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1

CHAPTER 2
Thermochemistry:
Energy Flow and Chemical Change

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2

Thermochemistry: Energy Flow and Chemical Change
 Forms of Energy and Their Interconversion
 Enthalpy: Chemical Change at Constant Pressure
 Calorimetry: Measuring the Heat of a Chemical or
Physical Change
 Stoichiometry of Thermochemical Equations
 Hess’s Law: Finding ∆H of Any Reaction
 Standard Enthalpies of Reaction (∆Hºrxn)

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3

Transfer and Interconversion of Energy
Thermodynamics is the study of
energy and its transformations
...

When energy is transferred from one object to
another, it appears as work and/or heat
...
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4

The system in this case is the contents of the reaction flask
...

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5

The System and Its Surroundings
A meaningful study of any transfer of energy requires that
we first clearly define both the system and its
surroundings
...

The total energy of the universe remains constant
...

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Figure 2

Energy diagrams for the transfer of internal energy (E)
between a system and its surroundings
...
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7

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Figure 4
8

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The two cases where energy is transferred
as work only
...

* For w: + means work done on system; – means work done by system
...
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Energy is conserved, and is neither created nor destroyed
...


∆Euniverse = ∆Esystem + ∆Esurroundings = 0

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Units of Energy

11

The SI unit of energy is the joule (J)
...

1 cal = 4
...

1 Btu is equivalent to 1055 J
...
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...


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Sample Problem 1
13

Determining the Change in Internal Energy
of a System

PROBLEM: When gasoline burns in a car engine, the heat released
causes the products CO2 and H2O to expand, which
pushes the pistons outward
...
If the expanding gases do 451 J of
work on the pistons and the system releases 325 J to the
surroundings as heat, calculate the change in energy (∆E)
in J, kJ, and kcal
...
Then ∆E = q + w
...


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Sample Problem 1
14
SOLUTION:
Heat is given out by a chemical reaction, so it makes sense to define
the system as the reactants and products involved
...

Heat is given out by the system, so q = –325 J
The gases expand to push the pistons, so the system does work on
the surroundings and w = –451 J
∆E = q + w =

–776 J x

-325 J + (-451 J) = –776 J

1 kJ
= –0
...
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185 kcal
–0
...
184 kJ

15

Enthalpy:
Chemical Change at Constant Pressure
 ∆E = q + w
 To determine ∆E, both heat and work must be measured
...


 Enthalpy (H) is defined as E + PV so
 ∆H = ∆E + ∆PV

 If a system remains at constant pressure and its volume
does not change much, then
 ∆H ≈ ∆E
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16

Figure 6

Two different paths for the energy change of a system
...


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Figure 7

Pressure-volume work
...

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∆H as a measure of ∆E

18

 ∆H is the change in heat for a system at constant
pressure
...

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19

Figure 8 Enthalpy diagrams for exothermic and endothermic processes
...
Permission required for reproduction or display
...

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B

H2O (s) → H2O (l)

Endothermic process
Heat is taken in
...
Calculate the work in J (1 atmL = 101
...


We are given the external pressure (988 torr) and initial (125 mL) and final
volumes (652 mL) and have to find the work done by the gas
...
We convert
the P to atm and use equation 6
...
Then we convert the answer
from atmL to J
...
5 J
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w (J)

Sample Problem 2

21

SOLUTION: Calculating ΔV:
ΔV(mL) = Vfinal – Vinitial = 652 mL – 125 mL = 527 mL
Converting ΔV from mL to L:
1L
= 0
...
30 atm

760 torr
Calculating w (using Equation 6
...
30 atm x 0
...
685 atmL
Converting P from atmL to J:

101
...
4 J

w (J) = –0
...
Energy is released as work, so the sign
should be negative
...
5 L = –0
...

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Sample Problem 3
22

Drawing Enthalpy Diagrams and Determining
the Sign of ΔH

PROBLEM: In each of the following cases, determine the sign of ∆H,
state whether the reaction is exothermic or endothermic,
and draw an enthalpy diagram
...
8 kJ
(b) 40
...
If heat is taken in as a “reactant”, the process is
endothermic
...

For the enthalpy diagram, the arrow always points from
reactants to products
...

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Sample Problem 3
SOLUTION: (a) H2 (g) + ½O2 (g) → H2O (l) + 285
...
The reactants are at a higher energy
than the products
...
8 kJ
H2O (l)

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(products)

EXOTHERMIC

Sample Problem 3
24
SOLUTION:

(b) 40
...
The reactants are at a lower energy
than the products
...
7 kJ

H2O (l)

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(reactants)

ENDOTHERMIC

Calorimetry

25

q = c x m x ∆T

q = heat lost or gained
c = specific heat capacity
m = mass in g
∆T = Tfinal – Tinitial

The specific heat capacity (c) of a substance is the
quantity of heat required to change the temperature of
1 gram of the substance by 1 K
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900

Wood

1
...
711

Cement

0
...
450

Glass

0
...
387

Granite

0
...
129

Steel

0
...
184

Ethyl alcohol, C2H5OH (l)

2
...
42

Carbon tetrachloride, CCl4 (l)

0
...
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How much heat is needed to raise the temperature
of the copper layer from 25ºC to 300
...
387 J/g∙K
...
387 J/g∙K) of Cu
and can find ∆T in ºC, which equals ∆T in K
...


SOLUTION:

∆T = Tfinal – Tinitial = 300
...
387 J x 125 g x 275 K
g∙K

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= 1
...


28

This device measures the heat transferred at constant pressure (qP)
...
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05 g solid is heated in a test-tube to 100
...
00 g of water in a coffee-cup calorimeter
...
10ºC to 28
...
Find
the specific heat capacity of the solid
...
In
addition, the heat given out by the solid (–qsolid) is equal to the
heat absorbed by the water (qwater)
...
49ºC – 25
...
39ºC = 3
...
49ºC – 100
...
51ºC = –71
...
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184 J/g∙K x 50
...
39 K
22
...
51 K)

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= 0
...
0 mL of 0
...
00ºC and 25
...
500 M HCl is carefully
added, also at 25
...
After stirring, the final temperature is
27
...
Calculate qsoln (in J) and the change in enthalpy, ∆H,
(in kJ/mol of H2O formed)
...
00 g/mL, and that c = 4
...
Since –qrxn = qsoln, we can find the heat of the reaction
by calculating the heat absorbed by the solution
...
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0 mL + 50
...
00 g/mL = 75
...
21ºC – 25
...
21ºC = 2
...
184 J/g∙K)(75
...
21 K) = 693 J
(b) To find ∆Hrxn we first need a balanced equation:
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

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Sample Problem 6
33
For HCl:
25
...
500 mol
103 mL
1L

x 1 mol H2O
1 mol HCl

= 0
...
0 mL NaOH x 1 L
x 0
...
0250 mol H2O
3 mL
1L
1 mol NaOH
10
HCl is limiting, and the amount of H2O formed is 0
...

∆Hrxn =

qrxn
mol H2O

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=

–693 J

x 1 kJ
0
...
4 kJ/mol H2O

Figure 10

A bomb calorimeter
...

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Sample Problem 7
35

Calculating the Heat of a Combustion
Reaction

PROBLEM: A manufacturer claims that its new dietetic dessert has
“fewer than 10 Calories per serving
...
The
initial temperature is 21
...
799ºC
...
151 kJ/K, is the manufacturer’s claim correct?
PLAN: When the dessert (system) burns, the heat released is
absorbed by the calorimeter:
–qsystem = qcalorimeter
To verify the energy provided by the dessert, we calculate
qcalorimeter
...
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799ºC – 21
...
937ºC = 4
...
151 kJ/K x 4
...
24 kJ
40
...
62 kcal or Calories

4
...


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37

Stoichiometry of Thermochemical Equations
 A thermochemical equation is a balanced equation that
includes ∆Hrxn
...

 The magnitude of ∆H is proportional to the amount of
substance
...


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38

Figure 11
The relationship between amount (mol) of substance and the
energy (kJ) transferred as heat during a reaction
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Its thermal decomposition can be
represented by the equation
Al2O3 (s) → 2Al (s) + 3 O2 (g)
2

∆Hrxn = 1676 kJ

If aluminum is produced this way, how many grams of
aluminum can form when 1
...

heat (kJ)
1676 kJ = 2 mol Al

mol of Al
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mass (g) of Al
multiply by M

Sample Problem 8
40
SOLUTION:
1
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98 g Al
1 mol Al

= 32
...

∆Hoverall = ∆H1 + ∆H2 + ………
...


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Calculating ∆H for an overall process

42



Identify the target equation, the step whose ∆H is
unknown
...


Manipulate each equation with known ∆H values so
that the target amount of each substance is on the
correct side of the equation
...

Multiply amount (mol) and ∆H by the same factor
...



All substances except those in the target equation must
cancel
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An environmental chemist is studying ways to
convert them to less harmful gases through the following
reaction:
CO (g) + NO (g) → CO2 (g) + ½N2 (g) ∆H = ?

Given the following information, calculate the unknown ∆H:
Equation A: CO (g) + ½O2 (g) → CO2 (g) ∆HA = –283
...
6 kJ
PLAN:

Manipulate Equations A and/or B and their ∆H values to get to
the target equation and its ∆H
...


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Sample Problem 9
44

SOLUTION:
Multiply Equation B by ½ and reverse it:
NO (g) → ½N2 (g) + ½O2 (g); ∆H = –90
...
0 kJ
∆H = –90
...
3 kJ

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Table 3 Selected Standard Enthalpies of Formation at 25°C (298K)

45

Formula

∆H°f (kJ/mol) Formula

Calcium
Ca(s)
CaO(s)
CaCO3(s)
Carbon
C(graphite)
C(diamond)
CO(g)
CO2(g)
CH4(g)
CH3OH(l)
HCN(g)
CSs(l)
Chlorine
Cl(g)

0
–635
...
9

0
1
...
5
–393
...
9
–238
...
9
121
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3

Silver
Ag(s)
AgCl(s)

Hydrogen
H(g)
H2(g)

218
...
9
90
...
8

Cl2(g)
HCl(g)

H2O(l)

–285
...
0

0
107
...
1

Sulfur
S8(rhombic)
0
S8(monoclinic) 0
...
8
SO3(g)

–396
...

(a) Silver chloride, AgCl, a solid at standard conditions
...

(c) Hydrogen cyanide, HCN, a gas at standard conditions
...
Make sure all substances are in their
standard states
...
3 or Appendix B
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Ag (s) + ½Cl2 (g) → AgCl (s) ∆Hºf = –127
...

3
Ca (s) + C(graphite) + 2O2 (g) → CaCO3 (s)

∆H°f = –1206
...

½H2 (g) + C(graphite) + ½N2 (g) → HCN (g)
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∆H°f = 135 kJ

Figure 12
48

The two-step process for determining ∆Hºrxn from ∆Hºf values
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The first step in the
industrial production process is the oxidation of ammonia:
Sample Problem 11

49

4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (g)
Calculate ∆H°rxn from ∆H°f values
...
3 or Appendix B and apply
the equation
∆Hrxn = Σm∆Hºf (products) – Σn∆Hºf (reactants)

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Sample Problem 11
50
SOLUTION:
∆Hrxn = Σm∆Hºf (products) – Σn∆Hºf (reactants)
∆Hrxn = [4(∆Hºf of NO(g) + 6(∆Hºf of H2O(g)]
– [4(∆Hºf of NH3(g) + 5(∆Hºf of O2(g)]
= (4 mol)(90
...
8 kJ/mol) –
[(4 mol)(-45
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Chemical Connections
52

Evidence for the enhanced greenhouse effect
...
Permission required for reproduction or display
...


Since the mid-19th
century, average global
temperature has risen
0
...

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53

Thermodynamics: Entropy and Free Energy

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Spontaneous Change

54

A spontaneous change is one that occurs without a
continuous input of energy from outside the system
...

A nonspontaneous change occurs only if the
surroundings continuously supply energy to the system
...

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55

The First Law of Thermodynamics Does
Not Predict Spontaneous Change
Energy is conserved
...

∆E = q + w
The total energy of the universe is constant:
∆Esys = −∆Esurr or ∆Esys + ∆Esurr = ∆Euniv = 0
The law of conservation of energy applies to all
changes, and does not allow us to predict the direction
of a spontaneous change
...
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...

Spontaneous exothermic processes include:
• freezing and condensation at low temperatures,
• combustion reactions,
Spontaneous iron and other metals
...

The sign of ∆H does not by itself predict the direction
of a spontaneous change
...
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Ba(OH)2·8H2O(s) + 2NH4NO3(s) → Ba2+(aq) + 2NO3–(aq) + 2NH3(aq) +
10H2O(l)
∆H°rxn = +62
...
The
reaction mixture absorbs heat from the surroundings so quickly that
the beaker freezes to a wet block
...
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solid → liquid → gas
crystalline solid + liquid → ions in solution
less freedom of particle motion
localized energy of motion

more freedom of particle motion
dispersed energy of motion

A change in the freedom of motion of particles in a system
is a key factor affecting the direction of a spontaneous
process
...
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 Each quantized energy state for a system of particles is
called a microstate
...


 At a given set of conditions, each microstate has the
same total energy as any other
...


 The larger the number of possible microstates, the larger
the number of ways in which a system can disperse its
energy
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...

S = k lnW
A system with fewer microstates has lower entropy
...

All spontaneous endothermic processes exhibit an
increase in entropy
...

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Figure 14

Spontaneous expansion of a gas
...

Increasing the volume increases the number of translational
energy levels the particles can occupy
...

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Figure 15

The entropy increase due to the expansion of a gas
...
More distributions of
particles are possible
...
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...


When the stopcock
opens, the number of
microstates is 2n, where
n is the number of
particles
...
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...


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Figure 17

Simulating a reversible process
...

We remove one grain of sand (an
“infinitesimal” decrease in pressure),
causing the gas to expand a tiny amount
...

This simulates a reversible process, since
it can be reversed by replacing the grain
of sand
...
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...

If we consider both the system and the surroundings, we
find that all real processes occur spontaneously in the
direction that increases the entropy of the universe
...

∆Suniv = ∆Ssys + ∆Ssurr > 0
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67

Comparing Energy and Entropy
The total energy of the universe remains constant
...

For enthalpy there is no zero point; we can only measure
changes in enthalpy
...

∆Suniv =∆Ssys +∆Ssurr > 0
For entropy there is a zero point, and we can determine
absolute entropy values
...
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...

Ssys = 0 at 0 K
A “perfect” crystal has flawless alignment of all its
particles
...

S = k lnW = k ln 1 = 0
To find the entropy of a substance at a given temperature, we cool it
as close to 0 K as possible
...
The sum of
these ∆S values gives the absolute entropy at the temperature of
interest
...
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...

The conventions for defining a standard state include:
•

1 atm for gases

•

1 M for solutions, and

•

the pure substance in its most stable form for
solids and liquids
...
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...

 For any substance, S° increases as temperature increases
...

 S° increases as the phase changes from solid to liquid to gas
...

 Entropy is related to atomic size and molecular
complexity
...


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Figure 18

Visualizing the effect of temperature on entropy
...
Adding heat increases T and the total energy, so the
particles have greater freedom of motion, and their energy is more
dispersed
...

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Figure 19

Visualizing the effect of temperature on entropy
...
Adding heat increases the total energy
(area under the curve), so the range of occupied energy levels
becomes greater, as does the number of microstates
...
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73

A system of 21 particles occupy energy levels (lines) in a box whose
height represents the total energy
...

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Figure 21
74

The increase in entropy during phase changes
from solid to liquid to gas
...
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...

Copyright © The McGraw-Hill Companies, Inc
...


pure solid
MIX

pure liquid
solution
The entropy of a salt solution is usually greater than that of the solid
and of water, but it is affected by the organization of the water molecules
around each ion
...
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...


Ethanol (A) and water (B) each have many H bonds between their own
molecules
...

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Figure 24

The entropy of a gas dissolved in a liquid
...
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...

Li

Na

Rb

Cs

186

Atomic radius (pm) 152

K
227

248

265

Molar mass (g/mol)

6
...
99

39
...
47

132
...
1

51
...
7

69
...
2

HF

HCl

Molar mass (g/mol)
S°(s)

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20
...
7

36
...
8

HBr
80
...
6

HI
127
...
3

Entropy and Structure

79

For allotropes, S° is higher in the form that allows the
atoms more freedom of motion
...
69 J/mol·K, whereas S° of diamond is 2
...


For compounds, S° increases with chemical
complexity
...
1

167

S°(g)

P4O10

NO

NO2

N 2 O4

211

240

304

229

These trends only hold for substances in the same
physical state
...
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...


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Sample Problem 12

Predicting Relative Entropy Values

81
PROBLEM: Choose the member with the higher entropy in each of the
following pairs, and justify your choice [assume constant
temperature, except in part (e)]:
(a) 1 mol of SO2(g) or 1 mol of SO3(g)
(b)

1 mol of CO2(s) or 1 mol of CO2(g)

(c)

3 mol of O2(g) or 2 mol of O3(g)

(d)

1 mol of KBr(s) or 1 mol of KBr(aq)

(e)

seawater at 2°C or at 23°C

(f) 1 mol of CF4(g) or 1 mol of CCl4(g)
PLAN: In general, particles with more freedom of motion have more
microstates in which to disperse their kinetic energy, so they have
higher entropy
...

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Sample Problem 12
82
SOLUTION:
(a) 1 mol of SO3(g)
...

(b) 1 mol of CO2(g)
...

(c) 3 mol of O2(g)
...
Although each
O3 molecule is more complex than each O2 molecule, the greater
number of molecules dominates because there are many more
microstates possible for 3 mol of particles than for 2
...
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...
The two samples have the same number of
ions, but their motion is more limited and their energy less
dispersed in the solid than in the solution
...

(e) Seawater at 23°C
...

(f)

1 mol of CCl4(g)
...


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84

Entropy Changes in the System
The standard entropy of reaction, ∆S°rxn, is the
entropy change that occurs when all reactants and
products are in their standard states
...


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85

Predicting the Sign of ∆S°rxn
We can often predict the sign of ∆S°rxn for processes
that involve a change in the number of moles of gas
...
g
...
g
...


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Sample Problem 13
86

Calculating the Standard Entropy of
Reaction, ∆S°rxn

PROBLEM: Predict the sign of ∆S°rxn and calculate its value for the
combustion of 1 mol of propane at 25°C
...
To find
∆S°rxn, we apply Equation 20
...

SOLUTION:
ΔS°rxn = [(3 mol CO2)(S° of CO2) + (4 mol H2O)(S° of H2O)]
– [(1 mol C3H8)(S° of C3H8) + (5 mol O2)(S° of O2)]
= [(3mol)(213
...
9J/K·mol)]
= –374 J/K
– [(1mol)(269
...
0 J/K·mol)]
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87

Entropy Changes in the Surroundings
A decrease in the entropy of the system is outweighed
by an increase in the entropy of the surroundings
...

In an exothermic process, the surroundings absorbs the
heat released by the system, and Ssurr increases
...

qsys > 0; qsurr < 0 and ∆Ssurr < 0
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88

Temperature at which Heat is Transferred
Since entropy depends on temperature, ∆S°surr is also
affected by the temperature at which heat is transferred
...
The heat transferred is
specific for the reaction and is the same regardless of the
temperature of the surroundings
...

∆Ssurr = – qsys
T
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∆Ssurr = – ∆Hsys for a process at constant P
T

Sample Problem 14

Determining Reaction Spontaneity

89
PROBLEM: At 298 K, the formation of ammonia has a negative
∆S°sys;
N2(g) + 3H2(g) → 2NH3(g); ∆S°sys = –197 J/K
Calculate ∆S°univ, and state whether the reaction occurs
spontaneously at this temperature
...
To find ∆S°surr we need ∆H°sys,
which is the same as ∆H°rxn
...

SOLUTION:
∆H°rxn = [(2 mol)(∆H°f of NH3)] – [(1 mol)(∆H°f of N2) + (3 mol)(∆H°f
of H2)]
= [(2 mol)(-45
...
8 kJ
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Sample Problem 14
90
∆Ssurr = -

∆Hsys
T

–91
...


Although the entropy of the system
decreases, this is outweighed by
the increase in the entropy of the
surroundings
...
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...


91

In this room-sized apparatus, a person exercises while respiratory gases,
energy input and output, and other physiological variables are monitored
...
For all
processes, however complex, the entropy of the university increases
...
edu
...

At equilibrium, there is no further net change, and ∆S°sys
is balanced by ∆S°surr
...


enazizah@unisza
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my

93

Figure 27 AB Components of ∆S°univ for spontaneous reactions
...
The
reaction will always be
spontaneous
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94

Components of ∆S°univ for spontaneous
reactions
...
∆Ssurr is always < 0 for
endothermic reactions
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edu
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The free energy change (∆G) is a measure of the
spontaneity of a process and of the useful energy available
from it
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edu
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∆G°rxn = Σm∆G°products – Σn∆G°reactants

enazizah@unisza
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Calculating ∆G°rxn from Enthalpy and
Entropy Values
PROBLEM: Potassium chlorate, a common oxidizing agent in
fireworks and matchheads, undergoes a solid-state
disproportionation reaction when heated
...
We use
∆Hf° to calculate ∆H°rxn, use S° values to calculate ∆S°rxn,
and then apply the Gibbs equation
...
7 kJ/mol)
= [(3 mol)(-432
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edu
...
7 kJ/mol)]

Sample Problem 15
98
∆S°rxn = ΣmS°(products) – ΣS°(reactants)
= [(3 mol of KClO4)(S° of KClO4) + (1 mol KCl)(S° of KCl)]
‒ [(4 mol KClO3)(S° + KClO3)]
= [(3 mol)(151
...
6 J/mol·K)
= –36
...
1 J/mol·K)]
∆G°sys = ∆H°sys – T ∆S°sys = –144 kJ – (298 K)(–36
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edu
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4:
4KClO3(s)

Δ

3KClO4(s) + KCl(s)

PLAN: We apply Equation 20
...

SOLUTION:
∆G°rxn = Σm∆G°products - Σn∆G°reactants
= [(3 mol KClO4)(∆Gf° of KClO4) + (1 mol KCl)(∆Gf° of KCl)]
‒ [(4 mol KClO3)(∆Gf° of KClO3)]
= [(3 mol)(-303
...
2 kJ/mol)]
‒ [(4 mol)(-296
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edu
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In practice, the maximum work is never done
...


∆G is the minimum work that must be done to a system
to make a nonspontaneous process occur at constant T
and P
...

enazizah@unisza
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my

101

Effect of Temperature on Reaction Spontaneity
∆Gsys = ∆Hsys - T ∆Ssys
Reaction is spontaneous at
all temperatures
If ∆H < 0 and ∆S > 0
∆G < 0 for all T
Reaction is nonspontaneous
at all temperatures
If ∆H > 0 and ∆S < 0
∆G > 0 for all T
enazizah@unisza
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102

Effect of Temperature on Reaction Spontaneity
Reaction becomes spontaneous
as T increases
If ∆H > 0 and ∆S > 0
∆G becomes more negative as T
increases
...

enazizah@unisza
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103

Table 20
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edu
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(a) What are the signs of ∆H and ∆S for this process? Explain
...

PLAN: (a) From the scenes, we determine any change in the amount
of gas, which indicates the sign of ∆S, and any change in
the freedom of motion of the particles, which indicates
whether heat is absorbed or released
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edu
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Sample Problem 16
105
SOLUTION:

(a)

The scene represents the condensation of water vapor, so the
amount of gas decreases dramatically, and the separated molecules
give up energy as they come closer together
...
In order for ∆G to
be < 0, the temperature must be low
...

enazizah@unisza
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my

Sample Problem 16
106

Determining the Effect of Temperature on ΔG

PROBLEM: A key step in the production of sulfuric acid is the oxidation
of SO2(g) to SO3(g):
2SO2(g) + O2(g) → 2SO3(g)
At 298 K, ∆G = –141
...
4 kJ; and ∆S = –187
...

(b) Assuming ∆H and ∆S are constant with increasing T (no phase
change occurs), is the reaction spontaneous at 900
...
We can then
calculate whether or not the reaction is spontaneous at the
higher temperature
...
edu
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With ∆S < 0, the term –T∆S > 0 and this term will become more
positive at higher T
...

(b)

∆G = ∆H – T∆S
Convert T to K: 900
...
15 = 1173 K
Convert S to kJ/K: –187
...
1879 kJ/K
∆G = –198
...
1879 kJ/K) = 22
...
°C
...
edu
...
9 kJ)
...

PLAN: We need to calculate the temperature at which ∆G crosses
over from a positive to a negative value
...

SOLUTION:
58
...
165 kJ/K
∆S
At any temperature above 352 K (= 79°C), the reaction becomes
spontaneous
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edu
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2 The cycling of metabolic free energy
...
edu
...


When ATP is hydrolyzed to ADP, the decrease in charge repulsion (A)
and the increase in resonance stabilization (B) causes a large
amount of energy to be released
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edu
...
edu
...
The reaction releases or absorbs a
large amount of free energy
...
The reaction releases or absorbs very
little free energy
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As ∆G° becomes more positive, K becomes smaller
...

To calculate ∆G for any conditions:
∆G = ∆G° + RT lnQ
enazizah@unisza
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my

Using Molecular Scenes to Find ∆G for a
Reaction at Nonstandard Conditions
PROBLEM: These molecular scenes represent three mixtures in which
A2 (black) and B2 (green) are forming AB
...
10 atm
...
4 kJ/mol
A2(g) + B2(g)
Sample Problem 18

114

(a) If mixture 1 is at equilibrium, calculate K
...
edu
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0 atm?
=P =P

Sample Problem 18
115
PLAN:
(a) Mixture 1 is at equilibrium, so we first write the expression for Q
and then find the partial pressure of each substance from the
number of molecules and calculate K
...
13
...

(c) Since 1
...

SOLUTION:
PAB2
Q=
(a) A2(g) + B2(g)
2AB(g)
PA2 x PB2
K=

(0
...
20)(0
...
edu
...
0

Sample Problem 18
116

(b) ∆G° = –RT lnK
–3
...
314 J T ln 4
...
4 kJ 1000 J
T=

mol·K

Mixture 1 is at equilibrium, so ∆G = 0;

1 kJ
mol
8
...
0
mol·K

∆G = ∆G° + RT lnQ
= –3400 J/mol + (8
...
0)
Mixture 2: J + 3400 J = 0
...
20)2
Q=
= 0
...
30)(0
...
314 J/mol·K)(295 K)(ln 0
...
edu
...
4x10-3 J

= 295 K

Sample Problem 18
117
Mixture 3:
Q=

(0
...
10)(0
...
314 J/mol·K)(295 K)(ln 36)
= –3400 J + 8800 J = +5
...

(c)

Under standard conditions, ∆G = ∆G° = –3
...


Yes, the reaction is spontaneous when the components are in
their standard states
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edu
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2SO2(g) + O2(g) → 2SO3(g)
(a) Calculate K at 298 K and at 973 K, (∆G°298 = −141
...
12 kJ/mol for reaction as written
...
500 atm of SO2, 0
...
100 atm of SO3; one is kept at 25°C and the other at
700
...
In which direction, if any, will the reaction proceed to
reach equilibrium at each temperature?
(c) Calculate ∆G for the system in part (b) at each temperature
...
edu
...

(b) To determine if a net reaction will occur, we find Q from the given
partial pressures and compare Q to each K from part (a)
...


SOLUTION:
(a) ∆G° = −RT lnK so K = e −(∆G°/RT)
−141
...
2
At 298 K, −(∆G°)/RT =
8
...
2 = 7x1024
K at 298 K = e
enazizah@unisza
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my

Sample Problem 19
120
−12
...
50
At 973 K, −(∆G°)/RT =
8
...
50 = 4
...
100)2

= 4
...
500)2(0
...

At 298 K, the system is far from equilibrium and will proceed far to
the right
...

enazizah@unisza
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my

Sample Problem 19
121
(c)
∆G298 = ∆G° + RT ln Q
= −141
...
314x10–3 kJ/mol·K x 298 K x ln 4
...
2 kJ/mol
∆G973 = ∆G° + RT ln Q
= −12
...
314x10–3 kJ/mol·K x 973 K x ln 4
...
9 kJ/mol

enazizah@unisza
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my



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