sum of ni is equal to N, so the
second and fourth terms in the second expression cancel
...
} dominates {N,0,0,
...
We shall see, in fact, that there is a configuration with so
great a weight that it overwhelms all the rest in importance to such an extent that the
system will almost always be found in it
...
This dominating configuration can be found by looking for the values of ni that lead to a maximum value of W
...
However, there are two
difficulties with this procedure
...
This requirement rules out
many configurations; {N,0,0,
...
}, for instance, have different
energies, so both cannot occur in the same isolated system
...
4)

i

where E is the total energy of the system
...
Thus,
increasing the population of one state by 1 demands that the population of another
state must be reduced by 1
...
2

(16
...
1 that the populations in the configuration of
greatest weight, subject to the two constraints in eqns 16
...
5, depend on the
energy of the state according to the Boltzmann distribution:

and is in error by less than 1 per cent
when x is greater than about 10
...
2 is
adequate
...
6a)

∑ e−βε

i

i

where ε0 ≤ ε1 ≤ ε2
...
6a is the justification of the remark that a single
parameter, here denoted β, determines the most probable populations of the states of
the system
...
3b that

β=

1

(16
...
In other
words, the thermodynamic temperature is the unique parameter that governs the most
probable populations of states of a system at thermal equilibrium
...
3, moreover, we see that β is a more natural measure of temperature than T itself
...
2 The molecular partition function
From now on we write the Boltzmann distribution as
pi =

e−βεi

(16
...
8]

i

i

The sum in q is sometimes expressed slightly differently
...
If, for example,
gi states have the same energy εi (so the level is gi-fold degenerate), we could write
q=

∑ gi e−βε

(16
...

Example 16
...

Method To use eqn 16
...
Whenever calculating
a partition function, the energies of the levels are expressed relative to 0 for the state
of lowest energy
...
5c
...
31, the energy levels of a linear rotor are hcBJ(J + 1), with

J = 0, 1, 2,
...
Each level consists of 2J + 1
degenerate states
...
For reasons explained in Section 17
...
2 THE MOLECULAR PARTITION FUNCTION

565

this expression applies only to unsymmetrical linear rotors (for instance, HCl,
not CO2)
...
2 Write the partition function for a two-level system, the lower state
(at energy 0) being nondegenerate, and the upper state (at an energy ε) doubly
degenerate
...
When T is close to zero, the parameter
β = 1/kT is close to infinity
...
The exception is the term with ε0 ≡ 0
(or the g0 terms at zero energy if the ground state is g0-fold degenerate), because then
ε0 /kT ≡ 0 whatever the temperature, including zero
...
10)

T→0

That is, at T = 0, the partition function is equal to the degeneracy of the ground state
...

Because e−x = 1 when x = 0, each term in the sum now contributes 1
...
11)

T→∞

In some idealized cases, the molecule may have only a finite number of states; then the
upper limit of q is equal to the number of states
...
The partition function for such a system can therefore be
2
expected to rise towards 2 as T is increased towards infinity
...
At T = 0, only
the ground level is accessible and q = g0
...


e
0
Fig
...
3 The equally spaced infinite array of
energy levels used in the calculation of the
partition function
...


Comment 16
...


10

q

5
Example 16
...
16
...
These levels can be thought of as the
vibrational energy levels of a diatomic molecule in the harmonic approximation
...
To evaluate eqn 16
...
16
...


0
0

5
kT /e

10

Fig
...
4 The partition function for the
system shown in Fig
...
3 (a harmonic
oscillator) as a function of temperature
...
How does q vary
with temperature when T is high, in the
sense that kT >> ε (or βε < 1)?
<

566

16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
1
...
2

1
...
5

1

10

kT / e

5
kT / e

10

Fig
...
5 The partition function for a two-level system as a function of temperature
...


Exploration Consider a three-level system with levels 0, ε, and 2ε
...


Low
temperature

High
temperature

Self-test 16
...

[q = 1 + e−βε, Fig
...
5]

It follows from eqn 16
...
2 for a uniform ladder of states of spacing ε,
q=

1

(16
...
0

1
...
7

0
...
05

1
...
99

3
...
16
...
16
...
2
...


Exploration To visualize the content
of Fig
...
6 in a different way, plot
the functions p0, p1, p2, and p3 against kT/ε
...
13)

Figure 16
...
At very low temperatures, where q
is close to 1, only the lowest state is significantly populated
...
At the same time, the partition function rises
from 1 and its value gives an indication of the range of states populated
...

The corresponding expressions for a two-level system derived in Self-test 16
...
14)

These functions are plotted in Fig
...
7
...
A common error is to suppose that all the molecules
2
2
in the system will be found in the upper energy state when T = ∞; however, we see

16
...
5

0
...
5
kT / e

0

5
kT / e

10

Fig
...
7 The fraction of populations of the two states of a two-level system as a function of
temperature (eqn 16
...
Note that, as the temperature approaches infinity, the populations
of the two states become equal (and the fractions both approach 0
...


Exploration Consider a three-level system with levels 0, ε, and 2ε
...


from eqn 16
...
The same
conclusion is true of multi-level systems too: as T → ∞, all states become equally
populated
...
3 Using the partition function to calculate a population

Calculate the proportion of I2 molecules in their ground, first excited, and second
excited vibrational states at 25°C
...
6 cm−1
...
9), so the partition function is given by eqn 16
...
13
...
At 298
...
226 cm−1
...
6 cm−1
207
...
036

Then it follows from eqn 16
...
645e−1
...
645, p1 = 0
...
081
...
The value of the
partition function, q = 1
...

Self-test 16
...
16
...
The upper curve shows that
variation of the entropy of a paramagnetic
system in the absence of an applied field
...

The isothermal magnetization step is from
A to B; the adiabatic demagnetization step
(at constant entropy) is from B to C
...
Common methods used to reach very low
temperatures include optical trapping and adiabatic demagnetization
...
Adiabatic demagnetization is based on the fact that, in the absence of a magnetic field, the unpaired electrons of a paramagnetic material are orientated at random, but in the presence of a
1
1
magnetic field there are more β spins (ms = − –) than α spins (ms = + –)
...
Even lower temperatures can be reached if nuclear spins (which also behave
like small magnets) are used instead of electron spins in the technique of adiabatic
nuclear demagnetization, which has been used to cool a sample of silver to about
280 pK
...
3)
...
2 Cooling a sample by adiabatic demagnetization

Consider the situation summarized by Fig
...
8
...
The sample is then exposed to a strong magnetic
field while it is surrounded by helium, which provides thermal contact with the
cold reservoir
...
Thermal contact between the sample and the surroundings is now broken
by pumping away the helium and the magnetic field is reduced to zero
...
At the end of this step the sample is the same as it was at A except that it
now has a lower entropy
...
That is, adiabatic demagnetization has cooled
the sample
...

However, closed approximate expressions can often be found and prove to be very
important in a number of chemical and biochemical applications (Impact 16
...
For
instance, the expression for the partition function for a particle of mass m free to move
in a one-dimensional container of length X can be evaluated by making use of the fact
that the separation of energy levels is very small and that large numbers of states are
accessible at normal temperatures
...
15)

This expression shows that the partition function for translational motion increases
with the length of the box and the mass of the particle, for in each case the separation
of the energy levels becomes smaller and more levels become thermally accessible
...


16
...
2 The partition function for a particle in a one-dimensional box

The energy levels of a molecule of mass m in a container of length X are given by
eqn 9
...


8mX 2

The lowest level (n = 1) has energy h2/8mX 2, so the energies relative to that level are

εn = (n2 − 1)ε

ε = h2/8mX 2

The sum to evaluate is therefore
qX =

∞

∑ e−(n −1)βε
n=1
2

The translational energy levels are very close together in a container the size of a typical laboratory vessel; therefore, the sum can be approximated by an integral:
∞

qX =

Ύ

∞

e−(n −1)βεdn ≈
2

1

Ύe

−n2βε

dn

0

The extension of the lower limit to n = 0 and the replacement of n2 − 1 by n2 introduces negligible error but turns the integral into standard form
...
For instance,
suppose the molecule we are considering is free to move in three dimensions
...
The total energy of a molecule ε is the sum of its translational energies in all three
directions:
(X
(Y
(Z
εn1n2n3 = ε n1 ) + ε n2 ) + ε n3 )

(16
...
Therefore, because ea+b+c = eaebec, the partition function factorizes as
follows:
q=

∑ e−βε

(X)
(Y)
(Z )
n1 −βε n2 −βε n3

all n

=

=

∑ e−βε

e−βε n e−βε n

(X)
n1

(Y)

(Z)

2

3

all n

A
(X)D A
(Y)D A
(Z)D
∑ e−βε n F C ∑ e−βε n F C ∑ e−βε n F
C n
n
n
1

1

2

2

3

(16
...


569

570

16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
Equation 16
...
The only change for the other two directions is to replace the length X by the
lengths Y or Z
...
18)

The product of lengths XYZ is the volume, V, of the container, so we can write
q=

V

Λ

3

A β D
Λ=h
C 2πm F

1/2

=

h

(16
...
The thermal wavelength decreases with increasing mass and temperature
...

Illustration 16
...
016 u; then

Λ=

6
...
016 × 1
...
38 × 10−23 J K−1) × (298 K)}1/2

= 7
...
Therefore,
q=

1
...
12 × 10−11 m)3

= 2
...
Many states are occupied if the thermal wavelength (which
in this case is 71
...

Self-test 16
...


[q = 7
...
19 can be expressed in terms
of the average separation of the particles in the container, d
...
Because q is the total number of accessible states, the
average number of states per molecule is q/N
...
However, V/N is the volume occupied by a single particle, and there>
fore the average separation of the particles is d = (V/N)1/3
...

>
>
That is, for eqn 16
...
For H2 molecules at 1 bar and 298 K, the average separation is 3 nm, which is significantly larger than their thermal wavelength
(71
...
3)
...
1 IMPACT ON BIOCHEMISTRY: THE HELIX–COIL TRANSITION IN POLYPEPTIDES
IMPACT ON BIOCHEMISTRY

I16
...
They are polypeptides formed from different amino
acids strung together by the peptide link, -CONH-
...
The unwinding of a helix
into a random coil is a cooperative transition, in which the polymer becomes increasingly more susceptible to structural changes once the process has begun
...

To calculate the fraction of polypeptide molecules present as helix or coil we need
to set up the partition function for the various states of the molecule
...
We
suppose that conformations hhhh and cccc contribute terms q0 and q4, respectively, to
the partition function q
...
Similarly, each of the six states with
two c amino acids contributes a term q2, and each of the four states with three c amino
acids contributes a term q3
...

That is, we suppose that the difference in energy between c ih4−i and c i+1h3−i has the
same value γ for all i
...
20)

where Γ = NAγ and s is called the stability parameter
...

q
q0

4

=

∑ C(4,i)s i
i=0

with C(4,i) =

4!
(4 − i)!i!

(16
...

The extension of this treatment to take into account a longer chain of residues is
now straightforward: we simply replace the upper limit of 4 in the sum by n:
q
q0

n

=

∑ C(n,i)s i

Comment 16
...
22)

i=0

A cooperative transformation is more difficult to accommodate, and depends on
building a model of how neighbours facilitate each other’s conformational change
...
Thus, the zipper model allows
a transition of the type
...
hhhcc
...
hhhch
...
The only exception to this rule is, of course, the very
first conversion from h to c in a fully helical chain
...
Each subsequent step is called a propagation step and has a stability
parameter s
...
24, you are invited to show that the partition function is:

0
...
82

n

q=1+

(16
...
1

∑ Z(n,i)σsi

where Z(n,i) is the number of ways in which a state with a number i of c amino acids
can be formed under the strictures of the zipper model
...
24),

1
...
05

i=1

∑ si − σ ∑ is i

(16
...
8 (͗i͘ = 1
...
0 (͗i͘ = 3
...
5 (͗i͘ = 15
...
0 × 10−3
...
5

0
...
0001
0
1
...
The
curves show the sigmoidal shape
characteristics of cooperative behaviour
...
16
...
25)

This is a general result that applies to any model of the helix–coil transition in which
the partition function q is expressed as a function of the stability parameter s
...
Calculations based on this more complete
Zimm–Bragg model give

0
...
Figure 16
...
0 × 10−3
...
When s = 1, there is a more widespread distribution of length of random coil segments
...
24) that

θ=

0
...
16
...
26)

Figure 16
...
The curves show the
sigmoidal shape characteristic of cooperative behaviour
...
That is,
the harder it is to get coil formation started, the sharper the transition from helix to coil
...
3 THE INTERNAL ENERGY

573

The internal energy and the entropy
The importance of the molecular partition function is that it contains all the information needed to calculate the thermodynamic properties of a system of independent
particles
...

0
...


E/Ne

16
...
2

The total energy of the system relative to the energy of the lowest state is
E=

∑ ni εi

(16
...
28)

0
...
4

d

N d

N dq

∑ dβ e−βε = − q dβ ∑ e−βε = − q dβ
i

q

5
kT /e

0
...
6

To manipulate this expression into a form involving only q we note that

E=−

0
0

i

i

(16
...
2
Illustration 16
...
16
...
Notice how the energy is zero at T = 0, when
1
only the lower state (at the zero of energy) is occupied, and rises to – Nε as T → ∞,
2
when the two levels become equally populated
...
29 that need to be made
...
Therefore, to obtain the conventional internal energy U, we must add the
internal energy at T = 0:
U = U(0) + E

(16
...
16
...
The graph at the top
shows the slow rise away from zero energy
at low temperatures; the slope of the graph
at T = 0 is 0 (that is, the heat capacity is
zero at T = 0)
...
5 as T → ∞ as both states
become equally populated (see Fig
...
7)
...
16
...


574

16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
Secondly, because the partition function may depend on variables other than the
temperature (for example, the volume), the derivative with respect to β in eqn 16
...
The complete
expression relating the molecular partition function to the thermodynamic internal
energy of a system of independent molecules is therefore
U = U(0) −

N A ∂q D

(16
...
31b)

These two equations confirm that we need know only the partition function (as a
function of temperature) to calculate the internal energy relative to its value at T = 0
...
To do so, we compare the equipartition expression for
the internal energy of a monatomic perfect gas, which from Molecular interpretation
2
...
32a)

with the value calculated from the translational partition function (see the following
Justification), which is
U = U(0) +

3N

(16
...
33)

(We have used N = nNA, where n is the amount of gas molecules, NA is Avogadro’s
constant, and R = NAk
...
1 and Further reading)
...
3 The internal energy of a perfect gas

To use eqn 16
...
19:
A ∂q D
A ∂ VD
d 1
V dΛ
B E =B
E =V
= −3 4
3
3
dβ Λ
Λ dβ
C ∂β F V C ∂β Λ F V
Then we note from the formula for Λ in eqn 16
...
4 THE STATISTICAL ENTROPY
Then, by eqn 16
...
32b
...
4 The statistical entropy
If it is true that the partition function contains all thermodynamic information, then
it must be possible to use it to calculate the entropy as well as the internal energy
...
2) that entropy is related to the dispersal of energy
and that the partition function is a measure of the number of thermally accessible
states, we can be confident that the two are indeed related
...
In Further information 16
...
34]

In this expression, W is the weight of the most probable configuration of the system
...

The statistical entropy behaves in exactly the same way as the thermodynamic
entropy
...
In the limit T → 0,
W = 1, so ln W = 0, because only one configuration (every molecule in the lowest level)
is compatible with E = 0
...
4)
...

To do so, we substitute the expression for ln W given in eqn 16
...
34 and,
as shown in the Justification below, obtain
S=

U − U(0)
T

+ Nk ln q

(16
...
4 The statistical entropy

The first stage is to use eqn 16
...
It follows from eqn 16
...
27
and 16
...
35 immediately follows
...
4 Calculating the entropy of a collection of oscillators

S /Nk

Calculate the entropy of a collection of N independent harmonic oscillators, and
evaluate it using vibrational data for I2 vapour at 25°C (Example 16
...

Method To use eqn 16
...
12
...
31a), and the two
expressions then combined to give S
...
12 is

q=

0
0

1
1 − e−βε

The internal energy is obtained by using eqn 16
...
16
...
The
entropy approaches zero as T → 0, and
increases without limit as T → ∞
...
16
...
Is there a
temperature at which this coefficient passes
through a maximum? If you find a
maximum, explain its physical origins
...
16
...
For I2 at 25°C, βε = 1
...
3), so
Sm = 8
...

Self-test 16
...
What is the entropy when the two states are equally thermally
accessible?
[S/Nk = βε /(1 + eβε ) + ln(1 + e−βε ); see Fig
...
13; S = Nk ln 2]

1

S /Nk

S /Nk

1

ln 2

ln 2

0
...
5

Fig
...
13 The temperature variation of the
entropy of a two-level system (expressed as
a multiple of Nk)
...


Exploration Draw graphs similar to
those in Fig
...
13 for a three-level
system with levels 0, ε, and 2ε
...
5
kT / e

1

0

0

5
kT / e

10

16
...
We shall also see how to obtain the molecular partition function from the more general form of the partition function developed here
...
5 The canonical ensemble
The crucial new concept we need when treating systems of interacting particles is the
‘ensemble’
...

(a) The concept of ensemble

To set up an ensemble, we take a closed system of specified volume, composition, and
temperature, and think of it as replicated Ñ times (Fig
...
14)
...
The total energy of all the systems is L and, because they are in thermal
equilibrium with one another, they all have the same temperature, T
...

The word ‘canon’ means ‘according to a rule’
...
In the microcanonical ensemble the condition of constant temperature is
replaced by the requirement that all the systems should have exactly the same energy:
each system is individually isolated
...
The number of members of the
ensemble in a state with energy Ei is denoted ñi, and we can speak of the configuration
of the ensemble (by analogy with the configuration of the system used in Section 16
...
Note that Ñ is unrelated to N, the number of molecules in the
actual system; Ñ is the number of imaginary replications of that system
...
1, some of the configurations of the ensemble will be very much
more probable than others
...
By analogy with the earlier discussion, we
can anticipate that there will be a dominating configuration, and that we can evaluate
the thermodynamic properties by taking the average over the ensemble using that
single, most probable, configuration
...

The quantitative discussion follows the argument in Section 16
...
The weight of a configuration {ñ0,ñ1,
...
16
...
The
individual replications of the actual system
all have the same composition and volume
...
Energy
may be transferred between them as heat,
and so they do not all have the same
energy
...


578

16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS

Energy

M=
Width of
range
Number of
states

(16
...


The configuration of greatest weight, subject to the constraints that the total energy of
the ensemble is constant at L and that the total number of members is fixed at Ñ, is
given by the canonical distribution:
ñi

The energy density of states is the
number of states in an energy range divided
by the width of the range
...
16
...
37)

i

i

The quantity Q, which is a function of the temperature, is called the canonical partition function
...
16
...
39, by the
number of states corresponding to that
energy (a steeply rising function)
...


The canonical distribution in eqn 16
...
We must appreciate that eqn 16
...
There may in fact be numerous states with almost identical energies
...
The density of states, the number of
states in an energy range divided by the width of the range (Fig
...
15), is a very
sharply increasing function of energy
...
37, a sharply decreasing function, multiplied by a sharply increasing
function (Fig
...
16)
...

We conclude that most members of the ensemble have an energy very close to the
mean value
...
6 The thermodynamic information in the partition function
Like the molecular partition function, the canonical partition function carries all the
thermodynamic information about a system
...
We can therefore use
Q to discuss the properties of condensed phases and real gases where molecular interactions are important
...
We use this quantity to calculate the internal energy of the system in the limit of Ñ (and L) approaching infinity:
U = U(0) + E = U(0) + L/Ñ

as

Ñ→∞

(16
...
7 as
"i =

e−βEi

(16
...
40)

16
...
31,
U = U(0) −

1 A ∂Q D

= U(0) −

Q C ∂β F V

A ∂ ln Q D
C ∂β F V

(16
...
Hence, we can calculate S from
S = k ln W = k ln M 1/Ñ =

k
Ñ

ln M

(16
...
4, that
S=

U − U(0)
T

+ k ln Q

(16
...
7 Independent molecules
We shall now see how to recover the molecular partition function from the more
general canonical partition function when the molecules are independent
...
44)

Justification 16
...
Therefore, we can write the total energy of a state i of the
system as
Ei = εi(1) + εi(2) + · · · + εi(N)
In this expression, εi(1) is the energy of molecule 1 when the system is in the state i,
εi(2) the energy of molecule 2 when the system is in the same state i, and so on
...

Therefore, instead of summing over the states i of the system, we can sum over all
the individual states i of molecule 1, all the states i of molecule 2, and so on
...
Suppose that molecule 1 is in some
state a, molecule 2 is in b, and molecule 3 is in c, then one member of the ensemble
has an energy E = εa + εb + εc
...
There are six such permutations in all, and N! in

579

580

16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
general
...
The detailed argument is quite
involved, but at all except very low temperatures it turns out that the correction factor
is 1/N!
...
45a)

Q = q /N!
N

(16
...
Their identity, however, is not the only
criterion
...
Identical molecules in a lattice can therefore be treated as distinguishable because their sites are distinguishable, and we use eqn 16
...
On the
other hand, identical molecules in a gas are free to move to different locations, and
there is no way of keeping track of the identity of a given molecule; we therefore use
eqn 16
...

(b) The entropy of a monatomic gas

An important application of the previous material is the derivation (as shown in the
Justification below) of the Sackur–Tetrode equation for the entropy of a monatomic
gas:
S = nR ln

A e5/2V D
C nNAΛ3 F

Λ=

h
(2πmkT)1/2

(16
...
Because the gas is perfect, we can
use the relation V = nRT/p to express the entropy in terms of the pressure as
S = nR ln

A e5/2kT D
C pΛ3 F

(16
...
6 The Sackur–Tetrode equation

For a gas of independent molecules, Q may be replaced by q N/N!, with the result
that eqn 16
...
2) to write
S=

U − U(0)
T

+ nR ln q − nR ln N + nR

The only mode of motion for a gas of atoms is translation, and the partition function is q = V/Λ3 (eqn 16
...
The internal energy
is given by eqn 16
...
46
...
5 Using the Sackur–Tetrode equation

Calculate the standard molar entropy of gaseous argon at 25°C
...
46b, divide both sides by
7
n
...
95 u
...
0 pm (by the same kind of calculation as in Illustration 16
...
Therefore,

1

7
S m = R ln 2

e5/2 × (4
...
60 × 10
5

5
6 = 18
...
60R at 298 K
...
7 Calculate the translational contribution to the standard molar
entropy of H2 at 25°C
...
2R]

Fig
...
17 As the width of a container is
increased (going from (a) to (b)), the
energy levels become closer together (as
1/L2), and as a result more are thermally
accessible at a given temperature
...


The Sackur–Tetrode equation implies that, when a monatomic perfect gas expands
isothermally from Vi to Vf , its entropy changes by
∆S = nR ln(aVf) − nR ln(aVi) = nR ln

Vf
Vi

(16
...
46a
...
1)
...
16
...


Checklist of key ideas
1
...
of the
energy levels ε0, ε1,
...

2
...

3
...

4
...

5
...


6
...

7
...

8
...

The canonical partition function, Q = ∑i e−βEi
...
The internal energy and entropy of an ensemble are,
respectively, U = U(0) − (∂ ln Q/∂β )V and S = {U − U(0)}/T
+ k ln Q
...
For distinguishable independent molecules we write Q = q N
...

11
...
46, is an expression for
the entropy of a monatomic gas
...
Chandler, Introduction to modern statistical mechanics
...

D
...
McQuarrie and J
...
Simon, Molecular thermodynamics
...


K
...
van Holde, W
...
Johnson, and P
...
Ho, Principles of physical
biochemistry
...

J
...
J
...
Educ
...


Further information
Further information 16
...
1 that ln W is easier to handle than W
...
Because ln W depends on all the ni, when a
configuration changes and the ni change to ni + dni, the function
ln W changes to ln W + d ln W, where
d ln W =

A ∂ lnW D
E dni
∂ni F

∑ dni = 0
i

d ln W =

=

∑
i

∑ dni − β ∑ εi dni
i

5
1 A ∂ ln W D
E + α − βεi 6 dni
∂ni F
7

∑2 B
i 3C

+ α − βεi = 0

i

(16
...

Differentiation of ln W as given in eqn 16
...
48)

The first constraint recognizes that the total energy must not change,
and the second recognizes that the total number of molecules must
not change
...

The way to take constraints into account was devised by the French
mathematician Lagrange, and is called the method of undetermined
multipliers
...
All we need
here is the rule that a constraint should be multiplied by a constant
and then added to the main variation equation
...

We employ the technique as follows
...
48 are multiplied by the constants −β and α, respectively (the
minus sign in −β has been included for future convenience), and then
added to the expression for d ln W:
A ∂ ln W D
B
E dn + α
C ∂ni F i

∂ni

∂ni

All this expression states is that a change in ln W is the sum of
contributions arising from changes in each value of ni
...
However, when the ni change, they do so
subject to the two constraints

∑ εidni = 0
i

∂ ln W

∂ ln W

∑B
C
i

All the dni are now treated as independent
...
The second term on the right in the second
line arises because ∂(ln N)/∂ni = (1/N)∂N/∂ni
...

For the derivative of the second term we first note that
∂ ln nj

=

∂ni

1 A ∂nj D
B
E
nj C ∂ni F

Morever, if i ≠ j, nj is independent of ni, so ∂nj /∂ni = 0
...
Then
∂(nj ln nj)

A ∂ ln nj D 5
E6
∂ni F 7

1 A ∂n D

∑ 2 B ∂nij E ln nj + nj B
F
C
j 3C

=

∑ 2 B ∂nij E ln nj + B ∂nij E 6
F
C
F7
j 3C

=

∑ B ∂nij E (ln nj + 1)
F
j C

=

∂ni

=

∑ δij(ln nj + 1) = ln ni + 1
j

A ∂n D 5

1 A ∂n D
A ∂n D

Heat

∑
j

and therefore
∂ ln W
∂ni

= −(ln ni + 1) + (ln N + 1) = −ln

ni

(a)

N

It follows from eqn 16
...
50)

∑ e−βε
j

j

(b)

and
ni
N

α −βεi

=e

α −βεi

=e e

=

1

∑ e−βε
j

Fig
...
18 (a) When a system is heated, the energy levels are
unchanged but their populations are changed
...

The levels in this case are the one-dimensional particle-in-a-box
energy levels of Chapter 9: they depend on the size of the container
and move apart as its length is decreased
...
6a
...
2 The Boltzmann formula

A change in the internal energy
U = U(0) +

∑ ni εi
i

(16
...
The most general change
is therefore
dU = dU(0) +

∑ ni dεi + ∑ εi dni
i
i

(16
...
16
...
43) that
under the same conditions
dU = dqrev = TdS
Therefore,
dS =

dU
T

= kβ

∑ εi dni
i

(16
...
49 to

βεi =

∂ ln W
∂ni

+α

16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS

and find that
dS = k

A ∂ ln W D
E dni + kα
∂ni F

∑B
i C

∑ dni
i

But because the number of molecules is constant, the sum over the
dni is zero
...
54)

where ε is the separation of the upper state N+ and the lower state N−
...
Indeed, for a general population,
T=

ε /k
ln(N− /N+)

(16
...

All the statistical thermodynamic expressions we have derived
apply to T < 0 as well as to T > 0, the difference being that states with
T < 0 are not in thermal equilibrium and therefore have to be
achieved by techniques that do not rely on the equalization of
temperatures of the system and its surroundings
...
However, it is possible to circumvent this
restriction in systems that have a finite number of levels or in
systems that are effectively finite because they have such weak
coupling to their surroundings
...
Pulse techniques in
NMR can achieve non-equilibrium populations (Section 15
...
5)
...

The expressions for q, U, and S that we have derived in this chapter
are applicable to T < 0 as well as to T > 0, and are shown in Fig
...
19
...
The entropy S is continuous at T = 0
...
16
...

Note that U → 0 as β → ∞ (that is, as T → 0, when only the lower
state is occupied) and U → Nε as β → −∞ (that is, as T → −0);

1

0
...
, S/Nk

Further information 16
...
34
...
5

-10

Fig
...
19 The partition function, internal energy, and entropy of a
two-level system extended to negative temperatures
...
The
entropy of the system is zero on either side of T = 0, and rises to
Nk ln 2 as T → ±∞
...

We get more insight into the dependence of thermodynamic
properties on temperature by noting the thermodynamic result
(Section 3
...
When S is plotted against U for a
two-level system (Fig
...
21), we see that the entropy rises as energy
is supplied to the system (as we would expect) provided that T > 0
(the thermal equilibrium regime)
...
This conclusion is consistent with
the thermodynamic definition of entropy, dS = dqrev /T (where, of
course, q denotes heat and not the partition function)
...
5

0

0
...
16
...


0
...
, S /Nk

585

0
...
16
...


to the shift towards population of the upper state alone as more
energy is packed into the system
...
The First Law (in essence, the
conservation of energy) is robust, and independent of how
populations are distributed over states
...

The efficiency of heat engines (Section 3
...

However, if the temperature of the cold reservoir is negative, then
the efficiency of the engine may be greater than 1
...

Alternatively, an efficiency greater than 1 implies that heat can be
converted completely into work provided the heat is withdrawn from
a reservoir at T < 0
...
The Third Law requires a slight amendment
on account of the discontinuity of the populations across T = 0: it is
impossible in a finite number of steps to cool any system down to
+0 or to heat any system above −0
...
1 Describe the physical significance of the partition function
...
2 Explain how the internal energy and entropy of a system composed of
two levels vary with temperature
...
3 Enumerate the ways by which the parameter β may be identified with

1/kT
...
4 Distinguish between the zipper and Zimm–Bragg models of the
helix–coil transition
...
5 Explain what is meant by an ensemble and why it is useful in statistical
thermodynamics
...
6 Under what circumstances may identical particles be regarded as
distinguishable?

586

16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS

Exercises
16
...
1b What is the temperature of a two-level system of energy separation

equivalent to 300 cm−1 when the population of the upper state is one-half that
of the lower state?

16
...
00 cm3
...
2b Calculate (a) the thermal wavelength, (b) the translational partition

function of an Ar atom in a cubic box of side 1
...

16
...

16
...


16
...
Deduce an expression for the partition function and
mean energy of the nucleus and sketch the variation of the functions with B
...
0 K, (b) 298 K
when B = 20
...

16
...
Calculate (a) the ratio of populations in the two states at (1) 1
...

16
...
0 K
...
00 K,
(2) 25
...
0 K,
(c) the molar energy at 25
...
0 K, (e) the
molar entropy at 25
...


16
...
Calculate the partition function of these electronic states at
1900 K
...
9a At what temperature would the population of the first excited
vibrational state of HCl be 1/e times its population of the ground state?

16
...
9b At what temperature would the population of the first excited

electronically excited level at 1250 cm−1, and a doubly degenerate level at
1300 cm−1
...


16
...
4a
...
5b Calculate the electronic contribution to the molar internal energy at

2000 K for a sample composed of the atoms specified in Exercise 16
...

16
...
At what temperature will 10 per cent
of the molecules be in the upper state?
16
...
At what temperature
will 15 per cent of the molecules be in the upper state?

16
...
Deduce an
expression for the partition function and mean energy of the electron and
sketch the variation of the functions with B
...
0 K, (b) 298 K when B = 1
...


rotational level of HCl be 1/e times its population of the ground state?
16
...
15 K
...
10b Calculate the standard molar entropy of xenon gas at (a) 100 K,
(b) 298
...

16
...


16
...

16
...

16
...


Problems*
16
...
00 × 1022 4He atoms in a box of dimensions 1
...
0 cm

Numerical problems
16
...
What is the number of configurations
available to the combined system? Also, compute the entropies S, S1, and S2
...
0 cm
...
0 mK, 2
...
0 K
...
What conclusions might you draw from the
results of your calculations?

* Problems denoted with the symbol ‡ were supplied by Charles Trapp and Carmen Giunta
...
3‡ By what factor does the number of available configurations increase
when 100 J of energy is added to a system containing 1
...
4‡ By what factor does the number of available configurations increase
when 20 m3 of air at 1
...
0010 per
cent at constant temperature?
16
...
00 cm
...

16
...
In an atomic
beam study of the atoms it was observed that 30 per cent of the atoms were in
the upper level, and the translational temperature of the beam was 300 K
...
7 (a) Calculate the electronic partition function of a tellurium atom at
(i) 298 K, (ii) 5000 K by direct summation using the following data:

Term

Degeneracy

Wavenumber/cm−1

Ground

5

0

1

1

4 707

2

3

4 751

3

5

10 559

(b) What proportion of the Te atoms are in the ground term and in the term
labelled 2 at the two temperatures? (c) Calculate the electronic contribution to
the standard molar entropy of gaseous Te atoms
...
8 The four lowest electronic levels of a Ti atom are: 3F2, 3F3, 3F4, and 5F1,
at 0, 170, 387, and 6557 cm−1, respectively
...
The boiling point of titanium is 3287°C
...
The degeneracies
of the levels are 2J + 1
...
9 The NO molecule has a doubly degenerate excited electronic level

121
...
Calculate
and plot the electronic partition function of NO from T = 0 to 1000 K
...
Calculate the electronic contribution
to the molar entropy of the NO molecule at 300 K and 500 K
...
10‡ J
...
Musgrove (J
...
Chem
...
Data 22, 1213 (1993))

have published tables of energy levels for germanium atoms and cations from
Ge+ to Ge+31
...
1

P0

E/cm−1

P1

3

1

1

1410
...
3

16 367
...
Hint
...

16
...
30, 425
...
27, 845
...
What proportion of I2 molecules are in the ground and
first two excited levels at the two temperatures? Calculate the vibrational
contribution to the molar entropy of I2 at the two temperatures
...
12‡ (a) The standard molar entropy of graphite at 298, 410, and 498 K is
5
...
03, and 11
...
If 1
...
00 mol C(graphite)

587

at 498 K, also insulated, how many configurations are there altogether for the
combined but independent systems? (b) If the same two samples are now
placed in thermal contact and brought to thermal equilibrium, the final
temperature will be 410 K
...
(c) Demonstrate that this process is
spontaneous
...
13 A sample consisting of five molecules has a total energy 5ε
...
(a) Calculate
the weight of the configuration in which the molecules are distributed evenly
over the available states
...
Calculate the weights of each configuration
and identify the most probable configurations
...
14 A sample of nine molecules is numerically tractable but on the verge of
being thermodynamically significant
...
13)
...
Go on to calculate the weights and
identify the most probable configuration
...
15 The most probable configuration is characterized by a parameter we
know as the ‘temperature’
...
13 and 16
...
(a) Show that
the temperature can be obtained by plotting pj against j, where pj is the
(most probable) fraction of molecules in the state with energy jε
...
14
...

16
...
The energy of the triplet exceeds that of the
singlet by ε
...

(b) Find expressions in terms of ε for the molar energy, molar heat capacity,
and molar entropy of such molecules and calculate their values at T = ε /k
...
17 Consider a system with energy levels εj = jε and N molecules
...
(b) Calculate the molecular partition function q for the
system when its mean energy is aε
...

16
...
What difference would it make if (a) a cruder
approximation, N! = N N, (b) the better approximation in Comment 16
...
19‡ For gases, the canonical partition function, Q, is related to the

molecular partition function q by Q = q N/N!
...


588

16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS

Applications: to atmospheric science, astrophysics,
and biochemistry
16
...
27) from the Boltzmann

distribution
...
Convert the barometric formula from pressure to
number density, N
...
0 km, a typical cruising altitude for commercial aircraft
...
21‡ Planets lose their atmospheres over time unless they are replenished
...
Prove that the atmosphere of planets cannot be in an equilibrium state
by demonstrating that the Boltzmann distribution leads to a uniform finite
number density as r → ∞
...
Recall that in a gravitational field the potential
energy is V(r) = −GMm/r, where G is the gravitational constant, M is the mass
of the planet, and m the mass of the particle
...
22‡ Consider the electronic partition function of a perfect atomic

hydrogen gas at a density of 1
...
These are the
mean conditions within the Sun’s photosphere, the surface layer of the Sun
that is about 190 km thick
...

(b) Develop a theoretical argument for truncating the sum and estimate the
maximum number of quantum states that contribute to the sum
...
Are there any general implications concerning electronic states
that will be observed for other atoms and molecules? Is it wise to apply these
calculations in the study of the Sun’s photosphere?
16
...
Show that the possible varieties (configurations) of the
species PL i (with PL 0 denoting P) are given by the binomial coefficients C(4,i)
...
24 Complete some of the derivations in the discussion of the helix–coil
transition in polypeptides (Impact I16
...
(a) Show that, within the tenets of
the zipper model,
n

q=1+

∑ Z(n,i)σs

i

i=1

and that Z(n,i) = n − i + 1 is the number of ways in which an allowed state with
a number i of c amino acids can be formed
...
Hint
...

16
...
1 to explore
the helix–coil transition in polypeptides
...
8, 1
...
5, with σ = 5
...
Discuss
the significance of any effects you discover
...
Use the results of the zipper model to calculate ͗i͘ for all the
combinations of s and σ used in Fig
...
10 and part (a)
...
First, we establish the relations between thermodynamic
functions and partition functions
...
These contributions can be calculated from spectroscopic
data
...
In the final section, we
see how to calculate the equilibrium constant of a reaction and through that calculation
understand some of the molecular features that determine the magnitudes of equilibrium
constants and their variation with temperature
...
1 The thermodynamic functions
17
...
3 Mean energies
17
...
5 Equations of state

A partition function is the bridge between thermodynamics, spectroscopy, and
quantum mechanics
...
It
also sheds light on the significance of these properties
...
6 Molecular interactions in

liquids
17
...
8 Equilibrium constants

Checklist of key ideas

Fundamental relations

Further reading
Discussion questions

In this section we see how to obtain any thermodynamic function once we know the
partition function
...

17
...
1)

where β = 1/kT
...
All the thermodynamic functions introduced in Part 1
are related to U and S, so we have a route to their calculation from Q
...
This relation implies that A(0) =
U(0), so substitution for U and S by using eqn 17
...
2)

(b) The pressure

By an argument like that leading to eqn 3
...
Therefore, on imposing constant temperature, the pressure and the
Helmholtz energy are related by p = −(∂A/∂V)T
...
2 that
p = kT

A ∂ ln Q D
C ∂V F T

(17
...
Because Q is in general a function of the volume,
temperature, and amount of substance, eqn 17
...

Example 17
...

Method We should suspect that the pressure is that given by the perfect gas law
...
45 and Table 17
...
3
...
The calculation shows that the equation of state of a gas
of independent particles is indeed the perfect gas law
...
1 Derive the equation of state of a sample for which Q = q Nf/N!, with

q = V/Λ3, where f depends on the volume
...
4)

17
...
32a), and have just shown that pV = nRT
...
5)°

(d) The Gibbs energy

One of the most important thermodynamic functions for chemistry is the Gibbs
energy, G = H − TS = A + pV
...
6)

This expression takes a simple form for a gas of independent molecules because pV in
the expression G = A + pV can be replaced by nRT:
G − G(0) = −kT ln Q + nRT

(17
...
8)°

with N = nNA
...

It will turn out to be convenient to define the molar partition function, qm = q/n
(with units mol−1), for then
G − G(0) = −nRT ln

qm
NA

(17
...
2 The molecular partition function
The energy of a molecule is the sum of contributions from its different modes of
motion:

εi = ε T + ε R + ε V + ε E
i
i
i
i

(17
...
The electronic contribution is not actually a ‘mode of motion’, but it is convenient to include it here
...
10 is only approximate
(except for translation) because the modes are not completely independent, but in
most cases it is satisfactory
...
The
separation of the vibrational and rotational modes is justified to the extent that the
rotational constant is independent of the vibrational state
...
2b):

591

592

17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
q=

∑ e−βε = ∑ e−βε −βε −βε −βε
T
i

i

i

R
i

V
i

E
i

i (all states)

∑

=

∑

∑

∑ e−βε −βε −βε −βε
T
i

R
i

V
i

E
i

(17
...

(a) The translational contribution

The translational partition function of a molecule of mass m in a container of volume
V was derived in Section 16
...
12)

(2πmkT)1/2

Notice that qT → ∞ as T → ∞ because an infinite number of states becomes accessible
as the temperature is raised
...

The thermal wavelength, Λ, lets us judge whether the approximations that led to the
expression for qT are valid
...
That will be so if Λ is small compared with the linear
dimensions of the container
...
For O2, a heavier molecule, Λ = 18 pm
...
2 that an
equivalent criterion of validity is that Λ should be much less than the average separation of the molecules in the sample
...
1, the partition function of a nonsymmetrical (AB)
linear rotor is
qR =

2

∑(2J + 1)e−βhcBJ( J+1)

(17
...


1

Example 17
...
17
...
The vertical axis is the value
of (2J + 1)e−βhcBJ( J+1)
...


Evaluate the rotational partition function of 1H35Cl at 25°C, given that B =
10
...

Method We use eqn 17
...
A useful relation is kT/hc =
207
...
15 K
...

Answer To show how successive terms contribute, we draw up the following table

by using kT/hcB = 0
...
17
...
0511J( J+1)

(2J + 1)e

1

2

3

4


...
71

3
...
79

3
...


0
...
2 THE MOLECULAR PARTITION FUNCTION
The sum required by eqn 17
...
9, hence qR = 19
...
Taking J up to 50 gives qR =
19
...
Notice that about ten J-levels are significantly populated but the number of
populated states is larger on account of the (2J + 1)-fold degeneracy of each level
...
6, in good agreement with the exact value and with much less
work
...
2 Evaluate the rotational partition function for HCl at 0°C
...
26]

At room temperature kT/hc ≈ 200 cm−1
...
2) and often smaller (though the very light H2 molecule,
for which B = 60
...
It follows that many rotational levels are
populated at normal temperatures
...
14a)

hcB
3/2

R

A π D
C ABC F

1/2

(17
...
However, before using
these expressions, read on (to eqns 17
...
16)
...
1 The rotational contribution to the molecular partition function

When many rotational states are occupied and kT is much larger than the separation
between neighbouring states, the sum in the partition function can be approximated by an integral, much as we did for translational motion in Justification 16
...
14a
...
First, we note that the energies of a symmetric rotor are
EJ,K,M = hcBJ(J + 1) + hc(A − B)K2
J

593

594

17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS

K = -J

K = +J

K = -5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5

J
5
4

with J = 0, 1, 2,
...
, −J, and MJ = J, J − 1,
...
Instead of considering these ranges, we can cover the same values by allowing K to range from −∞
to ∞, with J confined to |K|, |K| + 1,
...
17
...
Because
the energy is independent of MJ, and there are 2J + 1 values of MJ for each value of J,
each value of J is 2J + 1-fold degenerate
...

K = -1, J = 1, 2, 3,
...

K = +1, J = 1, 2, 3,
...


J = |K |

=

∞

∑ ∑ (2J + 1)e−hc {BJ( J+1)+(A−B)K }/kT

=

5
4

JKMJ /kT

can be written equivalently as

1

J

J

J= 0 K= −J MJ = −J

2

J = |K |

J

2

K=−∞ J=|K |
∞
∞
−{hc (A−B)/kT }K 2

∑e

∑ (2J + 1)e−hcBJ( J+1)/kT

J=|K |

K=−∞

Now we assume that the temperature is so high that numerous states are occupied
and that the sums may be approximated by integrals
...

and K = J, J − 1,
...
, ∞ for each value of K
...
17
...
Now we can
>
write

Synoptic table 17
...
4

309
1997

0
...
14b
...
053

ν1
ν2

CO2

=

hcB

∞

5
6
7

q=

kT

* For more values, see Table 13
...
439 K cm
...

Then ‘high temperature’ means T > θR and under these conditions the rotational
>
partition function of a linear molecule is simply T/θR
...
1
...


17
...
The large value of qR reflects the closeness in
energy (compared with kT) of the rotational levels in large, heavy molecules, and the
large number of them that are accessible at normal temperatures
...

For a homonuclear diatomic molecule or a symmetrical linear molecule (such as CO2
or HC
...
Hence, the number of thermally accessible states is only half the number
that can be occupied by a heteronuclear diatomic molecule, where rotation through
180° does result in a distinguishable state
...
15a)

2θ R

The equations for symmetrical and nonsymmetrical molecules can be combined into
a single expression by introducing the symmetry number, σ, which is the number of
indistinguishable orientations of the molecule
...
15b)

σθ R

For a heteronuclear diatomic molecule σ = 1; for a homonuclear diatomic molecule
or a symmetrical linear molecule, σ = 2
...
2 The origin of the symmetry number

01

J

Fig
...
3 The values of the individual terms
(2J + 1)e−βhcBJ( J+1) contributing to the
mean partition function of a 3:1 mixture of
ortho- and para-H2
...
At high
temperatures, the sum is approximately
equal to the sum of the terms over all
1
values of J, each with a weight of –
...


The quantum mechanical origin of the symmetry number is the Pauli principle,
which forbids the occupation of certain states
...
8, for example,
that H2 may occupy rotational states with even J only if its nuclear spins are
paired (para-hydrogen), and odd J states only if its nuclear spins are parallel (orthohydrogen)
...

To set up the rotational partition function we note that ‘ordinary’ molecular
hydrogen is a mixture of one part para-H2 (with only its even-J rotational states
occupied) and three parts ortho-H2 (with only its odd-J rotational states occupied)
...
17
...
From
the illustration we see that we would obtain approximately the same answer for the
partition function (the sum of all the populations) if each J term contributed half its
normal value to the sum
...

The same type of argument may be used for linear symmetrical molecules in
which identical bosons are interchanged by rotation (such as CO2)
...
8, if the nuclear spin of the bosons is 0, then only even-J states are
admissible
...
17
...


0 2

J

Fig
...
4 The relative populations of the
rotational energy levels of CO2
...
The full
line shows the smoothed, averaged
population of levels
...
2* Symmetry
numbers
s

Molecule
H2O

2

NH3

3

CH4

12

C6H6

q =
R

12

3/2

σ C hc F

A π D
C ABC F

1/2

(17
...
2
...
In NH3, there are three indistinguishable orientations around the axis shown in (1)
...
For benzene, any of six orientations
around the axis perpendicular to the plane of the molecule leaves it apparently
unchanged, as does a rotation of 180° around any of six axes in the plane of the
molecule (three of which pass along each C-H bond and the remaining three pass
through each C-C bond in the plane of the molecule)
...
17
...
2 in the
Data section
...
In a polyatomic molecule each normal mode
(Section 13
...
The overall vibrational partition function is
the product of the individual partition functions, and we can write qV = qV(1)qV(2)
...

If the vibrational excitation is not too great, the harmonic approximation may be
made, and the vibrational energy levels written as

H
1

10

qV

1
Ev = (v + – )hc#
2

v = 0, 1, 2,
...
17)

If, as usual, we measure energies from the zero-point level, then the permitted values
are εv = vhc# and the partition function is

5

qV =

∑ e−β vhc# = ∑ (e−β hc#)v
v

1
0
0

5
~
kT /hc n

10

(because eax = (ex)a)
...
2 (which is no accident: the
ladder-like array of levels in Fig
...
3 is exactly the same as that of a harmonic oscillator)
...

Note that the partition function is linearly
proportional to the temperature when the
temperature is high (T > θ V)
...
17
...
Estimate from
your plots the temperature above which
the harmonic oscillator is in the ‘high
temperature’ limit
...
18)

v

1
1 − e−β hc#

(17
...
17
...
In a polyatomic molecule, each normal mode
gives rise to a partition function of this form
...
3 Calculating a vibrational partition function

The wavenumbers of the three normal modes of H2O are 3656
...
8 cm−1,
and 3755
...
Evaluate the vibrational partition function at 1500 K
...
19 for each mode, and then form the product of the three con-

tributions
...
6 cm−1
...
2 THE MOLECULAR PARTITION FUNCTION
Answer We draw up the following table displaying the contributions of each mode:

Mode:

1

2

3

#/cm

3656
...
8

3755
...
507

1
...
602

1
...
276

1
...
031 × 1
...
028 = 1
...
However, there may be
so many normal modes in a large molecule that their excitation may be significant
even though each mode is not appreciably excited
...
14)
...
1 for the vibrational partition function of one normal mode, the
overall vibrational partition function is about qV ≈ (1
...
8, which indicates
significant vibrational excitation relative to a smaller molecule, such as H2O
...
3 Repeat the calculation for CO2, where the vibrational wavenumbers
are 1388 cm−1, 667
...

[6
...
For
example, the lowest vibrational wavenumber of CH4 is 1306 cm−1, so βhc# = 6
...
C-H stretches normally lie in the range 2850 to 2960 cm−1, so for
them βhc# ≈ 14
...
3 = 0
...
002 when βhc# = 6
...

Now consider the case of bonds so weak that βhc# < kT
...
20)

That is, for weak bonds at high temperatures,
qV =

1

β hc#

=

kT
hc#

(17
...
21 is valid can be expressed in terms of the
characteristic vibrational temperature, θ V = hc#/k (Table 17
...
The value for H2 is
abnormally high because the atoms are so light and the vibrational frequency is correspondingly high
...

(d) The electronic contribution

Electronic energy separations from the ground state are usually very large, so for most
cases q E = 1
...
1 cm-1

598

S

2
0

L

P1/2

2

S

L

Fig
...
6 The doubly degenerate ground
electronic level of NO (with the spin and
orbital angular momentum around the axis
in opposite directions) and the doubly
degenerate first excited level (with the spin
and orbital momenta parallel)
...


5
kT /e

10

Fig
...
7 The variation with temperature of
the electronic partition function of an NO
molecule
...
16
...


Exploration Plot the temperature
dependence of the electronic
partition function for several values of the
energy separation ε between two doubly
degenerate levels
...


electronically degenerate ground states, in which case qE = g E, where g E is the degeneracy
of the electronic ground state
...

Some atoms and molecules have low-lying electronically excited states
...
)
An example is NO, which has a configuration of the form
...
1)
...
17
...
The energy of the two states in which the orbital and spin momenta are
parallel (giving the 2Π3/2 term) is slightly greater than that of the two other states in
which they are antiparallel (giving the 2Π1/2 term)
...
8), is only 121 cm−1
...
If we denote the energies of the two levels as
E1/2 = 0 and E3/2 = ε, the partition function is
qE =

∑ gj e−βε = 2 + 2e−βε
j

(17
...
7 shows the variation of this function with temperature
...
At high temperatures,
q E approaches 4 because all four states are accessible
...
1
...
3 MEAN ENERGIES
(e) The overall partition function

The partition functions for each mode of motion of a molecule are collected in
Table 17
...
The overall partition function is the product of
each contribution
...
23)

Example 17
...
8778 cm−1,
−1
B = 14
...
2869 cm−1 and the information in Example 17
...


Method The starting point is eqn 17
...
For the standard value, we evaluate the

translational partition function at p7 (that is, at 105 Pa exactly)
...
3
...
3 for the other contributions
...
015 u, it follows that q m /NA = 1
...
For the vibra-

tional contribution we have already found that qV = 1
...
From Table 17
...
7
...
3145 J K−1 mol−1) × (1500 K)
× ln{(1
...
7 × 1
...
3 kJ mol−1
Comment 17
...
4 Repeat the calculation for CO2
...
3; B = 0
...


[−366
...
23 are approximate because they
assume that the rotational levels are very close together and that the vibrational levels
are harmonic
...


Using statistical thermodynamics
We can now calculate any thermodynamic quantity from a knowledge of the energy
levels of molecules: we have merged thermodynamics and spectroscopy
...

17
...
When the
molecular partition function can be factorized into contributions from each mode,
the mean energy of each mode M (from eqn 16
...
24)

The text’s web site contains links to
on-line databases of atomic and
molecular spectra
...
Then, if we note that Λ is a constant times β1/2,

1
...
25b)

Both conclusions are in agreement with the classical equipartition theorem (see
Molecular interpretation 2
...
Furthermore, the fact that the mean energy is independent of
2
the size of the container is consistent with the thermodynamic result that the internal
energy of a perfect gas is independent of its volume (Molecular interpretation 2
...


0
...
25a)

0

1
T /qR

2

The mean rotational energy of a
nonsymmetrical linear rotor as a function
of temperature
...

Fig
...
8

Exploration Plot the temperature

dependence of the mean rotational
energy for several values of the rotational
constant (for reasonable values of the
rotational constant, see the Data section)
...


(b) The mean rotational energy

The mean rotational energy of a linear molecule is obtained from the partition function given in eqn 17
...
When the temperature is low (T < θR), the series must be
summed term by term, which gives
qR = 1 + 3e−2βhcB + 5e−6βhcB + · · ·
Hence
͗ε R͘ =

hcB(6e−2βhcB + 30e−6βhcB + · · · )

(17
...
17
...
At high temperatures (T > θR), qR is given by
>
eqn 17
...
26b)

(qR is independent of V, so the partial derivatives have been replaced by complete
derivatives
...
(There is no rotation around the line of atoms
...

2
(c) The mean vibrational energy

The vibrational partition function in the harmonic approximation is given in eqn
17
...
Because qV is independent of the volume, it follows that
dqV
dβ

=

d A

D
hc#e−βhc#
=−
dβ C 1 − e−βhc# F
(1 − e−βhc#)2
1

(17
...
28)

17
...
29)

This result is in agreement with the value predicted by the classical equipartition
1
1
2
theorem, because the energy of a one-dimensional oscillator is E = – mv x + – kx 2 and
2
2
1
the mean energy of each quadratic term is – kT
...
4 Heat capacities
The constant-volume heat capacity is defined as CV = (∂U/∂T)V
...
30)

It follows that

A ∂U D
CV = −kβ
C ∂β F V
2

(17
...
The contribution of mode M is
M
CV = N

A ∂͗ε M͘ D
A ∂͗ε M͘ D
= −Nkβ 2
C ∂T F V
C ∂β F V

(17
...

2
Therefore, the molar constant-volume heat capacity is
T
CV,m = NA

3
d(– kT)
2

dT

3
= –R
2

(17
...
47 J K−1 mol−1
...
We saw in Section 2
...
33)°

When the temperature is high enough for the rotations of the molecules to be
highly excited (when T > θ R), we can use the equipartition value kT for the mean
>
rotational energy (for a linear rotor) to obtain CV,m = R
...
Only the lowest rotational state is occupied when the tempera>
2
ture is very low, and then rotation does not contribute to the heat capacity
...
The variation of the mean energy with temperature is illustrated in Fig
...
9
...
This approximation leads to

601

5

0
0

5
T /q V

10

Fig
...
9 The mean vibrational energy of a
molecule in the harmonic approximation
as a function of temperature
...


Exploration Plot the temperature
dependence of the mean vibrational
energy for several values of the vibrational
wavenumber (for reasonable values of the
vibrational wavenumber, see the Data
section)
...


602

17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS

Total

1

1

C V, m /R

Cm /R

0,1

0,2
1,2

0

0

0

2

T /q R

0

1,3

0,3

4
1
2
3
Temperature, T/q R

5

Fig
...
11 The rotational heat capacity of a
linear molecule can be regarded as the
sum of contributions from a collection of
two-level systems, in which the rise in
temperature stimulates transitions between
J levels, some of which are shown here
...
19
...
17
...


Exploration The Living graphs section

of the text’s web site has applets for
the calculation of the temperature
dependence of the rotational contribution
to the heat capacity
...


calculate the rotational heat capacity at intermediate temperatures by differentiating the equation for the mean rotational energy (eqn 17
...
The resulting (untidy)
expression, which is plotted in Fig
...
10, shows that the contribution rises from
zero (when T = 0) to the equipartition value (when T > θ R)
...
Problem 17
...
17
...

Molecular vibrations contribute to the heat capacity, but only when the temperature is high enough for them to be significantly excited
...
However, it is very unusual for the vibrations to be so highly excited that equipartition is valid, and it is more appropriate to use the full expression for the vibrational
heat capacity, which is obtained by differentiating eqn 17
...
2

Equation 17
...
7) with θ V
the Einstein temperature, θE
...


2

V
C V ,m = Rf

f=

A θ V D A e−θ V/2T D
C T F C 1 − e−θ V/T F

2

(17
...
The curve in Fig
...
12
shows how the vibrational heat capacity depends on temperature
...


17
...
17
...
34
...


Exploration The Living graphs section

of the text’s web site has applets
for the calculation of the temperature
dependence of the vibrational contribution
to the heat capacity
...

V,m

Dissociation

3
2

q V Temperature

Fig
...
13 The general features of the
temperature dependence of the heat
capacity of diatomic molecules are as
shown here
...
The heat capacity becomes very
large when the molecule dissociates
because the energy is used to cause
dissociation and not to raise the
temperature
...


(b) The overall heat capacity

The total heat capacity of a molecular substance is the sum of each contribution
(Fig
...
13)
...
In gases, all three translational modes
3
are always active and contribute – R to the molar heat capacity
...
If the temperature is high enough for ν * vibrational modes to be
V
2 *
active, the vibrational contribution to the molar heat capacity is ν * R
...
It follows that the total molar heat capacity is
V
1
CV,m = – (3 + ν R + 2ν * )R
*
V
2

(17
...
5 Estimating the molar heat capacity of a gas

Estimate the molar constant-volume heat capacity of water vapour at 100°C
...
3; the rotational constants of an
H2O molecule are 27
...
5, and 9
...


603

604

17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
Method We need to assess whether the rotational and vibrational modes are active

by computing their characteristic temperatures from the data (to do so, use hc/k =
1
...

Answer The characteristic temperatures (in round numbers) of the vibrations

are 5300 K, 2300 K, and 5400 K; the vibrations are therefore not excited at 373 K
...
The translational con3
tribution is – R = 12
...
Fully excited rotations contribute a further
2
−1
12
...
Therefore, a value close to 25 J K−1 mol−1 is predicted
...
1 J K−1 mol−1
...

Self-test 17
...
037 cm−1; see Table 13
...

[29 J K−1 mol−1]

17
...
3 is a very important route to the equations
of state of real gases in terms of intermolecular forces, for the latter can be built into
Q
...
1) that the partition function for a gas of independent particles leads to the perfect gas equation of state, pV = nRT
...
3 that their equations of state may be written
pVm
RT

=1+

B
Vm

+

C
2
Vm

+···

(17
...

The total kinetic energy of a gas is the sum of the kinetic energies of the individual
molecules
...
We therefore write
Q=

Z

(17
...
45 (Q = q N/N!, with q = V/Λ3), we see that for
a perfect gas of atoms (with no contributions from rotational or vibrational modes)
Z=

VN

(17
...
39)

where dτi is the volume element for atom i
...


17
...
1 Calculating a configuration integral

When the molecules do not interact with one another, EP = 0 and hence e−βEP = 1
...
This result coincides with
eqn 17
...


When we consider only interactions between pairs of particles the configuration
integral simplifies to

Ύ

1
Z = – e−βEPdτ1dτ2
2

(17
...
41)

2

The quantity f is the Mayer f-function: it goes to zero when the two particles are
so far apart that EP = 0
...
41 becomes
∞

Ύ fr dr

B = −2πNA

f = e−βEP − 1

2

(17
...

Intermolecular potential energies are discussed in more detail in Chapter 18,
where several expressions are developed for them
...
42 is used by considering the hard-sphere potential, which is infinite
when the separation of the two molecules, r, is less than or equal to a certain value σ,
and is zero for greater separations
...
43a)

f=0

=1

when r ≤ σ (and EP = ∞)
when r > σ (and EP = 0)

(17
...
42 that the second virial coefficient is
σ

Ύ r dr = –πN σ

B = 2πNA

2

2
3

A

3

(17
...

Such a potential can be found for weak attractive interactions (a < RT): it consists of
<
a hard-sphere repulsive core and a long-range, shallow attractive region (see Problem
17
...
A further point is that, once a second virial coefficient has been calculated for
a given intermolecular potential, it is possible to calculate other thermodynamic
properties that depend on the form of the potential
...
8), from the thermodynamic relation
lim µT = B − T

p→0

dB

(17
...
48
...
6 Molecular interactions in liquids
The starting point for the discussion of solids is the well ordered structure of a perfect
crystal, which will be discussed in Chapter 20
...
Liquids lie between these two extremes
...

(a) The radial distribution function

2

g (r )
100°C

1

25°C

4°C

0 200

600
r /pm

1000

The radial distribution function
of the oxygen atoms in liquid water at three
temperatures
...
(A
...
Narten, M
...

Danford, and H
...
Levy, Discuss
...

Soc
...
)
Fig
...
14

The average relative locations of the particles of a liquid are expressed in terms of
the radial distribution function, g(r)
...
In a perfect crystal, g(r) is a periodic array of sharp spikes, representing the
certainty (in the absence of defects and thermal motion) that molecules (or ions) lie at
definite locations
...
When the crystal melts, the long-range order is
lost and, wherever we look at long distances from a given molecule, there is equal
probability of finding a second molecule
...
It is still possible to detect a sphere of nearest neighbours at a distance r1,
and perhaps beyond them a sphere of next-nearest neighbours at r2
...

The radial distribution function of the oxygen atoms in liquid water is shown in
Fig
...
14
...
The form of g(r) at 100°C shows that the
intermolecular interactions (in this case, principally by hydrogen bonds) are strong
enough to affect the local structure right up to the boiling point
...
Infrared spectra show that about 90 per cent of hydrogen bonds are intact at
the melting point of ice, falling to about 20 per cent at the boiling point
...
dτN
(17
...
dτN

where β = 1/kT and VN is the N-particle potential energy
...


17
...
However, even a fluid of hard spheres without attractive interactions (a collection of ball-bearings in a container) gives a function that oscillates near the origin
(Fig
...
15), and one of the factors influencing, and sometimes dominating, the structure of a liquid is the geometrical problem of stacking together reasonably hard
spheres
...
The attractive part of the potential modifies this basic structure, but sometimes only quite weakly
...

There are several ways of building the intermolecular potential into the calculation
of g(r)
...
17
...
Then, whenever a particle leaves the box through one of its faces, its image arrives through the opposite face
...

In the Monte Carlo method, the particles in the box are moved through small but
otherwise random distances, and the change in total potential energy of the N particles in the box, ∆VN , is calculated using one of the intermolecular potentials discussed
in Section 18
...
Whether or not this new configuration is accepted is then judged from
the following rules:

Radial distribution function, g

(b) The calculation of g(r)
High
density

1
Low
density

0

1

2
R /d

3

4

Fig
...
15 The radial distribution function
for a simulation of a liquid using
impenetrable hard spheres (ball bearings)
...

If the potential energy is greater than before the change, then it is necessary to check if
the new configuration is reasonable and can exist in equilibrium with configurations
of lower potential energy at a given temperature
...
Because we are testing the viability of a configuration with a higher potential energy than the previous configuration in the calculation, ∆VN > 0 and the exponential factor varies between 0 and 1
...

The configurations generated with Monte Carlo calculations can be used to construct
g(r) simply by counting the number of pairs of particles with a separation r and averaging the result over the whole collection of configurations
...
To appreciate what is involved, we consider the motion of
a particle in one dimension
...
47)

where vi−1 is the velocity of the atom when it was at xi−1, its location at the start of the
interval
...
17
...


608

17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
dVN (x)

vi = vi−1 − m−1

dx

∆t

(17
...
The time interval ∆t is approximately 1 fs (10−15 s), which is shorter than the average time between
collisions
...
The time-consuming part of the calculation is the evaluation of the net force on
the molecule arising from all the other molecules present in the system
...
3 Particle trajectories according to molecular dynamics

Consider a particle of mass m moving along the x direction with an initial velocity
v1 given by
v1 =

∆x
∆t

If the initial and new positions of the atom are x1 and x2, then ∆x = x2 − x1 and
x2 = x1 + v1∆t
The particle moves under the influence of a force arising from interactions with
other atoms in the molecule
...
If the initial and new velocities
are v1 and v2, then ∆v = v2 − v1 and
v2 = v1 + a1∆t = v1 +

F1
m

∆t

Because F = −dV/dx, the force acting on the atom is related to the potential energy
of interaction with other nearby atoms, the potential energy VN(x), by
F1 = −

dVN(x)
dx

x1

where the derivative is evaluated at x1
...
48 for the calculation of a velocity vi from a
previous velocity vi−1
...
6 Consider a particle of mass m connected to a stationary wall with a
spring of force constant k
...

[vi = vi−1 + (k/m)(xi−1 − x0)]

A molecular dynamics calculation gives a series of snapshots of the liquid, and g(r)
can be calculated as before
...


(17
...
7 RESIDUAL ENTROPIES
(c) The thermodynamic properties of liquids

Once g(r) is known it can be used to calculate the thermodynamic properties of
liquids
...
50)

0

That is, U is essentially the average two-particle potential energy weighted by g(r)r 2dr,
which is the probability that the pair of particles have a separation between r and r + dr
...
51a)

The quantity v2 is called the virial (hence the term ‘virial equation of state’)
...
51b)

0

The first term on the right is the kinetic pressure, the contribution to the pressure
from the impact of the molecules in free flight
...
11, representing the contribution to the pressure from the intermolecular forces
...
The second term is therefore the average of this work over the range of
pairwise separations in the liquid as represented by the probability of finding two
molecules at separations between r and r + dr, which is g(r)r 2dr
...

17
...
3)
...
One possibility is that the
experimental determination failed to take a phase transition into account (and a contribution of the form ∆trsH/Ttrs incorrectly omitted from the sum)
...
The entropy at T = 0 is then
greater than zero and is called the residual entropy
...
There may be so little energy difference
between
...
AB BA BA AB
...
We can readily
calculate the entropy arising from residual disorder by using the Boltzmann formula
S = k ln W
...
Because the same energy can be achieved in 2N
different ways (because each molecule can take either of two orientations), the total
number of ways of achieving the same energy is W = 2N
...
52a)

609

610

17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
We can therefore expect a residual molar entropy of R ln 2 = 5
...
If s orientations are possible, the residual molar entropy will be
Sm = R ln s

(17
...
5 J K−1 mol−1 is in good agreement with the
experimental value (10
...
For CO, the measured residual entropy is
5 J K−1 mol−1, which is close to R ln 2, the value expected for a random structure of the
form
...

Fig
...
17 The possible locations of H atoms
around a central O atom in an ice crystal
are shown by the white spheres
...


Illustration 17
...
Each O atom is surrounded tetrahedrally by four H atoms, two of which are attached by short σ bonds, the other
two being attached by long hydrogen bonds (Fig
...
17)
...
17
...
However, not all
these arrangements are acceptable
...
Therefore, the number of permitted arrangements is
6
3
–
W = 22N(16 )N = (–)N
2

It then follows that the residual molar entropy is
3
3
3
Sm(0) ≈ k ln(–)NA = NAk ln(–) = R ln(–) = 3
...
4 J K−1 mol−1
...


17
...
9 in terms
of the molar partition function, qm = q/n
...
To calculate the
equilibrium constant, we need to combine these two equations
...


Fig
...
18 The six possible arrangements of
H atoms in the locations identified in
Fig
...
17
...


17
...
For these expressions,
we need the value of the molar partition function when p = p7 (where p 7 = 1 bar): we
7
denote this standard molar partition function q m
...
For a species J it follows that
7
7
G m(J) = G m(J,0) − RT ln

7
q J,m

NA

(17
...
By combining expressions
like this one (as shown in the Justification below), the equilibrium constant for the
reaction

aA+bB→cC+dD
is given by the expression
K=

7
7
(q C,m /NA)c(qD,m /NA)d

e−∆rE0/RT
7
7
(q A,m /NA)a(q B,m /NA)b

D0(reactants)

(17
...
17
...
In terms of the
stoichiometric numbers introduced in Section 7
...
54b)

Justification 17
...
55)

the reaction internal energy at T = 0 (a molar quantity)
...
17
...


611

612

17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS
ln K = −

∆r E0
RT

+ ln

7
7
(q C,m/NA)c(q D,m/NA)d
7
7
(q A,m/NA)a(q B,m/NA)b

This expression is easily rearranged into eqn 17
...


(b) A dissociation equilibrium

We shall illustrate the application of eqn 17
...
54 (with a = 1, b = 0, c = 2, and d = 0):
K=

7
(q X,m /NA)2
7
q X2,m /NA

e−∆rE0/RT =

7
(q X,m)2
7
q X2,mNA

e−∆rE0/RT

(17
...
56b)

where D0(X-X) is the dissociation energy of the X-X bond
...
The diatomic molecule X2 also has rotational and vibrational degrees of freedom, so its standard molar partition function is
7
q X2,m = gX2

7
A V m D R V RTgX2qR 2qV 2
X X
q q =
7 3
C Λ3 F X2 X2
p Λ X2
X2

where gX2 is the degeneracy of the electronic ground state of X2
...
54 that the equilibrium constant is
K=

kTg 2 Λ3 2
X X
p7gX2qR 2qV 2Λ6
X X X

e−D0 /RT

(17
...
All the quantities in this expression can be calculated
from spectroscopic data
...
3 and depend on the masses
of the species and the temperature; the expressions for the rotational and vibrational
partition functions are also available in Table 17
...

Example 17
...
1547 cm−1, # = 159
...
4 kJ mol−1
...

Method The partition functions required are specified in eqn 17
...
They are

evaluated by using the expressions in Table 17
...
For a homonuclear diatomic
molecule, σ = 2
...


17
...
14 pm
q R (Na2) = 2246
g(Na) = 2

Λ (Na) = 11
...
885
g(Na2 ) = 1

Then, from eqn 17
...
38 × 10−23 J K−1) × (1000 K) × 4 × (8
...
885 × (1
...
47

= 2
...

Self-test 17
...


[52]
P

(c) Contributions to the equilibrium constant

We are now in a position to appreciate the physical basis of equilibrium constants
...

Figure 17
...
The populations of the states are given by the Boltzmann distribution, and are independent of whether any given state happens to belong to R or
to P
...
If the spacings of R and P are similar (as in
Fig
...
20), and P lies above R, the diagram indicates that R will dominate in the
equilibrium mixture
...
17
...

It is quite easy to show (see the Justification below) that the ratio of numbers of R
and P molecules at equilibrium is given by
NP
NR

=

qP
qR

e−∆ rE 0 /RT

(17
...
58b)

just as would be obtained from eqn 17
...

Justification 17
...
The total number of R molecules is
the sum of these populations taken over the states belonging to R; these states we
label r with energies ε r
...
17
...
At equilibrium
all are accessible (to differing extents,
depending on the temperature), and the
equilibrium composition of the system
reflects the overall Bolzmann distribution
of populations
...


Comment 17
...
In the case of a more general
reaction, the conversion from q to q7
comes about at the stage of converting
the pressures that occur in K to numbers
of molecules
...

However, because ε p = ε p + ∆ε 0, where ∆ε 0 is the separation of zero-point energies
′
(as in Fig
...
21),
NP =

DE0

Fig
...
21 It is important to take into
account the densities of states of the
molecules
...
In classical thermodynamic terms,
we have to take entropies into account as
well as enthalpies when considering
equilibria
...
17
...
The
products P can dominate provided ∆E0 is
not too large and P has an appreciable
density of states
...

The equilibrium constant of the R 5 P reaction is proportional to the ratio of the
numbers of the two types of molecule
...
58b
...
58 can be seen most clearly by exaggerating the molecular
features that contribute to it
...
We also suppose that P has a large number of evenly, closely
spaced levels (Fig
...
22)
...
In this model
system, the equilibrium constant is
K=

e

N

kT

ε

e−∆ r E 0 /RT

(17
...
When ∆ r E 0 is small but still positive, K can
exceed 1 because the factor kT/ε may be large enough to overcome the small size of the
exponential term
...
At low temperatures K < 1 and the system con<
sists entirely of R
...
Hence P becomes dominant
...
This behaviour is what we saw, from
the outside, in Chapter 7
...
It shows that the density of states (and hence the
entropy) of each species as well as their relative energies controls the distribution of
populations and hence the value of the equilibrium constant
...
The molecular partition function can be written as q =
qTqRqVqE, with the contributions summarized in Table 17
...

2
...
4
...
The mean energy of a mode is ͗ε M͘ = −(1/q M)(∂q M/∂β )V , with
the contributions from each mode summarized in Table 17
...

4
...
5
...
The overall heat capacity is written as CV,m = –(3 + ν R + 2ν V
*
*)R
2
6
...
dτN for a real gas
...
In the virial equation of state, the second virial coefficient
can be written as B = −(NA /2V)∫fdτ1dτ2, where the Mayer
f-function is f = e−βEP − 1
...
The radial distribution function, g(r), where g(r)r 2dr, is the
probability that a molecule will be found in the range dr at a
distance r from another molecule
...
50 and 17
...

9
...

10
...
54)
...
Chandler, Introduction to modern statistical mechanics
...

K
...
Dill and S
...
Garland Publishing
(2002)
...
L
...
Dover, New
York (1986)
...
A
...
D
...

University Science Books, Sausalito (1999)
...
Widom, Statistical mechanics: a concise introduction for chemists
...


Table 17
...
561 × 10 −2(T/K)5/2(M/g mol−1)3/2
NA

Rotation
Linear molecules

qR =

kT
T
=
σ hcB θ R

Nonlinear molecules

qR =

3/2
1/2
1 A kT D A π D
B E B
E
σ C hc F C ABC F

Vibration

qV =

1
1
=
1 − e−hc#/kT 1 − e−θ V /T

θV =

hc# hν
=
k
k

θR =

hcB
k

kT
T
=
hc# θ V
q E = g0 [+ higher terms]
For T > θV, qV =
>

Electronic

where g0 is the degeneracy of the
electronic ground state
Note that β = 1/kT
...
6950
T/K
×
σ
(B/cm−1)

qR =

1
...
695 ×

T/K
#/cm−1

616

17 STATISTICAL THERMODYNAMICS 2: APPLICATIONS

Table 17
...


Table 17
...


C V = nR

(a)

(b)

EXERCISES

617

Discussion questions
17
...
5 Use concepts of statistical thermodynamics to describe the molecular
features that lead to the equations of state of perfect and real gases
...
2 Explain the origin of the symmetry number
...
6 Describe how liquids are investigated by using concepts of statistical
thermodynamics
...


17
...

17
...


17
...


Exercises
17
...

17
...

17
...


Do this calculation with and without the vibrational contribution to the
energy
...
2b Estimate the value of γ = Cp /CV for carbon dioxide
...
Which is closer to
the expected experimental value at 25°C?
17
...

17
...

17
...

17
...

17
...
878 cm−1, 14
...
287 cm−1
...
5b Calculate the rotational partition function of SO2 at 298 K from its

rotational constants 2
...
344 17 cm−1, and 0
...
Above
what temperature is the high-temperature approximation valid to within
10 per cent of the true value?
17
...
5a, calculate the rotational contribution
to the molar entropy of gaseous water at 25°C
...
9a Plot the molar heat capacity of a collection of harmonic oscillators
as a function of T/θ V, and predict the vibrational heat capacity of ethyne at
(a) 298 K, (b) 500 K
...

17
...
For data, see the preceding exercise
...
10a A CO2 molecule is linear, and its vibrational wavenumbers are 1388
...
4 cm−1, and 2349
...
The rotational constant of the molecule is 0
...

Calculate the rotational and vibrational contributions to the molar Gibbs
energy at 298 K
...
10b An O3 molecule is angular, and its vibrational wavenumbers are
1110 cm−1, 705 cm−1, and 1042 cm−1
...
553 cm−1, 0
...
3948 cm−1
...

17
...


Calculate the electronic contribution to the heat capacity of Cl atoms at
(a) 500 K and (b) 900 K
...
11b The first electronically excited state of O2 is 1∆ g and lies 7918
...
Calculate the electronic contribution to
the molar Gibbs energy of O2 at 400 K
...
12a The ground state of the Co2+ ion in CoSO4·7H2O may be regarded as

4
T9/2
...
Estimate the molar entropy of the solid at
these temperatures
...
6b From the results of Exercise 17
...
12b Estimate the contribution of the spin to the molar entropy of a solid
5
sample of a d-metal complex with S = –
...


17
...
7a Calculate the rotational partition function of CH4 (a) by direct

can adopt (a) three, (b) five, (c) six orientations of equal energy at T = 0
...
Take B = 5
...


17
...
7b Calculate the rotational partition function of CH3CN (a) by direct

summation of the energy levels at 298 K and 500 K, and (b) by the hightemperature approximation
...
28 cm−1 and B = 0
...

17
...
75 pm
...

17
...
1752 cm−1, 0
...
3505 cm−1
...


on account of the similarity of the H and F atoms
...

17
...
36 cm−1, B = 0
...
5422 eV
...


17
...
The Br2 molecule has a nondegenerate ground state, with no other electronic states nearby
...
33 cm−1
...
2 Explore whether a magnetic field can influence the heat capacity of a
paramagnetic molecule by calculating the electronic contribution to the heat
capacity of an NO2 molecule in a magnetic field
...
0 T magnetic field at (a) 50 K, (b) 298 K
...
9‡ Treat carbon monoxide as a perfect gas and apply equilibrium
statistical thermodynamics to the study of its properties, as specified below, in
the temperature range 100 –1000 K at 1 bar
...
8 cm−1, B =1
...
09 eV; neglect anharmonicity and centrifugal distortion
...
(b) Explore numerically the differences, if any, between
the rotational molecular partition function as calculated with the discrete
energy distribution and that calculated with the classical, continuous energy
distribution
...


17
...

What is the high-temperature contribution to the heat capacity and entropy
of such a freely rotating group at 25°C? The moment of inertia of CH3 about
its three-fold rotation axis (the axis that passes through the C atom and
the centre of the equilateral triangle formed by the H atoms) is 5
...


17
...
59; B(DCl)/cm−1 = 5
...
28; A(CHD3)/cm−1 = 2
...
63
...
4 Calculate the temperature dependence of the heat capacity of p-H2

17
...
Calculate the equilibrium constant at (a) 298 K and (b) 800 K
for the gas-phase exchange reaction H2O + DCl 5 HDO + HCl from the
following data: #(H2O)/cm−1 = 3656
...
8, 3755
...
7, 1402
...
5; A(H2O)/cm−1 = 27
...
51;
C(H2O)/cm−1 = 9
...
38; B(HDO)/cm−1 = 9
...
417; B(HCl)/cm−1 = 10
...
449;
#(HCl)/cm−1 = 2991; #(DCl)/cm−1 = 2145
...
1 The NO molecule has a doubly degenerate electronic ground state
and a doubly degenerate excited state at 121
...
Calculate the electronic
contribution to the molar heat capacity of the molecule at (a) 50 K, (b) 298 K,
and (c) 500 K
...
Use B = 60
...
The
experimental heat capacity of p-H2 does in fact show a peak at low
temperatures
...
5 The pure rotational microwave spectrum of HCl has absorption lines
at the following wavenumbers (in cm−1): 21
...
37, 63
...
75, 105
...
12 148
...
49, 190
...
87, 233
...
24, 275
...
62, 317
...
99, 360
...
36, 402
...
74, 444
...
11, 487
...
48
...

17
...
9987 cm−1 and its vibrational wavenumber
# = 2358 cm−1
...
1 J K−1 mol−1
...
7‡ J
...
Dojahn, E
...
M
...
E
...
Phys
...
100,
9649 (1996)) characterized the potential energy curves of the ground and
electronic states of homonuclear diatomic halogen anions
...
0 cm−1 and
equilibrium internuclear distance of 190
...
The first two excited states are
at 1
...
702 eV above the ground state
...

17
...
Negri,
G
...
Zerbetto (J
...
Chem
...
The wavenumber
for the single Au mode is 976 cm−1; wavenumbers for the four threefold
degenerate T1u modes are 525, 578, 1180, and 1430 cm−1; wavenumbers
for the five threefold degenerate T2u modes are 354, 715, 1037, 1190, and
1540 cm−1; wavenumbers for the six fourfold degenerate Gu modes are 345,
757, 776, 963, 1315, and 1410 cm−1; and wavenumbers for the seven fivefold
degenerate Hu modes are 403, 525, 667, 738, 1215, 1342, and 1566 cm−1
...


Theoretical problems
17
...

17
...
At temperatures above θ R = hcB/k, the
rotational contribution to the heat capacity becomes significant
...
But at this latter
temperature, dissociation of the molecule into the atoms must be considered
...
(b) Obtain an expression for the molar constant-pressure
heat capacity of hydrogen at all temperatures taking into account the
dissociation of hydrogen
...

17
...
Express the
results in terms of the vibrational temperature, θ V and plot graphs of each
property against T/θ V
...
15 Suppose that an intermolecular potential has a hard-sphere core of
radius r1 and a shallow attractive well of uniform depth ε out to a distance r2
...
42 and the condition ε < kT, that such a model is
<
approximately consistent with a van der Waals equation of state when
b < Vm, and relate the van der Waals parameters and the Joule–Thomson
<
coefficient to the parameters in this model
...


PROBLEMS
17
...
(b) Use this result to derive eqn 13
...
(c) Estimate the temperature at
which the spectrum of HCl shown in Fig
...
44 was obtained
...
17 A more formal way of arriving at the value of the symmetry number is
to note that σ is the order (the number of elements) of the rotational subgroup
of the molecule, the point group of the molecule with all but the identity and
the rotations removed
...

The rotational subgroup of NH3 is {E, 2C3}, so σ = 3
...
The
rotational subgroup of CH4 is obtained from the T character table as {E, 8C3,
3C2}, so σ = 12
...
(a) Estimate the rotational partition function of
′, ″},
ethene at 25°C given that A = 4
...
0012 cm−1, and C = 0
...

(b) Evaluate the rotational partition function of pyridine, C5H5N, at room
temperature (A = 0
...
1936 cm−1, C = 0
...

17
...
When presented with a table of
energy levels, it is often much more convenient to evaluate the following sums
directly:

q=

∑ e βε
−

≥=

j

j

∑ βε e βε
j

−

j

”=

j

∑ (βε ) e βε
j

2 −

1
S
1
0

3

3
P0
1
21 850

P1
3
21 870

3

P2
5
21 911

1

P1
3
35 051

S1
3
41 197

17
...
23 An average human DNA molecule has 5 × 108 binucleotides (rungs on

the DNA ladder) of four different kinds
...
24 It is possible to write an approximate expression for the partition
function of a protein molecule by including contributions from only two
states: the native and denatured forms of the polymer
...
It follows from Illustration 16
...

(a) Show that the constant-volume molar heat capacity is
R(ε m /RT)2e− ε m/RT
(1 + e−ε m /RT)2

Hint
...
(b) Plot the variation of CV,m with
temperature
...


3

sum:

ζ(β) =

Applications: to biology, materials science, environmental
science, and astrophysics

CV,m =

(a) Derive expressions for the internal energy, heat capacity, and entropy in
terms of these three functions
...
22 The heat capacity ratio of a gas determines the speed of sound in it
through the formula cs = (γ RT/M)1/2, where γ = Cp /CV and M is the molar
mass of the gas
...
Estimate the speed of
sound in air at 25°C
...

Then go on to show graphically that the total contribution to the heat capacity
of a linear rotor can be regarded as a sum of contributions due to transitions
0→1, 0→2, 1→2, 1→3, etc
...
17
...

17
...
19 to analyse the vibrational
contribution to the heat capacity in terms of excitations between levels and
illustrate your results graphically in terms of a diagram like that in Fig
...
11
...
21 Determine whether a magnetic field can influence the value of an
equilibrium constant
...
Data on the species are given in
4
Exercise 17
...
The electronic g value of the atoms is –
...


17
...
Viswanathan, R
...
Schmude, Jr
...
A
...
Phys
...
100, 10784 (1996)) studied thermodynamic properties of several
boron–silicon gas-phase species experimentally and theoretically
...
Among the computations they reported was
computation of the Gibbs energy of BSi(g) at several temperatures based on
a 4Σ − ground state with equilibrium internuclear distance of 190
...
Compute the standard molar Gibbs energy
7
7
G m(2000 K) − G m(0)
...
26‡ The molecule Cl2O2, which is believed to participate in the seasonal
depletion of ozone over Antarctica, has been studied by several means
...
Birk, R
...
Friedl, E
...
Cohen, H
...
Pickett, and S
...
Sander (J
...

Phys
...
4,
2409
...
7 MHz
...
J
...
Kronberg,
H
...
P
...
Willner (J
...
Chem
...
116, 1106 (1994)) report
its vibrational wavenumbers as 753, 542, 310, 127, 646, and 419 cm−1
...

17
...
Hutter, H
...
Lüthi, and F
...
Amer
...
Soc
...
Given that the ground state of C3, a molecule found
in interstellar space and in flames, is an angular singlet with moments of
inertia 39
...
032, and 0
...
4, 1224
...
00 K) − G m(0) and G m(1000 K) − G m(0) for C3
...
1 Electric dipole moments
18
...
3 Relative permittivities

Interactions between molecules
18
...
5 Repulsive and total

interactions
I18
...
All these properties reflect the degree to which the nuclei of atoms exert control over the
electrons in a molecule, either by causing electrons to accumulate in particular regions, or
by permitting them to respond more or less strongly to the effects of external electric fields
...
We begin our examination of molecular
interactions by describing the electric properties of molecules, which may be interpreted in terms of concepts in electronic structure introduced in Chapter 11
...
One result of this interaction is
the cohesion of molecules to form the bulk phases of matter
...
The interaction between ions is treated
in Chapter 5 (for solutions) and Chapter 20 (for solids)
...
6 Molecular interactions in gases
18
...
8 Condensation

Further information 18
...
The former competition
may result in an electric dipole moment
...


Further information 18
...
1 Electric dipole moments

Checklist of key ideas
Further reading

Discussion questions
Exercises
Problems

An electric dipole consists of two electric charges +q and −q separated by a distance R
...
The magnitude of m is
µ = qR and, although the SI unit of dipole moment is coulomb metre (C m), it is still
commonly reported in the non-SI unit debye, D, named after Peter Debye, a pioneer
in the study of dipole moments of molecules, where
1 D = 3
...
1)

The dipole moment of a pair of charges +e and −e separated by 100 pm is 1
...
8 D
...
The conversion factor in eqn 18
...
g
...
units: 1 D is the dipole moment of two equal and opposite
charges of magnitude 1 e
...
u
...


18
...
The permanent dipole moment stems from the partial charges on the atoms in the molecule
that arise from differences in electronegativity or other features of bonding (Section
11
...
Nonpolar molecules acquire an induced dipole moment in an electric field on
account of the distortion the field causes in their electronic distributions and nuclear
positions; however, this induced moment is only temporary, and disappears as soon
as the perturbing field is removed
...

The Stark effect (Section 13
...
In many cases microwave
spectroscopy cannot be used because the sample is not volatile, decomposes on vaporization, or consists of molecules that are so complex that their rotational spectra
cannot be interpreted
...
Computational
software is now widely available, and typically computes electric dipole moments by
assessing the electron density at each point in the molecule and its coordinates relative
to the centroid of the molecule; however, it is still important to be able to formulate
simple models of the origin of these moments and to understand how they arise
...

All heteronuclear diatomic molecules are polar, and typical values of µ include
1
...
42 D for HI (Table 18
...
Molecular symmetry is of the greatest
importance in deciding whether a polyatomic molecule is polar or not
...
Homonuclear polyatomic molecules may
be polar if they have low symmetry and the atoms are in inequivalent positions
...
Heteronuclear polyatomic molecules may be nonpolar if they have high symmetry, because individual
bond dipoles may then cancel
...

To a first approximation, it is possible to resolve the dipole moment of a polyatomic molecule into contributions from various groups of atoms in the molecule
and the directions in which these individual contributions lie (Fig
...
1)
...
1* Dipole moments
(µ) and polarizability volumes (α ′)
m/D

621

d-

m
d+ d+

–q

q
R

1 Electric dipole
Comment 18
...
Note that the direction of
the arrow is opposite to that of µ
...
57 D

D2h

(b) m obs = 0
mcalc = 0

C2v

(c) mobs = 2
...
7 D

d-

C2v

a¢/(10−30 m3)

2 Ozone, O3
CCl4

0

H2

0

10
...
819

H2O

1
...
48

HCl

1
...
63

HI

0
...
45

* More values are given in the Data section
...
48 D
m calc = 1
...
57 D), purple
...
18
...
2

Operations involving vectors are
described in Appendix 2, where eqn 18
...


In three dimensions, a vector m has
components µx, µy, and µz along the x-,
y-, and z-axes, respectively, as shown in
the illustration
...
For example, if
µx = −1
...
0 D
and points in the −x direction
...
2a)

When the two dipole moments have the same magnitude (as in the dichlorobenzenes), this equation simplifies to
1
µres ≈ 2µ1 cos –θ
2

Comment 18
...
1,2-Dichlorobenzene,
however, has a dipole moment which is approximately the resultant of two chlorobenzene dipole moments arranged at 60° to each other
...
2 Partial charges in
polypeptides
Atom

Partial charge/e

C(=O)

+0
...
06

H(-C)

+0
...
18

H(-O)

+0
...
36

O

– 0
...
2b)

Self-test 18
...


[µ(ortho)/µ(meta) = 1
...
These partial
charges are included in the output of many molecular structure software packages
...
3a)

J

Here qJ is the partial charge of atom J, x J is the x-coordinate of atom J, and the sum
is over all the atoms in the molecule
...
For an electrically neutral molecule, the origin of the coordinates
is arbitrary, so it is best chosen to simplify the measurements
...
3b)

Example 18
...
2 and the locations of the atoms
shown
...
3a to calculate each of the components of the dipole

moment and then eqn 18
...
Note that the partial charges are multiples of the fundamental charge, e = 1
...

Answer The expression for µx is

µ x = (−0
...
45e) × (0 pm) + (0
...
38e) × (−62
...
8e pm
= 8
...
609 × 10−19 C) × (10−12 m) = 1
...
1 ELECTRIC DIPOLE MOMENTS

623

(182, 87,0)

corresponding to µ x = 0
...
The expression for µy is:

H

µ y = (−0
...
45e) × (0 pm) + (0
...
6 pm)
+ (−0
...
1 × 10−30 C m

m

+0
...
36

(132,0,0)

C

N

+0
...
7 D
...
38
( 62,1 07,0)

µ = {(0
...
7 D)2}1/2 = 2
...
7 units of length to have x, y, and z components of 0
...
7, and 0 units; the orientation is superimposed on (6)
...
38
(0,118,0)

Self-test 18
...


[−3
...
4)

In the following pages we refer to the sample as a dielectric, by which is meant a
polarizable, nonconducting medium
...
In the presence of
a field, the dipoles become partially aligned because some orientations have lower
energies than others
...
We
show in the Justification below that, at a temperature T

µ2E

(18
...
Moreover, as we shall see, there is an
additional contribution from the dipole moment induced by the field
...
1 The thermally averaged dipole moment

The probability dp that a dipole has an orientation in the range θ to θ + dθ is given
by the Boltzmann distribution (Section 16
...

The average value of the component of the dipole moment parallel to the applied
electric field is therefore
π

Ύ

(94,–61,0)

H

The polarization, P, of a sample is the electric dipole moment density, the mean electric dipole moment of the molecules, ͗ µ͘, multiplied by the number density, N :

͗ µ z͘ =

C

Ύ

͗ µz͘ = µ cos θ dp = µ cos θ dp =

Ύe
Ύe

µ

x cos θ

0

cos θ sin θ dθ

π

x cos θ

0

sin θ dθ

H

+0
...
18

7

624

18 MOLECULAR INTERACTIONS
with x = µE /kT
...

1

Ύ ye dy
Ύ e dy

µ
͗ µz ͘ =

xy

−1
1

xy

−1

At this point we use
1

Ύ

−1

e xydy =

e x − e −x
x

1

Ύ

−1

ye xydy =

e x + e −x
x

−

e x − e −x
x2

It is now straightforward algebra to combine these two results and to obtain
͗ µz͘ = µL(x)
Comment 18
...


L(x) =

e x + e −x
e x − e −x

−

1
x

x=

µE
kT

(18
...

Under most circumstances, x is very small (for example, if µ = 1 D and T = 300 K,
then x exceeds 0
...
When x < 1, the exponentials in the
<
Langevin function can be expanded, and the largest term that survives is
1
L(x) = – x + · · ·
3

(18
...
6
...
2 Polarizabilities
An applied electric field can distort a molecule as well as align its permanent electric
dipole moment
...
8)

(See Section 20
...
8
...
The greater the polarizability, the larger is the induced
dipole moment for a given applied field
...

(a) Polarizability volumes

Polarizability has the units (coulomb metre)2 per joule (C2 m2 J−1)
...
5

When using older compilations of data,
it is useful to note that polarizability
volumes have the same numerical values
as the ‘polarizabilities’ reported using
c
...
s
...


α′ =

α
4πε0

[18
...
Because the units of 4πε0 are coulomb-squared
per joule per metre (C2 J−1 m−1), it follows that α ′ has the dimensions of volume
(hence its name)
...


18
...
1
...
The electron distribution can be distorted readily if the LUMO lies close to the HOMO in energy, so the polarizability is
then large
...
Molecules with
small HOMO–LUMO gaps are typically large, with numerous electrons
...
2 Polarizabilities and molecular structures

When an electric field is increased by dE, the energy of a molecule changes by −µdE,
and if the molecule is polarizable, we interpret µ as µ* (eqn 18
...
Therefore, the
change in energy when the field is increased from 0 to E is
E

E

Ύ µ*dE = −Ύ αE dE = − –αE

∆E = −

1

2

0

2

0

The contribution to the hamiltonian when a dipole moment is exposed to an electric field E in the z-direction is
H (1) = −µzE
Comparison of these two expressions suggests that we should use second-order
perturbation theory to calculate the energy of the system in the presence of the
field, because then we shall obtain an expression proportional toE 2
...
65b, the second-order contribution to the energy is

E

(2)

=

∑′

Ύψ *H
n

2

ψ0dτ

(1)

E (0) − E (0)
0
n

n

=E

∑′

2

n

Ύψ *µ ψ dτ
n z 0

E (0) − E (0)
0
n

2

|µ

|2

∑ ′ E(0)z,0nE (0)
0 − n

b= E 2

n

where µz,0n is the transition electric dipole moment in the z-direction (eqn 9
...
By
comparing the two expressions for the energy, we conclude that the polarizability of
the molecule in the z-direction is
|µ

|2

∑ ′ E(0)z,0nE(0)
n − 0

α=2

(18
...
10 can be appreciated by approximating the excitation
energies by a mean value ∆E (an indication of the HOMO–LUMO separation), and
supposing that the most important transition dipole moment is approximately
equal to the charge of an electron multiplied by the radius, R, of the molecule
...

If the excitation energy is approximated by the energy needed to remove an
electron to infinity from a distance R from a single positive charge, we can write
∆E ≈ e 2/4πε 0 R
...


For most molecules, the polarizability is anisotropic, by which is meant that its
value depends on the orientation of the molecule relative to the field
...
0067 nm3

625

626

18 MOLECULAR INTERACTIONS
and it is 0
...
The anisotropy
of the polarizability determines whether a molecule is rotationally Raman active
(Section 13
...

(b) Polarization at high frequencies

When the applied field changes direction slowly, the permanent dipole moment has
time to reorientate—the whole molecule rotates into a new direction—and follow the
field
...
Because
a molecule takes about 1 ps to turn through about 1 radian in a fluid, the loss of this
contribution to the polarization occurs when measurements are made at frequencies
greater than about 1011 Hz (in the microwave region)
...

The next contribution to the polarization to be lost as the frequency is raised is
the distortion polarization, the polarization that arises from the distortion of the
positions of the nuclei by the applied field
...
The time taken
for a molecule to bend is approximately the inverse of the molecular vibrational
frequency, so the distortion polarization disappears when the frequency of the radiation is increased through the infrared
...

At even higher frequencies, in the visible region, only the electrons are mobile
enough to respond to the rapidly changing direction of the applied field
...

Justification 18
...
2)
and is

α (ω) =

2
$

∑

′

n

ω n0| µz,0n |2

(18
...
As ω → 0,
n
0
the equation reduces to eqn 18
...
As ω becomes very
high (and much higher than any excitation frequency of the molecule so that the
2
ω n0 in the denominator can be ignored), the polarizability becomes

α(ω) = −

2
$ω 2

∑ ωn0 |µ0n |2 → 0

as

ω→∞

n

That is, when the incident frequency is much higher than any excitation frequency,
the polarizability becomes zero
...


18
...
3 Relative permittivities
When two charges q1 and q2 are separated by a distance r in a vacuum, the potential
energy of their interaction is (see Appendix 3):
V=

q1q2

(18
...
12b)

4πε r

where ε is the permittivity of the medium
...
13]

ε0

The relative permittivity can have a very significant effect on the strength of the interactions between ions in solution
...
Some of the consequences of this reduction for electrolyte solutions were explored in Chapter 5
...
The quantitative relation between the relative permittivity and the electric properties of the molecules is obtained by considering the polarization of a
medium, and is expressed by the Debye equation (for the derivation of this and the
following equations, see Further reading):

εr − 1
εr + 2

=

ρPm

(18
...
15]

The term µ2/3kT stems from the thermal averaging of the electric dipole moment in
the presence of the applied field (eqn 18
...
The corresponding expression without
the contribution from the permanent dipole moment is called the Clausius–Mossotti
equation:

εr − 1
εr + 2

=

ρNAα
3Mε 0

(18
...


627

628

18 MOLECULAR INTERACTIONS
Example 18
...
The relative permittivity of camphor (8) was measured at a series
of temperatures with the results given below
...


O

8 Camphor

θ /°C
0
20
40
60
80
100
120
140
160
200

ρ/(g cm −3 )
0
...
99
0
...
99
0
...
99
0
...
96
0
...
91

εr
12
...
4
10
...
0
9
...
90
8
...
60
7
...
21

Method Equation 18
...
The slope of the
graph is NA µ2/9ε 0 k and its intercept at 1/T = 0 is NAα /3ε 0
...

Answer For camphor, M = 152
...
We can therefore use the data to draw

up the following table:

θ /°C
0
20
40
60
80
100
120
140
160
200

122

-

P /(cm3 mol 1)
m

118

114

110

106
2

3
(103 K)/T

4

Fig
...
2 The plot of Pm /(cm3 mol−1)
against (103 K)/T used in Example 18
...


(103 K)/T
3
...
41
3
...
00
2
...
68
2
...
42
2
...
11

εr
12
...
4
10
...
0
9
...
90
8
...
60
7
...
21

(ε r − 1)/(ε r + 2)
0
...
776
0
...
750
0
...
725
0
...
688
0
...
634

Pm /(cm3 mol −1)
122
119
118
115
114
111
110
109
107
106

The points are plotted in Fig
...
2
...
7, so α ′ = 3
...

The slope is 10
...
46 × 10−30 C m, corresponding to 1
...
Because
the Debye equation describes molecules that are free to rotate, the data show that
camphor, which does not melt until 175°C, is rotating even in the solid
...

Self-test 18
...
71 at 20°C and 5
...
Assuming a constant density (1
...

[1
...
2 D]

18
...
17)

Therefore, the molar polarization, Pm, and the molecular polarizability, α , can be
measured at frequencies typical of visible light (about 1015 to 1016 Hz) by measuring
the refractive index of the sample and using the Clausius–Mossotti equation
...
6

The refractive index, nr, of the medium
is the ratio of the speed of light in a
vacuum, c, to its speed c¢ in the medium:
nr = c/c′
...
See
Appendix 3 for details
...
In addition, there are
interactions between ions and the partial charges of polar molecules and repulsive
interactions that prevent the complete collapse of matter to nuclear densities
...

18
...
12a)
...

(a) The potential energy of interaction

We show in the Justification below that the potential energy of interaction between a
point dipole µ1 = q1l and the point charge q2 in the arrangement shown in (9) is
µ1q2
V=−
(18
...
A
point dipole is a dipole in which the separation between the charges is much smaller
than the distance at which the dipole is being observed, l < r
...
18
...


q1

l

–q1

q2
r
9

Justification 18
...
Because l < r for a point dipole, this expression can be simplified by
<
expanding the terms in x and retaining only the leading term:
q1q2
2xq1q2
q1q2l
{−(1 + x + · · · ) + (1 − x + · · · )} ≈ −
=−
V=
4πε 0r 2
4πε 0r
4πε 0r
With µ1 = q1l, this expression becomes eqn 18
...
This expression should be multiplied by cos θ when the point charge lies at an angle θ to the axis of the dipole
...
18
...
The
potentials of the charges decrease (shown
here by a fading intensity) and the two
charges appear to merge, so their combined
effect approaches zero more rapidly than
by the distance effect alone
...
7

The following expansions are often
useful:
1
1+x
1
1−x

= 1 − x + x2 − · · ·

Example 18
...

Method We proceed in exactly the same way as in Justification 18
...


= 1 + x + x2 + · · ·

Answer The sum of the four contributions is

l

q1

l

-q1

q2

-q2

r

V=

1 A q1q2 q1q2 q1q2 q1q2 D
q1q2 A 1
1 D
−
+
+
−
=−
−2+
4πε 0 C r + l
1−xF
r
r
r−lF
4πε 0 r C 1 + x

with x = l/r
...
This step results in the expression

10

2xq1q2

V=−
-q2

q2
r

-q1

q1
11

4πε 0r

Therefore, because µ1 = q1l and µ 2 = q2l, the potential energy of interaction in the
alignment shown in the illustration is
V=−

µ1µ2
2πε 0r 3

This interaction energy approaches zero more rapidly (as 1/r 3) than for the previous case: now both interacting entities appear neutral to each other at large separations
...
1 for the general expression
...
4 Derive an expression for the potential energy when the dipoles are in

[V = µ1 µ 2 /4πε 0r 3]

the arrangement shown in (11)
...
3 summarizes the various expressions for the interaction of charges and
dipoles
...
18
...
Specifically, an n-pole is an array of point charges with an n-pole moment but no lower
moment
...
A dipole (n = 2), as we have seen, is an array
Table 18
...

The field arising from an arbitrary finite
charge distribution can be expressed as the
superposition of the fields arising from a
superposition of multipoles
...
6

Between rotating polar molecules

1/r 6

2

Between all types of molecules

Fig
...
4

London (dispersion)

The energy of a hydrogen bond A-H···B is typically 20 kJ mol−l and occurs on contact for A, B = O, N, or F
...


18
...
A quadrupole (n = 3) consists of an array of point charges that has neither net charge nor dipole moment (as for
CO2 molecules, 3)
...
The feature to remember is that the interaction energy falls off more
rapidly the higher the order of the multipole
...
In the cases
we consider, this will not be a
complication because the two charges of
the dipoles will be collinear and give rise
to fields in the same direction
...

Resultant

m

(18
...

For the point-dipole arrangement shown in Fig
...
5, the same procedure that was
used to derive the potential energy gives

µ

Fig
...
5 The electric field of a dipole is
the sum of the opposing fields from the
positive and negative charges, each
of which is proportional to 1/r 2
...


(18
...


q

The potential energy of interaction between two polar molecules is a complicated
function of their relative orientation
...
1)
4πε 0r 3

f (θ ) = 1 − 3 cos2θ

(18
...

In a fluid of freely rotating molecules, the interaction between dipoles averages to
zero because f(θ ) changes sign as the orientation changes, and its average value is zero
...

The interaction energy of two freely rotating dipoles is zero
...
In fact, the lower energy orientations are

-q2

r

-q1

q1

(c) Dipole–dipole interactions

µ1µ2 f(θ )

d+

Comment 18
...
We shall need this expression when we calculate the dipole
moment induced in one molecule by another
...
19)

(b) The electric field

E=

d+

12 CH4

1

The reason for the even steeper decrease with distance is the same as before: the array
of charges appears to blend together into neutrality more rapidly with distance the
higher the number of individual charges that contribute to the multipole
...


E=

631

13
Comment 18
...
Therefore the
average value of (1 − 3 cos2 θ) is
π

Ύ (1 − 3 cos θ) sin θ dθ = 0
...

We show in the following Justification that the average potential energy of two rotating molecules that are separated by a distance r is
͗V ͘ = −

C

C=

r6

2
2µ 1µ 2
2

3(4πε 0)2kT

(18
...

Justification 18
...
First, we note that the
average interaction energy of two polar molecules rotating at a fixed separation r is
given by
͗V͘ =

µ1 µ 2͗ f ͘
4πε 0r 3

where ͗ f ͘ now includes a weighting factor in the averaging that is equal to the probability that a particular orientation will be adopted
...
That is,
p ∝ e−V/kT

V=

µ1 µ 2 f
4πε 0r 3

When the potential energy of interaction of the two dipoles is very small compared
with the energy of thermal motion, we can use V < kT, expand the exponential
<
function in p, and retain only the first two terms:
p ∝ 1 – V/kT + · · ·
The weighted average of f is therefore
͗ f ͘ = ͗ f ͘0 −

µ1µ 2
4πε 0kTr 3

͗ f 2͘0 + · · ·

where ͗ · · · ͘0 denotes an unweighted spherical average
...
However, the average value of f 2 is nonzero because
f 2 is positive at all orientations, so we can write
͗V͘ = −

µ 2 µ 2 ͗ f 2͘0
1 2
(4πε0)2kTr 6

2
The average value ͗ f 2͘0 turns out to be – when the calculation is carried through in
3
detail
...
23
...
23 are its negative sign (the average interaction
is attractive), the dependence of the average interaction energy on the inverse sixth
power of the separation (which identifies it as a van der Waals interaction), and its
inverse dependence on the temperature
...
The inverse sixth power arises from the inverse third power of
the interaction potential energy that is weighted by the energy in the Boltzmann term,
which is also proportional to the inverse third power of the separation
...
07 kJ mol −1 when the separation is 0
...
This energy should be compared with

18
...
7 kJ mol−1 at the same temperature
...


(a)

(d) Dipole–induced-dipole interactions

A polar molecule with dipole moment µ1 can induce a dipole µ * in a neighbouring
2
polarizable molecule (Fig
...
6)
...
The average interaction
energy when the separation of the molecules is r is (for a derivation, see Further reading)
V=−

C
r6

C=

µ2 α 2
1 ′
4πε0

(18
...
Note that the C in this expression is different from the C in
eqn 18
...

The dipole–induced-dipole interaction energy is independent of the temperature
because thermal motion has no effect on the averaging process
...
For a molecule with µ = 1 D (such as HCl) near a
molecule of polarizability volume α ′ = 10 × 10−30 m3 (such as benzene, Table 18
...
8 kJ mol−1 when the separation is 0
...


(b)
Fig
...
6 (a) A polar molecule (purple
arrow) can induce a dipole (white arrow)
in a nonpolar molecule, and (b) the latter’s
orientation follows the former’s, so the
interaction does not average to zero
...
The abundant evidence for the
existence of interactions between them is the formation of condensed phases of nonpolar substances, such as the condensation of hydrogen or argon to a liquid at low
temperatures and the fact that benzene is a liquid at normal temperatures
...
To appreciate the origin of the interaction, suppose that the electrons in one
molecule flicker into an arrangement that gives the molecule an instantaneous dipole
moment µ *
...
The two dipoles
2
attract each other and the potential energy of the pair is lowered
...
18
...
Because of this correlation, the attraction between
the two instantaneous dipoles does not average to zero, and gives rise to an induceddipole–induced-dipole interaction
...

Polar molecules also interact by a dispersion interaction: such molecules also
possess instantaneous dipoles, the only difference being that the time average of each
fluctuating dipole does not vanish, but corresponds to the permanent dipole
...

The strength of the dispersion interaction depends on the polarizability of the first
molecule because the instantaneous dipole moment µ * depends on the looseness of
1

(a)

(b)
Fig
...
7 (a) In the dispersion interaction,
an instantaneous dipole on one molecule
induces a dipole on another molecule, and
the two dipoles then interact to lower the
energy
...


634

18 MOLECULAR INTERACTIONS
the control that the nuclear charge exercises over the outer electrons
...

The actual calculation of the dispersion interaction is quite involved, but a reasonable
approximation to the interaction energy is given by the London formula:
V=−

C
r

6

3
C = –α 1α 2
2 ′ ′

I1I2
I1 + I2

(18
...
4)
...
The dispersion interaction generally dominates all the interactions between
molecules other than hydrogen bonds
...
1 Calculating the strength of the dispersion interaction

For two CH4 molecules, we can substitute α ′ = 2
...
3 nm
...
2 kJ mol−1
...


(f ) Hydrogen bonding

H

B

Energy

A

The molecular orbital
interpretation of the formation of an AH···B hydrogen bond
...

Only the two lower energy orbitals are
occupied, and there may therefore be a net
lowering of energy compared with the
separate AH and B species
...
18
...
However, there is a type of
interaction possessed by molecules that have a particular constitution
...
Hydrogen bonding is conventionally regarded as being limited
to N, O, and F but, if B is an anionic species (such as Cl−), it may also participate in
hydrogen bonding
...

The formation of a hydrogen bond can be regarded either as the approach between
a partial positive charge of H and a partial negative charge of B or as a particular
example of delocalized molecular orbital formation in which A, H, and B each supply
one atomic orbital from which three molecular orbitals are constructed (Fig
...
8)
...
These three orbitals need to accommodate four electrons (two from the
original A-H bond and two from the lone pair of B), so two enter the bonding orbital
and two enter the nonbonding orbital
...

In practice, the strength of the bond is found to be about 20 kJ mol−1
...
4 INTERACTIONS BETWEEN DIPOLES

635

turned on when AH touches B and is zero as soon as the contact is broken
...
The properties
of liquid and solid water, for example, are dominated by the hydrogen bonding
between H2O molecules
...
The structural evidence for hydrogen bonding comes from noting that the
internuclear distance between formally non-bonded atoms is less than their van der
Waals contact distance, which suggests that a dominating attractive interaction is present
...
Moreover, the H···O distance is expected to be 260 pm but is found to be only 170 pm
...
In a symmetric hydrogen bond, the H atoms lies midway between the two other atoms
...
More common is
the unsymmetrical arrangement, where the A-H bond is shorter than the H···B bond
...
The experimental evidence from structural studies support a linear or
near-linear arrangement
...
18
...
To understand the
consequences of this effect, consider the thermodynamics of transfer of a nonpolar
hydrocarbon solute from a nonpolar solvent to water, a polar solvent
...
Therefore, it
is a large decrease in the entropy of the system (∆ transfer S < 0) that accounts for the
positive Gibbs energy of transfer
...
Substances characterized by a positive Gibbs energy of transfer from a
nonpolar to a polar solvent are called hydrophobic
...
26]

where S is the ratio of the molar solubility of the compound R-A in octanol, a nonpolar solvent, to that in water, and S0 is the ratio of the molar solubility of the compound H-A in octanol to that in water
...
It is observed experimentally that the π values of
most groups do not depend on the nature of A
...
For example, π for R = CH3, CH2CH3, (CH2)2CH3,
(CH2)3CH3, and (CH2)4CH3 is, respectively, 0
...
0, 1
...
0, and 2
...
This trend can be rationalized by ∆ transfer H becoming more positive
and ∆ transfer S more negative as the number of carbon atoms in the chain increases
...
18
...
As a result of this
acquisition of structure, the entropy of the
water decreases, so the dispersal of the
hydrocarbon into the water is entropyopposed; its coalescence is entropyfavoured
...
This is an
exothermic process and accounts for the negative values of ∆ transfer H
...
However, when many solute molecules cluster together, fewer (albeit
larger) cages are required and more solvent molecules are free to move
...
This
increase in entropy of the solvent is large enough to render spontaneous the association of hydrophobic molecules in a polar solvent
...
The hydrophobic interaction is an example of an ordering
process that is stabilized by a tendency toward greater disorder of the solvent
...
The total attractive interaction energy between rotating molecules is then the
sum of the three van der Waals contributions discussed above
...
) In a fluid phase, all three contributions to the potential energy vary as the inverse sixth power of the separation of
the molecules, so we may write
V=−

C6

(18
...

Although attractive interactions between molecules are often expressed as in eqn
18
...
First, we have
taken into account only dipolar interactions of various kinds, for they have the longest
range and are dominant if the average separation of the molecules is large
...
Secondly, the expressions have been derived by assuming that the molecules
can rotate reasonably freely
...

A different kind of limitation is that eqn 18
...
There is no reason to suppose that the energy of interaction of three (or
more) molecules is the sum of the pairwise interaction energies alone
...
28a)

(r ABr BCr CA)3

where
rBC
qC

C′ = a(3 cos θA cos θB cos θC + 1)

(18
...
The term in C′ (which represents the
non-additivity of the pairwise interactions) is negative for a linear arrangement of

18
...
5 Repulsive and total interactions
When molecules are squeezed together, the nuclear and electronic repulsions and the
rising electronic kinetic energy begin to dominate the attractive forces
...
18
...

In many cases, however, progress can be made by using a greatly simplified representation of the potential energy, where the details are ignored and the general features
expressed by a few adjustable parameters
...
29)

This very simple potential is surprisingly useful for assessing a number of properties
...
3), and r0, the separation at which
V = 0 (Table 18
...
The well minimum occurs at re = 21/6r0
...
An exponential function is more faithful to the
exponential decay of atomic wavefunctions at large distances, and hence to the overlap that is responsible for repulsion
...
These potentials can be used
to calculate the virial coefficients of gases, as explained in Section 17
...
The
potentials are also used to model the structures of condensed fluids
...
1), it has become
possible to measure directly the forces acting between molecules
...
18
...
At
long range the interaction is attractive, but
at close range the repulsions dominate
...
30)

rm

with n > m
...
The
Lennard-Jones potential is a special case of the Mie potential with n = 12 and m = 6
(Fig
...
11); it is often written in the form
1 A r D 12 A r D 6 5
0
0
6
V = 4ε 2
−
(18
...
It is found that the three-body term contributes about 10 per cent of the total
interaction energy in liquid argon
...
18
...
The green and purple
lines are the two contributions
...
4* Lennard-Jones
(12,6) parameters
(e/k)/K

r0 /pm

Ar

(18
...
244r0, and at
7
that distance is equal to −144(––)7/6ε /13r0, or −2
...
For typical parameters, the
26
magnitude of this force is about 10 pN
...
84

362
...
86

624
...
85

391
...
96

426
...


638

18 MOLECULAR INTERACTIONS
IMPACT ON MEDICINE

I18
...

Fig
...
12

A drug is a small molecule or protein that binds to a specific receptor site of a target
molecule, such as a larger protein or nucleic acid, and inhibits the progress of disease
...

Molecular interactions are responsible for the assembly of many biological structures
...
The binding of a ligand, or guest, to a biopolymer, or host, is also governed by molecular interactions
...
In all these cases, a site on the guest contains functional groups that can interact with complementary functional groups of the host
...
It is generally true that many specific intermolecular contacts must be made in a biological host–guest complex and, as a result, a guest
binds only hosts that are chemically similar
...

Interactions between nonpolar groups can be important in the binding of a guest to
a host
...
In addition to dispersion, repulsive, and hydrophobic
interactions, π stacking interactions are also possible, in which the planar π systems of
aromatic macrocycles lie one on top of the other, in a nearly parallel orientation
...
18
...
Some drugs with planar π systems, shown as a green rectangle in Fig
...
12,
are effective because they intercalate between base pairs through π stacking interactions, causing the helix to unwind slightly and altering the function of DNA
...

For example, at physiological pH, amino acid side chains containing carboxylic acid
or amine groups are negatively and positively charged, respectively, and can attract
each other
...
1)
...
Many effective drugs bind tightly and inhibit the
action of enzymes that are associated with the progress of a disease
...

There are two main strategies for the discovery of a drug
...
However, in many cases a number of so-called lead compounds are
known to have some biological activity but little information is available about the
target
...

In broad terms, the first stage of the QSAR method consists of compiling molecular descriptors for a very large number of lead compounds
...
1 IMPACT ON MEDICINE: MOLECULAR RECOGNITION AND DRUG DESIGN

639

mass, molecular dimensions and volume, and relative solubility in water and nonpolar
solvents are available from routine experimental procedures
...

In the second stage of the process, biological activity is expressed as a function of
the molecular descriptors
...

2

(18
...
The quadratic terms account for the fact that biological activity
can have a maximum or minimum value at a specific descriptor value
...
14
for details of membrane structure), or too hydrophobic (water-repelling), for then it
may bind too tightly to the membrane
...

In the final stage of the QSAR process, the activity of a drug candidate can be estimated from its molecular descriptors and the QSAR equation either by interpolation
or extrapolation of the data
...

The traditional QSAR technique has been refined into 3D QSAR, in which
sophisticated computational methods are used to gain further insight into the threedimensional features of drug candidates that lead to tight binding to the receptor site
of a target
...
The key assumption of
the method is that common structural features are indicative of molecular properties
that enhance binding of the drug to the receptor
...
An atomic probe,
typically an sp3-hybridized carbon atom, visits each grid point and two energies of
interaction are calculated: Esteric, the steric energy reflecting interactions between
the probe and electrons in uncharged regions of the drug, and Eelec, the electrostatic
energy arising from interactions between the probe and a region of the molecule carrying a partial charge
...
34)

r

where the c(r) are coefficients calculated by regression analysis, with the coefficients
cS and cE reflecting the relative importance of steric and electrostatic interactions,
respectively, at the grid point r
...
Figure 18
...
Indeed, we
see that the technique lives up to the promise of opening a window into the chemical
nature of the binding site even when its structure is not known
...
The ellipses
indicate areas in the protein’s binding site
with positive or negative electrostatic
potentials and with little or much steric
crowding
...
Also,
substituents that lead to the accumulation
of negative electrostatic potential at either
end of the drug are likely to show enhanced
affinity for the binding site
...
Krogsgaard-Larsen, T
...

Madsen (ed
...
)
Fig
...
13

640

18 MOLECULAR INTERACTIONS
dW

abundant
...


W

Gases and liquids
The definition of the solid angle,
dΩ, for scattering
...
18
...
18
...


RA

RB

b=0

The form of matter with the least order is a gas
...
In a
real gas there are weak attractions and repulsions that have minimal effect on the relative locations of the molecules but that cause deviations from the perfect gas law for
the dependence of pressure on the volume, temperature, and amount (Section 1
...

The attractions between molecules are responsible for the condensation of gases
into liquids at low temperatures
...
Second, although molecules attract each other when they are a few diameters
apart, as soon as they come into contact they repel each other
...
The molecules are held together by molecular interactions,
but their kinetic energies are comparable to their potential energies
...
6 that, although the molecules of a liquid are not free to escape completely from the bulk, the whole structure is very mobile and we can speak only of the
average relative locations of molecules
...


(a)

18
...
The beam is directed towards other molecules, and the scattering that occurs
on impact is related to the intermolecular interactions
...
The fraction is normally expressed in terms of dI, the rate at which molecules
are scattered into a cone that represents the area covered by the ‘eye’ of the detector
(Fig
...
14)
...
18
...
(The target molecule is
taken to be so heavy that it remains
virtually stationary
...
35)

The value of σ (which has the dimensions of area) depends on the impact parameter,
b, the initial perpendicular separation of the paths of the colliding molecules (Fig
...
15),
and the details of the intermolecular potential
...
18
...
If b = 0,
the lighter projectile is on a trajectory that leads to a head-on collision, so the only
scattering intensity is detected when the detector is at θ = π
...
Glancing blows,
with 0 < b ≤ RA + RB, lead to scattering intensity in cones around the forward direction
...
7 THE LIQUID–VAPOUR INTERFACE
The scattering pattern of real molecules, which are not hard spheres, depends on
the details of the intermolecular potential, including the anisotropy that is present when
the molecules are non-spherical
...
18
...
The variation
of the scattering cross-section with the relative speed of approach should therefore
give information about the strength and range of the intermolecular potential
...
The wave nature of the particles can be taken into account, at
least to some extent, by drawing all classical trajectories that take the projectile particle
from source to detector, and then considering the effects of interference between them
...
A particle with a certain
impact parameter might approach the attractive region of the potential in such a way
that the particle is deflected towards the repulsive core (Fig
...
18), which then repels
it out through the attractive region to continue its flight in the forward direction
...
The wavefunctions of the particles that take the two types of path interfere, and the intensity in the forward direction is modified
...
The same phenomenon
accounts for the optical ‘glory effect’, in which a bright halo can sometimes be seen
surrounding an illuminated object
...
)
The second quantum effect we need consider is the observation of a strongly
enhanced scattering in a non-forward direction
...

The origin of the phenomenon is illustrated in Fig
...
19
...
The
rainbow angle, θr, is the angle for which dθ/db = 0 and the scattering is strong
...
The vibrational temperature in supersonic beams is so low that
van der Waals molecules may be formed, which are complexes of the form AB in
which A and B are held together by van der Waals forces or hydrogen bonds
...
More recently, van der Waals clusters of water molecules
have been pursued as far as (H2O)6
...


641

Slow
molecule
Fast
molecule

Fig
...
17 The extent of scattering may
depend on the relative speed of approach
as well as the impact parameter
...


Interfering
paths

Fig
...
18 Two paths leading to the same
destination will interfere quantum
mechanically; in this case they give rise
to quantum oscillations in the forward
direction
...
7 The liquid–vapour interface
So far, we have concentrated on the properties of gases
...
6, we described
the structure of liquids
...
In this section we concentrate on the
liquid–vapour interface, which is interesting because it is so mobile
...

(a) Surface tension

Liquids tend to adopt shapes that minimize their surface area, for then the maximum
number of molecules are in the bulk and hence surrounded by and interacting with
neighbours
...
18
...
The rainbow angle, θr,
is the maximum scattering angle reached as
b is decreased
...


642

18 MOLECULAR INTERACTIONS

Synoptic table 18
...
88

Mercury

472

Methanol

22
...
75

shape with the smallest surface-to-volume ratio
...

Surface effects may be expressed in the language of Helmholtz and Gibbs energies
(Chapter 3)
...
The work
needed to change the surface area, σ, of a sample by an infinitesimal amount dσ is
proportional to dσ, and we write
dw = γ dσ

* More values are given in the Data section
...


[18
...
However, as
in Table 18
...
The work of surface formation at constant volume and temperature can
be identified with the change in the Helmholtz energy, and we can write
dA = γ dσ

(18
...
This is a more formal way of expressing
what we have already described
...
4 Using the surface tension

Calculate the work needed to raise a wire of length l and to stretch the surface of
a liquid through a height h in the arrangement shown in Fig
...
20
...


Force 2gl

l

Total area
2hl

h

Method According to eqn 18
...
Therefore,
all we need do is to calculate the surface area of the two-sided rectangle formed as
the frame is withdrawn from the liquid
...
The total increase is therefore 2lh and the work done is 2γ lh
...
This is why γ is called a tension and why its
units are often chosen to be newtons per metre (N m−1, so γ l is a force in newtons)
...
18
...


Self-test 18
...


[4πr 2γ ]

(b) Curved surfaces

The minimization of the surface area of a liquid may result in the formation of a
curved surface
...
What are widely called ‘bubbles’ in liquids are therefore strictly cavities
...
The treatments of both are similar, but a
factor of 2 is required for bubbles to take into account the doubled surface area
...


18
...
This relation is expressed by the Laplace equation,
which is derived in the following Justification:
2γ
r

(18
...
6 The Laplace equation

The cavities in a liquid are at equilibrium when the tendency for their surface area
to decrease is balanced by the rise of internal pressure which would then result
...
The
former has magnitude 4πr 2pout
...
The change in
surface area when the radius of a sphere changes from r to r + dr is
dσ = 4π(r + dr)2 − 4πr 2 = 8πrdr
(The second-order infinitesimal, (dr)2, is ignored
...
18
...


dw = 8πγ rdr
As force × distance is work, the force opposing stretching through a distance dr
when the radius is r is
F = 8πγ r
The total inward force is therefore 4πr 2pout + 8πγ r
...
18
...
Small
cavities have small radii of curvature, so the pressure difference across their surface
is quite large
...
10 mm in champagne implies a pressure difference of 1
...


P

P - 2g /r + rgh

which rearranges into eqn 18
...


h

P

(c) Capillary action

The tendency of liquids to rise up capillary tubes (tubes of narrow bore), which is
called capillary action, is a consequence of surface tension
...
The energy is lowest when a thin film covers as much of the glass
as possible
...
This curvature implies that the pressure just beneath the
curving meniscus is less than the atmospheric pressure by approximately 2γ /r, where
r is the radius of the tube and we assume a hemispherical surface
...
The excess external pressure presses the liquid up the tube until hydrostatic equilibrium (equal pressures at
equal depths) has been reached (Fig
...
22)
...
18
...
The pressure just
under the meniscus is less than that arising
from the atmosphere by 2γ /r
...


644

18 MOLECULAR INTERACTIONS
To calculate the height to which the liquid rises, we note that the pressure exerted
by a column of liquid of mass density ρ and height h is

Surface tension, g /(mN m-1)

78

p = ρgh

74

(18
...
Therefore, the height of the column at equilibrium is obtained by equating 2γ /r and ρgh,
which gives

70

h=
66

2γ

(18
...
Surface tension decreases with increasing temperature (Fig
...
23)
...
2 Calculating the surface tension of a liquid from its capillary rise

58
-20

0 20 40 60 80 100
Temperature, q /°C

Fig
...
23 The variation of the surface
tension of water with temperature
...
36 cm in a capillary of radius 0
...
1 kg m−3) × (9
...
36 × 10−2 m) × (2
...


gsg

qc

glg
gsl
The balance of forces that results
in a contact angle, θc
...
18
...
This retraction curves the surface with the
concave, high pressure side downwards
...
This compensation results in a capillary depression
...
If this contact angle is θc, then eqn 18
...
The origin of the contact angle can be traced to the balance
of forces at the line of contact between the liquid and the solid (Fig
...
24)
...
41)

This expression solves to
cos θc =

γsg − γsl
γ lg

(18
...
43)

eqn 18
...
44)

We now see that the liquid ‘wets’ (spreads over) the surface, corresponding to 0 <
θc < 90°, when 1 < wad/γ lg < 2 (Fig
...
25)
...
8 CONDENSATION
corresponding to 90° < θc < 180°, when 0 < wad /γ lg < 1
...
23, indicating a relatively low work of adhesion
of the mercury to glass on account of the strong cohesive forces within mercury
...
We saw in Section 4
...
Because curving a surface gives
rise to a pressure differential of 2γ /r, we can expect the vapour pressure above a curved
surface to be different from that above a flat surface
...
3 (p = p*eVm ∆P/RT, where p* is the vapour pressure when
the pressure difference is zero) we obtain the Kelvin equation for the vapour pressure
of a liquid when it is dispersed as droplets of radius r:
2γ Vm /rRT

p = p*e

(18
...
The pressure of the liquid outside the cavity is less than the pressure inside, so
the only change is in the sign of the exponent in the last expression
...
001 and 3, respectively
...
The first figure shows that the effect is usually small;
nevertheless it may have important consequences
...
Warm, moist air rises into the cooler
regions higher in the atmosphere
...
The initial step can be imagined as a
swarm of water molecules congregating into a microscopic droplet
...
Therefore, instead of growing
it evaporates
...
The vapour phase is then said to
be supersaturated
...

Clouds do form, so there must be a mechanism
...
The
first is that a sufficiently large number of molecules might congregate into a droplet so
big that the enhanced evaporative effect is unimportant
...
The more important process depends on the presence of
minute dust particles or other kinds of foreign matter
...

Liquids may be superheated above their boiling temperatures and supercooled
below their freezing temperatures
...
For example, superheating occurs because the vapour pressure
inside a cavity is artificially low, so any cavity that does form tends to collapse
...
Violent bumping often ensues as spontaneous nucleation leads to bubbles big enough to survive
...


1

cos qc

18
...
18
...


646

18 MOLECULAR INTERACTIONS

Checklist of key ideas
1
...

2
...
Orientation polarization is the polarization arising
from the permanent dipole moments
...

3
...
Electronic
polarizability is the polarizability due to the distortion of the
electron distribution
...
The permittivity is the quantity ε in the Coulomb potential
energy, V = q1q2/4πεr
...
The relative permittivity is given by εr = ε /ε0 and may be
calculated from electric properties by using the Debye
equation (eqn 18
...
16)
...
A van der Waals interaction between closed-shell molecules is
inversely proportional to the sixth power of their separation
...
The potential energy of the dipole–dipole interaction between
two fixed (non-rotating) molecules is proportional to µ 1µ 2/r 3
and that between molecules that are free to rotate is
2 2
proportional to µ 1µ 2/kTr 6
...
The dipole–induced-dipole interaction between two molecules
2
is proportional to µ 1α 2/r 6, where α is the polarizability
...
The potential energy of the dispersion (or London)
interaction is proportional to α1α 2/r 6
...
A hydrogen bond is an interaction of the form A-H···B,
where A and B are N, O, or F
...
A hydrophobic interaction is an interaction that favours
formation of clusters of hydrophobic groups in aqueous
environments and that stems from changes in entropy of
water molecules
...
The Lennard-Jones (12,6) potential, V = 4ε{(r0 /r)12 −(r0 /r)6},
is a model of the total intermolecular potential energy
...
A molecular beam is a collimated, narrow stream of molecules
travelling though an evacuated vessel
...

14
...

15
...

16
...

17
...

18
...


Further reading
Articles and texts

P
...
Atkins and R
...
Friedman, Molecular quantum mechanics
...

M
...
D
...
P
...
Chapman and Hall, London (1973)
...
N
...
Academic Press,
New York (1998)
...
A
...
Oxford University
Press (1997)
...
-J
...
Yatsimirsky, Principles and methods in
supramolecular chemistry
...

Sources of data and information

J
...
Jasper, The surface tension of pure liquid compounds
...
Phys
...
Ref
...

D
...
Lide (ed
...
CRC Press, Boca Raton (2000)
...
1 The dipole–dipole interaction

An important problem in physical chemistry is the calculation of the
potential energy of interaction between two point dipoles with
moments m1 and m2, separated by a vector r
...
46)

FURTHER INFORMATION
To calculate / 1, we consider a distribution of point charges qi located
at xi, yi, and zi from the origin
...
47)

i

Target gas

Oven
(source)

Solid
angle, dW

Comment 18
...
If there are several charges q2, q3,
...
The electric field
strength is the negative gradient of the electric potential: / = −—φ
...


where r is the location of the point of interest and the ri are the
locations of the charges qi
...
49)

The electric field strength is (see Comment 18
...
18
...
The
atoms or molecules emerge from a heated source, and pass through
the velocity selector, a rotating slotted cylinder such as that discussed
in Section 1
...
The scattering occurs from the target gas (which
might take the form of another beam), and the flux of particles
entering the detector set at some angle is recorded
...
48)

where the ellipses include the terms arising from derivatives with
respect to yi and zi and higher derivatives
...
Next
we note that ∑i qi xi = µx, and likewise for the y- and z-components
...
50)

Probability

φ(r) =

647

Super
-sonic
nozzle

Maxwell–
Boltzmann
distribution

Molecular speed

It follows from eqns 18
...
50 that
V=

m 1· m 2
4πε 0r 3

−3

(m 1· r)(m 2· r)
4πε 0r 5

(18
...
51 becomes:
V=

µ1µ 2 f(θ )
4πε 0r 3

f(θ ) = 1 − 3 cos 2θ

(18
...
22
...
2 The basic principles of molecular beams

The basic arrangement for a molecular beam experiment is shown in
Fig
...
26
...
The net effect of these collisions, which give
rise to hydrodynamic flow, is to transfer momentum into the

Fig
...
27 The shift in the mean speed and the width of the
distribution brought about by use of a supersonic nozzle
...
The molecules in the beam then travel with
very similar speeds, so further downstream few collisions take place
between them
...
Because the
spread in speeds is so small, the molecules are effectively in a state of
very low translational temperature (Fig
...
27)
...
Such jets are called supersonic
because the average speed of the molecules in the jet is much greater
than the speed of sound for the molecules that are not part of the jet
...
A skimmer consists of a conical nozzle
shaped to avoid any supersonic shock waves spreading back into the
gas and so increasing the translational temperature (Fig
...
28)
...
18
...


gas, and injecting molecules of interest into it in the hydrodynamic
region of flow
...
In

this context, a rotational or vibrational temperature means the
temperature that should be used in the Boltzmann distribution to
reproduce the observed populations of the states
...

The target gas may be either a bulk sample or another molecular
beam
...
The intensity of the incident beam is measured by the
incident beam flux, I, which is the number of particles passing
through a given area in a given interval divided by the area and the
duration of the interval
...
The state of the scattered
molecules may also be determined spectroscopically, and is of interest
when the collisions change their vibrational or rotational states
...
1 Explain how the permanent dipole moment and the polarizability of a
molecule arise
...
4 Account for the theoretical conclusion that many attractive interactions
between molecules vary with their separation as 1/r 6
...
2 Explain why the polarizability of a molecule decreases at high
frequencies
...
5 Describe the formation of a hydrogen bond in terms of molecular orbitals
...
3 Describe the experimental procedures available for determining the

electric dipole moment of a molecule
...
6 Account for the hydrophobic interaction and discuss its manifestations
...
7 Describe how molecular beams are used to investigate intermolecular
potentials
...
1a Which of the following molecules may be polar: CIF3, O3, H2O2?
18
...
2a The electric dipole moment of toluene (methylbenzene) is 0
...

Estimate the dipole moments of the three xylenes (dimethylbenzene)
...
2b Calculate the resultant of two dipole moments of magnitude 1
...
80 D that make an angle of 109
...

18
...
32 nm,
0), and −2e at an angle of 20° from the x-axis and a distance of 0
...

18
...

18
...
62 cm3 mol−1 at 351
...
47 cm3 mol−1 at 423
...

Calculate the polarizability and dipole moment of the molecule
...
4b The molar polarization of the vapour of a compound was found to vary
linearly with T −1, and is 75
...
0 K and 71
...
7 K
...

18
...
18 cm3 mol−1 and its density is 1
...
Calculate the relative
permittivity of the liquid
...
5b At 0°C, the molar polarization of a liquid is 32
...
92 g cm−3
...

Take M = 55
...


18
...
48 × 10−24 cm3; calculate the

dipole moment of the molecule (in addition to the permanent dipole
moment) induced by an applied electric field of strength 1
...

18
...
22 × 10−30 m3; calculate the
dipole moment of the molecule (in addition to the permanent dipole
moment) induced by an applied electric field of strength 15
...

18
...
732 for 656 nm light
...
32 g cm−3
...


PROBLEMS

649

18
...
622 for 643 nm light
...
99 g cm−3
...
Take M = 65
...


18
...
0 nm at 35
...
The vapour pressure of bulk water at that temperature is
5
...
0 kg m−3
...
8a The polarizability volume of H2O at optical frequencies is 1
...
11a The contact angle for water on clean glass is close to zero
...
The experimental value is 1
...
96 cm in a clean glass capillary tube of internal radius
0
...
The density of water at 20°C is 998
...


18
...
3 g mol−1 and
density 865 kg mol−1 at optical frequencies is 2
...

18
...
57 D and its polarizability
volume is 1
...
Estimate its relative permittivity at 25°C, when its
density is 1
...


18
...


18
...
17 × 10−30 C m and its
−29

18
...
Calculate
the surface tension of water at 30°C given that at that temperature water
climbs to a height of 9
...
320 mm
...
9956 g cm−3
...
5 × 10 m
...

3

18
...
12b Calculate the pressure differential of ethanol across the surface of a
spherical droplet of radius 220 nm at 20°C
...
39 mN m−1
...
The vapour pressure of bulk water at that temperature is
2
...
9982 g cm−3
...
1 Suppose an H2O molecule (µ = 1
...
What is
the favourable orientation of the molecule? Calculate the electric field (in volts
per metre) experienced by the anion when the water dipole is (a) 1
...
3 nm, (c) 30 nm from the ion
...
2 An H2O molecule is aligned by an external electric field of strength
1
...
66 × 10−24 cm3) is brought up slowly from
one side
...
1

3
...
0

6
...
0

5
...
0

ρ/(g cm−3)

1
...
64

1
...
61

1
...
53

1
...
Account for these results and
calculate the dipole moment and polarizability volume of the molecule
...
4 The relative permittivities of methanol (m
...
−95°C) corrected for density
variation are given below
...
791 g cm−3 at 20°C
...
2

3
...
0

5
...
5 In his classic book Polar molecules, Debye reports some early
measurements of the polarizability of ammonia
...


T/K

292
...
0

333
...
0

413
...
0

Pm /(cm3 mol−1)

57
...
01

51
...
99

42
...
6 Values of the molar polarization of gaseous water at 100 kPa as
determined from capacitance measurements are given below as a function of
temperature
...
3

420
...
7

484
...
0

Pm /(cm3 mol−1)

57
...
5

50
...
8

43
...

18
...
Luo, G
...
MeBane, O
...
F
...
R
...
Chem
...
3 The relative permittivity of chloroform was measured over a range of
temperatures with the following results:

θ /°C

and at 292
...
Combine the value calculated with the static molar
polarizability at 292
...


39
...
000 379 (for yellow
sodium light)
...
98, 3564 (1993)) reported experimental observation of the He2 complex,
a species that had escaped detection for a long time
...
51 × 10−23 J, hcD0 about 2 × 10−26 J, and R about 297 pm
...
(b) Plot the Morse potential given that
a = 5
...

18
...
D
...
T
...
Klemperer (Science 238, 1670 (1987))

examined several weakly bound gas-phase complexes of ammonia in search
of examples in which the H atoms in NH3 formed hydrogen bonds, but found
none
...
The permanent dipole moment of this complex is reported as
1
...
If the N and C atoms are the centres of the negative and positive
charge distributions, respectively, what is the magnitude of those partial
charges (as multiples of e)?
18
...
1 calculate the molar polarization, relative
permittivity, and refractive index of methanol at 20°C
...
7914 g cm−3
...


650

18 MOLECULAR INTERACTIONS

Theoretical problems
18
...

18
...
Go on to show how to deduce the polarizability
volume of a molecule from measurements of the refractive index of a gaseous
sample
...
12 Acetic acid vapour contains a proportion of planar, hydrogen-bonded
dimers
...
14 at 290 K and
increases with increasing temperature
...
What effect should isothermal dilution have on the relative
permittivity of solutions of acetic acid in benzene?
18
...
Hence, find a connection
between the van der Waals parameter a and C6, from n2alV 2 = (∂U/∂V)T
...
14 Suppose the repulsive term in a Lennard-Jones (12,6)-potential is
replaced by an exponential function of the form e−r/d
...

18
...
Show that
1
V = – N ∫V(R)dτ, where N is the number density of the molecules and V(R) is
2
their attractive potential energy and where the integration ranges from d to
infinity and over all angles
...


18
...
18)
...
0 nm
...
0 eV
...
What is the separation at which the force between the phenyl
groups (treated as benzene molecules) of two Phe residues is zero? Hint
...
At the end of the calculation, let
<
δr become vanishingly small
...
20 Molecular orbital calculations may be used to predict structures of
intermolecular complexes
...

Consider methyl-adenine (16, with R = CH3) and methyl-thymine (17, with
R = CH3) as models of two bases that can form hydrogen bonds in DNA
...
(b) Based on your tabulation of atomic charges, identify the
atoms in methyl-adenine and methyl-thymine that are likely to participate in
hydrogen bonds
...
For this step, you may want to
use your molecular modelling software to align the molecules properly
...
(e) Repeat parts (a)–(d) for
cytosine and guanine, which also form base pairs in DNA (see Chapter 19 for
the structures of these bases)
...
16 Consider the collision between a hard-sphere molecule of radius R1 and

mass m, and an infinitely massive impenetrable sphere of radius R2
...
Carry out the
calculation using simple geometrical considerations
...
17 The dependence of the scattering characteristics of atoms on the energy
of the collision can be modelled as follows
...
16, but that the
effective radius of the heavy atoms depends on the speed v of the light atom
...

1
1
Take R1 = – R2 for simplicity and an impact parameter b = – R2, and plot the
2
2
scattering angle as a function of (a) speed, (b) kinetic energy of approach
...
18 Phenylalanine (Phe, 15) is a naturally occurring amino acid
...
0 nm and treat the phenyl group as a benzene molecule
...
7 D and the polarizability volume
of benzene is α ′ = 1
...


18
...
(a) Using molecular modelling software and the
computational method of your choice, calculate the dipole moment of the
peptide link, modelled as a trans-N-methylacetamide (18)
...
0 nm (see eqn
18
...
(b) Compare the maximum value of the dipole–dipole interaction
energy from part (a) to 20 kJ mol−1, a typical value for the energy of a
hydrogen-bonding interaction in biological systems
...
22 This problem gives a simple example of a quantitative
structure–activity relation (QSAR)
...
(a) Consider a family of hydrocarbons R-H
...
5, 1
...
5, 2
...
5
...
(b) The equilibrium constants KI for the
dissociation of inhibitors (19) from the enzyme chymotrypsin were measured
for different substituents R:
R

CH3CO

CN

NO2

CH3

Cl

π

−0
...
025

0
...
5

0
...
73

−1
...
43

−2
...
40

Plot log KI against π
...

18
...
A
QSAR analysis of the activity A of a number of TIBO derivatives suggests the
following equation:

log A = b0 + b1S + b2W
where S is a parameter related to the drug’s solubility in water and W is a
parameter related to the width of the first atom in a substituent X shown in 20
...
Hint
...
To fit the data, you must use the mathematical procedure
of multiple regression, which can be performed with mathematical software or
an electronic spreadsheet
...
36 8
...
3

Cl

SCH3

7
...
25 6
...
52 7
...
53

S

3
...
24 4
...
45

2
...
89

4
...
03

3
...
00 1
...
70

1
...
60 1
...
95 1
...
60

(b) What should be the value of W for a drug with S = 4
...
60?

19
Determination of size and shape
19
...
2
19
...
4
19
...
1

19
...
7
19
...
9
I19
...
10
19
...
12

The different levels of
structure
Random coils
The structure and stability of
synthetic polymers
Impact on technology:
Conducting polymers
The structure of proteins
The structure of nucleic acids
The stability of proteins and
nucleic acids

Self-assembly
19
...
14 Micelles and biological

membranes
19
...
3 Impact on nanoscience:
Nanofabrication with selfassembled monolayers

Materials 1:
macromolecules
and aggregates
Macromolecules exhibit a range of properties and problems that illustrate a wide variety of
physical chemical principles
...
However, the molecules are so large and the solutions they form
depart so strongly from ideality, that techniques for accommodating these departures from
ideality need to be developed
...
We consider a range of influences in this chapter, beginning with a structureless random coil and ending with the structurally precise forces that
operate in polypeptides and nucleic acids
...
These assemblies, which include colloids and biological membranes,
exhibit some of the typical properties of molecules but have their own characteristic features
...
Some are natural:
they include polysaccharides such as cellulose, polypeptides such as protein enzymes,
and polynucleotides such as deoxyribonucleic acid (DNA)
...
Life in
all its forms, from its intrinsic nature to its technological interaction with its environment, is the chemistry of macromolecules
...
Natural macromolecules differ in certain respects from synthetic macromolecules, particularly in their composition and the resulting structure,
but the two share a number of common properties
...
Another level of complexity arises when small molecules aggregate
into large particles in a process that is called ‘self-assembly’ and give rise to aggregates
...

A similar type of aggregation gives rise to a variety of disperse phases, which include
colloids
...


Determination of size and shape

Checklist of key ideas
Further reading
Further information 19
...

However, there are several reasons why other techniques must also be used
...
1 MEAN MOLAR MASSES
place, the sample might be a mixture of molecules with different chain lengths and
extents of cross-linking, in which case sharp X-ray images are not obtained
...
Furthermore, although work on proteins and DNA has shown how immensely interesting and motivating the data can be, the information is incomplete
...
1 Mean molar masses
A pure protein is monodisperse, meaning that it has a single, definite molar mass
(although there may be small variations, such as one amino acid replacing another,
depending on the source of the sample)
...
The various techniques that are used to measure molar masses result in
different types of mean values of polydisperse systems
...
5) is the number-average molar mass, Jn, which is the value obtained by weighting
each molar mass by the number of molecules of that mass present in the sample:
Jn =

1
N

∑ Ni Mi

(19
...
Viscosity measurements give the viscosity-average molar mass, J v, lightscattering experiments give the weight-average molar mass, J w, and sedimentation
experiments give the Z-average molar mass, JZ
...
) Although
such averages are often best left as empirical quantities, some may be interpreted in
terms of the composition of the sample
...
2)

i

In this expression, mi is the total mass of molecules of molar mass Mi and m is the total
mass of the sample
...
3)

i

This expression shows that the weight-average molar mass is proportional to the
mean square molar mass
...
4)

653

654

19 MATERIALS 1: MACROMOLECULES AND AGGREGATES
Example 19
...
5

9
...
5

8
...
5

8
...
5

5
...
5

3
...
5

1
...
1 and 19
...
Calculate the two averages

by weighting the molar mass within each interval by the number and mass, respectively, of the molecule in each interval
...
Because the number of molecules is proportional to the amount of
substance (the number of moles), the number-weighted average can be obtained
directly from the amounts in each interval
...
5
Amount/mol

1
...
5

17
...
5

0
...
51

22
...
25

27
...
11

0
...
92

The number-average molar mass is therefore
Jn/(kg mol−1) =

1

(1
...
5 + 0
...
5 + 0
...
5 + 0
...
5
2
...
11 × 27
...
052 × 32
...
6 g:
Jw /(kg mol−1) =

1

(9
...
5 + 8
...
5 + 8
...
5 + 5
...
5
37
...
1 × 27
...
7 × 32
...
In this instance,
Jw /Jn = 1
...

Self-test 19
...


[19 kg mol −1]

The ratio Jw /Jn is called the heterogeneity index (or ‘polydispersity index’)
...
2 MASS SPECTROMETRY

655

same because the sample is monodisperse (unless there has been degradation)
...
Typical synthetic materials have J w /Jn ≈ 4
...
1; commercial polyethylene samples might be much more heterogeneous, with
a ratio close to 30
...
The spread of values is controlled
by the choice of catalyst and reaction conditions
...

Average molar masses may be determined by osmotic pressure of polymer solutions
...
A major problem for macromolecules of relatively low molar mass (less
than about 10 kg mol−1) is their ability to percolate through the membrane
...
Several techniques for the
determination of molar mass and polydispersity that are not so limited include mass
spectrometry, laser light scattering, ultracentrifugation, electrophoresis, and viscosity
measurements
...
2 Mass spectrometry
Mass spectrometry is among the most accurate techniques for the determination
of molar masses
...
Macromolecules
present a challenge because it is difficult to produce gaseous ions of large species
without fragmentation
...
We shall discuss MALDI-TOF mass spectrometry, so called because the
MALDI technique is coupled to a time-of-flight (TOF) ion detector
...
1 shows a schematic view of a MALDI-TOF mass spectrometer
...
This sample is then irradiated with a pulsed laser, such as a
nitrogen laser
...

The macromolecule is ionized by collisions and complexation with small cations, such
as H+, Na+, and Ag+
...
The time, t,
required for an ion of mass m and charge number z to reach the detector at the end of
the drift region is (see the Justification):

A m D
t=l
C 2zeEd F

1/2

(19
...
Because d, l, and E are fixed for a given experiment,
the time of flight, t, of the ion is a direct measure of its m/z ratio, which is given by:

A tD
= 2eEd
C lF
z

m

2

(19
...
19
...
A laser beam ejects
macromolecules and ions from the solid
matrix
...
Ions with
the smallest mass to charge ratio (m/z)
reach the detector first
...
1 The time of flight of an ion in a mass spectrometer

3

Relative intensity

n = 20

Consider an ion of charge ze and mass m that is accelerated from rest by an electric
field of strength E applied over a distance d
...
The drift region, l, and the time of flight, t, in the mass
spectrometer are both sufficiently short that we can ignore acceleration and write
v = l/t
...
6
...
19
...
,
J
...
Educ
...
)

Figure 19
...
The MALDI technique produces mostly singly charged
molecular ions that are not fragmented
...
Values of Jn, J w, and the
heterogeneity index can be calculated from the data
...


Example 19
...
19
...
The peak
at 4113 g mol−1 corresponds to the polymer for which n = 20
...

Method Because each peak corresponds to a different value of n, the molar mass

difference, ∆M, between peaks corresponds to the molar mass, M, of the repeating
unit (the group inside the brackets in 1)
...

Answer The value of ∆M is consistent with the molar mass of the repeating unit
shown in (1), which is 200 g mol−1
...

Self-test 19
...
3 LASER LIGHT SCATTERING

657

19
...
A familiar example is the light scattered
by specks of dust in a sunbeam
...
Unlike mass spectrometry, laser light scattering
measurements may be performed in nearly intact samples; often the only preparation
required is filtration of the sample
...
1)
...
The term
elastic refers to the fact that the incident and scattered photons have the same frequency and hence the same energy
...
If the medium is inhomogeneous,
as in an imperfect crystal or a solution of macromolecules, radiation is scattered into
other directions as well
...
19
...
This type of scattering
has several characteristic features
...

2 The intensity of scattered light is proportional to the molar mass of the particle
...
19
...
In
practice, data are collected at several angles to the incident laser beam (Example 19
...

4 For very dilute solutions excited by plane-polarized light, the Rayleigh ratio, Rθ ,
a measure of the intensity of scattered light at a given scattering angle θ, is defined as
Rθ =

I
I0

×

r2

(19
...
19
...

For a solution of a polymer of mass concentration cP, the Rayleigh ratio may be written as
Rθ = K Pθ cP J w, with K =

4π2n2 V(dnr /dcP)2
r,0

λ4NA

(19
...
6 and Appendix
3), (dn/dcP) is the change in refractive index of the solution with concentration of
polymer, V is the volume of the sample, and NA is Avogadro’s constant
...
When the molecule is much smaller than the
wavelength of incident radiation, Pθ ≈ 1
...
19
...
The intensity of
scattered light depends on the angle θ
between the incident and scattered beams
...
In a typical experimental
arrangement, φ = 90°
...
1 that
Pθ ≈ 1 −
(a)

(b)

Fig
...
4 (a) A spherical molecule and (b)
the hollow spherical shell that has the same
rotational characteristics
...


Synoptic table 19
...
98

Polystyrene

3
...

† In a poor solvent
...
9)

where Rg is the radius of gyration of the macromolecule, the radius of a thin hollow
spherical shell of the same mass and moment of inertia as the molecule (Fig
...
4 and
Section 19
...
Table 19
...

Illustration 19
...
9 that when the particles are very small and Pθ ≈ 1, the
medium scatters light of shorter wavelengths much more efficiently than light of
longer wavelengths
...
We also
see clouds because light scatters from them, but they look white, not blue
...
Although blue light scatters
more strongly, more molecules can contribute cooperatively when the wavelength
is longer (as for red light), so the net result is uniform scattering for all wavelengths:
white light scatters as white light
...
As a
result, the scattering intensity is distorted from the form characteristic of smallparticle, Rayleigh scattering, and the distortion is taken into account by values of Pθ
that differ from 1
...
However, eqn 19
...

In practice, even relatively dilute polymer dispersions can deviate considerably from
ideality
...
In thermodynamic
terms, the displacement and reorganization of solvent molecules implies that the
entropy change is especially important when a macromolecule dissolves
...
There are also significant contributions to the Gibbs energy from the
enthalpy of solution, largely because solvent–solvent interactions are more favourable
than the macromolecule–solvent interactions that replace them
...
8 as
KcP
Rθ

=

1
Pθ Jw

+ BcP

(19
...

For most solute–solvent systems there is a unique temperature (which is not always
experimentally attainable) at which the effects leading to non-ideal behaviour cancel
and the solution is virtually ideal
...
3 LASER LIGHT SCATTERING

659

temperature for real gases) is called the q-temperature (theta temperature)
...
As an example, for polystyrene in cyclohexane the θ-temperature
is approximately 306 K, the exact value depending on the average molar mass of the
polymer
...
Because a θ-solution
behaves nearly ideally, its thermodynamic and structural properties are easier to
describe even though the molar concentration is not low
...

Example 19
...

26
...
9

66
...
0

19
...
8

17
...
0

7

...
6
14
...
42 × 10−5 mol m5 kg−2
...
Assume that B is negligibly
small and that the polymer is small enough that eqn 19
...

Method Substituting the result of eqn 19
...
8 we obtain, after some

rearrangement:
1
Rθ

=

1
KcP J w

+

A 16π2R2 D A 1
D
g
1
sin2 –θ
2
C 3λ2 F C Rθ
F

6
...
5

1
Hence, a plot of 1/Rθ against (1/Rθ ) sin2 –θ should be a straight line with slope
2
16π2R2 /3λ2 and y-intercept 1/KcP J w
...
5

(Fig
...
5)
...
06

1
/Rθ ) sin2(–θ )/m−2
2

3

(10

5
...
83

6
...
33

17
...
3

48
...
391 and a y-intercept of
5
...
From these values and the value of K, we calculate Rg = 4
...
1 nm and J w = 987 kg mol−1
...

From analysis of the entire data set, the Rg, J w, and B values are obtained
...
3 The following data for an aqueous solution of a protein with cP =
2
...
0

45
...
0

85
...
0

23
...
9

21
...
7

20
...
40 × 10−2 mol m5 kg−2
...
Assume that B is negligibly small and that the protein is small enough that
eqn 19
...

[Rg = 39
...
96

2
...
19
...
3
...
2* Diffusion
coefficients in water at 20°C
M/(kg mol−1)
Sucrose

0
...
59 × 10 −10

Lysozyme

14
...
04 × 10 −10

Haemoglobin

68

6
...
9 × 10 −12

Collagen

* More values are given in the Data section
...
3* Frictional
coefficients and molecular geometry†
a/b

Prolate

Oblate

2

1
...
04

3

1
...
17

6

1
...
28

8

1
...
37

10

1
...
46

* More values and analytical expressions are
given in the Data section
...


A special laser scattering technique, dynamic light scattering, can be used to investigate the diffusion of polymers in solution
...
Suppose that at a time t the scattered waves from these particles interfere constructively at the detector, leading to a large signal
...
When this behaviour is extended to a very
large number of molecules in solution, it results in fluctuations in light intensity that
depend on the diffusion coefficient, D, which is a measure of the rate of molecular
motion and is given by the Stokes–Einstein relation (which is discussed further in
Section 21
...
11)

where f is the frictional coefficient, a measure of the forces that retard a molecule’s
motion
...
2 lists some typical values of D
...
6), the frictional coefficient is given by Stokes’s
relation:
f = 6πaη

(19
...
3
...
For dilute monodisperse
systems of random coils, it has been found empirically that D is related to the molar
mass M of the polymer by:
D = β D M −0
...
13)

The coefficient β D is obtained by determining D at fixed viscosity and temperature for
a variety of standard samples with known molar masses
...

19
...
The rate of sedimentation depends on the strength
of the field and on the masses and shapes of the particles
...

When the sample is at equilibrium, the particles are dispersed over a range of heights
in accord with the Boltzmann distribution (because the gravitational field competes
with the stirring effect of thermal motion)
...

Sedimentation is normally very slow, but it can be accelerated by ultracentrifugation, a technique that replaces the gravitational field with a centrifugal field
...
19
...

Modern ultracentrifuges can produce accelerations equivalent to about 105 that of
gravity (‘105 g’)
...


19
...
14)

where ρ is the solution density, vs is the partial specific volume of the solute (vs =
(∂V/∂mB)T , with mB the total mass of solute), and ρ vs is the mass of solvent displaced
per gram of solute
...
The
acceleration outwards is countered by a frictional force proportional to the speed, s, of
the particles through the medium
...
3)
...
The
forces are equal when
s=

=

bmr ω 2

Rotor
(a)

(19
...
16)

rω 2

bJn

(19
...
12), we obtain
S=

Solution
(b) 'Bottom'

s

Then, because the average molecular mass is related to the average molar mass Jn
through m = Jn/NA,
S=

w
r

The drift speed depends on the angular velocity and the radius, and it is convenient to
define the sedimentation constant, S, as
S=

661

bJn

(19
...
Again, if the molecules are not spherical, we use the appropriate value of f given in Table 19
...
As always when dealing with
macromolecules, the measurements must be carried out at a series of concentrations
and then extrapolated to zero concentration to avoid the complications that arise
from the interference between bulky molecules
...
4 Determining a sedimentation constant

The sedimentation of the protein bovine serum albumin (BSA) was monitored at
25°C
...
50 cm from the axis of rotation, and during centrifugation at 56 850 r
...
m
...
50

5
...
60

5
...
80

5
...
01

Calculate the sedimentation coefficient
...
16 can be interpreted as a differential equation for s = dr/dt
in terms of r; so integrate it to obtain a formula for r in terms of t
...


Fig
...
6 (a) An ultracentrifuge head
...
(b) Detail of the
sample cavity: the ‘top’ surface is the inner
surface, and the centrifugal force causes
sedimentation towards the outer surface; a
particle at a radius r experiences a force of
magnitude mrω 2
...
16 may be written

dr

8

= rω 2S

100 ln (r /r0)

dt

This equation integrates to

6

ln
4

r
r0

= ω 2St

It follows that a plot of ln(r/r0) against t should be a straight line of slope ω 2S
...
19
...
4
...
905

1
...
57

5
...
19

8
...
19
...
78 × 10−5; so ω 2S = 1
...

Because ω = 2π × (56 850/60) s−1 = 5
...
02 × 10−13 s
...
02 Sv
...
4 Calculate the sedimentation constant given the following data (the
other conditions being the same as above):

t/s

0

500

1000

2000

3000

4000

5000

r/cm

5
...
68

5
...
77

5
...
9

5
...
11 Sv]

At this stage, it appears that we need to know the molecular radius a to obtain the
molar mass from the value of S
...
11) between f and the diffusion
coefficient, D
...
19)

bD

where we are not specifying which mean molar mass because the average obtained
depends on technical details of the experiment
...
19 is independent
of the shape of the solute molecules
...

(b) Sedimentation equilibria

The difficulty with using sedimentation rates to measure molar masses lies in the
inaccuracies inherent in the determination of diffusion coefficients of polydisperse
systems
...
As we show in the Justification
below, the weight-average molar mass can be obtained from the ratio of concentrations of the macromolecules at two different radii in a centrifuge operating at angular
frequency ω :
Jw =

2RT
(r 2 − r 2)bω 2
2
1

ln

c2
c1

(19
...
5 ELECTROPHORESIS
An alternative treatment of the data leads to the Z–average molar mass
...
At these
slower speeds, several days may be needed for equilibrium to be reached
...
2 The weight-average molar mass from sedimentation experiments

The distribution of particles is the outcome of the balance between the effect of
the centrifugal force and the dispersing effect of diffusion down a concentration
gradient
...
The condition for equilibrium is that the chemical potential is
uniform, so
A ∂G D
A ∂µ D
B E = B E − Mω 2r = 0
C ∂r F T C ∂r F T
To evaluate the partial derivative of µ, we write
A ∂ ln c D
A ∂µ D
A ∂µ D A ∂p D
A ∂µ D A ∂c D
E
B E = B E B E + B E B E = Mvω 2rρ + RT B
C ∂r F T,p
C ∂r F T C ∂p F T,c C ∂r F T,c C ∂c F T,p C ∂r F T,p
The first result follows from the fact that (∂µ /∂p)T = Vm, the partial molar volume,
and Vm = Mv
...
The concentration term stems from the expression µ = µ 7 + RT ln c
...
20
...
5 Electrophoresis
Many macromolecules, such as DNA, are charged and move in response to an electric
field
...
Electrophoretic mobility is a result of a
constant drift speed, s, reached by an ion when the driving force zeE (where, as usual,
ze is the net charge and E is the field strength) is matched by the frictional force fs
...
7) is then:
s=

zeE
f

(19
...
The latter two factors are implied by
the dependence of s on f
...
According to eqn 19
...
However, there are limits to this strategy because
very large electric fields can heat the large surfaces of an electrophoresis apparatus
unevenly, leading to a non-uniform distribution of electrophoretic mobilities and
poor separation
...
The small size of the apparatus makes it easy to dissipate heat when large electric fields are applied
...
Each polymer fraction emerging from the capillary can be characterized further by other techniques, such as MALDI-TOF
...
1 Gel electrophoresis in genomics and proteomics

Advances in biotechnology are linked strongly to the development of physical techniques
...
However, decoding genomic DNA will not always lead to accurate predictions
of the amino acids present in biologically active proteins
...
Moreover, it is known that one piece of DNA may encode more than one
active protein
...

The procedures of genomics and proteomics, the analysis of the genome and proteome, of complex organisms are time-consuming because of the very large number
of molecules that must be characterized
...

Success in the characterization of the genome and proteome of any organism will
depend on the deployment of very rapid techniques for the determination of the order
in which molecular building blocks are linked covalently in DNA and proteins
...
Because the molecules must pass through the pores in the gel, the larger the
macromolecule the less mobile it is in the electric field and, conversely, the smaller the
macromolecule the more swiftly it moves through the pores
...
Two common gel materials for the study of proteins and nucleic acids
are agarose and cross-linked polyacrylamide
...

Polyacrylamide gels with varying pore sizes can be made by changing the concentration of acrylamide in the polymerization solution
...

The separation of very large pieces of DNA, such as chromosomes, by conventional
gel electrophoresis is not effective, making the analysis of genomic material rather
difficult
...
This problem can
be avoided with pulsed-field electrophoresis, in which a brief burst of the electric

19
...
6 Viscosity
The formal definition of viscosity is given in Section 21
...

The presence of a macromolecular solute increases the viscosity of a solution
...
At low concentrations the viscosity, η, of
the solution is related to the viscosity of the pure solvent, η0, by

η = η0(1 + [η]c + · · · )

(19
...
It follows from eqn 19
...
5

Speed, s /(mm s-1)

field is applied first along one direction and then along a perpendicular direction
...
In this way, the mobility of the
macromolecule can be related to its molar mass
...
For example, proteins
of the same size but different net charge travel along the slab at different speeds
...
Sodium dodecyl sulfate is an anionic detergent that is
very useful in this respect: it denatures proteins, whatever their initial shapes, into
rods by forming a complex with them
...
Under these conditions, different proteins in a mixture may be separated according to size only
...
However, molar masses obtained by this method, often referred to as SDS-PAGE when
polyacrylamide gels are used, are not as accurate as those obtained by MALDI-TOF or
ultracentrifugation
...
For instance, in acidic environments protons attach to basic
groups and the net charge is positive; in basic media the net charge is negative as a
result of proton loss
...
Consequently, the drift speed of a biopolymer depends on the pH
of the medium, with s = 0 at the isoelectric point (Fig
...
8)
...
In this
technique, a mixture of proteins is dispersed in a medium with a pH gradient along
the direction of an applied electric field
...
In this manner, the protein mixture can be separated into its components
...
However, the two techniques can be combined in two-dimensional (2D) electrophoresis
...
19
...
To improve the separation of closely spaced
bands, the first slab is attached to a second slab and SDS-PAGE is performed with the
electric field being applied in a direction that is perpendicular to the direction in
which isoelectric focusing was performed
...
19
...


665

0

-0
...
5

4

4
...
19
...
The isoelectric point of the
macromolecule corresponds to the pH at
which the drift speed in the presence of an
electric field is zero
...
(a)
Isoelectric focusing is performed on a thin
gel slab, resulting in separation along the
vertical direction of the illustration
...
The
dashed horizontal lines show how the
bands in the two-dimensional gel
correspond to the bands in the gel on
which isoelectric focusing was performed
...
19
...
4* Intrinsic viscosity
Solvent

Measuring
lines

K/(cm3 g−1)

a

Benzene

Polystyrene

q/°C
25

9
...
74

23

−2

0
...
2 × 10−3

0
...
3 × 10

* More values are given in the Data section
...
19
...
The
viscosity is measured by noting the time
required for the liquid to drain between
the two marks
...
In the Ostwald viscometer shown in
Fig
...
10, the time taken for a solution to flow through the capillary is noted, and
compared with a standard sample
...
23)

η0

=

t
t0

×

ρ

(19
...
) This ratio can be
used directly in eqn 19
...
Viscometers in the form of rotating concentric cylinders
are also used (Fig
...
11), and the torque on the inner cylinder is monitored while the
outer one is rotated
...

There are many complications in the interpretation of viscosity measurements
...

Some regularities are observed that help in the determination
...
The torque
on the inner drum is observed when the
outer container is rotated
...
19
...
25)

where K and a are constants that depend on the solvent and type of macromolecule
(Table 19
...

Example 19
...
58

6
...
74

7
...
98

8
...
25 with K = 3
...
63
...
7 THE DIFFERENT LEVELS OF STRUCTURE

667

0
...
23; therefore, form this ratio at
−1

the series of data points and extrapolate to c = 0
...
25
...
102

1
...
317

1
...
549

5
...
2

5
...
38

5
...
19
...
The extrapolated intercept at c = 0 is 0
...
0504 dm3 g−1
...
054

0
...
052

0
...
0 × 104 g mol−1

Self-test 19
...

[90 kg mol−1]

In some cases, the flow is non-Newtonian in the sense that the viscosity of the solution changes as the rate of flow increases
...
In some somewhat rare cases the
stresses set up by the flow are so great that long molecules are broken up, with further
consequences for the viscosity
...
050
0

2

6
8
4
c /(g dm-3)

10

Fig
...
12 The plot used for the
determination of intrinsic viscosity, which
is taken from the intercept at c = 0; see
Example 19
...


Structure and dynamics
The concept of the ‘structure’ of a macromolecule takes on different meanings at
the different levels at which we think about the arrangement of the chain or network
of monomers
...
Thus, the chains
-A-B-C- and -A-C-B- have different configurations
...

19
...
The residues may form either a chain, as in polyethylene, or
a more complex network in which cross-links connect different chains, as in crosslinked polyacrylamide
...
Thus, the repeating
unit of polyethylene is -CH2CH2-, and the primary structure of the chain is specified
by denoting it as -(CH2CH2)n-
...
For example, proteins are polypeptides formed
from different amino acids (about twenty occur naturally) strung together by the
peptide link, -CONH-
...
1

More rigorously, the repeating unit
in polyethylene is -CH2- and the
substance is polymethylene
...


668

19 MATERIALS 1: MACROMOLECULES AND AGGREGATES

=
(a)

(b)
Fig
...
13 (a) A polymer adopts a highly
organized helical conformation, an
example of a secondary structure
...
(b) Several
helical segments connected by short
random coils pack together, providing an
example of tertiary structure
...
The degradation of a polymer
is a disruption of its primary structure, when the chain breaks into shorter components
...
The secondary structure of an isolated molecule of polyethylene is a
random coil, whereas that of a protein is a highly organized arrangement determined
largely by hydrogen bonds, and taking the form of random coils, helices (Fig
...
13a),
or sheets in various segments of the molecule
...
When the hydrogen bonds in a protein are destroyed (for instance, by
heating, as when cooking an egg) the structure denatures into a random coil
...
For instance, the hypothetical protein shown in Fig
...
13b has helical regions
connected by short random-coil sections
...

The quaternary structure of a macromolecule is the manner in which large
molecules are formed by the aggregation of others
...
14 shows how four molecular subunits, each with a specific tertiary structure, aggregate together
...
For example, the oxygen-transport protein
haemoglobin consists of four subunits that work together to take up and release O2
...
8 Random coils
The most likely conformation of a chain of identical units not capable of forming
hydrogen bonds or any other type of specific bond is a random coil
...
The random coil model is a helpful starting point for estimating the
orders of magnitude of the hydrodynamic properties of polymers and denatured proteins in solution
...
19
...
We assume
that the residues occupy zero volume, so different parts of the chain can occupy the
same region of space
...
19
...
In a hypothetical one-dimensional freely jointed chain all the residues lie
in a straight line, and the angle between neighbours is either 0° or 180°
...


Fig
...
14 Several subunits with specific
tertiary structures pack together, providing
an example of quaternary structure
...
19
...


q

q

Arbitrary
angle

Fig
...
16 A better description is obtained
by fixing the bond angle (for example, at
the tetrahedral angle) and allowing free
rotation about a bond direction
...
8 RANDOM COILS
(a) Measures of size

1

As shown in the following Justification, we can deduce the probability, P, that the ends
of a one-dimensional freely jointed chain composed of N units of length l are a distance nl apart:
1/2

e−n /2N
2

(19
...
19
...
We
write this probability as f(r)dr, where
3

A a D
2 2
f (r) = 4π 1/2 r 2e−a r
Cπ F

A 3 D
a=
C 2Nl 2 F

1/2

Consider a one-dimensional freely jointed polymer
...
The distance between the ends of the chain is
(NR − NL)l, where l is the length of an individual bond
...

The number of ways W of forming a chain with a given end-to-end distance nl is
the number of ways of having NR right-pointing and NL left-pointing bonds
...
1 = N! ways of selecting whether a step should be to the right
or the left
...
However, we
end up at the same point for all NL! and NR! choices of which step is to the left and
which to the right
...
28)

The probability that the separation is nl is
P=
=

number of polymers with NR bonds to the right
total number of arrangements of bonds
N!/N R!(N − NR)!
2

N

=

1
{–(N +
2

N!
1
n)}!{–(N − n)}!2N
2

When the chain is compact in the sense that n < N, it is more convenient to
<
evaluate ln P: the factorials are then large and we can use Stirling’s approximation
(Section 16
...
4

0
...
3 The one-dimensional freely jointed chain

N!

0
...
27)

In some coils, the ends may be far apart whereas in others their separation is small
...

Although eqn 19
...

An alternative interpretation of eqn 19
...


W=

0
...
29)

0
-4

-2

0
n /N 1/2

2

4

Fig
...
17 The probability distribution
for the separation of the ends of a onedimensional random coil
...


670

19 MATERIALS 1: MACROMOLECULES AND AGGREGATES
where ν = n/N
...
26
...
6 Provide the algebraic steps that lead from eqn 19
...
29
...
The contour
length, Rc, is the length of the macromolecule measured along its backbone from
atom to atom
...
We show in the
following Justification that:

Rrms /l

40

Rrms = N 1/2l

4000

3000

2000

0

1000

20

0

(19
...
19
...


(19
...
19
...
The result must be multiplied by a factor when the chain is not freely jointed
(see below)
...
4 The root mean square separation of the ends of a freely
jointed chain

In Appendix 2 we see that the mean value ͗X͘ of a variable X with possible values x is
given by
+∞

͗X͘ =

Ύ

xf(x)dx

−∞

where the function f(x) is the probability density, a measure of the distribution of the
probability values over x, and dx is an infinitesimally small interval of x values
...
It follows that the general expression for
the mean nth power of the end-to-end separation (a positive quantity that can vary
from 0 to +∞) is
∞

͗Rn͘ =

Ύ r f(r)dr
n

0

To calculate Rrms, we first determine ͗R2͘ by using n = 2 and f (r) from eqn 19
...
8 RANDOM COILS
where we have used the standard integral
∞

Ύ xe

4 −a2x 2

dx =

0

3
2a2

When we use the expression for a in eqn 19
...
7 Calculate the mean separation of the ends of a freely jointed chain of
∞

N bonds of length l
...
You will need the standard integral

Ύ xe

3 −a2x 2

0

1
dx = – a4
...
3a
...
32)

where Rij is the separation of atoms i and j
...
33)

The radius of gyration may also be calculated for other geometries
...

The random coil model ignores the role of the solvent: a poor solvent will tend to
cause the coil to tighten so that solute–solvent contacts are minimized; a good solvent
does the opposite
...
The model is most reliable for a polymer in a
bulk solid sample, where the coil is likely to have its natural dimensions
...
Any stretching of the coil introduces order
and reduces the entropy
...
As shown in the Justification below, we can use the same model to deduce
that the change in conformational entropy, the statistical entropy arising from the
arrangement of bonds, when a coil containing N bonds of length l is stretched or compressed by nl is
1
∆S = − – kN ln{(1 + ν)1+ν(1 − ν)1−ν }
2

ν = n/N

(19
...
19
...


0

Justification 19
...
1

DS/kN

The conformational entropy of the chain is S = k ln W, where W is given by eqn 19
...

Therefore,
-0
...
3

-0
...
8 -0
...
Therefore, the maximum entropy is
0
n = n/N

0
...
8

Fig
...
19 The change in molar entropy of a
perfect elastomer as its extension changes;
ν = 1 corresponds to complete extension;
ν = 0, the conformation of highest entropy,
corresponds to the random coil
...
34
...
For long chains, we can simply take groups of neighbouring bonds
and consider the direction of their resultant
...
By concentrating on such groups rather
than individuals, it turns out that for long chains the expressions for the root mean
square separation and the radius of gyration given above should be multiplied by
F=

A 1 − cos θ D
C 1 + cos θ F

1/2

(19
...
5°), F = 21/2
...
36)

Illustration 19
...

Because l = 154 pm for a C-C bond, we find Rrms = 14 nm and Rg = 5
...
19
...
This value of Rg means that, on average, the coils rotate like hollow
spheres of radius 5
...


A random coil in three dimensions
...
The
root mean square distance between the
ends (Rrms), the mean radius, and the
radius of gyration (Rg) are indicated
...
19
...
Such self-avoidance tends to
swell the coil, so (in the absence of solvent effects) it is better to regard Rrms and Rg as
lower bounds to the actual values
...
9 THE STRUCTURE AND STABILITY OF SYNTHETIC POLYMERS
19
...
37a)

4
2

F /(kT/2l )

Synthetic polymers are classified broadly as elastomers, fibres, and plastics, depending
on their crystallinity, the degree of three-dimensional long-range order attained in
the solid state
...
Elastomers are polymers with numerous crosslinks that pull them back into their original shape when a stress is removed
...
We saw in Section 19
...
In the following Justification
we also see that the restoring force, F, of a one-dimensional perfect elastomer is
F=

Hooke’s
law

0

-2
-4
-6
-1

-0
...
This function is plotted in Fig
...
21
...
5

1

The restoring force, F, of a onedimensional perfect elastomer
...


Fig
...
21

(19
...
For small displacements, therefore, the whole coil
shakes with simple harmonic motion
...
6 Hooke’s law

The work done on an elastomer when it is extended through a distance dx is Fdx,
where F is the restoring force
...
The restoring force is
therefore
A ∂S D
E
F = −T B
C ∂x F T,V
If now we substitute eqn 19
...
37a
...

One example is nylon-66 (Fig
...
22)
...
19
...


19 MATERIALS 1: MACROMOLECULES AND AGGREGATES

Specific volume

674

Tg

Temperature
Fig
...
23 The variation of specific volume
with temperature of a synthetic polymer
...


Comment 19
...
A semiconductor is a substance
with an electrical conductivity that
increases as the temperature is raised
...


Oxidation

+


...

+

Fig
...
24 The mechanism of migration of a
partially localized cation radical, or
polaron, in polyacetylene
...

A plastic is a polymer that can attain only a limited degree of crystallinity and as a
result is neither as strong as a fibre nor as flexible as an elastomer
...
A sample of plastic
nylon-66 may be visualized as consisting of crystalline hydrogen-bonded regions of
varying size interspersed amongst amorphous, random coil regions
...
That is the case with nylon, poly(vinyl chloride), and the siloxanes
...
This change in crystallinity may be thought of as a
kind of intramolecular melting from a crystalline solid to a more fluid random coil
...
Thus,
polyethylene, which has chains that interact only weakly in the solid, has Tm = 414 K
and nylon-66 fibres, in which there are strong hydrogen bonds between chains, has
Tm = 530 K
...

All synthetic polymers undergo a transition from a state of high to low chain
mobility at the glass transition temperature, Tg
...
There is sufficient
energy available at normal temperatures for limited bond rotation to occur and the
flexible chains writhe
...
Glass transition temperatures well below 300 K
are desirable in elastomers that are to be used at normal temperatures
...
1)
...
19
...

IMPACT ON TECHNOLOGY

I19
...
Now we consider the electrical properties of synthetic polymers
...
However, a variety of newly
developed macromolecular materials have electrical conductivities that rival those
of silicon-based semiconductors and even metallic conductors
...
The Nobel Prize in
chemistry was awarded in 2000 to A
...
Heeger, A
...
McDiarmid, and H
...

One example of a conducting polymer is polyacetylene (Fig
...
24)
...
The product is a polaron, a partially localized
cation radical that does not delocalize but rather travels through the chain, as shown
in Fig
...
24
...
10 THE STRUCTURE OF PROTEINS

675

radicals that move independently
...

Conducting polymers are slightly better electrical conductors than silicon semiconductors but are far worse than metallic conductors
...

Recent studies of photon emission by conducting polymers may lead to new technologies for light-emitting diodes and flat-panel displays
...

19
...
For a protein to function correctly, it needs to have
a well defined conformation
...
The amino acid sequence of a protein
contains the necessary information to create the active conformation of the protein
from a newly synthesized random coil
...


R

H

Rotational
freedom

C

(a) The Corey–Pauling rules

The origin of the secondary structures of proteins is found in the rules formulated by
Linus Pauling and Robert Corey in 1951
...
The latter can act both as a
donor of the H atom (the NH part of the link) and as an acceptor (the CO part)
...
19
...

2 The N, H, and O atoms of a hydrogen bond lie in a straight line (with displacements of H tolerated up to not more than 30° from the N-O vector)
...

The rules are satisfied by two structures
...
The other, in which hydrogen bonding between peptide
links leads to a planar structure, is a sheet; this form is the secondary structure of the
protein fibroin, the constituent of silk
...
In the aqueous environment of biological cells, the outer
surface of a protein molecule is covered by a mobile sheath of water molecules, and its
interior contains pockets of water molecules
...

The simplest calculations of the conformational energy of a polypeptide chain ignore
entropy and solvent effects and concentrate on the total potential energy of all the

132
...
5°
124

O

1 The four atoms of the peptide link lie in a relatively rigid plane
...
5

Rotational
freedom

H

C

R

Fig
...
25 The dimensions that characterize
the peptide link (bonds in picometres)
...


676

19 MATERIALS 1: MACROMOLECULES AND AGGREGATES
interactions between nonbonded atoms
...

To calculate the energy of a conformation, we need to make use of many of the molecular interactions described in Chapter 18, and also of some additional interactions:
1 Bond stretching
...
If we liken the bond to a spring, then the potential energy takes the form of
Hooke’s law
1
Vstretch = –kstretch(R − Re)2
2

(19
...

2 Bond bending
...
If the equilibrium bond angle is θe, we write
1
Vbend = –kbend(θ − θe)2
2

(19
...

Self-test 19
...
5 kJ mol−1 to increase the angle to 30°
...
27 × 10−23 J deg−2 , equivalent to 37
...

In this case (an α-l-polypeptide) the chain
has been drawn in its all-trans form, with
ψ = φ = 180°
...
19
...
There is a barrier to internal rotation of one bond relative to
another (just like the barrier to internal rotation in ethane)
...
Figure 19
...
The
sign convention is that a positive angle means that the front atom must be rotated
clockwise to bring it into an eclipsed position relative to the rear atom
...
A helix is obtained when all the φ are equal and
when all the ψ are equal
...
For
a left-handed helix, both angles are positive
...
40)

in which A and B are constants of the order of 1 kJ mol−1
...

4 Interaction between partial charges
...
3):
VCoulomb =

qiqj
4πε r

(19
...
10 THE STRUCTURE OF PROTEINS

5 Dispersive and repulsive interactions
...
5):
VLJ =

C

D
− 6
r12 r

(19
...
In some models of structure, the interaction between partial
charges is judged to take into account the effect of hydrogen bonding
...
Charges
of −0
...
28e are assigned to N and H, respectively, and −0
...
39e to
O and C, respectively
...


677

Global
minimum

Parameter representing
conformation

(19
...
38–19
...
The procedure is known
as a molecular mechanics simulation and is automated in commercially available
molecular modelling software
...
19
...
The software packages include schemes for modifying the locations of the
atoms and searching for these minima systematically
...
In a molecular dynamics simulation, the molecule is set in motion by
heating it to a specified temperature, as described in Section 17
...
The possible trajectories of all atoms under the influence of the intermolecular potentials correspond
to the conformations that the molecule can sample at the temperature of the simulation
...
38–19
...
At high temperatures, more potential energy barriers can
be overcome and more conformations are possible
...


Fig
...
27 For large molecules, a plot of
potential energy against the molecular
geometry often shows several local minima
and a global minimum
...
19
...
Each turn of the helix contains 3
...

The pitch of a single turn (the distance between points separated by 360°) is 544 pm
...
All the R groups point away from the major axis of
the helix
...
29 shows the Ramachandran plots for the helical form of polypeptide
chains formed from the nonchiral amino acid glycine (R = H) and the chiral amino
acid l-alanine (R = CH3)
...
In contrast, the map for l-alanine
is unsymmetrical, and there are three distinct low-energy conformations (marked
I, II, III)
...
19
...
Carbon
atoms are shown in green, with nitrogen in
blue, oxygen in red, and hydrogen atoms in
grey
...
6 residues per turn, and a
translation along the helix of 150 pm per
residue, giving a pitch of 540 pm
...


678

19 MATERIALS 1: MACROMOLECULES AND AGGREGATES
180°

Fig
...
29 Contour plots of potential energy
against the torsional angles ψ and φ, also
known as Ramachandran plots, for (a) a
glycyl residue of a polypeptide chain and
(b) an alanyl residue
...

The glycyl diagram is symmetrical, but
regions I and II in the correspond to
right- and left-handed helices, are
unsymmetrical, and the minimum in
region I lies lower than that in region II
...
A
...
J
...
Mol
...
23, 47 (1967)
...

Fig
...
30

Comment 19
...
There are also
links to sites where you may visualize
the structures of proteins and nucleic
acids that have been obtained by
experimental and theoretical methods
...
This result is consistent
with the observation that polypeptides of the naturally occurring L-amino acids tend
to form right-handed helices
...

Some of the R groups point above and some point below the sheet
...

In an anti-parallel b-sheet (Fig
...
30a), φ = −139°, ψ = 113°, and the N-H····O
atoms of the hydrogen bonds form a straight line
...
Antiparallel β-sheets are very common
in proteins
...
19
...
This arrangement is a result of
the parallel arrangement of the chains: each N-H bond on one chain is aligned with
a N-H bond of another chain and, as a result, each C-O bond of one chain is aligned
with a C-O bond of another chain
...

Circular dichroism (CD) spectroscopy (Section 14
...
Consider a helical polypeptide
...
Therefore, we
expect the α-helix to have a unique CD spectrum
...
19
...

(d) Higher-order structures

Covalent and non-covalent interactions may cause polypeptide chains with well
defined secondary structures to fold into tertiary structures
...

Although we do not know all the rules that govern protein folding, a few general
conclusions may be drawn from X-ray diffraction studies of water-soluble natural
proteins and synthetic polypeptides
...
A wide variety of structures can result from these broad rules
...
19
...
The four nonpolar regions pack together to form a nonpolar interior
...
11 THE STRUCTURE OF NUCLEIC ACIDS
4

679

a helix
b sheet
Random
coil

2

eL - e R

0
-2
-4
-6
180

250
Wavelength, l /nm

Fig
...
31 Representative CD spectra of
polypeptides
...


Fig
...
32 A four-helix bundle forms from
the interactions between nonpolar
aminoacids on the surfaces of each helix,
with the polar aminoacids exposed to the
aqueous environment of the solvent
...
19
...
The retinol-binding protein of blood plasma, which is responsible
for transporting vitamin A, is an example of a β-barrel structure
...
4g)
...

19
...
Deoxyribonucleic acid (DNA) contains the instructions for protein synthesis, which is carried out by different forms of ribonucleic acid
(RNA)
...

Both DNA and RNA are polynucleotides (2), in which base–sugar–phosphate units
are linked by phosphodiester bonds
...
The most common bases are adenine (A, 3), cytosine (C, 4), guanine (G, 5), thymine (T, found in DNA only, 6), and uracil (U, found
in RNA only, 7)
...
Nonpolar aminoacids are in the
interior of the barrel
...
19
...
This charge distribution
leads to two important properties
...
The second is that the bases can interact by hydrogen bonding,
as shown for A-T (8) and C-G base pairs (9)
...

In DNA, two polynucleotide chains wind around each other to form a double
helix (Fig
...
34)
...

The structure is stabilized further by interactions between the planar π systems of
the bases
...
0 nm and a pitch of 3
...
Long stretches
of DNA can fold further into a variety of tertiary structures
...
19
...
Supercoiled DNA is found in the chromosome and can be visualized as the twisting of closed circular DNA (ccDNA), much like the twisting of a
rubber band
...
Therefore, RNA exists primarily as single chains that can fold into complex structures by formation of A-U and
G-C base pairs
...
19
...
Transfer RNAs help assemble polypeptide chains during
protein synthesis in the cell
...
19
...


Fig
...
35 A long section of DNA may form
closed circular DNA (ccDNA) by covalent
linkage of the two ends of the chain
...


Fig
...
36

(tRNA)
...
12 THE STABILITY OF PROTEINS AND NUCLEIC ACIDS
1

Self-assembly
Much of the material discussed in this chapter also applies to aggregates of particles
that form by self-assembly, the spontaneous formation of complex structures of
molecules or macromolecules held together by molecular interactions, such as

Fraction of unfolded protein

19
...
It can be achieved by changing the temperature or by adding chemical agents
...
When eggs are cooked, the protein
albumin is denatured irreversibly, collapsing into a structure that resembles a random
coil
...
Disulfide cross-links
between the chains of keratin render the protein and, hence hair fibres, inflexible
...
Oxidation re-forms the disulfide bonds and sets the new shape
...
Other means of chemically denaturing a protein include the
addition of compounds that form stronger hydrogen bonds than those within a helix
or sheet
...
The action of acids or bases, which can protonate or deprotonate groups
involved in hydrogen bonding or change the Coulombic interactions that determine
the conformation of a protein, can also result in denaturation
...
Thermal denaturation is similar
to the melting of synthetic polymers (Section 19
...
Denaturation is a cooperative
process in the sense that the biopolymer becomes increasingly more susceptible to
denaturation once the process begins
...
1)
...
5 (Fig
...
37)
...
Each G-C base pair has three hydrogen bonds
whereas each A-T base pair has only two
...

It follows that two factors render DNA sequences rich in C-G base pairs more stable
than sequences rich in A-T base pairs: more hydrogen bonds between the bases and
stronger stacking interactions between base pairs
...
For
example, Tm = 320 K for ribonuclease T1 (an enzyme that cleaves RNA in the cell),
which is not far above the temperature at which the enzyme must operate (close
to body temperature, 310 K)
...
0 and 298 K is only 19
...
Therefore,
unlike DNA, the stability of a protein does not increase in a simple way with the number of hydrogen bonding interactions
...
10
...
5

0

Tm
Temperature

Fig
...
37 A protein unfolds as the
temperature of the sample increases
...

The melting temperature, Tm, is the
temperature at which the fraction of
unfolded polymer is 0
...


682

19 MATERIALS 1: MACROMOLECULES AND AGGREGATES
Coulombic, dispersion, hydrogen bonding, or hydrophobic interactions
...
1), of protein quaternary structures from two or more polypeptide
chains, and of a DNA double helix from two polynucleotide chains
...

19
...
In this context, ‘small’ means something less than about 500 nm in diameter
(about the wavelength of visible light)
...
They pass through most filter papers, but can be detected by lightscattering and sedimentation
...
A sol is a dispersion of a solid in a liquid (such as clusters of gold atoms in water) or of a solid in a
solid (such as ruby glass, which is a gold-in-glass sol, and achieves its colour by light
scattering)
...
An emulsion is a dispersion of a liquid in a liquid (such as milk)
...
If the solvent is water, the terms hydrophilic and hydrophobic,
respectively, are used instead
...
Lyophilic
colloids generally have some chemical similarity to the solvent, such as -OH groups
able to form hydrogen bonds
...

The preparation of aerosols can be as simple as sneezing (which produces an
imperfect aerosol)
...
Material (for example, quartz) may be ground in the presence of the dispersion
medium
...
Arcing between electrodes immersed
in the support medium also produces a colloid
...
A precipitate (for example, silver iodide) already formed may be
dispersed by the addition of a peptizing agent (for example, potassium iodide)
...

Emulsions are normally prepared by shaking the two components together vigorously,
although some kind of emulsifying agent usually has to be added to stabilize the product
...
In milk, which is an emulsion of fats in water, the
emulsifying agent is casein, a protein containing phosphate groups
...
This coagulation may be prevented by ensuring that the emulsion is dispersed
very finely initially: intense agitation with ultrasonics brings this dispersion about, the
product being ‘homogenized’ milk
...
The
dispersal is aided if a charge is applied to the liquid, for then electrostatic repulsions

19
...
This procedure may also be used to produce emulsions, for the charged liquid phase may be directed into another liquid
...
2)
...
A membrane (for example, cellulose) is selected that is permeable to
solvent and ions, but not to the colloid particles
...

(b) Structure and stability

Colloids are thermodynamically unstable with respect to the bulk
...
7a), it
follows that dG < 0 if dσ < 0
...

At first sight, even the kinetic argument seems to fail: colloidal particles attract each
other over large distances, so there is a long-range force that tends to condense them
into a single blob
...
The energy of
attraction between two individual atoms i and j separated by a distance Rij, one in each
6
colloidal particle, varies with their separation as 1/R ij (Section 18
...
The sum of all
these pairwise interactions, however, decreases only as approximately 1/R2 (the precise variation depending on the shape of the particles and their closeness), where R is
the separation of the centres of the particles
...

Several factors oppose the long-range dispersion attraction
...
Thus the surface atoms of a
platinum sol in water react chemically and are turned into -Pt(OH)3H3, and this
layer encases the particle like a shell
...

(c) The electrical double layer

A major source of kinetic nonlability of colloids is the existence of an electric charge
on the surfaces of the particles
...
9)
...
First, there is a fairly immobile layer
of ions that adhere tightly to the surface of the colloidal particle, and which may
include water molecules (if that is the support medium)
...
The electric potential at the radius of shear relative to
its value in the distant, bulk medium is called the zeta potential, ζ, or the electrokinetic potential
...
The inner shell of charge and the outer ionic atmosphere is called the
electrical double layer
...
Derjaguin
and L
...
Verwey and J
...
G
...
It assumes that there is a balance between the repulsive interaction
between the charges of the electric double layers on neighbouring particles and the
attractive interactions arising from van der Waals interactions between the molecules
in the particles
...
44)

where A is a constant, ζ is the zeta potential,1 R is the separation of centres, s is the
separation of the surfaces of the two particles (s = R − 2a for spherical particles of
radius a), and rD is the thickness of the double layer
...
When the double layer is thin (rD < a),
<
<
the expression is replaced by
1
Vrepulsion = – Aaζ 2 ln(1 + e−s/rD)
2

(19
...
80):
rD =

A εRT D
C 2ρF 2Ib 7 F

1/2

(19
...
The
potential energy arising from the attractive interaction has the form
Vattraction = −

rD >> a

Potential energy, V

Coagulation

rD < a
<

Flocculation
Separation, s

Fig
...
38 The potential energy of interaction
as a function of the separation of the
centres of the two particles and its variation
with the ratio of the particle size to the
thickness a of the electric double layer rD
...


B
s

(19
...
The variation of the total potential energy with separation is shown in Fig
...
38
...
Aggregation of the particles arising from the
stabilizing effect of this secondary minimum is called flocculation
...
Coagulation, the irreversible aggregation of distinct particles into large particles, occurs when
the separation of the particles is so small that they enter the primary minimum of the
potential energy curve and van der Waals forces are dominant
...
This increase is the basis of the empirical Schulze–Hardy rule, that hydrophobic colloids are flocculated most efficiently by
ions of opposite charge type and high charge number
...
When river water containing colloidal clay flows into the sea, the salt water induces flocculation and coagulation, and is a major cause of silting in estuaries
...

The primary role of the electric double layer is to confer kinetic non-lability
...
This disruption may occur at high temperatures, which is one reason why sols precipitate
1
The actual potential is that of the surface of the particles; there is some danger in identifying it with the
zeta potential
...


19
...
The protective role of the double layer is the reason why it is
important not to remove all the ions when a colloid is being purified by dialysis, and
why proteins coagulate most readily at their isoelectric point
...
14 Micelles and biological membranes
Surfactant molecules or ions can cluster together as micelles, which are colloid-sized
clusters of molecules, for their hydrophobic tails tend to congregate, and their hydrophilic heads provide protection (Fig
...
39)
...
19
...
The hydrophilic groups are
represented by spheres and the
hydrophobic hydrocarbon chains are
represented by the stalks; these stalks are
mobile
...
The CMC is detected by noting a pronounced change in physical
properties of the solution, particularly the molar conductivity (Fig
...
40)
...
The hydrocarbon interior of
a micelle is like a droplet of oil
...
Micelles are
important in industry and biology on account of their solubilizing function: matter
can be transported by water after it has been dissolved in their hydrocarbon interiors
...

Non-ionic surfactant molecules may cluster together in clumps of 1000 or more,
but ionic species tend to be disrupted by the electrostatic repulsions between head
groups and are normally limited to groups of less than about 100
...
Spherical micelles do occur, but micelles are more commonly flattened
spheres close to the CMC
...
19
...
Liposomes may be used to carry nonpolar drug molecules in blood
...

These orderly arrangements of micelles are called lyotropic mesomorphs and, more
colloquially, ‘liquid crystalline phases’
...
Consequently, enthalpies of micelle formation display no readily
discernible pattern and may be positive (endothermic) or negative (exothermic)
...
That such micelles do form above the CMC indicates that the entropy
change accompanying their formation must then be positive, and measurements suggest a value of about +140 J K−1 mol−1 at room temperature
...
*) are important in the formation of micelles
...
19
...


(b) Membrane formation

Some micelles at concentrations well above the CMC form extended parallel sheets,
called lamellar micelles, two molecules thick
...
19
...


686

19 MATERIALS 1: MACROMOLECULES AND AGGREGATES

Comment 19
...


Fig
...
42 A depiction of the variation
with temperature of the flexibility of
hydrocarbon chains in a lipid bilayer
...

(b) At a specific temperature, the chains are
largely frozen and the bilayer is said to exist
as a gel
...
Such lamellar micelles show a close resemblance to
biological membranes, and are often a useful model on which to base investigations of
biological structures
...
The basic structural element of a
membrane is a phospholipid, such as phosphatidyl choline (10), which contains long
hydrocarbon chains (typically in the range C14–C24) and a variety of polar groups,
+
such as -CH2CH2N(CH3)3 in (10)
...
The lipid molecules form layers instead of
micelles because the hydrocarbon chains are too bulky to allow packing into nearly
spherical clusters
...
2)
...
It is better to think of the membrane
as a viscous fluid rather than a permanent structure, with a viscosity about 100 times
that of water
...
*), the
average distance a phospholipid molecule diffuses is proportional to the square-root
of the time; more precisely, for a molecule confined to a two-dimensional plane, the
average distance travelled in a time t is equal to (4Dt)1/2
...

All lipid bilayers undergo a transition from a state of high to low chain mobility at
a temperature that depends on the structure of the lipid
...
19
...

There is sufficient energy available at normal temperatures for limited bond rotation
to occur and the flexible chains writhe
...
19
...
At lower temperatures, the amplitudes of
the writhing motion decrease until a specific temperature is reached at which motion
is largely frozen
...
19
...
Biological membranes exist as liquid crystals at physiological temperatures
...
1)
...
For example, the melting temperature increases with the length of the hydrophobic chain of the lipid
...
It follows that stabilization of the gel
phase in membranes of lipids with long chains results in relatively high melting temperatures
...
Indeed, lipids
containing unsaturated chains, those containing some C=C bonds, form membranes

(a)

(b)

19
...

Interspersed among the phospholipids of biological membranes are sterols, such as
cholesterol (11), which is largely hydrophobic but does contain a hydrophilic -OH
group
...

Self-test 19
...
Integral proteins are proteins immersed in the mobile but viscous bilayer
...
There are two views of the motion of integral proteins in the bilayer
...
19
...
In the lipid raft model, a number of lipid and cholesterol molecules form ordered structures, or ‘rafts’, that envelop
proteins and help carry them to specific parts of the cell
...

Alternatively, material from the cell interior wrapped in cell membrane may coalesce
with the cell membrane itself, which then withdraws and ejects the material in the
process of exocytosis
...
By providing hydrophilic channels through an otherwise alien hydrophobic
environment, some proteins act as ion channels and ion pumps (Impact I21
...

19
...
The physical properties of surface films have also been investigated
...
When a monolayer has been transferred to a solid support, it is called
a Langmuir–Blodgett film, after Irving Langmuir and Katherine Blodgett, who
developed experimental techniques for studying them
...
19
...
This device consists of a shallow trough and a barrier that can
be moved along the surface of the liquid in the trough, and hence compress any
monolayer on the surface
...
Why do bacterial and plant cells grown at low temperatures synthesize more phospholipids with unsaturated chains than do cells grown at higher
temperatures?
[Insertion of lipids with unsaturated chains lowers the plasma membrane’s
melting temperature to a value that is close to the lower ambient temperature
...
19
...
In the
alternative lipid raft model, a number of
lipid and cholesterol molecules envelop
and transport the protein around the
membrane
...
19
...
The surfactant is spread on the
surface of the liquid in the trough, and then
compressed horizontally by moving the
compression barrier towards the mica float
...


Surface pressure, p/(mN m-1)

688

19 MATERIALS 1: MACROMOLECULES AND AGGREGATES

40

Stearic
acid

30

20

10

0
0

Isostearic
acid
Tri-p-cresyl
phosphate

0
...
4 0
...
8 1
Area per molecule/nm2

Fig
...
45 The variation of surface pressure
with the area occupied by each surfactant
molecule
...


tension of the pure solvent and the solution (π = γ * − γ ) is measured by using a
torsion wire attached to a strip of mica that rests on the surface and pressing against
one edge of the monolayer
...
In an actual experiment, a small amount (about 0
...

Some typical results are shown in Fig
...
45
...

This quantity is obtained from the extrapolation of the steepest part of the isotherm to
the horizontal axis
...
Neither, though, occupies as much area as the trip-cresyl phosphate molecule (14), which is like a wide bush rather than a lanky tree
...
19
...
This difference indicates
that the tri-p-cresyl phosphate film is more compressible than the stearic acid films,
which is consistent with their different molecular structures
...
When the monolayer is compressed beyond the point represented by the collapse
pressure, the monolayer buckles and collapses into a film several molecules thick
...
19
...

(b) The thermodynamics of surface layers

A surfactant is active at the interface between two phases, such as at the interface
between hydrophilic and hydrophobic phases
...
To establish the
relation between the concentration of surfactant at a surface and the change in surface
tension it brings about, we consider two phases α and β in contact and suppose that
the system consists of several components J, each one present in an overall amount nJ
...
However,
the components are not uniformly distributed because one may accumulate at the
interface
...
48)

Similarly, if it is supposed that the concentration of a species J is uniform right up to
the interface, then from its volume we would conclude that it contains an amount
n J (α) of J in phase α and an amount n J (β ) in phase β
...
This difference is expressed in terms of the
surface excess, ΓJ:

ΓJ =

n J(σ )

σ

(19
...


19
...
In the following Justification
we derive the Gibbs isotherm, between the changes in the chemical potentials of the
substances present in the interface and the change in surface tension:
dγ = −

∑ ΓJ dµJ

(19
...
7 The Gibbs isotherm

A general change in G is brought about by changes in T, p, and the n J:
dG = −SdT + Vdp + γ dσ +

∑ µJdnJ
J

When this relation is applied to G, G(α), and G(β ) we find
dG(σ ) = −S(σ )dT + γ dσ +

∑ µJdnJ(σ )
J

because at equilibrium the chemical potential of each component is the same in
every phase, µ J(α ) = µ J(β ) = µ J(σ )
...
1), the last equation integrates at constant temperature to
G(σ) = γ σ +

∑ µJnJ(σ )
J

We are seeking a connection between the change of surface tension dγ and the change
of composition at the interface
...
1 led
to the Gibbs–Duhem equation (eqn 5
...
50
...
This approximation implies that
only the surfactant, S, accumulates at the surface, and hence that Γoil and Γwater are
both zero
...
51)

For dilute solutions,
dµS = RT ln c

(19
...
It follows that
dc
dγ = RTΓS
c
at constant temperature, or

A ∂γ D
RTΓS
=−
C ∂c F T
c

(19
...
53 implies that (∂γ /∂c)T < 0
...
Conversely, if the concentration dependence of γ is known,
then the surface excess may be predicted and used to infer the area occupied by each
surfactant molecule on the surface
...
3 Nanofabrication with self-assembled monolayers

S

S

S

S

Nanofabrication is the synthesis of nanodevices, nanometre-sized assemblies of atoms
and molecules that can be used in nanotechnological applications, such as those discussed in Impact I9
...
Here we see how molecular self-assembly can be used as the
basis for nanofabrication on surfaces
...
To understand the formation of SAMs, consider exposing molecules such as
alkyl thiols RSH, where R represents an alkyl chain, to an Au(0) surface
...
19
...


If R is a sufficiently long chain, van der Waals interactions between the adsorbed RS
units lead to the formation of a highly ordered monolayer on the surface, as shown in
Fig
...
46
...

The atomic force microscope (Impact I9
...

In one application of the technique, enzymes were bound to patterned SAMs
...


Checklist of key ideas
1
...
Many proteins (specifically protein enzymes) are
monodisperse; a synthetic polymer is polydisperse
...
The number-average molar mass, Jn, is the value obtained by
weighting each molar mass by the number of molecules of
that mass present in the sample; the weight-average molar
mass, Jw, is the average calculated by weighting the molar
masses of the molecules by the mass of each one present in the
sample; the Z-average molar mass, JZ, is the average molar
mass obtained from sedimentation measurements
...
The heterogeneity index of a polymer sample is Jw /Jn
...
Techniques for the determination of the mean molar masses
of macromolecules include mass spectrometry (as MALDI),
ultracentrifugation, laser light scattering, and viscometry
...
The least structured model of a macromolecule is as a random
coil; for a freely jointed random coil of contour length Nl, the
root mean square separation is N1/2l and the radius of gyration
is Rg = (N/6)1/2l
...
The conformational entropy is the statistical entropy arising
from the arrangement of bonds in a random coil
...
The primary structure of a biopolymer is the sequence of its
monomer units
...
The secondary structure of a protein is the spatial arrangement
of the polypeptide chain and includes the α-helix and β -sheet
...
Helical and sheet-like polypeptide chains are folded into a
tertiary structure by bonding influences between the residues
of the chain
...
Some macromolecules have a quaternary structure as
aggregates of two or more polypeptide chains
...
Synthetic polymers are classified as elastomers, fibres, and
plastics
...
A perfect elastomer is a polymer in which the internal energy
is independent of the extension of the random coil; for small
extensions a random coil model obeys a Hooke’s law restoring
force
...
Synthetic polymers undergo a transition from a state of high
to low chain mobility at the glass transition temperature, Tg
...
The melting temperature of a polymer is the temperature at
which three-dimensional order is lost
...
A mesophase is a bulk phase that is intermediate in character
between a solid and a liquid
...
A disperse system is a dispersion of small particles of one
material in another
...
Colloids are classified as lyophilic (solvent attracting,
specifically hydrophilic for water) and lyophobic (solvent
repelling, specifically hydrophobic)
...
A surfactant is a species that accumulates at the interface of two
phases or substances and modifies the properties of the surface
...
The radius of shear is the radius of the sphere that captures the
rigid layer of charge attached to a colloid particle
...
The zeta potential is the electric potential at the radius of shear
relative to its value in the distant, bulk medium
...
The inner shell of charge and the outer atmosphere jointly
constitute the electric double layer
...
Many colloid particles are thermodynamically unstable but
kinetically non-labile
...
Flocculation is the reversible aggregation of colloidal particles;
coagulation is the irreversible aggregation of colloidal
particles
...
The Schultze–Hardy rule states that hydrophobic colloids are
flocculated most efficiently by ions of opposite charge type
and high charge number
...
A micelle is a colloid-sized cluster of molecules that forms at
the critical micelle concentration (CMC) and at the Krafft
temperature
...
A liposome is a vesicle with an inward pointing inner surface
of molecules surrounded by an outward pointing outer layer
...
A lamellar micelle is an extended layer of molecules two
molecules thick
...
A monolayer, a single layer of molecules on a surface
...

29
...

30
...


Further reading
Articles and texts

A
...
Adamson and A
...
Gast, Physical chemistry of surfaces
...

J
...
Wiley, New York (1999)
...
E
...
, Seymour/Carraher’s polymer chemistry, Marcel
Dekker (2000)
...
F
...
Wennerström, The colloidal domain : where
physics, chemistry, biology, and technology meet
...


C
...
Johnson and D
...
Gabriel, Laser light scattering
...

A
...
Leach, Molecular modelling: principles and applications
...

K
...
van Holde, W
...
Johnson, and P
...
Ho, Principles of physical
biochemistry
...

P
...
Heimenz and R
...
Marcel Dekker, New York (1997)
...
Flory, Principles of polymer chemistry
...


D
...
Lide (ed
...
CRC Press, Boca Raton (2000)
...
1 The Rayleigh ratio

Here we outline the key steps in the derivation of eqn 19
...
You are encouraged to consult sources in the Further
reading section for additional details
...
54)

where r is the distance between the sample and the detector and φ is
the angle of observation relative to the z-axis (φ = 90° in Fig
...
3)
...
7 that
Rθ =

π2α 2

ε 2 λ4
r

The relation between the polarizability and the refractive index, nr,
of a solution is (see Appendix 3 for a qualitative explanation and
Further reading for quantitative details)
2
n2 − n r,0 =
r

Nα

εr

(19
...
Because N = cP NA/M (where cP is the
mass concentration of the polymer and M is its molar mass), we have:

α=

εr M
cP NA

2
(nr − n2 )
r,0

For a dilute solution, nr differs little from nr,0 and we can write:
A dnr D
E c +···
nr = nr,0 + B
C dcP F P

692

19 MATERIALS 1: MACROMOLECULES AND AGGREGATES

It follows that

of atoms in each molecule
...
56)

The Rayleigh ratio for scattering by a single molecule now becomes
4π2n2 M 2 A dnr D
r,0
B
E
Rθ =
2
λ4N A C dcP F

2

(19
...
58)

Equation 19
...

Now we derive expressions for the structure factor
...
The scattering from all the particles is then calculated by
allowing for contributions from all possible orientations of each pair

1
N

2

∑
i,j

sin sRij
sRij

,

s=

4π

λ

1
sin –θ
2

(19
...

When the molecule is much smaller than the wavelength of the
incident radiation in the sense that sRij < 1 (for example, if R = 5 nm,
<
and λ = 500 nm, all the sRij are about 0
...
32)
...
9
...
1 Distinguish between number-average, weight-average, and Z-average
molar masses
...

19
...

19
...

19
...


19
...
Why are these methods generally more popular in the field of

polymer chemistry than the quantum mechanical procedures discussed in
Chapter 11?
19
...
Explain this effect
...
7 Explain the physical origins of surface activity by surfactant molecules
...
8 Discuss the physical origins of the surface Gibbs energy
...
9 Self-assembled monolayers (SAMs) are receiving more attention than
Langmuir–Blodgett (LB) films as starting points for nanofabrication
...
1a Calculate the number-average molar mass and the mass-average
molar mass of a mixture of equal amounts of two polymers, one having
M = 62 kg mol−1 and the other M = 78 kg mol−1
...
2b The radius of gyration of a long chain molecule is found to be 18
...


19
...


19
...
What average molar mass would be
obtained from measurement of (a) osmotic pressure, (b) light scattering?

19
...
3 nm
...
Assume the chain is randomly coiled and
estimate the number of links in the chain
...
3b A solution consists of 25 per cent by mass of a trimer with

The chain consists of links of length 450 pm
...


M = 22 kg mol−1 and its monomer
...
4a Evaluate the rotational correlation time, τ R = 4πa3η/3kT, for serum
albumin in water at 25°C on the basis that it is a sphere of radius 3
...

What is the value for a CCl4 molecule in carbon tetrachloride at 25°C?
(Viscosity data in Table 21
...
)
19
...
5 nm
...
5a What is the relative rate of sedimentation for two spherical particles of
the same density, but which differ in radius by a factor of 10?
19
...
10 g cm and 1
...
4, the former being the larger? Use ρ = 0
...


693

19
...
9 × 10−11 m2 s−1
...
1 Sv in a solution of density
997 kg m−3
...
721 cm3 g−1
...

19
...
The rotational rate
of the centrifuge was 50 000 r
...
m
...
61 cm3 g−1
...

19
...
The rotation rate of
the centrifuge was 1080 Hz
...
2 ×
10−4 m3 kg−1
...


19
...
749 × 10−3 m3 kg−1,
a sedimentation constant of 4
...
9 × 10−11 m2 s−1
...


19
...
0 cm from the centre of rotation in an ultracentrifuge operating at 80 000 r
...
m
...
6b A synthetic polymer has a specific volume of 8
...
10b Calculate the radial acceleration (as so many g) in a cell placed at 5
...
32 kHz
...
46 Sv, and a diffusion coefficient of
7
...
Determine its molar mass from this information
...
7a Find the drift speed of a particle of radius 20 µm and density
−3

1750 kg m which is settling from suspension in water (density
1000 kg m−3) under the influence of gravity alone
...
9 × 10−4 kg m−1 s−1
...
7b Find the drift speed of a particle of radius 15
...
The viscosity
of water is 8
...

19
...
3 × 10−11 m2 s−1
...
2 Sv in a solution of density
1
...
The specific volume of the macromolecule is 0
...

Determine the molar mass of the macromolecule
...
11a A polymer chain consists of 700 segments, each 0
...
If the

chain were ideally flexible, what would be the r
...
s
...
11b A polymer chain consists of 1200 segments, each 1
...
If the
chain were ideally flexible, what would be the r
...
s
...
12a Calculate the contour length (the length of the extended chain) and

the root mean square separation (the end-to-end distance) for polyethylene
with a molar mass of 280 kg mol−1
...
12b Calculate the contour length (the length of the extended chain) and
the root mean square separation (the end-to-end distance) for polypropylene
of molar mass 174 kg mol−1
...
1 In a sedimentation experiment the position of the boundary as a
function of time was found to be as follows:

t/min

15
...
1

36
...
2

r/cm

5
...
09

5
...
19

The rotation rate of the centrifuge was 45 000 r
...
m
...


The viscosity of the solvent is 0
...
What is the intrinsic viscosity of
the polymer?
19
...
From these data, calculate the
molar mass of the polystyrene samples
...

η(benzene) = 0
...
601 cP) at 25°C
...
22

5
...
00

10
...
2 Calculate the speed of operation (in r
...
m
...
2

248
...
4

371
...
3

to obtain a readily measurable concentration gradient in a sedimentation
equilibrium experiment
...
Use rtop = 5
...
0 cm, M ≈ 105 g mol−1, ρvs ≈ 0
...


19
...
3 The concentration dependence of the viscosity of a polymer solution is
found to be as follows:

c/(g dm−3)

1
...
89

5
...
17

η /(g m−1 s−1)

1
...
20

1
...
73

c/(g/102 cm3)

0

0
...
4

0
...
8

1
...
647

0
...
733

0
...
821

0
...
4 to deduce the molar mass of the polymer
...
6‡ Polystyrene in cyclohexane at 34
...
The following data

* Problems denoted with the symbol ‡ were supplied by Charles Trapp and Carmen Giunta
...
J
...
Hadjichristidis,
J
...
Lindner, and J
...
Mays (J
...
Chem
...
Data 23, 619 (1994)):
M/(kg mol−1) 10
...
8 106 249 359 860 1800 5470 9720 56 800
[η ]/(cm3 g−1) 8
...
9 28
...
0 51
...
6 113
...
What is the molar mass of a polystyrene
that forms a θ solution in cyclohexane with [η] = 100 cm3 g−1?
19
...
M
...
Janca (J
...
Sci
...
Chem
...
Their results for the intrinsic
viscosity, [η], as a function of average molar mass at 25°C are given in the
table below
...

(b) Compare your values to those in Table 19
...
5
...
0
[η ]/(cm3 g−1)

10
...
85

5
...
8

14
...
2 173

411

867

27
...
6 67
...
0

206
...
8 The concentration dependence of the osmotic pressure of solutions of a

macromolecule at 20°C was found to be as follows:
−3

c/(g dm )

1
...
72

5
...
60

Π /Pa

134

321

655

898

Determine the molar mass of the macromolecule and the osmotic virial
coefficient
...
9 The osmotic pressure of a fraction of poly(vinyl chloride) in a ketone
solvent was measured at 25°C
...
798 g cm−3
...
200

0
...
600

0
...
000

h/cm

0
...
2

1
...
76

3
...
10 The following table lists the glass transition temperatures, Tg, of several

polymers
...

Polymer

Poly(oxymethylene) Polyethylene

Poly(vinyl chloride) Polystyrene

Structure -(OCH2)n-

-(CH2CH2)n- -(CH2–CHCl)n-

-(CH2-CH(C6H5))n-

Tg /K

253

381

198

354

Theoretical problems
19
...
Present
arguments to show that the first polymer is a rigid rod whereas the second
is a random coil
...
12 The kinematic viscosity, ν, of a fluid is defined as η/ρ, where ρ is the
mass density
...
25) and hence
confirm eqn 19
...

19
...
What is the
number average molar mass when the distribution is narrow?
19
...
26 for a one-dimensional freely jointed chain can be
used to derive eqn 19
...
Hint
...
26, and similarly for the other two dimensions
...
Don’t count negative integers (that is, divide the volume of the
shell by 8, corresponding to the all-positive octant of values)
...
15 Use eqn 19
...
Evaluate these three quantities for a fully
flexible chain with N = 4000 and l = 154 pm
...
16 Construct a two-dimensional random walk by using a random number
generating routine with mathematical software or electronic spreadsheet
...
If there are many people working on the
problem, investigate the mean and most probable separations in the plots by
direct measurement
...
17 Evaluate the radius of gyration, Rg, of (a) a solid sphere of radius a,

(b) a long straight rod of radius a and length l
...
056902 × {(vs /cm3 g−1)(M/g mol−1)}1/3
...
750 cm3 g−1, and, in the
case of the rod, of radius 0
...

19
...
85
...
Evaluate B for l =
154 pm and N = 4000
...
Use B = – NAvP,
2
where vP is the excluded volume due to a single molecule
...
19 Radius of gyration is defined in eqn 19
...
Show that an equivalent
definition is that Rg is the average root mean square distance of the atoms or
groups (all assumed to be of the same mass), that is, that R2 = (1/N)∑j Rj2,
g
where Rj is the distance of atom j from the centre of mass
...
20 Consider the thermodynamic description of stretching rubber
...
Because dw = tdl, the basic equation is dU = TdS + tdl
...
) If G = U − TS − tl, find expressions for dG
and dA, and deduce the Maxwell relations

A ∂S D
A ∂t D
B E =−B E
C ∂l F T
C ∂T F l

A ∂S D
A ∂l D
B E =−B E
C ∂t F T
C ∂T F t

Go on to deduce the equation of state for rubber,
A ∂U D
A ∂t D
B
E =t− B E
C ∂l F T
C ∂T F l
19
...
Account for this result in terms of the
molecular nature of the sample
...
22 In this problem you will use molecular mechanics software of your
instructor’s choice to gain some appreciation for the complexity of the
calculations that lead to plots such as those in Fig
...
29
...
(a) Draw three initial conformers of the
dipeptide with R = H: one with φ = +75°, ψ = −65°, a second with φ = ψ =
+180°, and a third with φ = +65°, ψ = +35°
...
(Although any force
field will work satisfactorily, the AMBER force field is strongly recommended,
as it is optimized for calculations on biopolymers
...
9981 g cm−3
...
00 × 10−3 kg m−1 s−1 at 20°C
...
29 For some proteins, the isoelectric point must be obtained by
extrapolation because the macromolecule might not be stable over a very wide
pH range
...
5

final conformers represent? Rationalize any observed differences in total
potential energy of the final conformers
...
Rationalize any similarities and differences
between the final conformers of the dipeptides with R = H and R = CH3
...
23 Calculate the excluded volume in terms of the molecular volume on the
basis that the molecules are spheres of radius a
...
0 nm, and haemoglobin, a =
3
...
18)
...
00 g/(100 cm3) solutions of bushy stunt virus (M = 1
...
5 kg mol−1) from the ideal solution values
...
8, let Pθ = 1 and assume that both solutions have the same K value
...
24 Use the information below and the expression for Rg of a solid sphere

quoted in the Problem 19
...

M/(g mol −1)

vs /(cm3 g−1)

Serum albumin

66 × 10

0
...
6 × 10

3
6

2
...
741

12
...
556

4 × 106

DNA

Rg /nm

117
...
25 Suppose that a rod-like DNA molecule of length 250 nm undergoes a
conformational change to a closed-circular (cc) form
...
24 and an incident wavelength λ = 488 nm to calculate the ratio
of scattering intensities by each of these conformations, Irod/Icc, when θ = 20°,
45°, and 90°
...
Based on your
answer to part (a), at which angle would you conduct the experiments? Justify
your choice
...
26 In an ultracentrifugation experiment at 20°C on bovine serum albumin
the following data were obtained: ρ = 1
...
112 cm3 g−1,
ω /2π = 322 Hz,

r/cm

5
...
1

5
...
3

5
...
536

0
...
148

0
...
039

Evaluate the molar mass of the sample
...
27 Sedimentation studies on haemoglobin in water gave a sedimentation
constant S = 4
...
The diffusion coefficient is 6
...
Calculate the molar mass of haemoglobin using
vs = 0
...
998 g cm−3 for the
density of the solution
...
00 × 10−3 kg m−1 s−1
...
28 The rate of sedimentation of a recently isolated protein was monitored
at 20°C and with a rotor speed of 50 000 r
...
m
...
127

6
...
179

6
...
232

6
...
284

Calculate the sedimentation constant and the molar mass of the protein on
the basis that its partial specific volume is 0
...
62 × 10−11 m2 s−1 at 20°C, the density of the solution then being

5
...
5

6
...
10

Drift speed/(µm s−1)

−0
...
30

−0
...
30 Here we use concepts developed in Chapter 16 and this chapter to
enhance our understanding of closed-circular and supercoiled DNA
...
Initially, therefore, one
end of the polymer can be found anywhere within a sphere of radius N 1/2l
centred on the other end
...
What is the change in molar entropy? Plot
the function you derive as a function of N
...
For example, one twist (i = ±1)
makes ccDNA resemble the number 8
...
(ii) Plot the expression you
derived in part (a) for several values of the temperature
...
(iii) Calculate p0, p1, p5, and p10 at 298 K
...
31 The melting temperature of a DNA molecule can be determined by
differential scanning calorimetry (Impact I2
...
The following data were
obtained in aqueous solutions containing the specified concentration csalt of
an soluble ionic solid for a series of DNA molecules with varying base pair
composition, with f the fraction of G-C base pairs:

csalt = 1
...
375

0
...
589

0
...
750

Tm /K

339

344

348

351

354

−3

csalt = 0
...
375

0
...
589

0
...
750

Tm /K

359

364

368

371

374

(a) Estimate the melting temperature of a DNA molecule containing 40
...
Hint
...
(b) Do
the data show an effect of concentration of ions in solution on the melting
temperature of DNA? If so, provide a molecular interpretation for the effect
you observe
...
32 The fluidity of a lipid bilayer dispersed in aqueous solution depends on
temperature and there are two important melting transitions
...
The second
transition, which occurs at a higher temperature than the first, is from the
liquid crystalline state to a liquid state, in which the hydrophobic interactions
holding the aggregate together are largely disrupted
...
Explain these
observations
...

19
...
A batch of polydisperse polystyrene was prepared by initiating
the polymerization with t-butyl radicals
...
A sample
from this batch was embedded in an organic matrix containing silver
trifluoroacetate and the resulting MALDI-TOF spectrum consisted of a large
number of peaks separated by 104 g mol−1, with the most intense peak at
25 578 g mol−1
...

19
...
Solutions of these beads are studied by a physical
chemistry student by dilute solution viscometry with an Ostwald viscometer
in both the ‘good’ solvent toluene and the theta solvent cyclohexane
...
(a) Fit the data to the virial equation for viscosity,

η = η*(1 + [η]c + k′[η]2c 2 + · · · )
where k′ is called the Huggins constant and is typically in the range 0
...
40
...

(b) Use the empirical Mark–Kuhn–Houwink–Sakurada equation (eqn 19
...
For theta
solvents, a = 0
...
2 × 10−5 dm3 g−1 for cyclohexane; for the good
solvent toluene a = 0
...
15 × 10−5 dm3 g−1
...
84 × 1026 when
[η] is expressed in cubic decimetres per gram and the distance is in metres
...
(d) From the molar masses calculate

the average number of styrene (C6H5CH=CH2) monomer units, ͗n͘, (e)
Calculate the length of a fully stretched, planar zigzag configuration, taking
the C-C distance as 154 pm and the CCC bond angle to be 109°
...
33 to calculate the radius of gyration, Rg
...

Compare this result with that predicted by the Kirkwood–Riseman theory:
which gives the better fit? (g) Compare your values for M to the results of
Problem 19
...
Is there any reason why they should or should not agree? Is the
manufacturer’s claim valid?
c/(g dm−3 toluene)

0

1
...
0

t D /s

8
...
11

10
...
52

c/(g dm−3 cyclohexane)

0

1
...
5

2
...
32

8
...
85

9
...
0

19
...
S
...
Polym
...
, Polym
...
29, 1585 (1991)) has determined the molar mases
and Mark–Houwink constants for the electronically conducting polymer,
poly(3-hexylthiophene) (P3HT) in tetrahydrofuran (THF) at 25°C by
methods similar to those used for nonconducting polymers
...
Determine the constants in the Mark–Kuhn–Houwink–Sakurada
equation from these results and compare to the values obtained in your
solution to Problem 19
...

J v /(kg mol −1)
3 −1

[η]/(cm g )

3
...
1

15
...
8

6
...
44

23
...
28

Materials 2:
the solid state
First, we see how to describe the regular arrangement of atoms in crystals and the symmetry of their arrangement
...
X-ray diffraction leads to information about the structures of metallic,
ionic, and molecular solids, and we review some typical results and their rationalization in
terms of atomic and ionic radii
...


The solid state includes most of the materials that make modern technology possible
...
The properties of solids stem,
of course, from the arrangement and properties of the constituent atoms, and one of
the challenges of this chapter is to see how a wide range of bulk properties, including
rigidity, electrical conductivity, and optical and magnetic properties stem from the
properties of atoms
...


20
Crystal lattices
20
...
2 The identification of lattice

planes
20
...
1 Impact on biochemistry:

X-ray crystallography of
biological macromolecules
20
...
5 Metallic solids
20
...
7 Molecular solids and covalent

networks
The properties of solids
20
...
9 Electrical properties

Crystal lattices

I20
...
In this section we see
how to describe the arrangement of atoms inside crystals
...
10 Optical properties
20
...
12 Superconductors

20
...
A space lattice is the pattern
formed by points representing the locations of these motifs (Fig
...
1)
...
More formally, a
space lattice is a three-dimensional, infinite array of points, each of which is surrounded in an identical way by its neighbours, and which defines the basic structure
of the crystal
...

Fig
...
2

Fig
...
1 Each lattice point specifies the
location of a structural motif (for example,
a molecule or a group of molecules)
...


Fig
...
3 A unit cell can be chosen in a
variety of ways, as shown here
...

In this rectangular lattice, the rectangular
unit cell would normally be adopted
...
The crystal structure itself is obtained by associating
with each lattice point an identical structural motif
...
20
...
A unit cell can be thought
of as the fundamental region from which the entire crystal may be constructed by
purely translational displacements (like bricks in a wall)
...
20
...
Such
unit cells are called primitive
...
An infinite number of different unit cells can describe the same lattice, but the
one with sides that have the shortest lengths and that are most nearly perpendicular to
one another is normally chosen
...
20
...

Unit cells are classified into seven crystal systems by noting the rotational symmetry elements they possess
...
20
...
A monoclinic unit cell has one twofold axis; the unique

Comment 20
...
There is a
corresponding symmetry element for
each symmetry operation, which is the
point, line, or plane with respect to
which the symmetry operation is
performed
...
See Chapter 12
for a more detailed discussion of
symmetry
...
Note that the angle α lies in the
plane (b,c) and perpendicular to the axis a
...
20
...
20
...
The insert shows
the threefold symmetry
...
1 LATTICES AND UNIT CELLS
axis is by convention the b axis (Fig
...
6)
...
20
...
Table 20
...

There are only 14 distinct space lattices in three dimensions
...
20
...
It is conventional to portray these lattices by primitive unit
cells in some cases and by non-primitive unit cells in others
...
A body-centred unit cell (I) also
has a lattice point at its centre
...
A side-centred unit cell (A, B, or C) has
lattice points at its corners and at the centres of two opposite faces
...
20
...


a

Cubic P

Cubic I

Cubic F

a

a
c

Tetragonal P

Tetragonal I

b

a
c

Orthorhombic P

Orthorhombic C
a

Monoclinic P
a

Fig
...
7 A triclinic unit cell has no axes of
rotational symmetry
...
1 The seven crystal systems

Monoclinic C

a 120°

a

a

a

b
c

Triclinic

b

c

Hexagonal

Trigonal R

The fourteen Bravais lattices
...
P denotes a primitive unit cell (R is used for a trigonal lattice), I a bodycentred unit cell, F a face-centred unit cell, and C (or A or B) a cell with lattice points on two
opposite faces
...
20
...
20
...


centre of a molecule, as the location of a lattice point or the vertex of a unit cell, but
that is not a necessary requirement
...
2 The identification of lattice planes

c
b
a

The spacing of the planes of lattice points in a crystal is an important quantitative
aspect of its structure
...
20
...
Two-dimensional lattices are easier to visualize than
three-dimensional lattices, so we shall introduce the concepts involved by referring
to two dimensions initially, and then extend the conclusions by analogy to three
dimensions
...


20
...
20
...

Note that a 0 indicates that a plane is
parallel to the corresponding axis, and that
the indexing may also be used for unit cells
with non-orthogonal axes
...
20
...
Each plane in the illustration (except the plane passing through the
origin) can be distinguished by the distances at which it intersects the a and b axes
...
For example, we could denote the four sets in the illustration as
1 1
(1a,1b), (– a, –b), (−1a,1b), and (∞a,1b)
...
If the lattice in Fig
...
9 is the top view of a
2 3
three-dimensional orthorhombic lattice in which the unit cell has a length c in the zdirection, all four sets of planes intersect the z-axis at infinity
...

2 3
The presence of fractions and infinity in the labels is inconvenient
...
As we shall see, taking reciprocals
turns out to have further advantages
...
For example, the (1,1,∞)
1 1
planes in Fig
...
9a are the (110) planes in the Miller notation
...
Negative indices are written with a bar over the number,
and Fig
...
9c shows the (⁄10) planes
...
20
...
Figure 20
...

The notation (hkl) denotes an individual plane
...
Thus, we speak of the (110) plane in a lattice, and the set of all
{110} planes that lie parallel to the (110) plane
...
The same is true of k and the b axis
and l and the c axis
...
2 THE IDENTIFICATION OF LATTICE PLANES
planes are parallel to the a axis
...


701

a

(b) The separation of planes

1
2
d hk0

=

h2 + k 2

or dhk0 =

2

a

a

dhkl

(20
...
The separation of the {hk0} planes in the square lattice shown in Fig
...
11 is given by
2 1/2

By extension to three dimensions, the separation of the {hkl} planes of a cubic lattice
is given by
1
2
d hkl

=

h2 + k 2 + l 2

or dhkl =

2

a

a
(h + k + l )
2

2

2 1/2

(20
...
20
...


The corresponding expression for a general orthorhombic lattice is the generalization
of this expression:
1
2
d hkl

=

h2
a2

+

k2
b2

+

l2

(20
...
1 Using the Miller indices

{110}

{220}

Calculate the separation of (a) the {123} planes and (b) the {246} planes of an
orthorhombic unit cell with a = 0
...
94 nm, and c = 0
...

Method For the first part, simply substitute the information into eqn 20
...
For the

second part, instead of repeating the calculation, note that if all three Miller indices
are multiplied by n, then their separation is reduced by that factor (Fig
...
12):
1
2
d nh,nk,nl

=

(nh)2
a2

+

(nk)2
b2

+

(nl)2
c2

= n2

A h2 k 2 l 2 D
n2
+ 2+ 2 = 2
C a2 b c F d hkl

which implies that
dnh,nk,nl =

d hkl
n

Answer Substituting the indices into eqn 20
...
22 nm−2
(0
...
94 nm)2 (0
...
21 nm
...
11 nm
...

Self-test 20
...

[0
...
063 nm]

Fig
...
12 The separation of the {220} planes
is half that of the {110} planes
...


20 MATERIALS 2: THE SOLID STATE
Cooling
water

Metal
target

(a)

X-rays
(b)
Fig
...
13 When two waves are in the same
region of space they interfere
...
The component waves
are shown in green and blue and the
resultant in purple
...
20
...

Beryllium is transparent to X-rays (on
account of the small number of electrons in
each atom) and is used for the windows
...
3 The investigation of structure
A characteristic property of waves is that they interfere with one another, giving a greater
displacement where peaks or troughs coincide and a smaller displacement where peaks
coincide with troughs (Fig
...
13)
...
Therefore, the regions of constructive or destructive interference show
up as regions of enhanced or diminished intensities
...
Diffraction occurs when the
dimensions of the diffracting object are comparable to the wavelength of the radiation
...
Seventeen years later, Max von Laue
suggested that they might be diffracted when passed through a crystal, for by then he
had realized that their wavelengths are comparable to the separation of lattice planes
...
The bulk
of this section will deal with the determination of structures using X-ray diffraction
...
The analysis is aided by molecular
modelling techniques, which can guide the investigation towards a plausible structure
...
They
are typically generated by bombarding a metal with high-energy electrons (Fig
...
14)
...
1 Superimposed on the
1

Bremse is German for deceleration, Strahlung for ray
...
3 THE INVESTIGATION OF STRUCTURE

Ejected
electron

(a) NaCl
(b) KCl

X-ray

Energy

Electron
beam

Intensity

Ionization

L

K

Fig
...
16 The processes that contribute to
the generation of X-rays
...
Another electron (from
the L shell in this illustration) falls into the
vacancy and emits its excess energy as an
X-ray photon
...
20
...

The smaller number of lines in (b) is a
consequence of the similarity of the K+ and
Cl− scattering factors, as discussed later in
the chapter
...
20
...
These peaks arise from
collisions of the incoming electrons with the electrons in the inner shells of the atoms
...
20
...

If the electron falls into a K shell (a shell with n = 1), the X-rays are classified as Kradiation, and similarly for transitions into the L (n = 2) and M (n = 3) shells
...
Increasingly, X-ray diffraction makes use
of the radiation available from synchrotron sources (Further information 13
...

von Laue’s original method consisted of passing a broad-band beam of X-rays
into a single crystal, and recording the diffraction pattern photographically
...
There is currently a resurgence of interest in this approach because synchrotron
radiation spans a range of X-ray wavelengths
...
They used monochromatic radiation and a powdered
sample
...
In modern powder diffractometers the intensities
of the reflections are monitored electronically as the detector is rotated around the
sample in a plane containing the incident ray (Fig
...
17)
...
Powder diffraction data are also used to determine phase diagrams,
for different solid phases result in different diffraction patterns, and to determine the
relative amounts of each phase present in a mixture
...


703

Bremsstrahlung

Wavelength
Fig
...
15 The X-ray emission from a
metal consists of a broad, featureless
Bremsstrahlung background, with sharp
transitions superimposed on it
...


Comment 20
...


704

20 MATERIALS 2: THE SOLID STATE
W

c

f
Sample
2q
X-ray
beam

Detector

A four-circle diffractometer
...


Fig
...
18

The method developed by the Braggs (William and his son Lawrence, who later
jointly won the Nobel Prize) is the foundation of almost all modern work in X-ray
crystallography
...
There are many different sets of
planes in a crystal, so there are many angles at which a reflection occurs
...

Single-crystal diffraction patterns are measured by using a four-circle diffractometer (Fig
...
18)
...
The computer controls the settings, and moves the crystal and the detector for each one in
turn
...
Computing techniques are now available that lead not only to automatic indexing but also to the
automated determination of the shape, symmetry, and size of the unit cell
...

(b) Bragg’s law

q
d

q

A

C

An early approach to the analysis of diffraction patterns produced by crystals was to
regard a lattice plane as a semi-transparent mirror, and to model a crystal as stacks of
reflecting lattice planes of separation d (Fig
...
19)
...
It has also given rise to the name reflection to denote an intense
beam arising from constructive interference
...
20
...
One ray strikes point D on the upper plane
but the other ray must travel an additional distance AB before striking the plane
immediately below
...
The net path length difference of the two rays is then
AB + BC = 2d sin θ

q
B

Fig
...
19 The conventional derivation of
Bragg’s law treats each lattice plane as a
reflecting the incident radiation
...
Constructive
interference (a ‘reflection’) occurs when
AB + BC is equal to an integer number of
wavelengths
...
For many glancing angles the path-length difference
is not an integer number of wavelengths, and the waves interfere largely destructively
...
It follows
that a reflection should be observed when the glancing angle satisfies Bragg’s law:
nλ = 2d sin θ

(20
...
are called second-order, third-order, and so on; they
correspond to path-length differences of 2, 3,
...
In modern work it is
normal to absorb the n into d, to write the Bragg law as

λ = 2d sin θ

(20
...
1)
...


20
...
2 Using Bragg’s law

A first-order reflection from the {111} planes of a cubic crystal was observed at a
glancing angle of 11
...

What is the length of the side of the unit cell?
Method The separation of the planes can be determined from Bragg’s law
...
2, which may therefore be solved for a
...
5, the {111} planes responsible for the diffraction

have separation
d111 =

λ
2 sin θ

The separation of the {111} planes of a cubic lattice of side a is given by eqn 20
...
2°

= 687 pm

Self-test 20
...


[24
...

For example, in a cubic lattice of unit cell dimension a the spacing is given by eqn 20
...

2

1

2

3

4

5

6

8

9

9

10
...
) is missing because the sum of the squares of three integers
cannot equal 7 (or 15,
...

Self-test 20
...


A diffraction examination of the element polonium gave lines at the following
values of 2θ (in degrees) when 71
...
1, 17
...
0, 24
...
2, 29
...
7, 36
...
9, 40
...
8
...

[cubic P; a = 337 pm]

(c) Scattering factors

To prepare the way to discussing modern methods of structural analysis we need to note
that the scattering of X-rays is caused by the oscillations an incoming electromagnetic

705

706

20 MATERIALS 2: THE SOLID STATE
wave generates in the electrons of atoms, and heavy atoms give rise to stronger
scattering than light atoms
...
If the scattering factor is large, then
the atoms scatter X-rays strongly
...
6)

The value of f is greatest in the forward direction and smaller for directions away from
the forward direction (Fig
...
20)
...
We show in the Justification below that, in the forward direction (for
θ = 0), f is equal to the total number of electrons in the atom
...
1 The forward scattering factor

0
...
4 0
...
8 1
...
2

The variation of the scattering
factor of atoms and ions with atomic
number and angle
...


1
As θ → 0, so k → 0
...
20
...
It follows that in
the forward direction
∞

Ύ ρ(r)r dr
2

f = 4π

0

The integral over the electron density ρ (the number of electrons in an infinitesimal
region divided by the volume of the region) multiplied by the volume element
4πr 2dr is the total number of electrons, Ne, in the atom
...
For example, the scattering factors of Na+, K+, and Cl− are 8, 18,
and 18, respectively
...


(d) The electron density

The problem we now address is how to interpret the data from a diffractometer in
terms of the detailed structure of a crystal
...

If a unit cell contains several atoms with scattering factors fj and coordinates (xj a,
yj b, zj c), then we show in the Justification below that the overall amplitude of a wave
diffracted by the {hkl} planes is given by
Fhkl =

∑ fj eiφ

hkl( j)

where φhkl(j) = 2π(hxj + kyj + lzj)

(20
...
The quantity Fhkl is called the structure
factor
...
2 The structure factor

We begin by showing that, if in the unit cell there is an A atom at the origin and a
B atom at the coordinates (xa,yb,zc), where x, y, and z lie in the range 0 to 1, then
the phase difference, φ, between the hkl reflections of the A and B atoms is φhkl =
2π(hx + ky + lz)
...
3 THE INVESTIGATION OF STRUCTURE
Phase
difference = 2 ´ 2px

Phase
difference = 2px
A

A
a

xa

xa
B
A
(b)

Phase
difference = 2 ´ 2p

Diffraction from a crystal containing two kinds of atoms
...
(b) For a (200) reflection, the phase difference is 4π
...

Fig
...
21

Consider the crystal shown schematically in Fig
...
21
...
If there is a B atom at a fraction x of the distance between the two A planes, then
it gives rise to a wave with a phase difference 2πx relative to an A reflection
...
Now consider a (200) reflection
...
5 it would give
rise to a wave that differed in phase by 2π from the wave from the upper A layer
...
For a general (h00) reflection, the phase difference is therefore h × 2πx
...
7
...
For example,
1
if the unit cells are cubic I with a B atom at x = y = z = –, then the A,B phase difference
2
is (h + k + l)π
...
Hence the diffraction pattern for a cubic I lattice
can be constructed from that for the cubic P lattice (a cubic lattice without points at
the centre of its unit cells) by striking out all reflections with odd values of h + k + l
...
20
...

If the amplitude of the waves scattered from A is fA at the detector, that of the
waves scattered from B is fBeiφhkl, with φhkl the phase difference given in eqn 20
...
The
total amplitude at the detector is therefore

Primitive cubic

Phase
difference = 2p

Body-centred cubic
h + k + l = odd are
absent

A

Face-centred cubic
h,k,l all even or all
odd are present

B

(a)

707

(100)
(110)
(111)
(200)
(210)
(211)
(220)
(221), (300)
(310)
(311)
(222)
(320)
(321)
(400)
(410), (322)
(411), (330)
(331)
(420)
(421)
(332)
(422)
(502), (430)

Fhkl = fA + fBeiφhkl
Because the intensity is proportional to the square modulus of the amplitude of the
wave, the intensity, Ihkl, at the detector is
Ihkl ∝ F* Fhkl = (fA + fBe−iφhkl)(fA + fBeiφhkl)
hkl
2
2
Ihkl ∝ f A + f B

iφhkl

+ fA fB(e

The powder diffraction patterns
and the systematic absences of three
versions of a cubic cell
...
The
locations of the lines give the cell
dimensions
...
20
...
Hence, there is a variation
in the intensities of the lines with different hkl
...
3 Calculating a structure factor

Calculate the structure factors for the unit cell in Fig
...
23
...
7
...
20
...
Write f + for the Na+ scattering
factor and f − for the Cl − scattering factor
...
However, ions on faces are shared
1
1
between two cells (use – f ), those on edges by four cells (use – f ), and those at
2
4
1
corners by eight cells (use – f )
...
7, and summing over the coordinates of all 27 atoms in the

illustration:
The location of the atoms for the
structure factor calculation in Example
20
...
The purple circles are Na+; the green
circles are Cl−
...
20
...
For f + = f −, which
is the case for identical atoms in a cubic P arrangement, the hkl all-odd have zero
intensity, corresponding to the ‘systematic absences’ of cubic P unit cells
...
4 Which reflections cannot be observed for a cubic I lattice?

[for h + k + l odd, Fhkl = 0]
The intensity of the (hkl) reflection is proportional to |Fhkl |2, so in principle we can
determine the structure factors experimentally by taking the square root of the corresponding intensities (but see below)
...
8)

hkl

where V is the volume of the unit cell
...
8 is called a Fourier synthesis of
the electron density
...
3 THE INVESTIGATION OF STRUCTURE

709

Example 20
...
In
an X-ray analysis the structure factors were found as follows:
h:

0

1

2

3

4

5

6

7

8

9

Fh

16

−10

2

−1

7

−10

8

−3

2

−3

h:

10

11

12

13

14

15

Fh

6

−5

3

−2

2

−3

(and F−h = Fh)
...

Method Because F−h = Fh, it follows from eqn 20
...

Answer The results are plotted in Fig
...
24 (blue line)
...
The more terms there are included, the more
accurate the density plot
...


Electron density, r (x)

Vρ(x) =

0
0

0
...
5 Use mathematical software to experiment with different structure

factors (including changing signs as well as amplitudes)
...

[Fig
...
24 (purple line)]

(e) The phase problem

A problem with the procedure outlined above is that the observed intensity Ihkl is
proportional to the square modulus |Fhkl |2, so we cannot say whether we should use
+|Fhkl | or −|Fhkl | in the sum in eqn 20
...
In fact, the difficulty is more severe for noncentrosymmetric unit cells because, if we write Fhkl as the complex number |Fhkl |eiα,
where α is the phase of Fhkl and |Fhkl | is its magnitude, then the intensity lets us determine |Fhkl | but tells us nothing of its phase, which may lie anywhere from 0 to 2π
...
20
...
Some way must be found to assign phases to the structure
factors, for otherwise the sum for ρ cannot be evaluated and the method would be
useless
...
One
procedure that is widely used for inorganic materials with a reasonably small number
of atoms in a unit cell and for organic molecules with a small number of heavy atoms
is the Patterson synthesis
...
8:

Fig
...
24 The plot of the electron density
calculated in Example 20
...
5 (purple)
...
5
by using the interactive applets found in
the text’s web site
...
20
...
The distance and orientation
of each spot from the origin gives the
orientation and separation of one
atom–atom separation in (a)
...


1
V

∑ |Fhkl |2e−2πi(hx+ky+lz)

(20
...
Thus, if atom A is at the
coordinates (xA,yA,zA) and atom B is at (xB,yB,zB), then there will be a peak at (xA − xB,
yA − yB, zA − zB) in the Patterson map
...
The
height of the peak in the map is proportional to the product of the atomic numbers
of the two atoms, ZAZB
...
20
...
20
...

Heavy atoms dominate the scattering because their scattering factors are large, of
the order of their atomic numbers, and their locations may be deduced quite readily
...
To see why this is so, we have to note that a structure factor of a centrosymmetric cell has the form
F = (±)fheavy + (±)flight + (±)flight + · · ·

(20
...
The flight are all much smaller than fheavy, and their phases are more
or less random if the atoms are distributed throughout the unit cell
...
This phase can then be combined with the observed |F | (from the reflection intensity) to perform a Fourier synthesis of the full electron density in the unit
cell, and hence to locate the light atoms as well as the heavy atoms
...
Direct methods
are based on the possibility of treating the atoms in a unit cell as being virtually
randomly distributed (from the radiation’s point of view), and then using statistical
techniques to compute the probabilities that the phases have a particular value
...
For example,
the Sayre probability relation has the form
sign of Fh+h′,k+k′,l+l′ is probably equal to (sign of Fhkl) × (sign of Fh′k′l′)

(20
...

(f) Structure refinement

In the final stages of the determination of a crystal structure, the parameters describing the structure (atom positions, for instance) are adjusted systematically to give
the best fit between the observed intensities and those calculated from the model of
the structure deduced from the diffraction pattern
...
Not only does the procedure give accurate positions for all the atoms in
the unit cell, but it also gives an estimate of the errors in those positions and in the
bond lengths and angles derived from them
...


I20
...
1 X-ray crystallography of biological macromolecules

X-ray crystallography is the deployment of X-ray diffraction techniques for the
determination of the location of all the atoms in molecules as complicated as biopolymers
...
20
...
To interpret this image by using the Bragg law we have to be aware that it was obtained by
using a fibre consisting of many DNA molecules oriented with their axes parallel to
the axis of the fibre, with X-rays incident from a perpendicular direction
...

There are two principal features in Fig
...
26: the strong ‘meridional’ scattering
upward and downward by the fibre and the X-shaped distribution at smaller scattering angles
...
Because the
meridional pattern occurs at a distance of about 10 times that of the innermost spots
of the X-pattern, the large-scale structure is about 10 times bigger than the small-scale
structure
...
4 nm)
...
20
...
Each turn of the
helix defines two planes, one orientated at an angle α to the horizontal and the other
at −α
...
Thus, a DNA molecule
is like two arrays of planes, each set corresponding to those treated in the derivation of
Bragg’s law, with a perpendicular separation d = p cos α, where p is the pitch of the
helix, each canted at the angles ±α to the horizontal
...
The

a
a

(a)

(c)
(b)

The X-ray diffraction pattern
obtained from a fibre of B-DNA
...
(Adapted from an illustration
that appears in J
...
Glusker and K
...

Trueblood, Crystal structure analysis:
A primer
...
)

Fig
...
26

Fig
...
27 The origin of the X pattern
characteristic of diffraction by a helix
...
(b) The
diffraction spots from one set of planes
appear at an angle α to the vertical, giving
one leg of the X, and those of the other set
appear at an angle −α, giving rise to the
other leg of the X
...
(c) The
sequence of spots outward along a leg of
the X corresponds to first-, second-,
...


712

20 MATERIALS 2: THE SOLID STATE

Fig
...
28 The effect of the internal structure
of the helix on the X-ray diffraction
pattern
...

(b) Parallel planes passing through the
residues are perpendicular to the axis of the
molecule
...


Reservoir
solution

Drop of
biopolymer
solution

Fig
...
29 In a common implementation of
the vapour diffusion method of biopolymer
crystallization, a single drop of biopolymer
solution hangs above a reservoir solution
that is very concentrated in a non-volatile
solute
...
In the course
of evaporation (denoted by the downward
arrows), the biopolymer solution becomes
more concentrated and, at some point,
crystals may form
...
The sequence of spots outward along a leg corresponds to first-, second-,
...
in eqn 20
...
Therefore
from the X-ray pattern, we see at once that the molecule is helical and we can measure
the angle α directly, and find α = 40°
...
4 nm)/(4 tan 40°) = 1
...

To derive the relation between the helix and the cross-like pattern we have ignored
the detailed structure of the helix, the fact that it is a periodic array of nucleotide bases,
not a smooth wire
...
20
...
These planes give rise to the strong meridional
diffraction with an angle that allows us to determine the layer spacing from Bragg’s
law in the form λ = 2h sin θ as h = 340 pm
...
Most work is now done not on fibres but on crystals, in
which the large molecules lie in orderly ranks
...
The increase in the ionic strength of the solution
decreases the solubility of the protein to such an extent that the protein precipitates,
sometimes as crystals that are amenable to analysis by X-ray diffraction
...
2) or vapour diffusion
...
20
...
If the reservoir solution is more concentrated in a non-volatile solute (for example, a salt) than
is the biopolymer solution, then solvent will evaporate slowly from the drop until the
vapour pressure of water in the closed container reaches a constant, equilibrium value
...

Special techniques are used to crystallize hydrophobic proteins, such as those spanning the bilayer of a cell membrane
...
Dialysis or
vapour diffusion may then be used to induce crystallization
...
The three-dimensional structures of a very

20
...
20
...


large number of biological polymers have been determined in this way
...
This limitation stems from the fact that the Bragg rotation
method requires stable crystals that do not change structure during the lengthy data
acquisition times required
...

Time-resolved X-ray diffraction techniques make use of synchrotron sources,
which can emit intense polychromatic pulses of X-ray radiation with pulse widths
varying from 100 ps to 200 ps (1 ps = 10−12 s)
...
However, good
diffraction data cannot be obtained from a single X-ray pulse and reflections from
several pulses must be averaged together
...

An example of the power of time-resolved X-ray crystallography is the elucidation
of structural changes that accompany the activation by light of the photoactive yellow
protein of the bacterium Ectothiorhodospira halophila
...
20
...
A series of rearrangements
then follows, which includes the ejection of the ion from its binding site deep in the
protein, its return to the site, and re-formation of the cis conformation
...
Time-resolved X-ray diffraction studies in the nanosecond
to millisecond ranges identified a number of structural changes that follow electronic
excitation of the phenolate ion with a laser pulse: isomerization, ejection, protonation
of the exposed ion, and a number of amino acid motions
...
4 Neutron and electron diffraction
According to the de Broglie relation (eqn 8
...
Neutrons generated in a nuclear reactor and
then slowed to thermal velocities have wavelengths similar to those of X-rays and may
also be used for diffraction studies
...
In practice, a
range of wavelengths occurs in a neutron beam, but a monochromatic beam can be
selected by diffraction from a crystal, such as a single crystal of germanium
...
3

The text’s web site contains links to
databases of structures of biological
macromolecules
...
5 Calculating the typical wavelength of thermal neutrons

Calculate the typical wavelength of neutrons that have reached thermal equilibrium with their surroundings at 373 K
...
There are two link-

ing steps
...
Then the linear momentum can be expressed in terms of the
kinetic energy, the mean value of which is given in terms of the temperature by
the equipartition theorem (see Section 17
...

Answer From the equipartition principle, we know that the mean translational

kinetic energy of a neutron at a temperature T travelling in the x-direction is
1
E K = – kT
...
Hence, p = (mkT)1/2
...
626 × 10−34 J s
{(1
...
381 × 10−23 J K−1) × (373 K)}1/2
6
...
675 × 1
...
26 × 10−10 m = 226 pm
where we have used 1 J = 1 kg m2 s−2
...
6 Calculate the temperature needed for the average wavelength of the
neutrons to be 100 pm
...
90 × 10 3 K]

Fig
...
31 If the spins of atoms at lattice
points are orderly, as in this material,
where the spins of one set of atoms are
aligned antiparallel to those of the other
set, neutron diffraction detects two
interpenetrating simple cubic lattices on
account of the magnetic interaction of the
neutron with the atoms, but X-ray
diffraction would see only a single bcc
lattice
...
First, the
scattering of neutrons is a nuclear phenomenon
...
As a result, the intensity with which neutrons are scattered is independent of the number of electrons and neighbouring elements in the periodic table may scatter neutrons with markedly different intensities
...
A second difference is that neutrons possess a magnetic moment due to their
spin
...
One
consequence is that neutron diffraction is well suited to the investigation of magnetically ordered lattices in which neighbouring atoms may be of the same element but
have different orientations of their electronic spin (Fig
...
31)
...
However, their main
application is to the study of surfaces, and we postpone their discussion until Chapter 25
...
5 METALLIC SOLIDS

715

Crystal structure
The bonding within a solid may be of various kinds
...

20
...

(a) Close packing

Figure 20
...
A close-packed three-dimensional structure is obtained by stacking
such close-packed layers on top of one another
...

In all polytypes, the spheres of second close-packed layer lie in the depressions of
the first layer (Fig
...
33)
...
In one,
the spheres are placed so that they reproduce the first layer (Fig
...
34a), to give an
ABA pattern of layers
...
20
...
Two polytypes are formed if the two
stacking patterns are repeated in the vertical direction
...
, the spheres are hexagonally close-packed

Fig
...
32 The first layer of close-packed
spheres used to build a three-dimensional
close-packed structure
...
20
...

The two layers are the AB component of
the close-packed structure
...
20
...

(b) Alternatively, the third layer might lie
in the dips that are not above the spheres
in the first layer, resulting in an ABC
structure, which corresponds to cubic
close-packing

716

20 MATERIALS 2: THE SOLID STATE
Table 20
...


(b)
Fig
...
35 A fragment of the structure
shown in Fig
...
34 revealing the (a)
hexagonal (b) cubic symmetry
...
20
...


(hcp)
...
,
the spheres are cubic close-packed (ccp)
...
20
...
The ccp structure gives rise to a face-centred unit cell, so may
also be denoted cubic F (or fcc, for face-centred cubic)
...
Table 20
...

The compactness of close-packed structures is indicated by their coordination
number, the number of atoms immediately surrounding any selected atom, which is
12 in all cases
...
740 (see the following Justification)
...
0 per cent of the
volume is empty space
...

Justification 20
...
20
...


To calculate a packing fraction of a ccp structure, we first calculate the volume of a
unit cell, and then calculate the total volume of the spheres that fully or partially
occupy it
...

The second part involves counting the fraction of spheres that occupy the cell
...
20
...
Because a diagonal of any face passes completely through one
sphere and halfway through two other spheres, its length is 4R
...
Because each cell contains
1
1
4
the equivalent of 6 × – + 8 × – = 4 spheres, and the volume of each sphere is – πR3,
2
8
3
16
3
the total occupied volume is – πR
...
740
...
The packing fractions of structures that are
not close-packed are calculated similarly (see Exercises 20
...
17 and Problem 20
...


(b) Less closely packed structures

As shown in Table 20
...
The departure from close packing suggests that factors such as
specific covalent bonding between neighbouring atoms are beginning to influence the
structure and impose a specific geometrical arrangement
...
The coordination number of a bcc structure is
2
Strictly speaking, ccp refers to a close-packed arrangement whereas fcc refers to the lattice type of the
common representation of ccp
...


20
...
The packing fraction of 0
...
74), and shows that about two-thirds of the available space
is actually occupied
...
6 Ionic solids
Two questions arise when we consider ionic solids: the relative locations adopted by
the ions and the energetics of the resulting structure
...
The coordination number of an ion is the number of nearest neighbours of opposite charge; the structure
itself is characterized as having (n+ ,n−) coordination, where n+ is the coordination
number of the cation and n− that of the anion
...
As a result, ionic solids are generally less dense than metals
...
20
...
In this structure, an ion of one charge occupies the centre of a cubic unit cell with eight counter ions at its corners
...

When the radii of the ions differ more than in CsCl, even eight-coordinate packing
cannot be achieved
...
20
...
In this structure, each cation is surrounded by
six anions and each anion is surrounded by six cations
...
This structure is adopted by NaCl
itself and also by several other MX compounds, including KBr, AgCl, MgO, and ScN
...
12]

The two radii are those of the larger and smaller ions in the crystal
...
732 and that the rock-salt structure should be expected when 21/2 − 1 = 0
...
732
...
414, the most efficient packing leads to four-coordination of the
type exhibited by the sphalerite (or zinc blende) form of ZnS (Fig
...
39)
...
The deviation of
a structure from that expected on the basis of the radius-ratio rule is often taken to be
an indication of a shift from ionic towards covalent bonding; however, a major source
of unreliability is the arbitrariness of ionic radii and their variation with coordination
number
...

However, we need to apportion the total distance between the two ions by defining
the radius of one ion and then inferring the radius of the other ion
...
3)
...
20
...


Cl

Na

Fig
...
38 The rock-salt (NaCl) structure
consists of two mutually interpenetrating
slightly expanded face-centred cubic arrays
of ions
...


S

Zn

Fig
...
39 The structure of the sphalerite
form of ZnS showing the location of the Zn
atoms in the tetrahedral holes formed by
the array of S atoms
...
)

718

20 MATERIALS 2: THE SOLID STATE

Synoptic table 20
...
Because ionic radii are so arbitrary,
predictions based on them must be viewed cautiously
...
The lattice energy is always positive;
a high lattice energy indicates that the ions interact strongly with one another to give
a tightly bonded solid
...

† Coordination number
...
20
...


and its equivalent for other charge types and stoichiometries
...

Each ion in a solid experiences electrostatic attractions from all the other oppositely
charged ions and repulsions from all the other like-charged ions
...
Each cation is surrounded by anions, and there is a large negative contribution from the attraction of
the opposite charges
...
There is also
a negative contribution from the anions beyond those cations, a positive contribution
from the cations beyond them, and so on to the edge of the solid
...

First, consider a simple one-dimensional model of a solid consisting of a long line
of uniformly spaced alternating cations and anions, with d the distance between their
centres, the sum of the ionic radii (Fig
...
40)
...
The potential energy of the central ion is calculated by summing all the
terms, with negative terms representing attractions to oppositely charged ions and
positive terms representing repulsions from like-charged ions
...
Finally, we multiply EP by 2
2
3
4
to obtain the total energy arising from interactions on each side of the ion and then
multiply by Avogadro’s constant, NA, to obtain an expression for the lattice energy
per mole of ions
...
This energy is negative, corresponding to a net attraction
...
6 IONIC SOLIDS
EP = −A ×

|z1z2 |NAe 2

(20
...
4 Madelung constants
Structural type*

The factor A is a positive numerical constant called the Madelung constant; its value
depends on how the ions are arranged about one another
...
748
...
4 lists Madelung constants for
other common structures
...
These repulsions are taken into account by supposing that, because wavefunctions decay exponentially with distance at large distances
from the nucleus, and repulsive interactions depend on the overlap of orbitals, the
repulsive contribution to the potential energy has the form
E* = NAC′e−d/d *
P

(20
...
763

Fluorite

2
...
748

Rutile

2
...
638

Wurtzite

1
...


with C′ and d* constants; the latter is commonly taken to be 34
...
The total
potential energy is the sum of EP and E*, and passes through a minimum when
P
d(EP + E*)/dd = 0 (Fig
...
41)
...
21a):

4πε0d

C

1−

d* D
d F

A

(20
...
Provided we ignore zero-point
contributions to the energy, we can identify the negative of this potential energy with
the lattice energy
...

Experimental values of the lattice enthalpy (the enthalpy, rather than the energy)
are obtained by using a Born–Haber cycle, a closed path of transformations starting
and ending at the same point, one step of which is the formation of the solid compound from a gas of widely separated ions
...
20
...
It consists of the following steps (for convenience, starting at the
elements):

Repulsion

Potential energy

EP, min = −

NA |zAzB |e 2 A

Lattice
parameter
0

Attraction

Fig
...
41 The contributions to the total
potential energy of an ionic crystal
...
Ionization of K(g)

+418

[ionization enthalpy of K(g)]

4
...
Formation of solid from gas

−∆HL/(kJ mol−1)

6
...
Sublimation of K(s)
2
...
Some lattice enthalpies obtained in this way are listed in
Table 20
...
As can be seen from the data, the trends in values are in general accord
with the predictions of the Born–Mayer equation
...


-349
+418

K+(g) + Cl-(g)

K(g) + Cl(g)
+122
1
K(g) + - Cl2(g)
2
+89

-DHL

1
K(s) + - Cl2(g)
2

+437
KCl(s)
Fig
...
42 The Born–Haber cycle for KCl at
298 K
...


720

20 MATERIALS 2: THE SOLID STATE

Synoptic table 20
...


Comment 20
...
For example, oxygen has two
allotropes: O2 and O3 (ozone)
...
20
...
Each C atom is tetrahedrally
bonded to four neighbours
...


20
...
In
this section we can do no more than hint at the diversity of types of solids found when
molecules pack together or atoms link together in extended networks
...
The demands of directional bonding, which have only a small effect on the structures of many metals, now override the
geometrical problem of packing spheres together, and elaborate and extensive structures may be formed
...

Diamond and graphite are two allotropes of carbon
...
20
...
The network of
strong C-C bonds is repeated throughout the crystal and, as a result, diamond is the
hardest known substance
...
20
...
Because the
sheets can slide against each other when impurities are present, graphite is used widely
as a lubricant
...
2)
...
The simplest structural
motif is called a single-walled nanotube (SWNT) and is shown in Fig
...
45
...
The tubes have diameters between 1 and
2 nm and lengths of several micrometres
...
20
...
1)
...

Molecular solids, which are the subject of the overwhelming majority of modern
structural determinations, are held together by van der Waals interactions (Chapter
18)
...
20
...


(b)

Fig
...
44 Graphite consists of flat planes of hexagons of carbon atoms lying above one
another
...
When impurities are present, the planes can slide over one another
easily
...
8 MECHANICAL PROPERTIES

721

Fig
...
46 A fragment of the crystal structure
of ice (ice-I)
...
The central O atom is
attached by two short O-H bonds to two
H atoms and by two long hydrogen bonds
to the H atoms of two of the neighbouring
molecules
...


minimum Gibbs energy)
...
The problem is made more complicated by the role of hydrogen
bonds, which in some cases dominate the crystal structure, as in ice (Fig
...
46), but
in others (for example, in phenol) distort a structure that is determined largely by the
van der Waals interactions
...
The rational fabrication of modern materials depends crucially on
an understanding of this link
...
5

The web site contains links to databases
of properties of materials, such as metals
and polymers
...
8 Mechanical properties
The fundamental concepts for the discussion of the mechanical properties of solids
are stress and strain
...
The strain is the resulting distortion of the sample
...

Stress may be applied in a number of different ways
...
20
...
A pure
shear is a stress that tends to push opposite faces of the sample in opposite directions
...
For low stresses, the
strain is linearly proportional to the stress
...
Above a certain threshold, the strain becomes plastic
in the sense that recovery does not occur when the stress is removed
...
Brittle solids, such as ionic solids, exhibit sudden
fracture as the stress focused by cracks causes them to spread catastrophically
...
20
...

(a) Uniaxial stress, (b) shear stress,
(c) hydrostatic pressure
...
(b) Shear stress
...

Fig
...
48

[20
...
16c]

shear strain

where ‘normal stress’ refers to stretching and compression of the material, as shown in
Fig
...
48a and ‘shear stress’ refers to the stress depicted in Fig
...
48b
...
11
(eqn 2
...
A third ratio indicates how the sample changes its shape:

νP =

Poisson’s ratio:
(b)

[20
...
17]

normal strain

The moduli are interrelated:
G=

E
2(1 + νP)

K=

E

(20
...
Thus, in the Justification below, we show that, if
neighbouring molecules interact by a Lennard-Jones potential, then the bulk modulus
and the compressibility of the solid are related to the Lennard-Jones parameter ε (the
depth of the potential well) by
K=

8NAε

κ=

Vm

Vm

(20
...

Justification 20
...
45), to obtain
A ∂2U D
E
K=VB
C ∂V 2 F T
This expression shows that the bulk modulus (and through eqn 20
...

To develop this conclusion, we note that the variation of internal energy with volume
can be expressed in terms of its variation with a lattice parameter, R, such as the
length of the side of a unit cell
...
9 ELECTRICAL PROPERTIES

723

2

A ∂2U D A ∂R D
E
K=VB 2E B
C ∂R F T,0 C ∂V F T,0
where the 0 denotes that the derivatives are evaluated at the equilibrium dimensions
of the unit cell by setting R = R0 after the derivative has been calculated
...
Then, if the internal energy is given by a pairwise Lennard-Jones (12,6)-potential, eqn 18
...
6

(20
...
It then follows that
K=

72nNAε
3

9aR

=

8nNAε
V0

=

8NAε
Vm

To obtain the result in eqn 20
...
31)
...
19
...


20
...
Two types of solid are distinguished by the temperature
dependence of their electrical conductivity (Fig
...
50):
A metallic conductor is a substance with a conductivity that decreases as the temperature is raised
...

A semiconductor generally has a lower conductivity than that typical of metals, but
the magnitude of the conductivity is not the criterion of the distinction
...
We shall use this term, but it should be appreciated
that it is one of convenience rather than one of fundamental significance
...

(a) The formation of bands

The central aspect of solids that determines their electrical properties is the distribution of their electrons
...
In one, the nearly

Yield
point

Tensile
strength

Stress

The typical behaviour of a solid under stress is illustrated in Fig
...
49
...
For larger strains, though, dislocations begin to play a major
role and the strain becomes plastic in the sense that the sample does not recover its
original shape when the stress is removed
...
The slip planes of a ccp structure are the close-packed planes, and careful inspection of a unit cell shows that there are eight sets of slip planes in different
directions
...
In contrast, a
hexagonal close-packed structure has only one set of slip planes; and metals with
hexagonal close packing, like zinc or cadmium, tend to be brittle
...
20
...
At high strains, the body
is no longer elastic, may yield and become
plastic
...


724

20 MATERIALS 2: THE SOLID STATE
10

8

Conductivity/(S cm-1)

Metal
Superconductor

104

1

10-4

10

Semiconductor

-8

1

10

100

1000

T/K
Fig
...
50 The variation of the electrical
conductivity of a substance with
temperature is the basis of its classification
as a metallic conductor, a semiconductor,
or a superconductor
...


(a)

N=1

(b)

2

(c)

3

(d)

4

free-electron approximation, the valence electrons are assumed to be trapped in a
box with a periodic potential, with low energy corresponding to the locations of
cations
...
The latter model is more
in accord with the discussion in the foregoing chapters, and we confine our attention
to it
...
At first sight, this model may seem too restrictive and unrealistic
...

Suppose that each atom has one s orbital available for forming molecular orbitals
...
One
atom contributes one s orbital at a certain energy (Fig
...
51)
...

The third atom overlaps its nearest neighbour (and only slightly the next-nearest),
and from these three atomic orbitals, three molecular orbitals are formed: one is
fully bonding, one fully antibonding, and the intermediate orbital is nonbonding
between neighbours
...
At this stage, we can begin to see that the general effect of bringing up successive atoms is to spread the range of energies covered by the molecular orbitals,
and also to fill in the range of energies with more and more orbitals (one more for
each atom)
...
6) is

α−E
β
0
0
0
β
α−E
β
0
0
β
α−E
β
0
0
0
0
β
α−E
β
β
α−E
0
0
0
Ӈ
Ӈ
Ӈ
Ӈ
Ӈ
0
0
0
0
0

¥

The formation of a band of N
molecular orbitals by successive addition
of N atoms to a line
...

Fig
...
51

=0

where β is now the (s,s) resonance integral
...
, N

(20
...
22)

We can think of this band as consisting of N different molecular orbitals, the lowestenergy orbital (k = 1) being fully bonding, and the highest-energy orbital (k = N)
being fully antibonding between adjacent atoms (Fig
...
52)
...


20
...
20
...
In this case, the s and
p orbitals of the atoms are so widely spaced
that there is a band gap
...


Justification 20
...
Therefore, in this limit
E1 = α + 2β
When k has its maximum value of N,
EN = α + 2β cos

Nπ
N+1

As N approaches infinity, we can ignore the 1 in the denominator, and the cosine
term becomes cos π = −1
...


(b) The occupation of orbitals

Now consider the electronic structure of a solid formed from atoms each able to contribute one electron (for example, the alkali metals)
...
There are
N electrons to accommodate
...
20
...
However, unlike in molecules, there are empty
orbitals very close in energy to the Fermi level, so it requires hardly any energy to
excite the uppermost electrons
...


Unoccupied
levels
Fermi
level

Energy

The band formed from overlap of s orbitals is called the s band
...
20
...
If the atomic p orbitals lie higher in energy than the s orbitals, then the p
band lies higher than the s band, and there may be a band gap, a range of energies to
which no orbital corresponds
...


Occupied
levels

Fig
...
53 When N electrons occupy a band
of N orbitals, it is only half full and the
electrons near the Fermi level (the top of
the filled levels) are mobile
...
The population, P, of the orbitals is given by the Fermi–Dirac
distribution, a version of the Boltzmann distribution that takes into account the
effect of the Pauli principle:

1
...
8
P

P=

0
...
4

T=0

1/3
3

1

10

0
-6 -4

-2

0
2
(E - m )/m

4

e

+1

(20
...
The chemical potential in eqn 20
...

The shape of the Fermi–Dirac distribution is shown in Fig
...
54
...
2

1
(E−µ)/kT

6

The Fermi–Dirac distribution,
which gives the population of the levels at a
temperature T
...
The curves are
labelled with the value of µ /kT
...

Fig
...
54

Exploration Express the population
P as a function of the variables
(E − µ)/µ and µ/kT and then display the set
of curves shown in Fig
...
54 as a single
surface
...
24)

The population now resembles a Boltzmann distribution, decaying exponentially with
increasing energy
...

The electrical conductivity of a metallic solid decreases with increasing temperature
even though more electrons are excited into empty orbitals
...
That is, the electrons are scattered out of their paths through the solid,
and are less efficient at transporting charge
...
The Fermi level now lies at the top of the band (at T = 0), and there is a gap
before the next band begins (Fig
...
55)
...
As a consequence of electron promotion, positively charged ‘holes’ are left in in the valence band
...
In fact, it is a semiconductor,
because the electrical conductivity depends on the number of electrons that are promoted across the gap, and that number increases as the temperature is raised
...
Thus, the conventional distinction between an insulator and a semiconductor is related to the size of
the band gap and is not an absolute distinction like that between a metal (incomplete
bands at T = 0) and a semiconductor (full bands at T = 0)
...
55 depicts conduction in an intrinsic semiconductor, in which semiconduction is a property of the band structure of the pure material
...
A compound semiconductor is an
intrinsic semiconductor that is a combination of different elements, such as GaN,
CdS, and many d-metal oxides
...
If the dopants
can trap electrons, they withdraw electrons from the filled band, leaving holes which
allow the remaining electrons to move (Fig
...
56a)
...
An example is silicon doped with indium
...
9 ELECTRICAL PROPERTIES
(a) T = 0

727

(b) T > 0

Band
gap, Eg

Thermal
excitation

Energy

Energy

Conduction
band

Acceptor
band

Donor
band

Valence
band

(a)
Fig
...
55 (a) When 2N electrons are
present, the band is full and the material is
an insulator at T = 0
...


(b)

Fig
...
56 (a) A dopant with fewer electrons
than its host can form a narrow band that
accepts electrons from the valence band
...

(b) A dopant with more electrons than its
host forms a narrow band that can supply
electrons to the conduction band
...


semiconduction as arising from the transfer of an electron from a Si atom to a
neighbouring In atom
...
Alternatively, a dopant might carry
excess electrons (for example, phosphorus atoms introduced into germanium), and
these additional electrons occupy otherwise empty bands, giving n-type semiconductivity, where n denotes the negative charge of the carriers (Fig
...
56b)
...
2
...
Consider the application of a ‘reverse bias’ to the junction, in the
sense that a negative electrode is attached to the p-type semiconductor and a positive
electrode is attached to the n-type semiconductor (Fig
...
57a)
...
As a consequence, charge does not flow across the
junction
...
20
...
Now charge flows
across the junction, with electrons in the n-type semiconductor moving toward the
positive electrode and holes moving in the opposite direction
...
20
...


728

20 MATERIALS 2: THE SOLID STATE
through a material
...

As electrons and holes move across a p–n junction under forward bias, they recombine and release energy
...
In some solids, the energy
of electron–hole recombination is released as heat and the device becomes warm
...

IMPACT ON NANOSCIENCE

I20
...
1, I9
...
3) that research on nanometre-sized materials is motivated by the possibility that they will form the basis for
cheaper and smaller electronic devices
...
An important type of nanowire is based on carbon nanotubes, which,
like graphite, can conduct electrons through delocalized π molecular orbitals that
form from unhybridized 2p orbitals on carbon
...
The SWNT in Fig
...
45 is a semiconductor
...

Carbon nanotubes are promising building blocks not only because they have useful
electrical properties but also because they have unusual mechanical properties
...

Silicon nanowires can be made by focusing a pulsed laser beam on to a solid target
composed of silicon and iron
...
The phase diagram for this complex mixture shows that
solid silicon and liquid FeSin coexist at temperatures higher than 1473 K
...
It is observed that the silicon precipitate consists of nanowires with
diameters of about 10 nm and lengths greater than 1 µm
...
The result is formation of highly ordered structures
...

For example, Fig
...
58 shows an AFM image of germanium nanowires on a silicon
surface
...

Direct manipulation of atoms on a surface also leads to the formation of nanowires
...

Fig
...
58 Germanium nanowires fabricated
on to a silicon surface by molecular beam
epitaxy
...
Ogino et al
...
Chem
...
32, 447
(1999)
...
10 Optical properties
In this section, we explore the consequences of interactions between electromagnetic
radiation and solids
...


20
...
Now we consider the effects on the electronic absorption
spectrum of bringing atoms or molecules together into a solid
...
If the excitation corresponds to the removal of an electron from one orbital of a molecule and
its elevation to an orbital of higher energy, then the excited state of the molecule can
be envisaged as the coexistence of an electron and a hole
...
20
...

Exciton formation causes spectral lines to shift, split, and change intensity
...

A migrating excitation of this kind is called a Frenkel exciton
...
A migrating excitation
of this kind, which is now spread over several molecules (more usually ions), is a
Wannier exciton
...
Their migration implies
that there is an interaction between the species that constitute the crystal, for otherwise the excitation on one unit could not move to another
...
The strength of the interaction governs the rate at which
an exciton moves through the crystal: a strong interaction results in fast migration, and
a vanishingly small interaction leaves the exciton localized on its original molecule
...
Thus, an electric
dipole transition in a molecule is accompanied by a shift of charge, and the transient
dipole exerts a force on an adjacent molecule
...
This process continues and the excitation migrates through the crystal
...
An all-parallel arrangement of
the dipoles (Fig
...
60a) is energetically unfavourable, so the absorption occurs at a
higher frequency than in the isolated molecule
...
20
...

Illustration 20
...
4 that the potential energy of interaction between two parallel dipoles µ1 and µ 2 separated by a distance r is V = µ1 µ 2(1 − 3 cos2θ)/4πε0r 3,
where the angle θ is defined in (1)
...
It follows that V < 0 (an attractive interaction)
for 0° ≤ θ < 54
...
74° (for then 1 − 3 cos2θ = 0), and V > 0 (a
repulsive interaction) for 54
...
This result is expected on the basis of
qualitative arguments
...
By contrast, in a parallel arrangement,
the molecular interaction is repulsive because of the close approach of regions of
partial charge with the same sign
...
74°, the frequency of exciton
absorption is lower than the corresponding absorption frequency for the isolated
molecule (a red shift in the spectrum of the solid with respect to that of the isolated

Fig
...
59 The electron–hole pair shown on
the left can migrate through a solid lattice
as the excitation hops from molecule to
molecule
...


n

(a)

(b)

n

Fig
...
60 (a) The alignment of transition
dipoles (the yellow arrows) is energetically
unfavourable, and the exciton absorption is
shifted to higher energy (higher frequency)
...


m2
r
m1

q

730

20 MATERIALS 2: THE SOLID STATE
molecule)
...
74° < θ ≤ 90°, the frequency of exciton absorption
is higher than the corresponding absorption frequency for the isolated molecule (a
blue shift in the spectrum of the solid with respect to that of the isolated molecule)
...
74° the solid and the isolated molecule have absorption
lines at the same frequency
...
20
...
The
separation of the bands is the Davydov
splitting
...
The splitting between the bands is the Davydov splitting
...
20
...
Let the transition dipoles be along the length of the molecules
...
Within each unit cell the transition dipoles may
be arrayed in the two different ways shown in the illustration
...
The Davydov splitting is determined by the energy of
interaction between the transition dipoles within the unit cell
...
Again we
need to consider the consequences of interactions between particles, in this case
atoms, which are now so strong that we need to abandon arguments based primarily
on van der Waals interactions in favour of a full molecular orbital treatment, the band
model of Section 20
...

Consider Fig
...
53, which shows bands in an idealized metallic conductor
...
There is a near continuum of unoccupied energy levels above the Fermi level,
so we expect to observe absorption over a wide range of frequencies
...
Because this range of
absorbed frequencies includes the entire visible spectrum, we expect that all metals
should appear black
...

To explain the shiny appearance of a smooth metal surface, we need to realize that
the absorbed energy can be re-emitted very efficiently as light, with only a small fraction of the energy being released to the surroundings as heat
...
In essence, if the sample is excited with visible light, then visible
light will be reflected from the surface, accounting for the lustre of the material
...
Silver reflects light with nearly equal efficiency
across the visible spectrum because its band structure has many unoccupied energy
levels that can be populated by absorption of, and depopulated by emission of, visible
light
...

The material reflects at all wavelengths, but more light is emitted at lower frequencies
(corresponding to yellow, orange, and red) Similar arguments account for the colours
of other metals, such as the yellow of gold
...
We have already seen that promotion of electrons from the valence to the conduction band of a semiconductor can be the result of

20
...
In some materials, the band gap is very large and electron promotion
can occur only by excitation with electromagnetic radiation
...
20
...
Above this frequency threshold, a wide range of frequencies can be absorbed by
the material, as in a metal
...
2 Predicting the colour of a semiconductor

The semiconductor cadmium sulfide (CdS) has a band gap energy of 2
...
8 × 10−19 J)
...
8 × 10−19 J
6
...
8 × 1014 s−1

This frequency, of 5
...
1)
...


4

Pump

νmin =

F

1
...
7 Predict the colours of the following materials, given their band-gap

Thermal
decay

energies (in parentheses): GaAs (1
...
1 eV), and ZnS (3
...

[Black, red, and colourless]

Here we explore the further consequences of light emission in solids, focusing our
attention on ionic crystals and semiconductors used in the design of lasers and lightemitting diodes
...

The neodymium laser is an example of a four-level laser, in which the laser transition terminates in a state other than the ground state of the laser material (Fig
...
62)
...
The population inversion results from pumping a majority of the Nd3+ ions into an excited state
by using an intense flash from another source, followed by a radiationless transition
to another excited state
...
A
neodymium laser operates at a number of wavelengths in the infrared, the band at
1064 nm being most common
...
5) modes of operation
...
The electronic absorption spectrum of Ti3+ ion in sapphire is
very similar to that shown in Fig
...
13, with a broad absorption band centred at
around 500 nm that arises from vibronically allowed d–d transitions of the Ti3+ ion in
an octahedral environment provided by oxygen atoms of the host lattice
...
20
...
Therefore, the titanium sapphire laser is an
example of a vibronic laser, in which the laser transitions originate from vibronic
transitions in the laser medium
...
1),

Fig
...
62 The transitions involved in a
neodymium laser
...


2

E

Pump

(b) Light emission by solid-state lasers and light-emitting diodes

2

T2

Fig
...
63 The transitions involved in a
titanium sapphire laser
...
Monochromatic light from a
pump laser induces a 2E ← 2T2 transition
in a Ti3+ ion that resides in a site with
octahedral symmetry
...
As a result, the titanium sapphire
laser emits radiation over a broad spectrum
that spans from about 700 nm to about
1000 nm
...
7

The refractive index, nr, of the medium,
the ratio of the speed of light in a
vacuum, c, to its speed c′ in the medium:
nr = c/c′
...
See
Appendix 3 for details
...
Mode-locked titanium
sapphire lasers produce energetic (20 mJ to 1 J) and very short (20–100 fs, 1 fs = 10−15 s)
pulses
...

The unique electrical properties of p–n junctions between semiconductors can be
put to good use in optical devices
...
Practical
light-emitting diodes of this kind are widely used in electronic displays
...
Gallium arsenide
itself emits infrared light, but the band gap is widened by incorporating phosphorus,
and a material of composition approximately GaAs0
...
4 emits light in the red region
of the spectrum
...
In diode lasers, light emission due to electron–hole recombination is employed as the basis of laser action
...
One widely used material is Ga1−x Alx As, which
produces infrared laser radiation and is widely used in compact-disc (CD) players
...
One example is the
pumping of Nd:YAG lasers by Ga0
...
09As/Ga0
...
3As diode lasers
...
As a result, it
is now possible to construct a laser system for steady-state or time-resolved spectroscopy entirely out of solid-state components
...
Here we
explore two phenomena that not only can be studied conveniently with intense laser
beams but are commonly used in the laboratory to modify the output of lasers for
specific experiments, such as those described in Section 14
...

In frequency doubling, or second harmonic generation, an intense laser beam is
converted to radiation with twice (and in general a multiple) of its initial frequency as
it passes though a suitable material
...

We can account for frequency doubling by examining how a substance responds
nonlinearly to incident radiation of frequency ω = 2πν
...
At low light
intensity, most materials respond linearly, in the sense that µ = αE, where α is the
polarizability (see Section 18
...
To allow for nonlinear response by some materials at
high light intensity, we can write
1
µ = αE + –βE2 +
...
25)

where the coefficient β is the hyperpolarizability of the material
...
26)

20
...
Common
materials that can be used for frequency doubling in laser systems include crystals
of potassium dihydrogenphosphate (KH2PO4), lithium niobate (LiNbO3), and βbarium borate (β-BaB2O4)
...
Because a beam of light changes direction
when it passes from a region of one refractive index to a region with a different refractive index, changes in refractive index result in the self-focusing of an intense laser
pulse as it travels through the Kerr medium (Fig
...
64)
...
5)
...
The procedure makes use of the fact that the gain, the growth in intensity, of a frequency
component of the radiation in the cavity is very sensitive to amplification and, once a
particular frequency begins to grow, it can quickly dominate
...
A spontaneous fluctuation in intensity—a bunching of photons—
may begin to turn on the optical Kerr effect and the changes in the refractive index of
the Kerr medium will result in a Kerr lens, which is the self-focusing of the laser beam
...
The Kerr lens immediately disappears (if the medium is well chosen),
but is re-created when the intense pulse returns from the mirror at the far end
...
Sapphire is an example of a Kerr medium
that facilitates the mode locking of titanium sapphire lasers, resulting in very short
laser pulses of duration in the femtosecond range
...
Judicious use of nonlinear phenomena leads to more ways in
which the properties of optical signals, and hence the information they carry, can be
manipulated
...
11 Magnetic properties
The magnetic properties of metallic solids and semiconductors depend strongly on
the band structures of the material (see Further reading)
...
Much of the discussion applies to liquid and gas
phase samples as well as to solids
...
For instance, some molecules possess permanent magnetic dipole moments, and an applied
magnetic field can induce a magnetic moment, with the result that the entire solid
sample becomes magnetized
...
The magnetization induced by a field of
strength H is proportional to H, and we write
M = χH

[20
...

An intense laser beam is focused inside a
Kerr medium and passes through a small
aperture in the laser cavity
...


Fig
...
64

734

20 MATERIALS 2: THE SOLID STATE
where χ is the dimensionless volume magnetic susceptibility
...
28]

where Vm is the molar volume of the substance (we shall soon see why it is sensible to
introduce this quantity)
...
29]
−7

−2

−1 2

where µ0 is the vacuum permeability, µ0 = 4π × 10 J C m s
...
This density is increased if M adds to H (when χ > 0), but the density is
decreased if M opposes H (when χ < 0)
...
Those for which χ is negative are called diamagnetic
...
15), so molecules with a permanent
magnetic dipole moment of magnitude m contribute to the magnetization an amount
proportional to m2/3kT
...
An
applied field can also induce a magnetic moment by stirring up currents in the electron distribution like those responsible for the chemical shift in NMR (Section 15
...

The constant of proportionality between the induced moment and the applied field is
called the magnetizability, ξ (xi), and the magnetic analogue of eqn 18
...
30)

We can now see why it is convenient to introduce χm, because the product of the number density N and the molar volume is Avogadro’s constant, NA:
N Vm =

NVm
V

=

nNAVm
nVm

= NA

(20
...
32)

and the density dependence of the susceptibility (which occurs in eqn 20
...
The expression for χm is in agreement with the
empirical Curie law:

χm = A +

C
T

(20
...
As indicated above, and in contrast to electric
moments, this expression applies to solids as well as fluid phases
...
This
instrument consists of a sensitive balance from which the sample hangs in the form of a
narrow cylinder and lies between the poles of a magnet
...

A diamagnetic sample tends to be expelled from the field and appears to weigh less
when the field is turned on
...
11 MAGNETIC PROPERTIES
SQUID

Synoptic table 20
...
06

H2O(l)

−13
...
6

Cu(s)

+176

CuSO4·5H2O(s)

cm /(10−5 cm3 mol−1)

Current

−160
−38

Superconducting
wire

−6
...


known susceptibility
...
20
...
A SQUID takes
advantage of the quantization of magnetic flux and the property of current loops in
superconductors that, as part of the circuit, include a weakly conducting link through
which electrons must tunnel
...

Table 20
...
A typical paramagnetic volume susceptibility is about 10−3, and a typical diamagnetic volume susceptibility is about (−)10−5
...

(b) The permanent magnetic moment

The permanent magnetic moment of a molecule arises from any unpaired electron
spins in the molecule
...
8 that the magnitude of the magnetic
moment of an electron is proportional to the magnitude of the spin angular momentum, {s(s + 1)}1/2$
...
34)

where ge = 2
...
1)
...
It follows that the spin contribution to the molar magnetic susceptibility is

χm =

2
2
NAg e µ0 µBS(S + 1)

3kT

(20
...
The contribution decreases
with increasing temperature because the thermal motion randomizes the spin orientations
...

Illustration 20
...
24 g cm−3, and molar mass 200 g mol−1
...
3001 × 10−6 m3 K−1 mol−1

Current
Magnetic
field
Fig
...
65 The arrangement used to
magnetic susceptibility with a SQUID
...


736

20 MATERIALS 2: THE SOLID STATE
Consequently,

(a)

(b)

χm = 6
...
9 × 10−8 m3 mol−1
...
To obtain the volume magnetic susceptibility,
the molar susceptibility is divided by the molar volume Vm = M/ρ, where ρ is the
mass density
...
7 cm3 mol−1, so χ = 1
...


(c)
Fig
...
66 (a) In a paramagnetic material,
the electron spins are aligned at random in
the absence of an applied magnetic field
...
(c) In an
antiferromagnetic material, the electron
spins are locked into an antiparallel
arrangement
...


At low temperatures, some paramagnetic solids make a phase transition to a state
in which large domains of spins align with parallel orientations
...
20
...

In other cases, the cooperative effect leads to alternating spin orientations: the spins
are locked into a low-magnetization arrangement to give an antiferromagnetic phase
...
The ferromagnetic transition occurs at the Curie temperature,
and the antiferromagnetic transition occurs at the Néel temperature
...
These currents give rise to a magnetic field that usually opposes the applied field, so the substance is diamagnetic
...

The great majority of molecules with no unpaired electron spins are diamagnetic
...
In the few cases in which molecules are paramagnetic despite having no unpaired electrons, the induced electron currents flow in
the opposite direction because they can make use of unoccupied orbitals that lie close
to the HOMO in energy
...

We can summarize these remarks as follows
...
In a few cases (where there are low-lying excited
states) TIP is strong enough to make the molecules paramagnetic even though their
electrons are paired
...
12 Superconductors
The resistance to flow of electrical current of a normal metallic conductor decreases
smoothly with temperature but never vanishes
...
Following the discovery in 1911 that mercury is a superconductor below 4
...
Metals, such as tungsten,
mercury, and lead, tend to have Tc values below about 10 K
...
In 1986, high-temperature superconductors (HTSC) were discovered
...
12 SUPERCONDUCTORS

737

Cu
Ba

O

Y

Structure of the YBa2Cu3O7
superconductor
...

(b) The polyhedra show the positions of
oxygen atoms and indicate that the metal
ions are in square-planar and squarepyramidal coordination environments
...
20
...
For example, HgBa2Ca2Cu2O8 has Tc = 153 K
...
Some superconductors,
classed as Type I, show abrupt loss of superconductivity when an applied magnetic
field exceeds a critical value Hc characteristic of the material
...
36)

where Hc(0) is the value of Hc as T → 0
...
This complete exclusion of a magnetic field in a material is known as the
Meissner effect, which can be visualized by the levitation of a superconductor above
a magnet
...

There is a degree of periodicity in the elements that exhibit superconductivity
...
It is observed that, for simple metals, ferromagnetism and
superconductivity never coexist, but in some of the oxocuprate superconductors
ferromagnetism and superconductivity can coexist
...
20
...
The square-pyramidal CuO5 units arranged as two-dimensional layers and
the square planar CuO4 units arranged in sheets are common structural features of
oxocuprate HTSCs
...
The central concept of low-temperature superconduction is the existence of a
Cooper pair, a pair of electrons that exists on account of the indirect electron–electron
interactions fostered by the nuclei of the atoms in the lattice
...
20
...
Because that local distortion is rich in positive charge, it is
favourable for a second electron to join the first
...
The local distortion can
be easily disrupted by thermal motion of the ions in the solid, so the virtual attraction

-

e

Fig
...
68 The formation of a Cooper pair
...
These electron–lattice
interactions effectively bind the two
electrons into a pair
...
A Cooper pair undergoes less scattering than an
individual electron as it travels through the solid because the distortion caused by one
electron can attract back the other electron should it be scattered out of its path in a
collision
...

The Cooper pairs responsible for low-temperature superconductivity are likely
to be important in HTSCs, but the mechanism for pairing is hotly debated
...
It is believed that movement of electrons
along the linked CuO4 units accounts for superconductivity, whereas the linked CuO5
units act as ‘charge reservoirs’ that maintain an appropriate number of electrons in
the superconducting layers
...
However,
the potential uses of superconducting materials are not limited to the field to chemical instrumentation
...
The appropriate
technology is not yet available, but research in this area of materials science is active
...
Solids are classified as metallic, ionic, covalent, and molecular
...
A space lattice is the pattern formed by points representing the
locations of structural motifs (atoms, molecules, or groups of
atoms, molecules, or ions)
...
20
...

3
...
Unit cells are
classified into seven crystal systems according to their
rotational symmetries
...
Crystal planes are specified by a set of Miller indices (hkl) and
the separation of neighbouring planes in a rectangular lattice
2
is given by 1/d hkl = h2/a2 + k2/b2 + l 2/c 2
...
Bragg’s law relating the glancing angle θ to the separation of
lattice planes is λ = 2d sin θ, where λ is the wavelength of the
radiation
...
The scattering factor is a measure of the ability of an atom to
diffract radiation (eqn 20
...

7
...
7)
...
8)
...
A Patterson synthesis is a map of interatomic vectors
obtained by Fourier analysis of diffraction intensities
(eqn 20
...

9
...


10
...

11
...

12
...
12)
...
The lattice enthalpy is the change in enthalpy (per mole of
formula units) accompanying the complete separation of the
components of the solid
...
15)
...
A covalent network solid is a solid in which covalent bonds in
a definite spatial orientation link the atoms in a network
extending through the crystal
...

15
...

16
...
16a), the bulk modulus
(eqn 20
...
16c), and Poisson’s
ratio (eqn 20
...

17
...
An insulator is a semiconductor with a
very low electrical conductivity
...
According to the band theory, electrons occupy molecular
orbitals formed from the overlap of atomic orbitals: full
bands are called valence bands and empty bands are called
conduction bands
...
23)
...
Semiconductors are classified as p-type or n-type according
to whether conduction is due to holes in the valence band or
electrons in the conduction band
...
The spectroscopic properties of molecular solids can be
understood in terms of the formation and migration of
excitons, electron–hole pairs, from molecule to molecule
...
The spectroscopic properties of metallic conductors and
semiconductors can be understood in terms of the lightinduced promotion of electrons from valence bands to
conduction bands
...
Examples of solid state lasers include the neodymium laser,
the titanium sapphire laser, and diode lasers
...
Nonlinear optical phenomena arise from changes in the
optical properties of a material in the presence of an intense
field from electromagnetic radiation
...

24
...
When χ < 0,
the material is diamagnetic and moves out of a magnetic field
...

25
...

26
...

Antiferromagnetism results from alternating spin orientations
in a material and leads to weak magnetization
...
Temperature-independent paramagnetism arises from
induced electron currents within the orbitals of a molecule
that are occupied in its ground state
...
Superconductors conduct electricity without resistance below
a critical temperature Tc
...

They are also completely diamagnetic below Hc
...


Further reading
Articles and texts

O
...
Plenum, New York (1998)
...
D
...
, Magnetic materials
...
L
...
), 9, 1 VCH, New York (1994)
...
A
...
Oxford University Press (2000)
...
B
...
Chieh, Crystallography
...
L
...
), 4, 385 VCH, New York (1992)
...
D
...
, Materials science and engineering: an introduction
...

P
...
Cox, The electronic structure and chemistry of solids
...


Sources of data and information

D
...
Lide (ed
...

D
...
R
...


Discussion questions
20
...

20
...

20
...
4 Describe the phase problem and explain how it may be overcome
...
5 Describe the structures of elemental metallic solids in terms of the
packing of hard spheres
...
6 Describe the caesium-chloride and rock-salt structures in terms of the
occupation of holes in expanded close-packed lattices
...
7 Explain how X-ray diffraction can be used to determine the absolute

configuration of molecules
...
8 Explain how metallic conductors and semiconductors are identified and
explain their electrical and optical properties in terms of band theory
...
9 Describe the characteristics of the Fermi–Dirac distribution
...
10 To what extent are the electric and magnetic properties of molecules
analogous? How do they differ?

740

20 MATERIALS 2: THE SOLID STATE

Exercises
20
...
What points within a face-centred cubic unit cell are
1
equivalent to the point (– , 0, 0)?
2
20
...
What points within a body-centred cubic unit cell are
1
1
equivalent to the point (– , 0, – )?
2
2
20
...

20
...

20
...

20
...

20
...
3 pm is 20
...
Calculate the wavelength of the X-rays
...
4b The glancing angle of a Bragg reflection from a set of crystal planes
separated by 128
...
76°
...

20
...
5b What are the values of 2θ of the first three diffraction lines of fcc gold

{atomic radius 144 pm) when the X-ray wavelength is 154 pm?
20
...
433 pm and 154
...
Calculate the separation of the diffraction lines
arising from the two components in a powder diffraction pattern recorded in
a circular camera of radius 5
...
8 pm
...
6b A synchrotron source produces X-radiation at a range of wavelengths
...
4°
...
6° is known to be due to the (220) planes
...

20
...
7°, 13
...
7°, and
21
...
The reflection at 17
...
Index the
other reflections
...
11a Potassium nitrate crystals have orthorhombic unit cells of dimensions

a = 542 pm, b = 917 pm, and c = 645 pm
...

20
...
1 pm, b = 796
...
9 pm
...
42 pm (from aluminium)
...
12a Copper(I) chloride forms cubic crystals with four formula units per

unit cell
...
What is the (Bravais) lattice type of
the unit cell?
20
...
Identify the
(Bravais) lattice type of the unit cell
...
13a The coordinates, in units of a, of the atoms in a body-centred cubic

lattice are (0,0,0), (0,1,0), (0,0,1), (0,1,1), (1,0,0), (1,1,0), (1,0,1), and (1,1,1)
...

20
...
Calculate the structure factors Fhkl when all the atoms are identical
...
14a Calculate the packing fraction for close-packed cylinders
...
14b Calculate the packing fraction for equilateral triangular rods stacked
as shown in 2
...
401 and 96
...
Calculate
the separation of the diffraction lines arising from the two components in a
powder diffraction pattern recorded in a circular camera of radius 5
...
3 pm
...
7a The compound Rb3TlF6 has a tetragonal unit cell with dimensions
a = 651 pm and c = 934 pm
...

20
...
9 pm and c = 506
...

20
...
9 g cm−3
...


20
...
414
...
8b An orthorhombic unit cell of a compound of molar mass 135
...
15b Verify that the radius ratios for eightfold coordination is 0
...


has the dimensions a = 589 pm, b = 822 pm, and c = 798 pm
...
9 g cm−3
...

20
...
Calculate the spacing, d, of the (411) planes
...
9b An orthorhombic unit cell has dimensions a = 679 pm, b = 879 pm, and
c = 860 pm
...


2

20
...
3 determine the radius of the smallest

cation that can have (a) sixfold and (b) eightfold coordination with the
O2− ion
...
16b From the data in Table 20
...

20
...


20
...
17b Calculate the atomic packing factor for a side-centred (C) cubic unit

Kα radiation (wavelength 154 pm) at glancing angles 19
...
5°, 32
...


PROBLEMS
20
...
8 pm in hcp
but 142
...

20
...

20
...
Sketch the pattern
that would be obtained for a planar, triangular isolated BF3 molecule
...
19b In a Patterson synthesis, the spots correspond to the lengths and
directions of the vectors joining the atoms in a unit cell
...


741

20
...
41
...
0 dm3 is subjected to a uniaxial stress that
produces a strain of 2
...
25a Is arsenic-doped germanium a p-type or n-type semiconductor?
20
...
26a The promotion of an electron from the valence band into the

conduction band in pure TIO2 by light absorption requires a wavelength of
less than 350 nm
...


20
...
26b The band gap in silicon is 1
...
Calculate the minimum frequency
of electromagnetic radiation that results in promotion of electrons from the
valence to the conduction band
...
20b Calculate the wavelength of neutrons that have reached thermal
equilibrium by collision with a moderator at 300 K
...
27a The magnetic moment of CrCl3 is 3
...
How many unpaired
electrons does the Cr possess?

20
...
15) by calculating the

20
...
3µB
...
13 and
P
P
20
...

20
...
22a Cotton consists of the polymer cellulose, which is a linear chain of

glucose molecules
...
When
a cotton shirt is ironed, it is first moistened, then heated under pressure
...

20
...
These seals failed at 0°C, a temperature well above the crystallization
temperature of the rubber
...

20
...
2 GPa
...
0 kg is suspended from a
polyethylene thread of diameter 1
...
28a Calculate the molar susceptibility of benzene given that its volume

susceptibility is −7
...
879 g cm−3 at 25°C
...
28b Calculate the molar susceptibility of cyclohexane given that its volume
susceptibility is −7
...

20
...

However, experimentally it is found that χm /(m3 mol−1) = (1
...

Determine the number of unpaired spins in O2
...
29b Predict the molar susceptibility of nitrogen dioxide at 298 K
...
30a Data on a single crystal of MnF2 give χm = 0
...
53 K
...

20
...
00 × 10−8 m3 mol−1

at 298 K
...

20
...

20
...


20
...
What
strain will be produced when a mass of 10
...
10 mm?

20
...
24a Poisson’s ratio for polyethylene is 0
...
What change in volume takes

S = 1 in 15
...


place when a cube of polyethylene of volume 1
...
0 per cent?

orientational energy of an S = 1 system to be comparable to kT at 298 K?
20
...
1 In the early days of X-ray crystallography there was an urgent need to
know the wavelengths of X-rays
...
Another method was to estimate the
separation of lattice planes from the measured density of a crystal
...
17 g cm−3 and the (100) reflection using PdKα radiation occurred
at 6
...
Calculate the wavelength of the X-rays
...
2 The element polonium crystallizes in a cubic system
...
225, 0
...
388
from the (100), (110), and (111) sets of planes
...


742

20 MATERIALS 2: THE SOLID STATE

sixth and seventh lines observed in the powder diffraction pattern is larger
than between the fifth and sixth lines
...

20
...
Calculate the theoretical density of fully
crystalline polyethylene
...
92 to 0
...

20
...
8 pm, 627
...
2 pm,
and 658
...
In each case, anion and cation are in contact along
an edge of the unit cell
...
4 The powder diffraction patterns of (a) tungsten, (b) copper obtained in a

Fh +6
...
2 −4
...
2

camera of radius 28
...
20
...
Both were obtained with
154 pm X-rays and the scales are marked
...
Estimate the metallic radii of W and Cu
...
20
...
5 Elemental silver reflects X-rays of wavelength 154
...
076°, 22
...
256°
...
Assuming a cubic unit cell, determine its type and
dimension
...

20
...
The nucleus of a cultured
pearl is a piece of mother-of-pearl that has been worked into a sphere on a
lathe
...

Suggest an X-ray method for distinguishing between real and cultured pearls
...
7 In their book X-rays and crystal structures (which begins ‘It is now two

years since Dr
...
’) the Braggs give a number of
simple examples of X-ray analysis
...
If the side of the NaCl unit cell is 564 pm, what is
the side of the KCl unit cell? The densities of KCl and NaCl are 1
...
17 g cm−3 respectively
...
8 Calculate the coefficient of thermal expansion of diamond given that the

(111) reflection shifts from 22° 2′ 25″ to 21° 57′ 59″ on heating a crystal from
100 K to 300 K and 154
...

20
...
45 pm
...
516 g cm−3
...
1 +2
...
13 The scattering of electrons or neutrons from a pair of nuclei separated
by a distance Rij and orientated at a definite angle to the incident beam can be
calculated
...
0 +8
...
5 +4
...
5 −2
...
4 +3
...
0 +1
...
The electron scattering factor, f, is a measure of the intensity of the
electron scattering powers of the atoms
...
0 pm
...
0 keV electron
diffraction pattern of CCl4 with an (as yet) undertermined C-Cl bond length
but of known tetrahedral symmetry
...
Plot I/f 2 against positions of the maxima, which
occurred at 3° 0′, 5° 22′, and 7° 54′, and minima, which occurred at 1° 46′,
4° 6′, 6° 40′, and 9° 10′
...
14‡ B
...
Bovenzi and G
...
Pearse, Jr
...
Chem
...
Dalton Trans
...
The compound, which
they isolated from the reaction of the ligand with CuSO4 (aq), did not
contain a [Cu(C7H9N5O2)2]2+ complex cation as expected
...

The unit cell was primitive monoclinic with a = 1
...
8876 nm,
c = 1
...
254°
...
024 g cm−3
...
15‡ D
...
W
...
Donaubauer, and F
...
Heinemann

(Inorg
...
36, 1397 (1997)) describe the synthesis and reactivity of the
ruthenium nitrido compound [N(C4H9)4][Ru(N)(S2C6H4)2]
...
Compute the mass
density of the compound given that it crystallizes into an orthorhombic unit
cell with a = 3
...
9402 nm, and c = 1
...
Replacing the ruthenium with an osmium results in a
compound with the same crystal structure and a unit cell with a volume less
than 1 per cent larger
...


20
...
Naphthalene has

a monoclinic unit cell with two molecules per cell and sides in the ratio
1
...
436
...
152 g cm−3
...

20
...

Fully crystalline polyethylene has its chains aligned in an orthorhombic unit
cell of dimensions 740 pm × 493 pm × 253 pm
...
16 Aided by the Born–Mayer equation for the lattice enthalpy and a
Born–Haber cycle, show that formation of CaCl is an exothermic process
(the sublimation enthalpy of Ca(s) is 176 kJ mol−1)
...

20
...
It follows from the exponential form of the Fermi–Dirac distribution
that the conductance G, the inverse of the resistance (with units of siemens,
1 S = 1 Ω−1), of an intrinsic semiconductor should have an Arrhenius-like
temperature dependence, shown in practice to have the form G = G0e−Eg /2kT,
where Eg is the band gap
...
Estimate the value of Eg
...
0847

0
...
86

20
...
We
begin by considering a dimer, with each monomer having a single transition
with transition dipole moment µmon and wavenumber #mon
...
Now we write the hamiltonian matrix with diagonal elements
set to the energy between the excited and ground state of the monomer
(which, expressed as a wavenumber, is simply #mon), and off-diagonal
elements correspond to the energy of interaction between the transition
dipoles
...
1, we write this
interaction energy (as a wavenumber) as:

β=

2
µ mon

4πε0hcr 3

(1 − 3 cos2θ)

β D
E
#mon F

V
#mon
V
0

20
...
G
...
Marezio, M
...
de Brion, J
...
Tholence, Q
...
Santoro (Science 265, 380 (1994)) report the synthesis and
structure of a material that becomes superconducting at temperatures below
45 K
...
38606 nm and c = 2
...
The compound is made superconducting
by partially replacing Y by Ca, accompanied by a change in unit cell volume by
less than 1 per cent
...
55 given that the mass density of the compound is 7
...


20
...
3
...
23 Show that the volume of a triclinic unit cell of sides a, b, and c and

The eigenvalues of the matrix are the dimer transition wavenumbers #1 and
#2
...
(a) The intensity of absorption of incident
C ck F
radiation is proportional to the square of the transition dipole moment
(Section 9
...
The monomer transition dipole moment is µmon = ∫ψ * Nψ0dτ
1
= ∫ψ * Nψ0dτ , where ψ0 is the wavefunction of the monomer ground state
...
(b) Consider a dimer of monomers with µmon = 4
...
5 nm
...
How does this ratio vary
with the angle θ ? (c) Now expand the treatment given above to a chain of N
monomers (N = 5, 10, 15, and 20), with µmon = 4
...
5 nm
...
For example the
hamiltonian matrix for the case N = 4 is
⎛ #mon
⎜ V
@= ⎜
0
⎜ 0
⎝

20
...
19 K and Hc = 63 901 A m−1
...
For example, the
lowest-energy state of the five-ring compound C22H14 (4) is computed to have
S = 3, but the energies of S = 2 and S = 4 structures are each predicted to be
50 kJ mol−1 higher in energy
...
Estimate the molar susceptibility at
298 K if each level is present in proportion to its Boltzmann factor (effectively
assuming that the degeneracy is the same for all three of these levels)
...
19‡ J
...
Dannenberg, D
...
Halvick, and J
...
Rayez (J
...
Chem
...
The calculations

angles α, β, and γ is
V = abc(1 − cos2α − cos2 β − cos2γ + 2 cos α cos β cos γ )1/2
Use this expression to derive expressions for monoclinic and orthorhombic
unit cells
...

20
...

20
...
By calculating the phase of the I reflection in the
2 2
2 2 2
2
4
structure factor, show that the I atoms contribute no net intensity to the (114)
reflection
...
26 The coordinates, in units of a, of the A atoms, with scattering factor fA,

in a cubic lattice are (0,0,0), (0,1,0), (0,0,1), (0,1,1), (1,0,0), (1,1,0), (1,0,1),
1 1 1
and (1,1,1)
...

2 2 2
Calculate the structure factors Fhkl and predict the form of the powder
1
diffraction pattern when (a) fA = f, fB = 0, (b) fB = – fA, and (c) fA = fB = f
...
27 For an isotropic substance, the moduli and Poisson’s ratio may be
expressed in terms of two parameters λ and µ called the Lamé constants:

E=

µ (3λ + 2µ)
λ+µ

K=

3λ + 2µ
3

G=µ

νP =

λ
3(λ + µ)

Use the Lamé constants to confirm the relations between G, K, and E given in
eqn 20
...

20
...
(a) Use eqn 20
...
9
...

That is, show that the density of states increases towards the edges of the bands
in a one-dimensional metallic conductor
...
29 The treatment in Problem 20
...
In three dimensions, the variation of density of states is more like that
shown in Fig
...
70
...


20
...
Derive an expression
in terms of the equilibrium constant, K, for the dimerization to show how the
molar susceptibility varies with the pressure of the sample
...

20
...

It also has an unpaired electron, and so may be expected to be paramagnetic
...
The first excited state (at 121 cm−1) is paramagnetic
because the orbital magnetic moment adds to, rather than cancels, the spin
magnetic moment
...
Because
the upper state is thermally accessible, the paramagnetic susceptibility of NO
shows a pronounced temperature dependence even near room temperature
...


Energy

Applications: to biochemistry and nanoscience

p-band

s-band
Density of states, r
Fig
...
70
20
...
This
nonlinear optical phenomenon is known as frequency mixing and is used to
expand the wavelength range of lasers in laboratory applications, such as
spectroscopy and photochemistry
...
31 The magnetizability, ξ, and the volume and molar magnetic
susceptibilities can be calculated from the wavefunctions of molecules
...

Calculate ξ and χm for the ground state of a hydrogenic atom
...
34 Although the crystallization of large biological molecules may not be as
readily accomplished as that of small molecules, their crystal lattices are no
different
...
3 nm and a density of 1
...
Determine its molar
mass
...
35 What features in an X-ray diffraction pattern suggest a helical

conformation for a biological macromolecule? Use Fig
...
26 to deduce as
much quantitative information as you can about the shape and size of a DNA
molecule
...
36 A transistor is a semiconducting device that is commonly used either as
a switch or an amplifier of electrical signals
...
A useful starting point is the work summarized by Tans et al
...

20
...
The movement of atoms and ions depends on their ability to
leave one position and stick to another, and therefore on the energy changes
that occur
...
Calculate, by direct summation, its Coulombic
interaction when it is in an empty lattice point directly above an anion
...
We prepare the
ground for a discussion of the rates of reactions by considering the motion of
molecules in gases and in liquids
...
Characteristic physical and chemical
events take place at surfaces, including catalysis, and we see how to describe
them
...


21 Molecules in motion
22 The rates of chemical reactions
23 The kinetics of complex reactions
24 Molecular reaction dynamics
25 Processes at solid surfaces

This page intentionally left blank

Molecules in motion

One of the simplest types of molecular motion to describe is the random motion of
molecules of a perfect gas
...
Molecular mobility is
particularly important in liquids
...
Molecular and ionic motion have common features and, by considering them from a more general viewpoint, we derive expressions that govern the migration of properties through matter
...

Finally, we build a simple model for all types of molecular motion, in which the molecules
migrate in a series of small steps, and see that it accounts for many of the properties of
migrating molecules in both gases and condensed phases
...
1

The kinetic model of gases

I21
...
2

Collision with walls and
surfaces

21
...
4

Transport properties of a
perfect gas

Molecular motion in liquids
21
...
We set the scene by considering a simple type of motion, that of molecules in a perfect gas, and go on to see that
molecular motion in liquids shows a number of similarities
...
Four examples of transport properties are

Experimental results

21
...
7

The mobilities of ions

21
...
2 Impact on biochemistry: Ion

channels and ion pumps

Diffusion, the migration of matter down a concentration gradient
...

Electric conduction, the migration of electric charge along an electrical potential
gradient
...

It is convenient to include in the discussion effusion, the emergence of a gas from a
container through a small hole
...
9

The thermodynamic view

21
...
3 Impact on biochemistry:

Transport of non-electrolytes
across biological membranes
21
...
12 The statistical view

Molecular motion in gases
Here we present the kinetic model of a perfect gas as a starting point for the discussion
of its transport properties
...
The
kinetic model is one of the most remarkable—and arguably most beautiful—models
in physical chemistry for, from a set of very slender assumptions, powerful quantitative
conclusions can be deduced
...
1: The
transport characteristics of a
perfect gas
Discussion questions
Exercises
Problems

748

21 MOLECULES IN MOTION
21
...

2 The size of the molecules is negligible, in the sense that their diameters are much
smaller than the average distance travelled between collisions
...

An elastic collision is a collision in which the total translational kinetic energy of the
molecules is conserved
...
2]

Justification 21
...
21
...
In
an elastic collision of a molecule with a wall
perpendicular to the x-axis, the xcomponent of velocity is reversed but the
y- and z-components are unchanged
...
1)°

Area A
Volume = |vx Dt |A

Fig
...
2 A molecule will reach the wall on
the right within an interval ∆t if it is within
a distance vx ∆t of the wall and travelling to
the right
...
21
...
When a particle of mass m that is travelling
with a component of velocity vx parallel to the x-axis collides with the wall on the
right and is reflected, its linear momentum (the product of its mass and its velocity)
changes from mvx before the collision to −mvx after the collision (when it is travelling
in the opposite direction)
...
Many molecules
collide with the wall in an interval ∆t, and the total change of momentum is the
product of the change in momentum of each molecule multiplied by the number of
molecules that reach the wall during the interval
...
21
...
It follows that, if the wall has area
A, then all the particles in a volume A × vx ∆t will reach the wall (if they are travelling
towards it)
...

At any instant, half the particles are moving to the right and half are moving to
the left
...
The total momentum change in that interval is the product of
2
this number and the change 2mvx :
Momentum change =

nNAAvx ∆t
2V

× 2mvx =

2
nmANAv x ∆t

V

=

2
nMAv x ∆t

V

where M = mNA
...
1 THE KINETIC MODEL OF GASES
This rate of change of momentum is equal to the force (by Newton’s second law of
motion)
...
͘) of the quantity just calculated:
p=

2
nM͗v x ͘

V

This expression already resembles the perfect gas equation of state
...
Because the
2 1/2
root-mean-square speed, c, is defined as c = ͗v ͘ (eqn 21
...
It follows that c 2 = 3͗vx͘
...
1 follows immediately by substituting
1 2
2
2
͗v x ͘ = – c into p = nM͗vx ͘/V
...
1 is one of the key results of the kinetic model
...
2)
...
1 to be the
equation of state of a perfect gas, its right-hand side must be equal to nRT
...
3)°

We can conclude that the root mean square speed of the molecules of a gas is proportional
to the square root of the temperature and inversely proportional to the square root of the
molar mass
...
Sound waves are pressure waves, and for them to propagate the
molecules of the gas must move to form regions of high and low pressure
...
The root mean square speed of N2 molecules, for
instance, is found from eqn 21
...

Equation 21
...
However,
in an actual gas the speeds of individual molecules span a wide range, and the collisions in the gas continually redistribute the speeds among the molecules
...
The fraction of
molecules that have speeds in the range v to v + dv is proportional to the width of the
range, and is written f(v)dv, where f(v) is called the distribution of speeds
...
C
...
4)

749

21 MOLECULES IN MOTION

Relative numbers of molecules, f (v)

750

Low temperature or
high molecular mass
Intermediate
temperature
or molecular
mass
High
temperature or
low molecular
mass
Speed, v

Fig
...
3 The distribution of molecular
speeds with temperature and molar mass
...


Exploration (a) Plot different

distributions by keeping the molar
mass constant at 100 g mol−1 and varying
the temperature of the sample between
200 K and 2000 K
...
(c) Based on your observations,
provide a molecular interpretation of
temperature
...
Let’s consider its features, which are also shown pictorially in
Fig
...
3:
1 Equation 21
...
Its
presence implies that the fraction of molecules with very high speeds will be very small
2
because e−x becomes very small when x 2 is large
...

That is, heavy molecules are unlikely to be found with very high speeds
...
In other words, a greater fraction of the molecules can be expected to have
high speeds at high temperatures than at low temperatures
...
This factor goes to
zero as v goes to zero, so the fraction of molecules with very low speeds will also be
very small
...
4 and the 4π) simply
ensure that, when we add together the fractions over the entire range of speeds from
zero to infinity, then we get 1
...
4 to calculate the fraction of molecules in a given narrow range of
speeds, ∆v, we evaluate f(v) at the speed of interest, then multiply it by the width of the
range of speeds of interest; that is, we form f(v)∆v
...
5)

v1

This integral is the area under the graph of f as a function of v and, except in special
cases, has to be evaluated numerically by using mathematical software (Fig
...
4)
...
2 The Maxwell distribution of speeds

The Boltzmann distribution is a key result of physical chemistry and is treated fully
in Section 16
...
It implies that the fraction of molecules with velocity components
vx , vy , vz is proportional to an exponential function of their kinetic energy, which is
1
2 1
2 1
2
E = – mv x + – mv y + – mv z
2
2
2

Therefore, we can use the relation ax+y+z+ · · · = a xa ya z · · · to write
1

1

1

2
2
2
f = Ke−E /kT = Ke− (––mv x + ––mv y + ––mv z)/kT = Ke−mv x /2kTe−mv y /2kTe−mv z /2kT
2

2

2

2

2

2

where K is a constant of proportionality (at constant temperature) and fdvxdvydvz is
the fraction of molecules in the velocity range vx to vx + dvx, vy to vy + dvy, and vz to
vz + dvz
...

To determine the constant K, we note that a molecule must have a velocity somewhere in the range −∞ < vx < ∞, so

Ύ

∞

f(vx)dvx = 1
−∞

21
...
At this stage we know that
A M D
E
f(vx) = B
C 2πRT F

1/2

e−Mvx /2RT
2

(21
...
The probability f(v)dv that the molecules have a speed in the
range v to v + dv regardless of direction is the sum of the probabilities that the
velocity lies in any of the volume elements dvxdvydvz forming a spherical shell of
radius v and thickness dv (Fig
...
5)
...
Therefore,

A M D
E
f(v) = 4π B
C 2πRT F

Relative number of molecules

Substitution of the expression for f(vx) then gives

v1

Speed

v2

Fig
...
4 To calculate the probability that a
molecule will have a speed in the range v1
to v2, we integrate the distribution between
those two limits; the integral is equal to the
area of the curve between the limits, as
shown shaded here
...
4
...
1 Calculating the mean speed of molecules in a gas

v

What is the mean speed, K, of N2 molecules in air at 25°C?
Method We need to use the results of probability theory summarized in Appendix
2
...
A mean speed is calculated by multiplying each speed by the fraction of
molecules that have that speed, and then adding all the products together
...
To
employ this approach here, we note that the fraction of molecules with a speed in
the range v to v + dv is f (v)dv, so the product of this fraction and the speed is
vf (v)dv
...
4
...
21
...


752

21 MOLECULES IN MOTION
where we have used the standard result from tables of integrals (or software) that
∞

Ύ xe

3 −ax 2

dx =

0

1
2a2

Substitution of the data then gives
K=
1/e

A 8 × (8
...
02 × 10−3 kg mol−1)

1/2

= 475 m s−1

where we have used 1 J = 1 kg m2 s−2
...
1 Evaluate the root mean square speed of the molecules by integra-

c* = (2RT /M )

f (v)/4p(M/2pRT )1/2

1/2

c = (8RT /pM )1/2
c = (3RT /M )1/2

tion
...
1, we can use the Maxwell distribution to evaluate the
mean speed, K, of the molecules in a gas:
K=
0

1/2

1

v/(2RT /M )

(4/p)

1/2

(3/2)1/2

A summary of the conclusions that
can be deduced from the Maxwell
distribution for molecules of molar mass M
at a temperature T: c* is the most probable
speed, K is the mean speed, and c is the root
mean square speed
...
21
...
7)

We can identify the most probable speed, c*, from the location of the peak of the distribution:
c* =

A 2RT D
C M F

1/2

(21
...
6 summarizes these results
...
1

To find the location of the peak of the
distribution, differentiate f with respect
to V and look for the value of V at which
the derivative is zero (other than at 1 = 0
and 1 = ∞)
...
9)

This result is much harder to derive, but the diagram in Fig
...
7 should help to show
that it is plausible
...
10)

Note that the molecular masses (not the molar masses) and Boltzmann’s constant,
k = R/NA, appear in this expression; the quantity µ is called the reduced mass of the
molecules
...
10 turns into eqn 21
...

2
The Maxwell distribution has been verified experimentally
...
21
...
The spinning
cylinder has channels that permit the passage of only those molecules moving through
them at the appropriate speed, and the number of molecules can be determined by
collecting them at a detector
...
2
...
In particular,

21
...
21
...
When the molecules are
moving in the same direction, the mean
relative speed is zero; it is 2v when the
molecules are approaching each other
...
The last direction of approach is
the most characteristic, so the mean speed
of approach can be expected to be about
21/2v
...


Fig
...
8 A velocity selector
...
Only if the speed of a molecule is
such as to carry it along the channel that
rotates into its path will it reach the
detector
...


Synoptic table 21
...

We count a ‘hit’ whenever the centres of two molecules come within a distance d of
each other, where d, the collision diameter, is of the order of the actual diameters of
the molecules (for impenetrable hard spheres d is the diameter)
...
10
...
Some typical collision cross-sections are given in Table 21
...
6)
...
52

He

0
...
43

* More values are given in the Data section
...
11a)°
2

z=

0
...
11b)°

Justification 21
...
Then we note
what happens as one mobile molecule travels through the gas with a mean relative
speed Krel for a time ∆t
...
21
...
The number of stationary molecules with centres inside the collision tube is given by the

d
d
Area, s

Hit

Fig
...
9 In an interval ∆t, a molecule of
diameter d sweeps out a tube of radius d
and length Krel ∆t
...
In reality, the tube is not
straight, but changes direction at each
collision
...


754

21 MOLECULES IN MOTION
volume of the tube multiplied by the number density N = N/V, and is NσKrel∆t
...
The expression in terms of the pressure of the gas is obtained by using the perfect gas equation to write
N=

N
V

=

nNA
V

=

pNA
RT

=

p
kT

Equation 21
...
The reason for this increase is that the relative mean
speed increases with temperature (eqns 21
...
10)
...
11b shows that,
at constant temperature, the collision frequency is proportional to the pressure
...
For an N2 molecule in a
sample at 1 atm and 25°C, z ≈ 5 × 109 s−1, so a given molecule collides about 5 × 109
times each second
...

(c) The mean free path

Once we have the collision frequency, we can calculate the mean free path, λ
(lambda), the average distance a molecule travels between collisions
...
It follows that the mean free path is

λ=

K

(21
...
11b gives

λ=

kT
21/2σp

(21
...
A typical mean free path in
nitrogen gas at 1 atm is 70 nm, or about 103 molecular diameters
...
13, in a sample of constant volume, the pressure is proportional to T, so T/p remains constant when the temperature is increased
...
The distance between collisions is determined by the number of
molecules present in the given volume, not by the speed at which they travel
...
Each molecule
makes a collision within about 1 ns, and between collisions it travels about 103 molecular diameters
...

IMPACT ON ASTROPHYSICS

I21
...
It may seem absurd, therefore, to expect the kinetic model
and, as a consequence, the perfect gas law, to be applicable to the dense matter of
stellar interiors
...
50 times that of
liquid water and comparable to that of water about half way to its surface
...
2 COLLISIONS WITH WALLS AND SURFACES
we have to realize that the state of matter is that of a plasma, in which the electrons
have been stripped from the atoms of hydrogen and helium that make up the bulk of
the matter of stars
...
Therefore, a mean free path of only
0
...
We can therefore use pV = nRT as the equation of state for the stellar interior
...
Atoms are stripped of their electrons in the interior
of stars so, if we suppose that the interior consists of ionized hydrogen atoms, the
mean molar mass is one-half the molar mass of hydrogen, or 0
...
Half way to the centre of the
Sun, the temperature is 3
...
20 g cm−3 (slightly denser
than water); so the pressure there works out as 7
...

We can combine this result with the expression for the pressure from the kinetic
1
model (eqn 21
...
Because the total kinetic energy of the particles is E K = – Nmc 2, we
2
2
can write p = – EK /V
...
It follows that the kinetic energy density half way to
3
the centre of the Sun is about 0
...
In contrast, on a warm day (25°C) on
Earth, the (translational) kinetic energy density of our atmosphere is only 0
...

21
...
The collision flux, Z W, is the number of collisions with the area in a given time
interval divided by the area and the duration of the interval
...
We show in the Justification below that the collision flux is
ZW =

p

(21
...
00 bar) and T = 300 K, ZW ≈ 3 × 1023 cm−2 s−1
...
4 The collision flux

Consider a wall of area A perpendicular to the x-axis (as in Fig
...
2)
...
Therefore, all
molecules in the volume Avx ∆t, and with positive x-component of velocities, will
strike the wall in the interval ∆t
...

However, to take account of the presence of a range of velocities in the sample, we
must sum the result over all the positive values of vx weighted by the probability distribution of velocities (eqn 21
...
6,
∞

Ύ

vx f(vx)dvx =

0

A m D
C 2πkT F

1/2

Ύ

∞

vxe−mv x /2kTdvx =
2

0

A kT D
C 2πm F

1/2

where we have used the standard integral
∞

Ύ xe

−ax 2

dx =

0

1
2a

Therefore,
ZW = N

A kT D
C 2πm F

1/2
1
= – KN
4

(21
...
7 in the form K = (8kT/πm)1/2, which implies that – K =
4
1/2
(kT/2πm)
...
14
...
3 The rate of effusion
The essential empirical observations on effusion are summarized by Graham’s law of
effusion, which states that the rate of effusion is inversely proportional to the square
root of the molar mass
...
However, by using the
expression for the rate of collisions, we can obtain a more detailed expression for the
rate of effusion and hence use effusion data more effectively
...
14)
...
16)°

where, in the last step, we have used R = NAk and M = mNA
...

Equation 21
...
Thus, if the vapour pressure of a sample is p, and it is enclosed in a cavity
with a small hole, then the rate of loss of mass from the container is proportional to p
...
2 Calculating the vapour pressure from a mass loss

Caesium (m
...
29°C, b
...
686°C) was introduced into a container and heated to
500°C
...
50 mm was opened in the container for 100 s, a
mass loss of 385 mg was measured
...

Method The pressure of vapour is constant inside the container despite the effusion

of atoms because the hot liquid metal replenishes the vapour
...
16
...

Answer The mass loss ∆m in an interval ∆t is related to the collision flux by

∆m = Z W A0m∆t

21
...
It follows that
ZW =

∆m
A0m∆t

Because ZW is related to the pressure by eqn 21
...
9 g mol−1, substitution of the data gives p = 11 kPa (using 1 Pa =
1 N m−2 = 1 J m−1), or 83 Torr
...
2 How long would it take 1
...
4 Transport properties of a perfect gas
Transport properties are commonly expressed in terms of a number of ‘phenomenological’ equations, or equations that are empirical summaries of experimental observations
...

In the following sections, we introduce the equations for the general case and then
show how to calculate the parameters that appear in them
...
If matter is flowing (as in diffusion), we speak of a matter flux
of so many molecules per square metre per second; if the property is energy (as in
thermal conduction), then we speak of the energy flux and express it in joules per
square metre per second, and so on
...
For
example, the flux of matter diffusing parallel to the z-axis of a container is found to
be proportional to the first derivative of the concentration:
J(matter) ∝

dN
dz

(21
...

The SI units of J are number per metre squared per second (m−2 s−1)
...
There is no net flux if the concentration is uniform (dN /dz
= 0)
...
18)

The SI units of this flux are joules per metre squared per second (J m−2 s−1)
...
Because matter flows down a concentration gradient, from
high concentration to low concentration, J is positive if dN /dz is negative (Fig
...
10)
...
21
...
Fick’s first law
states that the flux of matter (the number of
particles passing through an imaginary
window in a given interval divided by the
area of the window and the duration of the
interval) is proportional to the density
gradient at that point
...
18 must be negative, and we
write it −D:

Synoptic table 21
...
0163

210

223

CO2

0
...
1442

187

196

N2

0
...

† 1 µP = 10−7 kg m−1 s−1
...
21
...
In this
illustration the fluid is undergoing laminar
flow, and particles bring their initial
momentum when they enter a new layer
...


dN

(21
...
Energy migrates down a temperature gradient, and the same reasoning leads to
J(energy) = −κ

dT

(21
...
The SI units of κ are joules per
kelvin per metre per second (J K−1 m−1 s−1)
...
2
...
21
...
The layer next to the wall of the vessel
is stationary, and the velocity of successive layers varies linearly with distance, z,
from the wall
...
A layer is
retarded by molecules arriving from a more slowly moving layer because they have a
low momentum in the x-direction
...
We interpret the net retarding effect as the fluid’s viscosity
...
The flux of the x-component of momentum is proportional to dvx /dz because there is no net flux when all the layers move at the same
velocity
...
21)

dz

The constant of proportionality, η, is the coefficient of viscosity (or simply ‘the
viscosity’)
...
Viscosities are
often reported in poise (P), where 1 P = 10−1 kg m−1 s−1
...
2
...
1 and summarized in Table 21
...

Table 21
...
4 TRANSPORT PROPERTIES OF A PERFECT GAS
The diffusion coefficient is
1
D = – λK
3

(21
...
13), so D
decreases with increasing pressure and, as a result, the gas molecules diffuse more slowly
...
7), so D also increases
with temperature
...

3 Because the mean free path increases when the collision cross-section of the
molecules decreases (eqn 21
...

Similarly, according to the kinetic model of gases, the thermal conductivity of a
perfect gas A having molar concentration [A] is given by the expression
1
κ = – λKCV,m[A]
3

(21
...
To interpret this expression, we note that:
1 Because λ is inversely proportional to the pressure, and hence inversely proportional to the molar concentration of the gas, the thermal conductivity is independent
of the pressure
...

The physical reason for the pressure independence of κ is that the thermal conductivity can be expected to be large when many molecules are available to transport the
energy, but the presence of so many molecules limits their mean free path and they
cannot carry the energy over a great distance
...
The thermal
conductivity is indeed found experimentally to be independent of the pressure, except
when the pressure is very low, when κ ∝ p
...
The flux is still
proportional to the number of carriers, but the length of the journey no longer
depends on λ, so κ ∝ [A], which implies that κ ∝ p
...
1):
1
η = – MλK[A]
3

(21
...

We can interpret this expression as follows:
1 Because λ ∝ 1/p (eqn 21
...

That is, the viscosity is independent of the pressure
2 Because K ∝ T 1/2 (eqn 21
...
That is, the viscosity of a gas increases with
temperature
...
The increase of
viscosity with temperature is explained when we remember that at high temperatures
the molecules travel more quickly, so the flux of momentum is greater
...
6, the viscosity of a liquid decreases with increase in temperature because intermolecular interactions must be overcome
...
One technique
depends on the rate of damping of the torsional oscillations of a disc hanging in the
gas
...
The other method is based
on Poiseuille’s formula for the rate of flow of a fluid through a tube of radius r:

105 h/(kg m-1 s-1)

3

2

dV

1

dt
(a) 300 K
0

-3 -2 -1 0 1
log (p/atm)

2

105 h/(kg m-1 s-1)

8

6

4

2

=

2
2
(p1 − p2)πr4

(21
...

Such measurements confirm that the viscosities of gases are independent of pressure over a wide range
...
21
...
01 atm to 20 atm
...
The blue line
in the illustration shows the calculated values using σ = 22 × 10−20 m2, implying a
collision diameter of 260 pm, in contrast to the van der Waals diameter of 335 pm
obtained from the density of the solid
...


Illustration 21
...
21
...

The blue line in the latter is the calculated
value
...


In a Poiseuille flow experiment to measure the viscosity of air at 298 K, the sample
was allowed to flow through a tube of length 100 cm and internal diameter
1
...
The high-pressure end was at 765 Torr and the low-pressure end was at
760 Torr
...
In 100 s, 90
...
The viscosity of air at 298 K is found by reorganizing
the Poiseuille formula, eqn 21
...
3 Pa):

η=

{(765 × 133
...
3 Pa)2} × π × (5
...
00 × 10−1 m) × (760 × 133
...
02 × 10−5 m3 D
C
F
100 s

= 1
...
The kinetic model expression gives η = 1
...
Viscosities are commonly
expressed in centipoise (cP) or (for gases) micropoise (µP), the conversion being
1 cP = 10−3 kg m−1 s−1; the viscosity of air at 20°C is about 180 µP
...
3 What volume would be collected if the pressure gradient were

doubled, other conditions remaining constant?

[181 cm3]

21
...
4* Viscosities of
liquids at 298 K

Molecular motion in liquids
We outlined what is currently known about the structure of simple liquids in Section 17
...
Here we consider a particularly simple type of motion through a liquid, that
of an ion, and see that the information that motion provides can be used to infer the
behaviour of uncharged species too
...
601

Mercury

1
...
26)

(Note the positive sign of the exponent
...
Such a variation is found experimentally, at least over reasonably small temperature ranges (Fig
...
13)
...

One problem with the interpretation of viscosity measurements is that the change
in density of the liquid as it is heated makes a pronounced contribution to the temperature variation of the viscosity
...
The intermolecular interactions between the molecules of the liquid govern the
magnitude of Ea, but the problem of calculating it is immensely difficult and still
largely unsolved
...
This behaviour is consistent with the rupture of hydrogen bonds
...
6 The conductivities of electrolyte solutions
Further insight into the nature of molecular motion can be obtained by studying the
motion of ions in solution, for ions can be dragged through the solvent by the application of a potential difference between two electrodes immersed in the sample
...

(a) Conductance and conductivity

The fundamental measurement used to study the motion of ions is that of the electrical
resistance, R, of the solution
...
891

* More values are given in the Data section
...
891 cP
...
0

1
...
Relaxation time measurements in NMR and EPR (Chapter 15) can be interpreted in terms of the mobilities of the molecules, and have been used to show that
big molecules in viscous fluids typically rotate in a series of small (about 5°) steps,
whereas small molecules in nonviscous fluids typically jump through about 1 radian
(57°) in each step
...
The same technique is used to examine
the internal dynamics of macromolecules
...
4)
...
The probability that a molecule has at least an energy Ea is
proportional to e−Ea /RT, so the mobility of the molecules in the liquid should follow
this type of temperature dependence
...
224

Water†

21
...
2

0
...
4

0

0

20

40 60
q/°C
q

80 100

Fig
...
13 The experimental temperature
dependence of the viscosity of water
...
A plot of ln η
against 1/T is a straight line (over a small
range) with positive slope
...
As resistance is expressed in ohms, Ω, the conductance of a
sample is expressed in Ω−1
...
The conductance of a sample decreases with its length l and increases with its cross-sectional area
A
...
27)

where κ is the conductivity
...

The conductivity of a solution depends on the number of ions present, and it is normal to introduce the molar conductivity, Λm, which is defined as

Λm =
390
...
8
KCl

L m /(S cm2 mol-1)
L

120

80

40
CH3COOH
0

0

20

40 60 80
-3
c/(mmol dm )

100

Fig
...
14 The concentration dependence of
the molar conductivities of (a) a typical
strong electrolyte (aqueous potassium
chloride) and (b) a typical weak electrolyte
(aqueous acetic acid)
...
28]

where c is the molar concentration of the added electrolyte
...

The molar conductivity is found to vary with the concentration
...
For instance, the concentration of ions in a solution
of a weak acid depends on the concentration of the acid in a complicated way, and
doubling the concentration of the acid added does not double the number of ions
...

The concentration dependence of molar conductivities indicates that there are two
classes of electrolyte
...
21
...
The characteristic of a weak
electrolyte is that its molar conductivity is normal at concentrations close to zero, but
falls sharply to low values as the concentration increases
...

(b) Strong electrolytes

Strong electrolytes are substances that are virtually fully ionized in solution, and
include ionic solids and strong acids
...

In an extensive series of measurements during the nineteenth century, Friedrich
Kohlrausch showed that at low concentrations the molar conductivities of strong
electrolytes vary linearly with the square root of the concentration:

Λm = Λ m − K c 1/2
°

(21
...
The constant Λm is the limiting molar con°
ductivity, the molar conductivity in the limit of zero concentration (when the ions
are effectively infinitely far apart and do not interact with one another)
...
) than on its specific identity
...


21
...
5* Limiting ionic conductivities in water at 298 K
l/(mS m2 mol−1)
H+

34
...
01

Cl

K+

7
...
56

19
...
63
7
...
00

* More values are given in the Data section
...
If the limiting molar conductivity
of the cations is denoted λ+ and that of the anions λ−, then his law of the independent
migration of ions states that

Λm = ν+λ+ + ν−λ−
°

(21
...
This
simple result, which can be understood on the grounds that the ions migrate independently in the limit of zero concentration, lets us predict the limiting molar conductivity of any strong electrolyte from the data in Table 21
...

Illustration 21
...
72 + 2 × 7
...
98 mS m2 mol−1
°

(c) Weak electrolytes

Weak electrolytes are not fully ionized in solution
...
The marked concentration dependence of
their molar conductivities arises from the displacement of the equilibrium
HA(aq) + H2O(l) 5 H3O+(aq) + A−(aq) Ka =

aH3O+aA−
aHA

(21
...

The conductivity depends on the number of ions in the solution, and therefore on
the degree of ionization, α, of the electrolyte; when referring to weak acids, we speak
instead of the degree of deprotonation
...
32)

If we ignore activity coefficients, the acidity constant, Ka, is approximately
Ka =

α 2c

(21
...
34)°

Comment 21
...
In eqn 21
...
The equilibrium
constant for the reaction between
HA(aq) and H2O(l) is the acidity
constant Ka of HA
...


764

21 MOLECULES IN MOTION
The acid is fully deprotonated at infinite dilution, and its molar conductivity is then
Λm
...
35)°

with α given by eqn 21
...

Illustration 21
...
0100 m CH3COOH(aq) at 298 K is Λm = 1
...
The degree of deprotonation, α, is calculated from eqn 21
...
05 mS cm2 mol−1 (Table 21
...
It follows that α = 0
...
The acidity constant,
Ka, is calculated by substitution of α into eqn 21
...
9 × 10−5
...
4 The molar conductivity of 0
...
61 mS m2

mol−1
...


1/L m
L

Once we know Ka, we can use eqns 21
...
35 to predict the concentration
dependence of the molar conductivity
...
21
...
More usefully, we can use the concentration dependence of Λm in
measurements of the limiting molar conductance
...
33 into
1

1/L °
m

α
L
cLm

=1+

αc

(21
...
35, we obtain Ostwald’s dilution law:
1

The graph used to determine the
limiting value of the molar conductivity of
a solution by extrapolation to zero
concentration
...
21
...
44]

Λm

=

1

Λm
°

+

Λ mc
°
Ka(Λ m)2

(21
...
21
...

°
21
...
The
central idea in this section is that, although the motion of an ion remains largely
random, the presence of an electric field biases its motion, and the ion undergoes net
migration through the solution
...
38)

l

In such a field, an ion of charge ze experiences a force of magnitude
F = zeE =

ze∆φ
l

(21
...
7 THE MOBILITIES OF IONS
Synoptic table 21
...
23
+

u/(10−8 m2 s−1 V−1)
OH−

20
...
19

Cl−

7
...
62

Br−

8
...
47

2−
SO4

8
...


(In this chapter we disregard the sign of the charge number and so avoid notational
complications
...
However, this acceleration is short-lived
...

If we assume that the Stokes formula (eqn 19
...
40)

The two forces act in opposite directions, and the ions quickly reach a terminal
speed, the drift speed, when the accelerating force is balanced by the viscous drag
...
41)

f

It follows that the drift speed of an ion is proportional to the strength of the applied
field
...
42]

where u is called the mobility of the ion (Table 21
...
Comparison of eqns 21
...
42 and use of eqn 21
...
43)

Illustration 21
...
For the viscosity, we use η = 1
...
0 × 10−3 kg m−1 s−1, Table 21
...

Then u ≈ 5 × 10−8 m2 V −1 s−1
...
That speed might seem slow, but not when expressed on
a molecular scale, for it corresponds to an ion passing about 104 solvent molecules
per second
...
21
...
Agmon (Chem
...
Letts
...
Proton transfer between
neighbouring molecules occurs when one
molecule rotates into such a position that
an O-H···O hydrogen bond can flip into
being an O···H-O hydrogen bond
...


Comment 21
...


+

+
(b)

(a)

+
(c)

Because the drift speed governs the rate at which charge is transported, we might
expect the conductivity to decrease with increasing solution viscosity and ion size
...
For example, the molar conductivities of the alkali metal ions increase
from Li+ to Cs+ (Table 21
...
The paradox is
resolved when we realize that the radius a in the Stokes formula is the hydrodynamic
radius (or ‘Stokes radius’) of the ion, its effective radius in the solution taking into
account all the H2O molecules it carries in its hydration sphere
...
Thus, an ion of small
ionic radius may have a large hydrodynamic radius because it drags many solvent
molecules through the solution as it migrates
...

The proton, although it is very small, has a very high molar conductivity (Table
21
...
5 ps, which is comparable to the time that
inelastic neutron scattering shows it takes a water molecule to reorientate through
about 1 rad (1 to 2 ps)
...
However, the actual mechanism is still highly contentious
...
This cluster of atoms is itself hydrated,
but the hydrogen bonds in the secondary sphere are weaker than in the primary
sphere
...
21
...
After this bond cleavage has taken place, and the released molecule has rotated through a few degrees (a
process that takes about 1 ps), there is a rapid adjustment of bond lengths and angles
+
in the remaining cluster, to form an H5O2 cation of structure H2O ···H+ ···OH2 (Fig
...
16b)
...
21
...
According to this model, there is no coordinated
motion of a proton along a chain of molecules, simply a very rapid hopping between
neighbouring sites, with a low activation energy
...
The mobility of NH4 is
also anomalous and presumably occurs by an analogous mechanism
...
As a
first step we establish in the Justification below the following relation between an ion’s
mobility and its molar conductivity:

λ = zuF
where F is Faraday’s constant (F = NAe)
...
44)°

21
...
5 The relation between ionic mobility and molar conductivity

To keep the calculation simple, we ignore signs in the following, and concentrate on
the magnitudes of quantities: the direction of ion flux can always be decided by
common sense
...
Let each formula unit give rise to ν+ cations of charge z+e and ν− anions of
charge z−e
...
The number of ions of one kind
that pass through an imaginary window of area A during an interval ∆t is equal to
the number within the distance s∆t (Fig
...
17), and therefore to the number in the
volume s∆tA
...
1 in the discussion
of the pressure of a gas
...
The flux through the window (the number of this type of ion passing
through the window divided by the area of the window and the duration of the
interval) is therefore
J(ions) =

s∆tAνcNA
A∆t

= sνcNA

Each ion carries a charge ze, so the flux of charge is
J(charge) = zsνceNA = zsνcF
Because s = uE, the flux is
J(charge) = zuνcFE
The current, I, through the window due to the ions we are considering is the charge
flux times the area:
I = JA = zuνcFEA
Because the electric field is the potential gradient, ∆φ/l, we can write
I=

zuνcFA∆φ

(21
...
27 in the form κ = Gl/A
...
4
...
Division by the molar concentration of ions, νc, then
results in eqn 21
...


Equation 21
...
Therefore, for the solution
itself in the limit of zero concentration (when there are no interionic interactions),

Λ m = (z+u+ν+ + z−u−ν−)F
°

(21
...
46b)°

Area, A
Cations

-

+
s- Dt
s+ Dt
Fig
...
17 In the calculation of the current,
all the cations within a distance s+ ∆t (that
is, those in the volume s+ A∆t) will pass
through the area A
...


768

21 MOLECULES IN MOTION
Illustration 21
...
The
experimental value for KCl, for instance, is 15 mS m2 mol−1
...
For a solution of two kinds of ion, the transport numbers of the
cations (t+) and anions (t−) are
t± =

I±

[21
...
Because the total current is the sum of the cation and anion currents, it follows that
t+ + t− = 1

The limiting transport number, t° , is defined in the same way but for the limit of zero
±
concentration of the electrolyte solution
...

The current that can be ascribed to each type of ion is related to the mobility of the
ion by eqn 21
...
Hence the relation between t° and u± is
±

M+XC

D

Time t

t° =
±
c

B

Time 0

Area, A
+

-

NX

In the moving boundary method
for the measurement of transport numbers
the distance moved by the boundary is
observed as a current is passed
...
One
procedure is to add bromothymol blue
indicator to a slightly alkaline solution of
the ion of interest and to use a cadmium
electrode at the lower end of the vertical
tube
...

Fig
...
18

z±ν±u±
z+ν+u+ + z−ν−u−

(21
...
49a simplifies to

l
A

(21
...
49b)°

u+ + u−

Moreover, because the ionic conductivities are related to the mobilities by eqn 21
...
49b that
t° =
±

ν±λ ±
ν+λ+ + ν−λ−

=

ν±λ ±
Λm
°

(21
...
51)°

Consequently, because there are independent ways of measuring transport numbers
of ions, we can determine the individual ionic conductivities and (through eqn 21
...

There are several ways to measure transport numbers (see Further reading)
...

Let MX be the salt of interest and NX a salt giving a denser solution
...
21
...
The MX solution, which is called the leading

21
...
There is a sharp boundary between the
two solutions
...
Thus, if any M ions
diffuse into the lower solution, they will be pulled upwards more rapidly than the N
ions around them, and the boundary will reform
...
That number is clANA, so the charge that the
M ions transfer through the plane is z+clAeNA
...
Therefore, the fraction due to the
motion of the M ions, which is their transport number, is
t+ =

z+clAF

(21
...

21
...
29)
...
9 we saw something similar: the activity coefficients of ions
at low concentrations also depend on c1/2 and depend on their charge type rather than
their specific identities
...

To accommodate the effect of motion, we need to modify the picture of an ionic
atmosphere as a spherical haze of charge
...
21
...
The
overall effect is the displacement of the centre of charge of the atmosphere a short
distance behind the moving ion
...
This reduction of the ions’ mobility is called the
relaxation effect
...

The ionic atmosphere has another effect on the motion of the ions
...
When the ionic atmosphere is present
this drag is enhanced because the ionic atmosphere moves in an opposite direction to
the central ion
...

The quantitative formulation of these effects is far from simple, but the Debye–
Hückel–Onsager theory is an attempt to obtain quantitative expressions at about the
same level of sophistication as the Debye–Hückel theory itself
...
53a)

with
A=

z 2eF 2 A 2 D
3πη C ε RT F

1/2

B=

qz 3eF A 2 D
24πε RT C ε RT F

1/2

(21
...
3) and q = 0
...
7)
...
21
...
The
attraction between the opposite charges
retards the motion of the central ion
...
7* Debye–Hückel–Onsager coefficients for (1,1)-electrolytes at 298 K

2
-1
(L m - L° )/(S cm mol )
m

-40

A/(mS m2 mol−1/(mol dm−3)1/2)

B/(mol dm−3)−1/2

15
...
923

Propanone

BaCl2

Solvent
Methanol

AgNO3

32
...
63
0
...


-80

-120

6
...
21
...
The agreement
is quite good at very low ionic strengths, corresponding to very low molar concentrations (less than about 10−3 M, depending on the charge type)
...
1

0
...
3

0
...
21
...


IMPACT ON BIOCHEMISTRY

I21
...
2)
...

Suppose that a membrane provides a barrier that slows down the transfer of
molecules or ions into or out of the cell
...
2 that the thermodynamic tendency to transport an ion through the membrane is partially determined
by a concentration gradient (more precisely, an activity gradient) across the membrane,
which results in a difference in molar Gibbs energy between the inside and the outside
of the cell, and a transmembrane potential gradient, which is due to the different potential energy of the ions on each side of the bilayer
...
It is also possible to move a species against these gradients, but now the flow
must be driven by an exergonic process, such as the hydrolysis of ATP
...

The transport of ions into or out of a cell needs to be mediated (that is, facilitated
by other species) because the hydrophobic environment of the membrane is inhospitable to ions
...
An example of a channel former is the polypeptide gramicidin A, which increases the membrane permeability to cations such as
H+, K+, and Na+
...
They are highly selective, so there is a channel protein for
Ca2+, another for Cl−, and so on
...

Ions such as H+, Na+, K+, and Ca2+ are often transported actively across membranes
by integral proteins called ion pumps
...
Because protein phosphorylation
requires dephosphorylation of ATP, the conformational change that opens or closes
the pump is endergonic and requires the use of energy stored during metabolism
...
2 IMPACT ON BIOCHEMISTRY: ION CHANNELS AND ION PUMPS
Let’s consider some of the experimental approaches used in the study of ion channels
...
Information about
the flow of ions across channels and pumps is supplied by the patch clamp technique
...
21
...
With mild
suction, a ‘patch’ of membrane from a whole cell or a small section of a broken cell can
be attached tightly to the tip of a micropipet filled with an electrolyte solution and
containing an electronic conductor, the so-called patch electrode
...
If the membrane is permeable to ions
at the applied potential difference, a current flows through the completed circuit
...

A detailed picture of the mechanism of action of ion channels has emerged from
analysis of patch clamp data and structural data
...
21
...
The pore through which ions move has a length of 3
...
2 nm and diameter of 1
...
2 nm and diameter of 0
...
The narrow region is
called the selectivity filter of the K+ ion channel because it allows only K+ ions to pass
...
Upon entering the selectivity
filter, the K+ ion is stripped of its hydrating shell and is then gripped by carbonyl
groups of the protein
...
The
Na+ ion, though smaller than the K+ ion, does not pass through the selectivity filter of
the K+ ion channel because interactions with the protein are not sufficient to compensate for the high Gibbs energy of dehydration of Na+ (∆dehydG 7 = +301 kJ mol−1)
...
In its hydrated form, the Na+ ion is too large (larger than a dehydrated K+ ion),
does not fit in the selectivity filter, and does not cross the membrane
...
For
example, K+ and Tl+ ions have similar radii and Gibbs energies of dehydration, so Tl+
can cross the membrane
...

The efficiency of transfer of K+ ions through the channel can also be explained by
structural features of the protein
...

An ion is lured into the channel by water molecules about halfway through the length
of the membrane
...
The
ion is ‘encouraged’ to leave the protein by electrostatic interactions in the selectivity
filter, which can bind two K+ ions simultaneously, usually with a bridging water
molecule
...

Now we turn our attention to a very important ion pump, the H+-ATPase responsible for coupling of proton flow to synthesis of ATP from ADP and Pi (Impact I7
...

Structural studies show that the channel through which the protons flow is linked in

771

Power supply
and current
measuring device
Patch
electrode

Micropipette
Intracellular
electrode
Cytosol

Ion channel

Cell

Fig
...
21 A representation of the patch
clamp technique for the measurement of
ionic currents through membranes in
intact cells
...

An intracellular electronic conductor is
inserted into the cytosol of the cell and the
two conductors are connected to a power
supply and current measuring device
...
2 nm

1
...
3 nm

1
...
21
...
The bulk of the
protein is shown in beige
...
2 nm and diameter of 1
...
2 nm and diameter of 0
...

The selectivity filter has a number of
carbonyl groups (shown in dark green)
that grip K+ ions
...


tandem to a unit composed of six protein molecules arranged in pairs of α and β subunits to form three interlocked αβ segments (Fig
...
23)
...
A protein at the centre of the interlocked structure, the γ subunit shown
as a white arrow, rotates and induces structural changes that cycle each of the three
segments between L, T, and O conformations
...
Then ADP and a Pi group migrate into the L site and, as it closes into
T, the earlier T site opens into O and releases its ATP
...

The proton flux drives the rotation of the γ subunit, and hence the conformational
changes of the α /β segments, as well as providing the energy for the condensation
reaction itself
...
For example, the rotation of the γ subunit has been observed directly by using
single-molecule spectroscopy (Section 14
...


Diffusion
We are now in a position to extend the discussion of ionic motion to cover the migration of neutral molecules and of ions in the absence of an applied electric field
...

21
...
In a system in which the chemical potential depends on the position x,
dw = dµ =

ATP

Pi
O L L ADP
T

ATP
Pi
L
ADP L T ATP ATP T O
L
O
ATP

ATP
ADP Pi

The mechanism of action of H+ATPase, a molecular motor that transports
protons across the mitochondrial
membrane and catalyses either the
formation or hydrolysis of ATP
...
21
...
54)

We also saw in Chapter 2 (Table 2
...
55)

By comparing these two expressions, we see that the slope of the chemical potential
can be interpreted as an effective force per mole of molecules
...
56]

There is not necessarily a real force pushing the particles down the slope of the chemical potential
...

(a) The thermodynamic force of a concentration gradient

In a solution in which the activity of the solute is a, the chemical potential is

µ = µ 7 + RT ln a

21
...
57)

If the solution is ideal, a may be replaced by the molar concentration c, and then
F=−

RT A ∂c D
c C ∂x F p,T

(21
...

Example 21
...
Calculate the thermodynamic force on the solute at 25°C given that the
concentration falls to half its value in 10 cm
...
58, the thermodynamic force is calculated by differ-

entiating the concentration with respect to distance
...

Answer The concentration varies with position as

c = c 0 e− x/λ
where λ is the decay constant
...
58 then implies that
F=

RT

λ

1
We know that the concentration falls to – c0 at x = 10 cm, so we can find λ from
2
−(10 cm)/λ
=e

...
It follows that

1
–
2

F = (8
...
0 × 10 −1 m) = 17 kN mol −1
where we have used 1 J = 1 N m
...
5 Calculate the thermodynamic force on the molecules of molar mass

M in a vertical tube in a gravitational field on the surface of the Earth, and evaluate
F for molecules of molar mass 100 g mol−1
...

[F = −Mg, −0
...
]

(b) Fick’s first law of diffusion

In Section 21
...
We shall now show that it can be deduced more generally and that it applies to
the diffusion of species in condensed phases too
...
The particles reach a steady drift
speed, s, when the thermodynamic force, F, is matched by the viscous drag
...
However, the
particle flux, J, is proportional to the drift speed, and the thermodynamic force is proportional to the concentration gradient, dc/dx
...

(c) The Einstein relation

If we divide both sides of eqn 21
...
59)

dx

In this expression, D is the diffusion coefficient and dc/dx is the slope of the molar
concentration
...
60)

This relation follows from the argument that we have used several times before
...
Hence, the amount of substance that can pass
through the window in that interval is s∆tAc
...
58, we find
D dc

s=−

c dx

=

DF
RT

(21
...

There is one case where we already know the drift speed and the effective force acting on a particle: an ion in solution has a drift speed s = uE when it experiences a force
ezE from an electric field of strength E (so F = NAezE = zFE)
...
61 gives
uE =

zFED
RT

and hence
u=

zFD
RT

(21
...
63)°

On inserting the typical value u = 5 × 10−8 m2 s−1 V−1, we find D ≈ 1 × 10−9 m2 s−1 at
25°C as a typical value of the diffusion coefficient of an ion in water
...
9 THE THERMODYNAMIC VIEW

775

(d) The Nernst–Einstein equation

The Einstein relation provides a link between the molar conductivity of an electrolyte
and the diffusion coefficients of its ions
...
44 and 21
...
64)°

RT

for each type of ion
...
65)°

which is the Nernst–Einstein equation
...

(e) The Stokes–Einstein equation

Equations 21
...
63 relate the mobility of an ion to the frictional force
and to the diffusion coefficient, respectively
...
66)

f

If the frictional force is described by Stokes’s law, then we also obtain a relation
between the diffusion coefficient and the viscosity of the medium:
D=

kT

(21
...
66 (and of its special case, eqn 21
...
Therefore, the equation also applies
in the limit of vanishingly small charge, that is, it also applies to neutral molecules
...
8)
...

Example 21
...

Method The starting point is the mobility of the ion, which is given in Table 21
...


The diffusion coefficient can then be determined from the Einstein relation, eqn
21
...
The ionic conductivity is related to the mobility by eqn 21
...
To estimate
the hydrodynamic radius, a, of the ion, use the Stokes–Einstein relation to find f
and the Stokes law to relate f to a
...
6, the mobility of SO 4 is 8
...
It follows

from eqn 21
...
8* Diffusion
coefficients at 298 K

= 1
...
31

I2 in hexane

4
...
33

Sucrose in water

0
...


776

21 MOLECULES IN MOTION
From eqn 21
...
891 cP (or 8
...
4):
a=

kT
6πηD

= 220 pm

2−
The bond length in SO 4 is 144 pm, so the radius calculated here is plausible and
consistent with a small degree of solvation
...
6 Repeat the calculation for the NH 4 ion
...
96 × 10−9 m2 s−1, 7
...
In particular, Walden’s rule is the empirical observation that the
product ηΛm is very approximately constant for the same ions in different solvents
(but there are numerous exceptions)
...
The usefulness
of the rule, however, is muddied by the role of solvation: different solvents solvate the
same ions to different extents, so both the hydrodynamic radius and the viscosity
change with the solvent
...
10 The diffusion equation

Volume Al
Area A

J(x + l )A

We now turn to the discussion of time-dependent diffusion processes, where we are
interested in the spreading of inhomogeneities with time
...
When the
source of heat is maintained and the bar can radiate, it settles down into a steady state
of nonuniform temperature
...
We shall focus
on the description of the diffusion of particles, but similar arguments apply to the
diffusion of physical properties, such as temperature
...

The central equation of this section is the diffusion equation, also called ‘Fick’s
second law of diffusion’, which relates the rate of change of concentration at a point
to the spatial variation of the concentration at that point:
∂c
∂t

x

x+l

J(x)A
Fig
...
24 The net flux in a region is the
difference between the flux entering from
the region of high concentration (on the
left) and the flux leaving to the region of
low concentration (on the right)
...
68)

We show in the following Justification that the diffusion equation follows from Fick’s
first law of diffusion
...
6 The diffusion equation

Consider a thin slab of cross-sectional area A that extends from x to x + l (Fig
...
24)
...
The amount (number of moles) of particles that enter the slab in the infinitesimal interval dt is JAdt, so the rate of increase
in molar concentration inside the slab (which has volume Al) on account of the flux
from the left is

21
...
The flux through that window is J′, and the rate of change of concentration that results is
∂c
∂t

=−

J′Adt
Aldt

=−

J′
l

The net rate of change of concentration is therefore
∂c
∂t

=

J − J′
l

Each flux is proportional to the concentration gradient at the window
...
68
...
If the concentration changes sharply from point to point (if
the distribution is highly wrinkled) then the concentration changes rapidly with
time
...
21
...
Where the curvature is negative (a heap), the change
in concentration is negative; the heap tends to spread
...
If the concentration decreases linearly with distance, then the concentration at any point is constant because the inflow of particles
is exactly balanced by the outflow
...
More succinctly: Nature abhors a wrinkle
...
If for the moment we ignore diffusion, then the flux of particles through an
area A in an interval ∆t when the fluid is flowing at a velocity v can be calculated in the
way we have used several times before (by counting the particles within a distance
v∆t), and is
J=

cA∆vt
A∆t

= cv

(21
...
The rate of change of concentration in a slab of
thickness l and area A is, by the same argument as before and assuming that the velocity does not depend on the position,
∂c J − J′ 1 G A ∂c D J 5 v
∂c
=
= 2c − Hc +
lK 6 = −v
(21
...
21
...
The
diffusion equation tells us that peaks in
distribution (regions of negative curvature)
spread and troughs (regions of positive
curvature) fill in
...
71)

∂x

A further refinement, which is important in chemistry, is the possibility that the
concentrations of particles may change as a result of reaction
...
71 (Section 24
...

(b) Solutions of the diffusion equation

The diffusion equation, eqn 21
...
Therefore, we
must specify two boundary conditions for the spatial dependence and a single initial
condition for the time-dependence
...
The single initial condition is that at t = 0 all N0 particles are concentrated
on the yz-plane (of area A) at x = 0
...
These requirements imply that the flux of particles is zero at the top and bottom surfaces of the system
...
5

c(x,t) =

0
...
5

1
0
...
0

0
...
72)

as may be verified by direct substitution
...
26 shows the shape of the concentration distribution at various times, and it is clear that the concentration spreads and
tends to uniformity
...
The concentration of diffused solute is spherically symmetrical and at a radius r is

0
...
5

n0

1
...
73)

Other chemically (and physically) interesting arrangements, such as transport of substances across biological membranes can be treated (Impact I21
...
In many cases the
solutions are more cumbersome
...
21
...
21
...
73, which
describes diffusion in three dimensions
...
In the capillary technique, a capillary tube, open at one end
and containing a solution, is immersed in a well stirred larger quantity of solvent,
and the change of concentration in the tube is monitored
...
In the
diaphragm technique, the diffusion occurs through the capillary pores of a sintered
glass diaphragm separating the well-stirred solution and solvent
...
Diffusion coefficients may also be measured by the dynamic
light scattering technique described in Section 19
...


The concentration profiles above
a plane from which a solute is diffusing
...
72 and are
labelled with different values of Dt
...
For
example, if x is in metres, Dt would be in
metres2; so, for D = 10−9 m2 s−1, Dt = 0
...


I21
...
3 Transport of non-electrolytes across biological membranes

We saw in Impact I21
...
Here
we use the diffusion equation to explore the way in which non-electrolytes cross the
lipid bilayer
...
To simplify the problem, we will assume that the concentration of A is
always maintained at [A] = [A]0 on one surface of the membrane and at [A] = 0 on
the other surface, perhaps by a perfect balance between the rate of the process that
produces A on one side and the rate of another process that consumes A completely
on the other side
...
7
...
74)

where D is the diffusion coefficient and the steady-state assumption makes partial
derivatives unnecessary
...
74 and the result, which may be verified by differentiation, is

A
C

[A](x) = [A]0 1 −

xD

(21
...
We now use Fick’s
first law to calculate the flux J of A through the membrane and the result is
J=D

[A]0

(21
...
This difference
arises from the significant difference in the solubility of A in an aqueous environment
and in the solution–membrane interface
...
77]

[A]s

where [A]s is the concentration of A in the bulk aqueous solution
...
78)

l

In spite of the assumptions that led to its final form, eqn 21
...

In many cases the flux is underestimated by eqn 21
...
However, the permeability increases
only for certain species and not others and this is evidence that transport can be mediated by carriers
...

A characteristic of a carrier C is that it binds to the transported species A and the
dissociation of the AC complex is described by
AC 5 A + C

K=

[A][C]
[AC]

(21
...
After writing [C]0 = [C] +
[AC], where [C]0 is the total concentration of carrier, it follows that

1

Flux, J/Jmax

[AC] =

[A][C]0

(21
...
80 and 21
...
5

J=

0
0

6
8
2
4
Concentration, [A]/K

10

Fig
...
27 The flux of the species AC
through a membrane varies with the
concentration of the species A
...


DKD[C]0

[A]

l

[A] + K

= Jmax

[A]

(21
...
We
see from Fig
...
27 that when [A] < K the flux varies linearly with [A] and that the
<
flux reaches a maximum value of Jmax = DKD[C]0 /l when [A] > K
...

21
...
We can also use them to calculate the net distance through which the particles diffuse in a given time
...
5 Calculating the net distance of diffusion

Calculate the net distance travelled on average by particles in a time t if they have a
diffusion constant D
...
In this case, we calculate the probability that a particle will be found at a certain
distance from the origin, and then calculate the average by weighting each distance
by that probability
...
The probability that any of the N0 = n0 NA
particles is in the slab is therefore cANAdx/N0
...
Therefore, the mean distance travelled by all
the particles is the sum of each x weighted by the probability of its occurrence:
∞

͗x͘ =

Ύ

0

xcANA
N0

dx =

1
(πDt)1/2

∞

Ύ xe

A Dt D
dx = 2
C πF

1/2

−x 2/4Dt

0

where we have used the same standard integral as that used in Justification 21
...

The average distance of diffusion varies as the square root of the lapsed time
...
7 Derive an expression for the root mean square distance travelled by
diffusing particles in a time t
...
12 THE STATISTICAL VIEW
As shown in Example 21
...
83)

The latter is a valuable measure of the spread of particles when they can diffuse in both
directions from the origin (for then ͗x͘ = 0 at all times)
...
21
...
The graph shows
that diffusion is a very slow process (which is why solutions are stirred, to encourage
mixing by convection)
...
12 The statistical view
An intuitive picture of diffusion is of the particles moving in a series of small steps and
gradually migrating from their original positions
...
The total distance travelled by a particle in a time t is therefore tλ /τ
...
The direction of each step may be
different, and the net distance travelled must take the changing directions into account
...
The same model was
used in the discussion of a one-dimensional random coil in Section 19
...

We show in the Justification below that the probability of a particle being at a distance x from the origin after a time t is
1/2

A 2τ D −x 2τ /2tλ2
P=
e
C πt F

(21
...
7 The one-dimensional random walk

Consider a one-dimensional random walk in which each step is through a distance
λ to the left or right
...

We write n = NR − NL and the total number of steps as N = NR + NL
...
82)

=

N!
1
1
{– (N + n)}!{– (N − n)}!2N
2
2

log ( x 2 1/2/m)

͗x͘ = 2

0

781

-4
-6

1 mm

-8
1 nm
-10
-10

1 ms 1 ms 1 s 1 h1 d1 y
-6

-2 0 2
log (t /s)

6

Fig
...
28 The root mean square distance
covered by particles with D = 5 ×
10−10 m2 s−1
...


782

21 MOLECULES IN MOTION
The use of Stirling’s approximation (Section 16
...
33)
A 2 D
ln P = ln B
E
C πN F

1/2

A
nD 1
A
nD
1
− – (N + n + 1)ln B 1 + E − – (N − n + 1)ln B 1 − E
2
2
NF
NF
C
C

1
For small net distances (n < N ) we can use the approximation ln(1 ± x) ≈ ±x − – x 2,
<
2
and so obtain

A 2 D
ln P ≈ ln B
E
C πN F

1/2

−

n2
2N

At this point, we note that the number of steps taken in a time t is N = t/τ and the net
distance travelled from the origin is x = nλ
...
84
...
72 and 21
...

Moreover, they can be found only at discrete points separated by λ instead of being
anywhere on a continuous line
...

We can now relate the coefficient D to the step length λ and the rate at which the
jumps occur
...
72 and 21
...
85)

Illustration 21
...
1 × 10−9 m2 s−1 and a = 210 pm
(as deduced from mobility measurements), it follows from λ = 2a that τ = 80 ps
...


The Einstein–Smoluchowski equation is the central connection between the microscopic details of particle motion and the macroscopic parameters relating to diffusion
(for example, the diffusion coefficient and, through the Stokes–Einstein relation, the
viscosity)
...
For if we
interpret λ /τ as K, the mean speed of the molecules, and interpret λ as a mean free
path, then we can recognize in the Einstein–Smoluchowski equation exactly the same
expression as we obtained from the kinetic model of gases, eqn 21
...
That is, the
diffusion of a perfect gas is a random walk with an average step size equal to the mean
free path
...
Diffusion is the migration of matter down a concentration
gradient; thermal conduction is the migration of energy down
a temperature gradient; electric conduction is the migration of
electric charge along an electrical potential gradient; viscosity
is the migration of linear momentum down a velocity
gradient
...
The kinetic model of a gas considers only the contribution
to the energy from the kinetic energies of the molecules
...

3
...

4
...

5
...

6
...

7
...
Graham’s law of effusion states that the rate of
effusion is inversely proportional to the square root of the
molar mass
...
Flux J is the quantity of a property passing through a given
area in a given time interval divided by the area and the
duration of the interval
...
Fick’s first law of diffusion states that the flux of matter
is proportional to the concentration gradient, J(matter)
= −DdN /dz, where D is the diffusion coefficient
...
The conductance, G, is the inverse of resistance
...

11
...
A weak electrolyte

is an electrolyte with a molar conductivity that is normal at
concentrations close to zero, but falls sharply to low values as
the concentration increases
...
Kohlrausch’s law for the concentration dependence of the
molar conductivity of a strong electrolyte is written as Λm =
Λ m − K c1/2, where the limiting molar conductivity, Λ m, is the
°
°
molar conductivity at zero concentration (Λ m = ν+λ+ + ν−λ−)
...
The drift speed s is the terminal speed when an accelerating
force is balanced by the viscous drag: s = uE, where u =
ze/6πηa is the ionic mobility and a is the hydrodynamic radius
(Stokes radius), the effective radius of a particle in solution
...
The ionic conductivity is the contribution of ions of one type
to the molar conductivity: λ = zuF
...
The transport number is the fraction of total current I carried
by the ions of a specified type: t± = I±/I
...
The Debye–Hückel–Onsager theory explains the
concentration dependence of the molar conductivity of a
strong electrolyte in terms of ionic interactions
...
The Einstein relation between the diffusion coefficient and the
ionic mobility is D = uRT/zF
...
The Nernst–Einstein relation between the molar conductivity
of an electrolyte and the diffusion coefficients of its ions is
2
Λm = (ν+z + D+ + ν −z 2 D−)(F 2/RT)
...
The Stokes–Einstein equation relates the diffusion coefficient
to the frictional force: D = kT/f
...
Walden’s rule states that the product ηΛm is very
approximately constant for the same ions in different solvents
...
The diffusion equation is a relation between the rate of change
of concentration at a point and the spatial variation of the
concentration at that point: ∂c/∂t = D∂2c/∂x 2
...
In a one-dimensional random walk, the probability P
that a molecule moves a distance x from the origin for a
period t by taking small steps with size λ and time τ is: P =
2
2
(2τ /π t)1/2e−x τ /2tλ
...
The Einstein–Smoluchowski equation relates the diffusion
coefficient to the parameters used in the formulation of the
random walk model, D = λ2/2τ
...
G
...
In
Encyclopedia of applied physics (ed
...
L
...

A
...
Bard and L
...
Faulkner, Electrochemical methods: fundamentals
and applications
...


R
...
Bird, W
...
Stewart, and E
...
Lightfoot, Transport phenomena
...

J
...
Murrell and A
...
Jenkins, Properties of liquids and solutions
...

K
...
van Holde, W
...
Johnson, and P
...
Ho, Principles of physical
biochemistry
...

A
...
Walton, Three phases of matter
...


784

21 MOLECULES IN MOTION

Sources of data and information

D
...
Lide (ed
...
CRC Press, Boca Raton (2000)
...
This material is on
CD-ROM
...
1 The transport characteristics of a
perfect gas

In this Further information section, we derive expressions for the
diffusion characteristics (specifically, the diffusion coefficient, the
thermal conductivity, and the viscosity) of a perfect gas on the basis
of the kinetic molecular theory
...
21
...
On average, the
molecules passing through the area A at z = 0 have travelled about
one mean free path λ since their last collision
...
This
number density is approximately
A dN D
E
N (−λ ) = N (0) − λ B
C dz F 0

(21
...

The average number of impacts on the imaginary window of area
1
A0 during an interval ∆t is ZW A0 ∆t, with Z W = –N K (eqn 21
...

4
Therefore, the flux from left to right, J(L→R), arising from the
supply of molecules on the left, is
J(L → R) =

1
–A 0N (−λ)K∆t
4

A 0 ∆t

N

1
= – N (−λ)K
4

(21
...
On average, the
molecules making the journey have originated from z = +λ where the
number density is N (λ)
...
88)

The average number density at z = +λ is approximately
A dN D
E
N (λ ) = N (0) + λ B
C dz F 0

(21
...
90)

A dN D
1
E
= − –Kλ B
2
C dz F 0
This equation shows that the flux is proportional to the first
derivative of the concentration, in agreement with Fick’s law
...
19 and 21
...
It must be remembered, however, that the calculation is
2
quite crude, and is little more than an assessment of the order of
magnitude of D
...
21
...


Fig
...
29

Long flight
(collides
in flight)
Fig
...
30 One approximation ignored in the simple treatment is that
some particles might make a long flight to the plane even though they
are only a short perpendicular distance away, and therefore they have
a higher chance of colliding during their journey
...
Because the path is long, the molecule is
likely to collide before reaching the window, so it ought to be added
to the graveyard of other molecules that have collided
...
The
3
modification results in eqn 21
...


Fast
layer

x
Slow
layer

Thermal conductivity

According to the equipartition theorem (Section 17
...
For monatomic particles, ν = –
...

We suppose that the number density is uniform but that the
temperature is not
...
Molecules also arrive from the
right after travelling a mean free path from a cooler region
...
91)

A dT D
E
B
C dz F 0

(21
...
93)

The energy flux is proportional to the temperature gradient, as we wanted
to show
...
20 shows that
1
κ = –νkλ KN
3

Fig
...
31 The calculation of the viscosity of a gas examines the net xcomponent of momentum brought to a plane from faster and slower
layers on average a mean free path away in each direction
...
95a)

Those arriving from the left bring a momentum

2
As before, we multiply by – to take long flight paths into account, and
3
so arrive at
1
Jz = − –ν kλ KN
3

-l

z

those travelling from the left transport mvx(−λ) to it
...
Those arriving
4
from the right on average carry a momentum

and the net flux is
A dT D
1
E
Jz = J(L → R) + J(L ← R) = − –νkλ KN B
2
C dz F 0

l
0

(21
...
23 then follows from CV,m = νkNA for a perfect gas,
where [A] is the molar concentration of A
...

Viscosity

A dvx D
E
mvx(−λ) = mvx(0) − mλ B
C dz F 0

(21
...
96)

The flux is proportional to the velocity gradient, as we wished to
show
...
21, and
2
multiplication by – in the normal way, leads to
3
1
η = –Nmλ K
3

(21
...
24 by using Nm = nM and
[A] = n/V
...
21
...
1 Use the kinetic theory to justify the following observations: (a) the rate
of a reaction in the gas phase depends on the energy with which two molecules
collide, which in turn depends on their speeds; (b) in the Earth’s atmosphere,
light gases, such as H2 and He, are rare but heavier gases, such as O2, CO2, and
N2, are abundant
...
2 Provide a molecular interpretation for each of the following processes:
diffusion, thermal conduction, electric conduction, and viscosity
...
3 Provide a molecular interpretation for the observation that the viscosity
of a gas increases with temperature, whereas the viscosity of a liquid decreases
with increasing temperature
...
4 Discuss the mechanism of proton conduction in liquid water
...
5 Limit the generality of the following expressions: (a) J = −D(dc/dx),

(b) D = kT/f, and (c) D = kT/6πηa
...
6 Provide a molecular interpretation for the observation that mediated
transport across a biological membrane leads to a maximum flux Jmax when
the concentration of the transported species becomes very large
...
7 Discuss how nuclear magnetic resonance spectroscopy, inelastic neutron
scattering, and dynamic light scattering may be used to measure the mobility
of molecules in liquids
...
1a Determine the ratios of (a) the mean speeds, (b) the mean kinetic
energies of H2 molecules and Hg atoms at 20°C
...
1b Determine the ratios of (a) the mean speeds, (b) the mean kinetic

energies of He atoms and Hg atoms at 25°C
...
2a A 1
...
0 × 10
3

21
...
00 mm
...
224 Pa at
450 K, by how much will the mass of the solid decrease in a period of 24
...
10a A manometer was connected to a bulb containing carbon dioxide

23

H2 molecules
...
2b The best laboratory vacuum pump can generate a vacuum of about

1 nTorr
...

21
...
36 nm2
...
3b At what pressure does the mean free path of argon at 25°C become

comparable to the diameters of the atoms themselves?
21
...
050 atm
...
43 nm2
...
4b At an altitude of 15 km the temperature is 217 K and the pressure

12
...
What is the mean free path of N2 molecules? (σ = 0
...
)

under slight pressure
...
When the experiment was repeated using nitrogen (for which
M = 28
...
Calculate the
molar mass of carbon dioxide
...
10b A manometer was connected to a bulb containing nitrogen under
slight pressure
...
1 cm to 42
...
5 s
...
3 s
...

21
...
0 m3 is struck by a meteor and a

hole of radius 0
...
If the oxygen pressure within the vehicle is
initially 80 kPa and its temperature 298 K, how long will the pressure take to
fall to 70 kPa?
21
...
0 m3 was punctured, and a hole of
radius 0
...
If the nitrogen pressure within the vehicle is
initially 122 kPa and its temperature 293 K, how long will the pressure take to
fall to 105 kPa?

21
...
0 s when the
temperature is 25°C and the pressure is (a) 10 atm, (b) 1
...
0 µatm?

21
...
5b How many collisions per second does an N2 molecule make at an
altitude of 15 km? (See Exercise 21
...
)

21
...
5 K m−1 in a sample of hydrogen in which the mean temperature is 260 K
...
6a Calculate the mean free path of molecules in air using σ = 0
...
13a Use the experimental value of the thermal conductivity of neon

21
...
13b Use the experimental value of the thermal conductivity of nitrogen

25°C and (a) 10 atm, (b) 1
...
0 µatm
...
52 nm2 at 25°C and (a) 15 atm, (b) 1
...
0 Torr
...
5 K m−1 in a sample of argon in which the mean temperature is 273 K
...
2) to estimate the collision cross-section of Ne atoms at 273 K
...
2) to estimate the collision cross-section of N2 molecules at 298 K
...
7a Use the Maxwell distribution of speeds to estimate the fraction of N2

21
...
0 cm
...
7b Use the Maxwell distribution of speeds to estimate the fraction of CO2

molecules at 300 K that have speeds in the range 200 to 250 m s−1
...
0 m2? What power of
heater is required to make good the loss of heat?

21
...
5 mm × 3
...
14b Two sheets of copper of area 1
...
0 cm
...


gas at 90 Pa and 500 K
...
What is the rate of loss of heat?

21
...
5 mm × 4
...
15a Use the experimental value of the coefficient of viscosity for neon

gas at 111 Pa and 1500 K
...
15b Use the experimental value of the coefficient of viscosity for nitrogen

21
...
50 mm
...
2) to estimate the collision cross-section of the molecules at 273 K
...
835 Pa at
400 K, by how much will the mass of the solid decrease in a period of 2
...
2) to estimate the collision cross-section of Ne atoms at 273 K
...
16a Calculate the inlet pressure required to maintain a flow rate of

9
...
50 m

EXERCISES
and diameter 1
...
The pressure of gas as it leaves the tube is 1
...

The volume of the gas is measured at that pressure
...
16b Calculate the inlet pressure required to maintain a flow rate of
8
...
5 m
and diameter 15 mm
...
00 bar
...

21
...
Take

σ ≈ 0
...
(The experimental values are 173 µP at 273 K, 182 µP at 20°C,
and 394 µP at 600°C
...
17b Calculate the viscosity of benzene vapour at (a) 273 K, (b) 298 K,
(c) 1000 K
...
88 nm2
...
18a Calculate the thermal conductivities of (a) argon, (b) helium at 300 K

787

21
...
01 × 10−8 m2 s−1 V−1

at 25°C
...
0 V
...
00 cm apart, what is the drift speed of the ion?
21
...
24b What fraction of the total current is carried by Cl− when current flows

through an aqueous solution of NaCl at 25°C?
21
...
99 mS m2 mol−1, 14
...
34 mS m2 mol−1, respectively
(all at 25°C)
...
0 mbar
...
What is the rate of flow of
energy as heat from one wall to the other in each case?

21
...
69 mS m2 mol−1, 9
...
78 mS m2
mol−1, respectively (all at 25°C)
...
18b Calculate the thermal conductivities of (a) neon, (b) nitrogen at 300 K

21
...
Each gas is confined in a cubic vessel of side 15 cm, one wall
being at 305 K and the one opposite at 295 K
...
87 mS m2 mol−1, 5
...
35 mS m2 mol−1, respectively
...
19a The viscosity of carbon dioxide was measured by comparing its rate of

m2 mol−1, 7
...
81 mS m2 mol−1, respectively
...
For the same pressure differential, the same volume of carbon dioxide
passed through the tube in 55 s as argon in 83 s
...

21
...
For the same pressure differential, the same
volume of the CFC passed through the tube in 72
...
0 s
...
Take M = 200 g mol−1
...
20a Calculate the thermal conductivity of argon (CV,m = 12
...
36 nm2) at room temperature (20°C)
...
20b Calculate the thermal conductivity of nitrogen (CV,m = 20
...
26b At 25°C the molar ionic conductivities of F−, Cl−, and Br− are 5
...
27a The mobility of a NO 3 ion in aqueous solution at 25°C is 7
...
Calculate its diffusion coefficient in water at 25°C
...
27b The mobility of a CH3CO 2 ion in aqueous solution at 25°C is

4
...
Calculate its diffusion coefficient in water at 25°C
...
28a The diffusion coefficient of CCl4 in heptane at 25°C is 3
...
Estimate the time required for a CCl4 molecule to have a root mean square
displacement of 5
...

21
...
05 × 10−9 m2 s−1
...
0 cm
...
29a Estimate the effective radius of a sucrose molecule in water 25°C given

mol−1, σ = 0
...


that its diffusion coefficient is 5
...
00 cP
...
21a Calculate the diffusion constant of argon at 25°C and (a) 1
...
29b Estimate the effective radius of a glycine molecule in water at 25°C

(b) 100 kPa, (c) 10
...
If a pressure gradient of 0
...
055 × 10−9 m2 s−1 and that the viscosity
of water is 1
...


21
...
0 Pa,

21
...
13 × 10−9 m2 s−1
...
0 MPa
...
20 bar m−1 is established
in a pipe, what is the flow of gas due to diffusion?

21
...
91 × 10−8 m2 s−1 V−1
...


21
...
24 × 10 m s V
...

21
...
92 × 10−8 m2 s−1 V−1

at 25°C
...
0 V
...
00 mm apart, what is the drift speed of the
Rb+ ion?

21
...
17 × 10−9 m2 s−1
...
31a What are the root mean square distances travelled by an iodine

molecule in benzene and by a sucrose molecule in water at 25°C in 1
...
31b About how long, on average, does it take for the molecules in Exercise
21
...
0 mm, (b) 1
...
00050

0
...
0050

0
...
020

0
...
1 Instead of the arrangement in Fig
...
8, the speed of molecules can also
be measured with a rotating slotted-disc apparatus, which consists of five
coaxial 5
...
0 cm, the slots in their rims being
displaced by 2
...
The relative intensities, I, of the
detected beam of Kr atoms for two different temperatures and at a series of
rotation rates were as follows:

R/Ω

3314

1669

342
...
1

89
...
14

ν/Hz

20

40

80

100

120

I (40 K)

0
...
513

0
...
015

0
...
592

0
...
217

0
...
057

Find the distributions of molecular velocities, f(vx), at these temperatures, and
check that they conform to the theoretical prediction for a one-dimensional
system
...
2 Cars were timed by police radar as they passed in both directions below a
bridge
...
What are
(a) the mean velocity, (b) the mean speed, (c) the root mean square speed?
21
...
80 (1), 1
...
84 (4), 1
...
88 (10), 1
...
92 (9), 1
...
96 (0), 1
...
What are (a) the
mean height, (b) the root mean square height of the population?
21
...
Be circumspect, and think about the
modes of motion that are thermally active at the two temperatures
...
5 A Knudsen cell was used to determine the vapour pressure of
germanium at 1000°C
...
50 mm amounted to 43 µg
...

21
...
Its half-life is 4
...
A sample of mass
1
...
0 cm3 that was impermeable to
α radiation, but there was also a hole of radius 2
...
What is the
pressure of helium at 298 K, inside the container after (a) 1
...
7 An atomic beam is designed to function with (a) cadmium, (b) mercury
...
0 cm × 1
...
The vapour pressure of cadmium is
0
...
What is the atomic
current (the number of atoms per unit time) in the beams?
21
...
The conductivity of water is 76 mS m−1 at
25°C and the conductivity of 0
...
1639 S m−1
...
21 Ω when filled with 0
...
0 Ω when filled with 0
...
What is the molar
conductivity of acetic acid at that concentration and temperature?
21
...
2063 cm−1
...
Determine the coefficient K
...
01 mS m2 mol−1 and λ(I−) =
7
...
010 mol dm−3 NaI(aq)
at 25°C
...
10 After correction for the water conductivity, the conductivity of a
saturated aqueous solution of AgCl at 25°C was found to be 0
...

What is the solubility of silver chloride at this temperature?
21
...
00-cm conductivity cell? How long
would it take an ion to move from one electrode to the other? In conductivity
measurements it is normal to use alternating current: what are the
displacements of the ions in (a) centimetres, (b) solvent diameters,
about 300 pm, during a half cycle of 1
...
12 The mobilities of H+ and Cl− at 25°C in water are 3
...
91 × 10−8 m2 s−1 V−1, respectively
...
0 mol dm−3 in
the salt? Note how concentration as well as mobility governs the transport
of current
...
13 In a moving boundary experiment on KCl the apparatus consisted
of a tube of internal diameter 4
...
021 mol dm−3
...
2 mA was passed,
and the boundary advanced as follows:

∆t/s

200

400

600

800

1000

x/mm

64

128

192

254

318

Find the transport number of K+, its mobility, and its ionic conductivity
...
14 The proton possesses abnormal mobility in water, but does it behave
normally in liquid ammonia? To investigate this question, a moving+
boundary technique was used to determine the transport number of NH4 in
liquid ammonia (the analogue of H3O+ in liquid water) at −40°C (J
...
Evans, and J
...
Gill, J
...
Soc
...
A steady current of
5
...
9 mm in a 0
...
03 mm in a
+
0
...
Calculate the transport number of NH4 at these
concentrations, and comment on the mobility of the proton in liquid
ammonia
...
146 mm and the density of liquid
ammonia is 0
...

21
...
The solution was in a horizontal tube of length 10 cm, and at first
there was a linear gradation of intensity of the purple solution from the left
(where the concentration was 0
...
050 mol dm−3)
...


* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady
...
16 Estimate the diffusion coefficients and the effective hydrodynamic radii
of the alkali metal cations in water from their mobilities at 25°C
...

Ionic radii are given Table 20
...

21
...
A set of measurements on methane in carbon
tetrachloride showed that its diffusion coefficient is 2
...
89 × 10−9 m2 s−1 at 25°C
...

21
...
0 cm
...
0 cm3 of water
...
0 dm3 of water is then poured very carefully on top of the layer, without
disturbing the layer
...
Find the concentration at 5
...
0 y
...
19 In a series of observations on the displacement of rubber latex spheres
of radius 0
...
2

113
...
Find the effective viscosity of water at the temperature
of this experiment (25°C)
...
20‡ A
...
Srivastava, R
...
Samant, and S
...
Patankar (J
...
Eng
...
They report the following conductances at 25°C in a solvent 80
per cent 1,3-dioxolan-2-one by mass:
NaI
c/(mmol dm−3)

32
...
28 12
...
64

2
...
24

0
...
26 51
...
01 55
...
99 58
...
67
KI
c/(mmol dm−3)

17
...
45

10
...
19 2
...
91 47
...
81

1
...
83

54
...
78

0
...
42

Calculate Λ m for NaI and KI in this solvent and λ°(Na) − λ°(K)
...
5
in the Data section
...
21‡ A
...
A
...
Vesovic, J
...
R
...
Millat,
and E
...
Phys
...
Ref
...
Deduce the effective molecular diameter of NH3 based on each of the
following vapour-phase viscosity coefficients: (a) η = 9
...
00 bar; (b) η = 1
...
0 bar
...
22‡ G
...
Lacmann, and W
...
Schmidt ( J
...
Chem
...
In cyclohexane at 22°C, the mobility is 1
...

−
Estimate the effective radius of the C 60 ion
...
93 × 10−3 kg m−1 s−1
...
The researchers interpreted the substantial
difference between this number and the van der Waals radius of neutral
C60 in terms of a solvation layer around the ion
...
23 Start from the Maxwell–Boltzmann distribution and derive an
expression for the most probable speed of a gas of molecules at a temperature

789

T
...

2
21
...
Calculate the distribution of speeds and determine the
mean speed of the molecules at a temperature T
...
25 A specially constructed velocity-selector accepts a beam of molecules
from an oven at a temperature T but blocks the passage of molecules with a
speed greater than the mean
...
26 What is the proportion of gas molecules having (a) more than, (b) less
than the root mean square speed? (c) What are the proportions having speeds
greater and smaller than the mean speed?
21
...
Evaluate the ratio for n = 3 and n = 4
...
28 Derive an expression that shows how the pressure of a gas inside an
effusion oven (a heated chamber with a small hole in one wall) varies with
time if the oven is not replenished as the gas escapes
...
Hint
...

21
...

21
...
72 is a solution of the diffusion equation with the
correct initial value
...
31 The diffusion equation is valid when many elementary steps are taken
in the time interval of interest, but the random walk calculation lets us discuss
distributions for short times as well as for long
...
84 to calculate the
probability of being six paces from the origin (that is, at x = 6λ) after (a) four,
(b) six, (c) twelve steps
...
32 Use mathematical software to calculate P in a one-dimensional random
walk, and evaluate the probability of being at x = nλ for n = 6, 10, 14,
...

Compare the numerical value with the analytical value in the limit of a large
number of steps
...
1 per cent?
21
...
7
...
34‡ A dilute solution of a weak (1,1)-electrolyte contains both neutral ion

pairs and ions in equilibrium (AB 5 A+ + B−)
...
29)
...
35 Calculate the escape velocity (the minimum initial velocity that will
take an object to infinity) from the surface of a planet of radius R
...
37 × 106 m, g = 9
...
38 × 106 m, mMars /mEarth = 0
...
At what temperatures do H2, He, and
O2 molecules have mean speeds equal to their escape speeds? What proportion
of the molecules have enough speed to escape when the temperature is
(a) 240 K, (b) 1500 K? Calculations of this kind are very important in
considering the composition of planetary atmospheres
...
36‡ Interstellar space is a medium quite different from the gaseous

environments we commonly encounter on Earth
...

Estimate the diffusion coefficient and thermal conductivity of H under these
conditions
...
Energy is in fact transferred much more effectively by
radiation
...
37 The principal components of the atmosphere of the Earth are diatomic

molecules, which can rotate as well as translate
...
15 J cm−3, what is the total kinetic
energy density, including rotation?
21
...
Nicholson, The sun
...
Both atoms
are completely ionized
...
4 × 10−15 A1/3 m, where A is the mass
number
...
(a) Calculate the excluded volume in 1
...
(b) The standard model suggests that the pressure in
the stellar interior is 2
...
Calculate the temperature of the Sun’s
interior based on the perfect gas model
...
(c) Would a van der Waals type of equation (with
a = 0) give a better value for T?

21
...
Hence, such
calculations are often called ‘Fermi calculations’
...
0 m a
part by diffusion in still air
...
40 The diffusion coefficient of a particular kind of t-RNA molecule is
D = 1
...
How long does it take
molecules produced in the cell nucleus to reach the walls of the cell at a
distance 1
...
41‡ In this problem, we examine a model for the transport of oxygen from

air in the lungs to blood
...
4) and the concentration c(x,t)
evolves by diffusion from the yz-plane of constant concentration, such as
might occur if a condensed phase is absorbing a species from a gas phase
...
10 × 10−9 m2 s−1)
on a spatial scale comparable to passage of oxygen from lungs through alveoli
into the blood
...

Hint
...


The rates of chemical
reactions
This chapter is the first of a sequence that explores the rates of chemical reactions
...
The results of such measurements show that reaction rates
depend on the concentration of reactants (and products) in characteristic ways that can be
expressed in terms of differential equations known as rate laws
...
The form of the rate law also provides insight into the series of elementary steps
by which a reaction takes place
...
Simple elementary
steps have simple rate laws, and these rate laws can be combined together by invoking
one or more approximations
...


This chapter introduces the principles of chemical kinetics, the study of reaction rates,
by showing how the rates of reactions may be measured and interpreted
...
The rate of a chemical reaction
might depend on variables under our control, such as the pressure, the temperature,
and the presence of a catalyst, and we may be able to optimize the rate by the appropriate choice of conditions
...


22
Empirical chemical kinetics
22
...
2 The rates of reactions
22
...
4 Reactions approaching

equilibrium
22
...
6 Elementary reactions
22
...
1 Impact on biochemistry: The

kinetics of the helix–coil
transition in polypeptides
22
...
1: The RRK
model of unimolecular reactions
Discussion questions

The first steps in the kinetic analysis of reactions are to establish the stoichiometry of
the reaction and identify any side reactions
...
The rates of most chemical reactions are sensitive to the temperature, so in conventional experiments the temperature of the reaction mixture
must be held constant throughout the course of the reaction
...
Gas-phase reactions, for instance,
are often carried out in a vessel held in contact with a substantial block of metal
...
Special efforts have to be made to study reactions at low temperatures,
as in the study of the kinds of reactions that take place in interstellar clouds
...
For work in the liquid phase and the solid phase, very low temperatures are often
reached by flowing cold liquid or cold gas around the reaction vessel
...
Non-isothermal conditions are sometimes employed
...

22
...
Many reactions reach equilibrium
over periods of minutes or hours, and several techniques may then be used to follow
the changing concentrations
...

Example 22
...

Method The total pressure (at constant volume and temperature and assuming

perfect gas behaviour) is proportional to the number of gas-phase molecules
...
To confirm this conclusion,
2
express the progress of the reaction in terms of the fraction, α, of N2O5 molecules
that have reacted
...
When a fraction α of the N2O5 molecules has decomposed, the
amounts of the components in the reaction mixture are:

N2O5
Amount:

NO2

n(1 − α)

2αn

O2
1
–αn
2

Total
3
n(1 + –α)
2

When α = 0 the pressure is p0, so at any stage the total pressure is
3
p = (1 + –α)p0
2
5
When the reaction is complete, the pressure will have risen to – times its initial
2
value
...
1 Repeat the calculation for 2 NOBr(g) → 2 NO(g) + Br2(g)
...
For example, the progress of the reaction

22
...
A reaction
that changes the number or type of ions present in a solution may be followed by
monitoring the electrical conductivity of the solution
...

Other methods of determining composition include emission spectroscopy, mass
spectrometry, gas chromatography, nuclear magnetic resonance, and electron paramagnetic resonance (for reactions involving radicals or paramagnetic d-metal ions)
...
Either a small sample is withdrawn or the bulk solution is monitored
...
22
...

The reaction continues as the thoroughly mixed solutions flow through the outlet tube,
and observation of the composition at different positions along the tube is equivalent
to the observation of the reaction mixture at different times after mixing
...
This makes the study of fast reactions particularly difficult because to
spread the reaction over a length of tube the flow must be rapid
...
22
...
The flow
ceases when the plunger of the syringe reaches a stop, and the reaction continues in
the mixed solutions
...
The technique allows for the study of reactions that occur on the millisecond to second timescale
...
1 later in the chapter)
...
Most work is now done with lasers with photolysis pulse widths that range from femtoseconds to nanoseconds (Section 14
...
The
apparatus used for flash photolysis studies is based on the experimental design for
time-resolved spectroscopy (Section 14
...
Reactions occurring on a picosecond or
femtosecond timescale may be monitored by using electronic absorption or emission,
infrared absorption, or Raman scattering
...
The laser pulse can initiate the reaction by forming a reactive species, such as an excited electronic state of a molecule, a radical, or an ion
...
An example of radical generation is the light-induced dissociation of Cl2(g) to yield Cl atoms that react with HBr
to make HCl and Br according to the following sequence:
Cl2 + hν → Cl + Cl
Cl + HBr → HCl* + Br
HCl* + M → HCl + M

Movable
spectrometer

Driving
syringes

Mixing
chamber

Fig
...
1 The arrangement used in the flow
technique for studying reaction rates
...
The location of
the spectrometer corresponds to different
times after initiation
...
22
...


794

22 THE RATES OF CHEMICAL REACTIONS
Here HCl* denotes a vibrationally excited HCl molecule and M is a body (an unreactive molecule or the wall of the container) that removes the excess energy stored in HCl
...

In contrast to real-time analysis, quenching methods are based on stopping, or
quenching, the reaction after it has been allowed to proceed for a certain time
...

These methods are suitable only for reactions that are slow enough for there to be
little reaction during the time it takes to quench the mixture
...
Different reaction times
can be selected by varying the flow rate along the outlet tube
...
Once the reaction has been quenched, the solution may be examined by
‘slow’ techniques, such as gel electrophoresis, mass spectrometry, and chromatography
...

22
...
The next few sections look at these observations in more detail
...
The
instantaneous rate of consumption of one of the reactants at a given time is −d[R]/dt,
where R is A or B
...
22
...
The rate of formation of
one of the products (C or D, which we denote P) is d[P]/dt (note the difference in
sign)
...

It follows from the stoichiometry for the reaction A + 2 B → 3 C + D that
d[D]
dt

Time, t
Reactant
(b)
Tangent,
rate = -slope
Time, t

Fig
...
3 The definition of (instantaneous)
rate as the slope of the tangent drawn
to the curve showing the variation of
concentration with time
...


1
=–
3

d[C]
dt

=−

d[A]
dt

1
= −–
2

d[B]
dt

so there are several rates connected with the reaction
...
1):
nJ − nJ,0
ξ=
(22
...
2]
V dt
It follows that
1 1 dnJ
v= ×
(22
...
2 THE RATES OF REACTIONS
(Remember that νJ is negative for reactants and positive for products
...
3b)

For a heterogeneous reaction, we use the (constant) surface area, A, occupied by the
species in place of V and use σJ = nJ /A to write
v=

1 dσJ

νJ dt

(22
...
With molar concentrations in moles per cubic decimetre and time in
seconds, reaction rates of homogeneous reactions are reported in moles per cubic
decimetre per second (mol dm−3 s−1) or related units
...
For heterogeneous reactions, rates are expressed in
moles per square metre per second (mol m−2 s−1) or related units
...
1 Rates of formation and consumption

If the rate of formation of NO in the reaction 2 NOBr(g) → 2 NO(g) + Br2(g) is
reported as 0
...
080 mmol
dm−3 s−1
...
16 mmol dm−3 s−1
...
16 mmol dm−3 s−1, or 9
...

Self-test 22
...
2 mol dm−3 s−1
under particular conditions
...
60 mol dm−3 s−1, (b) 0
...
For example, the rate of a reaction may be proportional to
the molar concentrations of two reactants A and B, so we write
v = k[A][B]

(22
...
The coefficient k is called the rate
constant for the reaction
...
An experimentally determined equation of this kind
is called the rate law of the reaction
...
)

[22
...
In this case, we write

795

796

22 THE RATES OF CHEMICAL REACTIONS
v = f(pA,pB,
...
5b]

The rate law of a reaction is determined experimentally, and cannot in general be
inferred from the chemical equation for the reaction
...
6)

In certain cases the rate law does reflect the stoichiometry of the reaction, but that is
either a coincidence or reflects a feature of the underlying reaction mechanism (see
later)
...
Moreover, as we shall see later, by knowing the rate law, we can go on to predict the composition of the reaction mixture at a later stage of the reaction
...

(c) Reaction order

Many reactions are found to have rate laws of the form
v = k[A]a[B]b · · ·

(22
...
A
reaction with the rate law in eqn 22
...
The overall order of a reaction with a rate law like that in eqn 22
...
The rate law in eqn 22
...

A reaction need not have an integral order, and many gas-phase reactions do not
...
8)

is half-order in A, first-order in B, and three-halves-order overall
...
Thus, the catalytic decomposition of phosphine (PH3) on hot tungsten at high pressures has the rate law
v=k

(22
...

Zero-order reactions typically occur when there is a bottle-neck of some kind in
the mechanism, as in heterogeneous reactions when the surface is saturated and the
subsequent reaction slow and in a number of enzyme reactions when there is a large
excess of substrate relative to the enzyme
...
7, the reaction does not have an overall order and may not even have definite orders with respect to each participant
...
6 shows that the reaction of hydrogen and bromine is first-order in
H2, the reaction has an indefinite order with respect to both Br2 and HBr and has no
overall order
...
First, we must see how to identify the rate
law and obtain the rate constant from the experimental data
...
Second, we must see how to construct reaction mechanisms
that are consistent with the rate law
...
2 THE RATES OF REACTIONS
this chapter and develop them further in Chapter 23
...
We shall see a
little of what is involved in this chapter, but leave the details until Chapter 24
...
If B is in large excess,
for example, then to a good approximation its concentration is constant throughout
the reaction
...
10)

which has the form of a first-order rate law
...
10 is
called a pseudofirst-order rate law
...

In the method of initial rates, which is often used in conjunction with the isolation
method, the rate is measured at the beginning of the reaction for several different
initial concentrations of reactants
...
Taking logarithms gives:
0
log v0 = log k + a log [A]0

(22
...

Example 22
...
The initial rates of reaction of 2 I(g) + Ar(g) → I2(g) + Ar(g) were as follows:
[I]0 /(10−5 mol dm−3)
v0 /(mol dm−3 s −1)

1
...
0

4
...
70 × 10 − 4 3
...
39 × 10 −2

6
...
13 × 10−2

(b) 4
...
74 × 10 −2 6
...
57 × 10−1

(c) 8
...
47 × 10 −2 1
...
13 × 10−1

The Ar concentrations are (a) 1
...
0 mmol dm−3, and (c) 10
...
Determine the orders of reaction with respect to the I and Ar atom
concentrations and the rate constant
...
The slopes of the two lines are the orders of reaction with respect to I and Ar,
respectively
...

Answer The plots are shown in Fig
...
4
...
The intercept corresponds to k = 9 × 109 mol−2 dm6 s−1
...
22
...


-2

0

0
...
4 0
...
8
log [I]0 + 5

1
...
2

0
...
6 0
...
0

A note on good practice The units of k come automatically from the calculation,

and are always such as to convert the product of concentrations to a rate in concentration/time (for example, mol dm−3 s−1)
...
3 The initial rate of a reaction depended on concentration of a sub-

stance J as follows:
[J]0 /(mmol dm−3)
v0 /(10

−7

−3 −1

mol dm s )

5
...
2

17

30

3
...
6

41

130

Determine the order of the reaction with respect to J and calculate the rate constant
...
4 × 10−2 dm3 mol−1 s−1]
The method of initial rates might not reveal the full rate law, for once the products
have been generated they might participate in the reaction and affect its rate
...
6 shows that the
full rate law depends on the concentration of HBr
...
The fitting may be done, in
simple cases at least, by using a proposed rate law to predict the concentration of any
component at any time, and comparing it with the data
...

22
...
Even the most complex rate laws may be
integrated numerically
...
We
examine a few of these simple cases here
...
12a)

22
...
1

Synoptic table 22
...
38 × 10−5

Br2(l)

25

4
...
51 h

700

5
...
6 min

C2H6 → 2 CH3

g

k/s

t1/2

The web site contains links to databases
of rate constants of chemical reactions
...
70 h

* More values are given in the Data section
...
12b)
0
...


0
...
1 First-order integrated rate law

First, we rearrange eqn 22
...
4

= −kdt

ksmall

This expression can be integrated directly because k is a constant independent of t
...
Then identify the time of interest (t)
and the corresponding concentration ([A]), and write these quantities as the upper
limits of their respective integrals
...
12b shows that, if ln([A]/[A]0) is plotted against t, then a first-order
reaction will give a straight line of slope −k
...
1
...
12b shows that in a firstorder reaction the reactant concentration decreases exponentially with time with a
rate determined by k (Fig
...
5)
...
3 Analysing a first-order reaction

The variation in the partial pressure of azomethane with time was followed at
600 K, with the results given below
...

0
10
...
12b is obtained immediately
...
2

1000
7
...
32

3000
3
...
59

1

ksmallt

2

3

Fig
...
5 The exponential decay of the
reactant in a first-order reaction
...


Exploration For a first-order reaction
of the form A → nB (with n possibly
fractional), the concentration of the
product varies with time as [B] = n[B]0
(1 − e−kt)
...
5, 1, and 2
...
Because the partial pressure of
a gas is proportional to its concentration, an equivalent procedure is to plot
ln(p/p0) against t
...


-0
...
360

−0
...
082

−1
...
6 shows the plot of ln(p/p0) against t
...
6 × 10−4
...
6 × 10−4 s−1
...
5
0

0

ln(p/p0)

t /s
Fig
...
6 The determination of the rate
constant of a first-order reaction: a straight
line is obtained when ln [A] (or, as here,
ln p) is plotted against t; the slope gives k
...
For a
graph of the form y = b + mx we can write

y = b + (m units)(x/units)
where ‘units’ are the units of x, and identify the (dimensionless) slope with ‘m
units’
...
4 In a particular experiment, it was found that the concentration of
N2O5 in liquid bromine varied with time as follows:

t/s

0
−3

[N2O5]/(mol dm )

200

400

600

1000

0
...
073

0
...
032

0
...

[k = 2
...
The time for [A] to decrease from [A]0 to – [A]0 in a first-order reaction is given
2
by eqn 22
...
13)

(ln 2 = 0
...
) The main point to note about this result is that, for a first-order reaction, the half-life of a reactant is independent of its initial concentration
...
Some half-lives are given in Table 22
...

2

22
...

From eqn 22
...
14)

k

(c) Second-order reactions

We show in the Justification below that the integrated form of the second-order rate
law
d[A]
dt

= −k[A]2

(22
...
15b)

[A]0

(22
...


0
...
15a we rearrange it into
d[A]
[A]2

= −kdt

The concentration of A is [A0] at t = 0 and [A] at a general time t later
...
6

ksmall
0
...
2

= k dt
0

Because the integral of 1/x 2 is −1/x, we obtain eqn 22
...
2 Second-order integrated rate law

1
[A]

−

1
[A]0

= kt

We can then rearrange this expression into eqn 22
...


Equation 22
...
The slope of the graph is k
...
2
...
15c, lets
us predict the concentration of A at any time after the start of the reaction
...
22
...

1
It follows from eqn 22
...
22
...
The grey lines are the
corresponding decays in a first-order
reaction with the same initial rate
...


Exploration For a second-order
reaction of the form A → nB (with
n possibly fractional), the concentration
of the product varies with time as [B] =
2
nkt[A]0 /(1 + kt[A]0)
...
5, 1, and 2
...
2* Kinetic data for second-order reactions
Reaction

Phase

q/°C

k/(dm3 mol−1 s−1)

2 NOBr → 2 NO + Br2

g

10

0
...
29 × 10−6

−

CH3Cl + CH3O

* More values are given in the Data section
...
16)

k[A]0

Therefore, unlike a first-order reaction, the half-life of a substance in a secondorder reaction varies with the initial concentration
...
In general,
for an nth-order reaction of the form A → products, the half-life is related to the rate
constant and the initial concentration of A by
t1/2 =

1
k[A]n−1

(22
...
12a
...
18)

Such a rate law cannot be integrated until we know how the concentration of B is related to that of A
...
19)

Therefore, a plot of the expression on the left against t should be a straight line from
which k can be obtained
...
3 Overall second-order rate law

It follows from the reaction stoichiometry that, when the concentration of A has
fallen to [A]0 − x, the concentration of B will have fallen to [B]0 − x (because each A
that disappears entails the disappearance of one B)
...
3 INTEGRATED RATE LAWS
dx
dt

Comment 22
...
19 by combining
the two logarithms by using ln y − ln z = ln(y/z) and noting that [A] = [A]0 − x and
[B] = [B]0 − x
...
3
...
3 Integrated rate laws
Order

Reaction

Rate law*

t1/2

0

A→P

v=k
kt = x for 0 ≤ x ≤ [A]0

[A]0 /2k

1

A→P

v = k[A]

(ln 2)/k

[A]0
kt = ln
[A]0 − x
2

A→P

v = k[A]2

1/k[A]0

x
kt =
[A]0([A]0 − x)
A+B→P

v = k[A][B]
1
[A]0([B]0 − x)
kt =
ln
[B]0 − [A]0
([A]0 − x)[B]0

A+2B→P

v = k[A][B]
1
[A]0([B]0 − 2x)
kt =
ln
[B]0 − 2[A]0
([A]0 − x)[B]0

A→P
with autocatalysis

3

A+2B→P

v = k[A][P]
1
[A]0([P]0 + x)
kt =
ln
[A]0 + [P]0
([A]0 − x)[P]0
v = k[A][B]2
kt =

n≥2

A→P

* x = [P] and v = dx /dt
...
The integral on the left is evaluated by using
the method of partial fractions and by using [A] = [A]0 and [B] = [B]0 at t = 0 to give:
x

803

2x
(2[A]0 − [B]0)([B]0 − 2x)[B]0
1
[A]0([B]0 − 2x)
+
ln
(2[A]0 − [B]0)2
([A]0 − x)[B]0

v = k[A]n
1 1
1
1 5
2
6
kt =
−
n−1
n − 1 3 ([A]0 − x)
[A]n−1 7
0

2n−1 − 1
(n − 1)k[A]n−1
0

1
(a − x)(b − x)

=

1 A 1
1 D
B
E
−
b − a C a − x b − xF

and integrate the expression on the
right
...
4 Reactions approaching equilibrium
Because all the laws considered so far disregard the possibility that the reverse reaction
is important, none of them describes the overall rate when the reaction is close to
equilibrium
...
In practice, however, most kinetic studies are made on
reactions that are far from equilibrium, and the reverse reactions are unimportant
...
The scheme we consider is
A→B

v = k[A]

B→A

v = k′[B]

(22
...
The net rate of change is therefore
d[A]
dt

= −k[A] + k′[B]

(22
...
Therefore,
d[A]
dt

[A] =

[J]/[J]0

k′ + ke−(k+k′)t
k′ + k

[A]/[A]0

[A]0

(22
...
8 shows the time dependence predicted by this equation
...
23 as:

0
...
4

[A]eq =
[B]/[B]0

K=
0

1

(k + k' )t

k′[A]0
k + k′

[B]eq = [A]0 − [A]∞ =

k[A]0
k + k′

(22
...
2

0

(22
...
8

= −k[A] + k′([A]0 − [A]) = −(k + k′)[A] + k′[A]0

2

3

Fig
...
8 The approach of concentrations to
their equilibrium values as predicted by
eqn 22
...


Exploration Set up the rate equations

and plot the corresponding graphs
for the approach to and equilibrium of the
form A 5 2 B
...
25)

(This expression is only approximate because thermodynamic equilibrium constants
are expressed in terms of activities, not concentrations
...
26)

This relation rearranges into eqn 22
...
The theoretical importance of eqn 22
...
Its practical importance is that, if one of the rate constants can be
measured, then the other may be obtained if the equilibrium constant is known
...
4 REACTIONS APPROACHING EQUILIBRIUM

805

K=

ka
k′
a

×

kb
k′
b

×···

(22
...

(b) Relaxation methods

The term relaxation denotes the return of a system to equilibrium
...
22
...
We shall
consider the response of reaction rates to a temperature jump, a sudden change in
temperature
...
4 that the equilibrium composition of a reaction depends on the temperature (provided ∆r H 7 is nonzero), so a shift in temperature acts as a perturbation on the system
...
Temperature jumps of between
5 and 10 K can be achieved in about 1 µs with electrical discharges
...
5) is sufficient to generate temperature jumps of
between 10 and 30 K within nanoseconds in aqueous samples
...

When a sudden temperature increase is applied to a simple A 5 B equilibrium that
is first-order in each direction, we show in the Justification below that the composition
relaxes exponentially to the new equilibrium composition:
x = x0 e−t/τ

1

τ

= ka + kb

(22
...

Justification 22
...
As the system is no longer at equilibrium, it readjusts to the
new equilibrium concentrations, which are now given by
ka[A]eq = kb[B]eq
and it does so at a rate that depends on the new rate constants
...
The
concentration of A then changes as follows:
d[A]
dt

= −ka[A] + kb[B]
= −ka([A]eq + x) + k b([B]eq − x)
= −(ka + k b)x

because the two terms involving the equilibrium concentrations cancel
...
12b and is given in eqn 22
...


Concentration, [A]

For a more general reaction, the overall equilibrium constant can be expressed in
terms of the rate constants for all the intermediate stages of the reaction mechanism:

Exponential
relaxation

T2

T1
Time, t

Fig
...
9 The relaxation to the new
equilibrium composition when a reaction
initially at equilibrium at a temperature T1
is subjected to a sudden change of
temperature, which takes it to T2
...
28 shows that the concentrations of A and B relax into the new equilibrium at a rate determined by the sum of the two new rate constants
...

Example 22
...
008 × 10−14 at 298 K
...
Given that the forward reaction is first-order and the reverse is secondorder overall, calculate the rate constants for the forward and reverse reactions
...
We can proceed as above, but it will be necessary to make the assumption that the deviation from equilibrium (x) is so small
that terms in x 2 can be neglected
...

Answer The forward rate at the final temperature is k1[H2O] and the reverse rate

is k2[H +][OH −]
...
It follows that
1

τ

= k1 + k2([H +]eq + [OH −]eq)

At this point we note that
Kw = a(H +)a(OH −) ≈ ([H +]eq /c 7)([OH −]eq)/c 7) = [H +]eq[OH −]eq/c 72
with c 7 = 1 mol dm−3
...
6 mol dm−3, so [H2O]eq /c 7 = 55
...
If
we write K = K w /55
...
81 × 10−16, we obtain
1

τ

= k2{K + 2K1/2}c 7
w

22
...
7 × 10−5 s) × (2
...
4 × 1011 dm3 mol−1 s−1

It follows that
k1 = k2Kc 7 = 2
...
6 × 1012 dm3 mol−1 s−1
...


Self-test 22
...

[1/τ = k([A] + [B])eq + k′([C] + [D])eq]

22
...
Many reactions in solution fall somewhere in the range spanned by the hydrolysis of methyl
ethanoate (where the rate constant at 35°C is 1
...
13)
...
This behaviour is normally expressed mathematically by introducing
two parameters, one representing the intercept and the other the slope of the straight
line, and writing the Arrhenius equation
Ea

(22
...
22
...

The parameter Ea, which is obtained from the slope of the line (−Ea /R), is called the
activation energy
...
4)
...
4* Arrhenius parameters
(1) First-order reactions

A/s−1

Ea /(kJ mol−1)

CH3NC → CH3CN

3
...
94 × 1013

103
...
0 ×1010

42

NaC2H5O + CH3I in ethanol

2
...


11

81
...
22
...
29)
...


808

22 THE RATES OF CHEMICAL REACTIONS
Example 22
...
Find Ea and A
...
011

760

790

810

840

910

1000

0
...
105

0
...
789

2
...
0

145

Method According to eqn 22
...
As explained in Example 22
...
The intercept at T = 0 is ln(A/dm3 mol−1 s−1)
...
43

1
...
32

1
...
23

1
...
51 −3
...
25 −1
...
24

1
...
00

0
...
00

4
...
22
...
The least-squares fit is to a line with slope
−22
...
7
...
7 × (8
...
7 dm3 mol−1 s−1 = 1
...
0

1
...
2 1
...
4

A note on good practice Note that A has the same units as k
...

Self-test 22
...
22
...
5
...
9 × 10

6

3
...
9 × 10

7

1
...
2 ×108

[8 ×1010 dm3 mol−1 s−1, 23 kJ mol−1]

The fact that Ea is given by the slope of the plot of ln k against 1/T means that, the
higher the activation energy, the stronger the temperature dependence of the rate
constant (that is, the steeper the slope)
...
If a reaction has zero activation energy, its
rate is independent of temperature
...
We shall see that
such behaviour is a signal that the reaction has a complex mechanism
...
However, it is still possible to define an activation energy at any temperature as
Ea = RT 2

A d ln k D
C dT F

[22
...
However, the definition in eqn 22
...
29, because it allows Ea to be obtained from the slope (at the

temperature of interest) of a plot of ln k against 1/T even if the Arrhenius plot is not a
straight line
...
7f)
...
29 as
k = Ae−Ea /RT

Potential energy

22
...
31)

To interpret Ea we consider how the molecular potential energy changes in the
course of a chemical reaction that begins with a collision between molecules of A and
molecules of B (Fig
...
12)
...
The reaction coordinate is the collection of motions, such
as changes in interatomic distances and bond angles, that are directly involved in the
formation of products from reactants
...
) The potential energy
rises to a maximum and the cluster of atoms that corresponds to the region close
to the maximum is called the activated complex
...
The climax of the reaction is at the peak of the potential energy, which
corresponds to the activation energy Ea
...
This crucial configuration is called the transition state of
the reaction
...

We also conclude from the preceding discussion that, for a reaction involving the
collision of two molecules, the activation energy is the minimum kinetic energy that reactants must have in order to form products
...
The fraction of collisions with a kinetic energy in excess of an
energy Ea is given by the Boltzmann distribution as e−Ea /RT
...
31 as the fraction of collisions that have enough kinetic
energy to lead to reaction
...
Hence, the product of A and the exponential factor, e−Ea /RT,
gives the rate of successful collisions
...


Accounting for the rate laws
We now move on to the second stage of the analysis of kinetic data, their explanation
in terms of a postulated reaction mechanism
...
6 Elementary reactions
Most reactions occur in a sequence of steps called elementary reactions, each of
which involves only a small number of molecules or ions
...
22
...
The height of the
barrier between the reactants and products
is the activation energy of the reaction
...
3

The terms actiVated complex and
transition state are often used as
synonyms; however, we shall preserve a
distinction
...
This equation, for instance, signifies that an H atom attacks
a Br2 molecule to produce an HBr molecule and a Br atom
...
In a unimolecular reaction, a single molecule shakes itself apart or
its atoms into a new arrangement, as in the isomerization of cyclopropane to propene
...
It is most important
to distinguish molecularity from order: reaction order is an empirical quantity, and
obtained from the experimental rate law; molecularity refers to an elementary reaction proposed as an individual step in a mechanism
...
32)

where P denotes products (several different species may be formed)
...
(Ten times as many decay in
the same interval when there are initially 1000 A molecules as when there are only
100 present
...

An elementary bimolecular reaction has a second-order rate law:
A+B→P

d[A]
dt

= −k[A][B]

(22
...

Therefore, if we have evidence that a reaction is a single-step, bimolecular process, we
can write down the rate law (and then go on to test it)
...
There is evidence that the mechanism of this
reaction is a single elementary step
CH3I + CH3CH2O− → CH3OCH2CH3 + I−
This mechanism is consistent with the observed rate law
v = k[CH3I][CH3CH2O−]

(22
...
For the present we emphasize
that, if the reaction is an elementary bimolecular process, then it has second-order kinetics,
but if the kinetics are second-order, then the reaction might be complex
...
Detailed analysis of this kind was one of the ways, for example, in which the
reaction H2(g) + I2(g) → 2 HI(g) was shown to proceed by a complex mechanism
...
7 CONSECUTIVE ELEMENTARY REACTIONS

811

many years the reaction had been accepted on good, but insufficiently meticulous
evidence as a fine example of a simple bimolecular reaction, H2 + I2 → HI + HI, in
which atoms exchanged partners during a collision
...
7 Consecutive elementary reactions
Some reactions proceed through the formation of an intermediate (I), as in the consecutive unimolecular reactions
k

k

a
b
A −→ I −→ P

Comment 22
...
5 min 239

U −−−−→

239

dy

2
...
) We can discover the characteristics of this type of reaction
by setting up the rate laws for the net rate of change of the concentration of each
substance
...
39 is a special case of this
standard form, with f(x) = constant
...
35)

1

and A is not replenished
...
The net rate of formation of I is therefore

0
...
36)

The product P is formed by the unimolecular decay of I:
d[P]

[A]/[A]0

= kb[I]

0
...
4

(22
...
2

We suppose that initially only A is present, and that its concentration is [A]0
...
35, is an ordinary first-order decay, so we can write
[A] = [A]0e−kat

(22
...
36, we obtain after rearrangement
d[I]
dt

+ kb[I] = ka[A]0e−kat

(22
...
40)

At all times [A] + [I] + [P] = [A]0, so it follows that
1

[P] = 2 1 +
3

kae−kbt − kbe−kat 5
k b − ka

6 [A]0
7

(22
...
22
...
The concentration of the product P rises from zero towards [A]0
...
5

1 1
...
5 3

Fig
...
13 The concentrations of A, I, and P
in the consecutive reaction scheme A → I
→ P
...
38,
22
...
41 with ka = 10kb
...
6
...
22
...
01
...
22
...


812

22 THE RATES OF CHEMICAL REACTIONS
Example 22
...
At what time will I be present in greatest concentration?
Method The time-dependence of the concentration of I is given by eqn 22
...
We

can find the time at which [I] passes through a maximum, tmax, by calculating
d[I]/dt and setting the resulting rate equal to zero
...
40 that

d[I]
dt

=−

ka[A]0(kae−k at − kbe−k bt )
k b − ka

This rate is equal to zero when
kae−kat = kbe−k bt
Therefore,
tmax =

1
ka − k b

ln

ka
kb

For a given value of ka, as k b increases both the time at which [I] is a maximum and
the yield of I decrease
...
7 Calculate the maximum concentration of I and justify the last remark
...
A reaction scheme involving many steps is nearly
always unsolvable analytically, and alternative methods of solution are necessary
...
An alternative
approach, which continues to be widely used because it leads to convenient expressions and more readily digestible results, is to make an approximation
...
22
...
22
...
It is supposed that the
concentrations of intermediates remain
small and hardly change during most of the
course of the reaction
...
42)

This approximation greatly simplifies the discussion of reaction schemes
...
36, which then becomes
ka[A] − kb[I] ≈ 0
Then
[I] ≈ (ka/k b)[A]

(22
...
7 CONSECUTIVE ELEMENTARY REACTIONS
For this expression to be consistent with eqn 22
...

On substituting this value of [I] into eqn 22
...
8

= k b[I] ≈ ka[A]

(22
...
We can write down the solution of this
equation at once by substituting the solution for [A], eqn 22
...
6

0
...
45)

0

This is the same (approximate) result as before, eqn 22
...
Figure 22
...


0
...
7 Using the steady-state approximation

Devise the rate law for the decomposition of N2O5,
2 N2O5(g) → 4 NO2(g) + O2(g)
on the basis of the following mechanism:
N2O5 → NO2 + NO3

ka

NO2 + NO3 → N2O5

ka
′

NO2 + NO3 → NO2 + O2 + NO

kb

NO + N2O5 → NO2 + NO2 + NO2

kc

A note on good practice Note that when writing the equation for an elementary

reaction all the species are displayed individually; so we write A → B + B, for
instance, not A → 2 B
...
Then, all net rates of change of the concentrations of intermediates
are set equal to zero and the resulting equations are solved algebraically
...
22
...
(The curve for [A]
is unchanged
...
8 Derive the rate law for the decomposition of ozone in the reaction

2 O3(g) → 3 O2(g) on the basis of the (incomplete) mechanism

Reactants

Products

Slow

ka
k′
a
kb

Fast

Fast

O3 → O2 + O
O2 + O → O3
O + O3 → O2 + O2

Slow

[d[O3]/dt = −kak b[O3]2/(k′ [O2] + k b[O3])]
a

(a)

(c) The rate-determining step

(b)
Fast

(c)

Slow

Fast

Potential energy

Fig
...
16 In these diagrams of reaction
schemes, heavy arrows represent fast steps
and light arrows represent slow steps
...


Equation 22
...
That is, the rate of formation of
P depends on the rate at which I is formed, not on the rate at which I changes into
P
...

Its existence has been likened to building a six-lane highway up to a single-lane bridge:
the traffic flow is governed by the rate of crossing the bridge
...

However, the rate-determining step is not just the slowest step: it must be slow and be
a crucial gateway for the formation of products
...
22
...

The rate law of a reaction that has a rate-determining step can often be written
down almost by inspection
...
Figure 22
...
Once over the initial barrier, the intermediates cascade into products
...

(d) Kinetic and thermodynamic control of reactions

RDS

Fast

Fast

Slow

Progress of reaction
Fig
...
17 The reaction profile for a
mechanism in which the first step (RDS) is
rate-determining
...
Suppose two products, P1 and P2, are produced by the following competing reactions:
A + B → P1

Rate of formation of P1 = k1[A][B]

A + B → P2

Rate of formation of P2 = k2[A][B]

The relative proportion in which the two products have been produced at a given
stage of the reaction (before it has reached equilibrium) is given by the ratio of the two
rates, and therefore of the two rate constants:

22
...
46)

k1

This ratio represents the kinetic control over the proportions of products, and is a
common feature of the reactions encountered in organic chemistry where reactants
are chosen that facilitate pathways favouring the formation of a desired product
...

(e) Pre-equilibria

From a simple sequence of consecutive reactions we now turn to a slightly more complicated mechanism in which an intermediate I reaches an equilibrium with the reactants A and B:
A+B5I→P

(22
...
This scheme involves a pre-equilibrium, in which an
intermediate is in equilibrium with the reactants
...
Because we assume that A, B, and I are in equilibrium, we can write
> a
K=

[I]
[A][B]

K=

ka

(22
...

The rate of formation of P may now be written:
d[P]
dt

= k b[I] = k bK[A][B]

(22
...
50)

Example 22
...

Method Begin by writing the net rates of change of the concentrations of the sub-

stances and then invoke the steady-state approximation for the intermediate I
...

Answer The net rates of change of P and I are

d[P]
dt
d[I]
dt

= k b[I]
= ka[A][B] − k′ [I] − k b[I] ≈ 0
a

815

816

22 THE RATES OF CHEMICAL REACTIONS
The second equation solves to
[I] ≈

ka[A][B]
k′a + k b

When we substitute this result into the expression for the rate of formation of P,
we obtain
d[P]
dt

≈ k[A][B]

k=

kak b
k′a + k b

This expression reduces to that in eqn 22
...

Self-test 22
...

[d[P]/dt = k bK[A]2[B]]

Potential energy

(f) The kinetic isotope effect

C-H
C-D

Ea(C-H)
Ea(C-D)

Reaction coordinate
Fig
...
18 Changes in the reaction profile
when a C-H bond undergoing cleavage is
deuterated
...
The only significant
change is in the zero-point energy of the
reactants, which is lower for C-D than for
C-H
...


The postulation of a plausible mechanism requires careful analysis of many experiments designed to determine the fate of atoms during the formation of products
...
A primary kinetic
isotope effect is observed when the rate-determining step requires the scission of a
bond involving the isotope
...
In both cases, the effect arises from the change in activation energy that accompanies the replacement of an atom by a heavier isotope on account of changes in the
zero-point vibrational energies (Section 13
...

First, we consider the origin of the primary kinetic isotope effect in a reaction in
which the rate-determining step is the scission of a C-H bond
...
22
...
On deuteration, the dominant change is the reduction of the
zero-point energy of the bond (because the deuterium atom is heavier)
...

We assume that, to a good approximation, a change in the activation energy arises
only from the change in zero-point energy of the stretching vibration, so
1
1
Ea(C-D) − Ea(C-H) = NA{– hc#(C-H) − – hc#(C-D)}
2
2

(22
...
From Section 13
...
It follows that
1/2
1
A µ CH D 5
1
6
Ea(C-D) − Ea(C-H) = – NAhc#(C-H) 2 1 −
(22
...
7 CONSECUTIVE ELEMENTARY REACTIONS
k(C-D)
k(C-H)

−λ

=e

A µ CH D
hc#(C-H) 1
21 −
with λ =
C µ CD F
2kT
3

817

1/2 5

6
7

(22
...


From infrared spectra, the fundamental vibrational wavenumber for stretching of
a C-H bond is about 3000 cm−1
...
538 and eqn 22
...
145 at 298 K
...
However, experimental values of k(C-D)/k(C-H) can differ
significantly from those predicted by eqn 22
...


In some cases, substitution of deuterium for hydrogen results in values of k(C-D)/
k(C-H) that are too low to be accounted for by eqn 22
...
Such abnormal kinetic isotope
effects are evidence for a path in which quantum mechanical tunnelling of hydrogen
atoms takes place through the activation barrier (Fig
...
19)
...
3
that the probability of tunnelling through a barrier decreases as the mass of the particle increases, so deuterium tunnels less efficiently through a barrier than hydrogen
and its reactions are correspondingly slower
...
We shall see in Chapter 23 that, because me is so small, tunnelling
is also a very important contributor to the rates of electron transfer reactions
...
The activation energy of the undeuterated compound is
Ea(H) = Ea + E ‡ (H) − E vib,0(H)
vib,0

Potential energy

Illustration 22
...
22
...

The effect is represented by drawing the
wavefunction of the proton near the
barrier
...


where Ea is the difference between the minima of the molecular potential energy
curves of the activated complex and the ground state of the reactant and E ‡ (H) and
vib,0
Evib,0(H) are the zero-point vibrational energies of the two states (Fig
...
20)
...
54)

where #‡ and # denote vibrational wavenumbers in the activated complex and reactant,
respectively
...


Fig
...
20

818

22 THE RATES OF CHEMICAL REACTIONS

Ea(D) − Ea(H) =

1A µ

1
– NAhc{#‡(C-H) − #(C-H)} 2
2

CH D

1/2

3 C µ CD F

5
− 16
7

(22
...
56)

Because µ CH /µ CD < 1, provided the vibrational wavenumber of the activated complex
is less than that of the reactant, λ > 1 and the deuterated form reacts more slowly than
the undeuterated compound
...
3 Assessing the secondary kinetic isotope effect

In the heterolytic dissociation CHCl3 → CHCl + + Cl− the activated complex
2
resembles the product CHCl +
...
Assuming that #‡(C-H) = 800 cm−1 on
2
account of the structural similarity between CHCl + and the activated complex,
2
it follows from µ CH / µ CD = 0
...
56 that k(D)/k(H) = 0
...

We predict that at room temperature the dissociation of CHCl3 should be about
40 per cent faster than dissociation of CDCl3
...
2 shows that the secondary kinetic isotope effect leads to higher
values of k(D)/k(H) than does the primary kinetic isotope effect
...


IMPACT ON BIOCHEMISTRY

I22
...
1 that a simple statistical model accounts for the thermodynamic
aspects of the helix–coil transition in polypeptides
...
Here we examine the kinetics of the helix–coil transition,
focusing primarily on experimental strategies and some recent results
...
In a typical stopped-flow
experiment, a sample of the protein with a high concentration of a chemical denaturant, such as urea or guanidinium hydrochloride, is mixed with a solution containing
a much lower concentration of the same denaturant
...
Unfolding is observed by
mixing a sample of folded protein with a solution containing a high concentration of
denaturant
...
However, the available data also indicate that, in a number of proteins, a significant portion of the folding process occurs in less than 1 ms, a time range not accessible
by the stopped-flow technique
...
For example, at ambient temperature the
formation of a loop between helical or sheet segments may be as fast as 1 µs and
the formation of tightly packed cores with significant tertiary structure occurs in the

I22
...
Among the fastest events are the formation and denaturation of
helices and sheets from fully unfolded peptide chains and we examine how the laserinduced temperature-jump technique has been used in the study of the helix–coil
transition
...
10)
...
Hence, a
temperature-jump experiment can be configured to monitor either folding or unfolding of a polypeptide, depending on the initial and final temperatures of the sample
...
A number of clever strategies have been employed
...
Another variation makes use of direct
excitation of overtones of the O-H or O-D stretching modes of H2O or D2O, respectively, with a pulsed infrared laser
...
Relaxation of the sample
can then be probed by a variety of spectroscopic techniques, including absorption,
emission, or Raman scattering
...

Much of the kinetic work on the helix–coil transition has been conducted in small
synthetic polypeptides rich in alanine, an aminoacid that is known to stabilize helical
structures
...
Using h and c to denote an aminoacid residue
belonging to a helical and coil region, respectively, the mechanism may be summarized as follows:
hhhh
...


very fast

chhh
...


rate-determining step

The rate-determining step is thought to account for the relaxation time of 160 ns
measured with a laser-induced temperature jump between 282
...
6 K in an
alanine-rich polypeptide containing 21 amino acids
...
7 kJ mol−1 associated with nucleation events of the form
...

→
...
in the middle of the chain
...
Models that use concepts of
statistical thermodynamics also suggest that a hhhh
...
transition at either
end of a helical segment has a significantly lower activation energy on account of the
converting aminoacid not being flanked by h regions
...

In apomyoglobin (myoglobin lacking the haem cofactor), the unfolding of the helices
appears to have a relaxation time of about 50 ns, even shorter than in synthetic peptides
...


819

820

22 THE RATES OF CHEMICAL REACTIONS
22
...
57)

The problem with the interpretation of first-order rate laws is that presumably a
molecule acquires enough energy to react as a result of its collisions with other
molecules
...
This term must be used with caution,
though, because the overall mechanism has bimolecular as well as unimolecular steps
...
22
...
The species A is
excited by collision with A, and the excited
A molecule (A*) may either be deactivated
by a collision with A or go on to decay by a
unimolecular process to form products
...
In the Lindemann–
Hinshelwood mechanism it is supposed that a reactant molecule A becomes energetically excited by collision with another A molecule (Fig
...
21):
d[A*]

A + A → A* + A

dt

= ka[A]2

(22
...
59)

Alternatively, the excited molecule might shake itself apart and form products P
...
60)

If the unimolecular step is slow enough to be the rate-determining step, the overall
reaction will have first-order kinetics, as observed
...
61)

This equation solves to
[A*] =

ka[A]2

(22
...
63)

At this stage the rate law is not first-order
...
8 UNIMOLECULAR REACTIONS

821

then we can neglect k b in the denominator and obtain
d[P]

= k[A]

dt

k=

kak b

(22
...
64 is a first-order rate law, as we set out to show
...
Thus, when k′ [A] < kb, the rate law in
<
a
eqn 22
...
65)

The physical reason for the change of order is that at low pressures the rate-determining
step is the bimolecular formation of A*
...
63 as
d[P]

= k[A]

dt

k=

kak b[A]
k b + k′ [A]
a

(22
...
67)

ka[A]

Hence, a test of the theory is to plot 1/k against 1/[A], and to expect a straight line
...
68a)

where s is the number of modes of motion over which the energy may be dissipated
and E* is the energy required for the bond of interest to break, we can write the Kassel
form of the unimolecular rate constant for the decay of A* to products as

A
C

1

s−1

EF

k b(E) = 1 −

2

(10-4 s-1)/k

Whereas the Lindemann–Hinshelwood mechanism agrees in general with the switch
in order of unimolecular reactions, it does not agree in detail
...
22 shows
a typical graph of 1/k against 1/[A]
...

An improved model was proposed in 1926 by O
...
Rice and H
...
Ramsperger and
almost simultaneously by L
...
Kassel, and is now known as the Rice–Ramsperger–
Kassel model (RRK model)
...
A
...
Here we outline Kassel’s original approach to the RRK model:
the details are set out in Further information 22
...
The essential feature of the model is that, although a molecule might have enough energy to
react, that energy is distributed over all the modes of motion of the molecule, and
reaction will occur only when enough of that energy has migrated into a particular
location (such as a bond) in the molecule
...
1 is

E* D
EF

s−1

kb

for E ≥ E*

where k b is the rate constant used in the original Lindemann theory
...
68b)

0

0

0
...
5
1
(10-3 mol dm-3)/[A]

2

Fig
...
22 The pressure dependence of the
unimolecular isomerization of transCHD=CHD showing a pronounced
departure from the straight line predicted
by eqn 22
...


822

22 THE RATES OF CHEMICAL REACTIONS
The energy dependence of the rate constant given by eqn 22
...
22
...
We see that the rate constant is smaller at a given
excitation energy if s is large, as it takes longer for the excitation energy to migrate
through all the oscillators of a large molecule and accumulate in the critical mode
...


1

5

kb(E)/kb

0
...
6

0
...
2

Although the rate of each step of a complex mechanism might increase with temperature and show Arrhenius behaviour, is that true of a composite reaction? To answer
this question, we consider the high-pressure limit of the Lindemann–Hinshelwood
mechanism as expressed in eqn 22
...
If each of the rate constants has an Arrheniuslike temperature dependence, we can use eqn 22
...
68 for three values
of s
...
22
...
69)

a
(A′ae−E′(a)/RT )

e−{Ea(a)+Ea(b)−E′a(a)}/RT

That is, the composite rate constant k has an Arrhenius-like form with activation energy
Ea = Ea(a) + Ea(b) − E′ (a)
a

(22
...
However, it is conceivable that Ea(a) + Ea(b) < E′ (a) (Fig
...
24), in
a
which case the activation energy is negative and the rate will decrease as the temperature is raised
...
The Lindemann–Hinshelwood mechanism is an unlikely
candidate for this type of behaviour because the deactivation of A* has only a small
activation energy, but there are reactions with analogous mechanisms in which a
negative activation energy is observed
...
63, it is clear that the temperature dependence may be difficult to predict because each rate constant in the

(a)

E' (a)
a
Ea(b)

Reaction coordinate

Ea(a)
Potential energy

Fig
...
24 For a reaction with a preequilibrium, there are three activation
energies to take into account, two referring
to the reversible steps of the preequilibrium and one for the final step
...


Potential energy

Ea(a)

(b)

E' (a)
a

Ea(b)

Reaction coordinate

FURTHER READING

823

expression for k increases with temperature, and the outcome depends on whether the
terms in the numerator dominate those in the denominator, or vice versa
...
65 rather than
eqn 22
...


Checklist of key ideas
1
...
Examples include real-time and
quenching procedures, flow and stopped-flow techniques,
and flash photolysis
...
The larger the activation energy, the
more sensitive the rate constant is to the temperature
...
The mechanism of reaction is the sequence of elementary
steps involved in a reaction
...
The instantaneous rate of a reaction is the slope of the tangent
ot the graph of concentration against time (expressed as a
positive quantity)
...
The molecularity of an elementary reaction is the number of
molecules coming together to react
...


3
...


13
...


4
...
, the rate constant
is k, the order with respect to A is a, and the overall order is
a+b+
...
An integrated rate law is an expression for the concentration
of a reactant or product as a function of time (Table 22
...

6
...
The
time constant τ is the time required for the concentration of a
reactant to fall to 1/e of its initial value
...

7
...

8
...

9
...

10
...
In the steady-state approximation, it is assumed that the
concentrations of all reaction intermediates remain constant
and small throughout the reaction
...
Provided a reaction has not reached equilibrium, the products
of competing reactions are controlled by kinetics, with
[P2]/[P1] = k2/k1
...
Pre-equilibrium is a state in which an intermediate is in
equilibrium with the reactants and which arises when the rates
of formation of the intermediate and its decay back into
reactants are much faster than its rate of formation of
products
...
The kinetic isotope effect is the decrease in the rate of a
chemical reaction upon replacement of one atom in a reactant
by a heavier isotope
...
A secondary kinetic isotope
effect is the reduction in reaction rate even though the bond
involving the isotope is not broken to form product
...
The Lindemann–Hinshelwood mechanism and the RRKM
model of ‘unimolecular’ reactions account for the first-order
kinetics of gas-phase reactions
...
Andraos, A streamlined approach to solving simple and complex
kinetic systems analytically
...
Chem
...
76, 1578 (1999)
...
H
...
F
...
G
...
), Comprehensive
chemical kinetics
...
1–38, Elsevier, Amsterdam (1969–2001)
...
N
...
M
...
Martinho, Integration of kinetic
rate equations by matrix methods
...
Chem
...
67, 375 (1990)
...
M
...
Chem
...
76, 275 (1999)
...
C
...
E
...
L
...
), Encyclopedia of
spectroscopy and spectrometry
...

S
...
Logan, Fundamentals of chemical kinetics
...

M
...
Pilling and P
...
Seakins, Reaction kinetics
...

J
...
Steinfeld, J
...
Francisco, and W
...
Hase, Chemical kinetics and
dynamics
...


Sources of data and information

NDRL/NIST solution kinetics database, NIST standard reference
database 40, National Institute of Standards and Technology,
Gaithersburg (1994)
...

NIST chemical kinetics database, NIST standard reference database
17, National Institute of Standards and Technology, Gaithersburg
(1998)
...


Further information
Further information 22
...
In
practice, of course, the vibrational modes of a molecule have different
frequencies, but assuming that they are all the same is a good first
approximation
...

We can represent the n quanta as follows:
ᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀᮀ
ᮀ
...
One such
distribution is

walls we used above)
...
For example, in a system of five oscillators
(other than the critical one) we might suppose that at least 6 quanta
out of the 28 available must be present in the critical oscillator, then
all the following partitions will result in dissociation:
ᮀᮀᮀᮀᮀᮀ|ᮀᮀᮀᮀᮀ|ᮀᮀᮀᮀᮀᮀᮀᮀ|ᮀᮀᮀᮀ||ᮀᮀᮀᮀᮀᮀ
ᮀᮀᮀᮀᮀᮀᮀ|ᮀᮀᮀᮀ|ᮀᮀᮀᮀᮀᮀᮀᮀ|ᮀᮀᮀᮀ||ᮀᮀᮀᮀᮀᮀ
ᮀᮀᮀᮀᮀᮀᮀᮀ|ᮀᮀᮀ|ᮀᮀᮀᮀᮀᮀᮀᮀ|ᮀᮀᮀᮀ||ᮀᮀᮀᮀᮀᮀ
···
(The leftmost partition is the critical oscillator
...
ᮀ|ᮀᮀ
The total number of arrangements of each quantum and wall
(of which there are n + s −1 in all) is (n + s − 1)! where, as usual,
x! = x(x − 1)!
...
However the n! arrangements of the n quanta are
indistinguishable, as are the (s −1)! arrangements of the s − 1 walls
...
It follows that
N=

(n + s − 1)!
n!(s − 1)!

(22
...
If we suppose
that a bond will break if it is excited to at least an energy E* = n*hν,
then the number of ways in which at least this energy can be localized
in one bond is
N* =

(n − n* + s − 1)!
(n − n*)!(s − 1)!

ᮀ|ᮀᮀᮀᮀ|ᮀᮀᮀᮀᮀᮀᮀᮀ|ᮀᮀᮀᮀ||ᮀᮀᮀᮀᮀ
ᮀᮀ|ᮀᮀᮀ|ᮀᮀᮀᮀᮀᮀᮀᮀ|ᮀᮀᮀᮀ||ᮀᮀᮀᮀᮀ
···

ᮀᮀᮀᮀᮀᮀ

and we see that we have the problem of permuting 28 − 6 = 22 (in
general, n − n*) quanta and 5 (in general, s − 1) walls, and therefore
a total of 27 (in general, n − n* + s − 1 objects)
...
The number N* is therefore
obtained from eqn 22
...

From the preceding discussion we conclude that the probability
that one specific oscillator will have undergone sufficient excitation to
dissociate is the ratio N*/N, which is
P=

N*
N

=

n!(n − n* + s − 1)!

(22
...
73 is still awkward to use, even when written out in
terms of its factors:

(22
...
1
(n − n*)(n − n* − 1)
...
1
(n + s − 1)(n + s − 2)
...
(n − n* + 1)
(n + s − 1)(n + s − 2)
...
(n − n*)s −1 factors
(n)(n)
...
68a
...
68b
...
1 Consult literature sources and list the observed timescales during which
the following processes occur: radiative decay of excited electronic states,
molecular rotational motion, molecular vibrational motion, proton transfer
reactions, the initial event of vision, energy transfer in photosynthesis, the
initial electron transfer events in photosynthesis, the helix-to-coil transition in
polypeptides, and collisions in liquids
...
5 Assess the validity of the following statement: the rate-determining step
is the slowest step in a reaction mechanism
...
2 Write a brief report on a recent research article in which at least one of
the following techniques was used to study the kinetics of a chemical reaction:
stopped-flow techniques, flash photolysis, chemical quench-flow methods,
freeze quench methods, temperature-jump methods, or pressure-jump
methods
...


22
...


22
...

22
...


22
...

22
...


22
...


Discuss how kinetic isotope effects in general can provide insight into the
mechanism of a reaction
...
10 Discuss the limitations of the generality of the expression k =
kakb[A]/(kb + k′ [A]) for the effective rate constant of a unimolecular reaction
a
A → P with the following mechanism: A + A 5 A* + A (ka, k′ ), A* → P (kb)
...


Exercises
22
...
0 mol dm−3 s−1
...

22
...
0 mol
dm−3 s−1
...

22
...
0 mol dm s
...

22
...
0 mol dm−3 s−1
...


22
...
1a was found to be v =

k[A][B]
...

22
...
1b was found to be
v = k[A][B]2
...

22
...
2a was reported as
d[C]/dt = k[A][B][C]
...
4b The rate law for the reaction in Exercise 22
...
Express the rate law in terms of the reaction rate; what are the
units for k in each case?

22
...
07 Torr s−1 when 5
...
76 Torr s−1 when 20
...
Determine
the order of the reaction
...
5b At 400 K, the rate of decomposition of a gaseous compound initially
at a pressure of 12
...
71 Pa s−1 when 10
...
67 Pa s−1 when 20
...
Determine the order of the
reaction
...
6a At 518°C, the half-life for the decomposition of a sample of gaseous
acetaldehyde (ethanal) initially at 363 Torr was 410 s
...
Determine the order of the reaction
...
6b At 400 K, the half-life for the decomposition of a sample of a gaseous
compound initially at 55
...
When the pressure was 28
...
Determine the order of the reaction
...
7a The rate constant for the first-order decomposition of N2O5 in the
reaction 2 N2O5(g) → 4 NO2(g) + O2(g) is k = 3
...
What is
the half-life of N2O5? What will be the pressure, initially 500 Torr, at (a) 10 s,
(b) 10 min after initiation of the reaction?
22
...
78 × 10−7 s−1 at 25°C
...
1 kPa, at (a) 10 s, (b) 10 min after
initiation of the reaction?

826

22 THE RATES OF CHEMICAL REACTIONS

22
...
050 mol dm−3 in A and 0
...
After
1
...
020 mol dm−3
...
(b) What is the half-life of the reactants?
22
...
075 mol dm−3 in A and 0
...
After
1
...
045 mol dm−3
...
(b) What is the half-life of the reactants?
22
...
9b If the rate laws are expressed with (a) concentrations in molecules per

metre cubed, (b) pressures in newtons per metre squared, what are the units
of the second-order and third-order rate constants?
22
...
11 dm3 mol−1 s−1
...
050 mol dm−3 and [CH3COOC2H5] =
0
...
12b Deduce an expression for the time it takes for the concentration of a
substance to fall to one-third its initial value in an nth-order reaction
...
13a The pKa of NH4 is 9
...
The rate constant at 25°C for the

+
reaction of NH4 and OH− to form aqueous NH3 is 4
...

Calculate the rate constant for proton transfer to NH3
...
15 mol dm−3 NH3(aq) at 25°C?

22
...
The measured
relaxation time is 3
...
The equilibrium constant for the system is 2
...
0 × 10−4 mol dm−3
...

22
...
80 × 10−3 dm3 mol−1 s−1 at 30°C and 1
...

Evaluate the Arrhenius parameters of the reaction
...
14b The rate constant for the decomposition of a certain substance is
1
...
01 × 10−2 dm3 mol−1 s−1 at 37°C
...

22
...
3 times more slowly than the bromination of
the undeuterated material
...
Use kf (C-H) = 450 N m−1
...
10b The second-order rate constant for the reaction A + 2 B → C + D is
0
...
What is the concentration of C after (a) 10 s, (b) 10 min
when the reactants are mixed with initial concentrations of [A] = 0
...
150 mol dm−3?

22
...
11a A reaction 2 A → P has a second-order rate law with k = 3
...
16a The effective rate constant for a gaseous reaction that has a

dm3 mol−1 s−1
...
260 mol dm−3 to 0
...


22
...
50 × 10−4 dm6

mol−2 s−1
...
077 mol dm−3 to 0
...

22
...


rates of displacement of (a) 1H and 3H, (b) 16O and 18O
...

Lindemann–Hinshelwood mechanism is 2
...
30 kPa and
2
...
Calculate the rate constant for the activation step in
the mechanism
...
16b The effective rate constant for a gaseous reaction that has a
Lindemann–Hinshelwood mechanism is 1
...
09 kPa and
2
...
Calculate the rate constant for the activation step
in the mechanism
...
1 The data below apply to the formation of urea from ammonium cyanate,
NH4CNO → NH2CONH2
...
9 g of ammonium cyanate was
dissolved in in enough water to prepare 1
...
Determine the
order of the reaction, the rate constant, and the mass of ammonium cyanate
left after 300 min
...
0

50
...
0

150

m(urea)/g

0

7
...
1

13
...
7

22
...
Determine the order of the reaction, the rate constant, and the molar
concentration of (CH3)3CBr after 43
...


t/h

0

[(CH3)3CBr]/(10 −2 mol dm−3) 10
...
00 4
...
00 8
...
10

0
...
67

0
...
00

0
...
32

12
...
25

0

Determine the order of the reaction and the rate constant
...
4 The following data have been obtained for the decomposition of
N2O5(g) at 67°C according to the reaction 2 N2O5(g) → 4 NO2(g) + O2(g)
...
It is
not necessary to obtain the result graphically, you may do a calculation using
estimates of the rates of change of concentration
...
000

0
...
497

0
...
246

0
...
15 6
...
00

18
...
80

22
...
Estimate the activation energy
...
96

3
...
07

k/(10 −3 s −1)

2
...
1

576

θ /°C

0

20
...
0

7
...
39

22
...


PROBLEMS
22
...
13 Show that the following mechanism can account for the rate law of the
reaction in Problem 22
...
74 s−1

HCl + HCl 5 (HCl)2

(2) CH3COOH → H2C=C=O + H2O

k2 = 4
...
7 Sucrose is readily hydrolysed to glucose and fructose in acidic solution
...
14 Consider the dimerization 2 A 5 A2, with forward rate constant ka and

The hydrolysis is often monitored by measuring the angle of rotation of planepolarized light passing through the solution
...
An experiment on the hydrolysis
of sucrose in 0
...
316 0
...
274 0
...
238 0
...
190 0
...
146

Determine the rate constant of the reaction and the half-life of a sucrose
molecule
...
8 The composition of a liquid-phase reaction 2 A → B was followed by a
spectrophotometric method with the following results:

t/min

0

10

20

30

40

∞

[B]/(mol dm−3)

0

0
...
153

0
...
230

0
...

22
...

The following data have been obtained:

t/(10−3 s)

0
...
62 0
...
60

[ClO]/(10− 6 mol dm−3) 8
...
09

7
...
79

3
...
00 5
...
20

4
...
95

Determine the rate constant of the reaction and the half-life of a ClO radical
...
10 Cyclopropane isomerizes into propene when heated to 500°C in the gas
phase
...

What are the order and rate constant for the reaction under these conditions?
22
...
In one study (M
...

Haugh and D
...
Dalton, J
...
Chem
...
97, 5674 (1975)), high pressures
of hydrogen chloride (up to 25 atm) and propene (up to 5 atm) were
examined over a range of temperatures and the amount of 2-chloropropane
formed was determined by NMR
...
In a series
of runs the ratio of [chloropropane] to [propene] was independent of
[propene] but the ratio of [chloropropane] to [HCl] for constant amounts of
propene depended on [HCl]
...
05, 0
...
01 for p(HCl) =
10 atm, 7
...
0 atm, respectively
...
12 Use mathematical software or an electronic spreadsheet to examine the
time dependence of [I] in the reaction mechanism A → I → P (k1, k2)
...
39 numerically (see Appendix 2) or use eqn 22
...
In all the following calculations, use [A]0 = 1 mol dm−3 and a time
range of 0 to 5 s
...
(b) Increase
the ratio k2/k1 steadily by decreasing the value of k1 and examine the plot of [I]
against t at each turn
...
(a) Derive the following expression for the relaxation
time in terms of the total concentration of protein, [A]tot, = [A] + 2[A2]:
1

τ2

= k 2 + 8kak b[A]tot
b

(b) Describe the computational procedures that lead to the determination of
the rate constants ka and kb from measurements of τ for different values of
[A]tot
...
500

0
...
251

0
...
101

τ /ns

2
...
7

3
...
0

5
...
15 In the experiments described in Problems 22
...
13 an inverse
temperature dependence of the reaction rate was observed, the overall rate of
reaction at 70°C being roughly one-third that at 19°C
...

22
...
Atkinson and J
...
Pitts,
J
...
Chem
...
In the reaction with benzene the rate constants
are 1
...
3 K, 3
...
2 K,
and 6
...
2 K
...

22
...
10 the isomerization of cyclopropane over a limited
pressure range was examined
...
These have been
obtained (H
...
Pritchard, R
...
Sowden, and A
...
Trotman-Dickenson, Proc
...
Soc
...
1

11
...
89

0
...
120

0
...
98

2
...
54

0
...
392

0
...

22
...
W
...
J
...
T
...
Gutman, and L
...


Krasnoperov (J
...
Chem
...
The reaction is bimolecular in both directions with Arrhenius
parameters A = 1
...
2 kJ mol−1 for the forward
reaction and k′ = 1
...
3 kJ mol−1 for the reverse
reaction
...

m
22
...
The activation energy for the reaction leading to
Product 1 is greater than that leading to Product 2
...
20 The reaction mechanism

A2 5 A + A
A+B→P

(fast)
(slow)

involves an intermediate A
...


828

22 THE RATES OF CHEMICAL REACTIONS

22
...
Derive an
expression for the concentration of A as a function of time when the initial
molar concentrations of A and B are [A]0 and [B]0
...

t/min

30

60

120

150

240

360

480

22
...


22
...


22
...
25 Show that the ratio t1/2 /t3/4, where t1/2 is the half-life and t3/4 is the time

3
for the concentration of A to decrease to – of its initial value (implying that
4
t3/4 < t1/2) can be written as a function of n alone, and can therefore be used
as a rapid assessment of the order of a reaction
...
26 Derive an equation for the steady-state rate of the sequence of reactions
A 5 B 5 C 5 D, with [A] maintained at a fixed value and the product D
removed as soon as it is formed
...
27‡ For a certain second-order reaction A + B → Products, the rate of
reaction, v, may be written

dx
dt

= k([A]0 − x)([B]0 + x)

where x is the decrease in concentration of A or B as a result of reaction
...
Draw a graph of v against x, and noting that v and x cannot be
negative, identify the portion of the curve that corresponds to reality
...
28 Consider the dimerization A ⇔ A2 with forward rate constant ka and
backward rate constant kb
...


v=

22
...
Comment
...


1
kb + 4ka[A]eq

Applications: to archaeology, biochemistry, and
environmental science
14

22
...
16 MeV)
...
What is its age?
22
...
This nuclide emits
β rays of energy 0
...
1 y
...
00 µg was
absorbed by a newly born child
...
31 Pharmacokinetics is the study of the rates of absorption and
elimination of drugs by organisms
...
A drug can be eliminated by many mechanisms, such as
metabolism in the liver, intestine, or kidney followed by excretion of
breakdown products through urine or faeces
...


B
drug dispersed
in blood

→

C
eliminated
drug

where the rate constants of absorption (A → B) and elimination are,
respectively, k1 and k2
...
Also, assume that elimination follows first-order kinetics
...
Also, write a
mathematical expression for the residual concentration of B, [R]n, which
we define to be the concentration of drug B immediately before the
administration of the (n + 1)th dose
...
Show that [P]∞ – [R]∞ = [B]0
...
0289 h−1
...

Prepare a graph that plots both [P]n /[B]0 and [R]n /[B]0 against n
...
Show that with the
initial concentration [B]0 = 0, the concentration of drug in the blood is
given by
[B] = [A]0(1 − e−k1t) − k2t
Plot [B]/[A]0 against t for the case k1 = 10 h−1, k2 = 4
...
1 mmol dm−3
...
(d) Using
the model from part (c), set d[B]/dt = 0 and show that the maximum value of
1 A k1[A]0 D
[B] occurs at the time tmax = ln B
E
...

22
...
5 hchh
...
5 cccc
...
1 that this type of nucleation is relatively slow, so neither
step may be rate-determining
...
(b) Apply the steady-state approximation and show that, under
these circumstances, the mechanism is equivalent to hhhh
...


PROBLEMS

829

(c) Use your knowledge of experimental techniques and your results from
parts (a) and (b) to support or refute the following statement: It is very
difficult to obtain experimental evidence for intermediates in protein folding
by performing simple rate measurements and one must resort to special flow,
relaxation, or trapping techniques to detect intermediates directly
...
C
...
L
...
R
...
Phys
...
A 101, 3125 (1997)) measured the rate constants for the elementary
bimolecular gas-phase reaction of methane with the hydroxyl radical over a
range of temperatures of importance to atmospheric chemistry
...
34 Propose a set of experiments in which analysis of the line-shapes of
NMR transitions (Section 15
...
What are the disadvantages and disadvantages of this
NMR method over methods that use electronic and vibrational spectroscopy?

T/K

22
...

22
...

To qualify, a reaction must proceed with favourable rates and equilibria
...
P
...
I
...
Amino acid analogues can be formed from HMU under prebiotic
conditions by reaction with various nucleophiles, such as H2S, HCN, indole,
imidazole, etc
...
75 −5488/(T/K) (at pH = 7), and log K = −1
...

For this reaction, calculate the rates and equilibrium constants over a range of
temperatures corresponding to possible prebiotic conditions, such as 0–50°C,
and plot them against temperature
...
Prebiotic conditions
are not likely to be standard conditions
...
Do you expect that the reaction would still be favourable?
22
...
Reaction with the hydroxyl radical OH is the main path by
which CH4 is removed from the lower atmosphere
...
Gierczak, R
...


295

223

k/(10 6 dm3 mol −1 s−1) 3
...
494

218

213

206

0
...
379 0
...
241 0
...
38‡ As we saw in Problem 22
...
T
...
K
...
C
...
L
...
R
...
Phys
...
A 101,
3125 (1997)) measured the rate constants for the bimolecular gas-phase
reaction CH4(g) + OH(g) → CH3(g) + H2O(g) and found A = 1
...
1 kJ mol−1 for the Arrhenius parameters
...
Take the average OH concentration to be 1
...
0 × 10−8 mol dm−3, and the temperature to
be −10°C
...

22
...
Gierczak, R
...
Talukdar, S
...
Herndon, G
...
Vaghjiani, and A
...


Ravishankara (J
...
Chem
...
From their data, the following Arrhenius
parameters can be obtained:
A/(dm3 mol −1 s−1)

Ea /(kJ mol−1)

CH4 + OH → CH3 + H2O

1
...
1

CD4 + OH → CD3 + DOH

6
...
5

CH4 + OD → CH3 + DOH

1
...
6

8

Compute the rate constants at 298 K, and interpret the kinetic isotope effects
...
40‡ The oxidation of HSO3 by O2 in aqueous solution is a reaction

of importance to the processes of acid rain formation and flue gas
desulfurization
...
E
...
-X
...
Lee, R
...
Chieng
−
(Inorg
...
34, 4543 (1995)) report that the reaction 2 HSO3 + O2 →
2−
−
2 SO4 + 2 H+ follows the rate law v = k[HSO3 ]2[H+]2
...
6
and an oxygen molar concentration of 2
...
6 × 106 dm9 mol−3 s−1, what is the initial rate of reaction?
−
How long would it take for HSO3 to reach half its initial concentration?

23
Chain reactions
23
...
2 Explosions

Polymerization kinetics
23
...
4 Chain polymerization

The kinetics of
complex reactions
This chapter extends the material introduced in Chapter 22 by showing how to deal with
complex reaction mechanisms
...
Under certain
circumstances, a chain reaction can become explosive, and we see some of the reasons
for this behaviour
...
There are two major classes of polymerization process
and the average molar mass of the product varies with time in distinctive ways
...
Finally, we describe the principles of photochemistry and apply them to problems
in environmental science, biochemistry, and medicine
...
5 Features of homogeneous

catalysis
23
...
Some
take place at a useful rate only after absorption of light or if a catalyst is present
...


Photochemistry
23
...
1 Impact on environmental

science: The chemistry of
stratospheric ozone
I23
...
8 Complex photochemical

processes
I23
...
1: The Förster
theory of resonance energy transfer
Discussion questions
Exercises
Problems

Many gas-phase reactions and liquid-phase polymerization reactions are chain reactions
...
The intermediates in a chain reaction are called chain carriers
...

Ions may also act as chain carriers
...

23
...
As a first example, consider the pyrolysis,
or thermal decomposition in the absence of air, of acetaldehyde (ethanal, CH3CHO),
which is found to be three-halves order in CH3CHO:
CH3CHO(g) → CH4(g) + CO(g)

v = k[CH3CHO]3/2

(23
...
The Rice–Herzfeld mechanism for this reaction is as
follows (the dot signifies an unpaired electron and marks a radical):
Initiation:
Propagation:
Propagation:
Termination:

CH3CHO → ·CH3 + ·CHO
v = ki[CH3CHO]
CH3CHO + ·CH3 → CH3CO· + CH4 v = kp[CH3CHO][·CH3]
CH3CO· → ·CH3 + CO
v = k′ [CH3CO·]
p
·CH3 + ·CH3 → CH3CH3
v = kt[·CH3]2

23
...
To simplify
the treatment, we shall ignore the subsequent reactions of ·CHO, except to note that
they give rise to the formation of CO and of the by-product H2
...
Radicals combine and end the chain in the termination step
...
According to the steady-state approximation (Section 22
...
The steady-state concentration of
·CH3 radicals is
1/2

A ki D
[CH3CHO]1/2
[·CH3] =
C 2k t F

(23
...
3)

which is in agreement with the three-halves order observed experimentally (eqn 23
...

However, this mechanism does not accommodate the formation of various known
reaction by-products, such as propanone (CH3COCH3) and propanal (CH3CH2CHO)
...
An example is the
hydrogen–bromine reaction:
H2(g) + Br2(g) → 2 HBr(g)

d[HBr]
dt

=

k[H2][Br2]3/2
[Br2] + k′[HBr]

The following mechanism has been proposed to account for this rate law (Fig
...
1):
Initiation:

Br2 + M → Br· + Br· + M

v = k i[Br2][M]

where M is either Br2 or H2
...

Propagation:
Retardation:
Termination:

Br· + H2 → HBr + H·
H· + Br2 → HBr + Br·
H· + HBr → H2 + Br·
Br· + Br· + M → Br2 + M*

HBr

(23
...
In this case, the chain
carrier H· attacks a molecule of HBr, the product
...
Other possible termination steps include the recombination of H atoms to form H2 and combination of H and Br atoms
...
The net rate of
formation of the product HBr is

Propagation
Br2
Initiation

Br2
Br

Termination

H
H2

Br2

Propagation
HBr

Fig
...
1 A schematic representation of the
mechanism of the reaction between
hydrogen and bromine
...
Similar
diagrams are used to depict the action of
catalysts
...
The following example illustrates the latter approach
...
1 Deriving the rate equation of a chain reaction

Derive the rate law for the formation of HBr according to the mechanism given above
...
Solve the resulting equations for the concentrations
of the intermediates, and then use the resulting expressions in the equation for the
net rate of formation of HBr
...
When we substitute these concentrations into the
expression for d[HBr]/dt, we obtain
d[HBr]

100

dt

=

2kp(k i /k t )1/2[H2][Br2]3/2
[Br2] + (k r /k′ )[HBr]
p

This equation has the same form as the empirical rate law (eqn 23
...
1, can be used to
explore how the concentration of HBr
changes with time
...


The rate law shows that the reaction slows down as HBr forms, or as the [HBr]/
[Br2] ratio increases
...
Numerical integration of the rate law with mathematical
software shows the predicted time dependence of the concentration of HBr for this
mechanism (Fig
...
2)
...
Find a
combination of rate constants that results
in steady states for these intermediates
...
1 Deduce the rate law for the production of HBr when the initiation
step is the photolysis, or light-induced decomposition, of Br2 into two bromine
atoms, Br·
...

[See eqn 23
...
23
...
2 EXPLOSIONS

833

23
...
The temperature of the system rises if the energy
released by an exothermic reaction cannot escape, and the reaction goes faster
...
A chain-branching explosion occurs when the number of chain centres grows exponentially
...
A chain reaction is involved, and the chain carriers include
H·, ·O·, and ·OH
...


5

Steady
reaction

4

of radicals, and for simplicity identify that concentration with the concentration of

Second
explosion
limit
Chain-branching
explosion

3

1000

900

800

2

700

First
explosion
limit

T /K
The explosion limits of the H2 + O2
reaction
...

Fig
...
3

Method Identify the onset of explosion with the rapid increase in the concentration

Thermal
explosion

Third
explosion
limit

600

Example 23
...
Recall that an O atom, with the ground-state configuration [He]2s 22p4, has two
unpaired electrons
...

u
The occurrence of an explosion depends on the temperature and pressure of
the system, and the explosion regions for the reaction, the conditions under which
explosion occurs, are shown in Fig
...
3
...
At these pressures the chain
carriers produced in the branching steps can reach the walls of the container where
they combine
...
The chain carriers react before reaching the walls and the branching
reactions are explosively efficient
...
The concentration of third-body M molecules is then so
high compared to the concentrations of chain carriers that the combination of H·
atoms with O2 molecules to form relatively unreactive HO2· molecules becomes faster
than the branching reaction between H· atoms and O2 molecules
...
When the pressure is increased to above the third explosion limit, diffusion of HO2· molecules to the walls becomes so slow that they can
react with H2 molecules (now at very high concentrations) to regenerate H atoms and
H2O2 molecules
...
Set
up the corresponding rate laws for the reaction intermediates and then apply the
steady-state approximation
...
Then,
d[H·]

Radical concentration, [X]Dk/ vinit

8

dt

There are two solutions
...
Then,

6

4

[H·] =
(b)

(a)

1
Time, | Dk |t

vinit
kterm − kbranch

(1 − e−(kterm−kbranch)t)

As can be seen from Fig
...
4a, in this regime there is steady combustion of hydrogen
...
Then,

2

0
0

= vinit + (k branch − k term)[H·]

2

The concentration of radicals in the
fuel-rich regime of the hydrogen–oxygen
reaction (a) under steady combustion
conditions, (b) in the explosive region
...

Fig
...
4

Exploration Using mathematical

software, an electronic spreadsheet,
or the interactive applets found in the
Living graphs section of the text’s web site,
explore the effect of changing the
parameter ∆k = kbranch − kterm on the shapes
of the curves in Figs
...
4a and 23
...


[H·] =

vinit
k branch − k term

(e(k branch−k term)t − 1)

There is now an explosive increase in the concentration of radicals (Fig
...
4b)
...

Self-test 23
...

[[H·] = vinitt]

Not all explosions are due to chain reactions
...


23
...
23
...
As a result, monomers are removed early in the reaction
and, as we shall see, the average molar mass of the product grows with time
...
The monomer is used up as it becomes
linked to the growing chains (Fig
...
6)
...

23
...
Stepwise polymerization is
the mechanism of production of polyamides, as in the formation of nylon-66:
H2N(CH2)6NH2 + HOOC(CH2)4COOH
→ H2N(CH2)6NHCO(CH2)4COOH + H2O
→ H-[NH(CH2)6NHCO(CH2)4CO]n-OH

(a)

(a)

(b)

(b)

(c)

(c)
Fig
...
5 In stepwise polymerization, growth
can start at any pair of monomers, and so
new chains begin to form throughout the
reaction
...
23
...
Chains grow as each chain
acquires additional monomers
...

A polyester, for example, can be regarded as the outcome of the stepwise condensation
of a hydroxyacid HO-M-COOH
...
Because the condensation reaction can occur
between molecules containing any number of monomer units, chains of many different lengths can grow in the reaction mixture
...
5a)

However, because there is one -OH group for each -COOH group, this equation is
the same as
d[A]
dt

= −k[A]2

(23
...
The solution of this rate law
is given by eqn 22
...
6)

1 + kt[A]0

The fraction, p, of -COOH groups that have condensed at time t is, after application
of eqn 23
...
7)

Next, we calculate the degree of polymerization, which is defined as the average
number of monomer residues per polymer molecule
...
For example, if there
were initially 1000 A groups and there are now only 10, each polymer must be 100
units long on average
...
7), the average number of monomers per polymer molecule, , is

20

15

10

=

5

0

[A]0 − [A]

0

0
...
4 0
...
8
Fraction condensed, p

1

The average chain length of a
polymer as a function of the fraction of
reacted monomers, p
...

Fig
...
7

[A]0
[A]

=

1
1−p

(23
...
23
...
When we express p in terms of the rate constant k
(eqn 23
...
8b)

The average length grows linearly with time
...

23
...
0 mol dm−3)
...
It results in the rapid growth of an individual polymer

23
...
Examples include the addition polymerizations of
ethene, methyl methacrylate, and styrene, as in
-CH2CHX· + CH2 =CHX → -CH2CHXCH2CHX·
and subsequent reactions
...
9)

Justification 23
...
We
have shown a reaction in which a radical is produced, but in some polymerizations the initiation step leads to the formation of an ionic chain carrier
...

(b) Propagation:
M + ·M1 → ·M2
M + ·M2 → ·M3
Ӈ
M + ·Mn−1 → ·Mn

vp = kp[M][·M]

If we assume that the rate of propagation is independent of chain size for sufficiently
large chains, then we can use only the equation given above to describe the propagation process
...

Because this chain of reactions propagates quickly, the rate at which the total concentration of radicals grows is equal to the rate of the rate-determining initiation
step
...
10)

where f is the fraction of radicals R· that successfully initiate a chain
...
In termination by
disproportionation a hydrogen atom transfers from one chain to another, corresponding to the oxidation of the donor and the reduction of the acceptor
...

Here we suppose that only mutual termination occurs
...
11)

Because the rate of propagation of the chains is the negative of the rate at which the
monomer is consumed, we can write vP = −d[M]/dt and
A fk i D
vP = kP[·M][M] = kp B E
C kt F

1/2

[I]1/2[M]

(23
...
9
...
13)

The kinetic chain length can be expressed in terms of the rate expressions in
Justification 23
...
To do so, we recognize that monomers are consumed at the rate that
chains propagate
...
Therefore, we can write the expression for the kinetic
chain length as

ν=

k p[·M][M]
2k t[M·]2

=

k p[M]
2k t[·M]

When we substitute the steady-state expression, eqn 23
...
14)

Consider a polymer produced by a chain mechanism with mutual termination
...
The
average number of units in each chain is ν
...
15)

with k given in eqn 23
...
We see that, the slower the initiation of the chain (the
smaller the initiator concentration and the smaller the initiation rate constant), the
greater the kinetic chain length, and therefore the higher the average molar mass of
the polymer
...


23
...
5 Features of homogeneous catalysis
We can obtain some idea of the mode of action of homogeneous catalysts by examining the kinetics of the bromide-catalysed decomposition of hydrogen peroxide:
2 H2O2(aq) → 2 H2O(l) + O2(g)
The reaction is believed to proceed through the following pre-equilibrium:
[H3O +]
2

H3O+ + H2O2 5 H3O + + H2O
2

K=

H3O+ + Br− → HOBr + H2O
2

v = k[H3O+][Br−]
2

HOBr + H2O2 → H3O+ + O2 + Br−

(fast)

[H2O2][H3O +]

where we have set the activity of H2O in the equilibrium constant equal to 1 and
assumed that the thermodynamic properties of the other substances are ideal
...
Therefore, we can obtain the rate law of the overall
reaction by setting the overall rate equal to the rate of the second step and using the
equilibrium constant to express the concentration of H3O + in terms of the reactants
...
The observed activation energy is that of the
effective rate constant kK
...
The catalyst lowers the activation energy of the reaction by providing an
alternative path that avoids the slow, rate-determining step of the uncatalysed reaction
(Fig
...
8)
...
When a little iodide ion is added, the activation energy falls
to 57 kJ mol−1 and the rate constant increases by a factor of 2000
...
For example, the enzyme catalase reduces the activation energy for the
decomposition of hydrogen peroxide to 8 kJ mol−1, corresponding to an acceleration
of the reaction by a factor of 1015 at 298 K
...
For
example, the decomposition of hydrogen peroxide in aqueous solution is catalysed by
bromide ion or catalase (Sections 23
...
6)
...
For example, the hydrogenation of
ethene to ethane, a gas-phase reaction, is accelerated in the presence of a solid catalyst
such as palladium, platinum, or nickel
...
We examine heterogeneous catalysis in Chapter 25 and consider
only homogeneous catalysis here
...
23
...
The result is
an increase in the rate of formation of
products
...
In base catalysis, a proton is transferred from the substrate to a base:
XH + B → X− + BH+

X− → products

Base catalysis is the primary step in the isomerization and halogenation of organic
compounds, and of the Claisen and aldol condensation reactions
...
6 Enzymes
S

Active
site

S

E

E
Lock
and
key

Induced
fit

ES

Fig
...
9 Two models that explain the
binding of a substrate to the active site of
an enzyme
...

In the induced fit model, binding of the
substrate induces a conformational change
in the active site
...


Enzymes are homogeneous biological catalysts
...
As is true of
any catalyst, the active site returns to its original state after the products are released
...
However, certain RNA molecules can also be biological
catalysts, forming ribozymes
...

The structure of the active site is specific to the reaction that it catalyses, with
groups in the substrate interacting with groups in the active site by intermolecular
interactions, such as hydrogen bonding, electrostatic, or van der Waals interactions
...
9 shows two models that explain the binding of a substrate to the active site
of an enzyme
...
Experimental evidence favours the induced fit model, in
which binding of the substrate induces a conformational change in the active site
...

Enzyme-catalysed reactions are prone to inhibition by molecules that interfere
with the formation of product
...
For example, an important strategy in the treatment of acquired
immune deficiency syndrome (AIDS) involves the steady administration of a specially
designed protease inhibitor
...
Without a properly formed envelope, HIV cannot replicate in
the host organism
...
Indeed, enzymes are such efficient catalysts that significant accelerations may be observed even when their concentration is more than three orders of
magnitude smaller than that of the substrate
...

2 For a given [E]0 and low values of [S]0, the rate of product formation is proportional to [S]0
...


23
...
According to this
mechanism, an enzyme–substrate complex is formed in the first step and either the
substrate is released unchanged or after modification to form products:
E + S 5 ES

ka, k′
a

ES → P

kb

(23
...
17)

1 + Km /[S]0

where KM = (k′ + k b)/ka is the Michaelis constant, characteristic of a given enzyme
a
acting on a given substrate
...
2 The Michaelis–Menten equation

The rate of product formation according to the Michaelis–Menten mechanism is
v = k b[ES]

(23
...
19)

vmax

KM =

k′ + k b
a
ka

=

[E][S]
[ES]

and note that KM has the same units as molar concentration
...
Moreover, because the substrate is typically in large excess relative to the
enzyme, the free substrate concentration is approximately equal to the initial substrate concentration and we can write [S] ≈ [S]0
...
17 when we substitute this expression for [ES] into that for the
rate of product formation (v = k b[ES])
...
17 shows that, in accord with experimental observations (Fig
...
10):
1 When [S]0 < KM, the rate is proportional to [S]0:
<
v=

ka
KM

Rate of reaction, v

where [E] and [S] are the concentrations of free enzyme and substrate, respectively
...
20a)

2 When [S]0 > KM, the rate reaches its maximum value and is independent of [S]0:
>
v = vmax = k b[E]0

(23
...
23
...
The approach to a
maximum rate, vmax, for large [S] is
explained by the Michaelis–Menten
mechanism
...


842

23 THE KINETICS OF COMPLEX REACTIONS
Substitution of the definitions of KM and vmax into eqn 23
...
23
...


1 + KM/[S]0

(23
...
22)

A Lineweaver–Burk plot is a plot of 1/v against 1/[S]0, and according to eqn 23
...
23
...
The value of k b is then calculated from the y-intercept
and eqn 23
...
However, the plot cannot give the individual rate constants ka and k′
a
that appear in the expression for KM
...
1b can give the additional data needed, because we can find the rate of
formation of the enzyme–substrate complex by monitoring the concentration after
mixing the enzyme and substrate
...

(b) The catalytic efficiency of enzymes

The turnover frequency, or catalytic constant, of an enzyme, kcat, is the number of
catalytic cycles (turnovers) performed by the active site in a given interval divided by
the duration of the interval
...
It follows
from the identification of kcat with k b and from eqn 23
...
23)
kcat = k b =
[E]0
The catalytic efficiency, ε (epsilon), of an enzyme is the ratio kcat /KM
...
We can think of the catalytic activity as
the effective rate constant of the enzymatic reaction
...
23, it follows that

ε=

Comment 23
...


kcat
Km

=

kak b
k′ + k b
a

(23
...
Because ka is the rate
constant for the formation of a complex from two species that are diffusing freely in
solution, the maximum efficiency is related to the maximum rate of diffusion of E
and S in solution
...
2) leads to rate
constants of about 108–109 dm3 mol−1 s−1 for molecules as large as enzymes at room
temperature
...
0 × 108 dm3 mol−1 s−1 and is said to have
attained ‘catalytic perfection’, in the sense that the rate of the reaction it catalyses is
controlled only by diffusion: it acts as soon as a substrate makes contact
...
3 Determining the catalytic efficiency of an enzyme

The enzyme carbonic anhydrase catalyses the hydration of CO2 in red blood cells
to give bicarbonate (hydrogencarbonate) ion:
−
CO2(g) + H2O(l) → HCO3 (aq) + H+(aq)

23
...
25

2
...
78 × 10 −2

5

20

5
...
33 × 10 −2

1
...
5 K
...
From eqn 23
...
24
...
800

1/(v/(mmol dm s ) )

0
...
200

0
...
0

−3 −1

20
...
0

6
...
12 shows the Lineweaver–Burk plot for the data
...
0 and the
y-intercept is 4
...
Hence,
vmax /(mmol dm−3 s−1) =

1
intercept

=

1
4
...
250

and
KM/(mmol dm−3) =

slope
intercept

=

40
...
00

= 10
...
5 × 10−4 mol dm−3 s−1
2
...
1 × 105 s−1

and

ε=

kcat
KM

=

1
...
0 × 10−2 mol dm−3

= 1
...

Self-test 23
...
Several solutions
containing the small peptide N-glutaryl-l-phenylalanine-p-nitroanilide at different concentrations were prepared and the same small amount of α-chymotrypsin
was added to each one
...
1, 273
...
3 nmol dm−3:

843

0
...
450

0
...
00

1
...
67

0
...
201

0
...
417

0
...
667

Determine the maximum velocity and the Michaelis constant for the reaction
...
80 mmol dm−3 s−1, KM = 5
...
2 0
...
6 0
...
23
...
3
...

The most general kinetic scheme for enzyme inhibition is then:
E + S 5 ES

1/v

a = a’ = 1
a

0

1/[S]

(b)

KI =

ESI 5 ES + I

a > 1, a’ = 1

kb

EI 5 E + I

(a)

ka, k′
a

ES → E + P

KS =
′

(23
...
25b)

[EI]

The lower the values of KI and K′ the more efficient are the inhibitors
...

As shown in the following Justification, the rate of reaction in the presence of an
inhibitor is
v=

vmax

(23
...
This equation is very similar to the
I
Michaelis–Menten equation for the uninhibited enzyme (eqn 23
...
27)

1/v

a = a’ = 1
a

Justification 23
...
25a and 23
...

Fig
...
13

Exploration Use eqn 23
...


[E]0 =

KM[ES]
[S]0

D
A α KM
α + [ES]α ′ = [ES]B
+ α ′E
F
C [S]0

The expression for the rate of product formation is then:
v = k b[ES] =

k b[E]0

α KM /[S]0 + α ′

which, upon rearrangement, gives eqn 23
...


There are three major modes of inhibition that give rise to distinctly different
kinetic behaviour (Fig
...
13)
...
This
condition corresponds to α > 1 and α ′ = 1 (because ESI does not form)
...
7 KINETICS OF PHOTOPHYSICAL AND PHOTOCHEMICAL PROCESSES

845

the Lineweaver–Burk plot increases by a factor of α relative to the slope for data on the
uninhibited enzyme (α = α ′ = 1)
...
23
...
In uncompetitive inhibition the inhibitor binds to
a site of the enzyme that is removed from the active site, but only if the substrate is
already present
...
In this case α = 1 (because EI does not form) and α ′ > 1
...
23
...
In non-competitive inhibition (also called mixed inhibition) the
inhibitor binds to a site other than the active site, and its presence reduces the ability
of the substrate to bind to the active site
...

This condition corresponds to α > 1 and α ′ > 1
...
Figure 23
...

In all cases, the efficiency of the inhibitor may be obtained by determining KM and
vmax from a control experiment with uninhibited enzyme and then repeating the
experiment with a known concentration of inhibitor
...
27), the mode of
inhibition, the values of α or α ′, and the values of KI or K′ may be obtained
...
The most important of all are the
photochemical processes that capture the radiant energy of the Sun
...
1)
...
2 and I23
...
Without photochemical processes,
the Earth would be simply a warm, sterile, rock
...
1 summarizes common
photochemical reactions
...
7 Kinetics of photophysical and photochemical processes
Photochemical processes are initiated by the absorption of radiation by at least one
component of a reaction mixture
...
Examples include fluorescence (Section 14
...
1, see also Impact I14
...
Products of a secondary process originate from intermediates that are formed directly from
the excited state of a reactant
...
8)
...
2)
...

(a) Timescales of photophysical processes

Electronic transitions caused by absorption of ultraviolet and visible radiation occur
within 10−16–10−15 s
...
Fluorescence is slower than

Comment 23
...


846

23 THE KINETICS OF COMPLEX REACTIONS
Table 23
...


Table 23
...


23
...
Therefore, the excited singlet state
can initiate very fast photochemical reactions in the femtosecond (10−15 s) to picosecond (10−12 s) timescale
...
1) and of photosynthesis
...
As a consequence, excited triplet states are photochemically important
...
The interplay between reaction rates and excited state lifetimes is a very important factor in the determination of
the kinetic feasibility of a photochemical process
...
1 Exploring the photochemical roles of excited singlet and triplet states

To estimate whether the excited singlet or triplet state of the reactant is a suitable
product precursor, we compare the emission lifetimes with the relaxation time, τ,
of the reactant due to the chemical reaction
...
7 × 104 s−1 and relaxation time τ = 1/(1
...
0 ns and an observed phosphorescence lifetime of 1
...

The excited singlet state is too short-lived and is not expected to be a major source
of product in this reaction
...


(b) The primary quantum yield

We shall see that the rates of deactivation of the excited state by radiative, non-radiative,
and chemical processes determine the yield of product in a photochemical reaction
...
It follows that the primary quantum
yield is also the rate of radiation-induced primary events divided by the rate of photon absorption
...
2), we write

φ=

number of events
number of photons absorbed

=

rate of process
intensity of light absorbed

=

v
Iabs

[23
...
Therefore, the total number of molecules deactivated by radiative
processes, non-radiative processes, and photochemical reactions must be equal to the
number of excited species produced by absorption of light
...
It follows that
v

∑ φi = ∑ I i
i

i

=1

(23
...
7(a), we write

φ f + φ IC + φ ISC + φ p = 1

847

848

23 THE KINETICS OF COMPLEX REACTIONS
where φ f, φ IC, φ ISC, and φp are the quantum yields of fluorescence, internal conversion,
intersystem crossing, and phosphorescence, respectively
...
If the excited singlet state also participates in a primary photochemical
reaction with quantum yield φR, we write

φ f + φIC + φISC + φ p + φ R = 1
We can now strengthen the link between reaction rates and primary quantum yield
already established by eqns 23
...
29
...
29 and rearranging, we obtain Iabs = ∑i vi
...
29 gives the general result

φi =

vi

∑

(23
...

(c) Mechanism of decay of excited singlet states

Consider the formation and decay of an excited singlet state in the absense of a chemical reaction:
Absorption:

S + hνi → S*

vabs = Iabs

Fluorescence:

S* → S + hνf

vf = kf[S*]

Internal conversion:

S* → S

vIC = kIC[S*]

Intersystem crossing:

S* → T*

vISC = kISC[S*]

in which S is an absorbing species, S* an excited singlet state, T* an excited triplet
state, and hνi and hνf are the energies of the incident and fluorescent photons, respectively
...
31)

where the observed fluorescence lifetime, τ 0, is defined as:

τ0 =

1
kf + kISC + kIC

(23
...
33)

23
...
4 The quantum yield of fluorescence

Most fluorescence measurements are conducted by illuminating a relatively dilute
sample with a continuous and intense beam of light
...
7) and write:
d[S*]
dt

= Iabs − kf[S*] − kISC[S*] − kIC[S*] = Iabs − (kf + kISC + kIC)[S*] = 0

Consequently,
Iabs = (kf + kISC + kIC)[S*]
By using this expression and eqn 23
...
33
...
12b)
...
Then, the exponential decay of the fluorescence intensity after the pulse is monitored
...
28, it follows that
1

τ0 =

kf + kISC + kIC

=

A
D 1 φf
kf
× =
C kf + kISC + kIC F kf kf

(23
...
2 Calculating the fluorescence rate constant of tryptophan

In water, the fluorescence quantum yield and observed fluorescence lifetime of
tryptophan are φ f = 0
...
6 ns, respectively
...
33 that
the fluorescence rate constant kf is
kf =

φf
τ0

=

0
...
6 × 10−9 s

= 7
...
Quenching may
be either a desired process, such as in energy or electron transfer, or an undesired side
reaction that can decrease the quantum yield of a desired photochemical process
...

The addition of a quencher, Q, opens an additional channel for deactivation of S*:
Quenching:

S* + Q → S + Q

vQ = kQ[Q][S*]

The Stern–Volmer equation, which is derived in the Justification below, relates the
fluorescence quantum yields φ f,0 and φ f measured in the absence and presence,
respectively, of a quencher Q at a molar concentration [Q]:

φf,0
φf

= 1 + τ 0 kQ[Q]

(23
...
Such a plot is called a Stern–Volmer plot (Fig
...
14)
...


f
ff / f

Justification 23
...

Fig
...
14

φf =

kf
kf + kISC + kIC + kQ[Q]

When [Q] = 0, the quantum yield is

φf,0 =

kf
kf + kISC + kIC

It follows that

φf,0
φf

A
kf
D A kf + kISC + kIC + kQ[Q] D
E ×B
E
=B
kf
kf + kISC + kIC F C
C
F
=

kf + kISC + kIC + kQ[Q]

=1+

kf + kISC + kIC
kQ
kf + kISC + kIC

[Q]

By using eqn 23
...
35
...
34, τ = φ f /kf), plots of If,0 /If and
τ0 /τ (where the subscript 0 indicates a measurement in the absence of quencher)
against [Q] should also be linear with the same slope and intercept as those shown for
eqn 23
...

Example 23
...

2+
Ruthenium(II) tris-(2,2′-bipyridyl), Ru(bipy)3 (2), has a strong metal-to-ligand
charge transfer (MLCT) transition (Section 14
...
The quenching of the
3+
*Ru(bipy)2+ excited state by Fe(H2O)6 in acidic solution was monitored by meas3
uring emission lifetimes at 600 nm
...
6

4
...
4

6

4
...
37

2
...
17

Method Re-write the Stern–Volmer equation (eqn 23
...

Answer Upon substitution of τ 0 /τ for φ0,f /φ f in eqn 23
...
7 KINETICS OF PHOTOPHYSICAL AND PHOTOCHEMICAL PROCESSES

τ

=

1

τ0

5
...
36)

Figure 23
...
36
...
8 × 109, so kQ = 2
...

This example shows that measurements of emission lifetimes are preferred
because they yield the value of kQ directly
...


4
...
0

2
...
4 The quenching of tryptophan fluorescence by dissolved O2 gas was

monitored by measuring emission lifetimes at 348 nm in aqueous solutions
...
3

5
...
6

1
...
92

0
...
57
−1 −1

[1
...
This fact may be used to determine the accessibility of amino acid
residues of a folded protein to solvent
...
Conversely, residues
in the hydrophobic interior of the protein are not quenched effectively by I−
...
For the system of Example 23
...
However, there are some criteria that govern the relative efficiencies of energy and electron transfer
...
The oscillating electric field of the
incoming electromagnetic radiation induces an oscillating electric dipole moment
in S
...
This is the ‘resonance condition’ for
absorption of radiation
...
If the
frequency of oscillation of the electric dipole moment in S is such that ν = ∆EQ /h then
Q will absorb energy from S
...
5
3+

-3

-3

[Fe ]/(10 mol dm )

10
...
0

[23
...
23
...
4
...
Förster in 1959, energy transfer is efficient when:

Table 23
...
2

Dansyl

ODR

4
...
9

IEDANS

FITC

4
...
2

Tryptophan

Haem (heme)

1 The energy donor and acceptor are separated by a short distance (of the order of
nanometres)
...
9

* Additional values may be found in J
...

Lacowicz in Principles of fluorescence spectroscopy,
Kluwer Academic/Plenum, New York (1999)
...


Intensity

Emission
spectrum
of S*
Absorption
spectrum
of Q

Frequency, n
n
Fig
...
16 According to the Förster theory,
the rate of energy transfer from a molecule
S* in an excited state to a quencher molecule
Q is optimized at radiation frequencies in
which the emission spectrum of S* overlaps
with the absorption spectrum of Q, as
shown in the shaded region
...

We show in Further information 23
...
38)

where R0 is a parameter (with units of distance) that is characteristic of each
donor–acceptor pair
...
38 has been verified experimentally and values of
R0 are available for a number of donor–acceptor pairs (Table 23
...

The emission and absorption spectra of molecules span a range of wavelengths,
so the second requirement of the Förster theory is met when the emission spectrum
of the donor molecule overlaps significantly with the absorption spectrum of the
acceptor
...
23
...

In many cases, it is possible to prove that energy transfer is the predominant mechanism of quenching if the excited state of the acceptor fluoresces or phosphoresces at a
characteristic wavelength
...

Equation 23
...
In a typical FRET experiment, a site on a biopolymer or membrane is labelled
covalently with an energy donor and another site is labelled covalently with an energy
acceptor
...
The distance
between the labels is then calculated from the known value of R0 and eqn 23
...

Several tests have shown that the FRET technique is useful for measuring distances
ranging from 1 to 9 nm
...
3 Using FRET analysis

As an illustration of the FRET technique, consider a study of the protein rhodopsin
(Impact I14
...
When an amino acid on the surface of rhodopsin was labelled
covalently with the energy donor 1
...
75 to 0
...
From eqn 23
...
68/0
...
093 and from eqn 23
...
4 nm for the 1
...
9 nm
...
9 nm to
be the distance between the surface of the protein and 11-cis-retinal
...
1 IMPACT ON ENVIRONMENTAL SCIENCE: THE CHEMISTRY OF STRATOSPHERIC OZONE
average distance travelled between collisions of donor and acceptor decreases
...

(f) Electron transfer reactions

According to the Marcus theory of electron transfer, which was proposed by R
...

Marcus in 1965 and is discussed fully in Section 24
...

2 The reaction Gibbs energy, ∆rG, with electron transfer becoming more efficient
as the reaction becomes more exergonic
...

3 The reorganization energy, the energy cost incurred by molecular rearrangements of donor, acceptor, and medium during electron transfer
...

Electron transfer can also be studied by time-resolved spectroscopy (Section 14
...

The oxidized and reduced products often have electronic absorption spectra distinct
from those of their neutral parent compounds
...

IMPACT ON ENVIRONMENTAL SCIENCE

I23
...
Indeed,
many of the natural trace constituents of our atmosphere participate in complex
chemical reactions that have contributed to the proliferation of life on the planet
...
The negative consequences of these changes for the environment are either
already being felt or, more disturbingly, are yet to be felt in the next few decades (see,
for example, the discussion of global warming in Impact I13
...
Careful kinetic analysis allows us to understand the origins of our complex atmosphere and point to ways
in which environmental problems can be solved or avoided
...
23
...
The pressure
decreases as altitude increases (see Problems 1
...
20), but the variation of
temperature with altitude is complex, owing to processes that capture radiant energy
from the Sun
...

In the troposphere, the region between the Earth’s surface and the stratosphere,
temperature decreases with increasing altitude
...
Then we know from
Section 2
...
23
...


The model predicts a decrease in temperature with increasing altitude because Cp,m/
7
CV,m ≈ – for air
...
The Chapman model
accounts for ozone formation and destruction in an atmosphere that contains only O2:
Initiation:

O2 + hν → O + O

185 nm < λ < 220 nm
−1

Termination:

O + O2 + M → O3 + M*

∆ r H = −106
...
9 kJ mol

v = k1[O2]
v = k2[O][O2][M]
v = k3[O3]
v = k4[O][O3]
v = k5[O]2[M]

where M is an arbitrary third body, such as O2 in an ‘oxygen-only’ atmosphere, which
helps to remove excess energy from the products of combination and recombination
reactions
...

Using values of the rate constants that are applicable to stratospheric conditions,
the Chapman model predicts a net formation of trace amounts of ozone, as seen in
Fig
...
18 (see also Problem 23
...
However, the model overestimates the concentration of ozone in the stratosphere because other trace species X contribute to catalytic
enhancement of the termination step O3 + O → O2 + O2 according to
X + O3 → XO + O2
XO + O → X + O2

(a)

855

2
O3
O

1

0
0

20
t /ms

40

[O3]/(1013 molecules cm-3)

4

[O]/(109 molecules cm-3)

[O],[O3]/(108 molecules cm-3)

23
...
23
...
The rate constants are consistent with reasonable stratospheric conditions
...
(b) Late reaction period, showing
that the concentration of O atoms begins to level off after about 4 hours of continuous
irradiation
...
For details of the calculation, see Problem 23
...
P
...
A
...
Oxford University Press (1999)
...
Chlorine atoms are produced by photolysis
of CH3Cl which, in turn, is a by-product of reactions between Cl− and decaying
vegetation in oceans
...
The hydroxyl radical is a product, along with the
methyl radical, of the reaction between excited oxygen atoms and methane gas, which
is a by-product of a number of natural processes (such as digestion of cellulose in
ruminant animals, anaerobic decomposition of organic waste matter) and industrial
processes (such as food production and fossil fuel use)
...

The chemistry outlined above shows that the photochemical reactions of the
Chapman model account for absorption of a significant portion of solar ultraviolet
radiation in the stratosphere
...
3)
...
For example,
chlorofluorocarbons (CFCs) have been used as propellants and refrigerants over
many years
...
For CF2Cl2, also known as CFC-12, the
reaction is:

O2

O

Cl

ClO

O3

O2

CF2Cl2 + hν → CF2Cl + Cl
We already know that the resulting Cl atoms can participate in the decomposition of
ozone according to the catalytic cycle shown in Fig
...
19
...

Ozone depletion has increased the amount of ultraviolet radiation at the Earth’s
surface, particularly radiation in the ‘UVB range’, 290–320 nm
...
23
...


856

23 THE KINETICS OF COMPLEX REACTIONS
thymine dimer (5) or a so-called 6–4 photoproduct (6)
...
There are several natural mechanisms for protection from
and repair of photochemical damage
...
Also, ultraviolet radiation can induce the production
of the pigment melanin (in a process more commonly known as ‘tanning’), which
shields the skin from damage
...

Consequently, there is concern that the depletion of stratospheric ozone may lead to
an increase in mortality not only of animals but also the plants and lower organisms
that form the base of the food chain
...

However, the temperature inversion shown in Fig
...
17 leads to trapping of gases in
the troposphere, so CFCs are likely to continue to cause ozone depletion over many
decades as the molecules diffuse slowly into the middle stratosphere, where they are
photolysed by intense solar UV radiation
...
2 Harvesting of light during plant photosynthesis

A large proportion of solar radiation with wavelengths below 400 nm and above
1000 nm is absorbed by atmospheric gases such as ozone and O2, which absorb ultraviolet radiation (Impact I23
...
2)
...
Plants use radiation in the wavelength range of 400–700 nm to drive the endergonic reduction of
CO2 to glucose, with concomitant oxidation of water to O2 (∆rG⊕ = +2880 kJ mol−1),
in essence the reverse of glycolysis and the citric acid cycle (Impact I7
...
The process takes place in the chloroplast, a special organelle of the plant cell, where chlorophylls a and b (7) and carotenoids (of
which β-carotene, 8, is an example) bind to integral proteins called light harvesting
complexes, which absorb solar energy and transfer it to protein complexes known as
reaction centres, where light-induced electron transfer reactions occur
...
Plants have two photosystems that drive the reduction of NADP+ (9) by water:
–
2 H2O + 2 NADP+ ––––––––––– ––––––→ O2 + 2 NADPH + 2 H+
lig ht, photosystems I and II

It is clear that energy from light is required to drive this reaction because, in the dark,
E ⊕ = −1
...
0 kJ mol−1
...
In photosystems I and II, absorption
of a photon raises a chlorophyll or carotenoid molecule to an excited singlet state
and within 0
...
7e)
...
There, a chlorophyll a dimer becomes electronically

I23
...
For example, the transfer of
an electron from the excited singlet state of P680, the chlorophyll dimer of the photosystem II reaction centre, to its immediate electron acceptor, a phaeophytin a molecule
(a chlorophyll a molecule where the central Mg2+ ion is replaced by two protons, which
are bound to two of the pyrrole nitrogens in the ring), occurs within 3 ps
...
The electrochemical reactions within the photosystem I
reaction centre also occur in this time interval
...
Photosynthesis
captures solar energy efficiently because the excited singlet state of chlorophyll is
quenched rapidly by processes that occur with relaxation times that are much shorter
than the fluorescence lifetime, which is typically about 1 ns in organic solvents at room
temperature
...
The electrons required
for this process come initially from P700 in its excited state
...
The net reaction is
–
–
NADP+ + 2 Cu+(Pc) + H+ –––– –––––––––→ NADPH + 2 Cu2+(Pc)
lig ht, photosystem I

Oxidized plastocyanin accepts electrons from reduced plastoquinone (PQ, 10)
...
2):
–– 6 –
PQH2 + 2 Cu2+(Pc) –– ––– ––––→ PQ + 2 H+ + 2 Cu+(Pc)
cyt b f complex

E ⊕ = +0
...
4 kJ mol−1

857

858

23 THE KINETICS OF COMPLEX REACTIONS

NADPH

hn
ATP

PSII
Fig
...
20 In plant photosynthesis, lightinduced electron transfer processes lead
to the oxidation of water to O2 and the
reduction of NADP+ to NADPH, with
concomitant production of ATP
...
The scheme
summarizes the general patterns of electron
flow and does not show all the intermediate
electron carriers in photosystems I and II,
the cytochrome b6 f complex, and
ferredoxin:NADP+ oxidoreductase
...

Plastoquinone is reduced by water in a process catalysed by light and photosystem
II
...
The resulting P680+ is then reduced ultimately by water
...
23
...
Experiments show that, for each molecule of NADPH
formed in the chloroplast of green plants, one molecule of ATP is synthesized
...
In the process, high energy molecules
(carbohydrates, such as glucose) are synthesized in the cell
...
During aerobic metabolism, the O2 released by
photosynthesis as a waste product is used to oxidize carbohydrates to CO2, driving
biological processes, such as biosynthesis, muscle contraction, cell division, and nerve
conduction
...

23
...
In this section, we consider two examples:
photochemical chain reactions and photosensitization
...
8 COMPLEX PHOTOCHEMICAL PROCESSES
(a) The overall quantum yield of a photochemical reaction

For complex reactions involving secondary processes, such as chain reactions initiated
by photolysis, many reactant molecules might be consumed as a result of absorption
of a single photon
...
In the photolysis
of HI, for example, the processes are
HI + hν → H· + I·
H· + HI → H2 + I·
I· + I· + M → I2 + M*
The overall quantum yield is 2 because the absorption of one photon leads to the
destruction of two HI molecules
...
In such cases the chain acts as a chemical amplifier of the
initial absorption step
...
5 Determining the quantum yield of a photochemical reaction

When a sample of 4-heptanone was irradiated for 100 s with 313 nm radiation with
a power output of 50 W under conditions of total absorption, it was found that
2
...
What is the quantum yield for the formation of ethene?
Method First, calculate the amount of photons generated in an interval ∆t: see

Example 8
...
Then divide the amount of ethene molecules formed by the amount
of photons absorbed
...
1, the amount (in moles) of photons absorbed is

n=

P∆t
(hc/λ)NA

If nC2H4 is the amount of ethene formed, the quantum yield is

Φ=
=

nC2H4
n

=

nC2H4NAhc

λP∆t
−3

(2
...
022 × 1023 mol−1) × (6
...
997 × 108 m s−1)
(3
...
21
Self-test 23
...
30
...
0 mol of molecules?
[3
...
We are assuming a

859

860

23 THE KINETICS OF COMPLEX REACTIONS
primary quantum yield of unity for the photodissociation of Br2
...
1 we can write
d[HBr]
dt

=

2kp(1/kt[M])1/2[H2][Br2]I 1/2
abs
[Br2] + (kr /k′ )[HBr]
p

(23
...

(c) Photosensitization

The reactions of a molecule that does not absorb directly can be made to occur if
another absorbing molecule is present, because the latter may be able to transfer its
energy to the former during a collision
...
3)
...
The Hg
atoms are excited (to Hg*) by resonant absorption of the radiation, and then collide
with H2 molecules
...

IMPACT ON MEDICINE

I23
...

The 1O2 molecules are very reactive and destroy cellular components and it is thought
that cell membranes are the primary cellular targets
...

Absorption:

P + hν → P*

Intersystem crossing:

P* → 3P

Photosensitization:

3

Oxidation reactions:

1

P + 3O2 → P + 1O2
O2 + reactants → products

The photosensitizer is hence a ‘photocatalyst’ for the production of 1O2
...
However, much effort is being expended to develop better
drugs with enhanced photochemical properties
...
From the point of view of pharmacological effectiveness, the drug must be soluble in tissue fluids, so it can be transported to the diseased organ through blood and secreted from the body through

CHECKLIST OF KEY IDEAS

861

urine
...
The drug must also have
unique photochemical properties
...
In practice, this means that the drug
should have a strong absorption band at λ > 650 nm
...
At
the same time, the quantum yield of triplet formation and of 1O2 formation must be
high, so many drug molecules can be activated and many oxidation reactions can
occur during a short period of laser irradiation
...


Checklist of key ideas
1
...
The chain ends in the termination
step
...

2
...

The longer a stepwise polymerization proceeds, the higher
the average molar mass of the product
...
In chain polymerization an activated monomer attacks
another monomer and links to it
...
The slower the initiation of the chain,
the higher the average molar mass of the polymer
...
Catalysts are substances that accelerate reactions but undergo
no net chemical change
...
A homogeneous catalyst is a catalyst in the same phase as the
reaction mixture
...

6
...

7
...

8
...

9
...


862

23 THE KINETICS OF COMPLEX REACTIONS

10
...


13
...
It is based on the
Stern–Volmer equation, φf,0 /φf = 1 + τ 0kQ[Q]
...
The primary quantum yield of a photochemical reaction is the
number of reactant molecules producing specified primary
products for each photon absorbed; the overall quantum yield
is the number of reactant molecules that react for each photon
absorbed
...
Collisional deactivation, electron transfer, and resonance
energy transfer are common fluorescence quenching
processes
...


12
...


15
...


Further reading
Articles and texts

K
...
van Holde, W
...
Johnson and P
...
Ho, Principles of physical
biochemistry
...


J
...

J
...
Educ
...


D
...
G
...
Wiley, New York (2004)
...
E
...
, Seymour/Carraher’s polymer chemistry, Marcel
Dekker (2000)
...
Fersht, Structure and mechanism in protein science: a guide to
enzyme catalysis and protein folding
...
H
...

A
...
Kuznetsov and J
...
Wiley, New York (1998)
...
I
...
S
...
L
...
Prentice Hall, Englewood Cliffs (1998)
...
J
...
University Science
Books, Sausalito (1991)
...
R
...
), CRC handbook of chemistry and physics, Section 5
...

S
...
Murov, I
...
L
...
Marcel Dekker, New York (1993)
...
The URL is available on the web site for this
book
...
The URL is available on the web site for this book
...
1 The Förster theory of resonance
energy transfer

From the qualitative description given in Section 23
...
From Section 18
...
We saw in Section 9
...
It follows that the rate of energy transfer,
w T, at a fixed distance R is given (using notation introduced in
Further information 9
...
In the last expression, the integrals are squares of transition
dipole moments at the radiation frequency ν, the first

EXERCISES
corresponding to emission of S* to S and the second to absorption
of Q to Q*
...
The rate of energy
transfer is proportional to R− 6, so it decreases sharply with increasing
separation between the energy donor and acceptor
...

Because the absorption and emission spectra of large molecules in
condensed phases are broad, it follows that the energy transfer rate is
optimal at radiation frequencies in which the emission spectrum of
the donor and the absorption spectrum of the acceptor overlap
significantly
...
In much the same way that we
defined the quantum yield as a ratio of rates, we can also define the
efficiency of energy transfer, E T, as the ratio

ET =

863

wT

w0 = (k f + k IC + k ISC)[S*]
(23
...
The efficiency, ET, may be expressed in terms of the
experimental fluorescence quantum yields ϕf,0 and ϕf of the donor in
the absence and presence of the acceptor, respectively
...
30 to write:
vf
vf
φf,0 =
and φf =
w0
w0 + w T
where vf is the rate of fluorescence
...
40 gives, after a little algebra, eqn 23
...

Alternatively, we can express w0 in terms of the parameter R0, the
characteristic distance at which w T = w0 for a specified pair of S and Q
(Table 23
...
By using wT ∝ R−6 and w0 ∝ R−6, we can rearrange the
0
expression for E T into eqn 23
...


Discussion questions
23
...
3 Discuss the features, advantages, and limitations of the
Michaelis–Menten mechanism of enzyme action
...
4 Distinguish between competitive, non-competitive, and uncompetitive
inhibition of enzymes
...


(2) A· → B· + C
(3) AH + B· → A· + D

23
...
Describe an experimental procedure for the
determination of the quantum yield
...
6 Discuss experimental procedures that make it possible to differentiate
between quenching by energy transfer, collisions, or electron transfer
...
2 Bearing in mind distinctions between the mechanisms of stepwise and

23
...
Then, discuss FRET in terms of Förster theory
...


Exercises
In the following exercises and problems, it is recommended that rate
constants are labelled with the number of the step in the proposed reaction
mechanism, and any reverse steps are labelled similarly but with a prime
...
1a Derive the rate law for the decomposition of ozone in the reaction
2 O3(g) → 3 O2(g) on the basis of the following proposed mechanism:

23
...
1b has also been proposed
...


(1) O3 5 O2 + O

k1, k′
1

23
...
1b On the basis of the following proposed mechanism, account for the
experimental fact that the rate law for the decomposition 2 N2O5(g) →
4 NO2(g) + O2(g) is v = k[N2O5]
...
Find the
dependence of the rate of decomposition of R2 on the concentration of R2
...
3a Refer to Fig
...
3 and determine the pressure range for a branching
chain explosion in the hydrogen–oxygen reaction at 800 K
...
5b Consider the following chain mechanism:

(1) A2 → A· + A·

23
...
23
...


(3) A· + P → B·

23
...
The carbanion then reacts with
a molecule of propanone to give the product
...

23
...
035 mol dm−3
...
15 × 10−3 mol dm−3 s−1
when the substrate concentration is 0
...
What is the maximum
velocity of this enzymolysis?
23
...
Use the steadystate approximation to find the concentration of the carbanion and derive the
rate equation for the formation of the product
...
4b Consider the acid-catalysed reaction

(1) HA + H+ 5 HAH+

k1k′ , both fast
1

(2) HAH+ + B → BH+ + AH

k2, slow

constant of 0
...
The rate of the reaction is 2
...
890 mol dm−3
...
7a In a photochemical reaction A → 2 B + C, the quantum efficiency with
500 nm light is 2
...
After
exposure of 300 mmol of A to the light, 2
...
How many
photons were absorbed by A?

Deduce the rate law and show that it can be made independent of the specific
term [H+]
...
7b In a photochemical reaction A → B + C, the quantum efficiency with
500 nm light is 1
...
After exposure of 200 mmol A to the
light, 1
...
How many photons were absorbed by A?

23
...
8a In an experiment to measure the quantum efficiency of a photochemical

(1) AH → A· + H·
(2) A· → B· + C
(3) AH + B· → A· + D

reaction, the absorbing substance was exposed to 490 nm light from a 100 W
source for 45 min
...
As a result of irradiation, 0
...
Determine the quantum efficiency
...
8b In an experiment to measure the quantum efficiency of a

(4) A· + B· → P
Use the steady-state approximation to deduce that the decomposition of AH is
first-order in AH
...
5 W source for 28
...
The intensity of the transmitted
light was 0
...
As a result of irradiation, 0
...
Determine the quantum efficiency
...
1 Studies of combustion reactions depend on knowing the concentrations
of H atoms and HO radicals
...
9 × 1010 dm3 mol−1 s−1

(2) OH + OH → H2O + O

k2 = 1
...
1 × 1010 dm3 mol−1 s−1

(J
...
Bradley, W
...
Hoyermann, and H
...
Wagner, J
...
Soc
...
I, 1889 (1973))
...
5 × 10−10 mol cm−3 and 5
...

23
...
N
...
A
...
A
...
Chem
...
Faraday Trans
...
Given the flow velocity as 6
...

l/cm

0

ln([O]0 /[O] ) 0
...
31

0
...
38

0
...
46

0
...
55

0
...
60

with [O]0 = 3
...
54 × 10−7 mol dm−3, p = 1
...

23
...
D
...
J
...
G
...
Electrochem
...

136, 2993 (1989)) proposed the following radical chain mechanism for the
initial stages of the gas-phase oxidation of silane by nitrous oxide:

(1) N2O → N2 + O
(2) O + SiH4 → SiH3 + OH
(3) OH + SiH4 → SiH3 + H2O
(4) SiH3 + N2O → SiH3O + N2

* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady
...
Use the steady-state approximation
to show that this mechanism predicts the following rate law for SiH4
consumption (provided k1 and k6 are in some sense small):
d[SiH4]
dt

= −k[N2O][SiH4]1/2

23
...
Despite the many studies there is not uniform
agreement on the mechanism
...
One such plausible mechanism is that of Example 23
...
Another is the
following:

fluorescence intensity of an aqueous solution of dansyl chloride with time
after excitation by a short laser pulse (with I0 the initial fluorescence intensity)
...
0

10
...
0

20
...
45

0
...
11

0
...

(b) The fluorescence quantum yield of dansyl chloride in water is 0
...
What
is the fluorescence rate constant?
23
...
This singlet changes rapidly into a triplet, which
phosphoresces
...
In an
experiment in methanol as solvent, the phosphorescence intensity varied
with amine concentration as shown below
...
What is the value of kq?

(2) H + O2 → OH + O

[Q]/(mol dm−3)

0
...
0050

0
...
41

0
...
16

23
...
Fluorescence
lifetime measurements of samples of Hg with and without N2 present are
summarized below (T = 300 K):
pN2 = 0
...
2, determine whether or not this
mechanism can lead to explosions under appropriate conditions
...
000

0
...
360

0
...
135

t/µs

0
...
0

10
...
0

20
...
5‡ For many years the reaction H2(g) + I2(g) → 2 HI(g) and its reverse

pN2 = 9
...
However, J
...
Sullivan
(J
...
Phys
...
Bodenstein (Z
...
Chem
...
000

0
...
342

0
...
117

t/µs

0
...
0

6
...
0

12
...
Under what conditions does this rate law reduce to the one for
the originally accepted mechanism?
23
...
The
decomposition of oxalic acid (COOH)2, in the presence of uranyl sulfate,
(UO2)SO4, proceeds according to the sequence

(1) UO2+ + hν → (UO2+)*
(2) (UO2+)* + (COOH)2 → UO2+ + H2O + CO2 + CO
with a quantum efficiency of 0
...
The amount of
oxalic acid remaining after exposure can be determined by titration (with
KMnO4) and the extent of decomposition used to find the number of incident
photons
...
232 g anhydrous oxalic acid, 25
...

After exposure for 300 s the remaining solution was titrated with 0
...
0 cm3 were required for complete oxidation of the
remaining oxalic acid
...

23
...
Tabulated below is the variation of the

You may assume that all gases behave ideally
...

23
...
L
...
P
...
Natl
...
Sci
...
2 nm to 4
...
2

1
...
8

1 − ET 0
...
94 0
...
8

3
...
4

3
...
0

0
...
74 0
...
40 0
...
3

4
...
24 0
...
38? If so, what is the value of R0
for the naphthyl-dansyl pair?

Theoretical problems
23
...
1, and it was noted there that it led to first-order
kinetics
...
How may the conditions be changed so that the
reaction shows different orders?
23
...

23
...

23
...

23
...

23
...
The rate law is d[A]/dt =
−k[A]2[OH]
...
17 Autocatalysis is the catalysis of a reaction by the products
...
The reaction gets
started because there are usually other reaction routes for the formation of
some P initially, which then takes part in the autocatalytic reaction proper
...
Hint
...
To integrate the resulting expression, the following relation
will be useful:
1
([A]0 − x)([P]0 + x)

=

A
1
1
1 D
B
E
+
[A]0 + [P]0 C [A]0 − x [P]0 + x F

(b) Plot [P]/[P]0 against at for several values of b
...
(c) Show that, for the autocatalytic process discussed in parts (a)
and (b), the reaction rate reaches a maximum at tmax = −(1/a) ln b
...
Solve the rate law for initial concentrations [A]0 and [P]0
...
(e) Another reaction with the
stoichiometry A → P has the rate law d[P]/dt = k[A][P]2; integrate the rate law
for initial concentrations [A]0 and [P]0
...

23
...
Thus the steady-state
concentration of products and reactants might differ significantly from
equilibrium values
...
What is the stationary state
concentration of B? Why does this ‘photostationary state’ differ from the
equilibrium state?
23
...

23
...
Devise
a
a mechanism that leads to this rate law when the chlorine pressure
is high
...
21 Photolysis of Cr(CO)6 in the presence of certain molecules M,

can give rise to the following reaction sequence:
(1) Cr(CO)6 + hν → Cr(CO)5 + CO
(2) Cr(CO)5 + CO → Cr(CO)6
(3) Cr(CO)5 + M → Cr(CO)5M
(4) Cr(CO)5M → Cr(CO)5 + M
Suppose that the absorbed light intensity is so weak that I < k4[Cr(CO)5M]
...
Show
that a graph of 1/f against [M] should be a straight line
...
22 Models of population growth are analogous to chemical reaction rate
equations
...

The numbers of births and deaths are proportional to the population, with
proportionality constants b and d
...
How well
does it fit the (very approximate) data below on the population of the planet
as a function of time?

Year

1750

1825

1922

1960

1974

1987

2000

N/109

0
...
23 Many enzyme-catalysed reactions are consistent with a modified
version of the Michaelis–Menten mechanism in which the second step is also
reversible
...

′[E]
′
′
max
M
a
(b) Find the limiting behaviour of this expression for large and small
concentrations of substrate
...
24 The following results were obtained for the action of an ATPase on
ATP at 20°C, when the concentration of the ATPase was 20 nmol dm−3:

[ATP]/(µmol dm−3)

0
...
80

1
...
0

3
...
81

0
...
30

1
...
69

Determine the Michaelis constant, the maximum velocity of the reaction, the
turnover number, and the catalytic efficiency of the enzyme
...
25 Enzyme-catalysed reactions are sometimes analysed by use of the
Eadie–Hofstee plot, in which v is plotted against v/[S]0
...

(b) Discuss how the values of KM and vmax are obtained from analysis of the
Eadie–Hofstee plot
...
23 by using an
Eadie–Hofstee plot to analyse the data
...
26 In general, the catalytic efficiency of an enzyme depends on the pH of

the medium in which it operates
...
This situation can be summarized by the following
mechanism:

PROBLEMS
EH + S ⇔ ESH

ka, k a
′

ESH → E + P

kb

EH 5 E− + H+

KE,a =

it, and the removed class, R, who have either had the disease and recovered,
are dead, or are immune or isolated
...
29 In light-harvesting complexes, the fluorescence of a chlorophyll
molecule is quenched by nearby chlorophyll molecules
...
6 nm, by what distance should two
chlorophyll a molecules be separated to shorten the fluorescence lifetime
from 1 ns (a typical value for monomeric chlorophyll a in organic solvents)
to 10 ps?
23
...
Justify this observation
with the help of molecular orbital theory
...
Show that a constant population is built into this system, namely, that
S + I + R = N, meaning that the timescales of births, deaths by other causes,
and migration are assumed large compared to that of the spread of the disease
...
(a) For the mechanism above,
show that
v=

867

KES,a
[H+]

[H+]
K E,b
[H+]
K ES,b

+
+

KE,a
[H+]
KES,a
[H+]

where vmax and KM correspond to the form EH of the enzyme
...
0 × 10−6 mol dm−3 s−1, KES,b = 1
...
0 × 10−8 mol dm−3
...
(c) Redraw the plot in part (b) by using the same
value of vmax, but KES,b = 1
...
0 × 10−10 mol dm−3
...

23
...
The following results were obtained when
the rate of the enzymolysis of carbobenzoxy-glycyl-D-phenylalanine (CBGP)
was monitored without inhibitor:

[CBGP]0 /(10 −2 mol dm−3)

1
...
84

5
...
13

Relative reaction rate

0
...
669

0
...
000

(All rates in this problem were measured with the same concentration of
enzyme and are relative to the rate measured when [CBGP]0 = 0
...
) When 2
...
25

2
...
00

5
...
172

0
...
344

0
...
0 × 10−2 mol dm−3 benzoate ion was
monitored and the results were:
[CBGP]0 /(10 −2 mol dm− 3)

1
...
50

5
...
00

Relative reaction rate

0
...
201

0
...
246

Determine the mode of inhibition of carboxypeptidase by the phenylbutyrate
ion and benzoate ion
...
28 Many biological and biochemical processes involve autocatalytic steps
(Problem 23
...
In the SIR model of the spread and decline of infectious
diseases the population is divided into three classes; the susceptibles, S, who
can catch the disease, the infectives, I, who have the disease and can transmit

23
...
In separate
experiments, it was observed that the electronic absorption spectrum of the
porphyrin sample showed bands at 420 nm and 550 nm, and the electronic
absorption spectrum of O2-saturated water showed no bands in the visible
range of the spectrum (and therefore no emission spectrum when excited in
the same range)
...
Propose additional experiments that test your
hypothesis
...
32‡ Ultraviolet radiation photolyses O3 to O2 and O
...
The quantum efficiency is 0
...
Data
from W
...
DeMore, S
...
Sander, D
...
Golden, R
...
Hampson, M
...
Kurylo,
C
...
Howard, A
...
Ravishankara, C
...
Kolb, and M
...
Molina, Chemical
kinetics and photochemical data for use in stratospheric modeling: Evaluation
Number 11, JPL Publication 94 –26 (1994)
...
33‡ Use the Chapman model to explore the behaviour of a model

atmosphere consisting of pure O2 at 10 Torr and 298 K that is exposed to
measurable frequencies and intensities of UV radiation
...
The rate constants k1 and k3
depend upon the radiation conditions; assume values of 1
...
016 s−1, respectively
...

(b) Write the rate expressions for the concentration of each chemical species
...
Examine relevant concentrations in the very early time period
t < 0
...
State all assumptions
...
What is the percentage
of ozone after 4
...
You will need a software package for
solving a ‘stiff’ system of differential equations
...
For help with using mathematical software
to solve systems of differential equations, see M
...
Cady and C
...
Trapp,
A Mathcad primer for physical chemistry
...

23
...
7 × 1010 dm3 mol−1 s−1)e−260/(T/K)
(W
...
DeMore, S
...
Sander, D
...
Golden, R
...
Hampson, M
...
Kurylo, C
...

Howard, A
...
Ravishankara, C
...
Kolb, and M
...
Molina, Chemical kinetics
and photochemical data for use in stratospheric modeling: Evaluation
Number 11, JPL Publication 94 –26 (1994))
...

23
...
The commonly accepted
mechanism has been that of H
...
F
...
Chem
...
22, 1724
(1952)):
(1) NO + NO → N2O + O

k1

(2) O + NO → O2 + N

k2

(3) N + NO → N2 + O

k3

(4) O + O + M → O2 + M

k4

(5) O2 + M → O + O + M

k′
4

(a) Label the steps of this mechanism as initiation, propagation, etc
...
What does this
expression for the rate become on the basis of the assumptions that v2 = v3
when [N] reaches its steady state concentration, that the rate of the
propagation step is more rapid than the rate of the initiation step, and that
oxygen atoms are in equilibrium with oxygen molecules? (c) Find an
expression for the effective activation energy, Ea,eff , for the overall reaction in
terms of the activation energies of the individual steps of the reaction
...
(e) It has been
pointed out by R
...
Wu and C
...
Yeh (Int
...
Chem
...
28, 89 (1996)) that
the reported experimental values of Ea,eff obtained by different authors have
varied from 253 to 357 kJ mol−1
...

Obtain the overall rate law based on the steady-state approximation and find
the forms that it assumes for low NO conversion (low O2 concentration)
...
Obtain the rate laws based
on these alternative mechanisms and again estimate the apparent activation
energies
...

Reactions in solution are classified into two types: diffusion-controlled and activationcontrolled
...
In
transition state theory, it is assumed that the reactant molecules form a complex that can
be discussed in terms of the population of its energy levels
...
This approach is useful for parametrizing the rates
of reactions in solution
...
As we shall see, such an
approach gives an intimate picture of the events that occur when reactions occur and is
open to experimental study
...


24
Reactive encounters
24
...
2

Diffusion-controlled
reactions

24
...
4

The Eyring equation

24
...
7

Potential energy surfaces

24
...
Here we examine the details of what happens to
molecules at the climax of reactions
...

As may be imagined, the calculation of the rates of such processes from first principles is very difficult
...
Only when we enquire more deeply do the complications emerge
...
Although a great deal of
information can be obtained from gas-phase reactions, many reactions of interest
take place in solution, and we shall also see to what extent their rates can be predicted
...
6

Some results from
experiments and calculations

24
...
10 The rates of electron transfer

processes
24
...
12 Experimental results
I24
...
Both
approaches are based on the view that reactant molecules must meet, and that reaction takes place only if the molecules have a certain minimum energy
...
5,

Electron transfer in and
between proteins
Checklist of key ideas
Further reading
Further information 24
...
In solution, the reactant molecules may simply
diffuse together and then acquire energy from their immediate surroundings while
they are in contact
...
1 Collision theory
We shall consider the bimolecular elementary reaction
A+B→P

v = k2[A][B]

(24
...

We can anticipate the general form of the expression for k2 by considering the
physical requirements for reaction
...
This requirement suggests that the
rate constant should also be proportional to a Boltzmann factor of the form e−Ea /RT
...
1, that
k2 ∝ σ (T/M)1/2e−Ea /RT
Not every collision will lead to reaction even if the energy requirement is satisfied,
because the reactants may need to collide in a certain relative orientation
...
2)

As we shall see in detail below, this expression has the form predicted by collision theory
...
24
...
If the
diameters of the two molecules are dA
and dB, the radius of the target area is
1
d = –(dA + dB) and the cross-section is πd 2
...
The collision density, ZAB, is the number of (A,B)
collisions in a region of the sample in an interval of time divided by the volume of
the region and the duration of the interval
...
1
...
3a)

where σ is the collision cross-section (Fig
...
1)

σ = πd 2

1
d = – (dA + dB)
2

(24
...
3c)

24
...
4)

Collision densities may be very large
...

Justification 24
...
11 that the collision frequency, z, for a single A molecule of
mass mA in a gas of other A molecules is
z = σ KrelN A

(24
...

As indicated in Section 21
...
6)

1
For future convenience, it is sensible to introduce µ = – m (for like molecules of mass
2
m), and then to write

A 8kT D
E
Krel = B
C πµ F

1/2

(24
...
5
...
8)

1
–
2

The factor of has been introduced to avoid double counting of the collisions
(so one A molecule colliding with another A molecule is counted as one collision
regardless of their actual identities)
...
9)
1
–
2

Note that we have discarded the factor of because now we are considering an A
molecule colliding with any of the B molecules as a collision
...
Equations 24
...
4 then follow
...
The latter condition can be incorporated by writing the
collision cross-section as a function of the kinetic energy of approach of the two colliding species, and setting the cross-section, σ (ε), equal to zero if the kinetic energy of
approach is below a certain threshold value, εa
...
Then, for a collision with a specific relative
speed of approach vrel (not, at this stage, a mean value),
d[A]
dt

= −σ (ε)vrel NA[A][B]

(24
...
1

See Further information 10
...
The
kinetic energy associated with the
relative motion of two particles takes the
1
2
form ε = – µvrel when the centre-of-mass
2
coordinates are separated from the
internal coordinates of each particle
...
2

To go from eqn 24
...
11, we
need to review concepts of probability
theory summarized in Appendix 2
...
The mean value of a function
g(X) is given by

1
...
95 eV

1

s E )/s
s(

2

0
...
11)

and hence recognize the rate constant as
∞

Ύ σ (ε)v

rel f(ε)dε

k2 = NA

(24
...
We show in
the Justification below that, above εa, σ (ε) varies as
(24
...
14)

Justification 24
...
2
0
0

5

E /Ea

10

15

Fig
...
2 The variation of the reactive crosssection with energy as expressed by
eqn 24
...
The data points are from
experiments on the reaction H + D2 →
HD + D (K
...
Katz, and R
...
Chem
...
84, 1934 (1986))
...
24
...
Intuitively we expect that a head-on collision between
2
A and B will be most effective in bringing about a chemical reaction
...
From trigonometry and the definitions of the distances a and d, and the angle θ given in Fig
...
3, it
follows that
A d 2 − a2 D
E
vrel,A–B = vrel cos θ = vrel B
C d2 F

εA–B = ε
B

q

a
vrel

A

The parameters used in the
calculation of the dependence of the
collision cross-section on the relative
kinetic energy of two molecules A and B
...
15)

We assume that only the kinetic energy associated with the head-on component of
the collision, εA–B, can lead to a chemical reaction
...
24
...
0 ´ 10 pm
0
...
24
...
Then, in the Justification below, we
show that

0
...
6

1
= −2
dt
3

d[A]

A
εa D
σ(ε) = 1 −
σ
C
εF

͗g(X)͘ = ∫g(x)f(x)dx
1
...
At this point we recognize that a wide range of approach energies is present
in a sample, so we should average the expression just derived over a Boltzmann distribution of energies f(ε), and write (see Comment 24
...
16)

The existence of an energy threshold, εa, for the formation of products implies that
there is a maximum value of a, amax, above which reactions do not occur
...
17)

2
Substitution of σ (ε) for πa max and σ for πd 2 in the equation above gives eqn 24
...

Note that the equation can be used only when ε > εa
...
1
...
1 COLLISION THEORY
1
in terms of the kinetic energy, ε, by writing ε = – µv2, then dv = dε /(2µε)1/2 and
2
eqn 21
...
18)

The integral we need to evaluate is therefore
∞

A 1 D
E
σ (ε)vrel f(ε)dε = 2π B
C πkT F
0

Ύ

A 8 D
E
=B
C πµ kT F

3/2 ∞

1/2

A 2ε D
σ (ε) B E
C µF
0

Ύ

A 1 D
B
E
C kT F

1/2

ε1/2e−ε /kTdε

∞

Ύ εσ (ε)e

−ε /kT

dε

0

To proceed, we introduce the approximation for σ (ε) in eqn 24
...
It follows that
∞

1/2

A 8kT D
E e−εa /kT
σ (ε)vrel f(ε)dε = σ B
C πµ F
0

Ύ

as in eqn 24
...


Equation 24
...
It follows that we can identify the activation energy, Ea, with
the minimum kinetic energy along the line of approach that is needed for reaction,
and that the pre-exponential factor is a measure of the rate at which collisions occur
in the gas
...
Table 24
...
5a)
...
1* Arrhenius parameters for gas-phase reactions
A/(dm3 mol−1 s−1)
Experiment

Theory

Ea /(kJ mol−1)

2 NOCl → 2 NO + 2 Cl

9
...
9 × 1010

102

2 ClO → Cl2 + O2

6
...
5 × 1010

0

2
...
24 × 106

7
...
7 × 10−6

K + Br2 → KBr + Br

1
...
1 × 10

* More values are given in the Data section
...
16

4
...
24
...


experiment, but for others there are major discrepancies
...
Moreover,
one reaction in the table has a pre-exponential factor larger than theory, which seems
to indicate that the reaction occurs more quickly than the particles collide!
We can accommodate the disagreement between experiment and theory by introducing a steric factor, P, and expressing the reactive cross-section, σ *, as a multiple
of the collision cross-section, σ * = Pσ (Fig
...
4)
...
19)

This expression has the form we anticipated in eqn 24
...
The steric factor is normally
found to be several orders of magnitude smaller than 1
...
1 Estimating a steric factor (1)

Estimate the steric factor for the reaction H2 + C2H4 → C2H6 at 628 K given that the
pre-exponential factor is 1
...

Method To calculate P, we need to calculate the pre-exponential factor, A, by using
eqn 24
...
Table 21
...
The best way to estimate the collision cross-section for dissimilar spherical species is to calculate the
collision diameter for each one (from σ = πd 2 ), to calculate the mean of the two
diameters, and then to calculate the cross-section for that mean diameter
...

Answer The reduced mass of the colliding pair is

µ=

m1m2
m1 + m2

= 3
...
016 u for H2 and m2 = 28
...
Hence

A 8kT D
C πµ F

1/2

= 2
...
1, σ (H2) = 0
...
64 nm2, giving a mean collision cross-section of σ = 0
...
Therefore,
1/2

A 8kT D
A=σ
NA = 7
...
24 × 106 dm3 mol−1 s−1, so it follows that P = 1
...
The
very small value of P is one reason why catalysts are needed to bring this reaction
about at a reasonable rate
...

Self-test 24
...
0 × 109

dm3 mol−1 s−1 at 298 K
...
42 nm2 and σ (Cl2) = 0
...

[0
...
1 COLLISION THEORY
An example of a reaction for which it is possible to estimate the steric factor is K +
Br2 → KBr + Br, with the experimental value P = 4
...
In this reaction, the distance of
approach at which reaction occurs appears to be considerably larger than the distance
needed for deflection of the path of the approaching molecules in a non-reactive
collision
...

This brilliant name is based on a model of the reaction that pictures the K atom
as approaching a Br2 molecule, and when the two are close enough an electron (the
harpoon) flips across from K to Br2
...
Under its influence the ions move together (the line is wound in), the
reaction takes place, and KBr + Br emerge
...

Example 24
...

Method We should begin by identifying all the contributions to the energy of

interaction between the colliding species
...
The first is the ionization energy, I, of K
...
The third is the Coulombic interaction energy between the ions when they have been formed: when their separation
is R, this energy is −e 2/4πε 0 R
...

Answer The net change in energy when the transfer occurs at a separation R is

E = I − Eea −

e2
4πε 0 R

The ionization energy I is larger than Eea, so E becomes negative only when R has
decreased to less than some critical value R* given by
e2
4πε 0R*

= I − Eea

When the particles are at this separation, the harpoon shoots across from K to Br2,
so we can identify the reactive cross-section as σ * = πR*2
...
With I = 420 kJ mol−1 (corresponding to 7
...
2 × 10−19 J), and d = 400 pm, we find P = 4
...
8)
...
2 Estimate the value of P for the harpoon reaction between Na and Cl2

for which d ≈ 350 pm; take Eea ≈ 230 kJ mol−1
...
2]

Example 24
...
First, the concept of a steric
factor is not wholly useless because in some cases its numerical value can be estimated
...
What we need is a more powerful theory that lets us calculate, and not merely guess, its value
...
4
...
2 Diffusion-controlled reactions
Encounters between reactants in solution occur in a very different manner from
encounters in gases
...
However, because a
molecule also migrates only slowly away from a location, two reactant molecules that
encounter each other stay near each other for much longer than in a gas
...
Such an encounter pair may accumulate enough
energy to react even though it does not have enough energy to do so when it first
forms
...

(a) Classes of reaction

The complicated overall process can be divided into simpler parts by setting up a simple kinetic scheme
...
20a)

As we shall see, kd (where the d signifies diffusion) is determined by the diffusional
characteristics of A and B
...
If we suppose that both processes are pseudofirst-order
reactions (with the solvent perhaps playing a role), then we can write
AB → A + B

v = kd
′[AB]

(24
...
20c)

The concentration of AB can now be found from the equation for the net rate of
change of concentration of AB:
d[AB]
dt

= kd[A][B] − k d
′[AB] − ka[AB] ≈ 0

(24
...
This expression solves to
[AB] =

kd[A][B]

(24
...
23)

Two limits can now be distinguished
...
2 DIFFUSION-CONTROLLED REACTIONS
Synoptic table 24
...
1 × 1016

100

Ethanol

3
...
4 × 104

45

Ethanol

CH3CH2Br

Ea /(kJ mol−1)

Water

(CH3)3CCl solvolysis

A/(dm3 mol−1 s−1)

4
...


k2 ≈

kakd
ka

= kd

(24
...
An indication that a reaction is
diffusion-controlled is that its rate constant is of the order of 109 dm3 mol−1 s−1 or
greater
...

An activation-controlled reaction arises when a substantial activation energy is
involved in the reaction AB → P
...
25)

where K is the equilibrium constant for A + B 5 AB
...
Some experimental data are given in Table 24
...

(b) Diffusion and reaction

The rate of a diffusion-controlled reaction is calculated by considering the rate at
which the reactants diffuse together
...
26)

where D is the sum of the diffusion coefficients the two reactant species in the solution
...
3 Solution of the radial diffusion equation

From the form of the diffusion equation (Section 21
...
For a spherically symmetrical system, ∇2
can be replaced by radial derivatives alone (see Table 8
...
27)

The general solution of this equation is
[B]r = a +

b
r

(24
...
8

as may be verified by substitution
...
One condition is that [B]r has its bulk value [B] as
r → ∞
...
It follows that a = [B] and b = −R*[B], and hence that
(for r ≥ R*)

R*

A
R* D
[B]r = B 1 + E [B]
r F
C

0
...
5 illustrates the variation of concentration expressed by this equation
...
4

0
...
29)

Rate of reaction = 4πR*2J

2

4

6
r /R *

8

10

The concentration profile for
reaction in solution when a molecule B
diffuses towards another reactant molecule
and reacts if it reaches R*
...
30)

From Fick’s first law (eqn 21
...
24
...
31)

(A sign change has been introduced because we are interested in the flux towards
decreasing values of r
...
32)

The rate of the diffusion-controlled reaction is equal to the average flow of B
molecules to all the A molecules in the sample
...
Because it is unrealistic to suppose that
all A are stationary; we replace DB by the sum of the diffusion coefficients of the two
species and write D = DA + DB
...
33)

Hence, the diffusion-controlled rate constant is as given in eqn 24
...


We can take eqn 24
...
66) relating the diffusion constant and the hydrodynamic radius RA and RB of each
molecule in a medium of viscosity η:
DA =

kT
6πηRA

DB =

kT
6πηRB

(24
...
35)

(The R in this equation is the gas constant
...
In this
approximation, the rate constant is independent of the identities of the reactants, and
depends only on the temperature and the viscosity of the solvent
...
3 THE MATERIAL BALANCE EQUATION
Illustration 24
...
326 cP (with 1 P = 10−1 kg m−1 s−1) is
kd =

8 × (8
...
26 × 10−4 kg m−1 s−1)

= 2
...
Because 1 m3 = 103 dm3, this result corresponds to 2
...
The experimental value is 1
...


24
...
We can have a glimpse of the kinds of calculations involved by considering the diffusion equation (Section 21
...

(a) The formulation of the equation

Consider a small volume element in a chemical reactor (or a biological cell)
...
71:
∂[J]
∂t

=D

∂2[J]
∂x

2

−v

∂[J]

(24
...
37)

if we suppose that J disappears by a pseudofirst-order reaction
...
38)

Loss
due to
reaction

Equation 24
...
If the rate constant is large,
then [J] will decline rapidly
...
The convection term,
which may represent the effects of stirring, can sweep material either into or out of the
region according to the signs of v and the concentration gradient ∂[J]/∂x
...
Some idea of how it is solved can be obtained by considering the special case in which there is no convective motion (as in an unstirred
reaction vessel):
∂[J]
∂t

=D

∂2[J]
∂x 2

− k[J]

(24
...
05
Concentration of reactant, [J]

t

[J]* = k

Ύ [J]e

−kt

dt + [J]e−kt

(24
...
71 is the solution for a system in which initially a layer of n0 NA molecules is
spread over a plane of area A:
n0e−x /4Dt
2

Dt = 0
...
24
...
In the absence
of reaction (grey lines) the concentration
profiles are the same as in Fig
...
26
...
41)

A(πDt)1/2

When this expression is substituted into eqn 24
...
24
...

Even this relatively simple example has led to an equation that is difficult to solve,
and only in some special cases can the full material balance equation be solved analytically
...


Exploration Use the interactive applet
found in the Living graphs section of
the text’s web site to explore the effect of
varying the value of the rate constant k on
the spatial variation of [J] for a constant
value of the diffusion constant D
...
3

24
...
3
at the end of the chapter
...
24
...
5b that an activated complex forms between reactants as they
collide and begin to assume the nuclear and electronic configurations characteristic
of products
...
We
now consider a more detailed calculation of rate constants which uses the concepts
of statistical thermodynamics developed in Chapter 17
...

Transition state theory is an attempt to identify the principal features governing the
size of a rate constant in terms of a model of the events that take place during the
reaction
...


A + B 5 C‡

K‡ =

pC‡ p7
pA pB

(24
...
43)

24
...
44)

k ‡K ‡

(24
...


Reactants

Products

(a) The rate of decay of the activated complex

An activated complex can form products if it passes through the transition state, the
arrangement the atoms must achieve in order to convert to products (Section 22
...

If its vibration-like motion along the reaction coordinate occurs with a frequency ν,
then the frequency with which the cluster of atoms forming the complex approaches
the transition state is also ν
...
For instance, the
centrifugal effect of rotations might also be an important contribution to the breakup of the complex, and in some cases the complex might be rotating too slowly, or
rotating rapidly but about the wrong axis
...
24
...
The horizontal
axis is the reaction coordinate, and the
vertical axis is potential energy
...


(24
...
In the absence of information to the contrary,
κ is assumed to be about 1
...
8 how to calculate equilibrium constants from structural data
...
54 of that section can be used directly, which in this case gives
K‡ =

7
NAq C‡
7 7
qAqB

e−∆E0 /RT

(24
...
48)

7
The q J are the standard molar partition functions, as defined in Section 17
...
Note
that the units of NA and the qJ are mol−1, so K ‡ is dimensionless (as is appropriate for
an equilibrium constant)
...
We have already assumed that a vibration of the
activated complex C‡ tips it through the transition state
...
49a)

where ν is its frequency (the same frequency that determines k‡)
...
24
...


Fig
...
8 In an elementary depiction of the
activated complex close to the transition
state, there is a broad, shallow dip in the
potential energy surface along the reaction
coordinate
...
However, this depiction is an
oversimplification, for in many cases there
is no dip at the top of the barrier, and the
curvature of the potential energy, and
therefore the force constant, is negative
...
We ignore this problem here,
but see Further reading
...
49b)

hν

We can therefore write
kT

q C‡ ≈

hν

’C ‡

(24
...
The
constant K ‡ is therefore
K‡ =

kT
hν

I‡

I‡ =

7
NA’ C‡

e−∆E0 /RT

7 7
q Aq B

(24
...

(c) The rate constant

We can now combine all the parts of the calculation into
k2 = k ‡

RT
7

p

K ‡ = κν

kT RT
hν p7

I‡

(24
...
53)

‡
‡
The factor I c is given by eqn 24
...

The partition functions for the reactants can normally be calculated quite readily,
using either spectroscopic information about their energy levels or the approximate
expressions set out in Table 17
...
The difficulty with the Eyring equation, however,
lies in the calculation of the partition function of the activated complex: C‡ is difficult
to investigate spectroscopically (but see Section 24
...
We shall illustrate what is involved in
one simple but significant case
...
Because the reactants J = A, B are structureless ‘atoms’, the only contributions to their partition functions are the translational terms:
q7 =
J

7
Vm

Λ3
J

ΛJ =

h
1/2

(2πmJkT)

7
Vm =

RT
p7

(24
...
It has one vibrational mode, but that mode corresponds to motion along the

24
...
It follows that the standard
molar partition function of the activated complex is
7
qC‡ =

7
A 2IkT D V m
C $2 F Λ 3 ‡
C

(24
...
56)

Finally, by identifying κ πr 2 as the reactive cross-section σ *, we arrive at precisely the
same expression as that obtained from simple collision theory (eqn 24
...

24
...
However, the concepts that it introduces, principally that of an
equilibrium between the reactants and the activated complex, have motivated a more
general, empirical approach in which the activation process is expressed in terms of
thermodynamic functions
...
57]

(All the ∆‡X in this section are standard thermodynamic quantities, ∆‡X 7, but we shall
omit the standard state sign to avoid overburdening the notation)
...
58)

Because G = H − TS, the Gibbs energy of activation can be divided into an entropy of
activation, ∆‡S, and an enthalpy of activation, ∆‡H, by writing
∆‡G = ∆‡H − T∆‡S

[24
...
59 is used in eqn 24
...
60)

The formal definition of activation energy, Ea = RT 2(∂ ln k/∂T), then gives Ea = ∆‡H +
2RT, so
k2 = e2Be∆ S/Re−Ea /RT
‡

(24
...
4

For reactions of the type A + B 5 P in
the gas phase, Ea = ∆‡H + 2RT
...


884

24 MOLECULAR REACTION DYNAMICS
from which it follows that the Arrhenius factor A can be identified as
D G(A)

Gibbs energy

‡

D G(B)
‡

DrG °(A)

A = e2Be∆ S/R
‡

(24
...
However, if there is a reduction in entropy below what would be
expected for the simple encounter of A and B, then A will be smaller than that expected
on the basis of simple collision theory
...
24
...
The approximate linear
correlation between ∆‡G and ∆rG 7 is the
origin of linear free energy relations
...
63)

Thus, the more complex the steric requirements of the encounter, the more negative
the value of ∆‡Ssteric, and the smaller the value of P
...
They are encountered when relationships between equilibrium constants
and rates of reaction are explored using correlation analysis, in which ln K (which is
equal to −∆ rG 7/RT) is plotted against ln k (which is proportional to −∆‡G/RT)
...
24
...
This linear correlation is the origin of the alternative name linear free energy relation (LFER; see
Further reading)
...
The statistical thermodynamic theory is very complicated to
apply because the solvent plays a role in the activated complex
...
64)

with the thermodynamic equilibrium constant
K=

aC‡
aAaB

= Kγ

[C‡]
[A][B]

Kγ =

γC‡
γAγB

(24
...
66)

If k° is the rate constant when the activity coefficients are 1 (that is, k° = k‡K), we can
2
2
write
k2 =

k°
2
Kγ

(24
...
9, particularly eqn 5
...
68)

with A = 0
...
Then
2
2
log k2 = log k° − A{z A + z B − (zA + zB)2}I1/2 = log k° + 2AzAzBI1/2
2
2

(24
...
5 THERMODYNAMIC ASPECTS
0
...
4

log (k /k °)

The charge numbers of A and B are zA and zB, so the charge number of the activated
complex is zA + zB; the zJ are positive for cations and negative for anions
...
69 expresses the kinetic salt effect, the variation of the rate constant
of a reaction between ions with the ionic strength of the solution (Fig
...
10)
...
The formation of a single, highly charged ionic complex from two less highly
charged ions is favoured by a high ionic strength because the new ion has a denser
ionic atmosphere and interacts with that atmosphere more strongly
...
Now the
charges cancel and the complex has a less favourable interaction with its atmosphere
than the separated ions
...
2

0

+

-0
...
0050

0
...
0150

0
...
0250

0
...
718

0
...
562

0
...
475

0
...
69, plot log(k/k°) against I1/2, when the slope will give

+

or -

+

-

-

22+

-0
...
What can be deduced about the charge of the activated
complex in the rate-determining stage?

2+
+

2+

Example 24
...
1
I 1/2

0
...
24
...

The ion types are shown as spheres, and
the slopes of the lines are those given by the
Debye–Hückel limiting law and eqn 24
...


1
...

Answer Form the following table:

0
...
0100

0
...
0200

0
...
0300

I

0
...
100

0
...
141

0
...
173

log(k/k°)

−0
...
20

−0
...
29

−0
...
35

1/2

These points are plotted in Fig
...
11
...
04, indicating that zAzB = −2
...
This analysis suggests that the pentaamminebromocobalt(III)
cation participates in the formation of the activated complex
...


0

-0
...
2

Self-test 24
...
Deduce the charge number of the other ion from the following data:

I

0
...
010

0
...
020

0
...
930

0
...
884

0
...
853

-0
...
030
0
...


0

0
...
2

Fig
...
11 The experimental ionic strength
dependence of the rate constant of a
hydrolysis reaction: the slope gives
information about the charge types
involved in the activated complex of the
rate-determining step
...
3
...
6 Reactive collisions
Molecular beams allow us to study collisions between molecules in preselected energy
states, and can be used to determine the states of the products of a reactive collision
...

(a) Experimental probes of reactive collisions
Detector

Source 1

Source 2
Fig
...
12 In a crossed-beam experiment,
state-selected molecules are generated in
two separate sources, and are directed
perpendicular to one another
...


v
7
6
5
4
3
2

1

0

Infrared chemiluminescence from
CO produced in the reaction O + CS →
CO + S arises from the non-equilibrium
populations of the vibrational states of CO
and the radiative relaxation to equilibrium
...
24
...
24
...
The detector for the products of the collision of two beams can be
moved to different angles, so the angular distribution of the products can be determined
...
9b, and with different orientations (by using
electric fields), it is possible to study the dependence of the success of collisions on
these variables and to study how they affect the properties of the outcoming product
molecules
...
By studying the intensities of the infrared emission spectrum, the populations of the vibrational states may be determined (Fig
...
13)
...
In this technique, a laser
is used to excite a product molecule from a specific vibration-rotation level; the intensity of the fluorescence from the upper state is monitored and interpreted in terms
of the population of the initial vibration-rotation state
...
16) can be used to monitor the progress of reaction
...
In MPI, the absorption of several photons by a molecule results in ionization if the total photon energy is greater than the ionization energy of the molecule
...
2)
...
In this
technique, product ions are accelerated by an electric field towards a phosphorescent
screen and the light emitted from specific spots where the ions struck the screen is
imaged by a charge-coupled device (CCD, Further information 13
...
An important
variant of MPI is resonant multiphoton ionization (REMPI), in which one or more
photons promote a molecule to an electronically excited state and then additional
photons are used to generate ions from the excited state
...

(b) State-to-state dynamics

The concept of collision cross-section was introduced in connection with collision
theory in Section 24
...
We shall write eqn 24
...
70)

24
...
Molecular beam studies provide a more sophisticated version of this quantity, for they provide the state-to-state
cross-section, σnn′, and hence the state-to-state rate constant, knn′:
knn′ = ͗σnn′vrel͘NA

(24
...
72)

n,n′

where fn(T) is the Boltzmann factor at a temperature T
...

Potential
energy

24
...
Potential energy surfaces may be constructed from experimental data,
with the techniques described in Section 24
...
7)
...

Special computational techniques are used to take into account electron correlation,
which arises from instantaneous interactions between electrons as they move closer to
and farther from each other in molecule or molecular cluster
...
An alternative
is to use semi-empirical methods, in which results of calculations and experimental
parameters are used to construct the potential energy surface
...
Detailed calculations show that the approach
of an atom along the H-H axis requires less energy for reaction than any other
approach, so initially we confine our attention to a collinear approach
...

At the start of the encounter RAB is infinite and RBC is the H2 equilibrium bond
length
...
The total energy of the three-atom system depends on
their relative separations, and can be found by doing a molecular orbital calculation
...
24
...
This surface is normally depicted
as a contour diagram (Fig
...
15)
...

A section through the surface at RAB = ∞, for example, is the same as the H2 bonding
potential energy curve drawn in Fig
...
16
...


RBC

RAB

Fig
...
14 The potential energy surface for
the H + H2 → H2 + H reaction when the
atoms are constrained to be collinear
...
24
...
24
...

Re marks the equilibrium bond length
of an H2 molecule (strictly, it relates to the
arrangement when the third atom is at
infinity)
...
24
...
24
...
Path A corresponds to a route
in which RBC is held constant as HA
approaches; path B corresponds to a route
in which RBC lengthens at an early stage
during the approach of HA; path C is the
route along the floor of the potential valley
...
However, we can obtain an initial idea of the paths available to the system for paths that correspond to least potential energy
...
If the HB-HC bond length is constant during the initial approach of HA, then
the potential energy of the H3 cluster rises along the path marked A in Fig
...
16
...
An alternative reaction path can be imagined (B) in which the HB-HC bond length increases
while HA is still far away
...

The path of least potential energy is the one marked C, corresponding to RBC
lengthening as HA approaches and begins to form a bond with HB
...
The
encounter of least potential energy is one in which the atoms take route C up the floor
of the valley, through the saddle point, and down the floor of the other valley as HC
recedes and the new HA-HB bond achieves its equilibrium length
...
4
...
In
terms of trajectories on potential surfaces, the transition state can be identified with a
critical geometry such that every trajectory that goes through this geometry goes on to
react (Fig
...
17)
...
8 Some results from experiments and calculations

RAB

RBC

Fig
...
17 The transition state is a set of
configurations (here, marked by the line
across the saddle point) through which
successful reactive trajectories must pass
...
Therefore, the shape of the surface can be explored experimentally by changing the relative speed of approach (by selecting the beam velocity) and the degree of
vibrational excitation and observing whether reaction occurs and whether the products emerge in a vibrationally excited state (Fig
...
18)
...
Thus, is trajectory C *, where the HBHC molecule is initially
2
vibrationally excited, more efficient at leading to reaction than the trajectory C 1 in
*,
which the total energy is the same but has a high translational kinetic energy?
(a) The direction of attack and separation

Figure 24
...
The potential barrier is least for collinear attack,
as we assumed earlier
...
) In contrast, Fig
...
20 shows the potential energy
changes that occur as a Cl atom approaches an HI molecule
...
The relevance
of this result to the calculation of the steric factor of collision theory should be noted:
not every collision is successful, because not every one lies within the reactive cone
...
8 SOME RESULTS FROM EXPERIMENTS AND CALCULATIONS

889

RBC
C2*

C1*

H

H

(b)

(a)

H

H

RBC
C3

C4

(c)
0

(d)

RAB

0

Fig
...
19 An indication of the anisotropy of
the potential energy changes as H approaches
H2 with different angles of attack
...
The surface indicates
the potential energy profile along the
reaction coordinate for each configuration
...
(a) C 1 corresponds to the path
*
along the foot of the valley
...
(c) C3 corresponds to A
approaching a non-vibrating BC molecule, but with insufficient translational kinetic energy
...

Fig
...
18

Cl
I
H
Unsuccessful
attack

memory of the approach direction has been lost
...
In the collision of K and I2, for
example, most of the products are thrown off in the forward direction
...
1c) because the
transition takes place at long range
...
In this mechanism, K
effectively bumps into a brick wall, and the KI product bounces out in the backward
direction
...

(b) Attractive and repulsive surfaces

Some reactions are very sensitive to whether the energy has been predigested into a
vibrational mode or left as the relative translational kinetic energy of the colliding
molecules
...
For F + HCl → Cl + HF, for example, the reaction is about five times as
efficient when the HCl is in its first vibrational excited state than when, although HCl
has the same total energy, it is in its vibrational ground state
...
In this case, successful
encounters occur only when Cl approaches
within a cone surrounding the H atom
...
24
...
5

In molecular beam work the remarks
we make in our discussion normally
refer to directions in a centre-of-mass
coordinate system
...
The way in which centre-of-mass
coordinates are constructed and the
events in them interpreted involves too
much detail for our present purposes,
but we should bear in mind that ‘forward’
and ‘backward’ have unconventional
meanings
...


890

24 MOLECULAR REACTION DYNAMICS

C*

C

C
‡

C*

Fig
...
21 An attractive potential energy
surface
...


Fig
...
22 A repulsive potential energy
surface
...
A reaction that is attractive in one
direction is repulsive in the reverse
direction
...
Figure 24
...
Figure 24
...
A surface that is attractive in one direction is repulsive in the
reverse direction
...
If the original molecule is vibrationally excited,
then a collision with an incoming molecule takes the system along C
...

If, however, the same amount of energy is present solely as translational kinetic
energy, then the system moves along C* and travels smoothly over the saddle point into
products
...

Moreover, the potential surface shows that once past the saddle point the trajectory
runs up the steep wall of the product valley, and then rolls from side to side as it falls
to the foot of the valley as the products separate
...

Now consider the repulsive surface (Fig
...
22)
...
As the reactants approach, the potential energy rises
...
This path corresponds to an unsuccessful encounter, even
though the energy is sufficient for reaction
...
This motion may be sufficient
to tip the system round the corner to the saddle point and then on to products
...

Reactions with repulsive potential surfaces can therefore be expected to proceed more
efficiently if the excess energy is present as vibrations
...

(c) Classical trajectories

A clear picture of the reaction event can be obtained by using classical mechanics to
calculate the trajectories of the atoms taking place in a reaction from a set of initial
conditions, such as velocities, relative orientations, and internal energies of the reacting

particles
...

Figure 24
...
This illustration shows clearly the vibration of
the original molecule and the approach of the attacking atom
...
The newly formed molecule shakes, but quickly settles down to steady, harmonic
vibration as the expelled atom departs
...
24
...
The reaction in the illustration is the exchange reaction KCl + NaBr → KBr +
NaCl
...


Internuclear distance, R /pm

24
...
The concept of trajectory
then fades and is replaced by the unfolding of a wavefunction that represents initially
the reactants and finally products
...
It is common to define a ‘channel’ as a group of molecules in
well-defined quantum mechanically allowed states
...
Furthermore, not every transition leads to a chemical reaction
...
What complicates a quantum mechanical
calculation of trajectories and rate constants even in this simple four-atom system is
that many reacting channels present at a given temperature can lead to the desired
products H2O + H, which themselves may be formed as many distinct channels
...
73)

i, j

Internuclear distance, R /pm

N(E) =

Fig
...
23 The calculated trajectories for a
reactive encounter between A and a
vibrating BC molecule leading to the
formation of a vibrating AB molecule
...
(M
...
N
...
D
...
Chem
...
, 43, 3258 (1965)
...
5

5
Time, t /ps

7

...
24
...
(P
...
Karplus, Faraday Disc
...
Soc
...
)

892

24 MOLECULAR REACTION DYNAMICS
where Pi,j(E) is the probability for a transition between a reacting channel i and a
product channel j and the summation is over all possible transitions that lead to product
...
74)

where Qr(T) is the partition function density (the partition function divided by the
volume) of the reactants at the temperature T
...
74 is that it
provides a direct connection between an experimental quantity, the rate constant, and
a theoretical quantity, N(E)
...
9 The investigation of reaction dynamics with ultrafast

laser techniques
The development of femtosecond pulsed lasers (Section 14
...
Pulsed-laser techniques can also be used to
control the outcome of chemical reactions
...
Also shown is movement
between a ‘covalent’ surface (in green) and
an ‘ionic’ surface (in purple) of the
wavepacket formed by laser excitation
...
24
...
In a typical experiment designed to detect an activated complex, a femtosecond
laser pulse is used to excite a molecule to a dissociative state, and then a second femtosecond pulse is fired at an interval after the dissociating pulse
...
For example,
when ICN is dissociated by the first pulse, the emergence of CN from the photoactivated state can be monitored by watching the growth of the free CN absorption (or,
more commonly, its laser-induced fluorescence)
...

Some sense of the progress that has been made in the study of the intimate mechanism of chemical reactions can be obtained by considering the decay of the ion pair
Na+I −
...
24
...

The system can be described with two potential energy surfaces, one largely ‘ionic’
and another ‘covalent’, which cross at an internuclear separation of 693 pm
...
Consequently, the electronically excited complex exists
as a superposition of states, or a localized wavepacket (Section 8
...
24
...

The complex can also dissociate, shown as movement of the wavepacket toward
very long internuclear separation along the dissociative surface
...
The dynamics of
the system is probed by a second laser pulse with a frequency that corresponds to the
absorption frequency of the free Na product or to the frequency at which Na absorbs
when it is a part of the complex
...


24
...
Femtosecond transition state spectroscopy has also been
used to study more complex reactions, such as the Diels–Alder reaction, nucleophilic
substitution reactions, and pericyclic addition and cleavage reactions
...
In other
experiments, the photoejection of carbon monoxide from myoglobin and the attachment of O2 to the exposed site have been studied to obtain rate constants for the two
processes
...
We already have at our disposal a
number of successful strategies for achieving this goal
...
However, this strategy is not very general, as a new catalyst needs to be developed for every
reaction type of interest
...
Here we examine two ways in
which the outcome of a chemical reaction can be affected by laser irradiation
...
Consider the gas-phase reaction between H and HOD
...

On the other hand, when the same stretching mode is excited to the v = 5 energy level,
HD and OH are the preferred products
...
However, the technique is limited to those cases in
which energy can be deposited and remains localized in the desired vibrational mode
of the reactant for a time that is much longer than the reaction time
...
24
...
The bound Na absorption intensity
shows up as a series of pulses that recur in about 1 ps, showing that the wavepacket
oscillates with about that period
...
The free Na
absorption also grows in an oscillating manner, showing the periodicity of wavepacket
oscillation, each swing of which gives it a chance to dissociate
...
25 ps, corresponding to a vibrational wavenumber of 27 cm−1
(recall that the activated complex theory assumes that such a vibration has a very low
frequency)
...
In contrast, although the
oscillation frequency of NaBr is similar, it barely survives one oscillation
...
Thus, a molecular beam can be used
to produce a van der Waals molecule (Section 18
...
The HI bond
can be dissociated by a femtosecond pulse, and the H atom is ejected towards the O
atom of the neighbouring CO2 molecule to form HOCO
...
24
...
The lower
curve is the absorption of the electronically
excited complex and the upper curve is the
absorption of free Na atoms (Adapted from
A
...
Zewail, Science 242, 1645 (1988))
...

A strategy that seeks to avoid the problem of vibrational relaxation uses ultrafast
lasers and is related closely to the techniques used for the spectroscopic detection of
transition states
...
The reaction can be initiated by exciting I2 to an electronic state at least 52 460 cm−1 above the ground state and
then followed by measuring the time dependence of the chemiluminescence of XeI*
...
The first pulse excites the I2 molecule to a low energy and
unreactive electronic state
...
In this case, the wavepacket does not have enough energy to
react, but excitation by another laser pulse with the proper wavelength can provide
the necessary additional energy
...
Because the reaction occurs
via the harpoon mechanism, the product yield is expected to be optimal if the second
pulse is applied when the wave packet is at a point where the Xe···I2 distance is just
right for electron transfer from Xe to I2 to occur (see Example 24
...
This type of
control of the I2 + Xe reaction has been demonstrated
...
Extension of these techniques to the controlled synthesis of materials in
routine laboratory work will require much more sophisticated knowledge of how
laser pulses may be combined to stimulate a specific molecular response in a complex
system
...
We begin by examining the features of a theory
that describes the factors governing the rates of electron transfer
...
We shall see that relatively simple expressions may be used to predict the
rates of electron transfer with reasonable accuracy
...
10 The rates of electron transfer processes
Consider electron transfer from a donor species D to an acceptor species A in solution
...
75)

In the first step of the mechanism, D and A must diffuse through the solution and
collide to form a complex DA, in which the donor and acceptor are separated by a
distance comparable to r, the distance between the edges of each species
...
10 THE RATES OF ELECTRON TRANSFER PROCESSES
D + A 5 DA

KDA =

ka
k′
a

=

[DA]
[D][A]

(24
...
Next, electron transfer occurs within the DA complex to
yield D+A−:
DA → D+A−

vet = ket[DA]

(24
...
The
D+A− complex has two possible fates
...
76c)

+ −

Second, D A can break apart and the ions diffuse through the solution:
D+A− → D+ + A−

vd = kd[D+A−]

(24
...
77)

kd F

Justification 24
...
75) to the rate of formation of separated ions, the reaction products (eqn 24
...
77, we divide the numerator and denominator on the right-hand
side of this expression by kdket and solve for the reciprocal of kobs
...
77 and the factors that determine the rate of electron
transfer reactions in solution, we assume that the main decay route for D +A− is dissociation of the complex into separated ions, or kd > kr
...
When
ket < k′ , we see that kobs ≈ (ka /k′ )ket or, after using eqn 24
...
78)

and the process is controlled by the activation energy of electron transfer in the DA
complex
...
4), we write
ket = κνe−∆ G/RT
‡

(24
...
Our first task is to write theoretical expressions for κν and ∆‡G by describing the motions of electrons and nuclei mathematically
...
11 Theory of electron transfer processes
Our discussion concentrates on the following two key aspects of the theory, which
was developed independently by R
...
Marcus, N
...
Hush, V
...
Levich, and R
...

Dogonadze:

DA

Gibbs energy

+

-

DA

l

DG
‡

1 Electrons are transferred by tunnelling through a potential energy barrier, the
height of which is partly determined by the ionization energies of the DA and D+A−
complexes
...

2 The complex DA and the solvent molecules surrounding it undergo structural
rearrangements prior to electron transfer
...


DrG °

(a) Electron tunnelling

R
0

P
0

q q* q
Displacement, q
Fig
...
27 The Gibbs energy surfaces of the
complexes DA and D+A− involved in an
electron transfer process are represented
by parabolas characteristic of harmonic
oscillators, with the displacement
coordinate q corresponding to the
changing geometries of the system
...
The
parabolas intersect at q = q*
...
11b)
...
2 that, according to the Franck–Condon principle, electronic
transitions are so fast that they can be regarded as taking place in a stationary nuclear
framework
...
We can represent
the potential energy (and the Gibbs energy) surfaces of the two complexes (the reactant complex, DA, and the product complex, D+A−) by the parabolas characteristic of
harmonic oscillators, with the displacement coordinate corresponding to the changing geometries (Fig
...
27)
...

According to the Franck–Condon principle, the nuclei do not have time to move
when the system passes from the reactant to the product surface as a result of the
transfer of an electron
...
24
...

The factor κν is a measure of the probability that the system will convert from
reactants (DA) to products (D+A−) at q* by electron transfer within the thermally
excited DA complex
...
11 THEORY OF ELECTRON TRANSFER PROCESSES

Ύ

͗HDA͘ = ψA@DAψBdτ
where HDA is a hamiltonian that describes the coupling of the electronic wavefunctions
...
80)

where r is the edge-to-edge distance between D and A, β is a parameter that measures
the sensitivity of the electronic coupling matrix element to distance, and ͗H° ͘2 is the
DA
value of the electronic coupling matrix element when D and A are in contact (r = 0)
...
80 is essentially the same as the
exponential decrease in transmission probability through a potential energy barrier
described in Section 9
...

(b) The expression for the rate of electron transfer

The full expression for ket turns out to be (see Further reading for a derivation)
ket =

2͗HDA͘2 A π3 D

C 4λRT F

h

1/2

e−∆ G /RT
‡

(24
...
80
...
1
that the Gibbs energy of activation ∆‡G is
∆‡G =

(∆rG 7 + λ)2
4λ

(24
...
These molecular rearrangements include the relative reorientation
of the D and A molecules in DA and the relative reorientation of the solvent molecules
surrounding DA
...
82 shows that ∆‡G = 0, with the implication that the
reaction is not slowed down by an activation barrier, when ∆rG 7 = −λ , corresponding
to the cancellation of the reorganization energy term by the standard reaction Gibbs
energy
...
24
...
Initially, the electron to be transferred occupies the HOMO of D, and the overall energy of DA is lower
than that of D+A− (Fig
...
28a)
...
24
...
Over reasonably short distances r, the main mechanism of electron transfer is tunnelling through the potential energy barrier depicted in Fig
...
28b
...
After an electron moves from the HOMO of D to the LUMO of A, the
system relaxes to the configuration represented by q P in Fig
...
28c
...

The tunnelling event responsible for electron transfer is similar to that described in
Section 9
...
We saw in
Section 9
...
24
...

(a) At the nuclear configuration denoted
by q R, the electron to be transferred in DA
0
is in an occupied electronic energy level
(denoted by a blue circle) and the lowest
unoccupied energy level of D+A− (denoted
by an unfilled circle) is of too high an
energy to be a good electron acceptor
...
(c) The system relaxes to
the equilibrium nuclear configuration of
D+A− denoted by q P, in which the lowest
0
unoccupied electronic level of DA is higher
in energy than the highest occupied
electronic level of D+A−
...
A
...
Sutin, Biochim
...
Acta 811, 265 (1985)
...
81 has some limitations
...
Weak coupling is observed
when the electroactive species are sufficiently far apart that the wavefunctions ψA
and ψD do not overlap extensively
...
7 nm apart
...
Examples of strongly coupled systems are mixed-valence, binuclear
d-metal complexes with the general structure LmMn+-B-Mp+Lm, in which the electroactive metal ions are separated by a bridging ligand B
...
0 nm
...
2)
...
At low temperatures,
thermal fluctuations alone cannot bring the reactants to the transition state and
transition state theory, which is at the heart of the theory presented in this section,
fails to account for any observed electron transfer
...
We saw in Section
9
...
Full quantum
‡
mechanical treatments of electron transfer reactions replace the (π3/4λRT)1/2e−∆ G/RT
term with Franck–Condon factors similar to those discussed in Section 14
...

24
...
In such cases, electron transfer occurs
after a donor–acceptor complex forms and it is not possible to exert control over r, the
edge-to-edge distance
...

‡
Under these conditions, the term e−∆ G/RT becomes a constant and, after taking the
natural logarithm of eqn 24
...
80, we obtain
ln ket = −βr + constant

(24
...
The
value of β depends on the medium through which the electron must travel from
donor to acceptor
...

The dependence of ket on the standard reaction Gibbs energy has been investigated
in systems where the edge-to-edge distance, the reorganization energy, and κν are
constant for a series of reactions
...
81 becomes
ln ket = −

1 A ∆rG 7 D
4λ C RT F

2
1
−–
2

A ∆rG 7 D
+ constant
C RT F

(24
...
Equation 24
...
Beyond that, the reaction enters the inverted
region, in which the rate constant decreases as the reaction becomes more exergonic
(∆ rG 7 becomes more negative)
...
12 EXPERIMENTAL RESULTS

899

1010
(c)

ket /s-1

109

(d)
(e)
(f)
(g)

108

(h)

7

10

(b)

(a)

special compounds in which the electron donor and acceptor are linked covalently to
a molecular spacer of known and fixed size (Fig
...
29)
...
84 and observed experimentally in Fig
...
29
can be explained by considering the dependence of the activation Gibbs energy on
the standard Gibbs energy of electron transfer
...
Now we let the minimum energy of the prod0
0
uct complex change while keeping q P constant, which corresponds to changing the
0
magnitude of ∆rG 7
...
30 shows the effect of increasing the exergonicity of the
process
...
24
...
As the process becomes more exergonic, the activation Gibbs
a
0
energy decreases and the rate constant increases
...
5
...
30b shows
that, when ∆‡G = 0 and q b = q R, the rate constant for the process reaches a maximum
*
0
as there is no activation barrier to overcome
...
81, this condition
occurs when −∆rG 7 = λ
...
24
...

c
0
The rate constant for the process decreases steadily as the activation barrier for the
process increases with decreasing ∆rG 7
...
24
...


106
0

0
...
5 2
-D rG °/eV

2
...
Kinetic measurements were
conducted in 2-methyltetrahydrofuran and
at 296 K
...
Each acceptor has a characteristic
standard reduction potential, so it follows
that the standard Gibbs energy for the
electron transfer process is different for
each compound in the series
...
84 and the
maximum of the parabola occurs at
−∆rG 7 = λ = 1
...
2 × 102 kJ mol−1
...
R
...
T
...
L
...
Amer
...
Soc
...
)
Fig
...
29

900

24 MOLECULAR REACTION DYNAMICS
D+A-

DA
R
q0

Energy

(a)

(b)

P
q0

(c)

q*(c) q*(b) q*(a)
Nuclear displacement

(a) ∆‡G > 0 and the transition
state is at q * > q R
...
(b) When ∆‡G = 0 and q b = q R
*
0
the rate constant for the process reaches a
maximum as there is no activation barrier
to overcome
...
The rate constant for the
c
0
process decreases steadily as the activation
barrier for the process increases with
decreasing ∆ rG 7
...
24
...

The basic theory presented in Section 24
...

IMPACT ON BIOCHEMISTRY

I24
...
2 and I21
...
Electron
transfer between protein-bound cofactors or between proteins also plays a role in
other biological processes, such as photosynthesis (Impact I23
...

Equation 24
...
2), which must form an encounter complex
before electron transfer can take place
...
0 nm, a relatively long distance on a
molecular scale, with the protein providing an intervening medium between donor
and acceptor
...
81
...
In that enzyme, bound copper ions
and haem groups work together to reduce O2 to water in the final step of respiration
...
81
...
More detailed work on the specific effect of secondary structure suggests
that 12
...
0 nm−1 when the intervening medium consists primarily of
α helices and 9
...
5 nm−1 when the medium is primarily β sheet
...

A value of β is not necessary for the prediction of the rate constants for electron
transfer processes between proteins if we take a different approach
...
78 and 24
...
85)

where Z = KDAκν
...
However, when λ >> |∆ rG 7 |, kobs may be estimated by a special case of the
Marcus cross-relation, which we derive in Further information 24
...
86)

where K is the equilibrium constant for the net electron transfer reaction (eqn 24
...
87a)
(24
...
1 IMPACT ON BIOCHEMISTRY: ELECTRON TRANSFER IN AND BETWEEN PROTEINS
The rate constants estimated by eqn 24
...

Example 24
...
5 × 10 2

0
...
6 × 10

0
...
7 × 104 dm3
mol−1 s−1
...
30 (ln K = νFE 7/RT) and 7
...
Then, we use eqn 24
...


Answer The two reduction half-reactions are

Right: cytochrome c(ox) + e− → cytochrome c(red)

7
E R = +0
...
286 V

The difference is
E 7 = (0
...
286 V) = −0
...
30 with ν = 1 and RT/F = 25
...
026V
−3

25
...
6
2
...
36
...
76 and the self-exchange rate constants, we
calculate
kobs = {(1
...
6 × 107 dm3 mol−1 s−1) × 0
...
0 × 104 dm3 mol −1 s−1
The calculated and observed values differ by only 25 per cent, indicating that the
Marcus relation can lead to reasonable estimates of rate constants for electron
transfer
...
4 Estimate kobs for the reduction by cytochrome c of plastocyanin, a
protein containing a copper ion that shuttles between the +2 and +1 oxidation
states and for which kAA = 6
...
350 V
...
8 × 103 dm3 mol−1 s−1]

901

902

24 MOLECULAR REACTION DYNAMICS
Table 24
...


Rate constants for individual steps

k′ , k′ ,
...
In collision theory, it is supposed that the rate is proportional
to the collision frequency, a steric factor, and the fraction of
collisions that occur with at least the kinetic energy Ea along
their lines of centres
...
A reaction in solution may be diffusion-controlled if its rate is
controlled by the rate at which reactant molecules encounter
each other in solution
...

3
...
The result is the
‡
Eyring equation, k2 = κ (kT/h)I c
...
The rate constant may be parametrized in terms of the
Gibbs energy, entropy, and enthalpy of activation,
k2 = (kT/h)e∆*S/Re−∆*H/RT
...
The kinetic salt effect is the effect of an added inert salt on the
rate of a reaction between ions, log k2 = log k° + 2AzAzBI1/2
...
Techniques for the study of reactive collisions include infrared
chemiluminescence, laser-induced fluorescence, multiphoton
ionization (MPI), reaction product imaging, and resonant
multiphoton ionization (REMPI)
...
A potential energy surface maps the potential energy as a
function of the relative positions of all the atoms taking part in
a reaction
...
In a
repulsive surface, the saddle point occurs late on the reaction
coordinate
...
Ultrafast laser techniques can be used to probe directly the
activated complex and to control the outcome of some
chemical reactions
...
The rate constant of electron transfer in a donor–acceptor
complex depends on the distance between electron donor
and acceptor, the standard reaction Gibbs energy, and the
‡
reorganization energy, λ: ket ∝ e−βre−∆ G/RT (constant T), with
∆‡G = (∆rG 7 + λ)2/4λ
...
The Marcus cross-relation predicts the rate constant for
electron transfer in solution from the reaction’s equilibrium
constant K and the self-exchange rate constants kii: kobs =
(kDDkAAK )1/2
...
J
...
N
...
J
...
Educ
...

G
...
Billing and K
...
Mikkelsen, Advanced molecular dynamics and
chemical kinetics
...

D
...
Clary, Quantum theory of chemical reaction dynamics
...

F
...
Crim, Vibrational state control of bimolecular reactions:
discovering and directing the chemistry
...
Chem
...
32, 877
(1999)
...
M
...
Taylor and Francis, London
(1985)
...
M
...
Ulstrup, Electron transfer in chemistry and
biology: an introduction to the theory
...

R
...
Levine and R
...
Bernstein, Molecular reaction dynamics and
chemical reactivity
...

J
...
Steinfeld, J
...
Francisco and W
...
Hase, Chemical kinetics and
dynamics
...

A
...
Zewail, Femtochemistry: atomic-scale dynamics of the chemical
bond using ultrafast lasers
...
Chem
...
Ed
...


Further information
Because Gm,R(q*) = Gm,P(q*), it follows from eqns 24
...
91, and
24
...
1 The Gibbs energy of activation of
electron transfer and the Marcus cross-relation

The simplest way to derive an expression for the Gibbs energy of
activation of electron transfer processes is to construct a model in
which the surfaces for DA (the ‘reactant complex’, denoted R) and
for D+A− (the ‘product complex’, denoted P) are described by
classical harmonic oscillators with identical reduced masses µ and
angular frequencies ω, but displaced minima, as shown in Fig
...
27
...
88a)

1
Gm,P(q) = – NA µω 2(q − q P)2 + Gm,P(q P)
0
0
2

(24
...
The standard
reaction Gibbs energy for the electron transfer process DA → D+A− is
∆rG 7 = Gm,P(q P) − Gm,R(q R), the difference in standard molar Gibbs
0
0
energy between the minima of the parabolas
...
24
...

We also note that q*, the value of q corresponding to the transition
state of the complex, may be written in terms of the parameter α, the
fractional change in q:
q* = q R + α (q P − q R)
0
0
0

(24
...
82 as
∆‡G =

∆ rG 72

+

4λ

∆ rG 7
2

+

λ
4

7

>
When λ > |∆ rG |, we obtain
∆ rG 7

(24
...
92)

(24
...
92 and 24
...
82
...

Equation 24
...
86) used in Impact I24
...
We begin by
using eqn 24
...
24
...
It follows from eqns 24
...
91 that

(24
...
89)

We see from Fig
...
27 that ∆‡G = Gm,R(q*) − Gm,R(q R)
...
88a, 24
...
89 that
1
1
∆‡G = –NA µω 2(q* − q R)2 = –NA µω 2{α(q P − q R)}2
0
0
0
2
2

1
α 2λ = –NAµω 2{(α − 1)(q P − q R)}2 + ∆rG 7
0
0
2
= (α − 1)λ + ∆rG 7

2

+

λ
4

This expression can be used without further elaboration to denote the
Gibbs energy of activation of the net reaction
...
17 (ln K = −∆rG 7/RT) to write

It follows that
kDD = ZDDe−λDD /4RT

kAA = ZAAe−λAA/4RT

To make further progress, we assume that the reorganization energy
of the net reaction is the arithmetic mean of the reorganization
energies of the self-exchange reactions:

λ DA =

λ DD + λ AA
2
∆ rG

Then, by combining this expression with the expressions for kDD and
kAA, and using the relations e x+y = e xe y and e x/2 = (e x )1/2, we obtain the
most general case of the Marcus cross-relation:

where

7

2

7/RT

kobs = (kDDkAAK )1/2f

It follows that the Gibbs energy of activation of the net reaction is
∆‡G =

K = e−∆rG

+

λ DD
8

+

λ AA
8

f=

Z
(ZAAZDD)1/2

In practice, the factor f is usually set to 1 and we obtain eqn 24
...


Therefore, the rate constant for the net reaction is
kobs = Ze−∆rG

7/2RT

e−λ DD /8RTe−λAA/8RT

Discussion questions
24
...

24
...

24
...

24
...

24
...


24
...
(b) Reactions with repulsive potential surfaces proceed
more efficiently if the excess energy is present as vibrations
...
7 A method for directing the outcome of a chemical reaction consists of
using molecular beams to control the relative orientations of reactants during
a collision
...
How should CH3I
molecules and Rb atoms be oriented to maximize the production of RbI?
24
...


Exercises
24
...
What is the percentage increase
when the temperature is raised by 10 K at constant volume?
24
...
What is the percentage
increase when the temperature is raised by 10 K at constant volume?
24
...
What is this
fraction when (a) Ea = 10 kJ mol−1, (b) Ea = 100 kJ mol−1 at (i) 300 K and
(ii) 1000 K?
24
...
What is this
fraction when (a) Ea = 15 kJ mol−1, (b) Ea = 150 kJ mol−1 at (i) 300 K and
(ii) 800 K?
24
...
2a

when the temperature is raised by 10 K
...
3b Calculate the percentage increase in the fractions in Exercise 24
...


24
...
The
collision cross-section is 0
...
32 × 10−27 kg, and
the activation energy is 171 kJ mol−1
...
4b Use the collision theory of gas-phase reactions to calculate the

theoretical value of the second-order rate constant for the reaction D2(g) +
Br2(g) → 2 DBr(g) at 450 K, assuming that it is elementary bimolecular
...
30 nm2, the reduced mass as 3
...

24
...
If the critical reaction distance is 0
...
5b Suppose that the typical diffusion coefficient for a reactant in aqueous

solution at 25°C is 4
...
If the critical reaction distance is 0
...
6a Calculate the magnitude of the diffusion-controlled rate constant
at 298 K for a species in (a) water, (b) pentane
...
00 × 10−3 kg m−1 s−1, and 2
...

24
...
The
viscosities are 3
...

24
...
89 cP
...
0 mmol dm−3 initially,
how long does it take for the concentration of the atoms to fall to half that
value? Assume the reaction is elementary
...
7b Calculate the magnitude of the diffusion-controlled rate constant at

298 K for the recombination of two atoms in benzene, for which η = 0
...

Assuming the concentration of the reacting species is 1
...

24
...
2 × 10−22 m2
...
95 and 0
...
Calculate the P-factor for the reaction
...
8b For the gaseous reaction A + B → P, the reactive cross-section obtained
from the experimental value of the pre-exponential factor is 8
...

The collision cross-sections of A and B estimated from the transport
properties are 0
...
40 nm2, respectively
...

24
...
37 × 10−3 kg m−1 s−1 at 40°C
...
150 mol dm−3 and 0
...

24
...
27 cP at 20°C
...
200 mol dm−3 and 0
...

24
...
Near 30°C the rate constant is given by the
empirical expression k2 = (2
...
Evaluate the
energy and entropy of activation at 30°C
...
10b The reaction A− + H+ → P has a rate constant given by the empirical

expression k2 = (8
...
Evaluate the energy and
entropy of activation at 25°C
...
11a When the reaction in Exercise 24
...
78 ×
1014)e−(9134 K)/T dm3 mol−1 s−1 near 30°C
...


905

24
...
45 × 1013)

e−(5375 K)/T dm3 mol−1 s−1 near 25°C
...


24
...
The energy of activation for the reaction is 58
...

At 65°C the rate constant is 7
...
Calculate the entropy of
activation at 65°C
...
12b A gas-phase recombination reaction is first-order in each of the
reactants
...
6 kJ mol−1
...
23 m3 s−1
...

24
...
40 nm2
...
13b Calculate the entropy of activation for a collision between two
structureless particles at 500 K, taking M = 78 g mol−1 and σ = 0
...

24
...
6 × 1012 dm3 mol−1 s−1 and its activation energy
is 10
...
What are (a) the entropy of activation, (b) the enthalpy of
activation, (c) the Gibbs energy of activation at 298 K?
24
...
3 × 1013 dm3 mol−1 s−1 and its activation energy is
30
...
What are (a) the entropy of activation, (b) the enthalpy
of activation, (c) the Gibbs energy of activation at 298 K?
24
...
At 25°C, k = 12
...
0525
...

24
...
55 dm6 mol−2 min−1 at an ionic strength of 0
...
Use the Debye–Hückel limiting law to estimate the rate
constant at zero ionic strength
...
16a For a pair of electron donor and acceptor, HAB = 0
...
182 eV and ket = 30
...
Estimate the value of the reorganization
energy
...
16b For a pair of electron donor and acceptor, ket = 2
...
665 eV
...
975 eV
when a substituent is added to the electron acceptor and the rate constant for
electron transfer changes to ket = 3
...
The experiments were
conducted at 298 K
...

24
...
02 × 105 s−1 when

r = 1
...
51 × 105 s−1 when r = 1
...
Assuming that ∆rG7
and λ are the same in both experiments, estimate the value of β
...
17b Refer to Exercise 24
...
Estimate the value of ket when r = 1
...


Problems*
Numerical problems
24
...
4 × 1010 dm3 mol−1 s−1
...
2 Nitrogen dioxide reacts bimolecularly in the gas phase to give 2 NO +
O2
...
What are the P factor and the
reactive cross-section for the reaction?

* Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady
...
6 × 102

9
...
3 × 105

3
...
60 nm2
...
3 The diameter of the methyl radical is about 308 pm
...
0dm3 sample of ethane at 298 K and 100 kPa is dissociated into methyl radicals
...
4 The rates of thermolysis of a variety of cis- and trans-azoalkanes have

been measured over a range of temperatures in order to settle a controversy
concerning the mechanism of the reaction
...
S
...
J
...
Amer
...
Soc
...
Calculate the enthalpy, entropy,
energy, and Gibbs energy of activation at –20°C
...
82
−1

10 × k/s
4

−20
...
02

−13
...
95

1
...
31

4
...
50

14
...
5 In an experimental study of a bimolecular reaction in aqueous solution,
the second-order rate constant was measured at 25°C and at a variety of ionic
strengths and the results are tabulated below
...
What is the charge on the other
ion involved?

I

0
...
0037

0
...
0065

0
...
05

1
...
16

1
...
26

24
...
Use the following data from 25°C to find the dependence of
log kr on the ionic strength:
I/(mol kg−1)

0
...
0525

0
...
1575

kr /(dm3 mol −1 min−1)

0
...
670

0
...
694

Evaluate the limiting value of kr at zero ionic strength
...
7 The total cross-sections for reactions between alkali metal atoms and

halogen molecules are given in the table below (R
...
Levine and R
...

Bernstein, Molecular reaction dynamics, Clarendon Press, Oxford, 72 (1974))
...

σ */nm2

Cl2

Br2

Na

1
...
16

0
...
54

1
...
27

Rb

1
...
97

1
...
96

2
...
95

I2

24
...
Cyfert, B
...
Wawrzeczyk (Int
...
Chem
...
28, 103
(1996)) examined the oxidation of tris(1,10-phenanthroline)iron(II) by
periodate in aqueous solution, a reaction that shows autocatalytic behaviour
...
2

0
...
462 0
...

(b) The experimentally determined Arrhenius parameters in the range
501–583 K are A = 7
...
935 × 104 K for k and and
A = 2
...
691 × 104 K for k′
...
41 kJ mol−1, D(F-F) = 160
...
2 kJ mol−1
...

24
...
07 × 105 dm3 mol−1 s−1 at 300 K and an activation
energy of 65
...
Calculate ∆‡S, ∆‡H, ∆‡U, and ∆‡G for the reaction
...
11‡ One of the most historically significant studies of chemical reaction

rates was that by M
...
physik
...
29, 295 (1899)) of the
gas-phase reaction 2 HI(g) → H2(g) + I2(g) and its reverse, with rate constants
k and k′, respectively
...
4 dm3 mol− 1 min−1)

0
...
588

1
...
10

6
...
9

k′/(22
...
0140 0
...
0659

0
...
375 3
...

24
...
86 with f = 1)
to estimate the rate constant for the reaction Ru(bpy)3+ + Fe(H2O)2+ →
3
6
2+
3+
Ru(bpy)3 + Fe(H2O)6 , where bpy stands for 4,4′-bipyridine
...
26 V

Fe(H2O)3+ + e− → Fe(H2O)2+
6
6

E 7 = 0
...
0 × 108 dm3 mol−1 s−1
*Fe(H2O)3+ + Fe(H2O) 2+ → *Fe(H2O)2+ + Fe(H2O)3+
6
6
6
6
kFe = 4
...
3 eV (Cl2), 1
...
7 eV
(I2), and ionization energies are 5
...
3 eV (K), 4
...
9 eV (Cs)
...
9‡ For the thermal decomposition of F2O by the reaction 2 F2O(g) →
2 F2(g) + O2(g), J
...
J
...
Phys
...
17,
235 (1972)) have suggested the following mechanism:

0
...
05

0
...
321

0
...
0125 0
...
283 0
...
224

What can be inferred about the charge of the activated complex of the ratedetermining step?

24
...
40 is a solution of eqn 24
...

24
...
40, and explore the effect of increasing reaction rate constant on the
spatial distribution of J
...
15 Estimate the orders of magnitude of the partition functions involved in
a rate expression
...
Check that in the collision of two structureless molecules the order
of magnitude of the pre-exponential factor is of the same order as that
predicted by collision theory
...

24
...
Consider the mechanism: H+(aq) + B(aq) → P, where H+
comes from the deprotonation of the weak acid, HA
...
First show that log [H+], derived from the ionization of HA,
depends on the activity coefficients of ions and thus depends on the ionic
strength
...

24
...
As an
example, consider the rate of diffusion of an atom stuck to the surface of a
solid
...
Find an expression for the rate of
diffusion, and evaluate it for W atoms on a tungsten surface (Ea = 60 kJ mol−1)
...
What is the value of the
diffusion coefficient D at 500 K? (Take the site separation as 316 pm and
ν = 1 × 1011 Hz
...
18 Suppose now that the adsorbed, migrating species treated in Problem
24
...

What effect does this have on the diffusion constant? Take the molecule to be
methane, for which B = 5
...

24
...


Applications: to environmental science and biochemistry
24
...
Atkinson (J
...
Chem
...
Data 26, 215 (1997)) has reviewed a

large set of rate constants relevant to the atmospheric chemistry of volatile
organic compounds
...
7 × 109 dm3 mol−1 s−1 for
R = C2H5 and 8
...
Assuming no energy
barrier, compute the steric factor, P, for each reaction
...
Obtain collision
diameters from collision cross-sections of similar molecules in the Data
section
...
24‡ The compound α-tocopherol, a form of vitamin E, is a powerful
antioxidant that may help to maintain the integrity of biological membranes
...
H
...
W
...
Amer
...
Soc
...
Once the duroquinone was photochemically excited, a bimolecular
reaction took place at a rate described as diffusion-limited
...
(b) The reported rate
constant was 2
...

24
...
It is important to characterize protein dimerization because the
process is considered to be the rate-determining step in the growth of crystals
of many proteins
...
21‡ Show that bimolecular reactions between nonlinear molecules are
much slower than between atoms even when the activation energies of both
reactions are equal
...
(1) All vibrational partition functions are close to unity;
(2) all rotational partition functions are approximately 1 × 101
...

24
...
It also demonstrates the
importance of femtosecond spectroscopy to our understanding of chemical
dynamics because direct experimental observation of the activated complex
removes much of the ambiguity of theoretical predictions
...
(a) Suppose that the H
approaches D2 from the side and forms a complex in the form of an isosceles
triangle
...
Let the critical coordinate
be the antisymmetric stretching vibration in which one H-D bond stretches
as the other shortens
...
Estimate k2
for this reaction at 400 K using the experimental activation energy of about
35 kJ mol−1
...
Use the same estimated molecular bond lengths and
vibrational frequencies to calculate k2 for this choice of model
...
Use mathematical software or write and run a program
that allows you to vary the structure of the complex and the parameters in
a plausible way, and look for a model (or more than one model) that gives
a value of k close to the experimental value, 4 × 105 dm3 mol−1 s−1
...
0100

0
...
0200

0
...
0300

0
...
20 In a molecular beam experiment to measure collision cross-sections it
was found that the intensity of a CsCl beam was reduced to 60 per cent of its
intensity on passage through CH2F2 at 10 µTorr, but that when the target was
Ar at the same pressure the intensity was reduced only by 10 per cent
...
10

13
...
50

27
...
10

52
...
26 A useful strategy for the study of electron transfer in proteins consists
of attaching an electroactive species to the protein’s surface and then
measuring ket between the attached species and an electroactive protein
cofactor
...
W
...
B
...
Rev
...
2) modified by replacement of the haem iron
by a zinc ion, resulting in a zinc-porphyrin (ZnP) moiety in the interior of the
protein, and by attachment of a ruthenium ion complex to a surface histidine
aminoacid
...
23 nm
...
For each ruthenium-modified
protein, either the Ru2+ → ZnP+ or the ZnP* → Ru3+, in which the electron
donor is an electronic excited state of the zinc-porphyrin formed by laser
excitation, was monitored
...
Use the following data to estimate the
reorganization energy for this system:

−∆ rG 7/eV

0
...
705

0
...
975

1
...
055

k et /(10 6 s−1)

0
...
52

1
...
99

5
...
1

24
...
The following table shows data
compiled by Moser et al
...
48

0
...
96

1
...
58 × 1012

3
...
00 × 109

1
...
35

2
...
98 × 107

63
...
Are these data in agreement with the behaviour predicted by eqn
24
...

24
...
27) decreases with decreasing
temperature in the range 300 K to 130 K
...
Account for these results
...
29 Azurin is a protein containing a copper ion that shuttles between the +2
and +1 oxidation states, and cytochrome c is a protein in which a haem-bound
iron ion shuttles between the +3 and +2 oxidation states
...
6 × 103
dm3 mol−1 s−1
...
5 × 10

0
...
304

2

Processes at solid
surfaces
In this chapter we see how solids grow at their surfaces and how the details of the structure
and composition of solid surfaces can be determined experimentally
...
This material is used to discuss how surfaces
affect the rate and course of chemical change by acting as the site of catalysis
...
Therefore, we
revisit in this chapter some of the topics treated in Chapter 7, but focus on the dynamics
of electrode processes rather than the equilibrium properties treated there
...


Processes at solid surfaces govern the viability of industry both constructively, as in
catalysis, and destructively, as in corrosion
...
The concept of a solid surface has been extended in recent years with the availability of microporous materials
as catalysts
...
In Chapter 24 we explored the dynamics
of electron transfer in homogeneous systems
...
A measure of this rate is the current density, j, the charge flux through a region (the electric current divided by the
area of the region)
...

We shall see that acronyms are widely used in surface studies; for convenience,
a list of the acronyms used in this chapter is given in Table 25
...


25
The growth and structure of
solid surfaces
25
...
2

The extent of adsorption

Physisorption and
chemisorption
25
...
5 The rates of surface processes
I25
...
3

Heterogeneous catalysis

Mechanisms of
heterogeneous catalysis
25
...
2 Impact on technology:
Examples of catalysis in the
chemical industry
25
...
8
25
...
10
25
...
12
I25
...
13
I25
...
The attachment of
particles to a surface is called adsorption
...
The reverse of adsorption is desorption
...
1: The relation
between electrode potential and the
Galvani potential
Discussion questions
Exercises
Problems

910

25 PROCESSES AT SOLID SURFACES
25
...
25
...
This primitive model is
largely supported by scanning tunnelling
microscope images (see Impact I9
...


Fig
...
2 Some of the kinds of defects that
may occur on otherwise perfect terraces
...


A simple picture of a perfect crystal surface is as a tray of oranges in a grocery store
(Fig
...
1)
...
The molecule loses energy as it bounces,
but it is likely to escape from the surface before it has lost enough kinetic energy to be
trapped
...

There is little energy advantage for an ion in solution to discard some of its solvating
molecules and stick at an exposed position on the surface
...
A common type of surface defect is a step between
two otherwise flat layers of atoms called terraces (Fig
...
2)
...
When an atom settles on a terrace it bounces
across it under the influence of the intermolecular potential, and might come to a step
or a corner formed by a kink
...
Likewise, when ions deposit from solution, the loss of the solvation interaction is
offset by a strong Coulombic interaction between the arriving ions and several ions at
the surface defect
...
As the process of settling
into ledges and kinks continues, there comes a stage when an entire lower terrace has
been covered
...
For continuing growth, a surface defect is needed that propagates as the crystal
grows
...
One reason for their formation may be that the crystal grows so quickly that its
particles do not have time to settle into states of lowest potential energy before being
trapped in position by the deposition of the next layer
...
25
...
Imagine a
cut in the crystal, with the atoms to the left of the cut pushed up through a distance of
one unit cell
...
A path encircling the screw axis spirals up to the top of
the crystal, and where the dislocation breaks through to the surface it takes the form
of a spiral ramp
...
The incoming particles lie in ranks on the ramp, and successive ranks reform the step at an angle to its initial position
...
Growth may therefore
continue indefinitely
...
Propagating spiral edges can also give rise
to flat terraces (Fig
...
4)
...
25
...
Successive tables
of atoms may form as counter-rotating defects collide on successive circuits, and
the terraces formed may then fill up by further deposition at their edges to give flat
crystal planes
...
This feature is explained in
Fig
...
6, where we see that, although the horizontal face grows forward most rapidly,
it grows itself out of existence, and the slower-growing faces survive
...
2 SURFACE COMPOSITION

Fig
...
3 A screw dislocation occurs where
one region of the crystal is pushed up
through one or more unit cells relative to
another region
...
As atoms lie along the step the
dislocation rotates round the screw axis
and is not annihilated
...
25
...
This accounts for the
appearance of this cadmium iodide crystal
...
M
...

Clarendon Press, Oxford (1978)
...
2 Surface composition
Under normal conditions, a surface exposed to a gas is constantly bombarded with
molecules and a freshly prepared surface is covered very quickly
...
14) for the
collision flux:
ZW =

p

(25
...
63 × 10 m s
24

(25
...
For air (M ≈ 29 g mol−1) at 1 atm and 25°C the
collision flux is 3 × 1027 m−2 s−1
...
Even if only a few collisions
leave a molecule adsorbed to the surface, the time for which a freshly prepared surface
remains clean is very short
...
When it is reduced
to 10−4 Pa (as in a simple vacuum system) the collision flux falls to about 1018 m−2 s−1,
corresponding to one hit per surface atom in each 0
...
Even that is too brief in most
experiments, and in ultrahigh vacuum (UHV) techniques pressures as low as 10−7 Pa
(when ZW = 1015 m−2 s−1) are reached on a routine basis and as low as 10−9 Pa (when
ZW = 1013 m−2 s−1) are reached with special care
...

The layout of a typical UHV apparatus is such that the whole of the evacuated part
can be heated to 150–250°C for several hours to drive gas molecules from the walls
...
The
sample is usually in the form of a thin foil, a filament, or a sharp point
...
Initial surface cleaning is achieved either by heating it electrically or by

Fig
...
5 Counter-rotating screw
dislocations on the same surface lead to the
formation of terraces
...
Subsequent
deposition can complete each terrace
...
25
...
Three successive stages of the
growth are shown
...
7 nm × 4
...
(a) The Si(001) surface before exposure to Si2H6(g)
...

(c) After 8 min, SiH3(surface) dissociates to SiH2(surface) and H(surface)
...
Wang, M
...
Bronikowski, and R
...
Hamers, Surface Science 64, 311
(1994)
...
25
...
The latter procedure demands care
because ion bombardment can shatter the surface structure and leave it an amorphous
jumble of atoms
...

We have already discussed three important techniques for the characterization of
surfaces: scanning electron microscopy (Impact I8
...
1), which reveals the atomic details of structure of the surface and of
adsorbates and can be used to visualize chemical reactions as they happen on surfaces
(Fig
...
7)
...

(a) Ionization techniques

Surface composition can be determined by a variety of ionization techniques
...
Their common feature is
that the escape depth of the electrons, the maximum depth from which ejected electrons
come, is in the range 0
...
0 nm, which ensures that only surface species contribute
...
4),
which in surface studies is normally called photoemission spectroscopy
...

In XPS, the energy of the incident photon is so great that electrons are ejected from
inner cores of atoms
...
Consequently, XPS gives lines characteristic of the
elements present on a surface
...
25
...
(M
...

Roberts and C
...
McKee, Chemistry of the
metal–gas interface, Oxford (1978)
...
25
...
This

25
...
The technique is very useful for studying the surface state of
heterogeneous catalysts, the differences between surface and bulk structures, and the
processes that can cause damage to high-temperature superconductors and semiconductor wafers
...
Its usefulness is its ability to reveal which orbitals of the
adsorbate are involved in the bond to the substrate
...
This difference is interpreted as meaning that the C6H6 molecules lie parallel to the surface and are attached to it by their π
orbitals
...
Among the advantages of SIMS are the ability to detect
adsorbed H and He atoms, which are not easily probed by XPS, and the high sensitivity of the mass spectrometer detector
...
However, it is possible to control the degree of
erosion to one or two monolayers by controlling the bombardment parameters
...
Measurement of transmitted radiation is not practical in surfaces, which are typically too opaque to infrared or visible
radiation
...

Raman spectroscopy is better suited for studies of surfaces because it involves the
detection of scattered radiation, but spectral bands are typically very weak
...
The effect is due in part to local accumulations of electron density
at the features of the roughened surface and at regions where bonding occurs
...

Disadvantages of SERS include weak enhancement observed on flat single crystal surfaces and the fact that the technique works for only certain metals
...
As in Raman spectroscopy, the spectrum of energy loss
can be interpreted in terms of the vibrational spectrum of the adsorbate
...
Very tiny amounts of adsorbate can be
detected, and one report estimated that about 48 atoms of phosphorus were detected
in one sample
...
25
...
The main peak arises

Fig
...
9 The electron energy loss spectrum
of CO adsorbed on Pt(111)
...

(Based on spectra provided by Professor
H
...
)

914

25 PROCESSES AT SOLID SURFACES
from CO attached perpendicular to the surface by a single Pt atom
...
This peak is due to CO
at a bridge site, attached to two Pt atoms, as in (1)
...
The Auger effect is the emission of a second
electron after high energy radiation has expelled another
...
The energy this
releases may result either in the generation of radiation, which is called X-ray fluorescence (Fig
...
10a) or in the ejection of another electron (Fig
...
10b)
...
The energies of the secondary electrons are
characteristic of the material present, so the Auger effect effectively takes a fingerprint
of the sample
...
In scanning Auger electron microscopy (SAM), the finely focused
electron beam is scanned over the surface and a map of composition is compiled; the
resolution can reach below about 50 nm
...
25
...
Alternatively, (b) the electron
falling into the orbital may give up its
energy to another electron, which is ejected
in the Auger effect
...
1)
...

These oscillations arise from a quantum mechanical interference between the wavefunction of a photoejected electron and parts of that electron’s wavefunction that are
scattered by neighbouring atoms
...
If the waves interfere constructively, then the photoelectron amplitude is higher,
and the photoelectron has a higher probability of appearing; correspondingly, the
X-ray absorption is greater
...
Such studies show that a solid’s
surface is much more plastic than had previously been thought, and that it undergoes
reconstruction, or structural modification, in response to adsorbates that are present
...
25
...

The electrons diffracted by the surface
layers are detected by the fluorescence they
cause on the phosphor screen
...
This technique
is like X-ray diffraction (Chapter 20) but uses the wave character of electrons, and the
sample is now the surface of a solid
...

The experimental arrangement is shown in Fig
...
11, and typical LEED patterns,
obtained by photographing the fluorescent screen through the viewing port, are
shown in Fig
...
12
...
By studying
how the diffraction intensities depend on the energy of the electron beam it is also
possible to infer some details about the vertical location of the atoms and to measure
the thickness of the surface layer, but the interpretation of LEED data is much more
complicated than the interpretation of bulk X-ray data
...
In practice, sharp patterns are obtained for surfaces ordered to depths of
about 20 nm and more
...
If the LEED pattern does not correspond to the pattern

25
...
25
...
CH
...
A
...
)

expected by extrapolation of the bulk surface to the surface, then either a reconstruction
of the surface has occurred or there is order in the arrangement of an adsorbed layer
...
As a general rule, it is found that metal surfaces are simply truncations of the bulk lattice, but the distance between the top layer
of atoms and the one below is contracted by around 5 per cent
...
Reconstruction occurs
in ionic solids
...
An actual example of the detail that can now
be obtained from refined LEED techniques is shown in Fig
...
13 for CH3C- adsorbed
on a (111) plane of rhodium
...
The importance of this type of measurement will emerge
later
...
25
...

The samples used were obtained by cleaving a crystal at different angles to a plane of

Fig
...
13 The structure of a surface close to
the point of attachment of CH3C- to the
(110) surface of rhodium at 300 K and the
changes in positions of the metal atoms
that accompany chemisorption
...
25
...

The photographs correspond to a platinum
surface with (top) low defect density,
(middle) regular steps separated by about
six atoms, and (bottom) regular steps
with kinks
...
A
...
)

916

25 PROCESSES AT SOLID SURFACES
atoms
...
The observation of additional structure
in the LEED patterns, rather than blurring, shows that the steps are arrayed regularly
...

One advantage is that the activity of specific crystal planes can be investigated by directing the beam on to an orientated surface with known step and kink densities (as
measured by LEED)
...
Another advantage is that the time of flight of a
particle may be measured and interpreted in terms of its residence time on the surface
...


The extent of adsorption
The extent of surface coverage is normally expressed as the fractional coverage, θ :

θ=

number of adsorption sites occupied
number of adsorption sites available

[25
...
The rate of adsorption, dθ/dt, is the rate of change of surface coverage,
and can be determined by observing the change of fractional coverage with time
...

One commonly used technique is therefore to monitor the rates of flow of gas into
and out of the system: the difference is the rate of gas uptake by the sample
...
In flash desorption the
sample is suddenly heated (electrically) and the resulting rise of pressure is interpreted
in terms of the amount of adsorbate originally on the sample
...
Gravimetry, in which the sample is weighed on a microbalance during the
experiment, can also be used
...

The key principle behind the operation of a QCM is the ability of a quartz crystal to
vibrate at a characteristic frequency when an oscillating electric field is applied
...
Masses as small
as a few nanograms (1 ng = 10−9 g) can be measured reliably in this way
...
3 Physisorption and chemisorption
Molecules and atoms can attach to surfaces in two ways
...
Van der
Waals interactions have a long range but are weak, and the energy released when a

25
...
Such small energies can be absorbed as vibrations of the lattice and dissipated
as thermal motion, and a molecule bouncing across the surface will gradually lose its
energy and finally adsorb to it in the process called accommodation
...
1)
...

In chemisorption (an abbreviation of ‘chemical adsorption’), the molecules (or
atoms) stick to the surface by forming a chemical (usually covalent) bond, and tend to
find sites that maximize their coordination number with the substrate
...
2)
...
A chemisorbed molecule may be torn apart at the demand of the
unsatisfied valencies of the surface atoms, and the existence of molecular fragments
on the surface as a result of chemisorption is one reason why solid surfaces catalyse
reactions
...
A spontaneous process
requires ∆G < 0
...
Therefore, in order for ∆G = ∆H − T∆S to be negative,
∆H must be negative (that is, the process is exothermic)
...
For example,
H2 adsorbs endothermically on glass because there is a large increase of translational
entropy accompanying the dissociation of the molecules into atoms that move quite
freely over the surface
...

The enthalpy of adsorption depends on the extent of surface coverage, mainly
because the adsorbate particles interact
...
Moreover, LEED studies show that such species settle
on the surface in a disordered way until packing requirements demand order
...
These adsorbates also show
order–disorder transitions when they are heated enough for thermal motion to overcome the particle–particle interactions, but not so much that they are desorbed
...
4 Adsorption isotherms
The free gas and the adsorbed gas are in dynamic equilibrium, and the fractional coverage of the surface depends on the pressure of the overlying gas
...

(a) The Langmuir isotherm

The simplest physically plausible isotherm is based on three assumptions:
1 Adsorption cannot proceed beyond monolayer coverage
...

3 The ability of a molecule to adsorb at a given site is independent of the occupation
of neighbouring sites (that is, there are no interactions between adsorbed molecules)
...
1* Maximum
observed enthalpies of physisorption
Adsorbate

Dad H 7/(kJ mol−1)

CH4

−21

H2

−84

H2O

−59

N2

−21

* More values are given in the Data section
...
2* Enthalpies of
chemisorption, ∆adH 7 /(kJ mol−1)
Adsorbate

Adsorbent (substrate)
Cr

CH4

Fe

Ni

−427

−285

−243

−192

CO
H2
NH3

−188

−134
−188

* More values are given in the Data section
...
The rate of change of surface coverage due to adsorption is proportional to the partial pressure p of A and the
number of vacant sites N(1 − θ ), where N is the total number of sites:
dθ
dt

= ka pN(1 − θ )

(25
...
3b)

At equilibrium there is no net change (that is, the sum of these two rates is zero), and
solving for θ gives the Langmuir isotherm:
Kp

θ=

K=

1 + Kp

ka

(25
...
1 Using the Langmuir isotherm

The data given below are for the adsorption of CO on charcoal at 273 K
...
In each case V has been corrected to 1
...
325 kPa)
...
3

26
...
0

53
...
7

80
...
3

3

10
...
6

25
...
5

36
...
6

46
...
4,

Kpθ + θ = Kp
With θ = V/V∞, where V∞ is the volume corresponding to complete coverage, this
expression can be rearranged into
p
V

=

p
V∞

+

1
KV∞

Hence, a plot of p/V against p should give a straight line of slope 1/V∞ and intercept
1/KV∞
...
3 26
...
0

53
...
7

80
...
3

1
...
44

1
...
69 1
...
92 2
...
25
...
The (least squares) slope is 0
...
The intercept at p = 0 is 1
...
25
...
1
...


K=

1
(111 cm ) × (1
...
51 × 10−3 kPa−1

25
...
1 Repeat the calculation for the following data:

p/kPa

13
...
7

40
...
3

66
...
0

93
...
3

19
...
3

34
...
0

45
...
0

V/cm

[128 cm3, 6
...
5a)

The rate of desorption is proportional to the frequency of encounters of atoms on the
surface, and is therefore second-order in the number of atoms present:
dθ
dt

= −kd(Nθ )2

(25
...
6)

1 + (Kp)1/2

The surface coverage now depends more weakly on pressure than for non-dissociative
adsorption
...
25
...
17
...
Different curves (and therefore different values of K) are obtained
at different temperatures, and the temperature dependence of K can be used to determine the isosteric enthalpy of adsorption, ∆adH 7, the standard enthalpy of adsorption
at a fixed surface coverage
...
23) to write:

A ∂ ln K D ∆ad H 7
=
C ∂T F θ RT 2

Fig
...
16 The Langmuir isotherm for
dissociative adsorption, X2(g) →
2 X(surface), for different values of K
...
4, generate
a family of curves showing the
dependence of 1/θ on 1/p for several
valuesof K
...
7)

Example 25
...
00 atm and 273 K) to be 10
...
1
...

T/K

200

210

220

230

240

250

p/kPa

4
...
95

6
...
20

8
...
85

Method The Langmuir isotherm can be rearranged to

Kp =

θ
1−θ

Therefore, when θ is constant,
ln K + ln p = constant

Fig
...
17 The Langmuir isotherm for nondissociative adsorption for different values
of K
...
6, generate
a family of curves showing the
dependence of 1/θ on 1/p for several
valuesof K
...


920

25 PROCESSES AT SOLID SURFACES
It follows from eqn 25
...

Answer We draw up the following table:

T/K

210

220

230

240

250

5
...
76

4
...
35

4
...
00

ln(p/kPa)
Fig
...
18 The isosteric enthalpy of
adsorption can be obtained from the slope
of the plot of ln p against 1/T, where p is the
pressure needed to achieve the specified
surface coverage
...
2
...
39

1
...
80

1
...
14

2
...
25
...
The slope (of the least squares fitted line) is
−0
...
904 × 103 K) × R = −7
...
The expression for
(∂ ln p/∂T)θ in this example is independent of the model for the isotherm
...
2 Repeat the calculation using the following data:

T/K

200

210

220

230

240

250

p/kPa

4
...
59

7
...
80

10
...
80
[−9
...
The most widely used isotherm dealing with multilayer adsorption was derived by Stephen Brunauer, Paul Emmett, and
Edward Teller, and is called the BET isotherm:
V
Vmon

Fig
...
19 Plots of the BET isotherm for
different values of c
...


Exploration Using eqn 25
...


=

cz
(1 − z){1 − (1 − c)z}

with z =

p
p*

(25
...
9)

Figure 25
...
They rise indefinitely as the
pressure is increased because there is no limit to the amount of material that may condense when multilayer coverage may occur
...


25
...
3 Using the BET isotherm

The data below relate to the adsorption of N2 on rutile (TiO2) at 75 K
...

p/kPa
V/mm

0
...
87

6
...
67

17
...
92

27
...
0 kPa
...
00 atm and 273 K and
refer to 1
...

Method Equation 25
...
The
results can then be combined to give c and Vmon
...
160 1
...
4

154

224

288

359

0
...
350 1
...
95

2
...
53

Fig
...
20 The BET isotherm can be tested,
and the parameters determined, by plotting
z/(1 − z)V against z = p/p*
...
3
...
67 17
...
92 27
...
11

103z

6
...
47

24
...
25
...
The least squares best line has an intercept at
0
...
98 × 10 −6 mm−3

The slope of the line is 1
...
23 × 10 −2) × 103 × 10 −4 mm−3 = 1
...
At 1
...
6 × 10−5 mol, or 2
...
Because each
atom occupies an area of about 0
...
5 m2
...
3 Repeat the calculation for the following data:

p/kPa

0
...
87

6
...
67

17
...
92

27
...
10)

This expression is applicable to unreactive gases on polar surfaces, for which c ≈ 102
because ∆desH 7 is then significantly greater than ∆vapH 7 (eqn 25
...
The BET isotherm

922

25 PROCESSES AT SOLID SURFACES
fits experimental observations moderately well over restricted pressure ranges, but it
errs by underestimating the extent of adsorption at low pressures and by overestimating it at high pressures
...
Deviations from the isotherm can often be traced to the failure
of these assumptions
...
Various attempts have been made to take these variations into account
...
11)

where c1 and c2 are constants, corresponds to supposing that the adsorption enthalpy
changes linearly with pressure
...
12)

corresponds to a logarithmic change
...
24)
...
Empirical, however, does not mean
useless for, if the parameters of a reasonably reliable isotherm are known, reasonably
reliable results can be obtained for the extent of surface coverage under various conditions
...

25
...
25
...
In each case, P is the enthalpy of
(non-dissociative) physisorption and C
that for chemisorption (at T = 0)
...


The rates of surface processes may be studied by techniques described in Section 25
...
1
...
We saw in Section 20
...
In addition to a number of crystals, surfaces are also suitable
materials for SHG
...

Figure 25
...
As the molecule approaches the surface its energy falls
as it becomes physisorbed into the precursor state for chemisorption
...
Even if the molecule does not
fragment, there is likely to be an initial increase of potential energy as the molecule
approaches the surface and the bonds adjust
...
This barrier, though, might be low, and
might not rise above the energy of a distant, stationary particle (as in Fig
...
21a)
...

Many gas adsorptions on clean metals appear to be non-activated
...
25
...
An example is H2 on copper, which has an
activation energy in the region of 20–40 kJ mol−1
...
5 THE RATES OF SURFACE PROCESSES

923

One point that emerges from this discussion is that rates are not good criteria for
distinguishing between physisorption and chemisorption
...
Physisorption is usually fast, but it can appear to be slow if adsorption is taking
place on a porous medium
...
If the energy is not dissipated quickly, the particle migrates over the
surface until a vibration expels it into the overlying gas or it reaches an edge
...
13]

The denominator can be calculated from the kinetic model, and the numerator can be
measured by observing the rate of change of pressure
...
For example, at room temperature CO has s in the range
0
...
0 for several d-metal surfaces, but for N2 on rhenium s < 10 −2, indicating
that more than a hundred collisions are needed before one molecule sticks successfully
...
74 on the (320) faces down to less than 0
...
The sticking probability decreases as the surface coverage
increases (Fig
...
22)
...
14)

where s0 is the sticking probability on a perfectly clean surface
...
The explanation is probably that the colliding molecule does not enter the chemisorbed state at
once, but moves over the surface until it encounters an empty site
...
A physisorbed particle vibrates in its shallow potential well, and
might shake itself off the surface after a short time
...
15)

Therefore, the half-life for remaining on the surface has a temperature dependence
t1/2 =

ln 2
kd

= τ0eE d /RT

τ0 =

ln 2
A

(25
...
) If we suppose that 1/τ 0 is approximately the
same as the vibrational frequency of the weak particle–surface bond (about 1012 Hz)
and Ed ≈ 25 kJ mol−1, then residence half-lives of around 10 ns are predicted at room
temperature
...
For chemisorption, with Ed = 100 kJ mol−1 and guessing that τ 0 = 10−14 s

Fig
...
22 The sticking probability of N2 on
various faces of a tungsten crystal and its
dependence on surface coverage
...
(Data provided by
Professor D
...
King
...
25
...
The three
peaks indicate the presence of three sites
with different adsorption enthalpies and
therefore different desorption activation
energies
...
W
...
D
...
Chem
...
51, 5352 (1969)
...

The desorption activation energy can be measured in several ways
...
Moreover, the transfer of concepts
such as ‘reaction order’ and ‘rate constant’ from bulk studies to surfaces is hazardous,
and there are few examples of strictly first-order or second-order desorption kinetics
( just as there are few integral-order reactions in the gas phase too)
...
A more
sophisticated technique is temperature programmed desorption (TPD) or thermal
desorption spectroscopy (TDS)
...
The TPD spectrum, the plot of desorption
flux against temperature, therefore shows a peak, the location of which depends on
the desorption activation energy
...
25
...

In many cases only a single activation energy (and a single peak in the TPD spectrum)
is observed
...
For instance, Cd atoms on tungsten
show two activation energies, one of 18 kJ mol −1 and the other of 90 kJ mol−1
...

Another example of a system showing two desorption activation energies is CO on
tungsten, the values being 120 kJ mol−1 and 300 kJ mol−1
...

(b) Mobility on surfaces

Fig
...
24 The events leading to an FIM
image of a surface
...
(The bouncing
motion is due to the intermolecular
potential, not gravity!)

A further aspect of the strength of the interactions between adsorbate and substrate is
the mobility of the adsorbate
...
The activation energy for diffusion over a surface need not be
the same as for desorption because the particles may be able to move through valleys
between potential peaks without leaving the surface completely
...
The defect
structure of the sample (which depends on the temperature) may also play a dominant role because the adsorbed molecules might find it easier to skip across a terrace
than to roll along the foot of a step, and these molecules might become trapped in
vacancies in an otherwise flat terrace
...

Diffusion characteristics of an adsorbate can be examined by using STM to follow
the change in surface characteristics or by field-ionization microscopy (FIM), which
portrays the electrical characteristics of a surface by using the ionization of noble gas
atoms to probe the surface (Fig
...
24)
...
A new image is then recorded,
and the new position of the atom measured (Fig
...
25)
...
1 IMPACT ON BIOCHEMISTRY: BIOSENSOR ANALYSIS

925

Fig
...
25 FIM micrographs showing the
migration of Re atoms on rhenium during
3 s intervals at 375 K
...
Ehrlich
...
The value of D for different
crystal planes at different temperatures can be determined directly in this way, and the
activation energy for migration over each plane obtained from the Arrhenius-like
expression
D = D0e−ED /RT

(25
...
Typical values for W atoms on tungsten
have ED in the range 57–87 kJ mol −1 and D0 ≈ 3
...
For CO on tungsten,
the activation energy falls from 144 kJ mol−1 at low surface coverage to 88 kJ mol−1
when the coverage is high
...
1 Biosensor analysis

Biosensor analysis is a very sensitive and sophisticated optical technique that is now
used routinely to measure the kinetics and thermodynamics of interactions between
biopolymers
...

The mobility of delocalized valence electrons accounts for the electrical conductivity of metals and these mobile electrons form a plasma, a dense gas of charged particles
...
Coulomb repulsion in the regions of high density causes electrons to move
away from each other, so lowering their density
...

Plasmons in the bulk may be visualized as waves that propagate through the solid
...

Biosensor analysis is based on the phenomenon of surface plasmon resonance
(SPR), the absorption of energy from an incident beam of electromagnetic radiation
by surface plasmons
...
It is common
practice to use a monochromatic beam and to vary the angle of incidence θ (Fig
...
26)
...
The angle corresponding to light absorption depends on the refractive index of
the medium in direct contact with the opposing side of the metallic film
...

As an illustration of biosensor analysis, we consider the association of two polymers,
A and B
...
Figure 25
...
25
...


Fig
...
27 The time dependence of a surface
plasmon resonance signal, R, showing
the effect of binding of a ligand to a
biopolymer adsorbed on to a surface
...
Passing
a solution containing no ligand over the
surface leads to dissociation and decrease
in R
...
The
system is normally allowed to reach equilibrium, which is denoted by the plateau in
Fig
...
27
...
Again, analysis of the decay of the SPR signal reveals the kinetics
of dissociation of the AB complex
...
25
...
Consider the equilibrium
A + B 5 AB

K=

kon
koff

where kon and koff are the rate constants for formation and dissociation of the AB complex, and K is the equilibrium constant for formation of the AB complex
...
18)

In a typical experiment, the flow rate of A is sufficiently high that [A] = a0 is essentially
constant
...
Finally, the SPR signal is often observed to be proportional to [AB]
...
We may then write
dR
dt

= kona 0(Rmax − R) − koff R = kona 0 Rmax − (kona 0 + koff)R

(25
...
It follows that (after some algebra)
Req = Rmax

A a0K D
C a0K + 1 F

(25
...

Biosensor analysis has been used in the study of thin films, metal–electrolyte surfaces, Langmuir–Blodgett films, and a number of biopolymer interactions, such as
antibody–antigen and protein–DNA interactions
...
For biological studies, the main disadvantage is the requirement for immobilization of at least one of the components of the system under study
...
3* Activation
energies of catalysed reactions
Reaction

Catalyst Ea /(kJ mol−1)

2 HI → H2 + I2

None

184

Au

105

Pt

59

2 NH3 →

None

350

N2 + 3 H2

W

162

* More values are given in the Data section
...
3)
...
In this section we consider heterogeneous catalysis, in which (as mentioned in the introduction to Section 23
...
For simplicity, we consider only gas/
solid systems
...

One consequence of the presence of a second species may be the modification of the
electronic structure at the surface of a metal
...
Such modifiers can act as promoters (to enhance
the action of catalysts) or as poisons (to inhibit catalytic action)
...
6 MECHANISMS OF HETEROGENEOUS CATALYSIS
25
...

This modification often takes the form of a fragmentation of the reactant molecules
...
Shape-selective catalysts, such as the zeolites
(Impact I25
...

The decomposition of phosphine (PH3) on tungsten is first-order at low pressures
and zeroth-order at high pressures
...
If the rate is supposed to be proportional to the surface
coverage and we suppose that θ is given by the Langmuir isotherm, we would write
v = kθ =

kKp

(25
...
When the pressure is so low that Kp < 1, we can
<
neglect Kp in the denominator and obtain
v = kKp

(25
...
When Kp > 1, we can neglect the 1 in the
>
denominator, whereupon the Kp terms cancel and we are left with
v=k

(25
...

Self-test 25
...

[v = k(Kp)1/2K′p′/(1 + K′p′)]

In the Langmuir–Hinshelwood mechanism (LH mechanism) of surface-catalysed
reactions, the reaction takes place by encounters between molecular fragments and
atoms adsorbed on the surface
...
23)

Insertion of the appropriate isotherms for A and B then gives the reaction rate in
terms of the partial pressures of the reactants
...
24)

then it follows that the rate law is
v=

kKAKB pA pB
(1 + KA pA + KB pB)2

(25
...


927

928

25 PROCESSES AT SOLID SURFACES
The Langmuir–Hinshelwood mechanism is dominant for the catalytic oxidation of
CO to CO2
...

The rate of formation of product is expected to be proportional to the partial pressure,
pB, of the non-adsorbed gas B and the extent of surface coverage, θA, of the adsorbed
gas A
...
26)

The rate constant, k, might be much larger than for the uncatalysed gas-phase reaction
because the reaction on the surface has a low activation energy and the adsorption
itself is often not activated
...
For example, if the adsorption of A follows a Langmuir isotherm
in the pressure range of interest, then the rate law would be
v=

kKpA pB
1 + KpA

(25
...
6 instead
...
27, when the partial pressure of A is high (in the sense KpA > 1)
there is almost complete surface coverage, and the rate is equal to kpB
...
When the pressure
of A is low (KpA < 1), perhaps because of its reaction, the rate is equal to kKpA pB; now
<
the extent of surface coverage is important in the determination of the rate
...
For example, the reaction between H(g)
and D(ad) to form HD(g) is thought to be by an ER mechanism involving the direct
collision and pick-up of the adsorbed D atom by the incident H atom
...

25
...
It has become possible to investigate how the
catalytic activity of a surface depends on its structure as well as its composition
...
The reaction
H2 + D2 → 2 HD has been studied in detail
...
Although the step itself might be
the important feature, it may be that the presence of the step merely exposes a more
reactive crystal face (the step face itself)
...
These observations suggest a reason why even small amounts of
impurities may poison a catalyst: they are likely to attach to step and kink sites, and so
impair the activity of the catalyst entirely
...

Molecular beam studies can also be used to investigate the details of the reaction
process, particularly by using pulsed beams, in which the beam is chopped into short

I25
...
4 Chemisorption abilities*
O2

C2H2

C2H4

CO

H2

CO2

N2

Ti, Cr, Mo, Fe

+

+

+

+

+

+

+

Ni, Co

+

+

+

+

+

+

−

Pd, Pt

+

+

+

+

+

−

−

Mn, Cu

+

+

+

+

±

−

−

Al, Au

+

+

+

+

−

−

−

Li, Na, K

+

+

−

−

−

−

−

Mg, Ag, Zn, Pb

+

−

−

−

−

−

−

* +, Strong chemisorption; ±, chemisorption; −, no chemisorption
...
The angular distribution of the products, for instance, can be used to assess
the length of time that a species remains on the surface during the reaction, for a long
residence time will result in a loss of memory of the incident beam direction
...
25
...

To be active, the catalyst should be extensively covered by adsorbate, which is the case if
chemisorption is strong
...
This pattern of behaviour suggests that the activity of a catalyst
should initially increase with strength of adsorption (as measured, for instance, by the
enthalpy of adsorption) and then decline, and that the most active catalysts should be
those lying near the summit of the volcano
...

Many metals are suitable for adsorbing gases, and the general order of adsorption
strengths decreases along the series O2, C2H2, C2H4, CO, H2, CO2, N2
...
Elements from the d block, such as
iron, vanadium, and chromium, show a strong activity towards all these gases, but
manganese and copper are unable to adsorb N2 and CO2
...
These trends are summarized in Table 25
...

IMPACT ON TECHNOLOGY

I25
...
5)
...
Other than the ones we
consider, these problems include the danger of the catalyst being poisoned by byproducts or impurities, and economic considerations relating to cost and lifetime
...
The alkene
(2) adsorbs by forming two bonds with the surface (3), and on the same surface there
may be adsorbed H atoms
...
The evidence for a two-stage reaction is the appearance of different isomeric
alkenes in the mixture
...
25
...
The lower curve refers to
the first series of d-block metals, the upper
curve to the second and third series d-block
metals
...


930

25 PROCESSES AT SOLID SURFACES
Table 25
...
The new alkene would not be formed if the two hydrogen atoms attached
simultaneously
...
Raw oils obtained from sources such as the
soya bean have the structure CH2(OOCR)CH(OOCR′)CH2(OOCR″), where R, R′,
and R″ are long-chain hydrocarbons with several double bonds
...
The geometrical configuration
of the chains is responsible for the liquid nature of the oil, and in many applications a
solid fat is at least much better and often necessary
...
The process, and the industry, is not made any easier by
the seasonal variation of the number of double bonds in the oils
...
Although
in some cases it is desirable to achieve complete oxidation (as in the production of
nitric acid from ammonia), in others partial oxidation is the aim
...
Likewise, the controlled oxidations of ethene to ethanol, ethanal (acetaldehyde), and (in the presence of acetic acid or chlorine) to chloroethene (vinyl chloride,
for the manufacture of PVC), are the initial stages of very important chemical industries
...

The physical chemistry of oxide surfaces is very complex, as can be appreciated by
considering what happens during the oxidation of propene to propenal on bismuth
molybdate
...
An O atom in the surface
can now transfer to this radical, leading to the formation of propenal and its desorption from the surface
...
The surface is left with vacancies and metal ions in
lower oxidation states
...
This sequence of events,
which is called the Mars van Krevelen mechanism, involves great upheavals of the
surface, and some materials break up under the stress
...
These small building blocks of polymers, perfumes,
and petrochemicals in general, are usually cut from the long-chain hydrocarbons
drawn from the Earth as petroleum
...
2 IMPACT ON TECHNOLOGY: CATALYSIS IN THE CHEMICAL INDUSTRY
long-chain hydrocarbons is called cracking, and is often brought about on silica–
alumina catalysts
...
These branched isomers burn
more smoothly and efficiently in internal combustion engines, and are used to produce
higher octane fuels
...
The platinum provides the metal function, and brings about
dehydrogenation and hydrogenation
...
The sequence of events in catalytic reforming
shows up very clearly the complications that must be unravelled if a reaction as
important as this is to be understood and improved
...
In this process first
one and then a second H atom is lost, and an alkene is formed
...
This carbocation can undergo several different reactions
...

Then the adsorbed molecule loses a proton, escapes from the surface, and migrates
(possibly through the gas) as an alkene to a metal part of the catalyst where it is
hydrogenated
...

The concept of a solid surface has been extended with the availability of
microporous materials, in which the surface effectively extends deep inside the
solid
...
25
...
Small neutral molecules,
such as CO2, NH3, and hydrocarbons (including aromatic compounds), can also
adsorb to the internal surfaces and we shall see that this partially accounts for the
utility of zeolites as catalysts
...
Examples include
the dehydration of methanol to form hydrocarbons such as gasoline and other fuels:
zeolite
x CH3OH −−−→ (CH2)x + xH2O

and the isomerization of m-xylene (8) to p-xylene (9)
...

Like enzymes, a zeolite catalyst with a specific compostion and structure is very selective
toward certain reactants and products because only molecules of certain sizes can enter

931

Fig
...
29 A framework representation of
the general layout of the Si, Al, and O
atoms in a zeolite material
...
Note the large central
pore, which can hold cations, water
molecules, or other small molecules
...
It is also possible that zeolites derive their
selectivity from the ability to bind and to stabilize only transition states that fit properly
in the pores
...


Processes at electrodes
A special kind of surface is an electrode and the special kind of process that occurs
there is the transfer of electrons
...
Indeed, the economic consequences of electron
transfer reactions are almost incalculable
...
3)
...
Each step of this wasteful
sequence could be improved by discovering more about the kinetics of electrochemical processes
...

As for homogeneous systems (Chapter 24), electron transfer at the surface of an
electrode involves electron tunnelling
...
Furthermore, specific interactions with the
electrode surface give the solute and solvent special properties that can be very different from those observed in the bulk of the solution
...
Then, we describe the kinetics of electrode processes by using a largely phenomenological (rather than strictly theoretical)
approach that draws on the thermodynamic language inspired by transition state theory
...
8 The electrode –solution interface

Fig
...
30 A simple model of the
electrode–solution interface treats it as two
rigid planes of charge
...

The plot shows the dependence of the
electric potential with distance from the
electrode surface according to this model
...


The most primitive model of the boundary between the solid and liquid phases is as
an electrical double layer, which consists of a sheet of positive charge at the surface of
the electrode and a sheet of negative charge next to it in the solution (or vice versa)
...
More sophisticated models for the electrode–solution interface attempt
to describe the gradual changes in the structure of the solution between two extremes:
the charged electrode surface and the bulk of the solution
...
In the Helmholtz layer
model of the interface the solvated ions arrange themselves along the surface of the
electrode but are held away from it by their hydration spheres (Fig
...
30)
...
In this simple model,
the electrical potential changes linearly within the layer bounded by the electrode surface on one side and the OHP on the other (see Exercise 25
...
In a refinement of this

25
...
The Helmholtz layer model ignores the disrupting effect of thermal
motion, which tends to break up and disperse the rigid outer plane of charge
...
9) with the latter’s single central
ion replaced by an infinite, plane electrode
...
31 shows how the local concentrations of cations and anions differ in
the Gouy–Chapman model from their bulk concentrations
...
The
modification of the local concentrations near an electrode implies that it might be
misleading to use activity coefficients characteristic of the bulk to discuss the thermodynamic properties of ions near the interface
...
Under such conditions, the activity coefficients are almost constant because the
inert ions dominate the effects of local changes caused by any reactions taking place
...

Neither the Helmholtz nor the Gouy–Chapman model is a very good representation of the structure of the double layer
...
The two are combined in
the Stern model, in which the ions closest to the electrode are constrained into a
rigid Helmholtz plane while outside that plane the ions are dispersed as in the Gouy–
Chapman model (Fig
...
32)
...


Fig
...
31 The Gouy–Chapman model of
the electrical double layer treats the outer
region as an atmosphere of counter-charge,
similar to the Debye–Hückel theory of ion
atmospheres
...


Fig
...
32 A representation of the Stern
model of the electrode–solution interface
...


933

934

25 PROCESSES AT SOLID SURFACES
(b) The electric potential at the interface

Fig
...
33 The variation of potential with
distance from an electrode that has been
separated from the electrolyte solution
without there being an adjustment of
charge
...


The potential at the interface can be analysed by imagining the separation of the electrode from the solution, but with the charges of the metal and the solution frozen in
position
...
25
...
As the test
charge approaches the electrode, which can be a metal or membrane electrode, it
enters a region where the potential varies more slowly
...
At
about 100 nm from the surface the potential varies only slightly with distance because
the closer the point of observation is to the surface, although the potential from a
given region of charge is stronger, a smaller area of surface is sampled (Fig
...
34)
...
As the test charge is taken
through the skin of electrons on the surface of the electrode, the potential it experiences changes until the probe reaches the inner, bulk metal environment, where the
potential is called the inner potential, φ
...

A similar sequence of changes of potential is observed as a positive test charge is
brought up to and through the solution surface
...

Now consider bringing the electrode and solution back together again but without
any change of charge distribution
...
Apart from a constant, this Galvani potential difference is the electrode potential that was discussed in
Chapter 7
...
1 for a quantitative treatment)
...
9 The rate of charge transfer

The origin of the distanceindependence of the outer potential
...

(b) Close to the electrode, the point charge
experiences a potential arising from a small
area but each contribution is strong
...

Fig
...
34

Because an electrode reaction is heterogeneous, it is natural to express its rate as the
flux of products, the amount of material produced over a region of the electrode surface in an interval of time divided by the area of the region and the duration of the
interval
...
28)

where [species] is the molar concentration of the relevant species in solution close to
the electrode, just outside the double layer
...
If the molar
concentrations of the oxidized and reduced materials outside the double layer are
[Ox] and [Red], respectively, then the rate of reduction of Ox, vOx, is
vOx = kc[Ox]

(25
...
29b)

(The notation kc and ka is justified below
...
For instance, in the deposition of

25
...

The net current density at the electrode is the difference between the current densities arising from the reduction of Ox and the oxidation of Red
...
Therefore, there is a cathodic current density of magnitude
jc = Fkc[Ox]

for

Ox + e− → Red

(25
...
There is also an opposing anodic current density of magnitude
ja = Fka[Red]

for

Red → Ox + e−

(25
...
The net current
density at the electrode is the difference
j = ja − jc = Fka[Red] − Fkc[Ox]

(25
...
25
...
25
...


Fig
...
35 The net current density is defined
as the difference between the cathodic and
anodic contributions
...
(b)
When jc > ja, the net current is cathodic,
and the net process is reduction
...
Likewise, a species already at
the inner plane must be detached and migrate into the bulk
...
4) as
k = Be−∆ G/RT
‡

(25
...

When eqn 25
...
30 we obtain
‡

j = FBa[Red]e−∆ Ga /RT − FBc[Ox]e−∆ Gc /RT
‡

‡

(25
...
That they are different is the central feature of the remaining
discussion
...
25
...

Consider the reduction reaction, Ox + e− → Red, and the corresponding reaction
profile
...
25
...
33a)

Thus, if the electrode is more positive than the solution, ∆φ > 0, then more work has
to be done to form an activated complex from Ox; in this case the activation Gibbs
energy is increased
...


Fig
...
36

936

25 PROCESSES AT SOLID SURFACES

Fig
...
37 When the transition state
resembles a species that has undergone
reduction, the activation Gibbs energy for
the anodic current is almost unchanged,
but the full effect applies to the cathodic
current
...


Fig
...
38 When the transition state
resembles a species that has undergone
oxidation, the activation Gibbs energy for
the cathodic current is almost unchanged
but the activation Gibbs energy for the
anodic current is strongly affected
...


Fig
...
39 When the transition state is
intermediate in its resemblance to reduced
and oxidized species, as represented here by
a peak located at an intermediate position
as measured by α (with 0 < α < 1), both
activation Gibbs energies are affected; here,
α ≈ 0
...
(a) Zero potential difference;
(b) nonzero potential difference
...
25
...
In a real system, the transition state has an intermediate resemblance to these extremes (Fig
...
39) and the activation Gibbs energy
for reduction may be written as
∆‡Gc = ∆‡Gc(0) + α F∆φ

(25
...
Experimentally, α is often found to be about 0
...

Now consider the oxidation reaction, Red + e− → Ox and its reaction profile
...
In this case, Red discards an electron to the electrode, so the
extra work is zero if the transition state is reactant-like (represented by a peak close to
the electrode)
...
In general, the activation Gibbs energy for this anodic process is
∆‡Ga = ∆‡Ga(0) − (1 − α )F∆φ

(25
...
32 with the result that

25
...
35)

This is an explicit, if complicated, expression for the net current density in terms of
the potential difference
...
35 can be simplified
...
36]

RT

Next, we identify the individual cathodic and anodic current densities:
‡
ja = FBa[Red]e−∆ Ga(0)/RTe(1−α)f∆φ 5
6 j = ja − jc
‡
jc = FBc[Ox]e−∆ Gc(0)/RTe−α f∆φ 7

(25
...
1 Calculating the current density 1

To calculate the change in cathodic current density at an electrode when the
potential difference changes from ∆φ′ and ∆φ, we use eqn 25
...
0 V, T = 298 K, and α = – (a typical value), we obtain
2

α f × (∆φ ′ − ∆φ) =

1
– × (9
...
0 V)
2

(8
...
6485 × 104 × 1
...
3145 × 298

Hence (after using 1 J = 1 VC),
j′
c
jc

−

=e

9
...
0
2×8
...
We can appreciate why the change is so
great by realizing that a change of potential difference by 1 V changes the activation
Gibbs energy by (1 V) × F, or about 50 kJ mol−1, which has an enormous effect on
the rates
...
5 Calculate the change in anodic current density under the same

[j′ /ja = 3 × 108]
a

circumstances
...
38)

When these equations apply, there is no net current at the electrode (as the cell is balanced), so the two current densities must be equal
...

When the cell is producing current (that is, when a load is connected between the
electrode being studied and a second counter electrode) the electrode potential changes

Comment 25
...

As explained earlier, they differ by a
constant amount, which may be
regarded as absorbed into the
constant B
...
39]

Hence, ∆φ changes to ∆φ = E + η and the two current densities become
ja = j0e(1−α)fη

jc = j0e−α fη

(25
...
32 we obtain the Butler–Volmer equation:
j = j0{e(1−α)fη − e−α fη}

(25
...

(d) The low overpotential limit

When the overpotential is so small that fη < 1 (in practice, η less than about 0
...
41 can be expanded by using ex = 1 + x + · · · to give
j = j0{1 + (1 − α)fη + · · · − (1 − αfη + · · · )} ≈ j0 fη

(25
...

When there is a small positive overpotential the current is anodic ( j > 0 when η > 0),
and when the overpotential is small and negative the current is cathodic ( j < 0 when
η < 0)
...
43)

Fj0

The importance of this interpretation will become clear below
...
2 Calculating the current density 2

The exchange current density of a Pt(s)|H2(g) |H+(aq) electrode at 298 K is
0
...
Therefore, the current density when the overpotential is +5
...
42 and f = F/RT = 1/(25
...
79 mA cm−2) × (5
...
69 mV

= 0
...
0 cm2 is therefore 0
...

Self-test 25
...
0, the other conditions being

the same?

[−18 mA (cathodic)]

(e) The high overpotential limit

When the overpotential is large and positive (in practice, η ≥ 0
...
41 is
much smaller than the first, and may be neglected
...
44)

ln j = ln j0 + (1 − α)fη

(25
...
9 THE RATE OF CHARGE TRANSFER

939

When the overpotential is large but negative (in practice, η ≤ −0
...
41 may be neglected
...
46)

ln(−j) = ln j0 − α fη

(25
...
The slope gives the value of α and the intercept at η = 0 gives the exchange
current density
...
25
...
A similar arrangement is typical of all kinds of electrochemical rate measurements
...
The current flowing through them is controlled externally
...
The potential
difference across the interface cannot be measured directly, but the potential of the
working electrode relative to a third electrode, the reference electrode, can be measured with a high impedance voltmeter, and no current flows in that half of the circuit
...
Changing the current flowing through the working circuit causes a change of potential of the working electrode, and that change is measured with the voltmeter
...


Fig
...
40 The general arrangement for
electrochemical rate measurements
...
No current flows in the reference
circuit
...
4 Interpreting a Tafel plot

The data below refer to the anodic current through a platinum electrode of area
2
...
Calculate the
exchange current density and the transfer coefficient for the electrode process
...
8

25
...
0

131

298

Method The anodic process is the oxidation Fe2+(aq) → Fe3+(aq) + e−
...
45)
...

Answer Draw up the following table:

η/mV

50

100

150

200

250

j/(mA cm−2)

4
...
5

29
...
5

149

ln( j/(mA cm−2))

1
...
53

3
...
18

5
...
25
...
The high overpotential region gives a straight
line of intercept 0
...
0165
...
88, so j0 = 2
...
From the latter,
(1 − α)

F
RT

= 0
...
25
...
The
data are from Example 25
...


940

25 PROCESSES AT SOLID SURFACES
so α = 0
...
Note that the Tafel plot is nonlinear for η < 100 mV; in this region
α fη = 2
...

Self-test 25
...
3

−1
...
4

−27
...
6

−510

[α = 0
...
041 mA cm−2]

Some experimental values for the Butler–Volmer parameters are given in Table
25
...
From them we can see that exchange current densities vary over a very wide
−
range
...
Exchange currents are generally large when the redox process involves
no bond breaking (as in the [Fe(CN)6]3−,[Fe(CN)6]4− couple) or if only weak bonds
are broken (as in Cl2,Cl−)
...

25
...

Voltammetry may also be used to identify species present in solution and to determine their concentration
...
Electrodes with potentials that change only slightly when
a current passes through them are classified as non-polarizable
...
From the linearized equation (eqn 25
...
The calomel and H2/Pt electrodes are both highly non-polarizable, which is one reason why they are so extensively
used as reference electrodes in electrochemistry
...
This assumption fails at high current
Synoptic table 25
...
9 × 10−4

Ni

6
...
0 × 10

Pt

2
...


0
...
58

25
...
A larger overpotential is then needed
to produce a given current
...

Consider a case for which the concentration polarization dominates all the rate processes and a redox couple of the type Mz+,M with the reduction Mz+ + z e− → M
...
29):
RT

E = E7 +

ln a

zF

(25
...
Therefore, the constant activity coefficient in a = γ c may be absorbed into E, and
we write the formal potential, E°, of the electrode as
RT

E° = E 7 +

zF

ln γ

[25
...
50)

When the cell is producing current, the active ion concentration at the OHP changes
to c′ and the electrode potential changes to
E′ = E ° +

RT
zF

ln c′

(25
...
52)

We now suppose that the solution has its bulk concentration, c, up to a distance
δ from the outer Helmholtz plane, and then falls linearly to c′ at the plane itself
...
25
...
The thickness of the Nernst
layer (which is typically 0
...

The concentration gradient through the Nernst layer is
dc
dx

=

c′ − c

δ

(25
...
The (molar) flux, J, is proportional to the concentration
gradient, and according to Fick’s first law (Section 21
...
54)

Therefore, the particle flux towards the electrode is
J=D

c − c′

δ

(25
...
25
...
Note that the diffusion layer is
much thicker relative to the OHP than
shown here
...
56)

For instance, for the couple [Fe(CN)6]2−/[Fe(CN)6]3−, z = 1, but for Fe3+/Fe, z = 3
...
This concentration occurs when an electron from an ion
that diffuses across the layer is snapped over the activation barrier and on to the electrode
...
57a)

By using the Nernst–Einstein equation (eqn 21
...
57b)

zFδ

Illustration 25
...
10 M Cu2+(aq) unstirred solution in which the thickness of the diffusion layer is about 0
...
With λ = 107 S cm2 mol−1 (Table 21
...
3 mm, c = 0
...
57b that
jlim = 5 mA cm−2
...

Self-test 25
...
010 mol dm−3 Ag+(aq) at 298 K
...
03 mm
...
56 that the concentration c′ is related to the current density
at the double layer by
c′ = c −

jδ

(25
...
However, this decline in concentration is small when the diffusion constant is
large, for then the ions are very mobile and can quickly replenish any ions that have
been removed
...
58 into eqn 25
...
59a)
(25
...
25
...
Initially, the absolute value of the potential is low, and the cathodic current is due to the

25
...
However, as the potential approaches the reduction
potential of the reducible solute, the cathodic current grows
...
57)
...
In differential pulse voltammetry the
current is monitored before and after a pulse of potential is applied, and the processed
output is the slope of a curve like that obtained by linear-sweep voltammetry (Fig
...
44)
...

In cyclic voltammetry the potential is applied in a sawtooth manner to the working electrode and the current is monitored
...
25
...
The shape of the curve is initially like that of a linear sweep experiment,
but after reversal of the sweep there is a rapid change in current on account of the
high concentration of oxidizable species close to the electrode that were generated on
the reductive sweep
...

When the reduction reaction at the electrode can be reversed, as in the case of the
[Fe(CN)6]3−/[Fe(CN)6]4− couple, the cyclic voltammogram is broadly symmetric about
the standard potential of the couple (as in Fig
...
45b)
...
As the
potential continues to change, the cathodic current begins to decline again because all

943

Fig
...
43 The change of potential with time
and the resulting current/potential curve in
a voltammetry experiment
...


Fig
...
44 A differential pulse voltammetry
experiment
...
The resulting current is shown as the
blue line and is sampled at the two points
shown
...


Fig
...
45 (a) The change of potential with
time and (b) the resulting current/potential
curve in a cyclic voltammetry experiment
...
The potential is now returned linearly to its initial value, and the reverse
series of events occurs with the [Fe(CN)6]4− produced during the forward scan now
undergoing oxidation
...

The overall shape of the curve gives details of the kinetics of the electrode process
and the change in shape as the rate of change of potential is altered gives information
on the rates of the processes involved
...
The appearance of the curve may also depend
on the timescale of the sweep for, if the sweep is too fast, some processes might not
have time to occur
...

Example 25
...

Method Decide which steps are likely to be reversible on the timescale of the

Fig
...
46 (a) When a non-reversible step in
a reaction mechanism has time to occur,
the cyclic voltammogram may not show
the reverse oxidation or reduction peak
...


potential sweep: such processes will give symmetrical voltammograms
...
However, at fast sweep rates, an intermediate might not have time
to react, and a reversible shape will be observed
...
25
...
At fast sweep rates, the second reaction does not have time
−
to take place before oxidation of the BrC6H4NO2 intermediate starts to occur
during the reverse scan, so the voltammogram will be typical of a reversible oneelectron reduction (Fig
...
46b)
...
9 Suggest an interpretation of the cyclic voltammogram shown in

Fig
...
47
...

[ClC6H4CN + e− 5 ClC6H4CN−,
−
+
−
ClC6H4CN + H + e → C6H5CN + Cl−, C6H5CN + e− 5 C6H5CN−]

25
...
25
...
9
...
The cell overpotential is the sum of the
overpotentials at the two electrodes and the ohmic drop (IR s, where R s is the internal

25
...
The additional
potential needed to achieve a detectable rate of reaction may need to be large when
the exchange current density at the electrodes is small
...
In this
section we see how to cope with both aspects of the overpotential
...

From eqn 25
...
60)

where j′ is the current density for electrodeposition and j is that for gas evolution, and
j′ and j0 are the corresponding exchange current densities
...
Note that η < 0 for a cathodic
process, so −η′ > 0
...
A
very crude criterion is that significant evolution or deposition occurs only if the overpotential exceeds about 0
...

Self-test 25
...


[j′/j = (δ j′ /cFD)e−αη′f]
0

A glance at Table 25
...
The most sluggish exchange currents occur for lead and
mercury, and the value of 1 pA cm−2 corresponds to a monolayer of atoms being
replaced in about 5 years
...
In contrast, the value for platinum (1 mA cm−2) corresponds to a monolayer being replaced in 0
...

The exchange current density also depends on the crystal face exposed
...
5 times the rate of the (111) face, for which
j0 = 0
...

25
...
Furthermore, we expect the cell potential to decrease as current is generated because it is then
no longer working reversibly and can therefore do less than maximum work
...
The potential of the cell is E′ = ∆φR − ∆φL
...

The cell potential is therefore
E′ = E + η R − η L

(25
...
61b)

with E the cell emf
...
61c)

The ohmic term is a contribution to the cell’s irreversibility—it is a thermal dissipation
term—so the sign of IRs is always such as to reduce the potential in the direction of zero
...
61 can be calculated from the Butler–Volmer equation for a given current, I, being drawn
...

Then from eqns 25
...
61c we find
4RT

E ′ = E − IRs −

F

ln

A I D
C AH F

H = ( j0L j0R)1/2

(25
...

The concentration overpotential also reduces the cell potential
...
59 as
RT 1 A
I DA
I D5
6
E′ = E −
ln 2 1 −
1−
(25
...
62 to obtain a full (but still very
approximate) expression for the cell potential when a current I is being drawn:
E ′ = E − IRs −

2RT

ln g(I)

(25
...
64b)

zF

with

Fig
...
48 The dependence of the potential
of a working galvanic cell on the current
density being drawn (blue line) and the
corresponding power output (purple line)
calculated by using eqns 25
...
65,
respectively
...


Exploration Using mathematical
software, and electronic spreadsheet,
or the interactive applets found in the
Living graphs section of the text’s web site,
confirm that the sharp decline in potential
and power observed in Fig
...
48 is true for
any value of Rs
...
25
...
Notice the very steep decline of working potential when the current is high and close to the limiting value for one of the electrodes
...
64 we can write
P = IE − I 2Rs −

2IRT
zF

ln g(I)

(25
...
The second term is the power generated uselessly as heat as a result of the resistance of the electrolyte
...

The general dependence of power output on the current drawn is shown in Fig
...
48
as the purple line
...
Information of this kind is
essential if the optimum conditions for operating electrochemical devices are to be
found and their performance improved
...
13 IMPACT ON TECHNOLOGY: FUEL CELLS

947

Electric storage cells operate as galvanic cells while they are producing electricity
but as electrolytic cells while they are being charged by an external supply
...
During charging the cathode reaction is the reduction of Pb2+ and its
deposition as lead on the lead electrode
...

The anode reaction during charging is the oxidation of Pb(II) to Pb(IV), which is
deposited as the oxide PbO2
...
Because
they have such high exchange current densities the discharge can occur rapidly, which
is why the lead battery can produce large currents on demand
...
3 Fuel cells

A fuel cell operates like a conventional galvanic cell with the exception that the reactants are supplied from outside rather than forming an integral part of its construction
...
25
...
One of the electrolytes used is
concentrated aqueous potassium hydroxide maintained at 200°C and 20–40 atm; the
electrodes may be porous nickel in the form of sheets of compressed powder
...
40 V

and the anode reaction is the oxidation
H2(g) + 2 OH−(aq) → 2 H2O(l) + 2 e−
For the corresponding reduction, E 7 = −0
...
Because the overall reaction
2 H2(g) + O2(g) → 2 H2O(l)

E 7 = +1
...
However, the
increased pressure compensates for the increased temperature, and E ≈ +1
...

One advantage of the hydrogen/oxygen system is the large exchange current density of the hydrogen reaction
...
1 nA cm−2, which limits the current available from the
cell
...
One type of highly developed fuel cell has phosphoric acid as the
electrolyte and operates with hydrogen and air at about 200°C; the hydrogen is
obtained from a reforming reaction on natural gas:
Anode:

2 H2(g) → 4 H+(aq) + 4 e−

Cathode:

O2(g) + 4 H+(aq) + 4 e− → 2 H2O(l)

This fuel cell has shown promise for combined heat and power systems (CHP systems)
...
Efficiency in a
CHP plant can reach 80 per cent
...
Although hydrogen gas is an attractive fuel, it has disadvantages for mobile applications: it is difficult to store and dangerous to handle
...
2)
...

Cells with molten carbonate electrolytes at about 600°C can make use of natural gas
directly
...
They include one version in which the

Fig
...
49 A single cell of a hydrogen/oxygen
fuel cell
...


948

25 PROCESSES AT SOLID SURFACES
electrolyte is a solid polymeric ionic conductor at about 100°C, but in current versions
it requires very pure hydrogen to operate successfully
...
Until these
materials have been developed, one attractive fuel is methanol, which is easy to handle
and is rich in hydrogen atoms:
Anode:

CH3OH(l) + 6 OH−(aq) → 5 H2O(l) + CO2(g) + 6 e−

Cathode:

O2(g) + 4 e− + 2 H2O(l) → 4 OH−(aq)

One disadvantage of methanol, however, is the phenomenon of ‘electro-osmotic
drag’ in which protons moving through the polymer electrolyte membrane separating
the anode and cathode carry water and methanol with them into the cathode compartment where the potential is sufficient to oxidize CH3OH to CO2, so reducing the
efficiency of the cell
...

A biofuel cell is like a conventional fuel cell but in place of a platinum catalyst it uses
enzymes or even whole organisms
...
One application will be as the
power source for medical implants, such as pacemakers, perhaps using the glucose
present in the bloodstream as the fuel
...
13 Corrosion
A thermodynamic warning of the likelihood of corrosion is obtained by comparing
the standard potentials of the metal reduction, such as
Fe2+(aq) + 2 e− → Fe(s)

E 7 = −0
...
23 V

In basic solution:
(c) 2 H2O(l) + O2(g) + 4 e− → 4 OH−(aq)

E 7 = +0
...
The electrode potentials we have quoted are standard values, and they change with the pH of the medium
...
059 V)pH
E(b) = E 7(b) + (RT/F)ln a(H+) = 1
...
059 V)pH
These expressions let us judge at what pH the iron will have a tendency to oxidize
(see Chapter 7)
...
If there is a thermodynamic tendency, we must
examine the kinetics of the processes involved to see whether the process occurs at a
significant rate
...
25
...
It can be taken to be a drop
of slightly acidic (or basic) water containing some dissolved oxygen in contact with
the metal
...
Those electrons are
replaced by others released elsewhere as Fe → Fe2+ + 2 e−
...
The droplet
acts as a short-circuited galvanic cell (Fig
...
50b)
...
4 IMPACT ON TECHNOLOGY: PROTECTING MATERIALS AGAINST CORROSION

949

The rate of corrosion is measured by the current of metal ions leaving the metal surface in the anodic region
...
We show in the justification below
that the corrosion current is related to the cell potential of the corrosion couple by
Icorr = H0 Ae fE /4

H0 = (j0 j0′)1/2

A = (AA′)1/2

(25
...
1 The corrosion current

Because any current emerging from the anodic region must find its way to the
cathodic region, the cathodic current, Ic, and the anodic current, Ia, must both be
equal to the corrosion current
...
67)

The Butler–Volmer equation is now used to express the current densities in terms of
overpotentials
...
46, j = −j0e−αfη) to apply, that polarization
overpotential can be neglected, that the rate-determining step is the transfer of a
1
single electron, and that the transfer coefficients are –
...
Moreover, because it is short-circuited by the
metal, the potential of the metal is the same in both regions, and so the potential
difference between the metal and the solution is the same in both regions too; it is
denoted ∆φcorr
...
66
...
Thermodynamically,
either hydrogen or oxygen reduction reaction (a) or (b) on p
...
However,
the exchange current density of reaction (b) on iron is only about 10−14 A cm−2,
whereas for (a) it is 10−6 A cm−2
...

For corrosion reactions with similar exchange current densities, eqn 25
...
That is, rapid corrosion can be expected
when the oxidizing and reducing couples have widely differing electrode potentials
...
4 Protecting materials against corrosion

Several techniques for inhibiting corrosion are available
...
66 we see
that the rate of corrosion depends on the surfaces exposed: if either A or A′ is zero,
then the corrosion current is zero
...
Paint also increases the effective solution resistance between the cathode and anode patches on the surface
...
The oxygen
then has access to the exposed metal and corrosion continues beneath the paintwork
...
Because the latter’s standard potential is −0
...
25
...
The oxidation of the
iron takes place in the region away from the
oxygen because the electrons are
transported through the metal
...


950

25 PROCESSES AT SOLID SURFACES
favoured and the iron survives (the zinc survives because it is protected by a hydrated
oxide layer)
...
14 V)
oxidizes the iron couple (E 7 = −0
...
Some oxides are inert kinetically in the sense
that they adhere to the metal surface and form an impermeable layer over a fairly wide
pH range
...
Thus, aluminium is inert in air even
though its standard potential is strongly negative (−1
...

Another method of protection is to change the electric potential of the object by
pumping in electrons that can be used to satisfy the demands of the oxygen reduction
without involving the oxidation of the metal
...
36 V)
...
25
...
A block of
magnesium replaced occasionally is much cheaper than the ship, building, or pipeline
for which it is being sacrificed
...
25
...

Table 25
...
(b) In impressed-current
cathodic protection electrons are supplied
from an external cell so that the object itself
is not oxidized
...

Fig
...
51

AES
AFM
BET isotherm
EELS
ER mechanism
ESCA
FIM
HREELS
IHP
LEED
LH mechanism
MBRS
MBS
OHP
QCM
RAIRS
SAM
SAM
SEM
SERS
SEXAFS
SHG
SIMS
SPM
SPR
STM
TDS
TPD
UHV
UPS
XPS

Auger electron spectroscopy
Atomic force microscopy
Brunauer–Emmett–Teller isotherm
Electron energy-loss spectroscopy
Eley–Rideal mechanism
Electron spectroscopy for chemical analysis
Field-ionization microscopy
High-resolution electron energy-loss spectroscopy
Inner Helmholtz plane
Low-energy electron diffraction
Langmuir–Hinshelwood mechanism
Molecular beam reactive scattering
Molecular beam scattering
Outer Helmholtz plane
Quartz crystal microbalance
Reflection–absorption infrared spectroscopy
Scanning Auger electron microscopy
Self-assembled monolayer
Scanning electron microscopy
Surface-enhanced Raman scattering
Surface-extended X-ray absorption fine structure spectroscopy
Second harmonic generation
Secondary ion mass spectrometry
Scanning probe microscopy
Surface plasmon resonance
Scanning tunnelling microscopy
Thermal desorption spectroscopy
Temperature programmed desorption
Ultra-high vacuum
Ultraviolet photoemission spectroscopy
X-ray photoemission spectroscopy

FURTHER READING

951

Checklist of key ideas
1
...
The reverse of
adsorption is desorption
...
The collision flux, Z W, of gas molecules bombarding a solid
surface is related to the gas pressure by Z W = p/(2πmkT)1/2
...
Techniques for studying surface composition and structure
include scanning electron microscopy (SEM), scanning probe
microscopy (STM), photoemission spectroscopy, sescondaryion mass spectrometry, surface-enhanced Raman scattering
(SERS), Auger electron spectroscopy (AES), low energy
electron diffraction (LEED), and molecular beam scattering
(MBS)
...
The fractional coverage, θ, is the ratio of the number of
occupied sites to the number of available sites
...
Techniques for studying the rates of surface processes
include flash desorption, biosensor analysis, second harmonic
generation (SHG), gravimetry by using a quartz crystal
microbalance (QCM), and molecular beam reactive
scattering (MRS)
...
Physisorption is adsorption by a van der Waals interaction;
chemisorption is adsorption by formation of a chemical
(usually covalent) bond
...
The Langmuir isotherm is a relation between the fractional
coverage and the partial pressure of the adsorbate:
θ = Kp/(1 + Kp)
...
The isosteric enthalpy of adsorption is determined from a plot
of ln K against 1/T
...
The BET isotherm is an isotherm applicable when multilayer
adsorption is possible: V/Vmon = cz/(1 − z){1 − (1 − c)z},
with z = p/p*
...
The sticking probability, s, is the proportion of collisions with
the surface that successfully lead to adsorption
...
Desorption is an activated process with half-life t1/2 = τ0eEd /RT;
the desorption activation energy is measured by temperature-

programmed desorption (TPD) or thermal desorption
spectroscopy (TDS)
...
In the Langmuir–Hinshelwood mechanism (LH mechanism)
of surface-catalysed reactions, the reaction takes place by
encounters between molecular fragments and atoms adsorbed
on the surface
...
In the Eley–Rideal mechanism (ER mechanism) of a surfacecatalysed reaction, a gas-phase molecule collides with another
molecule already adsorbed on the surface
...
An electrical double layer consists of a sheet of positive charge
at the surface of the electrode and a sheet of negative charge
next to it in the solution (or vice versa)
...
The Galvani potential difference is the potential difference
between the bulk of the metal electrode and the bulk of the
solution
...
Models of the double layer include the Helmholtz layer model
and the Gouy–Chapman model
...
The current density, j, at an electrode is expressed by the
Butler–Volmer equation, j = j0{e(1−α)fη − e−αfη}, where η is the
overpotential, η = E′ − E, α is the transfer coefficient, and j0 is
the exchange-current density
...
A Tafel plot is a plot of the logarithm of the current density
against the overpotential: the slope gives the value of α and
the intercept at η = 0 gives the exchange-current density
...
Voltammetry is the study of the current through an electrode
as a function of the applied potential difference
...

20
...

21
...


Further reading
Articles and texts

A
...
Adamson and A
...
Wiley, New
York (1997)
...
J
...
R
...
Wiley, New York (2000)
...
O’M
...
E
...
E
...
), Modern aspects of
electrochemistry
...
33
...

G
...
Knözinger, and J
...
VCH, Weinheim (1997)
...
G
...
W
...
), Advances in corrosion science
and technology
...

C
...
Hamann, W
...
Hammett, Electrochemistry
...

J
...
Lindon, G
...
Tranter, and J
...
Holmes (ed
...
Academic Press, San Diego (2000)
...
Mizuno and M
...
Chem
...
98,
199 (1998)
...
A
...
Chem
...
96, 1223 (1996)
...
D
...
Tuck, Modern battery construction
...

J
...
Wiley,
New York (1997)
...
M
...
Brett and A
...
O
...
Oxford
Chemistry Primers, Oxford University Press (1998)
...
Linden (ed
...
McGraw-Hill,
New York (1984)
...
1 The relation between electrode potential
and the Galvani potential

To demonstrate the relation between ∆φ and E, consider the cell
Pt|H2(g)|H+(g)|| M+(aq)|M(s) and the half-reactions
M+(aq) + e− → M(s)

1
H+(aq) + e− → – H2(g)
2

The Gibbs energies of these two half-reactions can be expressed in
terms of the chemical potentials, µ, of all the species
...
Thus, a cation in a region of positive
potential has a higher chemical potential (is chemically more active
in a thermodynamic sense) than in a region of zero potential
...
Because at constant
temperature and pressure the maximum electrical work can be
identified with the change in Gibbs energy (Section 7
...
The chemical potential of an ion in
the presence of an electric potential is called its electrochemical
potential, G
...
68]

where µ is the chemical potential of the species when the electrical
potential is zero
...

To express the Gibbs energy for the half-reactions in terms of the
electrochemical potentials of the species we note that the cations M+
are in the solution where the inner potential is φS and the electrons
are in the electrode where it is φM
...
Likewise, in the hydrogen half-reaction, the electrons
are in the platinum electrode at a potential φPt and the H+ ions are in
the solution where the potential is φS:
1
∆rGL = – G(H2) − {G(H+) + G(e−)}
2
1
= – µ(H2) − µ(H+) − µ(e−) + F∆φL
2

where ∆φL = φPt − φS is the Galvani potential difference at the lefthand electrode
...
When the cell is
balanced against an external source of potential the entire system is at
equilibrium
...
69)

If we compare this with the result established in Section 7
...
70)

This is the result we wanted to show, for it implies that the Galvani
potential difference at each electrode can differ from the electrode
potential by a constant at most; that constant cancels when the
difference is taken
...
1 (a) Distinguish between a step and a terrace
...

25
...
(b) Describe the advantages and

limitations of each of the microscopy, diffraction, and scattering techniques
designated by the acronyms AFM, FIM, LEED, MBRS, MBS, SAM, SEM,
and STM
...
3 Distinguish between the following adsorption isotherms: Langmuir,
BET, Temkin, and Freundlich
...
4 Consider the analysis of surface plasmon resonance data (as in biosensor
analysis) and discuss how a plot of a0 /Req against a0 may be used to evaluate
Rmax and K
...
5 Describe the essential features of the Langmuir–Hinshelwood,

Eley–Rideal, and Mars van Krevelen mechanisms for surface-catalysed
reactions
...
6 Account for the dependence of catalytic activity of a surface on the
strength of chemisorption, as shown in Fig
...
28
...
7 Discuss the unique physical and chemical properties of zeolites that

make them useful heterogeneous catalysts
...
8 (a) Discuss the main structural features of the electrical double layer
...

25
...

25
...

25
...
46
...
11 Discuss the principles of operation of a fuel cell
...
12 Discuss the chemical origins of corrosion and useful strategies for
preventing it
...
1a Calculate the frequency of molecular collisions per square centimetre
of surface in a vessel containing (a) hydrogen, (b) propane at 25°C when the
pressure is (i) 100 Pa, (ii) 0
...


energy for desorption
...
1b Calculate the frequency of molecular collisions per square centimetre

weakly so
...
What is the activation energy for chemisorption?

of surface in a vessel containing (a) nitrogen, (b) methane at 25°C when the
pressure is (i) 10
...
150 µTorr
...
2a What pressure of argon gas is required to produce a collision rate of

4
...
5 mm?

25
...
00 × 1019 s−1 at 525 K on a circular surface of diameter 2
...
3a Calculate the average rate at which He atoms strike a Cu atom in a
surface formed by exposing a (100) plane in metallic copper to helium gas at
80 K and a pressure of 35 Pa
...

25
...
Crystals of iron are body-centred cubic with
a cell edge of 145 pm
...
4a A monolayer of N2 molecules (effective area 0
...
00 g of an Fe/Al2O3 catalyst at 77 K, the boiling point of liquid
nitrogen
...
86 cm3 at 0°C and 760 Torr
...
4b A monolayer of CO molecules (effective area 0
...
00 g of an Fe/Al2O3 catalyst at 77 K, the boiling point of liquid
nitrogen
...
25 cm3 at 0°C and
1
...
What is the surface area of the catalyst?
25
...
00 g of a sample of silica at 0°C was 0
...
4 Torr and 1
...
What is the value of Vmon?
25
...
00 bar adsorbed on the surface of
3

3

1
...
60 cm at 52
...
73 cm at
104 kPa
...
6a The enthalpy of adsorption of CO on a surface is found to be −120 kJ
−1

mol
...

25
...
Estimate the mean lifetime of an NH3 molecule on the
surface at 500 K
...
7a The average time for which an oxygen atom remains adsorbed to a

tungsten surface is 0
...
49 s at 2362 K
...
7b The chemisorption of hydrogen on manganese is activated, but only

25
...
85 kPa−1 at 25°C
...
15, (b) 0
...

25
...
777 kPa−1 at 25°C
...
20, (b) 0
...


25
...
44 mg of CO when the pressure of the
gas is 26
...
The mass of gas adsorbed when
the pressure is 3
...
19 mg
...
Find the fractional coverage of
the surface at the two pressures
...
9b A certain solid sample adsorbs 0
...
0 kPa and the temperature is 300 K
...
0 kPa and the temperature is 300 K is 0
...
The Langmuir
isotherm is known to describe the adsorption
...

25
...
10 ps
...
10b For how long on average would an atom remain on a surface at 400 K
if its desorption activation energy were (a) 20 kJ mol−1, (b) 200 kJ mol−1? Take
τ0 = 0
...
For how long on average would the same atoms remain at 800 K?
25
...
5 mg of the

gas and obeys the Langmuir isotherm
...
00 mmol
of the adsorbed gas is desorbed is +10
...
What is the equilibrium pressure for
the adsorption of 2
...
11b A solid in contact with a gas at 8
...
67 mg
of the gas and obeys the Langmuir isotherm
...
00 mmol of the adsorbed gas is desorbed is +12
...
What is the equilibrium
pressure for the adsorption of the same mass of gas at 45°C?
25
...
Assume the adsorption follows the Langmuir isotherm
and predict the order of the HI decomposition reaction on each of the two
metal surfaces
...
12b Suppose it is known that ozone adsorbs on a particular surface in
accord with a Langmuir isotherm
...
13a Nitrogen gas adsorbed on charcoal to the extent of 0
...
2 MPa
...
13b Nitrogen gas adsorbed on a surface to the extent of 1
...
02 MPa
...
14a In an experiment on the adsorption of oxygen on tungsten it was

found that the same volume of oxygen was desorbed in 27 min at 1856 K and
2
...
What is the activation energy of desorption? How long
would it take for the same amount to desorb at (a) 298 K, (b) 3000 K?
25
...
44 s at
1012 K
...
15a The Helmholtz model of the electric double layer is equivalent to a

25
...
6 for the exchange current density and
transfer coefficient for the reaction Fe3+ + e−→ Fe2+ on platinum at 25°C to
determine what current density would be needed to obtain an overpotential of
0
...
Is the validity of the Tafel approximation affected at higher
overpotentials (of 0
...
20a Estimate the limiting current density at an electrode in which the

concentration of Ag+ ions is 2
...
The thickness of the
Nernst diffusion layer is 0
...
The ionic conductivity of Ag+ at infinite
dilution and 25°C is 6
...

25
...
5 mmol dm−3 at 25°C
...
32 mm
...
60 mS m2 mol−1
...
21a A 0
...
00 mA cm−2
...
60 V
...

25
...
10 M FeSO4(aq) solution is electrolysed between a magnesium

cathode and a platinum anode with a current density of 1
...
The
hydrogen overpotential is 0
...
What will be the concentration of Fe2+ ions
when evolution of H2 just begins at the cathode? Assume all activity
coefficients are unity
...
Hence the potential difference across the double layer
is given by ∆ϕ = σd/ε, where d is the distance between the plates and σ is the
surface charge density
...
0 M NaCl(aq) if the surface charge density is 0
...


is 0
...
What is the current density at an electrode when its
overpotential is (a) 10 mV, (b) 100 mV, (c) −5
...
5
...
15b Refer to the preceding exercise
...
22b The exchange current density for a Pt|Fe3+,Fe2+ electrode is 2
...
5 M NaCl(aq) if the surface charge density is
0
...

25
...
39
...
0 mA cm−2 when the overvoltage is 125 mV
...
16b The transfer coefficient of a certain electrode in contact with M2+ and
3+

M in aqueous solution at 25°C is 0
...
The current density is found to be
17
...
What is the overvoltage
required for a current density of 72 mA cm−2?
25
...
16a
...
17b Determine the exchange current density from the information given

in Exercise 25
...


25
...
The standard potential of the electrode is +0
...
Calculate the current
flowing through an electrode of surface area 1
...
Take unit activity for both ions
...
23a Suppose that the electrode potential is set at 1
...
The exchange

current density is 6
...
50
...
1 to 10
...


25
...
50 V
...
1 to
10
...

25
...
10?
The surface area of the electrode is 1
...

25
...
010?
25
...
18a To a first approximation, significant evolution or deposition occurs in

electrolysis only if the overpotential exceeds about 0
...
To illustrate this
criterion determine the effect that increasing the overpotential from 0
...
60 V has on the current density in the electrolysis of 1
...
0 mA cm−2 at 0
...
Take α = 0
...

25
...
50 V to
0
...
0 M NaOH(aq), which
is 1
...
50 V and 25°C
...
50
...
19a Use the data in Table 25
...
20 V as calculated from (a) the Butler–Volmer equation, and (b) the Tafel
equation
...
4 V and more)?

layer in each second when the Pt,H2 |H+, Pt|Fe3+,Fe2+, and Pb,H2 |H+
electrodes are at equilibrium at 25°C? Take the area as 1
...

Estimate the number of times each second a single atom on the surface takes
part in a electron transfer event, assuming an electrode atom occupies about
(280 pm)2 of the surface
...
25b How many electrons or protons are transported through the double
layer in each second when the Cu,H2 |H+ and Pt|Ce4+,Ce3+ electrodes are at
equilibrium at 25°C? Take the area as 1
...
Estimate the
number of times each second a single atom on the surface takes part in a
electron transfer event, assuming an electrode atom occupies about (260 pm)2
of the surface
...
26a What is the effective resistance at 25°C of an electrode interface when

the overpotential is small? Evaluate it for 1
...


PROBLEMS
25
...
0 cm2 (a) Pb,H2 |H+, (b) Pt|Fe2+,Fe3+ electrodes
...
27a State what happens when a platinum electrode in an aqueous solution

955

25
...
76 V at 25°C
...
79 mA cm−2
...
)

containing both Cu2+ and Zn2+ ions at unit activity is made the cathode of an
electrolysis cell
...
30a Can magnesium be deposited on a zinc electrode from a unit activity

25
...


25
...
28a What are the conditions that allow a metal to be deposited from

nickel–cadmium cell, and the maximum possible power output when 100 mA
is drawn at 25°C
...
28b The overpotential for hydrogen evolution on cadmium is about 1 V at

current densities of 1 mA cm−2
...
29a The exchange current density for H discharge at zinc is about 50 pA

cm−2
...
31a Calculate the maximum (zero-current) potential difference of a

25
...

25
...
0 A m−2
...

25
...
0 A m−2
...


Problems*
Numerical problems
25
...
As an illustration, consider a two-dimensional square lattice of
univalent positive and negative ions separated by 200 pm, and consider a
cation on the upper terrace of this array
...
Now consider a high step in the same lattice, and let the cation move
into the corner formed by the step and the terrace
...

25
...
Calculate the
collision frequency per square centimetre of surface made by O2 molecules at
(a) 100 kPa, (b) 1
...
Estimate the number of collisions made
with a single surface atom in each second
...
Take the nearestneighbour distance as 291 pm
...
3 Nickel is face-centred cubic with a unit cell of side 352 pm
...
10 µTorr
...
4 The data below are for the chemisorption of hydrogen on copper

powder at 25°C
...
Then find the value of K for the adsorption equilibrium and the
adsorption volume corresponding to complete coverage
...
042

0
...
221

0
...
411

0
...
5 The data for the adsorption of ammonia on barium fluoride are
reported below
...


(a) θ = 0°C, p* = 429
...
0

37
...
6

79
...
7

100
...
4

V/cm3

11
...
5

14
...
0

15
...
3

16
...
6°C, p* = 819
...
3

8
...
4

29
...
1

74
...
1

102
...
2

9
...
3

11
...
9

13
...
4

14
...
6 The following data have been obtained for the adsorption of H2 on the

surface of 1
...
The volume of H2 below is the volume that
the gas would occupy at STP (0°C and 1 atm)
...
050

0
...
150

0
...
250

V/cm3

1
...
33

1
...
36

1
...
The density of liquid hydrogen is
0
...

25
...
Check the applicability of this isotherm to the following
data for the adsorption of acetic acid on charcoal at 25°C and find the values
of the parameters c1 and c2
...
05

0
...
50

1
...
5

wa /g

0
...
06

0
...
16

0
...

25
...
This is the case, for instance, in the catalytic decomposition of
ammonia on platinum at 1000°C
...


956

25 PROCESSES AT SOLID SURFACES

this type of process, show that the rate of ammonia decomposition should
follow
dpNH3
dt

= −kc

pNH3
pH2

in the limit of very strong adsorption of hydrogen
...
Solve the rate equation for the
catalytic decomposition of NH3 on platinum and show that a plot of F(t) =
(1/t) ln(p/p0) against G(t) = (p − p0)/t, where p is the pressure of ammonia,
should give a straight line from which kc can be determined
...

t/s

0

30

60

100

160

200

250

p/kPa

13
...
7

11
...
7

10
...
9

9
...
9‡ A
...
Zardkoohi (J
...
Eng
...
They fitted their observations to a Freundlich isotherm of the form
cads = Kc1/n, where cads is the concentration of adsorbed phenol and csol is
sol
the concentration of aqueous phenol
...
26

15
...
43

31
...
00

c ads /(mg g −1)

4
...
2

35
...
0

67
...
What further information would be
necessary in order to express the data in terms of fractional coverage, θ ?
25
...
Huang and W
...
Cheng (J
...
188, 270 (1997))

examined the adsorption of the hexacyanoferrate(III) ion, [Fe(CN)6]3−, on γAl2O3 from aqueous solution
...
5:
T/K

283

298

308

318

10 −11K

2
...
078

1
...
085

Determine the isosteric enthalpy of adsorption, ∆adsH 7, at this pH
...

Determine ∆adsG 7
...
11‡ M
...
Olivier and R
...
Chem
...
Data 42, 230 (1997))
studied the adsorption of butane on silica gel
...
00

38
...
03

76
...
97

n/(mol kg −1)

1
...
17

1
...
04

2
...
47

165
...
41

205
...
91

n/(mol kg −1)

2
...
22

3
...
35

3
...
12‡ The following data were obtained for the extent of adsorption, s, of

acetone on charcoal from an aqueous solution of molar concentration, c,
at 18°C
...
0 23
...
0

s/(mmol acetone/g charcoal) 0
...
75 1
...
0 165

390

800

1
...
15 3
...
10

Which isotherm fits this data best, Langmuir, Freundlich, or Temkin?
25
...
Evaluate α and j0 for the
electrode
...
66

8
...
9

100

−
25
...
9 µA cm−2 when the concentration of KI is 6
...
The diffusion coefficient of I3 is
−9 2 −3
1
...
What is the thickness of the diffusion layer?

25
...
65 summarizes, in an
approximate way, some of the parameters involved
...
Express E in terms of the concentration and
conductivities of the ions present in the cell
...
Take electrodes of area 5 cm2 separated
by 5 cm
...
Take the concentration as 1 mol dm−3, the temperature 25°C, and
neglect activity coefficients
...
On the
same graph, plot the power output of the cell
...
17 Consider a cell in which the current is activation-controlled
...
Carry through this analysis for the cell
in Problem 25
...

25
...
Kanya (J
...

Chem
...
The values in the table below are based on the data
obtained with an electrode of surface area 9
...
70 µmol dm−3 in Fe2+
...
(b) Calculate the
cathodic current density, jc, from the rate of deposition of Fe2+ for each value
of E
...

v/(pmol s−1)

1
...
18

3
...
26

−E /mV

702

727

752

812

25
...


−3

25
...

What should their relative activities be in order to ensure simultaneous
deposition from a mixture?

335

How would the current density at this electrode depend on the overpotential
of the same set of magnitudes but of opposite sign?

Gouy–Chapman model is given by eqn 19
...
Use this equation to calculate
and plot the thickness as a function of concentration and electrolyte type at
25°C
...
1 to 100 mmol dm−3
...
20‡ V
...
Losev and A
...
Pchel’nikov (Soviet Electrochem
...
388

0
...
350

0
...
590

1
...
507

Use these data to calculate the transfer coefficient and the exchange current
density
...
365 V?
25
...
In a study
of methone (1,1-dimethyl-3,5-cyclohexanedione) by E
...
Hermolin,
and E
...
Acta 16, 1437 (1971)), the following
current–voltage data were obtained for the reduction of the quinone in
anhydrous butanol on a mercury electrode:
−E/V

1
...
58

1
...
72

1
...
98

≥2
...
Does this mechanism help to explain the
current–voltage data?
25
...
Bowden and
T
...
Roy
...
A120, 59 (1928)), who measured the overpotential
for H2 evolution with a mercury electrode in dilute aqueous solutions of
H2SO4 at 25°C
...
9

η /V

0
...
65 0
...
3

28

957

the interface show if it varies linearly and periodically (as a sawtooth
1
waveform) between η− and η+ around η0? Take α = –
...
30 Derive an expression for the current density at an electrode where the
rate process is diffusion-controlled and ηc is known
...
What changes occur if anion currents are involved?

Applications: to chemical engineering and environmental
science
25
...
As a first step in
the investigation they determined the form of the adsorption isotherm
...
Is the Langmuir isotherm suitable at this pressure?

p/kPa

13
...
7

40
...
3

66
...
0

17
...
0

47
...
8

75
...
3

1650

3300

V/cm 3

0
...
84 0
...
93

0
...
At 15°C, p*( butadiene) = 200 kPa
...


100

250

630

Explain any deviations from the result expected from the Tafel equation
...
32‡ In a study relevant to automobile catalytic converters, C
...
Wartnaby,

Theoretical problems
25
...
Confirm this assertion
...
Estimate the equilibrium distance of an atom
above the surface
...
24 Use the Gibbs adsorption isotherm (another name for eqn 19
...

25
...
50, can be
integrated to give a relation between Va and p, as in a normal adsorption
isotherm
...


A
...
Y
...
A
...
Phys
...
100, 12483 (1996))
measured the enthalpy of adsorption of CO, NO, and O2 on initially clean
platinum 110 surfaces
...
How
much more strongly adsorbed is NO at 500°C than at 400°C?
25
...

Activated carbon has long been used as an adsorbent in this process, but the
presence of moisture in the stream reduces its effectiveness
...
-S
...
-H
...
Envir
...
ASCE, 123, 437(1997)) have studied the effect of
moisture content on the adsorption capacities of granular activated carbon
(GAC) for normal hexane and cyclohexane in air streams
...
The following table gives values of qVOC,RH=0 for
cyclohexane:
c/ppm

33
...
5°C

57
...
4°C

99°C

200

0
...
069

0
...
042

0
...
093

0
...
072

0
...
042

25
...
Find

1000

0
...
088

0
...
063

0
...


2000

0
...
092

0
...
068

0
...
112

0
...
087

0
...
058

25
...
25
...

Then, derive an expression for R(t) that applies to the dissociation part of the
surface plasmon resonance experiment in Fig
...
27
...
28 If α = – , an electrode interface is unable to rectify alternating current
2
1
because the current density curve is symmetrical about η = 0
...
Suppose that
the overpotential varies as η = η0 cos ω t
...
In each case work in the limit of small η0
2
but to second order in η0F/RT
...
0 cm hydrogen–platinum electrode with α = 0
...


25
...
What waveform will the current across

(a) By linear regression of 1/q VOC, RH=0 against 1/cVOC, test the goodness of fit
and determine values of a and b
...
Test
the goodness of fit of the data to these equations and obtain values for ka, kb,
∆adsH, and ∆bH
...
34‡ M
...
Chou and J
...
Chiou (J
...
Engrg
...
The
following table shows the adsorption capacities (qwater = mwater /mGAC) of GAC
for pure water from moist air streams as a function of relative humidity (RH)
in the absence of VOCs at 41
...

RH

0
...
26

0
...
57

0
...
00

q water

0
...
026

0
...
091

0
...
229

958

25 PROCESSES AT SOLID SURFACES

The authors conclude that the data at this and other temperatures obey a
Freundlich type isotherm, qwater = k(RH)1/n
...
5°C and determine the constants k and n
...
5°C
...
10

36
...
40

6
...
21

2
...
13

4
...
80

1
...
55

0
...
00

0
...
25

0
...
53

0
...
81

rVOC

1
...
98

0
...
84

0
...
67

0
...
36 Calculate the thermodynamic limit to the zero-current potential of
fuel cells operating on (a) hydrogen and oxygen, (b) methane and air, and
(c) propane and air
...


The authors propose that these data fit the equation rVOC = 1 − qwater
...
Suggest reasons for any differences
...
35‡ The release of petroleum products by leaky underground storage

tanks is a serious threat to clean ground water
...
D
...
Kershaw, B
...
Kulik, and
S
...
Geotech
...
Engrg
...
Though sorption involves more than surface interactions, sorption
data are usually found to fit one of the adsorption isotherms
...
(a) Determine the units
of the empirical constants
...


(c) Compare the sorption efficiency of ground rubber to that of granulated
activated charcoal, which for benzene has been shown to obey the Freundlich
isotherm in the form q = 1
...
6 with coefficient of determination R2 = 0
...

eq

25
...
Take as a criterion of corrosion a
metal ion concentration of at least 10−6 mol dm−3
...
38 Estimate the magnitude of the corrosion current for a patch of zinc of
area 0
...
Take the exchange current densities as 1 µA cm−2 and the local ion
concentrations as 1 µmol dm−3
...
39 The corrosion potential of iron immersed in a de-aerated acidic
solution of pH = 3 is −0
...
2802 V
...
10 µA cm−2
...


Appendix 1
Quantities, units,
and notational
conventions

A1
Names of quantities
Units

The result of a measurement is a physical quantity (such as mass or density) that is
reported as a numerical multiple of an agreed unit:
physical quantity = numerical value × unit
For example, the mass of an object may be reported as m = 2
...
010 kg dm−3 where the units are, respectively, 1 kilogram (1 kg) and 1 kilogram
per decimetre cubed (1 kg dm−3)
...
Thus, the expression (physical quantity)/unit
is simply the numerical value of the measurement in the specified units, and hence
is a dimensionless quantity
...
5 and the density as d/(kg dm−3) = 1
...

Physical quantities are denoted by italic or (sloping) Greek letters (as in m for mass
and Π for osmotic pressure)
...


Names of quantities
A substance is a distinct, pure form of matter
...
This number is found experimentally to be approximately 6
...
If a sample contains N entities, the amount
of substance it contains is n = N/NA, where NA is the Avogadro constant: NA = 6
...
Note that NA is a quantity with units, not a pure number
...
Two examples are mass and volume
...
Examples are temperature,
mass density (mass divided by volume), and pressure
...
A molar property is
intensive
...
The one exception to the
notation Xm is the molar mass, which is denoted M
...
The molar mass of a molecular compound is the

Notational conventions
Further reading

960

Appendix 1 QUANTITIES, UNITS, AND NOTATIONAL CONVENTIONS
mass per mole of molecules, and the molar mass of an ionic compound is the mass per
mole of formula units
...
The names atomic weight and molecular weight
are still widely used in place of molar mass (often with the units omitted), but we shall
not use them in this text
...
Molar concentration
is usually expressed in moles per decimetre cubed (mol dm−3 or mol L−1; 1 dm3
is identical to 1 L)
...
Such a solution is widely called a ‘1 molar’ solution and
denoted 1 m
...
Its units are typically moles of solute
per kilogram of solvent (mol kg−1)
...
1
...
Thus, volume is (length)3 and may be reported as
a multiple of 1 metre cubed (1 m3), and density, which is mass/volume, may be
reported as a multiple of 1 kilogram per metre cubed (1 kg m−3)
...
The names of units
derived from names of people are lower case (as in torr, joule, pascal, and kelvin), but
their symbols are upper case (as in Torr, J, Pa, and K)
...
2
...
In a perfect world, Greek prefixes of units are upright (as in µm) and sloping for
physical properties (as in µ for chemical potential), but available typefaces are not
always so obliging
...
3
...
1 The SI base units
Physical quantity

Symbol for quantity

Base unit

Length

l

metre, m

Mass

M

kilogram, kg

Time

t

second, s

Electric current

I

ampere, A

Thermodynamic temperature

T

kelvin, K

Amount of substance

n

mole, mol

Luminous intensity

I

candela, cd

UNITS

Table A1
...


Table A1
...
Powers of units apply to the
prefix as well as the unit they modify:
1 cm3 = 1 (cm)3 = 1 (10−2 m)3 = 10−6 m3
Note that 1 cm3 does not mean 1 c(m3)
...

There are a number of units that are in wide use but are not a part of the
International System
...
These include
the litre (L), which is exactly 103 cm3 (or 1 dm3) and the atmosphere (atm), which is
exactly 101
...
Others rely on the values of fundamental constants, and hence
are liable to change when the values of the fundamental constants are modified by
more accurate or more precise measurements
...
602 177 33 × 10−19 J
...
4 gives
the conversion factors for a number of these convenient units
...
The default numbering of equations is (C
...
n] is used to denote a definition and {C
...
A subscript r

961

962

Appendix 1 QUANTITIES, UNITS, AND NOTATIONAL CONVENTIONS
Table A1
...
325 kPa

electronvolt

eV

1
...
485 31 kJ mol−1
* All values in the final column are exact, except for the definition of 1 eV, which depends on the measured
value of e
...
A superscript ° indicates that the equation is valid for an ideal system,
such as a perfect gas or an ideal solution
...

For numerical calculations, we take special care to use the proper number of significant figures
...
, 100
...
, etc)
...
M
...
), Quantities, units, and symbols in physical chemistry
...


A2

Appendix 2
Mathematical
techniques
Basic procedures

Basic procedures
A2
...
1 Logarithms and exponentials
The natural logarithm of a number x is denoted ln x, and is defined as the power to
which e = 2
...
must be raised for the result to be equal to x
...
1)

ln x − ln y = ln(x/y)

(A2
...
2

Complex numbers and
complex functions

A2
...
3)

We also encounter the common logarithm of a number, log x, the logarithm compiled with 10 in place of e
...
Common and natural logarithms are related by

Calculus
A2
...
5

Power series and Taylor
expansions

A2
...
4)

A2
...
The following properties are important:

Functionals and functional
derivatives

A2
...
9

Differential equations

ln x = ln 10 log x ≈ 2
...
=e

(A2
...
6)

(ex)a = eax

(A2
...
10 Random selections

A2
...
11 Some results of probability

theory

Complex numbers have the form
z = x + iy

(A2
...
The real numbers x and y are, respectively, the real and imaginary
parts of z, denoted Re(z) and Im(z)
...
9)

Matrix algebra
A2
...
13 Simultaneous equations
A2
...
10)

The following rules apply for arithmetic operations involving complex numbers:
1 Addition
...
11)

Further reading

964

Appendix 2 MATHEMATICAL TECHNIQUES
2 Multiplication
...
12)

3 Division
...
A2
...


=

z(z′)*
|z′| 2

(A2
...
We write the complex conjugate, f *, of a complex function, f, by replacing i wherever it occurs by −i
...

Complex exponential functions may be written in terms of trigonometric functions
...
14)

which implies that
v

v

q

u

u

u+
v

u

18
0°
-

q

q

q
(a)

v
(b)

(c)

(a) The vectors V and u make an
angle θ
...
(c) To finish the process, we
draw the resultant vector by joining the tail
of u to the head of V
...
A2
...
A2
...
Comparison with the result
shown in Fig
...
2c for the addition of u
to V shows that reversing the order of
vector addition does not affect the result
...
A2
...
16)

A2
...
The vector shown in Fig
...
1
has components on the x, y, and z axes with magnitudes vx, vy, and vz, respectively
...
17)

where i, j, and k are unit vectors, vectors of magnitude 1, pointing along the positive
directions on the x, y, and z axes
...
18)

Using this representation, we can define the following vector operations:
1 Addition and subtraction
...
15)

(A2
...
Consider two vectors V and u making an angle θ
(Fig
...
2a)
...
A2
...
In the second step, we draw a vector Vres, the
resultant vector, originating from the tail of V to the head of u, as shown in Fig
...
2c
...
That is, we obtain the same
Vres whether we add u to V (Fig
...
2c) or V to u (Fig
...
3)
...
A2
...
To calculate the magnitude of the third side, vres, we
make use of the law of cosines, which states that:
For a triangle with sides a, b, and c, and angle C facing side c:
c 2 = a2 + b 2 − 2ab cos C
This law is summarized graphically in Fig
...
4 and its application to the case shown
in Fig
...
2c leads to the expression

A2
...

Fig
...
4

u

z

The graphical method for
subtraction of the vector u from the vector
V (as shown in Fig
...
2a) consists of two
steps: (a) reversing the direction of u to
form −u, and (b) adding −u to V
...
A2
...
20)

The subtraction of vectors follows the same principles outlined above for addition
...
A2
...
We note that subtraction of u from V
amounts to addition of −u to V
...
A2
...
A2
...


x

uv sin q v x u

Fig
...
6 The direction of the crossproducts of two vectors u and V with an
angle θ between them: (a) u × V and (b) V × u
...
21, are perpendicular to
both u and V but the direction depends on
the order in which the product is taken
...
There are two ways to multiply vectors
...
21a)

where θ is the angle between the two vectors and l is a unit vector perpendicular to
both u and V, with a direction determined as in Fig
...
6
...
21b)

where the structure in the middle is a determinant (see below)
...
22)

f (x)

As its name suggests, the scalar product of two vectors is a scalar
...


Fig
...
7

A2
...
The slope of a function, like the slope of a hill, is obtained
by dividing the rise of the hill by the horizontal distance (Fig
...
7)
...
In fact, we let it become infinitesimally small
—hence the name infinitesimal calculus
...
Therefore, the slope of the function f at
x is the vertical distance, which we write δf, divided by the horizontal distance, which
we write δx:
Slope =

rise in value
horizontal distance

=

δf
δx

=

f(x + δx) − f(x)
δx

(A2
...
In this limit, the δ is replaced by a d, and we write
Slope at x =

df
dx

= lim

f(x + δx) − f(x)
δx

δx→0

(A2
...
Some important derivatives are given inside
the front cover of the text
...
25)

Rule 2 (the product rule) For two functions f and g:
d(fg) = f dg + g df

(A2
...
27)

Rule 4 (the chain rule) For a function f = f(g), where g = g(t),
df
dt

=

df dg

(A2
...

For instance, the area under the graph of the function f drawn in Fig
...
8 can be written as the value of f evaluated at a point multiplied by the width of the region, δx, and
then all those products f(x)δx summed over all the regions:

dx

Area between a and b =

∑ f(x)δx

When we allow δx to become infinitesimally small, written dx, and sum an infinite
number of strips, we write:

f (x)

Ύ f(x)dx
b

Area between a and b =

(A2
...
A2
...


The elongated S symbol on the right is called the integral of the function f
...
When written with limits
(as in eqn A2
...
The definite integral is the
indefinite integral evaluated at the upper limit (b) minus the indefinite integral evaluated at the lower limit (a)
...
5 POWER SERIES AND TAYLOR EXPANSIONS
Average value of f (x) from a to b =

Ύ f(x)dx
b

1
b−a

(A2
...

Integration is the inverse of differentiation
...
Some important integrals
are given on the front back cover of the text
...
Two integration techniques are useful:
Technique 1 (integration by parts) For two functions f and g:

Ύf dx dx = fg − Ύg dx dx
dg

Ύ

df

(A2
...
It follows that

G

dx J

Ύ (a − x)(b − x) = b − a HI Ύ a − x − Ύ b − x KL
dx

1

=

dx

A
1
1 D
ln
− ln
+ constant
b−a C a−x
b − xF
1

(A2
...
5 Power series and Taylor expansions
A power series has the form
∞

c0 + c1(x − a) + c2(x − a)2 + · · · + cn(x − a)n + · · · =

∑ cn(x − a)n

(A2
...
It is often useful to express a function f(x) in the vicinity
of x = a as a special power series called the Taylor series, or Taylor expansion, which
has the form:
f(x) = f(a) +
∞

=

A df D
1 A d2f D
1 A dnf D
(x − a) +
(x − a)2 + · · · +
(x − a)n
C dx F
2! C dx 2 F a
n! C dx n F a

1 A dnf D

∑ n! C dxn F (x − a)n

n= 0

(A2
...
1
By definition 0! = 1
...

<

(A2
...
6 Partial derivatives
A partial derivative of a function of more than one variable, such as f(x,y), is the slope
of the function with respect to one of the variables, all the other variables being held
constant (see Fig
...
21)
...
Thus, if f is a function of x and y, then when x and y change by dx and dy, respectively, f changes by
df =

A ∂f D
A ∂f D
dx +
dy
C ∂x F y
C ∂y F x

(A2
...
The quantity
df is also called the differential of f
...
37)

∂y∂x

For the function f given above, it is easy to verify that
A ∂ A ∂f D D
A ∂ A ∂f D D
B
E = 3ax 2
B
E = 3ax 2
∂y C ∂x F y F
∂x C ∂y F x F
C
C
x
y
In the following, z is a variable on which x and y depend (for example, x, y, and z
might correspond to p, V, and T)
...
38)

Relation 2

A ∂y D
1
=
C ∂x F z (∂x/∂y)z

(A2
...
40)

By combining this relation and Relation 2 we obtain the Euler chain relation:

A ∂y D A ∂x D A ∂z D
= −1
C ∂x F z C ∂z F y C ∂y F x

(A2
...

df = g(x,y)dx + h(x,y)dy

is exact if

A ∂g D A ∂h D
=
C ∂y F x C ∂x F y

If df is exact, its integral between specified limits is independent of the path
...
42)

A2
...
7 Functionals and functional derivatives
Just as a function f can be regarded as a set of mathematical procedures that associates
a number f(x) to a specified value of a variable x, so a functional G gives a prescription
for associating a number G[f ] to a function f(x) over a specified range of the variable
x
...
Functionals are important in
quantum chemistry
...

To make the following discussion more concrete, consider the functional

Ύ f(x) dx
1

G[f ] =

2

(A2
...
However, if f(x) = sin πx, then
3
1
G[f ] = – over the range 0 ≤ x ≤ 1
...
By analogy with eqn
A2
...
44)

δf

δf→0

However, this equation does not give us a simple method for calculating the functional derivative
...
45)

where the integral is evaluated in the range over which x varies
...
45 is used to calculate a functional derivative, consider the functional given by eqn A2
...
We begin by writing

Ύ
Ύ {f(x) + 2f(x)δf(x) + δf(x) }dx
= Ύ {f(x) + 2f(x)δf(x)}dx = G[f ] + Ύ 2f(x)δf(x)dx
1

G[ f + δf ] =

1

{f(x) + δf(x)}2dx =

0

2

2

0

1

1

2

0

0

where we have ignored the minute contribution from δf 2 to arrive at the penultimate
expression and then used eqn A2
...
It follows that

Ύ 2f(x)δf(x)dx
1

G[f + δf ] − G[f ] =

0

By comparing this expression with eqn A2
...
8 Undetermined multipliers
Suppose we need to find the maximum (or minimum) value of some function f that
depends on several variables x1, x2,
...
When the variables undergo a small
change from xi to xi + δxi the function changes from f to f + δf, where

969

970

Appendix 2 MATHEMATICAL TECHNIQUES
n

δf =

A ∂f D

∑ C ∂x F δxi
i

(A2
...
47)

i

If the xi were all independent, all the δxi would be arbitrary, and this equation could
be solved by setting each (∂f/∂xi) = 0 individually
...
We proceed
as follows
...
For example, in Chapter 16, one constraint was n0 + n1 + · · · = N, which can be written
g = 0,

with g = (n0 + n1 + · · ·) − N

The constraint g = 0 is always valid, so g remains unchanged when the xi are varied:
δg =

A ∂g D

∑ C ∂x F δxi = 0
i

(A2
...
47:
n 1
A ∂f D
A ∂g D 5
2
6 δx = 0
+λ
(A2
...
All those other δxi (i = 1, 2,
...
But here is the trick: λ is arbitrary; therefore we can choose
it so that the coefficient of δxn in eqn A2
...
That is, we choose λ so that

A ∂f D
A ∂g D
+λ
=0
C ∂xn F
C ∂xn F

(A2
...
49 becomes
n−1 1
A ∂f D
A ∂g D 5
2
6 δx = 0
+λ
C ∂xi F 7 i
i 3 C ∂xi F

∑

(A2
...
, n − 1

(A2
...
50 has exactly the same form as this equation, so the maximum or
minimum of f can be found by solving

A ∂f D
A ∂g D
+λ
=0
C ∂xi F
C ∂xi F

i = 1, 2,
...
53)

The use of this approach was illustrated in the text for two constraints and therefore
two undetermined multipliers λ1 and λ2 (α and −β )
...
One approach is to solve
eqn A2
...
In Chapter 16 we
used the alternative procedure of keeping λ undetermined until a property was calculated for which the value was already known
...


A2
...
9 Differential equations
(a) Ordinary differential equations

An ordinary differential equation is a relation between derivatives of a function of
one variable and the function itself, as in
a

d2y
dx

2

+b

dy
dx

+ cy = 0

(A2
...
may be functions of x
...
54 is a second-order equation
...

A solution of a differential equation is an expression for y as a function of x
...
A general solution of a
differential equation is the most general solution of the equation and is expressed in
terms of a number of constants
...
A first–order differential equation requires the specification
of one boundary (or initial) condition; a second–order differential equation requires
the specification of two such conditions, and so on
...
For
example, the equation
dy
dx

= axy

with a constant may be rearranged into
dy
y

= axdx

and then integrated to
1
ln y = – ax 2 + A
2

where A is a constant
...
For example, it is sensible to try the substitution y = sx, and to change the
variables from x and y to x and s
...

Solutions to complicated differential equations may also be found by referring to
tables (see Further reading)
...
55)

971

972

Appendix 2 MATHEMATICAL TECHNIQUES
appear in the study of chemical kinetics
...
56)

Mathematical software is now capable of finding analytical solutions of a wide variety
of differential equations
...
One powerful approach commonly used to lay siege to
second-order differential equations is to express the solution as a power series:
∞

y=

∑ cn x n

(A2
...

This approach results, for instance, in the Hermite polynomials that form part of the
solution of the Schrödinger equation for the harmonic oscillator (Section 9
...
All the
second-order differential equations that occur in this text are tabulated in compilations of solutions or can be solved with mathematical software, and the specialized
techniques that are needed to establish the form of the solutions may be found in
mathematical texts
...
In such cases, we resort to
numerical methods, in which approximations are made in order to integrate the
differential equation
...
The general form of such programs to solve
df/dx = g(x), for instance, replaces the infinitesimal quantity df = g(x)dx by the small
quantity ∆f = g(x)∆x, so that
f (x + ∆x) ≈ f(x) + g(x)∆x
and then proceeds numerically to step along the x-axis, generating f (x) as it goes
...
Among the simple numerical methods, the fourth-order Runge–
Kutta method is one of the most accurate
...
Here we illustrate the procedure with a first-order
differential equation of the form:
dy
dx

= f(x,y)

(A2
...
36, which
describes the time dependence of the concentration of an intermediate I in the reaction sequence A → I → P
...
58, we proceed by rewriting it in terms of finite differences instead of differentials:
∆y
∆x

= f(x,y)

where ∆y may also be written as y(x + ∆x) − y(x)
...
59)

A2
...
60a)

1
k2 = f(x + – ∆x, y +
2
1
k3 = f(x + – ∆x, y +
2

1
– k1)∆x
2
1
– k2)∆x
2

(A2
...
60c)

k4 = f(x + ∆x, y + k3)∆x

(A2
...
60(a–d) to calculate values of y for a range of x values
...
The
accuracy of the calculation increases with decreasing values of the increment ∆x
...
An example is
∂2y
∂t 2

=a

∂2y

(A2
...
In certain cases, partial differential equations may be separated into ordinary differential equations
...
2) may be separated by writing the wavefunction, ψ (x,y), as the product X(x)Y(y), which results in
the separation of the second-order partial differential equation into two second-order
differential equations in the variables x and y
...


Statistics and probability
Throughout the text, but especially in Chapters 16, 17, 19, and 21, we use several elementary results from two branches of mathematics: probability theory, which deals
with quantities and events that are distributed randomly, and statistics, which provides tools for the analysis of large collections of data
...

A2
...
Consider a simple coin-toss problem
...
62)

A nD
The numbers N(n,i), which are sometimes denoted
, are also called binomial
C iF
coefficients
...
For example, there are six possible results for the roll of a die
...
nm!

,

n=

∑ ni

(A2
...
In Chapter 16 we use the multinomial coefficient to determine the number of ways to configure a system of identical particles
given a specific distribution of particles into discrete energy levels
...
We can simplify factorials
of large numbers by using Stirling’s approximation:
2
n! ≈ (2π)1/2nn+ –e−n
1

(A2
...

For very large values of n, it is possible to use another form of the approximation:
ln n! ≈ n ln n − n

(A2
...
11 Some results of probability theory
Here we develop two general results of probability theory: the mean value of a variable
and the mean value of a function
...

The mean value (also called the expectation value) ͗X͘ of a variable X is calculated
by first multiplying each discrete value xi that X can have by the probability pi that xi
occurs and then summing these products over all possible N values of X:
N

͗X͘ =

∑ xi pi
i= 1

When N is very large and the xi values are so closely spaced that X can be regarded
as varying continuously, it is useful to express the probability that X can have a value
between x and x + dx as
Probability of finding a value of X between x and x + dx = f(x)dx
where the function f(x) is the probability density, a measure of the distribution of the
probability values over x, and dx is an infinitesimally small interval of x values
...
66)

−∞

This expression is similar to that written for the case of discrete values of X, with
f (x)dx as the probability term and integration over the closely spaced x values replacing summation over widely spaced xi
...
12 MATRIX ADDITION AND MULTIPLICATION
The mean value of a function g(X) can be calculated with a formula similar to that
for ͗X͘:
+∞

͗g(X)͘ =

Ύ

g(x)f(x)dx

(A2
...
Matrices may be combined together by addition or
multiplication according to generalizations of the rules for ordinary numbers
...

Consider a square matrix M of n2 numbers arranged in n columns and n rows
...
Each element is therefore denoted Mrc
...
This is an example of a 2 × 2
matrix
...
Thus, the matrix
⎛ 1 0 0⎞
D = ⎜ 0 2 0⎟
⎜ 0 0 1⎟
⎝
⎠
is diagonal
...
68)

where δrc is the Kronecker delta, which is equal to 1 for r = c and to 0 for r ≠ c
...
The unit matrix, 1 (and occasionally I), is
a special case of a diagonal matrix in which all nonzero elements are 1
...
69)

Thus, for the matrix M we have been using,
⎛ 1 3⎞
MT = ⎜
⎟
⎝ 2 4⎠
Matrices are very useful in chemistry
...

A2
...
70)

975

976

Appendix 2 MATHEMATICAL TECHNIQUES
(that is, corresponding elements are added)
...
71)

n

For example, with the matrices given above,
⎛ 1 2⎞ ⎛ 5 6⎞ ⎛ 1 × 5 + 2 × 7 1 × 6 + 2 × 8⎞ ⎛ 19 22⎞
P= ⎜
⎟⎜
⎟ =⎜
⎟ =⎜
⎟
⎝ 3 4⎠ ⎝ 7 8⎠ ⎝ 3 × 5 + 4 × 7 3 × 6 + 4 × 8⎠ ⎝ 43 50⎠
It should be noticed that in general MN ≠ NM, and matrix multiplication is in general
non-commutative
...
72)

The inverse of a matrix can be constructed by using mathematical software, but in
simple cases the following procedure can be carried through without much effort:
1 Form the determinant of the matrix
...

⎛ 1 3⎞
2 Form the transpose of the matrix
...

⎝ 2 4⎠
3 Form *′, where *′ is the cofactor of the element Mrc, that is, it is the determinrc
ant formed from M with the row r and column c struck out
...
For example,
⎛ −2 1 ⎞
1
M −1 = –– ⎛ 4 −2⎞ = ⎜ 3
⎟
1
−2 ⎜
⎝ −3 1 ⎠ ⎝ 2 − 2 ⎟
⎠
A2
...
73)

an1x1 + an2x 2 + · · · + annxn = bn
can be written in matrix notation if we introduce the column vectors x and b:
⎛ x1 ⎞
⎜x ⎟
x = ⎜ 2⎟
⎜ M⎟
⎜x ⎟
⎝ n⎠

⎛ b1 ⎞
⎜b ⎟
b = ⎜ 2⎟
⎜ M⎟
⎜b ⎟
⎝ n⎠

A2
...
74)

The formal solution is obtained by multiplying both sides of this matrix equation by
a−1, for then
x = a−1b

(A2
...
14 Eigenvalue equations
An eigenvalue equation is a special case of eqn A2
...
76)

where λ is a constant, the eigenvalue, and x is the eigenvector
...
77)
(i)

There are n corresponding eigenvectors x
...
77 has a solution only if the
determinant of the coefficients is zero
...
78)

The n eigenvalues the secular equation yields may be used to find the n eigenvectors
...

Thus, because
(
⎛ x11)⎞
⎜ (1)⎟
x(1) = ⎜ x2 ⎟
⎜ M ⎟
⎜ x (1)⎟
⎝ n ⎠

(
⎛ x12)⎞
⎜ (2)⎟
x(2) = ⎜ x2 ⎟
⎜ M ⎟
⎜ x (2)⎟
⎝ n ⎠

etc
...
, x(n)) = ⎜ x2
⎜ M
⎜ x (1)
⎝ n

(
(
x12) L x1n)⎞
(
(
x22) L x2n)⎟
⎟
M
M ⎟
(
(
xn2) L xnn)⎟
⎠

so that Xrc = x (c)
...
, λn along the diagonal, then all the eigenvalue equations ax(i) =
li x(i) may be confined into the single equation
aX = XL

(A2
...
Therefore, if we form X −1 from X, we construct a similarity transformation
L = X −1aX

(A2
...
It follows that if the matrix X that
causes X−1aX to be diagonal is known, then the problem is solved: the diagonal matrix
so produced has the eigenvalues as its only nonzero elements, and the matrix X used
to bring about the transformation has the corresponding eigenvectors as its columns
...


Further reading
P
...
Atkins, J
...
de Paula, and V
...
Walters, Explorations in physical
chemistry
...
H
...


R
...
Mortimer, Mathematics for physical chemistry
...


J
...
Barrante, Applied mathematics for physical chemistry
...


D
...
McQuarrie, Mathematical methods for scientists and engineers
...


M
...
Cady and C
...
Trapp, A Mathcad primer for physical chemistry
...
H
...


D
...
), CRC standard mathematical tables and formulae
...


A3

Appendix 3
Essential concepts
of physics
Energy

Energy

The central concept of all explanations in physical chemistry, as in so many other
branches of physical science, is that of energy, the capacity to do work
...
Although energy can be transferred from one location to another, the total energy is constant
...
For a body of mass m travelling at a speed v,
(A3
...
The zero of potential energy is arbitrary
...
No universal expression for the potential energy can be given because it depends on the type
of interaction the body experiences
...
If the body is at a height
h above the surface of the Earth, then its potential energy is mgh, where g is a constant
called the acceleration of free fall, g = 9
...

The total energy is the sum of the kinetic and potential energies of a particle:
E = EK + EP

Kinetic and potential energy

A3
...
2)

A3
...
4

A3
...
1

Newton’s second law

A3
...
6

The harmonic oscillator

Waves
A3
...
8

Features of electromagnetic
radiation

A3
...
10 Optical activity

Electrostatics
A3
...
12 The Coulomb potential
A2
...
2 Energy units

A2
...
3)

Calories (cal) and kilocalories (kcal) are still encountered in the chemical literature: by
definition, 1 cal = 4
...
An energy of 1 cal is enough to raise the temperature of 1 g
of water by 1°C
...
4)

Further reading

980

Appendix 3 ESSENTIAL CONCEPTS OF PHYSICS

Classical mechanics
Classical mechanics describes the behaviour of objects in terms of two equations
...


pz
p

A3
...
5)

dt

The velocity is a vector, with both direction and magnitude
...
The linear momentum, p, of a particle of mass m is related to
its velocity, V, by
p = mV

The linear momentum of a particle
is a vector property and points in the
direction of motion
...
A3
...
6)

Like the velocity vector, the linear momentum vector points in the direction of travel
of the particle (Fig
...
1)
...
7)

This equation can be used to show that a particle will have a definite trajectory, or
definite position and momentum at each instant
...
Because v = dx/dt, it follows from eqns A3
...
7 that

A 2EK D
=
dt C m F

dx

1/2

(A3
...
9)

Potential energy, V

The linear momentum is a constant:
p(t) = mv(t) = m

Force

dt

= (2mEK)1/2

(A3
...

Force

A3
...
The force points in
the direction of lower potential energy
...
A3
...
11a)

This relation implies that the direction of the force is towards decreasing potential
energy (Fig
...
2)
...
5 ROTATIONAL MOTION
F = −∇V

∇=i

∂
∂x

+j

∂
∂y

+k

∂
∂z

981

(A3
...
In one dimension:
dp
dt

=F

(A3
...
12b)

The second derivative, d2x/dt 2, is the acceleration of the particle, its rate of change of
velocity (in this instance, along the x-axis)
...
12 will also give the trajectory
...
For example, it can be used to show that, if a particle of mass m is initially
stationary and is subjected to a constant force F for a time τ, then its kinetic energy increases from zero to
EK =

F 2τ 2
2m

(A3
...
Because the applied force,
F, and the time, τ, for which it acts may be varied at will, the solution implies that the
energy of the particle may be increased to any value
...
5 Rotational motion
The rotational motion of a particle about a central point is described by its angular
momentum, J
...
A3
...

The magnitude of the angular momentum, J, is given by the expression
J = Iω

(A3
...
The analogous roles of m and I,
of v and ω , and of p and J in the translational and rotational cases, respectively, should
be remembered, because they provide a ready way of constructing and recalling equations
...
15)

To accelerate a rotation it is necessary to apply a torque, T, a twisting force
...
16)

If a constant torque is applied for a time τ, the rotational energy of an initially stationary body is increased to

Fig
...
3 The angular momentum of a
particle is represented by a vector along the
axis of rotation and perpendicular to the
plane of rotation
...
The direction of motion is
clockwise to an observer looking in the
direction of the vector
...
17)

2I

The implication of this equation is that an appropriate torque and period for which it
is applied can excite the rotation to an arbitrary energy
...
6 The harmonic oscillator
A harmonic oscillator consists of a particle that experiences a restoring force proportional to its displacement from its equilibrium position:
F = −kx

0
Displacement, x
Fig
...
4 The force acting on a particle that
undergoes harmonic motion
...
The
corresponding potential energy is parabolic
(proportional to x2)
...
18)

An example is a particle joined to a rigid support by a spring
...
The negative sign in F signifies that the direction of the force is opposite to
that of the displacement (Fig
...
4)
...
18, into Newton’s equation, eqn A3
...
The
resulting equation is
m

d2x
dt 2

= −kx

A solution is
x(t) = A sin ω t

p(t) = mω A cos ω t

ω = (k/m)1/2

(A3
...
They also show that the particle is stationary (p = 0)
when the displacement, x, has its maximum value, A, which is called the amplitude of
the motion
...
To confirm this remark we note that the kinetic energy is
EK =

p2
2m

=

(mω A cos ω t)2
2m

1
= – mω 2A2 cos 2ω t
2

(A3
...
21)

The force on the oscillator is F = −kx, so it follows from the relation F = −dV/dx that
the potential energy of a harmonic oscillator is
1
1
V = – kx 2 = – kA2 sin2ω t
2
2

(A3
...
23)

(We have used cos2ω t + sin2ω t = 1
...
It follows that the energy of an oscillating particle can be raised to any value by stretching
the spring to any desired amplitude A
...
In other words, the particle will oscillate at the same
frequency regardless of the amplitude of its motion
...
8 FEATURES OF ELECTROMAGNETIC RADIATION

Waves

983

Wavelength, l

Waves are disturbances that travel through space with a finite velocity
...
Waves can be characterized by a wave equation, a differential equation that describes the motion of the wave in space and time
...

These concepts are used in classical physics to describe the wave character of electromagnetic radiation, which is the focus of the following discussion
...
7 The electromagnetic field
In classical physics, electromagnetic radiation is understood in terms of the electromagnetic field, an oscillating electric and magnetic disturbance that spreads as a harmonic wave through empty space, the vacuum
...
As its name suggests, an electromagnetic field has two components, an electric field that acts on charged particles
(whether stationary of moving) and a magnetic field that acts only on moving charged
particles
...
A3
...
The frequency is measured in hertz, where 1 Hz = 1 s−1
...
24)

Therefore, the shorter the wavelength, the higher the frequency
...
25)

A wavenumber can be interpreted as the number of complete wavelengths in a given
length
...
The
classification of the electromagnetic field according to its frequency and wavelength is
summarized in Fig
...
6
...
8 Features of electromagnetic radiation
Consider an electromagnetic disturbance travelling along the x direction with wavelength λ and frequency ν
...
26a)

B(x,t) = B 0cos{2πν t − (2π/λ)x + φ}

(A3
...
If two waves, in the same region of
space, with the same wavelength are shifted by φ = π or −π (so the peaks of one wave
coincide with the troughs of the other), then the resultant wave will have diminished

(b)
Fig
...
5 (a) The wavelength, λ, of a wave is
the peak-to-peak distance
...

At a given location, the instantaneous
amplitude of the wave changes through a
complete cycle (the four dots show half a
cycle) as it passes a given point
...

Wavelength and frequency are related by
λν = c
...
The waves are said to interfere destructively
...

Equations A3
...
26b represent electromagnetic radiation that is planepolarized; it is so called because the electric and magnetic fields each oscillate in a single plane (in this case the xy-plane, Fig
...
7)
...
A3
...
An alternative mode of polarization is circular polarization, in which the electric and magnetic fields rotate around the direction of propagation in either a clockwise or a counterclockwise sense but remain perpendicular
to it and each other
...
26a and A3
...

Fig
...
6

∂2

4π2

∂x

λ2

ψ (x,t) = −
2

ψ (x,t)

∂2
∂t 2

ψ (x,t) = −4π2ν 2ψ (x,t)

(A3
...

According to classical electromagnetic theory, the intensity of electromagnetic
radiation is proportional to the square of the amplitude of the wave
...
1 are based on the interaction
between the electric field of the incident radiation and the detecting element, so light
intensities are proportional to E 2
...
9 Refraction
A beam of light changes direction (‘bends’) when it passes from one transparent
medium to another
...
28]

It follows from the Maxwell equations (see Further reading), that the refractive index
at a (visible or ultraviolet) specified frequency is related to the relative permittivity εr
(discussed in Section 20
...
29)

Table A3
...

Because the relative permittivity of a medium is related to its polarizability by
eqn 20
...
To see why this is so, we
need to realize that propagation of light through a medium induces an oscillating
dipole moment, which then radiates light of the same frequency
...
Because photons of high-frequency light carry more
energy than those of low-frequency light, they can distort the electronic distributions
of the molecules in their path more effectively
...
This dependence on frequency is
the origin of the dispersion of white light by a prism: the refractive index is greater for
blue light than for red, and therefore the blue rays are bent more than the red
...
10 OPTICAL ACTIVITY
term dispersion is a term carried over from this phenomenon to mean the variation
of the refractive index, or of any property, with frequency
...
10 Optical activity
The concept of refractive index is closely related to the property of optical activity
...
To understand this effect, it is useful to regard the incident plane-polarized beam as a superposition of two oppositely rotating circularly
polarized components
...
A3
...
On entering the medium, one component propagates faster than the
other if their refractive indices are different
...
In terms of the
refractive indices, the difference is
∆t = (nR − nL)

Fig
...
7 Electromagnetic radiation consists
of a wave of electric and magnetic fields
perpendicular to the direction of
propagation (in this case the x-direction),
and mutually perpendicular to each other
...


l
c

The phase difference between the two components when they emerge from the sample is therefore
∆θ = 2πν∆t =

z

2πc∆t

λ

= (nR − nL)

Synoptic table A3
...
524

1
...
497

CS2 (l)

1
...
628

1
...
340

1
...
331

KI(s)

1
...
666

1
...
The two rotating electric vectors have a different
phase when they leave the sample from the value they had initially, so their superposition gives rise to a plane-polarized beam rotated through an angle ∆θ relative
to the plane of the incoming beam
...
A sample in which these two
refractive indices are different is said to be circularly birefringent
...
One interpretation
is that, if a molecule is helical (such as a polypeptide α-helix described in Section 19
...
1), its polarizability depends on whether or not the electric field
of the incident radiation rotates in the same sense as the helix
...
This difference is
known as circular dichroism, which is explored in Chapter 14
...


Resultant

∆t =

x

Left

Right

Electrostatics
Electrostatics is the study of the interactions of stationary electric charges
...
60 × 10−19 C
...
65 × 104 C mol−1
...
A3
...


986

Appendix 3 ESSENTIAL CONCEPTS OF PHYSICS
A3
...
30)

4πε 0r

The constant ε0 is the vacuum permittivity, a fundamental constant with the value
8
...
This very important relation is called the Coulomb potential
energy and the interaction it describes is called the Coulomb interaction of two
charges
...

It follows from eqns A3
...
30 that the electrical force, F, exerted by a charge
q1 on a second charge q2 has magnitude
F=

q1q2

(A3
...
With charge
in coulombs and distance in metres, the force is obtained in newtons
...
3)
...
12 The Coulomb potential
The potential energy of a charge q1 in the presence of another charge q2 can be
expressed in terms of the Coulomb potential, φ :
V = q1φ

φ=

q2

(A3
...
The combination joules per coulomb
occurs widely in electrostatics, and is called a volt, V:
1 V = 1 J C−1

(A3
...
present in the system, the total potential experienced by the charge q1 is the sum of the potential generated by each charge:

φ = φ2 + φ3 + · · ·

(A3
...
With charge in
coulomb and length in metres, the charge density is expressed in coulombs per
metre-cubed (C m−3)
...
35)

where ∇ = (∂ /∂x + ∂ /∂y + ∂ /∂z )
...
35 reduces to the form used in Further information 5
...

2

2

2

2

2

2

2

A3
...
The electric
field strength (which, like the force, is actually a vector quantity) is the negative gradient of the electric potential:
/ = −∇φ

(A3
...
14 Electric current and power
The motion of charge gives rise to an electric current, I
...
37)

If the current flows from a region of potential φi to φf, through a potential difference
∆φ = φf − φi, the rate of doing work is the current (the rate of transfer of charge)
multiplied by the potential difference, I∆φ
...
38)

With current in amperes and the potential difference in volts, the power works out in
watts
...
39)

The energy is obtained in joules with the current in amperes, the potential difference
in volts, and the time in seconds
...
P
...
B
...
Sands, The Feynman lectures on
physics
...
Addison–Wesley, Reading (1966)
...
S
...
Academic Press,
San Diego (2000)
...
A
...
Ritchie and D
...
Sivia, Foundations of physics for chemists
...


R
...
M
...

Benjamin Cummings, San Francisco (1999)
...
The
remainder will be found on the pages indicated
...
1
1
...
3
1
...
5
1
...
7

Pressure units (4)
The gas constant in various units (9)
The composition of dry air at sea level (11)
Second virial coefficients*
Critical constants of gases*
van der Waals coefficients*
Selected equations of state (19)

2
...
2
2
...
4 Enthalpies of transition [notation] (51)
2
...
6 Thermochemical properties of some fuels (53)
2
...
7a Standard enthalpies of hydration at infinite dilution*
2
...
8 Expansion coefficients and isothermal
compressibilities*
2
...
1

3
...
6

Standard entropies (and temperatures) of phase
transitions*
Standard entropies of vaporization of liquids*
Standard Third-Law entropies [see Tables 2
...
7]*
Standard Gibbs energies of formation [see Tables 2
...
7]*
The Maxwell relations (104)
The fugacity of nitrogen*

5
...
2
5
...
4
5
...
6

Henry’s law constants for gases*
Freezing point and boiling point constants*
Standard states [summary of definitions] (158)
Ionic strength and molality (164)
Mean activity coefficients in water*
Relative permitivities (dielectric constants)*

3
...
3
3
...
1
7
...
3
7
...
1
8
...
1
9
...
3
9
...
1
10
...
3
10
...
1
11
...
3
11
...
5

Some hybridization schemes (368)
Bond lengths*
Bond dissociation enthalpies*
Pauling and Mulliken electronegativities*
Ab initio calculations and spectroscopic data (398)

12
...
2
12
...
1
13
...
3

Moments of inertia [formulae] (440)
Properties of diatomic molecules*
Typical vibrational wavenumbers*

14
...
2
14
...
4

15
...
2
15
...
1

Rotational and vibrational temperatures [see also
Table 13
...
2
17
...
4
17
...
1

Symmetry numbers [see also Table 13
...
2
18
...
4
18
...
1
19
...
3
19
...
1
20
...
3
20
...
5
20
...
1
21
...
3

Collision cross-sections*
Transport properties of gases*
Transport properties of perfect gases [summary of
formulae] (758)
Viscosities of liquids*
Limiting ionic conductivities in water*
Ionic mobilities in water*

21
...
5
21
...
7

989

21
...
1
22
...
3
22
...
1
23
...
3

Photochemical processes (846)
Common photophysical processes (846)
Values of R0 for donor–acceptor pairs (852)

24
...
2
24
...
4]*
Summary of uses of k [notation] (902)

25
...
2
25
...
4
25
...
6
25
...
1 The SI base units (960)
A1
...
3 Common SI prefixes (961)
A1
...
1 Refractive indices relative to air*
Character tables

The following tables reproduce and expand the data given in the short tables in the text, and follow their numbering
...
The general references are as follows:
AIP: D
...
Gray (ed
...
McGraw Hill, New York (1972)
...
Abramowitz and I
...
Stegun (ed
...
Dover, New York (1963)
...
Emsley, The elements
...

HCP: D
...
Lide (ed
...
CRC Press, Boca Raton (2000)
...
M
...
P
...
Macmillan, London (1992)
...
W
...
Kaye and T
...
Laby (ed
...
Longman, London (1973)
...
N
...
Randall, resived by K
...
Pitzer and L
...
McGraw-Hill, New York (1961)
...
Phys
...
Reference Data, 11, Supplement 2 (1982)
...
A
...
H
...
Butterworth, London (1959)
...
B
...
D
...
P
...
Chapman & Hall, London (1986)
...
698

Argon(g)

1
...
340

Bromine(l)

3
...
260

Carbon(s, d)

1
...
960

Fluorine(g)

1
...
5

2740

Tb /K

CaCO3(s, calcite)

2
...
320

2573
265
...
3
3931
331
...
2
1357
53
...
2
2840
85
...
125

4
...
071

14
...
3

Iodine(s)

4
...
7

7
...
413

116
...
284

383(–H2O)

423(–5H2O)

2
...
3

206
...
187

159
...
1

2
...
4

237
...
997

273
...
2

D2O(l)

1
...
0

374
...
817

195
...
8

KBr(s)

2
...
984

1049

1773s

NaCl(s)

2
...
841

283
...
2

Organic compounds
Acetaldehyde, CH3CHO(l, g)

0
...
5

Iron(s)

CuSO4·5H2O(s)
HBr(g)

H2O(l)

83
...
513

Chlorine(g)

Gold(s)

Tf /K

Inorganic compounds

Elements
Aluminium(s)

r/(g cm−3 )
at 293 K†

Tb /K

1808

3023
120
...
350

600
...
049

289
...
534

453
...
787

178

329

Magnesium(s)

1
...
0

1363

Aniline, C6H5NH2(l)

1
...
243

490

615
353
...
546

234
...
7

Neon(g)

1
...
5

27
...
879

278
...
880

63
...
4

Carbon tetrachloride, CCl 4(l)

1
...
9

90
...
499

209
...
789

156

351
...
0

1
...
6

111
...
791

179
...
6

Naphthalene, C10H8(s)

1
...
4

491

Octane, C8H18(l)

0
...
4

398
...
073

314
...
0

Sucrose, C12H22O11(s)

1
...
140

54
...
820

317
...
862
10
...
971

Sulfur(s, α)

2
...
950

Xenon(g)
Zinc(s)

2
...
133

336
...
0
386
...
3
692
...
8
4018
166
...
† For gases, at their boiling points
...
4 Second virial coefficients, B/(cm3 mol−1)
100 K

Nuclide

m/u

991

273 K

373 K

600 K

Abundance/%
Air

−167
...
5

3
...
0

Ar

−187
...
7

−4
...
9

1

1
...
985

2

2
...
015

CH4

3

3
...
000 13

CO2

4

4
...
0

13
...
0151

7
...
4

12
...
3

7

7
...
58

Kr

−62
...
7

1
...
0129

19
...
5

6
...
7

11

11
...
22

Ne

−6
...
4

12
...
8

12

12*

98
...
5

−22
...
7

12
...
0034

1
...
7

−81
...
6

14

14
...
63

15

15
...
37

16

15
...
76

17

16
...
9992

F

19

18
...
9738

100

S

32

31
...
9715

34

33
...
22

Ar

35

34
...
53

37

Br2

36
...
4

79

C2H4

78
...
54

81

80
...
46

H

H
H

He

He
He

Li

Li
Li

B

B
B

C

C
C

N

N
N

O

O
O
O
F
P
S
S
S

Cl

Cl
Cl

Br

Br
Br

I

127

I

* Exact value
...
9045

100

−53
...
2

−160
...
1

−72
...
4

15
...
4

Data: AIP, JL
...
22 of Section 1
...
21 using
B′ = B/RT
...


0
...
204

Table 1
...
0
pc /atm

0
...
5

150
...
292

584

0
...
50

124

283
...
270

C2H6

48
...
4

0
...
6

260

562
...
274

CH4

45
...
6

0
...
1

417
...
276

CO2

72
...
2

0
...
8

F2

55

0
...
0

102

75
...
00

Vc /(cm3 mol −1 )

98
...
0

510
...
8

64
...
3

55
...
0

363
...
5

81
...
26
80
...
76

33
...
4

0
...
7

0
...
21

0
...
64

423
...
27

92
...
39

0
...
0

N2

33
...
10

126
...
292

327
...
307

122
...
86
111
...
74

44
...
5

405
...
242

O2

50
...
0

154
...
308

405
...
0

118
...
75

0
...
0

Data: AIP, KL
...
6 van der Waals coefficients
a/(atm dm6 mol−2 )

b/(10−2 dm3 mol−1 )

a/(atm dm6 mol−2 )

b/(10 −2 dm3 mol−1)
4
...
337

3
...
484

C2H4

4
...
82

He

0
...
38

C2H6

5
...
51

Kr

5
...
06

11
...
352

3
...
57

CH4

2
...
31

Ne

0
...
67

Cl2

6
...
42

NH 3

4
...
71

CO

1
...
95

O2

1
...
19

CO2

3
...
29

SO2

6
...
68

H2

0
...
65

Xe

4
...
16

H2O

5
...
05

Data: HCP
...
2 Temperature variation of molar heat capacities†
b/(10−3 K−1)

a

c/(105 K2 )

Monatomic gases
20
...
32

0
...
03

0
...
85

CO2

44
...
79

−8
...
56

2
...
51

H2

27
...
26

0
...
40

0
...
71

N2

28
...
77

−0
...
75

O2

29
...
1
4
...
26

−1
...
67

Liquids (from melting to boiling)
C10H8, naphthalene

79
...
4075

0

I2

80
...
29

0

0

Solids
Al

20
...
38

C (graphite)

16
...
77

C10H8, naphthalene

−115
...
920 × 103

0
−8
...
64

6
...
12

49
...
94

16
...
13

11
...
96

† For Cp,m /(J K−1 mol−1) = a + bT + c/T 2
...


DATA SECTION

993

Table 2
...
30

2436

Tf /K

250
...
81

1
...
29

8
...
6

25
...
2

4
...
4

26
...
506

217
...
15

6
...
15

40
...
016 at 298 K

Br2

265
...
57

332
...
45

Cl2

172
...
41

239
...
41

H2S

187
...
377

F2

53
...
26

85
...
16

H2SO4

283
...
56

H2

13
...
117

20
...
916

NH3

195
...
5

0
...
22

0
...
3

2
...
8

N2
Na
O2

63
...
0
54
...
7
458
...
52
0
...
35

2
...
444

161

2
...
4

2
...
18

2
...
0

89
...
86

184
...
7

278
...
59

353
...
8

178

13
...
1

28
...
80

490
...
51

CH3OH

80
...
3
CCl4

111
...
6

K

0
...
820

Xe

23
...
01

90
...
7

C2H6

5
...
652

CCl4

41
...
67

Organic compounds
CH4
90
...
30

212
...
3

175
...
16

337
...
27
37
...
47

349
...
00

158
...
60

352

43
...


Table 2
...
011

C(s) (diamond)

12
...
040

D f H 7/(kJ mol−1)
0
+1
...
51

D f G 7/(kJ mol−1)
0
+2
...
740

8
...
51

2
...
113

−395
...
36

213
...
11

Hydrocarbons
CH4(g), methane

16
...
81

−50
...
26

35
...
04

+145
...
92

194
...
70

−890

C2H2(g), ethyne

26
...
73

+209
...
94

43
...
05

+52
...
15

219
...
56

−1411

C2H6(g), ethane

30
...
68

−32
...
60

52
...
08

+20
...
78

267
...
89

−2058

C3H6(g), cyclopropane

42
...
30

+104
...
55

55
...
10

−103
...
49

269
...
5

−2220

C4H8(g), 1-butene

56
...
13

+71
...
71

85
...
11

−6
...
95

300
...
91

−2710

C4H8(g), trans-2-butene

56
...
17

+63
...
59

87
...
13

−126
...
03

310
...
45

−2878

C5H12(g), pentane

72
...
44

−8
...
40

120
...
15

−173
...
12

+49
...
3

136
...
3

994

DATA SECTION

Table 2
...
12

C6H12(l), cyclohexane

84
...
18

D f G 7/(kJ mol−1)

−198
...
93

7
Sm /(J K−1 mol−1)†

+129
...
31

+26
...
4

C 7 /(J K −1 mol −1)
p,m

81
...
5

D c H 7/(kJ mol −1)

−3302
−3920
−4163

204
...
14

+50
...
0

320
...
6

C7H16(l), heptane

100
...
4

+1
...
6

224
...
4

361
...
23

−249
...
23

−255
...
18

+78
...
04

−238
...
27

126
...
6

CH3OH(g)

32
...
66

−161
...
81

43
...
07

−277
...
78

160
...
46

−1368

C2H5OH(g)

46
...
10

−168
...
70

65
...
12

−165
...
9

146
...
03

−424
...
35

128
...
5

−389
...
8

60
...
05

−485
...
46

59
...
01

−369
...
6

(COOH)2(s), oxalic

90
...
2

122
...
1

−3054

99
...
3

−255
−875

178
...
3
117

−254

−245
...
6

146
...
1

−2231

CH3CH(OH)COOH(s), lactic

90
...
0

CH3COOC2H5(l), ethyl acetate

88
...
0

−332
...
4

Alkanals and alkanones
HCHO(g), methanal

30
...
57

−102
...
77

CH3CHO(l), ethanal

44
...
30

−128
...
2

CH3CHO(g)

44
...
19

−128
...
3

57
...
08

−248
...
4

200
...
7

−1790

−910

212

Sugars
C6H12O6(s), α-d-glucose

180
...
16

−1268

35
...
16

−1266

C12H22O11(s), sucrose

342
...
06

−333
...
33

104
...
14

−632

CH3NH2(g), methylamine

31
...
97

+32
...
41

53
...
13

+31
...
07

−532
...
5

99
...
2

−3393
−373
...
† Standard entropies of ions may be either positive or negative because the values are relative to the entropy of the hydrogen ion
...
7 Thermodynamic data for elements and inorganic compounds (all values relate to 298 K)
M/(g mol −1)

D f H 7/(kJ mol −1)

D f G 7/(kJ mol−1)

7
S m /(J K −1 mol −1)†

C 7 , m /(J K−1 mol−1)
p

Aluminium (aluminum)
Al(s)

26
...
98

+10
...
98

+326
...
98

+5483
...
98

−531

−485

Al2O3(s, α)

101
...
7

−1582
...
92

79
...
24

−704
...
8

110
...
84

154
...
786

3+

Argon
Ar(g)

39
...
20
+285
...
33

24
...
55

24
...
54

21
...
7

Antimony
Sb(s)

121
...
69

25
...
77

+145
...
75

232
...
05

Arsenic
As(s, α)

74
...
1

24
...
21

20
...
92

+302
...
0

As4(g)

299
...
9

+92
...
95

+66
...
93

222
...
07

Barium
Ba(s)

137
...
8

28
...
24

20
...
34

+180

+146

Ba2+(aq)

137
...
64

−560
...
34

−553
...
1

70
...
78

BaCl2(s)

208
...
6

−810
...
68

75
...
01

Be(g)

9
...
98

Bi(g)

208
...
82

Br2(g)

159
...
3

0
+207
...
907

0
+286
...
2

0

+9
...
50

16
...
27

20
...
74

25
...
00

20
...
23

75
...
110

245
...
02

+82
...
02

20
...
91

+111
...
91

−219
...
91

−121
...
96

+82
...
92

−36
...
45

198
...
8
29
...
40

0

0

51
...
98

Cd(g)

112
...
01

+77
...
75

20
...
40

−75
...
612

−73
...
7 (Continued)
M/(g mol −1)

D f H 7/(kJ mol −1)

D f G 7/(kJ mol−1)

7
S m /(J K −1 mol −1)†

C 7 , m /(J K−1 mol−1)
p

Cadmium (Continued)
CdO(s)

128
...
2

−228
...
8

CdCO3(s)

172
...
6

−669
...
5

Caesium (cesium)
Cs(s)

132
...
23

Cs(g)

132
...
06

+49
...
60

Cs+(aq)

132
...
28

−292
...
05

Calcium
Ca(s)

40
...
08

Ca2+(aq)

40
...
08

0

32
...
79
−10
...
42

25
...
3

154
...
786

−542
...
58

−53
...
09

−604
...
2

0

43
...
75

42
...
09

−1206
...
8

92
...
88

CaCO3(s) (aragonite)

100
...
1

−1127
...
7

81
...
08

−1219
...
3

CaCl2(s)

110
...
8

−748
...
6

CaBr2(s)

199
...
8

−663
...
5)
C(s) (graphite)
12
...
895

0
+2
...
87

5
...
527

C(s) (diamond)

12
...
011

C2(g)

24
...
90

+775
...
42

43
...
011

−110
...
17

197
...
14

CO2(g)

44
...
51

−394
...
74

37
...
010

−413
...
98

117
...
03

−699
...
08

187
...
02

−691
...
77

+91
...
68

+671
...
377

67
...
59

158
...
113
20
...
01

−677
...
81

−56
...
82

−135
...
21

216
...
14

+89
...
27

151
...
7

HCN(g)

27
...
1

+124
...
78

35
...
03
26
...
87
+150
...
97
+172
...
84
+94
...
63

Chlorine
Cl2(g)

70
...
07

33
...
45

+121
...
68

165
...
840

Cl−(g)

34
...
13

Cl−(aq)

35
...
16

−131
...
5

HCl(g)

36
...
31

−95
...
91

HCl(aq)

36
...
16

−131
...
5

Chromium
Cr(s)

52
...
00

CO2−(aq)
3

0
+396
...
8

131
...
4
29
...
4

23
...
35

174
...
79

DATA SECTION

997

Table 2
...
99

Cr2O2−(aq)
7

215
...
15
−1490
...
75
−1301
...
21
+261
...
54

0

0

Cu(g)

63
...
32

+298
...
38

Cu+(aq)

63
...
67

+49
...
6

+65
...
6

2+

+64
...
150

24
...
79

Cu (aq)

63
...
08

−168
...
0

93
...
64

CuO(s)

79
...
3

−129
...
63

42
...
60

−771
...
8

109

100
...
11

146
...
4

280

CuSO4(s)
CuSO4 ·H2O(s)

177
...
8

CuSO4 ·5H2O(s)

249
...
7

Deuterium
D2(g)

4
...
022

0
+0
...
7

0
−1
...
96

29
...
80

29
...
27

D2O(g)

20
...
20

−234
...
34

D2O(l)

20
...
60

−243
...
94

84
...
022

−245
...
11

199
...
81

HDO(l)

19
...
89

−241
...
29

Fluorine
F2(g)

38
...
78

31
...
00

+78
...
91

158
...
74

F−(aq)

19
...
63

−278
...
8

HF(g)

20
...
1

−273
...
78

Gold
Au(s)

196
...
97

Helium
He(g)

0
+366
...
3

−106
...
13

47
...
42

180
...
79

4
...
15

20
...
016

0

0

130
...
824
20
...
008

+217
...
25

114
...
008

0

0

0

H+(g)

1
...
015

0

+1536
...
99

H2O(l)

18
...
83

−237
...
91

H2O(g)

18
...
82

−228
...
83

33
...
015

−187
...
35

109
...
1

75
...
81

0

0

116
...
44

I2(g)

253
...
44

+19
...
69

36
...
7 (Continued)
M/(g mol −1)

D f H 7/(kJ mol −1)

D f G 7/(kJ mol−1)

7
S m /(J K −1 mol −1)†

C 7 , m /(J K−1 mol−1)
p

Iodine (Continued)
I(g)

126
...
84

+70
...
90

−55
...
57

HI(g)

127
...
48

+1
...
85

Fe(g)

55
...
3

Fe2+(aq)

55
...
1

−78
...
7

Fe3+(aq)

55
...
5

−4
...
9

Fe3O4(s) (magnetite)

231
...
4

−1015
...
4

Fe2O3(s) (haematite)

159
...
2

−742
...
40

103
...
91

−100
...
4

60
...
54

119
...
2

−166
...
93

62
...
08

20
...
80

0

0

Lead
Pb(s)

207
...
19

+195
...
19

−1
...
7

0

0
+161
...
43

180
...
3
206
...
786
−142
...
158

27
...
10

180
...
68

143
...
81

26
...
37

20
...
5

PbO(s, yellow)

223
...
32

−187
...
70

45
...
19

−218
...
93

66
...
81

PbO2(s)

239
...
4

−217
...
6

64
...
94

0

0

29
...
77

Li(g)

6
...
37

+126
...
77

20
...
94

−278
...
31

+13
...
6

Magnesium
Mg(s)

24
...
68

24
...
31

+147
...
10

148
...
786

Mg2+(aq)

24
...
85

−454
...
31

−601
...
43

MgCO3(s)

84
...
22

Mercury
Hg(l)

200
...
02

27
...
59

+61
...
82

174
...
786

Hg2+(aq)

200
...
1

+164
...
2

Hg2+(aq)
2

401
...
4

+153
...
5

−1095
...
32

−1012
...
79

−138
...
94

37
...
7

75
...
62

71
...
59

−90
...
54

Hg2Cl2(s)

472
...
22

−210
...
5

HgCl2(s)

271
...
3

−178
...
0

HgS(s, black)

232
...
6

−47
...
3

70
...
06
102

DATA SECTION

999

Table 2
...
18

Nitrogen
N2(g)

28
...
61

29
...
007

+472
...
56

153
...
786

NO(g)

30
...
25

+86
...
76

29
...
01

+82
...
20

219
...
45

NO2(g)

46
...
18

+51
...
06

37
...
1

+9
...
89

304
...
28

0

0

146
...
786

N2O5(s)

108
...
1

+113
...
2

N2O5(g)

108
...
3

+115
...
7

84
...
01

−174
...
71

155
...
87

HNO3(aq)

143
...
01

−207
...
25

146
...
6

−
NO3 (aq)

62
...
0

−108
...
4

−86
...
03

−46
...
45

192
...
03

−80
...
50

111
...
04

−132
...
31

+113
...
03

−114
...
03

+264
...
3

140
...
68

HN3(g)

43
...
1

+328
...
97

98
...
05

+149
...
21

139
...
04

−365
...
87

151
...
1

NH4Cl(s)

53
...
43

−202
...
6

Oxygen
O2(g)

31
...
138

29
...
999

+249
...
73

161
...
912

O3(g)

47
...
7

+163
...
93

OH−(aq)

17
...
99

−157
...
75

Phosphorus
P(s, wh)

30
...
09

23
...
97

+314
...
25

163
...
786

P2(g)

61
...
3

+103
...
13

32
...
90

+58
...
44

279
...
15

PH3(g)

34
...
4

+13
...
23

37
...
33

−287
...
8

311
...
84

PCl3(l)

137
...
7

−272
...
1

PCl5(g)

208
...
9

−305
...
6

112
...
24

−443
...
00

−964
...
1

110
...
06

H3PO3(s)

+50
...
00
94
...
0

H3PO4(l)

94
...
9

H3PO4(aq)

94
...
4

79
...
20
−148
...
8

H3PO4(s)

35
...
7

−222

1000

DATA SECTION

Table 2
...
8

94
...
4

−1018
...
89

−2984
...
0

P4O6(s)

219
...
1

Potassium
K(s)

39
...
86

211
...
18

29
...
59

160
...
786

K(g)

39
...
24

K+(g)

39
...
26

K+(aq)

39
...
38

−283
...
5

21
...
11

−424
...
08

78
...
9

KF(s)

58
...
27

−537
...
57

49
...
56

−436
...
14

82
...
30

KBr(s)

119
...
80

−380
...
90

52
...
01

−327
...
89

106
...
93

Silicon
Si(s)

28
...
83

20
...
09

+455
...
3

167
...
25

SiO2(s, α)

60
...
94

−856
...
84

44
...
87

0

0

42
...
351

Ag(g)

107
...
55

+245
...
00

20
...
87

+105
...
11

+72
...
8

Silver
Ag(s)

0

0

AgBr(s)

187
...
37

−96
...
1

52
...
32

−127
...
79

96
...
79

Ag2O(s)

231
...
05

−11
...
3

65
...
88

−129
...
41

140
...
05

Sodium
Na(s)

22
...
21

28
...
99

+107
...
76

153
...
79

Na+(aq)

22
...
12

−261
...
0

46
...
00

−425
...
49

64
...
54

NaCl(s)

58
...
15

−384
...
13

50
...
90

−361
...
98

86
...
38

NaI(s)

149
...
78

−286
...
53

52
...
06

S(s, β) (monoclinic)

32
...
33

S(g)

32
...
80

22
...
1

32
...
6

+278
...
25

167
...
673

64
...
37

+79
...
18

32
...
06

+33
...
8

−14
...
06

−296
...
19

248
...
87

80
...
72

−371
...
76

50
...
7 (Continued)
M/(g mol −1)

D f H 7/(kJ mol −1)

D f G 7/(kJ mol−1)

7
S m /(J K −1 mol −1)†

C 7 , m /(J K−1 mol−1)
p

Sulfur (Continued)
H2SO4(l)
H2SO4(aq)
2−
SO4 (aq)
−
HSO 4 (aq)
H2S(g)
H2S(aq)
HS−(aq)
SF6(g)

98
...
08
96
...
07
34
...
08
33
...
05

−813
...
27
−909
...
34
−20
...
7
−17
...
00
−744
...
53
−755
...
56
−27
...
08
−1105
...
90
20
...
1
+131
...
79
121
+62
...
82

138
...
23

Tin
Sn(s, β)

118
...
69

Sn2+(aq)

118
...
8

−27
...
69

−285
...
9

56
...
31

SnO2(s)

150
...
7

−519
...
3

52
...
30

169
...
786

Zinc
Zn(s)

65
...
63

25
...
37

+130
...
14

160
...
79

Zn2+(aq)

65
...
89

−147
...
37

−348
...
30

0

0

+302
...
55

26
...
49

+267
...
28

20
...
1

46

43
...
25

Source: NBS
...


Table 2
...

Data: Principally J
...
Bockris and A
...
N
...
1
...


Table 2
...

Data: Principally J
...
Bockris and A
...
N
...
1
...


1002

DATA SECTION

Table 2
...
9 Inversion temperatures, normal freezing and boiling
points, and Joule–Thomson coefficients at 1 atm and 298 K

kT /(10 −6 atm−1 )

Liquids

TI /K
Air

Tf /K

m JT /(K atm−1)

Tb /K

603

Benzene

12
...
1

Argon

Carbon tetrachloride

12
...
5

Carbon dioxide

Ethanol

11
...
8

Helium

Mercury

1
...
7

Hydrogen

Water

2
...
6

0
...
8

1500

194
...
3
1
...
22

−0
...
501

0
...
030

0
...
354

0
...
861

2
...

Data: AIP(α), KL(κT)
...
0

Krypton

1090

116
...
8

Methane
Solids
Copper

20
...
062

968

90
...
6

Neon

231

24
...
1

Nitrogen

621

63
...
4

0
...
8

90
...
31

s: sublimes
...
W
...
McGraw-Hill,
New York (1957)
...
1 Standard entropies (and temperatures) of phase transitions, ∆ trsS 7/(J K −1 mol−1)
Fusion (at Tf )

Vaporization (at T b )

Ar

14
...
8 K)

74
...
3 K)

Br2

39
...
9 K)

88
...
4 K)

C6H6

38
...
6 K)

87
...
2 K)

CH3COOH

40
...
8 K)

61
...
4 K)

CH3OH

18
...
2 K)

104
...
2 K)

Cl2

37
...
1 K)

85
...
0 K)

H2

8
...
0 K)

44
...
38 K)

H2O

22
...
2 K)

H2S

12
...
6 K)

He

4
...
8 K and 30 bar)

109
...
2 K)
87
...
0 K)
19
...
22 K)

N2

11
...
2 K)

75
...
4 K)

NH3

28
...
4 K)

97
...
73 K)

O2
Data: AIP
...
17 (at 54
...
63 (at 90
...
2 Standard entropies of vaporization of liquids at their normal boiling point
D vap H 7/(kJ mol−1)

q b /°C

D vap S 7/(J K−1 mol−1)

Benzene

30
...
1

+87
...
74

46
...
7

Carbon tetrachloride

30
...
7

+85
...
1

80
...
1

Decane

38
...
7

Dimethyl ether

21
...
6

78
...
0

Hydrogen sulfide

18
...
4

+87
...
3

Methane

356
...
2

−161
...
18

+73
...
21

65
...
1

Water

40
...
0

+109
...


Table 3
...
5 and 2
...
4 Standard Gibbs energies of formation at 298 K: see Tables 2
...
7

Table 3
...
999 55

300

1
...
9956

400

1
...
9912

600

1
...
9703

800

1
...
9672

1000

1
...
9721

Data: LR
...
1 Henry’s law constants for gases at 298 K, K/(kPa kg mol−1)
Water
CH4
CO2

Benzene

7
...
1 × 10

3

44
...
90 × 102

H2

1
...
79 × 104

N2

1
...
87 × 104

O2

7
...
J
...
A
...
Wiley, New York (2001)
...
2 Freezing-point and boiling-point constants
K f /(K kg mol −1)

K b /(K kg mol −1)

Acetic acid

3
...
07

Benzene

5
...
53

Camphor

40

Carbon disulfide

3
...
37

30

4
...
94

5
...
27

3
...
86

0
...


Table 5
...
001

0
...
966

0
...
830

0
...
005

0
...
927

0
...
639

0
...
16

0
...
905

0
...
732

0
...
560

0
...
05

0
...
816

0
...
340

0
...
035

0
...
798

0
...
524

0
...
356

0
...
50

0
...
652

0
...
155

0
...
014

1
...
811

0
...
725

0
...
387

2
...
011

0
...
554

0
...
954

In2 (SO4 )3

Data: RS, HCP, and S
...
Van Nostrand (1942)
...
6 Relative permittivities (dielectric constants) at 293 K
Nonpolar molecules
Methane (at −173°C)

Polar molecules
1
...
54 (at 298 K)
80
...
238

Ammonia

Cyclohexane

2
...
9 (at 298 K)
22
...
26 at −85°C
5
...
0

Ethanol

25
...


2
...
6

DATA SECTION

1005

Table 7
...
(a) In electrochemical order
Reduction half-reaction

E 7/V

Cu2+ + e− → Cu+

Reduction half-reaction

E 7/V

+0
...
0
+2
...
07
+2
...
98
+1
...
78
+1
...
67
+1
...
61

Sn4+ + 2e− → Sn2+
AgBr + e− → Ag + Br−
Ti4+ + e− → Ti3+
2H+ + 2e− → H2
Fe3+ + 3e− → Fe
−
O2 + H2O + 2e− → HO2 + OH−
2+
−
Pb + 2e → Pb
In+ + e− → In
Sn2+ + 2e− → Sn
AgI + e− → Ag + I−
Ni2+ + 2e− → Ni

+0
...
07
0
...
04
−0
...
13
−0
...
14
−0
...
23

+1
...
51
+1
...
40
+1
...
33
+1
...
23
+1
...
23
+1
...
97
+0
...
92
+0
...
86
+0
...
80
+0
...
77
+0
...
62
+0
...
56
+0
...
52
+0
...
49
+0
...
40
+0
...
36
+0
...
27
+0
...
20

Co2+ + 2e− → Co
In3+ + 3e− → In
Tl+ + e− → Tl
2−
PbSO4 + 2e− → Pb + SO4
3+
−
2+
Ti + e → Ti
Cd2+ + 2e− → Cd
In2+ + e− → In+
Cr3+ + e− → Cr 2+
Fe2+ + 2e− → Fe
In3+ + 2e− → In+
S + 2e− → S2−
In3+ + e− → In2+
U4+ + e− → U3+
Cr3+ + 3e− → Cr
Zn2+ + 2e− → Zn
Cd(OH)2 + 2e− → Cd + 2OH−
2H2O + 2e− → H2 + 2OH−
Cr2+ + 2e− → Cr
Mn2+ + 2e− → Mn
V2+ + 2e− → V
Ti2+ + 2e− → Ti
Al3+ + 3e− → Al
U3+ + 3e− → U
Sc3+ + 3e− → Sc
Mg2+ + 2e− → Mg
Ce3+ + 3e− → Ce
La3+ + 3e− → La
Na+ + e− → Na
Ca2+ + 2e− → Ca
Sr2+ + 2e− → Sr
Ba2+ + 2e− → Ba
Ra2+ + 2e− → Ra
Cs+ + e− → Cs
Rb+ + e− → Rb
K+ + e− → K
Li+ + e− → Li

−0
...
34
− 0
...
36
−0
...
40
− 0
...
41
− 0
...
44
−0
...
49
−0
...
74
−0
...
81
−0
...
91
−1
...
19
−1
...
66
−1
...
09
−2
...
48
−2
...
71
−2
...
89
−2
...
92
−2
...
93
−2
...
05

Strongly oxidizing
+

−

H4XeO6 + 2H + 2e → XeO3 + 3H2O
F2 + 2e− → 2F −
O3 + 2H+ + 2e− → O2 + H2O
2−
2−
S2O8 + 2e− → 2SO4
2+
−
+
Ag + e → Ag
Co3+ + e− → Co2+
H2O2 + 2H+ + 2e− → 2H2O
Au+ + e− → Au
Pb4+ + 2e− → Pb2+
2HClO + 2H+ + 2e− → Cl2 + 2H2O
Ce4+ + e− → Ce3+
2HBrO + 2H+ + 2e− → Br2 + 2H2O
−
MnO4 + 8H+ + 5e− → Mn2+ + 4H2O
3+
−
2+

Mn + e → Mn
Au3+ + 3e− → Au
Cl2 + 2e− → 2Cl −
2−
Cr2O7 + 14H+ + 6e− → 2Cr 3+ + 7H2O
O3 + H2O + 2e− → O2 + 2OH−
O2 + 4H+ + 4e− → 2H2O
−
−
ClO4 + 2H+ + 2e− → ClO3 + H2O
+
−
MnO2 + 4H + 2e → Mn2+ + 2H2O
Br2 + 2e− → 2Br−
Pu4+ + e− → Pu3+
−
NO3 + 4H+ + 3e− → NO + 2H2O
2+
2Hg 2+ + 2e− → Hg 2
−
−
ClO + H2O + 2e → Cl− + 2OH−
Hg 2+ + 2e− → Hg
−
NO3 + 2H+ + e− → NO2 + H2O
+
Ag + e− → Ag
Hg2+ + 2e− → 2Hg
2
Fe3+ + e− → Fe2+
BrO− + H2O + 2e− → Br− + 2OH−
2−
Hg2SO4 + 2e− → 2Hg + SO4
2−
−
MnO4 + 2H2O + 2e → MnO2 + 4OH−
−
2−
MnO4 + e− → MnO4
−
−
I2 + 2e → 2I
CU+ + e− → Cu
−
I3 + 2e− → 3I−
NiOOH + H2O + e− → Ni(OH)2 + OH−
2−
Ag2CrO4 + 2e− → 2Ag + CrO4
−
−
O2 + 2H2O + 4e → 4OH
−
−
ClO4 + H2O + 2e− → ClO3 + 2OH−
3−
−
[Fe(CN)6] + e → [Fe(CN)6]4−
Cu2+ + 2e− → Cu
Hg2Cl2 + 2e− → 2Hg + 2Cl−
AgCl + e− → Ag + Cl−
Bi3+ + 3e− → Bi

1006

DATA SECTION

Table 7
...
(b) In electrochemical order
Reduction half-reaction

E 7/V

Reduction half-reaction

E 7/V

Ag+ + e− → Ag
Ag2+ + e− → Ag+
AgBr + e− → Ag + Br−
AgCl + e− → Ag + Cl−
2−
Ag2CrO4 + 2e− → 2Ag + CrO4
−
−
AgF + e → Ag + F
AgI + e− → Ag + I−
Al3+ + 3e− → Al
Au+ + e− → Au
Au3+ + 3e− → Au
Ba2+ + 2e− → Ba
Be2+ + 2e− → Be
Bi3+ + 3e− → Bi
Br2 + 2e− → 2Br−
BrO− + H2O + 2e− → Br− + 2OH−
Ca2+ + 2e− → Ca
Cd(OH)2 + 2e− → Cd + 2OH−
Cd2+ + 2e− → Cd
Ce3+ + 3e− → Ce
Ce4+ + e− → Ce3+
Cl2 + 2e− → 2Cl−
ClO− + H2O + 2e− → Cl− + 2OH−
−
−
ClO4 + 2H+ + 2e− → ClO3 + H2O
−
−
−
ClO4 + H2O + 2e → ClO3 + 2OH−
2+
−
Co + 2e → Co
Co3+ + e− → Co2+
Cr2+ + 2e− → Cr
2
Cr2O7 − + 14H+ + 6e− → 2Cr 3 + + 7H2 O
Cr3+ + 3e− → Cr
Cr3+ + e− → Cr2+
Cs+ + e− → Cs
Cu+ + e− → Cu
Cu2+ + 2e− → Cu
Cu2+ + e− → Cu+
F2 + 2e− → 2F −
Fe2+ + 2e− → Fe
Fe3+ + 3e− → Fe
Fe3+ + e− → Fe2+
[Fe(CN)6]3− + e− → [Fe(CN)6]4−
2H+ + 2e− → H2
2H2O + 2e− → H2 + 2OH−
2HBrO + 2H+ + 2e− → Br2 + 2H2O
2HClO + 2H+ + 2e− → Cl2 + 2H2O
H2O2 + 2H+ + 2e− → 2H2O
H4XeO6 + 2H+ + 2e− → XeO3 + 3H2O
2+
Hg2 + 2e− → 2Hg
Hg2Cl2 + 2e− → 2Hg + 2Cl−
Hg2+ + 2e− → Hg
2+
2Hg2+ + 2e− → Hg2
2−
−
Hg2SO4 + 2e → 2Hg + SO4

+0
...
98
+0
...
22
+0
...
78
−0
...
66
+1
...
40
+2
...
85
+0
...
09
+0
...
87
−0
...
40
−2
...
61
+1
...
89
+1
...
36
−0
...
81
−0
...
33
−0
...
41
−2
...
52
+0
...
16
+2
...
44
−0
...
77
+0
...
83
+1
...
63
+1
...
0
+0
...
27
+0
...
92
+0
...
54
+0
...
14
−0
...
44
−0
...
49
−2
...
52
−3
...
36
−1
...
51
+1
...
51
+0
...
60
−2
...
23
+0
...
80
+0
...
10
+0
...
23
−0
...
08
+2
...
24
−0
...
67
−0
...
20
+0
...
92
−2
...
48
+2
...
09
−0
...
15
−2
...
63
−0
...
00
−0
...
79
−0
...
19
−0
...
76

DATA SECTION

1007

Table 7
...
(a) In order of acid strength
Acid

HA

A−

Ka

pKa

Hydriodic

HI

I−

1011

−11

−

109

−9
−7

Hydrobromic

HBr

Br

Hydrochloric

HCl

Cl −

107

Sulfuric

H2SO4

−
HSO4

102

Perchloric*

HClO4

−
ClO4

4
...
0

Oxalic

(COOH)2

−
HOOCCO2

5
...
25

Sulfurous

H2SO3

−
HSO3

1
...
85

Hydrogensulfate ion

−
HSO4

2−
SO4

1
...
99

Phosphoric

H3PO4

−
H2PO4

6
...
16

Glycinium ion

+

NH2CH2COOH

4
...
35

Hydrofluoric

HF

F−

6
...
20

Formic

HCOOH

1
...
75

Hydrogenoxalate ion

−
HOOCCO2

−
HCO2
2−
C2O 4

1
...
81

Lactic

CH3CH(OH)COOH

−
CH3CH(OH)CO2

1
...
86

Acetic (ethanoic)

CH3COOH

−
CH3CO2

1
...
76

Butanoic

CH3CH2CH2COOH

1
...
83

Propanoic

CH3CH2COOH

−
CH3CH2CH2CO2
−
CH3CH2CO2

1
...
87

Anilinium ion

+
C6H5NH 3

C6H5NH2

1
...
87

Pyridinium ion

C5H5NH+

C6H5N

5
...
23

Carbonic

H2CO3

4
...
35

Hydrosulfuric

H2S

−
HCO3
−

HS

8
...
05

Dihydrogenphosphate ion

−
H2PO 4

2−
HPO4

6
...
21

Hypochlorous

HClO

ClO−

4
...
40

Hydrazinium ion

+
NH2NH 3

NH2NH2

8 × 10

Hypobromous

HBrO

BrO−

2
...
55

Hydrocyanic

HCN

CN−

6
...
21

Ammonium ion

NH +
4

NH3

5
...
25

Boric*

B(OH)3

−
B(OH)4

5
...
27

Trimethylammonium ion

(CH3)3NH+

(CH3)3N

1
...
80

Phenol

C6H5OH

C6H5O −

1
...
99

Hydrogencarbonate ion

−
HCO 3

2−
CO3

4
...
33

Hypoiodous

HIO

IO−

3 × 10 −11

10
...
2 × 10 −11

10
...
2 × 10 −11

10
...
9 × 10 −11

10
...
75

Methylammonium ion
Dimethylammonium ion

NH3CH2COOH

+

−2
1

−9

−1
...
1

Triethylammonium ion

(CH3CH2)3NH

(CH3CH2)3N

1
...
4 × 10 −11

10
...
1 × 10 −12

11
...
8 × 10 −13

12
...
00

Hydrogensulfide ion
* At 293 K
...
0 × 10

1008

DATA SECTION

Table 7
...
(b) In alphabetical order
Acid

HA

A−

Acetic (ethanoic)

CH3COOH

−
CH3CO 2

1
...
76

Ammonium ion

+
NH4

NH3

5
...
25

Anilinium ion

+
C6H5NH3

C6H5NH2

1
...
87

Boric*

B(OH)3

−
B(OH) 4

5
...
27

Butanoic

CH3CH2CH2COOH

1
...
5 × 10−7

6
...
4 × 10−11

10
...
2 × 10−8

7
...
83

(CH3)2NH

1
...
73

CH3CH2NH2

2
...
65

HCOOH

−
HCO 2

1
...
75

Glycinium ion

+

NH2CH2COOH

4
...
35

Hydrazinium ion

NH2NH +
3

NH2NH2

Hydriodic

HI

I−

1011

−11

Hydrobromic

HBr

Br−

109

−9

Hydrochloric

HCl

Cl−

107

NH3CH2COOH

−

−9

8 × 10

8
...
2 × 10

Hydrofluoric

HF

F−

6
...
20

Hydrogenarsenate ion

2−
HAsO 4

3−
AsO4

5
...
29

Hydrogencarbonate ion

−
HCO 3

2−
CO 3

4
...
33

Hydrogenoxalate ion

Hydrogensulfide ion

−
HOOCCO 2
2−
HPO 4
−
HSO4
−

HS

2−
C2O 4
3−
PO 4
2−
SO 4
2−

Hydronium ion

H3O+

H2O

1

0
...
9 × 10−8

7
...
8 × 10−9

8
...
0 × 10−8

7
...
5 × 10

4
...
21

3
...
32

1
...
99

1
...
00

−11

Hypoiodous

HIO

IO

3 × 10

Lactic

CH3CH(OH)COOH

−
CH3CH(OH)CO 2

1
...
86

Methylammonium ion

CH3NH +
3

CH3NH2

2
...
66

Oxalic

(COOH)2

−
HOOCCO2

5
...
25

Perchloric*

HClO4

−
ClO 4

4
...
0 × 10−10

9
...
9 × 10−3

2
...
4 × 10−5

4
...
9 × 10

10
...
6

Pyridinim ion

C5H5NH

C6H5N

Sulfuric

H2SO4

−
HSO 4

Sulfurous

H2SO3

−
HSO 3

1
...
85

Triethylammonium ion

(CH3CH2)3NH+

(CH3CH2)3N

1
...
75

Trimethylammonium ion

(CH3)3NH+

(CH3)3N

1
...
80

* At 293 K
...
23
−2

1009

DATA SECTION

Table 9
...
45

0
...
01

0
...
50

0
...
02

0
...
55

0
...
03

0
...
60

0
...
04

0
...
65

0
...
05

0
...
70

0
...
06

0
...
75

0
...
07

0
...
80

0
...
08

0
...
85

0
...
09

0
...
90

0
...
10

0
...
95

0
...
15

0
...
00

0
...
20

0
...
20

0
...
25

0
...
40

0
...
30

0
...
60

0
...
35

0
...
80

0
...
40

0
...
00

0
...


Table 10
...
6875

1s

Li
2
...
6848

2s

1
...
9120

B
4
...
6727

N
6
...
6579

F
8
...
6421

2
...
2166

3
...
4916

5
...
7584

2
...
1358

3
...
4532

5
...
7584

P
14
...
5409

Cl
16
...
5075

1s

Na
10
...
6089

Al
12
...
5745

2s

6
...
3920

8
...
0200

9
...
6288

11
...
2304

2p

6
...
8258

8
...
9450

10
...
9770

12
...
0082

3s

2
...
3075

4
...
9032

5
...
3669

7
...
7568

4
...
2852

4
...
4819

6
...
7641

3p

Data: E
...
L
...

IBM Res
...
J
...
Phys
...


1010

DATA SECTION

Table 10
...
0

2372
...
4

Li
513
...
4

B
800
...
2

N
1402
...
9

F
1681

Ne
2080
...
0

1757
...
1

3388
...
2

Na
495
...
7

Al
577
...
5

P
1011
...
1

Ar
1520
...
4

1450
...
6

1577
...
2

2251

2297

2665
...
6

999
...
8

Ca
589
...
8

Ge
762
...
0

Se
940
...
9

Kr
1350
...
4

1145

1979

1537

1798

2044

2104

2350

2963

2735

Rb
403
...
5

In
558
...
6

Sb
833
...
2

I
1008
...
4

2632

1064
...
6

1411
...
9

2046

2704

2943
...
8

Tl
589
...
5

Bi
703
...
1

1971
...
4

1610

2878

3081
...
5
2420

Data: E
...
4 Electron affinities, Eea/(kJ mol−1)
H

He

72
...
8

Be
≤0

B
23

C
122
...
7

Ar
−35

−844
Na
52
...
6

P
71
...
4
−532

K
48
...
37

Ga
36

Ge
116

As
77

Se
195
...
5

Kr
−39

Rb
46
...
03

In
34

Sn
121

Sb
101

Te
190
...
3

Xe
−41

Cs
45
...
95

Tl
30

Pb
35
...


DATA SECTION

Table 11
...
3

Cl2

198
...
81

F2

141
...
138

HBr

141
...
45

HF

91
...
92

N2

109
...
75

(b) Mean bond lengths from covalent radii*
H
37
C

77(1)

N

74 (1)

67(2)

O

66(1)

65 (2)

F

64

Cl

99

57(2)

60(3)
Si

118

P

110

S

104(1)
95(2)

Ge

122

As

121

Se

104

Br

114

Sb

141

Te

137

I

133

* Values are for single bonds except where indicated otherwise (values in parentheses)
...


Table 11
...
N

945

H-O

428

H-F

565

H-Cl

431

Br-Br

193

I-I

151

H-Br

366

H-I

299

H-C6H5

469

I-CH3

237

Polyatomic molecules
H-CH3
435

H-NH2

460

H-OH

492

H3C-CH3

368

H2C=CH2

720

HC
...


54

1011

1012

DATA SECTION

Table 11
...

Data: HCP and L
...
Cornell University Press (1960)
...
4 Pauling (italics) and Mulliken electronegativities
H

He

2
...
06
Li
0
...
57

B
2
...
55

N
3
...
44

F
3
...
28

1
...
83

2
...
08

3
...
43

4
...
93

Mg
1
...
61

Si
1
...
19

S
2
...
16

Ar

1
...
63

1
...
03

2
...
65

3
...
36

K
0
...
00

Ga
1
...
01

As
2
...
55

Br
2
...
0

1
...
30

1
...
95

2
...
51

3
...
98

Rb
0
...
95

In
1
...
96

Sb
2
...
10

I
2
...
6

0
...
21

1
...
83

2
...
34

2
...
59

Cs
0
...
89

Tl
2
...
33

Bi
2
...
L
...
Inorg
...
Chem
...
C
...
E
...
, 42,
1523 (1980)
...
C
...
Am
...
Soc
...
The Mulliken values have been
scaled to the range of the Pauling values
...
2 Properties of diatomic molecules
§0 /cm−1

qV /K

B/cm−1

qR /K

r/pm

1

2321
...
8

42
...
8

2

1

4400
...
864

87
...
138

574
...
1

2

2

3118
...
442

43
...
154

577
...
6

2

1

4138
...
956

30
...
680

965
...
4

1

35

2990
...
593

15
...
45

516
...
7

1

1

H81Br

2648
...
465

12
...
44

411
...
7

1

1

2308
...
511

9
...
92

313
...
9

1

14

2358
...
9987

2
...
76

2293
...
7

2

16

1580
...
4457

2
...
75

1176
...
5

2

19

891
...
8828

1
...
78

445
...
4

2

35

Cl2

559
...
2441

0
...
75

322
...
3

2

12 16

2170
...
9313

2
...
81

1903
...
8

1

323
...
0809

10
...
3

245
...
2

1

+
H2

H2
H2
H19F

1

H Cl
H127I
N2
O2
F2
C O

79

Br81Br

k/(N m−1)

D/(kJ mol −1)

s

Data: AIP
...
3 Typical vibrational
wavenumbers, #/cm−1

Table 14
...
C stretch

2100 –2260

O-H stretch

3590 –3650

H-bonds

3200 –3570

C=O stretch

1640 –1780

C
...
J
...
Chapman and Hall
...
00

<1
...
24

<120

Red

700

4
...
43

1
...
84

1
...
00

193

Yellow

580

5
...
72

2
...
66

1
...
34

226

Blue

470

6
...
13

2
...
14

2
...
95

285

Near ultraviolet
Far ultraviolet

300

10
...
33

4
...
0

>5
...
20

>598

Data: J
...
Calvert and J
...
Pitts, Photochemistry
...


1014

DATA SECTION

Table 14
...
10

163

1
...
73

174

5
...
7–3
...
9

350

15

>3
...
6

280

4
...
9

255

5
...
3 × 103

5
...
0 × 105

-NO2
C6H5-

2+

Strong
10
1
...
2

810

10

[Cu(NH3)4]2+(aq)

1
...
0

167

7
...
2 Nuclear spin properties
Nuclide

Spin I

Magnetic
moment m/m N

g-value

g/(107 T −1 s−1)

NMR frequency at
1 T, n /MHz

1
–
2

Natural
abundance %

−1
...
8260

−18
...
164

1

99
...
792 85

5
...
752

42
...
0156

1

0
...
857 45

1
–
2

2
...
1067

6
...
2553

−20
...
6

3

1
...
6002

2
...
575

11

80
...
6886

1
...
5841

13
...
7024

1
...
7272

10
...
414

13

1
...
635

1

0
...
037

5
–
2

−1
...
7572

−3
...
774

C
N
O

0
...
9328

3
...
628 87

5
...
177

40
...
1316

2
...
840

17
...
6438

0
...
054

3
...
4

3
–
2

0
...
5479

2
...
176

37

24
...
6841

0
...
184

3
...
74

* Radioactive
...

Data: KL and HCP
...
3 Hyperfine coupling constants for atoms, a/mT
Nuclide

Spin

1

1
–
2

50
...
8(1s)

13

1
–
2

113
...
2(2s)

19

1
–
2

1720(2s)

108
...
6(3p)

35

3
–
2
3
–
2

168(3s)

10
...
4(3p)

H
H
C
N
F
P
Cl

37

Cl

Isotropic
coupling

Anisotropic
coupling

6
...
8(2p)

Data: P
...
Atkins and M
...
R
...
Elsevier, Amsterdam (1967)
...
1 Dipole moments, polarizabilities, and polarizability volumes
m/(10−30 C m)

m/D

a/(10− 40 J−1 C2 m2 )

Ar

0

0

C2H5OH

5
...
69

C6H5CH3

1
...
36

C6H6

0

0

10
...
6

CCl4

0

0

10
...
7

CH2Cl2

5
...
57

6
...
57

CH3Cl

6
...
87

4
...
04

CH3OH

5
...
71

3
...
59

CH4

0

0

2
...
89

CHCl 3

3
...
01

8
...
46

CO

0
...
117

1
...
20

CO2

0

0

2
...
93

H2

0

0

0
...
911

H2O

6
...
85

1
...
65

HBr

2
...
80

3
...
01

HCl

3
...
08

2
...
93

He

0

0

0
...
22

HF

6
...
91

0
...
57

HI

1
...
42

5
...
06

N2

0

0

1
...
97

NH3

4
...
47

2
...
47

1,2-C6H4(CH3)2

2
...
62

1
...
85

Data: HCP and C
...
F
...
Bordewijk, Theory of electric polarization
...


1015

1016

DATA SECTION

Table 18
...
5 Surface tensions of liquids at 293 K
g /(mN m−1)

r0 /pm

Ar

111
...
3

Benzene

C2H2

209
...
5

Carbon tetrachloride

28
...
0

C2H4

200
...
9

Ethanol

22
...
12

478
...
4

C6H6

377
...
4

Mercury

CCl4

378
...
1

Methanol

22
...
27

448
...
75

CO2

201
...
4

72
...
29

357
...
0 at 100°C

Kr

154
...
5

N2

91
...
9

O2

113
...
4

Xe

213
...
0

472

Data: KL
...
Cuadros, I
...
Ahamuda, Molec
...


Table 19
...
2 Diffusion coefficients of macromolecules in water at
20°C

Rg /nm

M/(kg mol −1)

D/(10 −10 m2 s−1)

2
...
8

493

Polystyrene

3
...
9 × 10 4

3

Sucrose

50 (in poor solvent)

Ribonuclease

13
...
19

Lysozyme

14
...
04

Serum albumin

65

0
...
69

Urease

480

0
...
069

Myosin

493

0
...
0
92
...
Tanford, Physical chemistry of macromolecules
...


0
...
586

Data: C
...
Wiley, New York (1961)
...
3 Frictional coefficients and molecular geometry
Major axis/Minor axis

Table 19
...
04

1
...
11

1
...
18

1
...
25

1
...
31

1
...
38

1
...
43

1
...
49

1
...
54

1
...
95

2
...
07

2
...
5
81

a
0
...
50

Benzene

23†

83

0
...
70

Amylose

0
...
50

Various
proteins‡

Guanidine
hydrochloride +
HSCH2CH2OH

7
...
66

† The θ temperature
...

Data: K
...
Van Holde, Physical biochemistry
...


Data: K
...
Van Holde, Physical biochemistry
...

Sphere; radius a, c = af0
Prolate ellipsoid; major axis 2a, minor axis 2b, c = (ab2)1/3

1

f= 2

5
6f
ln{[1 + (1 − b /a ) ]/(b/a)} 7 0
(1 − b 2/a2)1/2

2/3

3 (b/a)

2

2 1/2

Oblate ellipsoid; major axis 2a, minor axis 2b, c = (a 2b)1/3

1

f= 2

5
6f
arctan[(a /b − 1) ] 7 0
(a 2/b 2 − 1)1/2

2/3

3 (a/b)

2

2

1/2

Long rod; length l, radius a, c = (3a 2/4)1/3

1

5
6 f0
3 (3/2) {2 ln(l/a) − 0
...


Table 20
...
Values for ions without a coordination number stated are estimates
...
D
...
T
...
B25, 925 (1969)
...
5 Lattice enthalpies, ∆H L/(kJ mol−1)

F

Cl

Table 20
...
0

Li

1037

852

815

761

Benzene

−7
...
4

Na

926

787

752

705

Cyclohexane

−7
...
5

K

821

717

689

649

Carbon tetrachloride

−8
...
4

Rb

789

695

668

632

NaCl(s)

−13
...
75

Cs

750

676

654

620

Cu(s)

−96

−6
...
9

−2
...
5

−4
...
79 × 103
+600
+1
...
2

Pt(s)

+262

+22
...

Data: Principally D
...
Chem
...
31, 1646 (1959)
...
1 Collision cross-sections,
σ /nm2

2832

+7
...
7

K(s)

+5
...
5

Data: KL and χm = χM/ρ
...
2 Transport properties of gases at 1 atm
k/(J K −1 m−1 s−1)

h/mP

273 K

273 K

Ar

0
...
64

C6H6

0
...
0241

173

182

CH4

0
...
0163

210

223

Cl2

0
...
0164

97

103

CO2

0
...
0302

103

110

0
...
079

123

132

He

0
...
0145

136

147

N2

0
...
1682

84

88

Ne

0
...
1442

187

196

O2

0
...
0087

234

250

0
...
0240

166

176

H2

SO2

293 K

Ne
Data: KL
...
0465

298

313

O2

0
...
0052

212

228

Data: KL
...
5 Limiting ionic conductivities in water at 298 K, λ /(mS m2 mol−1)

Table 21
...
601

Carbon tetrachloride

0
...
06

Mercury

1
...
553

Pentane

0
...
81

Ca

11
...
09

Cs+

7
...
635

Cu2+

10
...
73

34
...
86
14
...
891

7
...
87

F−

Mg2+

30
...
010

[Fe(CN)6]4 −

44
...
26

−
HCO2

5
...
49

I−

7
...
35

−
NO3

+

Rb

7
...
91

Sr2+

11
...
00

Zn2+

10
...
370 23(t − 20) + 8
...

Data: AIP, KL
...
146

Data: KL, RS
...
6 Ionic mobilities in water at 298 K, u/(10 −8 m2 s−1 V −1)

+

5
...
60

Na+

† The viscosity of water over its entire liquid
range is represented with less than 1 per cent
error by the expression

B = 109 + t

12
...
7 Debye–Hückel–Onsager coefficients for (1,1)electrolytes at 25°C

Anions
Solvent

−

Ag

6
...
09

Ca2+

6
...
56

Cl−

7
...
24

Cu2+

A/(mS m2 mol−1/
(mol dm−3)1/2 )

H+

36
...
46

−

7
...
70

4
...
5

5
...
4

[Fe(CN)6]

+
NH4

7
...
65

−
NO3

7
...
92

OH−

20
...
47

2
SO4 −

8
...
4 and u = λ /zF
...
96

Acetone (propanone)

3
...
63

Acetonitrile

2
...
716

Ethanol

8
...
83

15
...
923

4
...
776

Methanol
Nitrobenzene
Nitromethane
Water

111
6
...
708
0
...
O’M
...
K
...
Reddy, Modern electrochemistry
...


1020

DATA SECTION

Table 21
...
05

9
...
96

Br−

2
...
13

N2 in CCl4(l)

3
...
31

Cl −

2
...
17

O2 in CCl4(l)

3
...
03

F−

1
...
055

Ar in CCl4(l)

3
...
33

I−

Dextrose in water

0
...
89

Sucrose in water

0
...
26

CH3OH in water

5
...
58

C2H5OH in water

OH

2
...
24

Data: AIP and (for the ions) λ = zuF in conjunction with Table 21
...


Table 22
...
38 × 10 −5

HNO3(l)

25

1
...
27 × 10 −5

4
...
70 h
131 h

C2H6 → 2 CH3

g

700

5
...
6 min

Cyclopropane → propene

g

500

6
...
2 min

CH3N2CH3 → C2H6 + N2

g

Sucrose → glucose + fructose

aq(H+)

3
...
0 × 10 −5

25

3
...

Data: Principally K
...
Laidler, Chemical kinetics
...
J
...
W
...
Oxford University Press (1995);
J
...
Harper & Row, New York (1976)
...


Table 22
...
80

2 NO2 → 2 NO + O2

g

300

0
...
42 × 10−2

D2 + HCl → DH + DCl

g

600

0
...
8 × 1010

−

methanol

20

2
...
23 × 10 − 6

water

25

1
...
6 × 1012

CH3Cl + CH3O

CH3Br + CH3O
+

−

H + OH → H2O

ice

Data: Principally K
...
Laidler, Chemical kinetics
...
J
...
W
...
Oxford University Press (1995); J
...
Harper & Row,
New York (1976)
...
4 Arrhenius parameters
First-order reactions

A/s−1

E a /(kJ mol−1)

Cyclopropane → propene

1
...
98 × 1013

160

cis-CHD=CHD → trans-CHD=CHD

3
...
98 × 1013

261

C2H5I → C2H4 + HI

2
...
51 × 107

384

2 N2O5 → 4 NO2 + O2

4
...
94 × 1011

250

1
...
0

NO + Cl2 → NOCl + Cl

10

315
42

4
...
25

Second-order, solution

8

A/(dm3 mol−1 s−1)

E a /(kJ mol−1)

C2H5ONa + CH3I in ethanol

2
...
6

−

C2H5Br + OH in water

4
...
5

C2H5I + C2H5O− in ethanol

1
...
6

CH3I + C2H5O in ethanol

2
...
6

C2H5Br + OH− in ethanol

4
...
5

−

CO2 + OH− in water

1
...
19 × 1012

Sucrose + H2O in acidic water

1
...
7
107
...
1 × 1016

100

in methanol

2
...
0 × 10

13

112

in acetic acid

4
...
4 × 104

45

C6H5NH2 + C6H5COCH2Br
in benzene

91

34

Data: Principally J
...
Harper & Row, New York (1976) and A
...
Frost and R
...

Pearson, Kinetics and mechanism
...


1021

1022

DATA SECTION

Table 24
...
4 × 109

5
...
0

0
...
0 × 109

4
...
0

5
...
3 × 10

0
...
5 × 10 −3

H2 + C2H4 → C2H6
K + Br2 → KBr + Br

2
...
24 × 10

7
...
0 × 10

2
...
7 × 10 −6

180
0
...
8

Data: Principally M
...
Pilling and P
...
Seakins, Reaction kinetics
...


Table 24
...
See Table 22
...
1 Maximum observed enthalpies of physisorption, ∆adH 7/(kJ mol−1)
C2H2

−38

H2

−84

C2H4

−34

H2O

−59

CH4

−21

N2

−21

Cl2

−36

NH3

−38

CO

−25

O2

−21

CO2

−25

Data: D
...
Haywood and B
...
W
...
Butterworth (1964)
...
2 Enthalpies of chemisorption, ∆adH 7/(kJ mol−1)
Adsorbate

Adsorbent (substrate)
Ti

Ta

H2

W

Cr

Mo

−188

−188

N2

Nb

−167

Mn

−586

−71

Co

Ni

−134

−682

−494
−192

−703

−552

−456

−339

−301

NH3
−577

Pt

−117

−720
−640

CO2

Rh

−293

O2
CO

C2H4

Fe

−427

−372

−222

−176

−225

−146

−184

−188
−427

Data: D
...
Haywood and B
...
W
...
Butterworth (1964)
...
6 Exchange current densities and transfer coefficients at
298 K

Table 25
...
9 × 10 −4

a

None
Au(s)

105

Pt(s)

59

Cu

1 × 10 −6

None

350

Ni

6
...
58

162

Hg

7
...
50

None

245

Pb

5
...
5 × 10 −3

0
...
0 × 10 −5

0
...
O’M
...
K
...
Reddy, Modern electrochemistry
...


Data: G
...
Bond, Heterogeneous catalysis
...


Table A3
...
5236

Benzene

589 nm
1
...
4965

Carbon tetrachloride

1
...
4676

1
...
6748

1
...
6182

Ethanol

1
...
3618

1
...
5050

1
...
4973

Kl(s)

1
...
6664

1
...
3362

1
...
3277

Methylbenzene

1
...
4955

1
...
3404

1
...
3312

Data: AIP
...


The groups Dn
h=4

D2, 222

E

z
C2

y
C2

x
C2

A1

1

1

1

1

B1

1

1

−1

−1

z, xy

Rz

B2

1

−1

1

−1

y, xz

Ry

B3

1

−1

−1

1

x, yz

Rx

D3, 32

E

2C3

1

1

1

A2

1

1

−1

2

−1

h=6

3C′
2

A1
E

x 2, y 2, z 2

z 2, x 2 + y 2
z

Rz

(x, y), (xz, yz), (xy, x − y )
2

0

2

D4, 422

E

C2

2C4

2C′
2

2C 2
″

A1

1

1

1

1

1

A2

1

1

1

−1

−1

z

B1

1

1

−1

1

−1

x2 − y2

B2

1

1

−1

−1

1

xy

E

2

−2

0

0

0

(x, y), (xz, yz)

(Rx, Ry)

h=8
z 2, x 2 + y 2
Rz

(Rx, Ry)

The groups Dnh
D3h, %2m

E

σh

2C3

2S3

3C′
2

3σv

A′
1

1

1

1

1

1

1

h = 12
z 2, x 2 + y 2

A′
2

1

1

1

1

−1

−1

A1
″

1

−1

1

−1

1

−1

A2
″

1

−1

1

−1

−1

1

z

E′

2

2

−1

−1

0

0

(x, y), (xy, x 2 − y 2)

E″

2

−2

−1

1

0

0

(xz, yz)

Rz

(Rx, Ry)

1025

1026

DATA SECTION

2S4

σh

2σv

2σd

1

1

1

1

1

1

1

−1

−1

1

−1

1

1

−1

1

−1

1

−1

1

xy
(xz, yz)

D4h, 4/mmm

E

2C4

C2

2C′
2

2C 2
″

A1g

1

1

1

1

1

1

A2g

1

1

1

−1

−1

B1g

1

−1

1

1

−1

B2g

1

−1

1

−1

1

i

Eg

2

0

−2

0

0

2

0

−2

0

1

1

1

1

1

−1

−1

−1

−1

1

1

1

−1

−1

−1

−1

−1

1

1

B1u

1

−1

1

1

−1

−1

1

−1

−1

1

B2u

1

−1

1

−1

1

−1

1

−1

1

−1

Eu

2

0

−2

0

0

−2

0

2

0

Rz
x2 − y2

−1

A2u

x 2 + y 2, z 2

0

A1u

h = 16

0

(Rx, Ry)

z

(x, y)

5C2

σh

2S5

3
2S5

5σv

1

1

1

1

1

1

1

1

−1

1

1

1

−1

2 cos α

2 cos 2α

0

2

2 cos α

2 cos 2α

0

(x, y)

2

2 cos 2α

2 cos α

0

2

2 cos 2α

2 cos α

0

(x 2 − y 2, xy)

A″
1

1

1

1

1

−1

−1

A″
2

1

1

1

−1

−1

−1

−1

1

z

E″
1

2

2 cos α

2 cos 2α

0

−2

−2 cos α

−2 cos 2α

0

(xz, yz)

E″
2

2

2 cos 2α

2 cos α

0

−2

−2 cos 2α

−2 cos α

0

D5h

E

2C5

2
2C 5

A′
1

1

1

A′
2

1

E′
1

2

E′
2

−1

•C′
2

D•h

E

2Cf

+
A1g(Σ g )

1

1

1

+
A1u (Σ u )

1

1

1

−
A2g(Σ g )
−
A2u(Σ u )

1

1

1

1

i

h = 20
x 2 + y 2, z 2

Rz

−1

2iC•

iC′
2

1

1

1

−1

−1

−1

−1

1

1

−1

−1

−1

1

(Rx, Ry)

h=•

1

z 2, x 2 + y 2
z
Rz

E1g(Πg)

2

2 cos φ

0

2

−2 cos φ

0

(xz, yz)

E1u(Πu)

2

2 cos φ

0

−2

2 cos φ

0

(x, y)

E2g(∆g)

2

2 cos 2φ

0

2

2 cos 2φ

0

(xy, x 2 − y 2)

E2u(∆u)

2

2 cos 2φ

0

−2

−2 cos 2φ

0

Ӈ

a = 72°

(Rx, Ry)

DATA SECTION

1027

The cubic groups
Td, ∞3m

E

8C3

3C2

6σd

6S4

A1

1

1

1

1

1

A2

1

1

1

−1

−1

E

2

−1

2

0

0

T1

3

0

−1

−1

1

T2

3

0

−1

1

−1

h = 24
x2 + y2 + z2
(3z 2 − r 2, x 2 − y 2 )
(Rx, Ry, Rz)
(x, y, z), (xy, xz, yz)

2
3C2 (= C 4 )

Oh (m3m)

E

8C3

6C2

6C2

A1g

1

1

1

1

1

6S4

8S6

3σh

6σd

1

1

1

1

1

i

A2g

1

1

−1

−1

1

1

−1

1

1

2

−1

0

0

2

2

0

−1

2

x2 + y2 + z2

−1

Eg

h = 48

0

T1g

3

0

−1

1

−1

3

1

0

−1

−1

T2g

3

0

1

−1

−1

3

−1

0

−1

(2z 2 − x 2 − y 2, x 2 − y 2)

1

A1u

1

1

1

1

1

−1

−1

−1

−1

1

1

−1

−1

1

−1

1

−1

−1

1

Eu

2

−1

0

0

2

−2

0

1

−2

0

T1u

3

0

−1

1

−1

−3

−1

0

1

1

T2u

3

0

1

−1

−1

−3

1

0

1

−1

(xy, yz, xy)

−1

A2u

(Rx, Ry, Rz)

The icosahedral group
2
12C 5

I

E

A

1

1

1

3

1
–(1 +
2
1
–(1 −
2

1
–(1 −
2
1
–(1 +
2

T1

12C5

5)

h = 60

20C3

15C2

1

1

5)

0

−1

(x, y, z)

5)

(Rx, Ry, Rz)

T2

3

0

−1

G

4

−1

−1

1

0

G

5

0

0

−1

1

z2 + y2 + z2

5)

(2z 2 − x 2 − y 2, x 2 − y 2, xy, yz, zx)

Further information: P
...
Atkins, M
...
Child, and C
...
G
...
Oxford University
Press (1970)
...
18
2
...
1
1
...
3
1
...
5
1
...
7
1
...
9
1
...
11
1
...
13
11
...
15
1
...
17
1
...
19
1
...
21

1
...
5 bar, (b) 10
...

(a) 1
...

120 kPa
...
67 × 103 kg
...
5 × 103 Pa
...

R = 0
...
9987 g mol−1
...

2
...

(a) 3
...
2 kPa
...
14 g mol−1
...

(a) (i) 1
...

(b) (i) 0
...

a = 1
...
36 × 10−5 m3 mol−1
...
12, repulsive; (b) 2
...

(a) 0
...
112 dm3 mol−1
...
7 cm3; (b) −0
...

(a) xN = 0
...
37;
(b) pN = 2
...
5 atm;
(c) 4
...

b = 0
...
94 × 10−10 m, a = 3
...

(a) 1259 K; (b) 0
...

(a) p = 2
...
2 atm, T = 718 K;
(c) p = 1
...

b = 1
...
67
...
20
2
...
22
2
...
24
2
...
26
2
...
28
2
...
30
2
...
32
2
...
1
3
...
3
3
...
5
3
...
7
3
...
9

Chapter 2
2
...
2
2
...
4
2
...
6
2
...
8
2
...
10
2
...
12
2
...
14
2
...
16
2
...

−91 J
...
62 × 103 J;
(b) ∆U = ∆H = 0, q = −w = 1
...

p2 = 143 kPa, w = 0, q = 3
...
28 × 103 J
...
8 J
...
6 kJ, w = 5
...
0 kJ
...

(a) ∆H = q = 14
...
1 kJ
...
9 kJ, w = 0, ∆U = q = 14
...

200 K
...

8
...

Cp,m = 53 J K−1 mol−1, CV,m = 45 J K−1 mol−1
...
0 × 103 J mol−1, ∆U = 1
...

q = 0, w = −3
...
5 × 103 J
...
0205 m3, Tf = 279 K, w = −6
...

q = ∆H = 24
...
6 kJ, ∆U = 22
...

−3053
...


−1152 kJ mol−1
...
3 J K−1, ∆T = +64
...

+84
...

+1
...

(a) ∆r H 7 = −589
...
13 kJ mol−1
...
48 kJ mol−1, ∆ f H 7(H2O) = −241
...

−760
...

+52
...

−566
...

(a) ∆ r H 7(298 K) = −175 kJ mol−1, ∆ rU 7(298 K) = −173 kJ mol−1;
(b) ∆ r H 7(348 K) = −176 mol−1
...
49 kJ mol−1
...

0
...

∆Um = +129 J mol−1, q = +7
...
62 kJ mol−1
...
27 × 10−3 K−1
...
6 × 102 atm
...
2 J atm−1 mol−1, q (supplied) = 27
...


3
...
11
3
...
13

3
...
15
3
...
17
3
...
19
3
...
21
3
...
8 × 102 J K−1; (b) 1
...

152
...

−7
...

∆S = q = 0, w = ∆U = +2
...
58 kJ
...

(a) 0; (b) −230 J; (c) −230 J; (d) −5
...
2 J K−1
...
6 J K−1; (b) −104
...

(a) −21
...
0 J K−1 mol−1
...
40 kJ mol−1; (b) −5798 kJ mol−1
...
55 kJ mol−1; (b) −5798 kJ mol−1
...
2 kJ mol−1
...

(a) ∆S(gas) = +3
...
0 J K−1, ∆S(total) = 0;
(b) ∆S(gas) = +3
...
0 J K−1;
(c) ∆S(gas) = 0, ∆S(surroundings) = 0, ∆S(total) = 0
...
11 kJ mol−1
...
500; (b) 0
...
5 kJ
...
0 J
...
8 J K−1
...
0 kJ
...
71 kJ mol−1
...
93 kJ mol−1
...

+2
...


Chapter 4
4
...
2
4
...
4

296 K = 23°C
...
5 J K−1 mol−1, ∆fusH = +2
...

25
...

(a) 31
...
9 K
...
5
4
...
7
4
...
9
4
...

3
...

Frost will sublime, 0
...

(a) 29
...
22 atm = 168 Torr; At 60°C,
p1 = 0
...

272
...

6
...
73 per cent
...
3
7
...
5
7
...
7
7
...
1
5
...
3
5
...
5
5
...
7
5
...
9
5
...
11
5
...
13
5
...
15
5
...
17
5
...
19
5
...
21

843
...

18 cm3
...
2 × 103 kPa
...
5 × 102 kPa
...

178 g mol−1
...
077°C
...
3 J, ∆mixS = 6
...

∆mixG = −3
...
5 J K−1, ∆mix H = 0
...
7358
...
51 mmol kg−1, O2: 0
...

0
...

11 kg
...
0 kg mol−1
...
9701, xA = 0
...

−3536 J mol−1, 212 Torr
...
436, aB = 0
...
98, γ B = 0
...

0
...

(a) 45
...
8 g Ba(NO3)2
...
661
...
3
...
10
7
...
12
7
...
14

7
...
1
6
...
3
6
...
5
6
...
7
6
...
11
6
...
14

6
...
5, yB = 0
...

xA = 0
...
347, p = 73
...

(a) the solution is ideal; (b) yA = 0
...
5418
...
4 kPa; (b) yB = 0
...
23; (c) p(final) = 4
...

(a) yA = 0
...
67, yA = 0
...

C = 3
...

(a) C = 2, P = 2; (b) F = 2
...
53, T = T2, xB = 0
...

(a) xB ≈ 0
...
8; (c) xAB2 ≈ 0
...

A solid solution with x(ZrF4) = 0
...
The solid
solution continues to form, and its ZrF4 content increases until it
reaches x(ZrF4) = 0
...
At that temperature, the entire
sample is solid
...
47, a second liquid phase appears
...
314, only one liquid
remains
...


7
...
17
7
...
1
7
...
16841; (c) ∆rG 7 = 4
...

(a) K = 0
...
96
...
3 × 1054, ∆rG 7 = 308
...
5 × 1049, ∆rG 7 = −306
...

(a) Mole fractions: A: 0
...
0302, C: 0
...
6742,
Total: 0
...
6; (c) K = 9
...
6 kJ mol−1
...
4 × 103 K
...
191 kJ mol−1, ∆r S 7 = −21 J K−1 mol−1
...
0 kJ mol−1
...
6 × 10−2
...

(a) At 427°C, K = 9
...
08;
(b) ∆rG 7 = −12
...

T = 397 K
...
8 kJ mol−1
...
45 V
4
L: Cl2(g) + 2e− → 2Cl−(aq)
+1
...
91 V
4
(b) R: Sn4+(aq) + 2e− → Sn2+(aq)
+ 0
...
77 V
Overall (R – L): Sn4+(aq) + 2Fe2+(aq)
→ Sn2+(aq) + 2Fe3+(aq)
−0
...
23 V
L: Cu2+(aq) + 2e− → Cu(s)
+0
...
89 V
(a) R: 2H2O(1) + 2e− → 2OH−(aq) + H2(g)
−0
...
71 V
and the cell is Na(s)|Na+(aq), OH−(aq)|H2(g)Pt
+1
...
54 V
L: 2H+(aq) + 2e− → H2(g)
0
...
54 V
(c) R: 2H+(aq) + 2e− → H2(g)
0
...
083 V
and the cell is Pt|H2(g) |H+(aq), OH−(aq)|H2(g) |Pt
0
...
89 kJ mol−1;
F
(c) E 7 = +0
...

(a) K = 1
...
2 × 10−7
...
4 × 10−20; (b) 5
...


Chapter 8
8
...
2
8
...
4

8
...
6
8
...
8

Chapter 7

1029

8
...
10
8
...
3 × 10−5 m s−1
...
89 × 10−27 kg m s−1, ν = 0
...

∆x = 5
...

(a) E = 0
...
32 × 10−15 J, E (per mole) = 7
...
99 × 10−23 J, E (per mole) = 0
...

(a) ν = 0
...
98 × 10−6 m s−1
...

(a) 3
...
52 × 1018 s−1
...
84 × 10−19 J; ν = 1
...

(a) E = 2
...
00 × 10−19 J, or
181 kJ mol−1; (c) E = 6
...
0 × 10−10 kJ mol−1
...
23 × 10−10 m; (b) λ = 3
...
88 × 10−1 m
...
8 × 105 m s−1
...
13
8
...
67 × 10−16 J
...


3
1
3
1
1
10
...

2
2
2
2
2
5 3 1
10
...
(b) S′ = – , – , – ;
2 2 2

multiplicities = 6,4,2 respectively
...

10
...

(c) J = 9 7
2 2
10
...

10
...
1
9
...
3

9
...
14 × 10−19 J, 1
...
08 × 104 cm−1, 129 kJ mol−1;
(b) 3
...
17 eV, 1
...

(a) P = 0
...
029
...

L
A 3 D 1/2 h
A 3 D 1/2
L=B E
= B E λc
...
1

L 3L L 7L 9L
, , , ,
...
2

9
...
6

6
...
26 × 1010, ∆E = 1
...
5 pm, the particle behaves
classically
...
92 × 10−21 J
...

λ = 13
...

λ = 18
...

(a) ∆E = 2
...
14 × 10−20 J
...
96α, or ±2
...

E0 = 2
...

Magnitude = 2
...
0546 ×
10−34 J s and ±2
...


9
...
8
9
...
10
9
...
12
9
...
15
9
...
1
10
...
3
10
...
5

I = 12
...

r = 11
...
53a0 /Z, r = 0
...
382a0, r = 3
...

1
N=

...
7
10
...
9
10
...
11
10
...
13
10
...
4
11
...
6

A
D 1/2
1
N=B
E
...
145S + 0
...
844A − 0
...

0
...
844S
11
...

−µe4
11
...

12π 3ε 2 $2
0
−16
11
...
39 × 10 J
...
12 (a) a2ue4 e1 , E = 7α + 7β; (b) a2ue 1g , E = 5α + 7β
...
13 (a) 19
...
44824β
...
7

Chapter 12
12
...


12
...
6

11
...

(a) 1σ 2 2σ *2 3σ 2 1π 4 2π *4; (b) 1σ 2 2σ *2 1π 4 3σ 2;
(c) 1σ 2 2σ *2 3σ 2 1π 4 2π *3
...

BrCl is likely to have a shorter bond length than BrCl−; it has a bond
order of 1, while BrCl− has a bond order of 1/2
...

2
2

A 1 D A 1 D A ZD
E × B E × (6 − 6ρ + ρ 2)2e−ρ,
P3s = 4πr 2 B E × B
C 4π F C 243 F C α 0 F
r = 0
...
19 a0 /Z and 13
...

r = 1
...

1
(a) angular momentum = 6– h = 2
...
49 × 10−34 J s, angular nodes = 1,
2
radial nodes = 0;
1
(c) angular momentum = 2– h = 1
...

1 3
9 11
–
(a) j = – , – ; (b) j = – ,–
...

(a) g = 1; (b) g = 64; (c) g = 25
...

(a) r = 110 pm, r = 20
...
4 pm
...


12
...
4
12
...
8
12
...
10
12
...
12
12
...
14
12
...

(a) CH3Cl
...

Forbidden
...

(a) C∞V; (b) D3; (c) C4V, C2V; (d) Cs
...

(a) C∞V; (b) D5h; (c) C2V; (d) D3h; (e) Oh; (f) Td
...

NO− : px and py, SO3: all d orbitals except d 2
...

(a) B3U(x-polarized), B2U(y-polarized), B1U(z-polarized); (b) A1U or
E1U
...


Chapter 13
13
...
2
13
...
73 × 10−32 J m−3 s; (b) ν = 6
...

s = 6
...

(a) 1
...
48 ps
...
4
13
...
6
13
...
8
13
...
10
13
...
12
13
...
14
13
...
16
13
...
18
13
...
20
13
...
22
13
...
24
13
...
26

(a) 160 MHz; (b) 16 MHz
...
4754 × 1011 s−1
...
307 × 10−47 kg m2; (b) R = 141 pm
...
420 × 10−46 kg m2; (b) R = 162
...

R = 116
...

RCO = 116
...
9 pm
...

R = 141
...

(a) H2O, C2V; (b) H2O2, C2; (c) NH3, C3V; (d) N2O, C∞V
...

k = 0
...

28
...

k = 245
...

(a) 0
...
561
...
3 cm−1, DCl: # = 2143
...
8 cm−1, DI:
# = 1640
...

# = 2374
...
087 × 10−3
...
235 × 104 cm−1 = 4
...

#R = 2347
...

(a) CH3CH3; (b) CH4; (c) CH3Cl
...

(a) IR active = A2 + E′, Raman active = A1 + E′;
″
(b) IR active = A1 + E, Raman active = A1 + E
...

A1g + A2g + E1u
...

22
...

14
...
9 × 105 cm2 mol−1
...
4 1
...

14
...
56 × 108 dm3 mol−1 cm−2
...
6 Rise
...
7 ε = 522 dm3 mol−1 cm−1
...
8 ε = 128 dm3 mol−1 cm−1, T = 0
...

14
...
010 cm; (b) 0
...

14
...
39 × 108 dm3 mol−1 cm−2; (b) 1
...

14
...

14
...
2

15
...
11
15
...
15
15
...
17
15
...
19

15
...
22

1031

(a) 4
...
63 × 10−5 T
...

2
...

(a) The H and F nuclei are both chemically and magnetically
equivalent
...

B1 = 9
...
25 µs
...
3 T
...
0022
...
2 mT, g = 1
...

Eight equal parts at ±1
...
435 ± 1
...
865, 330
...
735, 331
...
845, 333
...
625 and 336
...

(a) 332
...

I = 1
...

(a) 15
...
04 pm; (b) 2
...
82 × 1027
...
3
= 187
...

qHe
16
...
006
...
5 E = 7
...

16
...

16
...
997, 0
...
999 99, 0
...

n2
n3
16
...
39 × 10−11, = 1
...
368, = 0
...
779, = 0
...
503; (c) Um = 88
...
53 J K−1 mol−1; (e) Sm = 6
...

16
...
26 K
...
10 (a) 147 J K−1 mol−1; (b) 169
...

16
...
7 J K−1 mol−1
...
12 (a)
16
...
2

Chapter 17
Chapter 15
15
...
2
15
...
4
15
...
6

17
...

EmI = −2
...
35 × 10−26 J
...
3 MHz
...
88 × 10−26 J; (b) ∆E = 5
...

3
...

B/T
gI
(a) 300 MHz
(b) 750 MHz

14

N

0
...
5
244

17
...
3
17
...
5
19

31

F

5
...
49
18
...
2634
17
...
5

17
...
7
17
...
9
17
...
7
15
...
3 × 10−7; (b) 2
...
34 × 10−5
...
(b)
= 13
...
11
17
...
14

(a) O3 : 3R [experimental = 3
...
3R]
5
(c) CO2 : –R [experimental = 4
...
29
...

(a) 2; (b) 2; (c) 6; (d) 24; (e) 4
...
8479 K, T = 0
...

R
Sm = 84
...

(a) At 298 K, qR = 2
...
At 500 K, qR = 5
...
50 × 103
...
43 × 103
...
97 × 103; (b) At 100°C, qR = 1
...

(a) At 298 K, Sm = 5
...

(b) At 500 K, Sm = 16
...

R
R
V
V
Gm − Gm(0) = −20
...
110 kJ mol−1
...
65 kJ mol
...
9 J mol−1 K−1
...
25
...
14 –
...

18
...
4 D
...
3 µ = 9
...
0°
...
4 µ = 3
...
55 × 10−39 C2 m2 J−1
...
5 εr = 8
...

18
...
71 × 10−36 C m
...
7 α = 3
...

18
...
10
...
9 εr = 16
...
10 p = 5
...

18
...
12 × 10−2 N m−1
...
12 pin − pout = 2
...


20
...
1

20
...
17
20
...
20
20
...
23
20
...
25
20
...
27
20
...
29

Chapter 19
−1

−1

•n = 68 kg mol , •w = 69 kg mol
...
2 Rg = 1
...

19
...
8 kg mol−1; (b) •w = 11 kg mol−1
...
4 τ = 9
...

19
...

19
...

19
...
47 × 10− 4 m s−1
...
8 • = 56 kg mol−1
...
9 •w = 3
...

19
...
86 × 105
...
11 Rrms = 38
...

19
...
26 × 10−6 m, Rrms = 1
...

19
...
30
20
...
32

21
...
2
21
...
4
21
...
6
21
...
9

20
...

2 2
2 2

21
...
2

21
...
4

(3 1 3) and (6 4 3)
...
2 pm
...
7 pm
...
5

hkl

sin θ

θ/°

2θ/°

111

0
...
1

38
...
378

22
...
4

220

0
...
3

64
...
3

21
...
13
21
...
15
21
...
17

21
...
054 cm
...
2582 nm3
...
8 d = 5, d = 2
...

20
...

20
...

20
...
19

20
...
20
21
...
22

100

574
...
166

21
...
8

3
...
5

7
...
11 hkl

= 0
...

R
(a) 57 pm; (b) 111 pm
...
370
...

λ = 252 pm
...

strain = 5
...

∆V = 0
...

p-type
...
71 × 104 H
...

−8
...

χm = 1
...

2
...

χm = 1
...

r = 0
...


Chapter 21

21
...
23
21
...
26

20
...


21
...
13 Fhkl = 2f for h + k + l even; 0 for h + k + l odd
...
28

(a) 7
...

(a) c = 4
...
01 s−1
...
4 × 107 Pa
...
1 × 10−7 m
...
9 × 108 s−1
...
7 × 10−9 m; (b) λ = 5
...
1 × 10−5 m
...
6 × 10−2
...
3 × 1021
...
98 × 10−4 kg
...

t = 1
...

0
...

1
...

22 J s−1
...
00 × 10−19 m2
...
00 × 105 Pa
...
95 × 10−5 kg m−1 s−1;
(b) At 298 K: η = 0
...
81 × 10−5 kg m−1 s−1
...
0114 J m−1 s−1 K−1, 0
...
0 × 10−3 J m−1 s−1 K−1, 0
...

52
...

κ = 9
...

(a) D = 0
...
87 mol m−2 s−1;
(b) D = 1
...
7 × 10−5 mol m−2 s−1;
(c) D = 7
...
8 × 10−7 mol m−2 s−1
...
09 × 10−3 S m2 mol−1
...
81 × 10−5 m V−1 s−1
...
604
...
96 mS m2 mol−1
...
74 × 10−8 m2 V−1 s−1;
Cl−: u = 7
...
09 × 10−8 m2 V−1 s−1
...
09 × 10−9 m2 s−1
...
1 × 103 s
...
29 207 pm
...
8
−11

21
...


21
...
8 × 103 s
...
2 × 102 s; (b) 3
...


24
...
10
24
...
12
24
...
1
22
...
3
22
...
5
22
...
7
22
...
9

22
...
11
22
...
13
22
...
15
22
...
0 mol dm−3 s−1; B = 3
...
0 mol dm−3 s−1; D = 2
...

Rate of consumption of B = 1
...

Rate of reaction = 0
...

Rate of formation of C = 0
...

Rate of formation of D = 0
...

Rate of consumption of A = 0
...

d[A]
d[C]
K: dm3 mol−2 s−1
...

dt
dt
v = k[A][B][C]−1, K: s−1
...
00
...

t1/2 = 1
...
5 kPa; (b) p = 29
...

(a) k = 3
...
4 h; t1/2(B) = 0
...

(a) Second-order units: m3 molecule−1 s−1, Third-order units: m6
molecule−2 s−1; (b) Second-order units; Pa−1 s−1, Third-order units;
Pa−2 s−1
...
5 × 10−3 mol dm−3; (b) 0
...

1
...

3n−1 − 1
t1/3 =
[A]1−n
...
7 × 10 s , kr = 8
...

Ea = 9
...
94 dm3 mol−1 s−1
...
06; (b) k18 /k16 ≈ 0
...

ka = 9
...
9 s−1 MPa−1
...
2
23
...
4
23
...
6
23
...
8

A k1 D 1/2
= −k1[R2] − k2 B E [R2]3/2
...
(b) p = 1
...

k1k2K1/2
a
[HA]3/2[B]
...

dt
υmax = 2
...

1
...

Φ = 1
...

d[R2]

24
...
15
24
...
17

25
...
2
25
...
4
25
...
6
25
...
8
25
...
10
25
...
12

25
...
14
25
...
16
25
...
18
25
...
20
25
...
22

Chapter 24
24
...
3
24
...
5
24
...
7

z = 6
...
07 × 1034 m−3 s−1, 1
...

(a) (i) 2
...
10; (b) (i) 7
...
6 × 10−10
...
2, (ii) 1
...
4, (ii) 1
...

k = 1
...

kd = 3
...
2 × 1010 dm3 mol−1 s−1
...
97 × 106 m3 mol−1 s−1; (b) kd = 2
...

kd = 1
...
10 × 1010 dm3 mol−1 s−1,
t1/2 = 5
...


P = 1
...

v = 1
...

∆‡H = 48
...
2 J K−1 mol−1
...
8 kJ mol−1
...

∆‡S = −80
...

(a) ∆‡S = −24
...
5 kJ mol−1;
(c) ∆‡G = 34
...

k° = 1
...

2
λ = 1
...
39 × 10−24 J
...
4 × 103 s−1
...
23

24
...
24
25
...
26
25
...
30
25
...
32

(a) (i) 2
...
75 × 1013 cm−2 s−1;
(b) (i) 3
...
60 × 1013 cm−2 s−1
...
3 × 102 Pa
...
6 × 104 s−1
...
8 m2
...
7 cm3
...

Ed = 3
...

(a) 0
...
9 kPa
...
75, θ2 = 0
...

(a) At 400 K: 4
...
4 × 10−12 s;
(b) At 400 K: 1
...
4 s
...
50 kPa
...
A plot of θ
1 + (Kp)1/3
versus p at low pressures (where the denominator is approximately 1)
would show progressively weaker dependence on p for dissociation
into two or three fragments
...
40 kJ mol−1
...
85 × 105 J mol−1
...
48 × 1036 s; (b) t = 1
...

ε = 2
...

167 mV
...
6 mA cm−2
...
5 mA cm−2
...
34 A cm−2;
(b) j = 0
...
The validity of the Tafel equation improves as the
overpotential increases
...
3 A m−2
...

j = (2
...
42)E ′/f × (3
...
58)E′/f × (3
...

At r = 0
...
5 A cm−2,
At r = 1: j/j0 = 4
...

0
...

For the Cu, H2 |H+ electrode: N = 6
...
2 × 10−3 s−1
...
5 × 1014 s−1 cm−2, f = 0
...

(a) 5
...

Deposition would not occur
...

E 7 = 1
...
180 W
...
0 mm y−1
...
1
1
...
5
1
...
9

1
...
13
1
...
17
1
...
21
1
...
25
1
...
31

−233°N
...
95°C
...
0245 kPa; (b) p = 9
...
0245 kPa
...
5 dm3 mol−1; (b) Vm = 12
...

(a) 0
...
69 dm3 mol−1, 2
...
11 dm3 mol−1
(a) 0
...
6957; (c) 0
...

b = 59
...
649 dm6 atm mol−2, p = 21 atm
...
6 cm3 mol−1, a = 1
...

RT
3C
B2
B3
1
Vc =
, Tc =
, pc =
, Zc =
2
B
3RC
27C
3
B′ = 0
...
0 dm3 mol−1
...

0
...

4
...
00; (b) −0
...
5 km
...
0 × 10−3 bar
...
1

2
...
5

2
...
9
2
...
13
2
...
17
2
...
21

T1 = 273 K = T3, T2 = 546 K
Step 1 → 2: w = −2
...
40 × 103 J
q = +5
...
67 × 103 J
Step 2 → 3: w = 0
qv = ∆U = −3
...
67 × 103 J
Step 3 → 1: ∆U = ∆H = 0
−q = w = +1
...
70 × 103 J
w = 0, ∆U = +2
...
03 kJ,
(a) w = 0, ∆U = +6
...
19 kJ, ∆H = +8
...
19 kJ, ∆H = −8
...
19 kJ;
(c) ∆U = ∆H = 0, −q = w = +4
...

(a) w0 = −1
...
8 kJ; (c) w = −1
...

−87
...

∆ r H 7 = +17
...
0 kJ mol−1
...
903, k = −73
...

∆c H 7 = 25 968 kJ mol−1, ∆ f H 7(C60) = 2357 kJ mol−1
...

41
...

3
...

(a) dz = (2x − 2y + 2)dx + (4y − 2x − 4)dy
A
1D
(c) dz = B y + E dx + (x − 1)dy
xF
C

2
...
27
2
...
37

(a) −1
...
6 kJ
...

aT 2
(a) µ =
Cp
2
A
2apT D
E
(b) Cv = Cp − R B 1 +
R F
C

2
...
41
2
...
47

7
...

(a) −25 kJ; (b) 9
...

∆T = 2°C, ∆h = 1
...
8 m, ∆T = 3
...
8 m
...
5 K MPa−1; (b) 14
...


Chapter 3
3
...
3

3
...
3 J K−1 mol−1, ∆Ssur = +21
...
4 J K−1 mol−1
...
7 J K−1 mol−1, ∆Ssur = −111
...
5 J K−1 mol−1
...
9 kJ, q(H2O) = −43
...
1 J K−1,
∆S(Cu) = 145
...

(b) θ = 49
...
1 K, q(Cu) = 38
...
8 J K−1, ∆S(Cu) = 129
...

Step 1

Step 2

Step 3

Step 4

Cycle

q
+11
...
74 kJ
0
−5
...
5 kJ
−3
...
74 kJ
+3
...
8 kJ
∆U
0
−3
...
74 kJ
0
∆H
0
−6
...
23 kJ
0
∆S +19
...
1 J K−1
0
0
∆Stot
0
0
0
0
0
∆G
−11
...
5 kJ Indeterminate
0

3
...
9
3
...
13
3
...
17
3
...
31
3
...
7 J K−1 mol−1; (b) 232
...

Tf
Tf
∆S = nCp,m ln
+ nCp,m ln , ∆S = +22
...

Th
Tc
(a) 63
...
08 J K−1 mol−1
...
1 kJ mol−1
...
60 J K−1 mol−1
...

p2 ∆B
πT ≈
×
R ∆T
(a) 3
...
30 atm
...
50 kJ

SOLUTIONS TO ODD PROBLEMS

3
...
37
3
...
43

3
...
9991 atm
−21 kJ mol−1
...

A VB D
ε=1− B
E
C VA F

+
...
13

1/c

6
...
21

ε = 0
...
00 kJ; (b) 8
...


4
...
1
7
...
3
4
...
7
4
...
11
4
...
17

4
...
23

7
...
0 K, 11
...

(a) 5
...
5 per cent
...
63 cm3 mol−1 ; (b) +30
...
6 kJ mol−1
...

(a) Tb = 227
...
18 K; (c) T = 383
...
0 kJ mol−1
...
8 Torr
...

T
∞
2pr(Tr,Vr)
(a) ∆Ur(Tr,Vr) = −
dVr ; (c) 0
...
90
...
9
7
...
13
7
...
17

7
...
21

∆ vap H = 1
...


5
...
5
5
...
11

5
...
15
5
...
19
5
...
29

kA = 15
...
03 kPa
...
4 cm3 mol−1, VH20 = 18
...

VE = 57
...
8 cm3, ∆V = +0
...

4 ions
...
63 cm3 mol−1, V2 = 99
...

KH = 371 bar, γCO2 = 1
...
99(at 20p/bar), 1
...
99 (at 40p/bar), 0
...
94 (at 80p/bar)
...
6 kJ
...
36 cm mol
...
0 atm = 56 µg, pN2 at 1
...
7 × 102 µg N2
...
1 × 105 g mol−1,
(d) B′ = 21
...
0 cm3 g−1, C′ = 196 cm6 g−2
...
1
6
...
7
6
...
532
...
877 (78 K), 1
...
039 (82 K), 0
...
993 (86 K), 0
...
987 (90
...

MgCu2: 16 per cent mg by mass, Mg2Cu: 43 per cent mg by mass
...
2 at per cent Si at 1268°C; 69
...
Congruent melting compounds: Ca2Si = 1314°C; CaSi =
1324°C
...
0 K; (c)

A ∂E D
∆rV
(a) B E = −
;
∂p F T ,n
νF
C

A ∂E D
(c) the linear fit and constancy of B E are very good;
C ∂p F
7
...
25
7
...
33
7
...
2 × 10−7 atm−1
...
15 V
...

(b) +0
...

(iv) HNO3·3H2O is most stable
...
1
8
...
5
8
...
15

Chapter 6

(a) ∆rG 7 = +4
...
101 atm
...
48 R
...
161 × (T/K)
...
740, second experiment, K = 5
...

∆H 7 = +158 kJ mol−1
...
2 × 108; (b) At 700 K: 2
...

(a) CuSO4: 4
...
2 × 10−2; (b) γ± (CuSO4) = 0
...
60; (c) Q = 5
...
102 V; (e) E = +1
...

pH = 2
...

E 7 = +0
...
9659 (1
...
9509 (3
...
9367 (5
...
9232 (7
...
9094 (10
...
533
...
80 × 10−3 mV atm−1;

Chapter 5
5
...
Relative amounts, nCa /nliq = 2
...
53 at slightly above 1030°C, (ii) nSi/nCaSi2 = 0
...

(i) Below a denaturant concentration of 0
...

(a) 2150°C (b) y(MgO) = 0
...
35, (c) c = 2640°C
...
85
...
3

Chapter 4

1035

8
...
6 × 1033 J m−3; (b) ∆E = 2
...

CV
CV
(a) θ E = 2231 K,
= 0
...
897
...
0 × 10−6; (b) 1
...

A 2 D 1/2
1
1
1
(a) N = B E ; (b) N =
; (c) N =
; (d) N =

...

(a) Yes, eigenvalue = −k2;
(b) Yes, eigenvalue = −k2;
(c) Yes, eigenvalue = 0;
(d) Yes, eigenvalue = 0;
(e) No
...

dx
dx

1036
8
...
21
8
...
29

SOLUTIONS TO ODD PROBLEMS
$2k2

10
...
4 cm−1,


...

0
0
(a) λrelativistic = 5
...

(a) Methane is unstable above 825 K; (b) λ max (1000 K) = 2880 nm;
(c) Excitance ratio = 7
...
8 × 10−3;
(d) 2
...


He = 60 954
...
The wavenumbers for n = 2 → n = 1: 4He =
329 170 cm−1, 3He = 329 155 cm−1
...
27 (a) receeding; s = 3
...

3

Chapter 11
R = 2
...

(a) P = 8
...
0 × 10− 6;
(b) P = 8
...
0 × 10− 6;
(c) P = 3
...
9 × 10−7/ P = 5
...

11
...
15 (a) C2H4: −3
...
623, C6H8: −5
...
873;
(b) 8
...

11
...
122 V; (c) E 7 = −0
...

11
...
7

Chapter 9
9
...
24 × 10−39 J, n = 2
...
8 × 10−30 J
...
30 × 10−22 J, minimum angular momentum = ±η
...
5

(1)
(a) E 1 =

9
...
13

͗T͘ =

9
...
1

εa
L

N2
2κ

+

ε
π

A πa D
A πD
ε ε
sin B E ; (b) E (1) = + sin B E = 0
...

1
10 π
C LF
C 10 F

; (b) ͗x͘ =

N2


...

2C
2F

Chapter 12
12
...
19
9
...
31
9
...
35
9
...


ml

, 54°44′
...
3 × 10−19 J; (b) v = 4
...

ω = 2
...

(a) lz = 5
...
39 × 10−24 J; (b) v = 9
...

F = 5
...


Chapter 10
n2 → 6, transitions occur at 12 372 nm, 7503 nm, 5908 nm, 5129 nm,

...

10
...
5 eV
...
5 2P1/2 and 2P3/2, of which the former has the lower energy, 2P3/2 and
2
P5/2 of which the former has the lower energy, the ground state will
be 2P3/2
...
7 3
...
000 272
...
9 (a) ∆# = 0
...

10
...

10
...

2
Z
Z
Z
10
...

a0
4a0
4a0
10
...
3

−
iσz

1

2

(a) E = 0, angular momentum = 0; (b) E =

θ = arccos


...

C2σ h = i
...
7

1
͗T͘ = − –͗V͘
...
25

1/2

The matrices do not form a group since the products iσz, iσy, iσx and
their negatives are not among the four given matrices
...
9 All five d orbitals may contribute to bonding
...

12
...

12
...

Lanthanide ion (a) tetrahedral complex: f → A1 + T1 + T2 in Td
symmetry, and there is one nondegenerate orbital and two sets of
triply degenerate orbitals
...

12
...
1

T/K
(a) 1500
(b) 2500
(c) 5800

E/J m−3

Eclass/J m−3

2
...
884 × 10−4
3
...
206
3
...
528

SOLUTIONS TO ODD PROBLEMS
1

1/2
kT A πm D
E , δν ≈ 700 MHz, below 1 Torr
...
3

τ=

13
...
83 pm, R1 = 123
...

I = 2
...
5 pm, hence we expect lines at 10
...
11, 31
...
cm−1
...

B = 14
...

linear
...
15 eV; (b) 5
...

(a) # = 152 m−1, k = 2
...
93 × 10−46 kg m2,
B = 95
...

(b) xe = 0
...

(a) C3v; (b) 9; (c) 2A1 + A2 + 3E
...
(e) All but the A2 mode may be Raman active
...

HgCl2: 230, HgBr2: 240, HgI2: 250 pm
...
5, O2 : 1;
(d) Fe3+O2−; (e) Structures 6 and 7 are consistent with this
2
2
observation, but structures 4 and 5 are not
...
0768 c, T = 8
...

B = 2
...
35 K
...
7
13
...
11
13
...
15
13
...
19
13
...
25
13
...
29
13
...
7

49 364 cm−1
...

A = 1
...

5
...


14
...
7506
−8
...
8352
−8
...
2489
−8
...
1
14
...
5

n
= 1
...
0 × 102
...

(a) 3 + 1, 3 + 3; (b) 4 + 4, 2 + 2
...
4 × 103
...
24 × 105 dm3 mol−1 dm−2
...
24 × 105 dm3 mol−1 cm−2; (b) A = 0
...

V1 − V0 = 3
...
538 cm−1, #0 = 2034
...


14
...
15
14
...
23
14
...
27
14
...
3

15
...
13
15
...
17

15
...

Width of the CH3 spectrum is 3aH = 6
...
The width of the CD3
spectrum is 6aD
...
1 mT
...
10, P(N2pz) = 0
...
48, P(O) =
0
...
8, Φ = 131°
...
78 × 10−5 Z
...

1
Aτ
I(ω) ≈

...
58 mT
...
282 × 10−21 J K−1, S1 = 0
...
645 × 10−21 J K−1
...
3
≈ 2
...

W
16
...
5 × 10−15 K, q = 7
...

16
...
00; (ii) q = 6
...
00 at 298 K, p0 = 0
...
5 × 10−11 at 298 K,
p2 = 0
...

(c) (i) Sm = 13
...
07 J K−1 mol−1
...
9 (a) p0 = 0
...
36; (b) 0
...
At 300 K,
Sm = 11
...
4 J K−1 mol−1
...
11 (a) At 100 K: q = 1
...
953, p1 = 0
...
002;
Um − Um(0) = 123 J mol−1, Sm = 1
...

(b) At 298 K: q = 1
...
645, p1 = 0
...
083,
Um − Um(0) = 1348 J mol−1, Sm = 8
...

16
...

16
...

16
...
351; (b) 0
...
029
...
2 J K−1 mol−1, Sm = 15 J K−1 mol−1
...
5 q = 19
...

7
17
...
4 J mol−1 K−1
...
11 At 298 K: K = 3
...
At 800 K: K = 2
...

17
...
55 K, θ V = 6330 K
...
1
17
...

2
C 2hcB F

17
...
17 (a) q R = 660
...
26 × 104
...
23 S = 9
...


7
7
17
...
5 kJ mol−1
...
27 At 10 K, G m − G m(0) = 660
...


7
7
At 1000 K, G m − G m(0) = 241
...


18
...
42 × 10−5, β state lies lower
...
29 s
...
3 T,

15
...
1

15
...
3
18
...
7

(a) ε = 1
...

α ′ = 1
...
86 D
...
24 × 10−24 cm3, µ = 1
...
66 cm3 mol−1, µ = 1
...

m
(a) ε = 1
...


1038
18
...
14 cm3 mol−1, εr = 1
...
33
...
19 (a) V = −39 J mol−1, (b) The force approaches zero as the distance

becomes very large
...
21 (a) µ = 1
...
55 × 10−23 J
...
3
21
...
7
21
...
1
19
...
5
19
...
9

21
...
97 × 10−13 s or 5
...

[η] = 0
...

M = 158 kg mol−1
...
0117 cm3 g−1 and a = 0
...

•n = 155 kg mol−1, B = 13
...


A 2γ D 1/2
19
...

C πF

21
...
15 (a) Rrms = lN1/2, Rrms = 9
...
17
21
...
13

l, Rmean = 8
...
21
21
...
95 nm;

21
...
When M = 100 kg mol−1,
C 5F
2 3

Irod / Icc
0
...
876
(b) 90°
...
27 •n = 69 kg mol−1, a = 3
...

19
...
85
...
514

Chapter 20
20
...
3
20
...
7
20
...
11
20
...
17
20
...
21
20
...
25
20
...

Yes, the data support
...
55 pm, ρ = 10
...

a(KCl) = 628 pm, are broadly conistent
...
654 g cm−3
...
01 g cm−3
...
385 g cm−3, ρos = 1
...

0
...

For S = 2, χm = 0
...
254 ×
10−6 m3 mol−1, For S = 4, χm = 0
...

x = 0
...

For a monoclinic cell, V = abc sin β
...

Fhkl ∝ 1 + e5iπ + e6iπ + e7iπ = 1 − 1 + 1 −1 = 0
...

2me
2me

f(nc*)
f(c*)
f(3c*)

19
...
40
...
50 nm, Rg = 46 nm
...
23 vp = 8vmol
For BSV, B = 28 m3 mol−1
...
33 m3 mol−1
...
6 × 10−2 corresponding to 2
...

Π°
Π − Π°
For Hb,
= 5
...

Π°
19
...
89 m; (b) ͗h2͘ = 1
...

p = 7
...
3 mPa
...

Λm = 12
...
30 mS m2 mol−1 M−1/2
...
96 mS m2 mol−1; (b) κ = 119
...
5 Ω
...
0 × 10−3 cm s−1, s(Na+) = 5
...
6 × 10−3 cm s−1
...

(a) d(Li+) = 1
...
7 × 10−6 cm; d(K+) = 2
...

(b) 43, 55 and 81 solvent molecule diameters respectively
...
48 and t− = 0
...
u+ = 7
...
λ+ = 72 S cm2 mol−
1

...
1 × 10−20 N molecule−1; (b) 2
...
1 × 10−20 N molecule−1
...
3 kJ mol−1
...
2 × 10−3 kg m−1 s−1
...
68 × 10−10 m; (b) 3
...

͗vx͘ = 0
...


f(c*)

(nc*)2 e−mn c* /2kT
2

=

2

c*2e−mc* /2kT
2

= 3
...
4] = n2e−(n −1)mc* /2kT = n2e(1−n ),

f(4c*)
f(c*)

2

2

2

= 4
...


21
...
016; (c) p = 0
...

21
...
25 J cm−3
...
39 t = 108 s
...
1
22
...
5
22
...
9
22
...
13
22
...
17
22
...
0594 dm3 mol−1 min−1, m = 2
...

First-order, k = 1
...

9
...

k = 3
...

k = 2
...
98 × 10−3 s
...

rate = kK1K2[HCl]3[CH3CH=CH2]; look for evidence of proposed
intermediates, e
...
using infrared spectroscopy to search for (HCl)2
...

There are marked deviations at low pressures, indicating that the
Lindemann theory is deficient in that region
...


22
...
27 vmax = k B
22
...


22
...
00765 min−1 = 0
...
5 h = 91 min
...
35 v = k[A][B], k =

k1k2


...
37 Ea = 13
...
03 × 109 dm3 mol−1 s−1
...
39 k1 = 3
...
1 × 105 dm3 mol−1 s−1,
k2
k3 = 4
...
13
...
3

24
...
0
...

d[SiH4]
dt

A k1k4k5 D 1/2
=B
E [N2O][SiH4]1/2
C k6 F

2kbka[I2][H2]
=

...
This is equivalent to kb[H2] = ka
′
d[HI]
and hence,
= 2kbK[I2][H2]
...
7 (a) τ0 = 6
...
105 ns−1
...
9 kq = 1
...

p1/2M
23
...

1−p
23
...

23
...
19

d[HI]

24
...
4 nm−1
...
1

25
...

dt
C k3 F

d[A]

k2k4[CO]

...
27 Uncompetitive
...
29 R = 2
...

23
...
39 kJ mol−1;
23
...
5
25
...
9
25
...
13
25
...
21
25
...
33

A ka D 1/2
E [NO]2, where ke is the rat constant
= −2kb B
C 2kd[M] F

for NO + O2 → O + NO2
...

25
...
5

(a) σ * = 4
...
15
...
7 × 1011 M−1 s−1, t = 3
...

2−
...
9

(a) −

24
...
3

d[F2O]
dt

A k1 D 1/2
= k1[F2O]2 + k2 B E [F2O]3/2; (b) ∆H(FO-F) ≈
C k4 F

E1 = 160
...
4 kJ mol−1, E2 ≈ 60 kJ mol−1
...
11 Linear regression analysis of ln(rate constant) against 1/T yields the

following results: R = 0
...
998 48, which indicate that
the data are a good fit
...
15 P = 5
...

v3
24
...
7 × 10−15 m2 s−1, (b) D = 1
...

v2
24
...
6 × 10−3; For O2 with cyclohexyl: P = 1
...


For a cation above a flat surface, the energy is 0
...
For a cation at the
foot of a high cliff, the energy is −0
...
The latter is the more likely
settling point
...
61 × 1015 cm−2; (b) 1
...
86 × 1015 cm−2
...
37

25
...
8 × 105
9
...
9 × 105

8
...
2 × 10−1
7
...
4 × 105
2
...
2 × 105

1
...
7 × 10−2
1
...
1 cm3; (b) c = 264, Vmon12
...

c2 = 2
...
16
...
138 mg g−1, n = 0
...

n∞ = 5
...
02 Pa−1
...
78, α = 0
...

δ = 2
...
25 mm
...

BET isotherm is a much better representation of the data
...
4 cm3, c = 3
...

(a) R values in the range 0
...
991, the fit is good at all
temperatures
...
68 × 10−3, ∆adH = −8
...
62 × 10−5 ppm−1,
∆bH = −15
...

(c) ka may be interpreted to be the maximum adsorption capacity at
an adsorption enthalpy of zero, while kb is the maximum affinity
in the case for which the adsorbant–surface bonding enthalpy is
zero
...

(b) R (Linear) = 0
...
9682, R (Langmuir) =
0
...
The Langmuir isotherm can be eliminated as it gives
a negative value for KL : the fit to the Freudlich isotherm has a
large standard deviation
...

(c) qrubber /qcharcoal = 0
...
46, hence much worse
...

(b) Ni: corrodes, Cd: corrodes, Mg: corrodes, Ti: corrodes, Mn:
corrodes
...
28 mg cm−2d−1
...


A
A2 spectrum 531
ab initio method 394
absolute value 963
absorbance 432
absorption spectroscopy 431
abundant-spin species 541
acceleration 981
acceleration of free fall 979
acceptable wavefunction 272
accommodation 917
acetaldehyde pyrolysis 830
achiral molecule 412
acid 763
acid catalysis 839
acidity constant 763, (T) 1007
acronyms 950
activated complex 809, 881
activated complex theory 880
activation, enthalpy of 51
activation energy 807, (T) 1023
composite reaction 822
negative 822
activation Gibbs energy (electrode)
935
activation-controlled reaction 877
active site 840
active transport 770
activity 158, 204
ion 163
activity coefficient 159
determination 228
additional work 34, 99
adiabat 48
adiabatic bomb calorimeter 38
adiabatic boundary 6
adiabatic demagnetization 568
adiabatic expansion 47, 48, 69
adiabatic flame calorimeter 42
adiabatic nuclear demagnetization
568
adiabatic process, entropy change
80
ADP 224, 225
adsorbate 909
adsorbent 909
adsorption 909
adsorption isotherm 917
adsorption rate 923
aerobic metabolism 225
aerosol 682
AES 914
AFM 289, 637
air, composition 11

Airy radius 466
all-trans-retinal 490
allowed transition 335, 435
alloy 175
microstructure 191
alpha-helix 677
alveoli 147
amount of substance 959
amplitude 982
anaerobic metabolism 225
angstrom 961
angular momentum 297, 981
commutator 307
magnitude 305
operator 307
orbital 326
quantization 298
summary of properties 309
total 349, 352
z-component 306
angular velocity 981
anharmonic 455
anharmonicity constant 455
anode 217
anodic current density 935
anti-parallel beta-sheet 678
anti-Stokes radiation 431
anti-bonding orbital 371
anticyclone 12
antiferromagnetic phase 736
antioxidant 386
antisymmetric stretch 461
antisymmetric wavefunction 338
apomyoglobin 819
aragonite 43
area 966
argon viscosity 760
argon-ion laser 507
aromatic stability 392
array detector 470
Arrhenius equation 807
Arrhenius parameters 807, 873, (T)
1021
ascorbic acid 386
asymmetric rotor 442
asymmetry potential 230
asymptotic solution 323
atmosphere 11, 462
temperature 463
temperature profile 854
atmosphere (unit) 4, 961
atmosphere composition 853
atmospheric ozone 855
atom 320
configuration 337
many-electron 336
selection rule 335
term symbol 352

atomic force microscopy 289, 637
atomic level 349
atomic orbital 326
atomic weight 959
atomization, enthalpy of 51
ATP 224, 858
attractive surface 890
Aufbau principle, see building-up
principle 340
Auger electron spectroscopy 914
autocatalysis 803
autoprotolysis rate 806
avalanche photodiode 473
average molar mass 653
average value 528, 966
Avogadro’s principle 7
AX energy levels 524
AX2 spectrum 526
AX3 spectrum 526
Axilrod–Teller formula 636
axis of improper rotation 406
axis of symmetry 405
azeotrope 184
azimuth 301

B
Balmer series 320
band formation 724
band gap 725
band head 487
band spectra 457
band width 725
bar 4, 961
barometer 4
barometric formula 12
barrier penetration 286
barrier transmission 287
base 763
base catalysis 840
base pairs 680
base-stacking 681
base unit 960
basis set 380
Bayard–Alpert pressure gauge 5
Beer–Lambert law 432
bends, the 147
benzene, MO description 391
Berthelot equation of state 19
BET isotherm 920
beta-barrel 679
beta-pleated sheet 678
beta-sheet 678
bilayer 686
bimolecular reaction 810
binomial coefficient 973
binomial expansion 571, 973
biochemical cascade 491

biofuel cell 948
biological standard state 161, 209
biosensor analysis 925
bipolaron 674
Birge–Sponer plot 456
bivariant 176
black body 245
black-body radiation 245
block-diagonal matrix 414
Blodgett, K
...
81
bond 362
bond dissociation energy 377
bond dissociation enthalpy 377, (T)
1011
bond enthalpy 55
bond length (T) 1011
determination 448
bond order 376
bond order correlations 377
bond torsion 676
bonding orbital 370
Born equation 102, 110
Born interpretation 256, 272
Born, M
...
and L
...
920
bubble 642
buckminsterfullerene 410
building-up principle 340
bulk modulus 722
bumping 645
butadiene, MO description 389

C
caesium-chloride structure 717
cage effect 876
calamitic liquid crystal 189
calcite 43
calorie 31, 979
calorimeter 38
calorimeter constant 38
calorimetry 38
camphor 628
candela 960
canonical distribution 578
canonical ensemble 577
canonical partition function 578
capacitance manometer 5
capillary action 643
capillary electrophoresis 664
capillary technique 778
carbon dioxide
isotherm 15
phase diagram 120
supercritical 119
vibrations 461
carbon dioxide laser 507
carbon monoxide, residual entropy
610
carbon nanotube 728
carbonyl group 489
Carnot cycle 82
carotene 281, 398, 857
CARS 465
casein 682
catalyst 839
catalyst properties 930
catalytic constant 842
catalytic efficiency 842
catalytic hydrogenation 930
catalytic oxidation 930
cathode 217
cathodic current density 935
cathodic protection 950
cavity 642
CCD 473
ccp 716
CD spectra 491
cell, electrochemical 216
cell emf 219

cell notation 218
cell overpotential 944
cell potential 219
cell reaction 218
Celsius scale 6
centre of symmetry 406
centrifugal distortion 446
centrifugal distortion constant 446
centrifugal effect 323
ceramic 736
cesium, see caesium 717
CFC 855
chain carrier 830
chain polymerization 835, 836
chain reaction 830
chain relation 68, 968
chain rule 966
chain transfer 837
chain-branching explosion 833
channel former 770
Chapman model 855
character 413
character table 413, (T) 1023
characteristic rotational temperature
594
characteristic vibrational
temperature 597
charge-coupled device 473
charge transfer rate 934
charge-transfer transition 489
Charles’s law 7
chemical equilibrium 208
Boltzmann distribution 208, 212
chemical exchange 532, 533
chemical kinetics 791
chemical potential 122
chemical equilibrium 201
general definition 138
significance 139
standard 141
variation with pressure 123
variation with temperature 123
chemical potential (band theory)
726
chemical potential gradient 772
chemical quench flow method 794
chemical shift 519
electronegativity 522
typical 520
chemiluminescence 886
chemiosmotic theory 227
chemisorption 917
chemisorption ability 929
chiral molecule 412, 491
chlorofluorocarbon 855
chlorophyll 856, 857
chloroplast 254, 856
cholesteric phase 189
cholesterol 687
chorine atom ozone decomposition
855
CHP system 947
chromatic aberration 490

chromatography 119
chromophore 487, (T) 1014
chromosphere 346
chronopotentiometry 940
circular dichroism 491
circular polarization 491
circularly birefringent 985
circularly polarized 984
circumstellar space 480
cis-retinal 490, 853
citric acid cycle 225, 856
Clapeyron equation 126
class 415, 416
classical mechanics 243
clathrate 635
Clausius inequality 86, 95
Clausius–Clapeyron equation 128
Clausius–Mossotti equation 627
Clebsch–Gordan series 352
close-packed 715
closed shell 339
closed system 28
cloud colour 658
cloud formation 645
CMC 685
CNDO 394
co-adsorption 926
coagulation 684
COBE 438
coefficient of performance 85
coefficient of thermal conductivity
758
coefficient of viscosity 758, 759, 785
cofactor (matrix) 976
coherence length 497
coherent anti-Stokes Raman
spectroscopy 465
coherent radiation 497
colatitude 301
cold denaturation 819
collapse pressure 688
colligative property 150
collision 9, 753
elastic 748
reactive 886
collision cross-section 753, 870, (T)
1018
collision density 870
collision diameter 753
collision flux 755
collision frequency 753, 755
collision theory 809, 870
collision-induced emission 846
collisional deactivation 438, 846
collisional lifetime 438
colloid 682
colloid stability 683
colour 481, (T) 1013
column vector 976
combination difference 458
combination principle 321
combinatorial function 973
combined gas law 10

1041

combined heat and power system
947
combustion, enthalpy of 51
common logarithm 963
commutator 271
angular momentum 307
commute 271
competitive inhibition 844
complementary observable 271
complete neglect of differential
overlap 394
complete set 267
complete shell 339
complex conjugate 256, 963
complex mode process 891
complex number 256, 963
component 175
compound semiconductor 726
compressibility 722
compression factor 14, 111
Compton wavelength 316
computational chemistry 393
concentration cell 218
concentration polarization 941
concentration profile 878
condensation 17, 645
conductance 761
conducting polymer 674
conductivity 762
thermal 758
configuration
atom 337
macromolecule 667
statistical 561
configuration integral 604
confocal microscopy 504
conformation 667
conformational conversion 532
conformational energy 675
conformational entropy 671
congruent melting 192
conjugated polyene 401
consecutive reactions 811
consolute temperature 186
constant
acidity 763
anharmonicity 455
boiling point 151
calorimeter 38
centrifugal distortion 446
critical 16
dielectric 110
equilibrium 203
Faraday’s 985
force 452, 982
freezing point 153
gas 8
Lamé 743
Madelung 719
Michaelis 841
normalization 255
Planck’s 246
rotational 443

1042

INDEX

Rydberg 320, 327
scalar coupling 524
second radiation 275
spin–orbit coupling 350
time 801
constituent 175
constrained chain 672
constructive interference 370
contact angle 644
continuum generation 503
contour diagram (reaction) 887
contour length 670
convection 12, 777
convective flux 777
cooling 86
cooling curve 178, 191
Cooper pair 737
cooperative transition 572
coordination 717
coordination number 716
core hamiltonian 393
Corey–Pauling rules 675
corona 346
correlation analysis 884
correlation diagram 355
correlation spectroscopy 543
corresponding states 21
corrosion 948
corrosion current 949
Cosmic Background Explorer 438
cosmic ray 244, 984
COSY 543
Coulomb integral 380
Coulomb interaction 986
Coulomb operator 393
Coulomb potential 110, 986
shielded 167
Coulomb potential energy 986
counter electrode 939
covalent bond 362
covalent network solid 720
cracking 931
cream 682
critical compression factor 21
critical constant 16, 17
critical field 737
critical isotherm 15
critical micelle concentration 685
critical molar volume 17
critical point 17
critical pressure 17, 118
critical solution temperature 186
critical temperature 17, 118
cross-peaks 546
cross-product 965
cross-relation 900, 903
cross-section 753, 870
state-to-state 887
crossed-beam technique 648
crossed molecular beams 886
cryogenics 568
crystal diode 473
crystal structure 697, 715

crystal system 698
crystallinity 673
crystallographic point group 408
crystallography 711
cubic close packed 716
cubic F 716
cubic group 410
cubic unit cell 698
cumulative reaction probability 891
Curie law 734
Curie temperature 736
current 987
current density 909
curvature 264
curved surface 643
CW spectrometer 520
cyclic boundary condition 301
cyclic voltammetry 943
cyclone 12
cytochrome 228
cytosol 771

D
d block 342
D lines 351
d orbital 334
d orbital hybridization 367
d–d transition 484, 487
d-metal complex 488
Dalton’s law 13, 179
Daniell cell 218
dark current 473
Davisson, C
...
252
Debye equation 627
Debye extrapolation 91
Debye formula 248
Debye length 168
Debye T 3 law 91
Debye temperature 248
Debye, P
...
683
deshielded nucleus 519
desorption 909
desorption rate 923
destructive interference 372
detection period 543
detector 471, 473
determinant 382, 975
deuterium lamp 470
DFT 395
diagonal peaks 546
dialysis 155, 712
diamagnetic 377, 734
diamagnetic contribution 521
diamond structure 720
diamond-anvil cell 178
diathermic boundary 5
diatomic molecule (T) 1013
diatomic molecule spectra 482
dielectric 619
dielectric constant 110, (T) 1004
Dieterici equation of state 19
differential 968
differential equation 811, 971
differential overlap 394
differential pulse voltammetry 943
differential scanning calorimeter 46
differential scattering cross-section
640
differentiation 966
diffraction 702
diffraction grating 471
diffraction limit 466
diffraction order 471
diffraction pattern 702
diffractometer 703
diffuse double layer 933
diffusion 747, 757, 772, 776
reaction 877, 879
relation to curvature 777
relation to mobility 774
diffusion coefficient 758, 759, 784,
(T) 1016
viscosity 775
diffusion equation 877
diffusion-controlled limit 877
dihelium 373
dilute-spin species 541
diode laser 732
dioxygen, electronic states 483

dipole 620
dipole moment 620, (T) 1015
induced 624
measurement 446
dipole–charge interaction 629
dipole–dipole interaction 631, 646
dipole–dipole interaction (EPR) 553
dipole–induced dipole interaction
633
Dirac bracket notation 313
direct method 710
direct mode process 891
direct product decomposition 420
discotic liquid crystal 189
dismutation 385
disorder 81
disperse phase 682
dispersing element 432, 471
dispersion 81, 175, 985
dispersion interaction 633, 677
disproportionation 837
dissociation 495
degree of 207, 211
dissociation energy 363
determination 456
dissociation equilibrium 612
dissociation limit 495
distillation 182
partially miscible liquids 187
distinguishable molecules 580
distortion polarization 626
distribution of speeds 749
disulfide bond 681
DNA 652, 680
analysis 664
damage 855
intercalation 638
structure from X-rays 711
Dogonadze, R
...
896
donor–acceptor pair 852
dopant 726
doping 191
Doppler broadening 436
Doppler effect 361, 436
dot product 350, 514, 965
drift speed 661, 765, 774
droplet 642
drug design 638
dry air 11
DSC 42
duality 253
Dulong and Petit law 247
Dulong, P
...
247
dust grain 438
DVLO theory 684
dye laser 508
dynamic light scattering 660
dynode 473

E
Eadie–Hofstee plot 867
Earth surface temperature 463

INDEX
eddy 11
edible fat 930
EELS 913
effect 769
cage 876
centrifugal 323
Doppler 361, 436
electrophoretic 769
Joule–Thomson 64
kinetic isotope 816
kinetic salt 885
Meissner 737
photoelectric 250
relaxation 769
salting-in 173
salting-out 173
Stark 446
effective mass 453
effective nuclear charge 339, (T)
1009
effective potential energy 323
effective transverse relaxation time
538
efficiency 83, 585
effusion 747
Ehrenfest classification 129
Ehrenfest equations 134
eigenfunction 262
eigenvalue 262
eigenvalue equation 261, 977
eigenvector 977
Einstein, A
...
920
emulsification 683
emulsion 682
end separation (polymer) 669

endergonic 202
endothermic process 29
energy 29, 979
conformational 675
electron in magnetic field 514
harmonic oscillator 291
multipole interaction 630
nucleus in magnetic field 515
particle in a box 280
quantization 246, 260
rotational 443
zero-point 281
energy density 755
energy dispersal 77
energy flux 757
energy pooling 846
ensemble 577
enthalpy 40
electron gain 343
ionization 343
partition function 590
reaction 212
variation with temperature 46
enthalpy and entropy, relation
between 44
enthalpy density 53
enthalpy of activation 51, 883
enthalpy of atomization 51
enthalpy of chemisorption 917, (T)
1022
enthalpy of combustion 51, 52
enthalpy of electron gain 51, 343
enthalpy of formation 51
enthalpy of fusion 50, (T) 993
enthalpy of hydration 51
enthalpy of ionization 51, 343
enthalpy of mixing 51, 143
enthalpy of physisorption 917, (T)
1022
enthalpy of reaction 51
enthalpy of solution 51
enthalpy of sublimation 51
enthalpy of transition 50
notation 51
enthalpy of vaporization 49, 50, (T)
993
entropy
Boltzmann formula 575
conformational 671
excess 149
from Q 579
harmonic oscillator 576
internal energy 589
measurement 91
partial molar 94
partition function 589
reaction 93
residual 93, 609
statistical 575
statistical definition 80
thermodynamic definition 78
Third-Law 93, 575
two-level system 576

1043

units 79
variation with temperature 89
entropy change
adiabatic process 80
heating 89
perfect gas expansion 79
phase transition 87
surroundings 79
entropy determination 91
entropy of activation 883
entropy of mixing 143
entropy of transition (T) 1002
entropy of vaporization 88, (T) 1003
enzyme 839, 840
epifluorescence 504
EPR 516, 549
EPR spectrometer 549
equation
Arrhenius 807
Born 102, 110
Born–Mayer 719
Clapeyron 126
Clausius–Clapeyron 128
Clausius–Mossotti 627
Debye 627
differential 971
diffusion 776
eigenvalue 261, 977
Einstein–Smoluchowski 782
Eyring 882
fundamental 139
generalized diffusion 777
Gibbs–Duhem 140
Gibbs–Helmholtz 105
Hartree–Fock 393
Karplus 528
Margules 162
Mark–Kuhn–Houwink–Sakurada
666
material balance 879
McConnell 552
Michaelis–Menten 841
Nernst 221
Nernst–Einstein 775
partial differential 973
Poisson’s 168, 986
Roothaan 393
Sackur–Tetrode 580
secular 380, 977
Stern–Volmer 849
Stokes–Einstein 775, 878
Thomson 127
transcendental 186
van der Waals 17
van ’t Hoff 156, 212, 919
virial 16
Wierl 742
equation of state 3
partition function 604
thermodynamic 104
equilibrium 35
approach to 804
Boltzmann distribution 208, 212

1044

INDEX

chemical 201
effect of compression 211
effect of temperature 211
mechanical 4
response to pressure 210
sedimentation 662
thermal 6
thermodynamic criterion 122
equilibrium bond length 363
equilibrium constant 203
contributions to 613
electrochemical prediction 228
molecular interpretation 208
partition function 611
relation between 208
relation to rate constant 804
standard cell emf 221
standard Gibbs energy of reaction
206
thermodynamic 205
equilibrium table 207
equipartition theorem 31, 247, 600
equivalent nuclei 526, 530
ER mechanism 928
error function 297
error function (T) 1009
ESCA 913
escape depth 912
ESR 516, 549
essential symmetry 699
ethanal pyrolysis 830
ethanol 396
ethene, MO description 387
Euler chain relation 68, 968
eutectic 191
eutectic halt 192
evanescent wave 925
evolution period 543
exact differential 58, 968
criterion for 103
excess entropy 149
excess function 149
exchange–correlation energy 395
exchange–correlation potential 395
exchange current density 937, (T)
1023
exchange operator 393
exchange process 532
excimer formation 846
exciplex laser 507
excited state absorption 846
excited state decay 848
exciton 304, 729
exciton band 730
excluded volume 18
exclusion principle 337
exclusion rule 464
exercise 53
exergonic 202
exocytosis 687
exothermic process 29
exp-6 potential 637
expansion coefficient 62, (T) 1002

expansion work 33
expectation value 267, 974
explosion 833
exponential decay 799
exponential function 963
extended Debye–Hückel law 165
extensive property 31, 959
extent of reaction 201, 794
extinction coefficient 432
extra work, see additional work 34
extrinsic semiconductor 726
eye 490
Eyring equation 882

F
f block 342
face-centred cubic 716
face-centred unit cell 699
factorial 967
far infrared 244
far-field confocal microscopy 504
fat 53
fcc 716
FEMO 401
femtochemistry 892
femtosecond spectroscopy 893
Fermi calculation 790
Fermi contact interaction 528
Fermi level 725
fermion 309, 338
Fermi–Dirac distribution 726
ferrocene 411
ferromagnetism 736
fibre 673
Fick’s first law of diffusion 757,
773
Fick’s second law of diffusion 776
FID 535, 554
field 244
electric 244, 983
electromagnetic 243, 983
magnetic 244, 983
field-ionization microscopy 924
FIM 924
fine structure
atomic 351
fine structure (NMR) 524
finite barrier 288
first ionization energy 342
First Law of thermodynamics 32
first-order correction 310
first-order differential equation
811
first-order phase transition 129
first-order reaction 796, 798
first-order spectra 532
flash desorption 916
flash photolysis 793
flocculation 684
flow method 793
fluctuations 578
fluid mosaic model 687

fluorescence 492, 846
laser-induced 886
solvent effect 493
fluorescence lifetime 848
fluorescence microscopy 494
fluorescence quantum yield 848
fluorescence resonance energy
transfer 852
flux 757
toward electrode 941
Fock operator 393
Fock, V
...
852
four-centred integral 395
four-circle diffractometer 704
four-helix bundle 678
Fourier transform 554
Fourier transform technique 432,
471
Fourier-transform NMR 533
fractional coverage 916
fractional distillation 183
fracture 723
framework representation 931
Franck–Condon factor 486
Franck–Condon principle 484, 493
Franklin, R
...
702
frontier orbital 388
FT-NMR 533
fuel, thermochemical properties 53
fuel cell 947

fuel-rich regime 834
fugacity 111
fugacity coefficient 111
full rotation group 411
functional 395, 969
functional derivative 395, 969
functional MRI 541
fundamental equation 103, 139
fundamental transition 455
fusion, enthalpy of 50

G
g subscript 372
g-value 514, 550
Galileo 4
Galvani potential 952
Galvani potential difference 932,
934
galvanic cell 216
working 945
galvanizing 949
gamma-ray region 244, 984
gas 3
kinetic model 9
gas constant 8
gas laser 506
gas laws 7
gas mixture 12
gas solubility 147
gas solvation 124
gas-sensing electrode 230
gauss 514
Gaussian function 292
Gaussian-type orbital 395
gel 682
gel electrophoresis 664
general solution 971
generalized diffusion equation 777
generalized displacement 34
generalized force 34
genomics 664
gerade symmetry 372
Gerlach, W
...
248
GFP 494
Gibbs energy 96
formation 100
maximum non-expansion work
99
mixing 142
mixing (partial miscibility) 186
partial molar 138
partition function 591
perfect gas 107
properties 105
reaction 100, 201
solvation 110
standard reaction 202, 220
surface 688
variation with pressure 106
variation with temperature 105
Gibbs energy of activation 883

INDEX
Gibbs energy of activation (electron
transfer) 903
Gibbs energy of formation 204
Gibbs energy of mixing, ideal
solution 148
Gibbs energy of reaction 100
Gibbs isotherm 689
Gibbs, J
...
176
Gibbs–Duhem equation 140
Gibbs–Helmholtz equation 105
glancing angle 704
glass electrode 229
glass transition temperature 674
global warming 462
globar 470
glucose oxidation 226
glycolysis 225
Gouy balance 734
Gouy–Chapman model 933
Grahame model 933
Graham’s law of effusion 756
grand canonical ensemble 577
graphical representation 396
graphite structure 720
gravimetry 916
green fluorescent protein 494
greenhouse effect 462
gross selection rule 436
Grotrian diagram 336
Grotthuss mechanism 766
group theory 404
GTO 395
Gunn oscillator 550

H
haematoporphyrin 861
hair 681
half-life 800, (T) 1020
summary 803
half-reaction 216
hamiltonian
core 393
Hückel method 388
hydrogen molecule-ion 368
hamiltonian matrix 389
hamiltonian operator 261
hard sphere packing 716
hard-sphere potential 637
harmonic motion 290
harmonic oscillator 291
energy 291
entropy 576
penetration 296
wavefunction 291
harmonic oscillator (classical) 982
harmonic wave 983
Harned cell 222
harpoon mechanism 875
Hartree, D
...
344
Hartree–Fock equations 393
Hartree–Fock self-consistent field
344

hcp 715, 716
heat 29
heat and work equivalence of 32
heat at constant pressure 41
heat capacity 39, 247, (T) 992
constant pressure 45
constant volume 38, 39
contributions summary 616, 620
molar 45
partition function 601
phase transition 130
relation between 63, 69
relation between (perfect gas) 47
rotational transitions 602
specific 39
variation with temperature 46
vibrational contribution 602
heat capacity ratio 48
heat engine 76
efficiency 83
Heisenberg uncertainty principle
269, 272
Heisenberg, W
...
145
Henry’s law 145
Henry’s law constant (T) 1003
Hermann–Mauguin system 406
Hermite polynomial 292
hermitian operator 264, 283
hermiticity 313
Hertz 244
Hess’s law 53
heterogeneous catalysis 927
rate law 927
heterogeneous catalyst 839
heterogeneous reaction rate 795
heteronuclear diatomic molecule,
MO description 368, 379
heteronuclear spin system 532
hexagonal unit cell 699
hexagonally close-packed 715
HF-SCF 344
high-energy phosphate bond 225
high-performance liquid
chromatography 119
high-temperature superconductor
736
highest occupied molecular orbital
388
Hinshelwood, C
...
820
HOMO 388
homogeneity index 654
homogeneous catalyst 839
homogenized milk 682

homonuclear diatomic molecule
MO description 368
molecular orbital diagram 375
VB description 363
homonuclear spin system 532
Hooke’s law 673
HPLC 119
HREELS 913
HTSC 736
Hückel approximations 387
Hückel method 387
Hückel, E
...
703
Humphreys series 359
Hund’s maximum multiplicity rule
341
Hush, N
...
896
hybrid orbital 366
hybridization 366
hybridization schemes 368
hydration, enthalpy of 51
hydrodynamic flow 647
hydrodynamic radius 766
hydrofluorocarbon 856
hydrogen atom
bound state 326
energies 324
wavefunction 324
hydrogen bond 634, 677
hydrogen–bromine reaction 831,
860
hydrogen electrode 222
hydrogen fluoride, MO description
379, 384
hydrogen ion
conduction by 766
enthalpy of formation 55
Gibbs energy of formation 100
standard entropy 94
hydrogen molecule
MO description 373
VB description 364
hydrogen molecule ion 368
hydrogen–oxygen reaction 833
hydrogen peroxide decomposition
839
hydrogen storage 947
hydrogen/oxygen fuel cell 947
hydrogenation 929
hydrogenic atom 320
hydrogenic orbital, mean radius
330
hydronium ion 766
hydrophilic 682
hydrophobic 635, 682
hydrophobic interaction 636
hydrophobicity constant 635
hydrostatic pressure 5
hydrostatic stress 721
hyperbola 7
hyperfine coupling constant 551, (T)
1015
hyperfine structure 551

1045

hyperpolarizability 732
hypertonic 155

I
IC 495, 846
ice 121
phase diagram 121
residual entropy 93, 610
structure 121, 721
icosahedral group 410
ideal gas, see perfect gas 8
ideal solution 144
Gibbs energy of mixing 148
ideal–dilute solution 146
identity operation 405
IHP 933
immiscible liquids 184
impact parameter 640
impressed-current cathodic
protection 950
improper rotation 406
incident beam flux 648
incongruent melting 193
indefinite integral 966
independent migration of ions 763
independent molecules 579
indicator diagram 35
indicator solution 768
indistinguishable molecules 580
induced dipole moment 624
induced fit model 840
induced magnetic moment 736
induced-dipole–induced-dipole
interaction 633
induction period 812
inelastic neutron scattering 761
inexact differential 58
infectious disease kinetics 867
infinite temperature 567, 584
infrared 244
infrared active 454
infrared activity 467
infrared chemiluminescence 886
infrared inactive 454
infrared region 984
inhibition 844
inhomogeneous broadening 538
initial condition 971
initiation step 831
inner Helmholtz plane 933
inner potential 934
insulator 723
integral 966
integral protein 687
integrated absorption coefficient 433
integrated rate law 798
summary 803
integrated signal 521
integration 966
integration by parts 967
intensive property 959
interference 251, 370

1046

INDEX

interferogram 472
interferometer 471
intermolecular interaction 14
internal conversion 495
internal energy 30
fluid 609
from Q 578
general changes in 59
heat at constant volume 37
molecular contributions 31
partition function 573
properties 103
statistical 573
internal pressure 60
International System (point groups)
406
International System (units) 960
interstellar cloud 439
intersystem crossing 494
intrinsic semiconductor 726
intrinsic viscosity 665, (T) 1017
inverse matrix 976
inversion operation 406
inversion recovery technique 538
inversion symmetry 372
inversion temperature 66, (T) 1002
inverted region 899
iodine, metallic 178
ion
activity 163
Gibbs energy of formation 101
standard entropy 94
ion channel 770
ion mobility 765, 774
ion pump 770
ion-selective electrode 229
ionic atmosphere 164, 683, 769
ionic bond 362
ionic mobility (T) 1019
ionic radius (T) 1017
ionic solid 717
ionic strength 164
ionization, enthalpy of 51, 343
ionization energy 327, 342, (T) 1010
periodicity 343
spectroscopic measurement 325
ion–ion interaction (conductivity)
769
irreducible representation 415
irrep 415
ISC 494, 846
isenthalpic process 64
isobar 10, 12
isobaric calorimeter 41
isochore 10
isodensity surface 396
isoelectric focusing 665
isoelectric point 665
isolated system 29
isolation method 797
isopleth 181
isosteric enthalpy of adsorption 919
isotherm 7, 10, 15

isothermal compressibility 62, (T)
1002
isothermal expansion 79
isothermal Joule–Thomson
coefficient 65
isothermal reversible expansion 36
isotope abundance (T) 991
isotope separation 501
isotopomer 501

J
Jablonski diagram 495
Jeans, J
...
P
...
S
...
702
Knudsen method 756
Kohlrausch’s law 762
Kohn–Sham equations 395
Koopmans’ theorem 378
Krafft temperature 685
Kronecker delta 311, 975
krypton-ion laser 507

L
Lagrange method 582, 970
Laguerre polynomial 324

Lamb formula 521
lambda line 121, 122
lambda-transition 130
Lamé constants 743
lamellar micelle 685
laminar flow 758
lamp 470
Landau, L
...
687
Langmuir–Blodgett film 687
Langmuir–Hinshelwood mechanism
927
Laplace equation 643
laplacian 168, 255, 301
Laporte selection rule 483
Larmor frequency 515, 534
laser 732
laser action 496
laser radiation characteristics 500
laser-induced fluorescence 886
lattice energy 718
lattice enthalpy (T) 1018
lattice point 698
law
Beer–Lambert 432
Boyle’s 7
Charles’s 7
combined gas 10
cosines 370
Curie 734
Dalton’s 13, 179
Debye T3 91
Debye–Hückel limiting 164
Dulong and Petit 247
extended Debye-Hückel 165
Fick’s first 757, 773
Fick’s second 776
First 32
gas 7
Graham’s 756
Henry’s 145
Hess’s 53
Hooke’s 673, 723
Kirchhoff’s 56
Kohlrausch’s 762
limiting 7
motion 981
Newton’s second 981
Ostwald’s dilution 764
Raoult’s 144, 179
Rayleigh–Jeans 245
Second 76
Stefan–Boltzmann 275
Stokes’ 775
Third 93
Wien’s 275
Zeroth 6
law of cosines 370, 964
LCAO-MO 369, 374, 386
symmetry considerations 424
LCAO-MO (solids) 724

Le Chatelier, H
...
N
...
(Lord Cherwell) 820
Lindemann–Hinshelwood
mechanism 820
line alternation 451
line broadening (NMR) 532
line intensity 517
line shape 436
linear combination 267
degenerate orbital 334
linear combination of atomic
orbitals 369, 374, 386
linear free energy relation 884, 899
linear momentum, wavefunction
261
linear rotor 442, 445
linear-sweep voltammetry 942
Lineweaver–Burk plot 842
lipid bilayer 779
lipid raft model 687
liposome 685
liquid, molecular motion 761
liquid crystal 191, 685
phase diagram 192
liquid crystal display 189
liquid junction potential 218
liquid structure 606
liquid viscosity 761
liquid–liquid phase diagram 185
liquid–solid phase diagram 189

INDEX
liquid–vapour boundary 127
lithium atom 337
litre 961
LMCT 489
local contribution to shielding 521
local density approximation 396
local minima 677
lock-and-key model 840
logarithm 963
London formula 635
London interaction 633
long-range order 606
longitudinal relaxation time 536
low energy electron diffraction 914
low overpotential limit 938
low temperature 85
lower critical solution temperature
186
lowest occupied molecular orbital
388
Luggin capillary 939
LUMO 388
lung 147
Lyman series 320
lyophilic 682
lyophobic 682
lyotropic liquid crystal 189
lyotropic mesomorph 685

M
macromolecule 652
macular pigment 490
Madelung constant 719
magic-angle spinning 549
magnetic field 244, 983
magnetic flux density 734
magnetic induction 514
magnetic levitation 737
magnetic moment 514, 735
magnetic quantum number 302
magnetic resonance imaging 540
magnetic susceptibility 522, (T) 1018
magnetically equivalent nuclei 530
magnetizability 734
magnetization 733
magnetization vector 534
magnetogyric ratio 514
MALDI 655
MALDI-TOF 655
manometer 5, 24
many-electron atom 320, 336
Marcus cross-relation 903
Marcus inverted region 899
Marcus theory 853, 896
Marcus, R
...
821, 896
Margules equation 162
Mark–Kuhn–Houwink–Sakurada
equation 666
Mars van Kreelen mechanism 930
MAS 549
mass spectrometry 655
material balance equation 879

matrix addition 975
matrix algebra 975
matrix diagonalization 389
matrix element 310, 313, 975
matrix-assisted laser desorption/
ionization 655
matter flux 757
matter, nature of 309
maximum multiplicity 341
maximum velocity 840
maximum work 96
Maxwell construction 20
Maxwell distribution 750
Maxwell relation 104
Mayer f-function 605
MBE 728
MBRS 928
MBS 916
McConnell equation 552
mean activity coefficient 163, (T)
1004
mean bond enthalpy 55, (T) 1012
mean cubic molar mass 653
mean displacement 294
mean distance diffused 781
mean energies summary 616
mean energy 599
mean free path 754
mean molar mass 653
mean radius, hydrogenic orbital 330
mean rotational energy 600
mean speed 751, 752
mean square displacement 294
mean square molar mass 653
mean translational energy 600
mean value 528, 974
mean value theorem 967
mean vibrational energy 600
measurement, interpretation 267
mechanical equilibrium 4
mechanical property 721
mechanism of reaction 791
Meissner effect 737
melting, response to pressure 123
melting point (T) 990
melting temperature 119
melting temperature (polymer) 674
membrane
formation 685
transport across 779
mercury photosensitization 860
meridional scattering 711
meso-tartaric acid 407
mesopause 854
mesophase 189
mesosphere 854
metal extraction 215
metal-to-ligand transition 489
metallic conductor 723
metallic lustre 730
metastable excited state 496
metastable phase 118
methane, VB description 365

method of initial rates 797
method of undetermined multipliers
582
mho 762
micelle 685
Michaelis constant 841
Michaelis–Menten equation 841
Michaelis–Menten mechanism 841
Michelson interferometer 432, 471
microcanonical ensemble 577
microporous material 931
microstructure 191
microwave background radiation
438
microwave region 244, 984
Mie potential 637
milk 682
Miller indices 700
mirror plane 406
Mitchell, P
...
T
...
683
Overhauser effect spectroscopy 548
overlap, symmetry relation 421
overlap density 370
overlap integral 371, 375
overpotential 938
overtone 456
oxidant 216
oxidation 216
oxidative phosphorylation 225, 227
oxygen
electronic states 483
molecular properties 483
ozone 853

P
p band 725
P branch 458
p orbital 332
real form 333
p-type semiconductivity 726
P680 857
P700 858
packing fraction 716
PAGE 664
para-hydrogen 452

parabolic potential 291, 452
parallel band 461
parallel beta-sheet 678
parallel spins 347
paramagnetic 377, 734
paramagnetic contribution 521
paramagnetism 376
parcel (of air) 12
parity 372, 482
parity selection rule 483
partial charge 379
partial derivative 39, 968
partial differential equation 973
partial fraction 967, 803
partial molar entropy 94
partial molar Gibbs energy 138
partial molar quantity 136
partial molar volume 137
partial pressure 12
partial vapour pressure 124
partially miscible 149
partially miscible liquids 185
distillation 187
particle in a box 278
partition function 568
quantum number 280
particle in a sphere 304
particle on a ring 297
particle on a sphere 301
particular solution 971
partition function
canonical 578
contributions to 615
electronic 597
enthalpy 590
entropy 575, 589
equally spaced levels 563
equation of state 604
equilibrium constant 611
factorization 569
Gibbs energy 591
heat capacity 601
Helmholtz energy 590
internal energy 573, 589
molar 591
molecular 564, 591
overall 599
particle in a box 568
pressure 590
rate constant 882
rotational 592
second virial coefficient 605
standard molar 611
thermodynamic functions from
616
thermodynamic information 578
translational 568, 592
two-level system 564
vibrational 596
partition ratio 779
pascal 4, 961
Pascal’s triangle 526
Paschen series 320

INDEX
passive transport 770
patch clamp technique 771
patch electrode 771
path function 57
Patterson synthesis 709
Pauli exclusion principle 337
Pauli principle 338, 451
Pauling electronegativity 379, (T)
1012
PDT 861
PEMD pulse sequence 543
penetration 286, 296, 340
peptide link 571, 667, 675
peptizing agent 682
perfect elastomer 673
perfect gas 8
enthalpy of mixing 143
entropy change 79, 87
entropy of mixing 143
equilibria 202
Gibbs energy 107
Gibbs energy of mixing 142
internal energy 574
isothermal expansion 87
molar volume 11
statistical entropy 580
transport properties 757, 784
perfect-gas temperature scale 6
periodicity 341
peripheral protein 687
permanent waving 681
permittivity 110, 627, 984, 986
perpendicular band 461
persistence length 75
perturbation theory 310, 313
polarizability 625
time-dependent 311
time-independent 310
Petit, A
...
247
phaeophytin 857
phase 117, 174
phase (wave) 983
phase boundary 118, 126
phase diagram 118
carbon dioxide 120
helium 121
ice 121
liquid crystal 190
liquid–liquid 185
liquid–solid 189
sodium and potassium 192
water 120, 177
phase problem 709
phase rule 176
phase separation 185
phase transition 117, 129
entropy of 87
phase-sensitive detection 550
phosphatidyl choline 686
phosphine decomposition 927
phospholipid 199, 686
phosphorescence 492, 494, 846
photocatalyst 861

photocathode 473
photochemical processes 845
photochemistry 845
photodeflection 502
photodiode 473
photodissociation 502
photodynamic therapy 860
photoelectric effect 250
photoelectron 378
photoelectron spectroscopy 378, 912
photoemission spectroscopy 912
photoionization 501
photoisomerization 502
photomultiplier tube 473
photon 250
photophosphorylation 858
photosensitization 860
photosphere 346
photosynthesis 856
photosystem I and II 856
phototactic response 713
photovoltaic cell detector 473
physical quantity 959
physical state 3
physisorption 916
pi bond 365
pi orbital 374
pi-bond formation energy 390
pi-electron bonding energy 390
pi pulse 538
pi-stacking interaction 638
pi*-n transition 489
pi*-pi transition 489
pi/2 pulse 534
Planck distribution 246
Planck, M
...
C
...
144
Raoult’s law 144, 179
rate
charge transfer 934
surface process 922
rate constant (T) 1020
diffusion controlled 878
electron transfer 895
Kassel form 821
partition function 882
state-to-state 887
rate law 795
heterogeneous catalysis 927
rate of adsorption 916
rate of formation 794
rate of reaction 794
rate-determining step 814
Rayleigh radiation 431
Rayleigh ratio 657, 691
Rayleigh scattering 657
Rayleigh, Lord 245
Rayleigh–Jeans law 245
RDS 814
reaction centre 856
reaction coordinate 809
reaction enthalpy 51
from enthalpy of formation 55
measurement 212
temperature dependence 56
reaction entropy 93
reaction Gibbs energy 100, 201, 220
reaction mechanism 791
reaction order 796
reaction product imaging 886
reaction profile 809
reaction quotient 202
reaction rate 794
collision theory 809, 870
temperature dependence 807
reactive collision 886
reactive cross-section 871, 874
read gradient 540
real gas 8, 14
real-time analysis 793
reciprocal identity 70
recursion relation 292
redox couple 216
redox reaction 216
reduced mass 322, 752
reduced representation 414
reduced variable 21
reducing agent 216
reductant 216
reference electrode 939
reference state 54
refinement 710
reflected wave 287
reflection 406
reflection (X-ray) 704
reflection-absorption infrared
spectroscopy 913
reflection symmetry 483

reforming 931
refraction 984
refractive index 732, 984, (T) 1023
refrigeration 85
regular solution 162, 186
relation between Q and q 579
relative mean speed 752
relative motion 357
relative permittivity 110, 627
relativistic effect 276
relaxation effect 769
relaxation method 805
relaxation time 536, 539
REMPI 886
reorganization energy 897
representation 414
representative matrix 413
repulsion 637
repulsive surface 890
residual entropy 93, 609
resolution (microscopy) 466
resolution (spectroscopy) 473
resonance 513
resonance condition 516
resonance energy transfer 851, 863
resonance integral 380
resonance Raman spectroscopy 465
resonant mode (laser) 497
resonant multiphoton ionization 886
respiratory chain 226
restoring force elastomer 673
resultant vector 964
retardation step 831
retinal 490, 853
retinol 491
reversible change 35
reversible expansion 36
rheology 721
rheometer 666
rhodamine 6G 508
rhodopsin 490
ribosome 840
ribozyme 840
Rice, O
...
821
Rice–Herzfeld mechanism 830
ridge (atmospheric) 12
rigid rotor 442
Rise–Ramsperger–Kassel model 821
Ritz combination principles 321
RNA 680, 840
rock-salt structure 717
rods and cones 490
Röntgen, W
...
320

S
s band 725
S branch 459
s orbital 328
Sackur–Tetrode equation 580
sacrificial anode 950
saddle point 888
SALC 422
salt bridge 216
salting-in effect 173
salting-out effect 173
SAM 690, 914
SATP 11
saturable absorber 498
Sayre probability relation 710
scalar coupling constant 524
scalar product 350, 514, 965
scanning Auger electron microscopy
914
scanning electron microscopy 254
scanning probe microscopy 289
scanning tunnelling microscopy 289
Scatchard equation 156
scattering factor 706
scattering theory 891
SCF 119, 344, 393
Scherrer, P
...
307
Stern–Gerlach experiment 307
Stern–Volmer equation 849
Stern–Volmer plot 850
steroid binding 639
sticking probability 923
stimulated absorption 434
stimulated emission 434, 846
stimulated Raman spectroscopy 501
Stirling’s approximation 563, 974
STM 289
stoichiometric coefficient 203
stoichiometric number 203
Stokes formula 765
Stokes radiation 431
Stokes radius 766
Stokes’ law 775
Stokes–Einstein equation 775, 878
Stokes–Einstein relation 660
stopped-flow technique 793
STP 11

1052

INDEX

strain 721
stratosphere 853
stress 721
strong electrolyte 762
strongly coupled spectra 532
structure factor (X-ray) 706
structure factor (light scattering) 657
structure refinement 710
sublimation
enthalpy of 51
sublimation vapour pressure 118
subshell 328
subshell energies 340
substance 959
substrate 840, 909
sulfur dioxide spectrum 484
Sun 754
supercoiled DNA 680
superconducting magnet 517
superconducting quantum
interference device 735
superconductor 723, 736
supercooled 645
supercritical carbon dioxide 119
supercritical fluid 17, 118
supercritical fluid chromatography
119
supercritical water 119
superfluid 121
superheated 645
superoxide ion 385
superposition 267, 364
superradiant 507
supersaturated 645
supersonic beam 647
supersonic nozzle 648
supertwist 189
surface composition 688, 911
surface defect 910
surface excess 688
surface film balance 687
surface Gibbs energy 688
surface growth 910
surface plasmon resonance 925
surface potential 934
surface pressure 687
surface reconstruction 915
surface tension 642, 689, (T) 1016
surface-enhanced Raman scattering
913
surface-extended X-ray absorption
fine structure spectroscopy
914
surfactant accumulation 689
surroundings 28
entropy change 79
susceptibility 522
sweating 53
SWNT 720, 728
symmetric rotor 442, 444
symmetric stretch 461
symmetry and degeneracy 286
symmetry axis 405

symmetry element 404
symmetry number 595
symmetry operation 404
symmetry species 415, 416
symmetry-adapted linear
combination 422
synchrotron radiation 470, 713
synchrotron storage ring 470
system 28
one-component 177
systematic absences 707
Système International 4

T
T1-weighted image 540
T2-weighted image 541
T 3 law 91
Tafel plot 939
Taylor expansion 967
Taylor series 967
TDS 924
Teller, E
...
D
...
683
vibration 452
vibrational microscopy 466
vibrational modes 460
vibrational motion 290
vibrational partition function 596
vibrational progression 485
vibrational Raman spectra 459
vibrational term 453
vibrational wavenumber (T) 1013
vibration–rotation spectra 457
vibronic laser 731
vibronic transition 484
virial 609
virial coefficient 16
virial equation of state 16, 19

virial theorem 296
viscosity 665, 747, 758, 759, 761, 785,
(T) 1019
diffusion coefficient 775
viscosity-average molar mass 653
visible region 984
vision 490
vitamin C 386
volcano curve 929
voltammetry 940
volume magnetic susceptibility 734
von Laue, M
...
711
watt 961, 979
wave 983
wave equation 983
wave packet 269
wavefunction 253, 272
acceptability 259
acceptable 272
antisymmetric 338
constraints 259
harmonic oscillator 291
hydrogen 324
interpretation 256
linear momentum 261
particle in a box 280
particle on a rectangular surface
285
radial 323
separation 322
trial 380
wavelength 244
wavenumber 244, 983
wavepacket 892
wave–particle duality 253
weak acid 763
weak electrolyte 763, 764
weather 11
weather map 12

1053

weight (configuration) 562
weight-average molar mass 652
wet 644
white paper 658
wide-field epifluorescence method
504
Wierl equation 742
Wilkins, M

---

Title: Physical Chemistry
Description: A full book of Physical Chemistry with simple problem solvings for beginners.

Buy These NotesPreview