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Title: Physical Chemistry
Description: A full book of Physical Chemistry with simple problem solvings for beginners.
Description: A full book of Physical Chemistry with simple problem solvings for beginners.
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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: 0716787598
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 endofchapter 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 nanometresized 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 partialdiﬀerential 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
workedout 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 Selftests as inchapter 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 buildingup 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 semiempirical 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 (0716785862), through use of an activation code card
...
whfreeman
...
Key features of the eBook include:
• Easy access from any Internetconnected 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 online
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
(0716785676)
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 (0716762064) provides full solutions to the ‘a’ exercises and the
oddnumbered 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
...
GarzaLópez, 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
NilsOla 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, EspooVantaa 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 temperaturedependence 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
Halfreactions 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
Timeindependent perturbation theory
Timedependent 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
Valencebond 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 moleculeion
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
Twodimensional NMR
Solidstate 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
Selfassembly
19
...
14
19
...
3
Colloids
Micelles and biological membranes
Surface ﬁlms
Impact on nanoscience: Nanofabrication with
selfassembled 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
Diffusioncontrolled 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 metresquared:
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
Selftest 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 highpressure gas will fall as it expands
and that of the lowpressure 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 crosssectional area A; then its volume is Ah and its
mass is m = ρAh
...
3)
Note that the pressure is independent of the shape and crosssectional area of the
column
...
(a)
Selftest 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 redhot 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 perfectgas 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 lowpressure 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 righthand 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 sealevel 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, largescale 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 shortrange 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 hardsphere 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 nonspherical 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 waterlike liquid of density
1
...
6(a) A manometer consists of a Ushaped 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 crosssectional 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 midlatitude
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 nineteenthcentury 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 constanttemperature 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
...
Selftest 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 constantvolume 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 partialdiﬀerential 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 constantvolume 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 constantvolume 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
...
Selftest 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 gasphase reactions
In the reaction 2 H2(g) + O2(g) → 2 H2O(l), 3 mol of gasphase molecules is
replaced by 2 mol of liquidphase 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
Selftest 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 constantpressure 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 threedimensional 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 longchain 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 pathindependence 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
Selftest 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(AB), the average enthalpy change associated with the breaking of a speciﬁc
AB bond,
AB(g) → A(g) + B(g)
∆H(AB)
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 temperaturedependence 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
Selftest 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 pathdependence 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)
...
Selftest 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
...
Selftest 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 constantvolume 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 crosssectional 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 constantpressure 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 dribose 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 constantpressure 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 constantvolume 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 constantvolume 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 doublestranded 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 tertbutyl radical, namely,
(a) tertC4H9 → secC4H9, (b) tertC4H9 → C3H6 + CH3, (c) tertC4H9 →
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 2,2dichloro1,1,1triﬂuoroethane (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 HFC134a)
...
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 constantpressure 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 nonspontaneous, 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
...
Selftest 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 highenergy states
as well as in lowenergy 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 noncancelling 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 icecold 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 nonexpansion 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
constantvolume 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
...
Selftest 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 OH 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
ThirdLaw 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) ThirdLaw entropies
Entropies reported on the basis that S(0) = 0 are called ThirdLaw 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
...
Selftest 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 nonexpansion
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)
...
Selftest 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 nonexpansion 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 nonexpansion 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 nonexpansion 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 nonexpansion 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)
Selftest 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 gasphase 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
...
Selftest 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 (nonexpansion) 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
Selftest 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 (nonexpansion) 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 constantvolume 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 crosssection 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 nonexpansion 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 constantpressure 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 constantpressure 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 ThirdLaw
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 nonexpansion 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 nonexpansion
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 constantvolume 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 liquidlike 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 highperformance 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, scCO2based 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 siloxanebased polymeric stabilizers, which allow polymerization reactions to proceed in scCO2
...
An alternative
and much cheaper approach is poly(ethercarbonate) 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 nonaqueous 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 snowlike 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 helium4 has two liquid phases
...
1
0
...
17
4
...
20
(Tb) (Tc)
(Tl)
Temperature, T /K
The phase diagram for helium
(4He)
...
HeliumII 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
bodycentred 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 λline (lambda line) in Fig
...
7
...
Helium3 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
Selftest 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 nonzero 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 ﬁrstorder phase transition
...
At a ﬁrstorder 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 secondorder 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 λtransition is applied to a phase transition that is not ﬁrstorder 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 Secondorder phase transitions and λtransitions
One type of secondorder 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 λcurve 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 lowtemperature 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 ﬁrstorder phase transition, a secondorder phase
transition, and a λtransition 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 freon12 (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
freon12 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
...
DupontPavlovsky (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 secondorder
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
Selftest 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 idealdilute 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 AB interactions in the mixture is the same as
the average energy of AA and BB interactions in the pure liquids
...
5
...
5
...
Real solutions are composed of particles for which AA, AB, and BB 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 idealdilute 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 lefthand 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 boilingpoint 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 viriallike 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 g1 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
...
Selftest 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 idealdilute, 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
...
Selftest 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 timeaveraged, 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 1,1electrolyte), but it fails at higher
molalities
...
50 and
C = 0 as a representation of experimental
data for a certain 1,1electrolyte
...
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 idealdilute 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 1,2epoxybutane 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 onecomponent 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)
...
Selftest 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 onecomponent 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 onephase,
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 ﬁrstorder 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 diamondanvil 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 ﬁrstorder 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
Twocomponent 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 highboiling azeotrope
...
Fig
...
16
a4
a2
a3
Mole fraction of A, zA
a1
a
1
Fig
...
17 A lowboiling 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 twophase 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 onephase 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 hexanerich phase than nitrobenzenerich
phase
...
Because the phase diagram has been constructed experimentally, these conclusions are not based on any assumptions about ideality
...
Selftest 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 βparameter 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 singlephase 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 lowboiling 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 onephase 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 twophase 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 twophase 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 Xray
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 threephase 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 nonspherical shapes, such as being
long and thin (1), or disklike (2)
...
Calamitic liquid crystals (from the Greek word for reed) are made from long and
thin molecules, whereas discotic liquid crystals are made from disklike 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
...
Disklike
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) Twocomponent, 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) Twocomponent, temperature–composition, solid–liquid diagram,
one compound of formula AB2 that melts incongruently, negligible
solid–solid solubility; (b) twocomponent, 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
1,2dimethylbenzene 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 2methyl1propanol (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 tetraﬂuoromethanerich
PROBLEMS
solution changes from solid methane to a methanerich 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 1butanol 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 1butanol 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 twophase
region are given below as a function of temperature:
T/K
(a) Plot the phase diagram
...
750 mol of N,Ndimethylacetamide 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 pazoxyanisole 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 2methylbutane and the
conversion of lalanine to dalanine
...
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 nonspontaneous
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 Ushaped 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
...
Selftest 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 righthand 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
Selftest 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
Selftest 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 nonspontaneous reaction is
driven by an external source of current
...
5 Halfreactions 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 halfreactions
...
In general we write a couple as Ox/Red and the corresponding reduction halfreaction as
Ox + ν e− → Red
(7
...
1 Varieties of electrode
Electrode type
Designation
Redox couple
Halfreaction
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 halfreactions
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 halfreactions:
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 halfreactions
...
This quotient is
deﬁned like the reaction quotient for the overall reaction, but the electrons are ignored
...
8 Writing the reaction quotient of a halfreaction
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 halfreaction 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 reenter 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 Utube 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 righthand electrode is the cathode, and hence that the spontaneous reaction is one in which reduction is taking place in the righthand compartment
...
If the lefthand electrode turns out to
be the cathode, then the reverse of the corresponding cell reaction is spontaneous
...
Then we subtract from it the lefthand reduction halfreaction (for, by implication, that
7
...
Thus, in the cell Zn(s)  ZnSO4(aq)  CuSO4(aq)  Cu(s)
the two electrodes and their reduction halfreactions are
Righthand electrode: Cu2+(aq) + 2 e− → Cu(s)
Lefthand 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 halfreactions 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 nonexpansion (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 righthand electrode and the standard
hydrogen electrode is the lefthand 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
higherorder 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 halfreaction 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
...
Selftest 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 ADPphosphate
bond has been called a ‘highenergy 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 membranebound protein complexes and two mobile electron carriers before reducing O2 to H2O
...
The respiratory chain begins in complex I (NADHQ oxidoreductase), where
NADH is oxidized by coenzyme Q (Q, 4) in a twoelectron 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
Lefthand 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) Speciesselective 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 gassensing 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 (hydrocarbonattracting) membrane is attached to
a small reservoir of a hydrophobic (waterrepelling) 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 noncalorimetric 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
...
Selftest 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 nonspontaneous 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 BiBi2S3(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 ﬂuorideselective
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 halfcell for a ﬂuorideselective
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 ﬂuorideselective 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 gasphase 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 oneelectron reaction between cytochrome c, cyt, and
2,6dichloroindophenol, 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 costeﬀective 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
blackbody 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 blackbody 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, Xrays, and even γrays) 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 blackbody 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 blackbody
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 PierreLouis 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 constantvolume 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 lowtemperature
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
lowtemperature 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 wavelike—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 cm1, 551 nm)
2
...
86 ´ 104 cm1, 539 nm)
2
...
69 ´ 104 cm1, 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 longestablished 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 particlelike 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 nanometresized
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
hcross or hbar) 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
quantummechanical postulate
...
1
...
8
...
1 The Schrödinger equation
For onedimensional 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
...
Selftest 8
...
3
...
The principal constraint is that ψ must not be inﬁnite anywhere
...
17 would be inﬁnite (in other words, ψ would not be
squareintegrable) 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
singlevalued; 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 secondorder diﬀerential equation,
the second derivative of ψ must be welldeﬁned if the equation is to be applicable
everywhere
...
8
...
At this stage we see that ψ must be continuous, have a continuous slope, be
singlevalued, and be squareintegrable
...
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 lefthand 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
...
Selftest 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 xaxis is multiplication (of the wavefunction)
by x and the operator for linear momentum parallel to the xaxis 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$
...
Selftest 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
...
Selftest 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
...
Selftest 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
welldeﬁned 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 illdeﬁned 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
onedimensional 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, nonnegative
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 noncommuting 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 singlevalued, and be squareintegrable
...
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 (Xray),
(c) 1
...
8
...
4a
...
5b Calculate the speed to which a stationary 4He atom (mass 4
...
4b
...
6a A glowworm of mass 5
...
10 W entirely in the backward direction
...
6b A photonpowered 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 Â + iU and Â − iU are not hermitian
if Â 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 blackbody 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 blackbody
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 groundstate 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 groundstate 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 reradiates it into space as blackbody 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
...
Gasphase 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)
@=−
$2
theory
d2
2m dx 2
(9
...
1 are
ψk = Aeikx + Be−ikx
Timeindependent
perturbation theory
9
...
9
Further information 9
...
2)
Note that we are now labelling both the wavefunctions and the energies (that is,
the eigenfunctions and eigenvalues of @) with the index k
...
2: Perturbation
theory
Discussion questions
Exercises
Problems
278
9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
¥
Potential energy, V
¥
0
0
Wall
L
x
Wall
A particle in a onedimensional
region with impenetrable walls
...
Fig
...
1
functions are solutions by substituting ψk into the lefthand side of eqn 9
...
In this case, all values of k, and therefore all
values of the energy, are permitted
...
We saw in Section 8
...
That
is, eikx is an eigenfunction of the operator Yx with eigenvalue +k$, 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 gasphase molecule that is free to move in a onedimensional
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 onedimensional 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?
Selftest 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 zeropoint 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
βCarotene (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 lefthand 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
...
Selftest 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 twodimensional 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
xaxis, and likewise for EY and motion parallel to the yaxis
...
3 The separation of variables technique applied to the particle in a
twodimensional 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 righthand 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 twodimensional
versions of the wavefunctions shown in Fig
...
3
...
We treat a particle in a threedimensional 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 isotopedependence 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 enzymecatalysed reactions
...
A problem related to the one just considered is that of a particle in a squarewell
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 constantcurrent 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 constantz
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 zeropoint energy is that v cannot take negative
values, for if it did the wavefunction would be illbehaved
...
We can picture this zeropoint 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 XH 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) × (bellshaped 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 onedimensional 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
...
Selftest 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 xyplane (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 zaxis 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 xyplane is represented by
a vector J with the single nonzero
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 zaxis 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
zcomponent 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 zaxis 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 zaxis 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 xyplane 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 zaxis
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 zaxis 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 zaxis 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 zaxis 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 ycomponents
...
The vector model of angular momentum uses pictures like that in Fig
...
40b
...
Each cone has a deﬁnite projection (of ml units) on the zaxis,
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 lightemitting 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 nonnegative number) and ms, the
spin magnetic quantum number, for the projection on the zaxis
...
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 groundstate 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 spin1 particles
(as are some atomic nuclei), but for our purposes the most important spin1 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 zaxis, ml $
Spin
s
1
–
2
Magnitude, {s(s + 1)}1/2$
Spin magnetic
ms
1
±–
2
Component on zaxis, ms $
Total
j
l + s, l + s − 1,
...
, −j
Component on zaxis, 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 timeindependent perturbation theory, the perturbation is always present and
unvarying
...
In timeindependent 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 ‘ﬁrstorder correction’ to the energy, a contribution proportional to
@ (1), and E (2) is the ‘secondorder 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 TIMEDEPENDENT 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 secondorder energy
is a similar average, but over the distortion
induced by the perturbation
...
5 Using perturbation theory
Find the ﬁrstorder correction to the groundstate 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 ﬁrstorder 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 secondorder 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 timedependent
...
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 timedependent 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 zcomponent of the
angular momentum is ml $, ml = l, l − 1,
...
15
...
16
...
A fermion is a particle with a halfintegral 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 secondorder
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 timeindependent perturbation theory, in
which a system is subjected to a perturbation that does not vary with
9 QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
1 Timeindependent perturbation theory
To develop expressions for the corrections to the wavefunction and
energy of a system subjected to a timeindependent 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 ﬁrstorder 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 timedependent 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 Timedependent perturbation theory
To cope with a perturbed wavefunction that evolves with time, we
need to solve the timedependent 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 timeindependent
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
timeindependent 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 onedimensional 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 zeropoint energy
...
9
...
Why is
9
...
Exercises
9
...
0 nm
...
8b Calculate the zeropoint 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 squarewell 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 squarewell potential
...
11a Refer to Exercise 9
...
9
...
What would be the
length of the box such that the zeropoint 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 zeropoint 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 ﬁrstorder correction to the groundstate 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 secondorder 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 cofactors 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 halfangle 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 onedimensional 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 threedimensional 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 onedimensional 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 manyelectron
atoms
10
...
5 Selfconsistent 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 oneelectron 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 manyelectron atom
...
They also provide a set of
concepts that are used to describe the structures of manyelectron 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
Selftest 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 righthand 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 oneelectron 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 zcomponent 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 highenergy 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 manyelectron atoms (see Section 10
...
Selftest 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 n2fold 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
Selftest 10
...
19
...
For example, the 1s, 2s, and 3s orbitals have 0, 1, and 2 radial nodes, respectively
...
Selftest 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 constantvolume 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
...
Selftest 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 zaxis
...
The
wavefunction of a 2porbital 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 xyplane,
where z = 0, so the xyplane 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 nonzero angular momentum
about the zaxis: e+iφ corresponds to clockwise rotation when viewed from below, and
e−iφ corresponds to counterclockwise 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 zaxis, 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 zaxis (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 nonzero
...
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 cm1 (Ha)
20 571 cm1 (Hb)
23 039 cm1 (Hg)
1
24 380 cm (Hd)
102 824 cm1
97 492 cm1
82 259 cm1
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 ycomponents, 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)
...
Selftest 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 manyelectron atoms
The Schrödinger equation for a manyelectron 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 manyelectron 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 groundstate
conﬁguration 1s22s1, with the central nucleus surrounded by a complete heliumlike
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 groundstate conﬁguration of
C is 1s22s22p2, or more succinctly [He]2s22p2, with [He] the heliumlike 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 closedshell 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 buildingup 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 constantpressure 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 blackbody 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 blackbody emission is strong from the Xray 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 manyelectron 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 groundstate 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 oneelectron 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 ﬁrstorder 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
lowj level lies below the highj 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 /cm1
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 manyelectron 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 closedshell
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 twoelectron 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
Selftest 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 jjcoupling
...
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 jjcoupling 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 twoelectron 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 jjcoupling
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 oneelectron 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 manyelectron atom
in the gas phase; the second ionization energy I2 is the
minimum energy necessary to remove an electron from an
ionized manyelectron atom in the gas phase
...
The electron aﬃnity Eea is the energy released when an
electron attaches to a gasphase 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 manyelectron 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 longestwavelength 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 manyelectron 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 crosssection, 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 Earthbound 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
Valencebond 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 valencebond 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)
...
Valencebond 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 groundstate 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 zeropoint 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 zaxis 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
...
Selftest 11
...
[A trigonal pyramidal molecule with each NH 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
valencebond 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 valencebond description of the CH4 molecule leads to
a tetrahedral molecule containing four equivalent CH bonds
...
For
example, the (unnormalized) 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 sp2hybridized 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 valencebond theory to suggest possible shapes for the
hydrogen peroxide molecule, H2O2
...
8°); rotation around the OO bond is possible, so the molecule
interconverts between planar and nonplanar 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
+
moleculeion, 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 oneelectron wavefunctions obtained by solving the Schrödinger equation
Hψ = Eψ are called molecular orbitals (MO)
...
3 THE HYDROGEN MOLECULEION
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 LCAOMO
...
(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 MOLECULEION
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 LCAOMO 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 socalled 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 buildingup principle
to deduce the ground electronic conﬁgurations of manyelectron 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 offaxis
...
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 groundstate 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, HH, 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 singlybonded
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 Xray 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
Selftest 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 valencebond 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 groundstate 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 AB 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 buildingup 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 fourdimensional for this fouratom 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, · CH2CH=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 πbond energy the energy of two πbonds
...
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 valencebond and molecular orbital terms, with typically valencebond 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 groundstate conﬁguration a 2ue1g
...
The πelectron 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 πelectron energy of only 3(2α + 2β ) = 6α + 6β
...
The πbond 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 semiempirical
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 Nelectron closedshell 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) Semiempirical 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
(ABCD) = 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 twocentre integrals of the form
395
2
A e2 D
(ABCD) = 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 fourcentre 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 groundstate energy of an nelectron 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 localdensity 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 ballandstick 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 pbenzoquinone
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 lowestenergy 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 βcarotene 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 valencebond 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 nonneighbours are set equal to zero,
and all remaining resonance integrals are set equal (to β)
...
The πelectron 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 selfconsistent 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 semiempirical 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 groundstate 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 groundstate 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 πelectron 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 onedimensional 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 πelectron 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 xzplane 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 electronsensitive 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 bondlength 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 moleculeion in its ground state
...
Do the same for the
moleculeion the instant after the electron has been excited into the
antibonding LCAOMO
...
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 semiempirical, 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 semiempirical
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
+
LCAOMOs for the H2 moleculeion (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 nonbonding 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 LCAOMOs 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 LCAOMOs 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 (semiempirical, ab initio, or density functional theory
methods), calculate ELUMO, the energy of the LUMO of each substituted
1,4benzoquinone, and plot ELUMO against E 7
...
2)
...
(c) The 1,4benzoquinone
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 nfold rotation (the operation) about an nfold 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 nfold 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 socalled ‘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 transCHCl=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 nfold 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 BF 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 endtoend rotation and endtoend 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
Selftest 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 OH bond in H2O is
cancelled by an equal but opposite component of the dipole of the second OH 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 mirrorimage 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 LAlanine, 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 zaxis 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 threedimensional 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 blockdiagonal form:
⎛[n] 0 0 ⎞
D = ⎜ 0 [n] [n]⎟
⎜ 0 [n] [n]⎟
⎝
⎠
The blockdiagonal 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 onedimensional 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 onedimensional
representation ‘spanned’ by pS, and a twodimensional representation spanned by
(pA, pB)
...
4 CHARACTER TABLES AND SYMMETRY LABELS
orbital plays a role diﬀerent from the other two
...
5)
The onedimensional 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 onedimensional 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 onedimensional representation; we shall denote it Γ (1)′
...
2)
...
An A or a B is used to denote a onedimensional 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 twodimensional irreducible representation and T a threedimensional
irreducible representation; all the irreducible representations of C2v are onedimensional
...
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 blockdiagonal 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 symmetryadapted 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 counterclockwise 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
...
Selftest 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 symmetryadapted 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 righthand 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 symmetryadapted 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
−
Selftest 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 symmetryadapted 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 NH 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′
...
Selftest 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 groundstate conﬁguration is likely to be
...
) that
have a particular symmetry
...
), as input and generates combinations
of the speciﬁed symmetry
...
Symmetryadapted 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
...
Selftest 12
...
The zcomponent of this vector is deﬁned
through
Ύ
µz,ﬁ = −e ψ f*zψi dτ
A1
A2
B1
ypolarized
xpolarized
xpolarized
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:
xcomponent
Forbidden
B2
ycomponent
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 zpolarized
...
Selftest 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 threebladed propellor, (c) a
fourlegged 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) cisCHBr=CHBr, (f) transCHCl=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) cisdichloroethene and (b) transdichloroethene 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 yaxis
(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 zcomponent 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 [transAg(CF3)2(CN)2]−, the AgCN 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 symmetryadapted
linear combinations
...
16 Construct the symmetryadapted 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 lowestlying 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 βcarotene belong? (b) Classify the irreducible representations spanned by
429
the carbon 2pz orbitals and ﬁnd their symmetryadapted linear combinations
...
(d) What states of this model of βcarotene 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 lowestlying 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 lowerfrequency 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 crosssection 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 blackbody 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 timedependent 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 electricdipole 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 Dopplershifted 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 halfheight 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 halfheight 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 linebroadening 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)
...
Earthbound 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
...
Selftest 13
...
The CCl 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 endoverend 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 endoverend rotation (K = 0)
corresponds to 19
...
1 GHz)
...
2 A CH335Cl molecule has a CCl bond length of 178 pm, a CH
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, laboratoryﬁxed 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 spin1 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
...
Suc