Student Solutions
Manual to
Accompany Atkins’
Physical Chemistry
INTERNATIONAL EDITION
Peter Bolgar
Haydn Lloyd
Aimee North
Vladimiras Oleinikovas
Stephanie Smith
and
James Keeler
Department of Chemistry
University of Cambridge
UK
Table of contents
Preface
1
2
3
4
5
6
vii
The properties of gases
1
1A The perfect gas
1
1B The kinetic model
12
1C Real gases
23
Internal energy
39
2A Internal energy
39
2B Enthalpy
45
2C Thermochemistry
48
2D State functions and exact differentials
57
2E Adiabatic changes
63
The second and third laws
69
3A Entropy
69
3B Entropy changes accompanying specific processes
75
3C The measurement of entropy
86
3D Concentrating on the system
96
3E Combining the First and Second Laws
102
Physical transformations of pure substances
113
4A Phase diagrams of pure substances
113
4B Thermodynamic aspects of phase transitions
115
Simple mixtures
131
5A The thermodynamic description of mixtures
131
5B The properties of solutions
142
5C Phase diagrams of binary systems: liquids
159
5D Phase diagrams of binary systems: solids
166
5E Phase diagrams of ternary systems
172
5F
177
Activities
Chemical equilibrium
193
6A The equilibrium constant
193
iv
TABLE OF CONTENTS
7
8
9
6B The response of equilibria to the conditions
202
6C Electrochemical cells
215
6D Electrode potentials
221
Quantum theory
237
7A The origins of quantum mechanics
237
7B Wavefunctions
243
7C Operators and observables
247
7D Translational motion
256
7E Vibrational motion
271
7F
281
Rotational motion
Atomic structure and spectra
291
8A Hydrogenic Atoms
291
8B Many-electron atoms
300
8C Atomic spectra
303
Molecular Structure
311
9A Valence-bond theory
311
9B Molecular orbital theory: the hydrogen molecule-ion
315
9C Molecular orbital theory: homonuclear diatomic molecules
319
9D Molecular orbital theory: heteronuclear diatomic molecules
323
9E Molecular orbital theory: polyatomic molecules
329
10 Molecular symmetry
345
10A Shape and symmetry
345
10B Group theory
355
10C Applications of symmetry
366
11 Molecular Spectroscopy
377
11A General features of molecular spectroscopy
377
11B Rotational spectroscopy
387
11C Vibrational spectroscopy of diatomic molecules
400
11D Vibrational spectroscopy of polyatomic molecules
413
11E Symmetry analysis of vibrational spectroscopy
416
11F Electronic spectra
418
11G Decay of excited states
428
TABLE OF CONTENTS
12 Statistical thermodynamics
439
12A The Boltzmann distribution
439
12B Partition functions
443
12C Molecular energies
453
12D The canonical ensemble
461
12E The internal energy and entropy
463
12F Derived functions
476
13 Molecules in motion
487
13A Transport properties of a perfect gas
487
13B Motion in liquids
494
13C Diffusion
499
14 Chemical kinetics
509
14A The rates of chemical reactions
509
14B Integrated rate laws
515
14C Reactions approaching equilibrium
531
14D The Arrhenius equation
535
14E Reaction mechanisms
539
14F Examples of reaction mechanisms
545
14G Photochemistry
551
15 Reaction dynamics
569
15A Collision theory
569
15B Diffusion-controlled reactions
574
15C Transition-state theory
577
15D The dynamics of molecular collisions
589
15E Electron transfer in homogeneous systems
591
16 Magnetic resonance
599
16A General principles
599
16B Features of NMR spectra
602
16C Pulse techniques in NMR
612
16D Electron paramagnetic resonance
620
17 Molecular Interactions
17A Electric properties of molecules
625
625
v
vi
TABLE OF CONTENTS
17B Interactions between molecules
637
17C Liquids
643
17D Macromolecules
646
17E Self-assembly
657
18 Solids
663
18A Crystal structure
663
18B Diffraction techniques
666
18C Bonding in solids
673
18D The mechanical properties of solids
678
18E The electrical properties of solids
679
18F The magnetic properties of solids
681
18G The optical properties of solids
684
19 Processes at solid surfaces
689
19A An introduction to solid surfaces
689
19B Adsorption and desorption
694
19C Heterogeneous catalysis
706
19D Processes at electrodes
710
Preface
�is manual provides detailed solutions to the (a) Exercises and the odd-numbered Discussion questions and Problems from the international edition of Atkins’ Physical Chemistry.
Conventions used is presenting the solutions
We have included page-speci�c references to equations, sections, �gures and other features
of the main text. Equation references are denoted [��B.�b–���], meaning eqn ��B.�b located
on page ��� (the page number is given in italics). Other features are referred to by name,
with a page number also given.
Generally speaking, the values of physical constants (from the �rst page of the main text)
are used to � signi�cant �gures except in a few cases where higher precision is required.
In line with the practice in the main text, intermediate results are simply truncated (not
rounded) to three �gures, with such truncation indicated by an ellipsis, as in 0.123...; the
value is used in subsequent calculations to its full precision.
�e �nal results of calculations, generally to be found in a box , are given to the precision
warranted by the data provided. We have been rigorous in including units for all quantities
so that the units of the �nal result can be tracked carefully. �e relationships given on
the back of the front cover are useful in resolving the units of more complex expressions,
especially where electrical quantities are involved.
Some of the problems either require the use of mathematical so�ware or are much easier
with the aid of such a tool. In such cases we have used Mathematica (Wolfram Research,
Inc.) in preparing these solutions, but there are no doubt other options available. Some of
the Discussion questions relate directly to speci�c section of the main text in which case we
have simply given a reference rather than repeating the material from the text.
Acknowledgements
In preparing this manual we have drawn on the equivalent volume prepared for the ��th edition of Atkins’ Physical Chemistry by Charles Trapp, Marshall Cady, and Carmen Giunta. In
particular, the solutions which use quantum chemical calculations or molecular modelling
so�ware, and some of the solutions to the Discussion questions, have been quoted directly
from the solutions manual for the ��th edition, without signi�cant modi�cation. More
generally, we have bene�ted from the ability to refer to the earlier volume and acknowledge,
with thanks, the in�uence that its authors have had on the present work.
�is manual has been prepared by the authors using the LATEX typesetting system, in
the implementation provided by MiKTEX (miktex.org); the vast majority of the �gures
and graphs have been generated using PGFPlots. We are grateful to the community who
maintain and develop these outstanding resources.
Finally, we are grateful to the editorial team at OUP, Jonathan Crowe and Roseanna
Levermore, for their invaluable support in bringing this project to a conclusion.
viii
PREFACE
Errors and omissions
In such a complex undertaking some errors will no doubt have crept in, despite the authors’
best e�orts. Readers who identify any errors or omissions are invited to pass them on to us
by email to pchem@ch.cam.ac.uk.
1
1A
The properties of gases
The perfect gas
Answers to discussion questions
D�A.�
�e partial pressure of gas J, p J , in a mixture of gases is given by [�A.�–�], p J =
x J p, where p is the total pressure and x J is the mole fraction of J.
If the gases are perfect, the partial pressure is also the pressure the gas would
exert if it occupied on its own the same container as the mixture at the same
temperature. �is leads to Dalton’s law, which is that the pressure of a mixture
of gases is the sum of the pressures that each one would exert if it occupied the
container alone.
Dalton’s law is a limiting law because it holds exactly only in the limit that there
are no interactions between the molecules, which for real gases will be in the
limit of zero pressure.
D세에 1.146kg 있음
Solutions to exercises
E�A.�(a)
.
3
Consider � m of air: the mass of gas is therefore �.��� kg. If perfect gas behaviour is assumed, the amount in moles is given by n = pV �RT
n=
pV
(0.987 × 105 Pa) × (1 m3 )
=
= 39.5... mol
RT (8.3145 J K−1 mol−1 ) × ([27 + 273.15] K)
(i) �e total amount in moles is n = n O2 +n N2 . �e total mass m is computed
from the amounts in moles and the molar masses M as
m = n O2 × M O2 + n N2 × M N2
�ese two equations are solved simultaneously for n O2 to give the following expression, which is then evaluated using the data given
n O2 =
=
m − M N2 n
M O2 − M N2
702
=
m-MN.HN
2
T
(1146 g) − (28.02 g mol−1 ) × (39.5... mol)
= 9.50... mol
(32.00 g mol−1 ) − (28.02 g mol−1 )
�e mole fractions are therefore
x O2 =
n O2 9.50... mol
=
= �.���
n
39.5... mol
x N2 = 1 − x O2 = �.���
2
1 THE PROPERTIES OF GASES
�e partial pressures are given by p i = x i p tot
p O2 = x O2 p tot = 0.240(0.987 bar) = �.��� bar
p N2 = x N2 p tot = 0.760(0.987 bar) = �.��� bar
(ii) �e simultaneous equations to be solved are now
n = n O2 + n N2 + n Ar
m = n O2 M O2 + n N2 M N2 + n Ar M Ar
Because it is given that x Ar = 0.01, it follows that n Ar = n�100. �e two
unknowns, n O2 and n N2 , are found by solving these equations simultaneously to give
n N2 =
100m − n(M Ar + 99M O2 )
100(M N2 − M O2 )
100×(1146 g)−(39.5... mol)×[(39.95 g mol−1 )+99×(32.00 g mol−1 )]
100 × [(28.02 g mol−1 ) − (32.00 g mol−1 )]
= 30.8... mol
=
From n = n O2 + n N2 + n Ar it follows that
n O2 = n − n Ar − n N2
= (39.5... mol) − 0.01 × (39.5... mol) − (30.8... mol) = 8.31... mol
�e mole fractions are
30.8... mol
nN
x N2 = 2 =
= �.���
n
39.5... mol
�e partial pressures are
x O2 =
n O2 8.31... mol
=
= �.���
n
39.5... mol
p N2 = x N2 p tot = 0.780 × (0.987 bar) = �.��� bar
p O2 = x O2 p tot = 0.210 × (0.987 bar) = �.��� bar
Note: the �nal values are quite sensitive to the precision with which the intermediate results are carried forward.
E�A.�(a)
�e vapour is assumed to be a perfect gas, so the gas law pV = nRT applies. �e
task is to use this expression to relate the measured mass density to the molar
mass.
First, the amount n is expressed as the mass m divided by the molar mass M to
give pV = (m�M)RT; division of both sides by V gives p = (m�V )(RT�M).
�e quantity (m�V ) is the mass density ρ, so p = ρRT�M, which rearranges
to M = ρRT�p; this is the required relationship between M and the density.
M=
=
ρRT
p
(1.23 kg m−3 ) × (8.3145 J K−1 mol−1 ) × (330 K)
20.0 × 103 Pa
= �.��� kg mol−1
�e relationships 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have been used.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�A.�(a)
Charles’ law [�A.�b–�] states that V ∝ T at constant n and p, and p ∝ T at
constant n and V . For a �xed amount the density ρ is proportional to 1�V , so it
follows that 1�ρ ∝ T. At absolute zero the volume goes to zero, so the density
goes to in�nity and hence 1�ρ goes to zero. �e approach is therefore to plot
1�ρ against the temperature (in ○ C) and then by extrapolating the straight line
�nd the temperature at which 1�ρ = 0. �e plot is shown in Fig �.�.
θ�○ C
−85
0
100
ρ�(g dm−3 )
1.877
1.294
0.946
(1�ρ)�(g−1 dm3 )
0.532 8
0.772 8
1.057 1
(1�ρ)�(g−1 dm3 )
1.0
0.5
0.0
−300
Figure 1.1
−200
−100
θ�○ C
0
100
�e data are a good �t to a straight line, the equation of which is
(1�ρ)�(g−1 dm3 ) = 2.835 × 10−3 × (θ�○ C) + 0.7734
�e intercept with 1�ρ = 0 is found by solving
0 = 2.835 × 10−3 × (θ�○ C) + 0.7734
E�A.�(a)
�is gives θ = −273 ○ C as the estimate of absolute zero.
(i) �e mole fractions are
x H2 =
n H2
2.0 mol
=
= 2
n H2 + n N2 2.0 mol + 1.0 mol 3
x N2 = 1 − x H2 = 13
(ii) �e partial pressures are given by p i = x i p tot . �e total pressure is given
3
4
1 THE PROPERTIES OF GASES
by the perfect gas law: p tot = n tot RT�V
2 (3.0 mol) × (8.3145 J K−1 mol−1 ) × (273.15 K)
×
3
22.4 × 10−3 m3
= 2.0 × 105 Pa
p H2 = x H2 p tot =
1 (3.0 mol) × (8.3145 J K−1 mol−1 ) × (273.15 K)
×
3
22.4 × 10−3 m3
5
= 1.0 × 10 Pa
p N2 = x N2 p tot =
Expressed in atmospheres these are �.� atm and �.� atm, respectively.
(iii) �e total pressure is
(3.0 mol) × (8.3145 J K−1 mol−1 ) × (273.15 K)
= 3.0 × 105 Pa
22.4 × 10−3 m3
or �.�� atm.
Alternatively, note that � mol at STP occupies a volume of ��.� dm3 , which is
the stated volume. As there are a total of �.� mol present the (total) pressure
must therefore be �.� atm.
E�A.�(a)
From the inside the front cover the conversion between pressure units is: � atm
≡ 101.325 kPa ≡ ��� Torr; � bar is 105 Pa exactly.
(i) A pressure of ��� kPa is converted to Torr as follows
108 kPa ×
E�A.�(a)
1 atm
760 Torr
×
= ��� Torr
101.325 kPa
1 atm
(ii) A pressure of �.��� bar is 0.975 × 105 Pa, which is converted to atm as
follows
1 atm
0.975 × 105 Pa ×
= �.��� atm
101.325 kPa
�e perfect gas law [�A.�–�], pV = nRT, is rearranged to give the pressure,
p = nRT�V . �e amount n is found by dividing the mass by the molar mass of
Xe, ���.�� g mol−1 .
��� � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � ��
�8.2057 × 10−2 dm3 atm K−1 mol−1 � × (298.15 K)
(131 g)
p=
(131.29 g mol−1 )
1.0 dm3
= 24.4 atm
n
So no , the sample would not exert a pressure of �� atm, but ��.� atm if it were
a perfect gas.
E�A.�(a)
Because the temperature is constant (isothermal) Boyle’s law applies, pV =
const. �erefore the product pV is the same for the initial and �nal states
p f Vf = p i Vi
hence
p i = p f Vf �Vi
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e initial volume is �.�� dm3 greater than the �nal volume so Vi = 4.65+2.20 =
6.85 dm3 .
pi =
Vf
4.65 dm3
× pf =
× (5.04 bar) = 3.42 bar
Vi
6.85 dm3
(i) �e initial pressure is �.�� bar
(ii) Because a pressure of � atm is equivalent to �.����� bar, the initial pressure
expressed in atm is
E�A.�(a)
1 atm
× 3.40 bar = �.�� atm
1.01325 bar
If the gas is assumed to be perfect, the equation of state is [�A.�–�], pV = nRT.
In this case the volume and amount (in moles) of the gas are constant, so it
follows that the pressure is proportional to the temperature: p ∝ T. �e ratio
of the �nal and initial pressures is therefore equal to the ratio of the temperatures: p f �p i = Tf �Ti . �e pressure indicated on the gauge is that in excess
of atmospheric pressure, thus the initial pressure is 24 + 14.7 = 38.7 lb in−2 .
Solving for the �nal pressure p f (remember to use absolute temperatures) gives
Tf
× pi
Ti
(35 + 273.15) K
=
× (38.7 lb in−2 ) = 44.4... lb in−2
(−5 + 273.15) K
pf =
E�A.�(a)
�e pressure indicated on the gauge is this �nal pressure, minus atmospheric
pressure: 44.4... − 14.7 = 30 lb in−2 . �is assumes that (i) the gas is behaving
perfectly and (ii) that the tyre is rigid.
�e perfect gas law pV = nRT is rearranged to give the pressure
p=
nRT
V
��� � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � �
255 × 10−3 g (8.3145 × 10−2 dm3 bar K−1 mol−1 ) × (122 K)
=
×
20.18 g mol−1
3.00 dm3
n
= �.���� bar
Note the choice of R to match the units of the problem. An alternative is to
use R = 8.3154 J K−1 mol−1 and adjust the other units accordingly, to give a
pressure in Pa.
[(255 × 10−3 g)�(20.18 g mol−1 )] × (8.3145 J K−1 mol−1 ) × (122 K)
3.00 × 10−3 m3
5
= 4.27 × 10 Pa
p=
where 1 dm3 = 10−3 m3 has been used along with 1 J = 1 kg m2 s−2 and 1 Pa =
1 kg m−1 s−2 .
5
6
1 THE PROPERTIES OF GASES
E�A.��(a)
�e vapour is assumed to be a perfect gas, so the gas law pV = nRT applies. �e
task is to use this expression to relate the measured mass density to the molar
mass.
First, the amount n is expressed as the mass m divided by the molar mass M to
give pV = (m�M)RT; division of both sides by V gives p = (m�V )(RT�M).
�e quantity (m�V ) is the mass density ρ, so p = ρRT�M, which rearranges
to M = ρRT�p; this is the required relationship between M and the density.
M=
ρRT (3.710 kg m−3 ) × (8.3145 J K−1 mol−1 ) × ([500 + 273.15] K)
=
p
93.2 × 103 Pa
= 0.255... kg mol−1
where 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have been used. �e molar mass
of S is 32.06 g mol−1 , so the number of S atoms in the molecules comprising
the vapour is (0.255... × 103 g mol−1 )�(32.06 g mol−1 ) = 7.98. �e result is
expected to be an integer, so the formula is likely to be S8 .
E�A.��(a)
�e vapour is assumed to be a perfect gas, so the gas law pV = nRT applies; the
task is to use this expression to relate the measured data to the mass m. �is
is done by expressing the amount n as m�M, where M is the the molar mass.
With this substitution it follows that m = MPV �RT.
�e partial pressure of water vapour is 0.60 times the saturated vapour pressure
M pV
RT
(18.0158 g mol−1 ) × (0.60 × 0.0356 × 105 Pa) × (400 m3 )
=
(8.3145 J K−1 mol−1 ) × ([27 + 273.15] K)
m=
= 6.2 × 103 g = �.� kg
Solutions to problems
P�A.�
(a) �e expression ρgh gives the pressure in Pa if all the quantities are in
SI units, so it is helpful to work in Pa throughout. From the front cover,
��� Torr is exactly � atm, which is 1.01325×105 Pa. �e density of ��.�� g cm−3
is equivalent to 13.55 × 103 kg m−3 .
p = p ex + ρgh
= 1.01325 × 105 Pa + (13.55 × 103 kg m−3 ) × (9.806 m s−2 )
× (10.0 × 10−2 m) = 1.15 × 105 Pa
(b) �e calculation of the pressure inside the apparatus proceeds as in (a)
p = 1.01325 × 105 Pa + (0.9971 × 103 kg m−3 ) × (9.806 m s−2 )
× (183.2 × 10−2 m) = 1.192... × 105 Pa
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e value of R is found by rearranging the perfect gas law to R = pV �nT
R=
= �.��� J K−1 mol−1
�e perfect gas law pV = nRT implies that pVm = RT, where Vm is the molar
volume (the volume when n = 1). It follows that p = RT�Vm , so a plot of p
against T�Vm should be a straight line with slope R.
However, real gases only become ideal in the limit of zero pressure, so what is
needed is a method of extrapolating the data to zero pressure. One approach is
to rearrange the perfect gas law into the form pVm �T = R and then to realise
that this implies that for a real gas the quantity pVm �T will tend to R in the limit
of zero pressure. �erefore, the intercept at p = 0 of a plot of pVm �T against p
is an estimate of R. For the extrapolation of the line back to p = 0 to be reliable,
the data points must fall on a reasonable straight line. �e plot is shown in
Fig �.�.
p�atm
0.750 000
0.500 000
0.250 000
(pVm �T)�(atm dm3 mol−1 K−1 )
P�A.�
pV
(1.192... × 105 Pa) × (20.000 × 10−3 m3 )
=
nT [(1.485 g)�(4.003 g mol−1 )] × ([500 + 273.15] K)
Vm �(dm3 mol−1 ) (pVm �T)�(atm dm3 mol−1 K−1 )
29.8649
0.082 001 4
44.8090
0.082 022 7
89.6384
0.082 041 4
0.08206
0.08204
0.08202
0.08200
0.0
0.2
0.4
p�atm
0.6
0.8
Figure 1.2
�e data fall on a reasonable straight line, the equation of which is
(pVm �T)�(atm dm3 mol−1 K−1 ) = −7.995 × 10−5 × (p�atm) + 0.082062
�e estimate for R is therefore the intercept, 0.082062 atm dm3 mol−1 K−1 .
�e data are given to � �gures, but they do not fall on a very good straight line
so the value for R has been quoted to one fewer signi�cant �gure.
7
1 THE PROPERTIES OF GASES
P�A.�
For a perfect gas pV = nRT which can be rearranged to give p = nRT�V . �e
amount in moles is n = m�M, where M is the molar mass and m is the mass of
the gas. �erefore p = (m�M)(RT�V ). �e quantity m�V is the mass density
ρ, and hence
p = ρRT�M
It follows that for a perfect gas p�ρ should be a constant at a given temperature.
Real gases are expected to approach this as the pressure goes to zero, so a
suitable plot is of p�ρ against p; the intercept when p = 0 gives the best estimate
of RT�M. �e plot is shown in Fig. �.�.
ρ�(kg m−3 )
0.225
0.456
0.664
1.062
1.468
1.734
p�kPa
12.22
25.20
36.97
60.37
85.23
101.30
(p�ρ)�(kPa kg−1 m3 )
54.32
55.26
55.68
56.85
58.06
58.42
58
(p�ρ)�(kPa kg−1 m3 )
8
56
54
0
20
40
60
p�kPa
80
100
Figure 1.3
�e data fall on a reasonable straight line, the equation of which is
(p�ρ)�(kPa kg−1 m3 ) = 0.04610 × (p�kPa) + 53.96
�e intercept is (p�ρ)lim p→0 , which is equal to RT�M.
M=
RT
(8.3145 J K−1 mol−1 ) × (298.15 K)
=
= 4.594×10−2 kg mol−1
(p�ρ)lim p→0
53.96 × 103 Pa kg−1 m3
�e estimate of the molar mass is therefore ��.�� g mol−1 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P�A.�
(a) For a perfect gas pV = nRT so it follows that for a sample at constant
volume and temperature, p 1 �T1 = p 2 �T2 . If the pressure increases by
∆p for an increase in temperature of ∆T, then with p 2 = p 1 + ∆p and
T2 = T1 + ∆T is follows that
p 1 p 1 + ∆p
=
T1 T1 + ∆T
hence
∆p =
For an increase by �.�� K, ∆T = 1.00 K and hence
∆p =
p 1 ∆T
T1
p 1 ∆T (6.69 × 103 Pa) × (1.00 K)
=
= 24.5 Pa
T1
273.16 K
Another way of looking at this is to write the rate of change of pressure
with temperature as
∆p p 1 6.69 × 103 Pa
=
=
= 24.5... Pa K−1
∆T T1
273.16 K
(b) A temperature of ���.�� ○ C is equivalent to an increase in temperature
from the triple point by 100.00 + 273.15 − 273.16 = 99.99 K
∆p′ = ∆T ′ × �
∆p
6.69 × 103 Pa
� = (99.99 K) ×
= 2.44... × 103 Pa
∆T
273.16 K
�e �nal pressure is therefore 6.69 + 2.44... = 9.14 kPa .
(c) For a perfect gas ∆p�∆T is independent of the temperature so at ���.� ○ C
a �.�� K rise in temperature gives a pressure rise of 24.5 Pa , just as in (a).
P�A.�
�e molar mass of SO� is 32.06+2×16.00 = 64.06 g mol−1 . If the gas is assumed
to be perfect the volume is calculated from pV = nRT
��� � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � ��
nRT
200 × 106 g (8.3145 J K−1 mol−1 ) × ([800 + 273.15] K)
V=
=�
�
p
1.01325 × 105 Pa
64.06 g mol−1
n
= 2.7 × 105 m3
Note the conversion of the mass in t to mass in g; repeating the calculation for
��� t gives a volume of 4.1 × 105 m3 .
�e volume of gas is therefore between �.�� km3 and �.�� km3 .
P�A.��
Imagine a column of the atmosphere with cross sectional area A. �e pressure
at any height is equal to the force acting down on that area; this force arises
from the gravitational attraction on the gas in the column above this height –
that is, the ‘weight’ of the gas.
Suppose that the height h is increased by dh. �e force on the area A is reduced
because less of the atmosphere is now bearing down on this area. Speci�cally,
the force is reduced by that due to the gravitational attraction on the gas contained in a cylinder of cross-sectional area A and height dh. If the density of
9
10
1 THE PROPERTIES OF GASES
the gas is ρ, the mass of the gas in the cylinder is ρ × A dh and the force due to
gravity on this mass is ρgA dh, where g is the acceleration due to free fall. �e
change in pressure dp on increasing the height by dh is this force divided by
the area, so it follows that
dp = −ρgdh
�e minus sign is needed because the pressure decreases as the height increases.
�e density is related to the pressure by starting from the perfect gas equation,
pV = nRT. If the mass of gas is m and the molar mass is M, it follows that
n = m�M and hence pV = (m�M)RT. Taking the volume to the right gives
p = (m�MV )RT. �e quantity m�V is the mass density ρ, so p = (ρ�M)RT;
this is rearranged to give an expression for the density: ρ = M p�RT.
�is expression for ρ is substituted into dp = −ρgdh to give dp = −(M p�RT)gdh.
Division by p results in separation of the variables (1�p) dp = −(M�RT)gdh.
�e le�-hand side is integrated between p 0 , the pressure at h = 0 and p, the
pressure at h. �e right-hand side is integrated between h = 0 and h
�
p 1
p0
p
dp = �
h
0
−
Mg
dh
RT
Mg
p
h
[ln p] p 0 = −
[h]0
RT
p
M gh
ln
=−
p0
RT
�e exponential of each side is taken to give
p = p 0 e−h�H
with
It is assumed that g and T do not vary with h.
H=
RT
Mg
(a) �e pressure decrease across such a small distance will be very small because h�H � 1. It is therefore admissible to expand the exponential and
retain just the �rst two terms: ex ≈ 1 + x
p = p 0 (1 − h�H)
�is is rearranged to give an expression for the pressure decrease, p − p 0
p − p 0 = −p 0 h�H
If it is assumed that p 0 is one atmosphere and that H = 8 km,
p − p 0 = −p 0 h�H = −
(1.01325 × 105 Pa) × (15 × 10−2 m)
= −2 Pa
8 × 103 m
(b) �e pressure at �� km is calculated using the full expression
p = p 0 e−h�H = (1 atm) × e−(11 km)�(8 km) = 0.25 atm
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P�A.��
Imagine a volume V of the atmosphere, at temperature T and pressure p tot .
If the concentration of a trace gas is expressed as X parts per trillion (ppt), it
means that if that gas were con�ned to a volume X × 10−12 × V at temperature
T is would exert a pressure p tot . From the perfect gas law it follows that n =
pV �RT, which in this case gives
n trace =
p tot (X × 10−12 × V )
RT
Taking the volume V to the le� gives the molar concentration, c trace
c trace =
n trace X × 10−12 × p tot
=
V
RT
An alternative way of looking at this is to note that, at a given temperature and
pressure, the volume occupied by a gas is proportional to the amount in moles.
Saying that a gas is present at X ppt implies that the volume occupied by the
gas is X × 10−12 of the whole, and therefore that the amount in moles of the gas
is X × 10−12 of the total amount in moles
n trace = (X × 10−12 ) × n tot
�is is rearranged to give an expression for the mole fraction x trace
x trace =
n trace
= X × 10−12
n tot
�e partial pressure of the trace gas is therefore
p trace = x trace p tot = (X × 10−12 ) × p tot
�e concentration is n trace �V = p trace �RT, so
c trace =
(a) At �� ○ C and �.� atm
n trace X × 10−12 × p tot
=
V
RT
X CCl3 F × 10−12 × p tot
RT
261 × 10−12 × (1.0 atm)
=
(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × ([10 + 273.15] K)
c CCl3 F =
= 1.1 × 10−11 mol dm−3
X CCl2 F2 × 10−12 × p tot
RT
509 × 10−12 × (1.0 atm)
=
(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × ([10 + 273.15] K)
c CCl2 F2 =
= 2.2 × 10−11 mol dm−3
11
12
1 THE PROPERTIES OF GASES
(b) At ��� K and �.��� atm
X CCl3 F × 10−12 × p tot
RT
261 × 10−12 × (0.050 atm)
=
(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (200 K)
c CCl3 F =
= 8.0 × 10−13 mol dm−3
X CCl2 F2 × 10−12 × p tot
RT
509 × 10−12 × (0.050 atm)
=
(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (200 K)
c CCl2 F2 =
= 1.6 × 10−12 mol dm−3
1B The kinetic model
Answer to discussion questions
D�B.�
For an object (be it a space cra� or a molecule) to escape the gravitational
�eld of the Earth it must acquire kinetic energy equal in magnitude to the
gravitational potential energy the object experiences at the surface of the Earth.
�e gravitational potential between two objects with masses m 1 and m 2 when
separated by a distance r is
V =−
Gm 1 m 2
r
where G is the (universal) gravitational constant. In the case of an object of
mass m at the surface of the Earth, it turns out that the gravitational potential
is given by
GmM
V =−
R
where M is the mass of the Earth and R its radius. �is expression implies that
the potential at the surface is the same as if the mass of the Earth were localized
at a distance equal to its radius.
As a mass moves away from the surface of the Earth the potential energy increases (becomes less negative) and tends to zero at large distances. �is change
in potential energy must all be converted into kinetic energy if the mass is to
escape. A mass m moving at speed υ has kinetic energy 12 mυ 2 ; this speed will
be the escape velocity υ e when
�
GmM
2GM
2
1
mυ e =
hence
υe =
2
R
R
�e quantity in the square root is related to the acceleration due to free fall,
g, in the following way. A mass m at the surface of the Earth experiences
a gravitational force given GMm�R 2 (note that the force goes as R−2 ). �is
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
force accelerates the mass towards the Earth, and can be written mg. �e two
expressions for the force are equated to give
GMm
= mg
R2
hence
GM
= gR
R
�is expression for GM�R is substituted into the above expression for υ e to give
�
2GM �
υe =
= 2Rg
R
�e escape velocity is therefore a function of the radius of the Earth and the
acceleration due to free fall.
�e radius of the Earth is 6.37×106 m and g = 9.81 m s−2 so the escape velocity
is 1.11×104 m s−1 . For comparison, the mean speed of He at ��� K is ���� m s−1
and for N� the mean speed is ��� m s−1 . For He, only atoms with a speed in
excess of eight times the mean speed will be able to escape, whereas for N� the
speed will need to be more than twenty times the mean speed. �e fraction of
molecules with speeds many times the mean speed is small, and because this
2
fraction goes as e−υ it falls o� rapidly as the multiple increases. A tiny fraction
of He atoms will be able to escape, but the fraction of heavier molecules with
su�cient speed to escape will be utterly negligible.
D�B.�
�e mean free path is given by [�B.��–��], λ = kT�σ p. In a container of constant volume, the mean free path is directly proportional to temperature and
inversely proportional to pressure. �e former dependence can be rationalized
by noting that the faster the molecules travel, the farther on average they go
between collisions. �e latter also makes sense in that the lower the pressure,
the less frequent are collisions, and therefore the further the average distance
between collisions.
Perhaps more fundamental than either of these considerations is the dependence on the size of the container and on the size of the molecules. �e ratio
T�p is directly proportional to volume for a perfect gas, so the average distance
between collisions is directly proportional to the size of the container holding a
given number of gas molecules. Finally, the mean free path is inversely proportional to the size of the molecules as given by the collision cross section (and
therefore inversely proportional to the square of the radius of the molecule).
Solutions to exercises
E�B.�(a)
�e most probable speed is given by [�B.��–��], υ mp = (2RT�M)1�2 , the mean
speed is given by [�B.�–��], υ mean = (8RT�πM)1�2 , and the mean relative
speed
between two molecules of the same mass is given by [�B.��a–��], υ rel =
√
2υ mean .
M CO2 = 12.01 + 2 × 16.00 = 44.01 g mol−1 .
2RT 1�2
2 × (8.3145 J K−1 mol−1 ) × (293.15 K)
υ mp = �
� =�
�
M
44.01 × 10−3 kg mol−1
1�2
= 333 m s−1
13
14
1 THE PROPERTIES OF GASES
υ mean = �
8RT 1�2
8 × (8.3145 J K−1 mol−1 ) × (293.15 K)
� =�
�
πM
π × (44.01 × 10−3 kg mol−1 )
υ rel =
E�B.�(a)
√
√
2υ mean = 2 × (376 m s−1 ) = 531 m s−1
= 376 m s−1
�e collision frequency is given by [�B.��b–��], z = σ υ rel p�kT, with the
√ relative
speed for two molecules of the same type given by [�B.��a–��], υ rel = 2υ mean .
�e mean speed is given by [�B.�–��], υ mean = (8RT�πM)1�2 . From the Resource section the collision cross-section σ is �.�� nm2 .
z=
σ υ rel p σ p √
8RT 1�2
=
× 2×�
�
kT
kT
πM
(0.27 × 10−18 m2 ) × (1.01325 × 105 Pa) √
=
× 2
(1.3806 × 10−23 J K−1 ) × (298.15 K)
�
= 1.7 × 1010 s−1
E�B.�(a)
1�2
8 × (8.3145 J K−1 mol−1 ) × (298.15 K)
�
π × (2 × 1.0079 × 10−3 kg mol−1 )
1�2
where 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have been used. Note the
conversion of the collision cross-section σ to m2 : � nm2 = (1 × 10−9 )2 m2 =
1 × 10−18 m2 .
�e mean speed is given by [�B.�–��], υ mean = (8RT�πM)1�2 . �e collision
frequency is given by [�B.��b–��], z = σ υ rel p�kT, with the
√ relative speed for
two molecules of the same type given by [�B.��a–��], υ rel = 2υ mean . �e mean
free path is given by [�B.��–��], λ = kT�σ p
(i) �e mean speed is calculated as
υ mean = �
8RT 1�2
8 × (8.3145 J K−1 mol−1 ) × (298.15 K)
� =�
�
πM
π × (2 × 14.01 × 10−3 kg mol−1 )
1�2
= 475 m s−1
(ii) �e collision cross-section σ is calculated from the collision diameter d
as σ = πd 2 = π × (395 × 10−9 m)2 = 4.90... × 10−19 m2 . With this value
the mean free path is calculated as
λ=
kT
(1.3806 × 10−23 J K−1 ) × (298.15 K)
=
= 82.9×10−9 m = ��.� nm
σ p (4.90... × 10−19 m2 ) × (1.01325 × 105 Pa)
where 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have been used.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(iii) �e collision rate is calculated as
z=
σ υ rel p σ p √
8RT 1�2
=
× 2×�
�
kT
kT
πM
(4.90... × 10−19 m2 ) × (1.01325 × 105 Pa) √
=
× 2
(1.3806 × 10−23 J K−1 ) × (298.15 K)
8 × (8.3145 J K−1 mol−1 ) × (298.15 K)
�
�
π × (2 × 14.01 × 10−3 kg mol−1 )
1�2
= 8.10 × 109 s−1
E�B.�(a)
An alternative for the calculation of z is to use [�B.��–��], λ = υ rel �z,
rearranged to z = υ rel �λ
√
√
υ rel
2υ mean
2 × (475 m s−1 )
z=
=
=
= 8.10 × 109 s−1
λ
λ
82.9 × 10−9 m
�e container is assumed to be spherical with radius r and hence volume V =
4
πr 3 . �is volume is expressed in terms the the required diameter d = 2r as
3
V = 16 πd 3 . Rearrangement of this expression gives d
d=�
6V 1�3
6 × 100 cm3
� =�
�
π
π
1�3
= 5.75... cm
�e mean free path is given by [�B.��–��], λ = kT�σ p. �is is rearranged to
give the pressure p with λ equal to the diameter of the vessel
p=
kT
(1.3806 × 10−23 J K−1 ) × (298.15 K)
=
= �.�� Pa
σ d (0.36 × 10−18 m2 ) × (5.75... × 10−2 m)
Note the conversion of the diameter from cm to m.
E�B.�(a)
�e mean free path is given by [�B.��–��], λ = kT�σ p.
λ=
E�B.�(a)
kT
(1.3806 × 10−23 J K−1 ) × (217 K)
=
σ p (0.43 × 10−18 m2 ) × (0.05 × 1.01325 × 105 Pa)
= 1.4 × 10−6 m = �.� µm
1�2
(i) �e
� mean speed is given by [�B.�–��], υ mean = (8RT�πM) , so υ mean ∝
1�M. �e ratio of the mean speeds therefore depends on the ratio of
the molar masses
M Hg
υ mean,H2
=�
�
υ mean,Hg
M H2
1�2
=�
200.59 g mol−1
�
2 × 1.0079 g mol−1
1�2
= �.���
(ii) �e mean translational kinetic energy �E k � is given by 12 m�υ 2 �, where
�υ 2 � is the mean square speed, which is given by [�B.�–��], �υ 2 � = 3RT�M.
�e mean translational kinetic energy is therefore
1
1
3RT
�E k � = m�υ 2 � = m �
�
2
2
M
15
16
1 THE PROPERTIES OF GASES
�e molar mass M is related to the mass m of one molecule by M = mN A ,
where N A is Avogadro’s constant, and the gas constant can be written R =
kN A , hence
1
3RT
1
3kN A T
3
�E k � = m �
� = m�
� = kT
2
M
2
mN A
2
�e mean translational kinetic energy is therefore independent of the
identity of the gas, and only depends on the temperature: it is the same
for H� and Hg.
�is result is related to the principle of equipartition of energy: a molecule
has three translational degrees of freedom (x, y, and z) each of which
contributes 12 kT to the average energy.
E�B.�(a)
�e rms speed is given by [�B.�–��], υ rms = (3RT�M)1�2 .
3RT
υ rms,H2 = �
�
M H2
1�2
3 × (8.3145 J K−1 mol−1 ) × (293.15 K)
=�
�
2 × 1.0079 × 10−3 kg mol−1
1�2
= 1.90 km s−1
where 1 J = 1 kg m2 s−2 has been used. Note that the molar mass is in kg mol−1 .
υ rms,O2 = �
E�B.�(a)
3 × (8.3145 J K−1 mol−1 ) × (293.15 K)
�
2 × 16.00 × 10−3 kg mol−1
1�2
= 478 m s−1
�e Maxwell–Boltzmann distribution of speeds, f (υ), is given by [�B.�–��].
�e fraction of molecules with speeds between υ 1 and υ 2 is given by the integral
�
υ2
υ1
f (υ) dυ
If the range υ 2 − υ 1 = δυ is small, the integral is well-approximated by
f (υ mid ) δυ
where υ mid is the mid-point of the velocity range: υ mid = 12 (υ 2 + υ 1 ). In this
exercise υ mid = 205 m s−1 and δυ = 10 m s−1 .
fraction = f (υ mid ) δυ = 4π × �
−Mυ 2mid
M 3�2 2
� υ mid exp �
� δυ
2πRT
2RT
2 × 14.01 × 10−3 kg mol−1
= 4π × �
�
2π × (8.3145 J K−1 mol−1 ) × (400 K)
× exp �
3�2
× (205 m s−1 )2
−(2 × 14.01 × 10−3 kg mol−1 ) × (205 m s−1 )2
� × (10 m s−1 )
2 × (8.3145 J K−1 mol−1 ) × (400 K)
= 6.87 × 10−3
where 1 J = 1 kg m2 s−2 has been used. �us, �.���% of molecules have velocities in this range.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�B.�(a)
�e mean relative speed is given by [�B.��b–��], υ rel = (8kT�πµ)1�2 , where
µ = m A m B �(m A + m A ) is the e�ective mass. Multiplying top and bottom of
the expression for υ rel by N A and using N A k = R gives υ rel = (8RT�πN A µ)1�2
in which N A µ is the molar e�ective mass. For the relative motion of N� and H�
this e�ective mass is
NA µ =
υ rel = �
M N2 M H2
(2 × 14.01 g mol−1 ) × (2 × 1.0079 g mol−1 )
=
= 1.88... g mol−1
M N2 + M H2 (2 × 14.01 g mol−1 ) + (2 × 1.0079 g mol−1 )
8RT
�
πN A µ
1�2
=�
8 × (8.3145 J K−1 mol−1 ) × (298.15 K)
�
π × (1.88... × 10−3 kg mol−1 )
1�2
= 1832 m s−1
�e value of the e�ective mass µ is dominated by the mass of the lighter molecule,
in this case H� .
Solutions to problems
P�B.�
A rotating slotted-disc apparatus consists of a series of disks all mounted on a
common axle (sha�). Each disc has a narrow radial slot cut into it, and the slots
on successive discs are displaced from one another by a certain angle. �e discs
are then spun at a constant angular speed.
Detector
Source
Selector
Imagine a molecule moving along the direction of the axle with a certain velocity such that it passes through the slot in the �rst disc. By the time the molecule
reaches the second disc the slot in that disc will have moved around, and the
molecule will only pass through the slot if the speed of the molecule is such
that it arrives at the second disc at just the time at which the slot appears in
the path of the molecule. In this way, only molecules with a speci�c velocity
(or, because the slot has a �nite width, a small range of velocities) will pass
through the second slpt. �e velocity of the molecules which will pass through
the second disc is set by the angular speed at which the discs are rotated and
the angular displacement of the slots on successive discs.
�e angular velocity of the discs is 2πv rad s−1 so in time t the discs move
through an angle θ = 2πvt. If the spacing of the discs is d, a molecule with
velocity υ x will take time t = d�υ x to pass from one disc to the next. If the
second slit is set at an angle α relative to the �rst, such a molecule will only pass
through the second slit if
2πv �
d
�=α
υx
hence
υx =
2πvd
α
17
18
1 THE PROPERTIES OF GASES
If the angle α is expressed in degrees, α = π(α ○ �180○ ), this rearranges to
υx =
2πvd
360○ vd
=
○
○
π(α �180 )
α○
With the values given the velocity of the molecules is computed as
υx =
360○ vd 360○ v(0.01 m)
=
= 180v(0.01 m)
α○
2○
�e Maxwell–Boltzmann distribution of speeds in one dimension is given by
[�B.�–��]
m 1�2 −mυ 2x �2k T
f (υ x ) = �
� e
2πkT
�e given data on the intensity of the beam is assumed to be proportional to
f (υ x ): I ∝ f (υ x ) = Af (υ x ). Because the constant of proportionality is not
known and the variation with υ x is to be explored, it is convenient to take
logarithms to give
ln I = ln[Af (υ x )] = ln A + ln �
m 1�2 mυ 2x
� −
2πkT
2kT
A plot of ln I against υ 2x is expected to be a straight line with slope −m�2kT;
such a plot is shown in Fig. �.�.
ν�Hz υ x �m s−1 υ 2x �(104 m2 s−2 ) I(40 K) ln I(40 K) I(100 K) ln I(100 K)
20
36
0.13
0.846
−0.167
0.592
−0.524
40
72
0.52
0.513
−0.667
0.485
−0.724
80
144
2.07
0.069
−2.674
0.217
−1.528
100
180
3.24
0.015
−4.200
0.119
−2.129
120
216
4.67
0.002
−6.215
0.057
−2.865
At both temperatures the data fall on reasonable straight lines, with slope −1.33
at �� K and −0.516 at ��� K.
If the Maxwell–Boltzmann distribution applies the expected slope at �� K is
computed as
−
m
M
83.80 × 10−3 kg mol−1
=−
=−
= −1.26 × 10−4 m−2 s2
2kT
2RT
2 × (8.3145 J K−1 mol−1 ) × (40 K)
where R = N A k has been used. �e expected slope of the above graph is therefore −1.26, which compares reasonably well with that found experimentally.
At ��� K the expected slope is
−
83.80 × 10−3 kg mol−1
= −5.04 × 10−5 m−2 s2
−1
−1
2 × (8.3145 J K mol ) × (100 K)
Again, the expected slope −0.504 compares reasonably well with that found
experimentally.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
ln I
0.0
T = 40 K
T = 100 K
−2.0
−4.0
−6.0
0
1
2
Figure 1.4
P�B.�
3
υ 2x �(104 m2 s−2 )
4
5
�e Maxwell–Boltzmann distribution of speeds in one dimension (here x) is
given by [�B.�–��]
m 1�2 −mυ 2x �2k T
f (υ x ) = �
� e
2πkT
�e �rst task is to �nd an expression for the mean speed, which is found using
∞
[�B.�–��], �υ n � = ∫0 υ n f (υ) dυ. In this case
�υ x � = �
0
∞
υx �
m 1�2 −mυ 2x �2k T
� e
dυ
kT
�e required integral is of the form of G.� from the Resource section
�
0
∞
xe−ax dx =
2
With a = m�2kT the mean speed is
1
2a
m 1�2
1
kT 1�2
� �
�=�
�
kT
2(m�2kT)
2πm
υmean = �υ x � = �
A�er the beam emerges from the velocity selector, f (υ x ) is zero for υ x > υmean .
�e probability distribution is therefore changed and so needs to be re-normalized
such that
υ mean
2
Kx �
e−mυ x �2k T dυ x = 1
0
�is integral is best evaluated using mathematical so�ware which gives
�
υ mean
0
e−mυ x �2k T dυ = �
2
πkT 1�2
1
� erf( √ )
2m
2 π
where erf(x) is the error function. �e normalized distribution is therefore
f new (υ x ) = �
2
2m 1�2
1
�
e−mυ x �2k T
1
πkT
erf( 2√π )
19
20
1 THE PROPERTIES OF GASES
�e new mean speed is computed using this distribution; again this intergral is
best evaluated using mathematical so�ware. Note that the integral extends up
to υmean
υmean, new = �
υ mean
2
2m 1�2
1
�
υ x e−mυ x �2k T dυ x
�
1
πkT
erf( 2√π ) 0
= (1 − e1�4π ) �
2kT
�
πm
= (1 − e1�4π )2υmean
P�B.�
1�2
��� � � � � � � � � ��� � � � � � � � � � �
1
kT 1�2
1
1�4π
=
(1
−
e
)2
�
�
2πm
erf( 2√1 π )
erf( 2√1 π )
υ mean
1
erf( 2√1 π )
�e error function is evaluated numerically to give υmean, new ≈ 0.493 υmean .
�e Maxwell–Boltzmann distribution of speeds in three dimensions is given
by [�B.�–��]
M 3�2 2 −M υ 2 �2RT
f (υ) = 4π �
� υ e
2πRT
with M the molar mass. �e most probable speed is given by [�B.��–��], υmp =
(2RT�M)1�2 . If the interval of speeds, ∆υ is small, the fraction of molecules
with speeds in this range, centred at speed υmp is well-approximated by f (υmp )∆υ.
�e required fraction of molecules with speeds in the range ∆υ around n × υmp
compared to that centred around υmp is given by
2
2
f (n × υmp )∆υ (n × υmp )2 e−M(nυmp ) �2RT
=
= n 2 e−Mυmp (n −1)�2RT
2 �2RT
2
−M
υ
mp
f (υmp )∆υ
υmp
e
2
In taking the ratio, with the exception of the term υ 2 , all of the terms in f (υ)
which multiply the exponential cancel. In this expression the term υmp is replaced by (2RT�M)1�2 to give
2
2
2
2
f (n × υmp )∆υ
= n 2 e−M υmp (n −1)�2RT = n 2 e−M(2RT�M)(n −1)�2RT = n 2 e(1−n )
f (υmp )∆υ
P�B.�
For n = 3 this expression evaluates to 3.02 × 10−3 and for n = 4 it evaluates
to 4.89 × 10−6 . �ese numbers indicate that very few molecules have speeds
several times greater than the most probable speed.
�e key idea here is that for an object to escape the gravitational �eld of the
Earth it must acquire kinetic energy equal in magnitude to the gravitational
potential energy the object experiences at the surface of the Earth. �e gravitational potential energy between two objects with masses m 1 and m 2 when
separated by a distance r is
V =−
Gm 1 m 2
r
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
where G is the (universal) gravitational constant. In the case of an object of
mass m at the surface of the Earth, it turns out that the gravitational potential
energy is given by
GmM
V =−
R
where M is the mass of the Earth and R its radius. �is expression implies that
the potential at the surface is the same as if the mass of the Earth were localized
at a distance equal to its radius.
As a mass moves away from the surface of the Earth the potential energy increases (becomes less negative) and tends to zero at large distances. If the mass
is to escape its kinetic energy must be greater than or equal to this change in
potential energy. A mass m moving at speed υ has kinetic energy 12 mυ 2 ; this
speed will be the escape velocity υ e when
GmM
1
mυ 2e =
2
hence
R
υe = �
2GM 1�2
�
R
�e quantity in the square root is related to the acceleration due to free fall,
g, in the following way. A mass m at the surface of the Earth experiences
a gravitational force given GMm�R 2 (note that the force goes as R −2 ). �is
force accelerates the mass towards the Earth, and can be written mg. �e two
expressions for the force are equated to give
GMm
= mg
R2
hence
GM
= gR
R
(�.�)
�is expression for GM�R is substituted into the above expression for υ e to give
υe = �
2GM 1�2
� = (2Rg)1�2
R
�e escape velocity is therefore a function of the radius of the Earth and the
acceleration due to free fall.
�e quoted values for the Earth give
�
�
υ e = 2Rg = 2 × (6.37 × 106 m) × (9.81 m s−2 ) = 1.12 × 104 m s−1
For Mars, data is not given on the acceleration due to free fall. However, it
follows from eqn �.� that g = GM�R 2 , and hence
gMars
MMars REarth 2
=
�
�
gEarth MEarth RMars
�e acceleration due to freefall on Mars is therefore computed as
gMars = gEarth
MMars REarth 2
�
�
MEarth RMars
= (9.81 m s−2 ) × (0.108) × �
6.37 × 106 m
� = 3.76... m s−2
3.38 × 106 m
2
21
22
1 THE PROPERTIES OF GASES
�e escape velocity on Mars is therefore
�
�
υ e = 2Rg = 2 × (3.38 × 106 m) × (3.76... m s−2 ) = 5.04 × 103 m s−1
�e mean speed is given by [�B.�–��], υmean = (8RT�πM)1�2 . �is expression
is rearranged to give the temperature T at which the mean speed is equal to the
escape velocity
υ 2 πM
T= e
8R
For H� on the Earth the calculation is
T=
(1.12 × 104 m s−1 )2 × π × (2 × 1.0079 × 10−3 kg mol−1 )
= 1.19 × 104 K
8 × (8.3145 J K−1 mol−1 )
�e following table gives the results for all three gases on both planets
planet υ e �m s−1
Earth 1.12 × 104
Mars 5.04 × 103
T�104 K (H2 )
1.19
0.242
T�104 K (He)
2.36
0.481
T�104 K (O2 )
18.9
3.84
�e fraction of molecules with speed greater than υ e is found by integrating the
Maxwell–Boltzmann distribution from this speed up to in�nity:
fraction with speed ≥ υ e = F = �
∞
υe
4π �
M 3�2 2 −M υ 2 �2RT
� υ e
dυ
2πRT
�is integral is best computed using mathematical so�ware, to give the following results for the fraction F; an entry of zero indicates that the calculated
fraction is zero to within the machine precision.
planet
Earth
Mars
T�K
���
����
���
����
F(H2 )
�
1.49 × 10−4
1.12 × 10−5
�.���
F(He)
�
9.52 × 10−9
5.09 × 10−11
4.31 × 10−2
F(O2 )
�
�
�
4.61 × 10−14
�ese results indicate that the lighter molecules have the greater chance of
escaping (because they are moving faster on average) and that increasing the
temperature increases the probability of escaping (again becuase this increases
the mean speed). Escape from Mars is easier than from the Earth because of
the lower escape velocity, and heavier molecules are seemingly very unlikely to
escape from the Earth.
P�B.�
�e Maxwell–Boltzmann distribution of speeds in three dimensions is given
by [�B.�–��]
M 3�2 2 −M υ 2 �2RT
f (υ) = 4π �
� υ e
2πRT
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e fraction with speed between υ 1 and υ 2 is found by integrating the distribution between these speeds; this is best done using mathematical so�ware
fraction with speed between υ 1 and υ 2 = �
P�B.��
υ2
υ1
4π �
M 3�2 2 −Mυ 2 �2RT
� υ e
dυ
2πRT
At ��� K and with M = 2 × 16.00 g mol−1 the fraction is �.���� and at ���� K
the fraction is �.���� .
Two hard spheres will collide if their line of centres approach within 2r of one
another, where r is the radius of the sphere. �is distance de�nes the collision
diameter, d = 2r, and the collision cross-section is the area of a circle with this
radius, σ = πd 2 = π(2r)2 . �e pressure is computed from the other parameters
using the perfect gas law: p = nRT�V .
�e collision frequency is given by [�B.��b–��], z = σ υ rel p�kT, with the
√ relative
speed for two molecules of the same type given by [�B.��a–��], υ rel = 2υ mean .
�e mean speed is given by [�B.�–��], υ mean = (8RT�πM)1�2 .
Putting this all together gives
σ υ rel p π(2r)2 √
8RT 1�2 nRT
=
× 2×�
� ×
kT
kT
πM
V
1�2
√
8RT
nN A
= π(2r)2 × 2 × �
� ×
πM
V
z=
where to go to the second line R = N A k has been used. �e expression is
evaluated to give
z = π(2×(0.38 × 10
−9
×
1�2
√
8×(8.3145 J K−1 mol−1 )×(298.15 K)
m)) × 2×�
�
π×(16.0416 × 10−3 kg mol−1 )
2
(0.1 mol) × (6.0221 × 1023 mol−1 )
= 9.7 × 1010 s−1
1 × 10−3 m3
1C Real gases
Answer to discussion questions
D�C.�
�e critical constants represent the state of a system at which the distinction
between the liquid and vapour phases disappears. �is situation is usually
described by saying that above the critical temperature the liquid phase cannot
be produced by the application of pressure alone. �e liquid and vapour phases
can no longer coexist, though supercritical �uids have both liquid and vapour
characteristics.
D�C.�
�e van der Waals equation is a cubic equation in the volume V . Every cubic
equation has some values of the coe�cients for which the number of real roots
passes from three to one. In fact, any equation of state of odd degree n > 1 can
in principle account for critical behavior because for equations of odd degree
23
24
1 THE PROPERTIES OF GASES
in V there are necessarily some values of temperature and pressure for which
the number of real roots of V passes from n to �. �at is, the multiple values of
V converge from n to � as the temperature approaches the critical temperature.
�is mathematical result is consistent with passing from a two phase region
(more than one volume for a given T and p) to a one phase region (only one V
for a given T and p), and this corresponds to the observed experimental result
as the critical point is reached.
Solutions to exercises
E�C.�(a)
�e relation between the critical constants and the van der Waals parameters
is given by [�C.�–��]
Vc = 3b
pc =
a
27b 2
Tc =
8a
27Rb
All three critical constants are given, so the problem is over-determined: any
pair of the these expressions is su�cient to �nd values of a and b. It is convenient to use R = 8.2057 × 10−2 dm3 atm K−1 mol−1 and volumes in units of
dm3 .
If the expressions for Vc and p c are used, a and b are found in the following
way
Vc = 3b
hence b = Vc �3 = (0.0987 dm3 mol−1 )�3 = 0.0329 dm3 mol−1
a
a
pc =
=
hence a = 27(Vc �3)2 p c
2
27b
27(Vc �3)2
a = 27(Vc �3)2 p c = 27([0.0987 dm3 mol−1 ]�3)2 × (45.6 atm)
= 1.33 atm dm6 mol−2
�ere are three possible ways of choosing two of the expressions with which to
�nd a and b, and each choice gives a di�erent value. For a the values are �.��,
�.��, and �.��, giving an average of �.�� atm dm6 mol−2 . For b the values are
�.����, �.����, and �.����, giving an average of �.���� dm3 mol−1 .
In Section �C.�(a) on page �� it is argued that b = 4Vmolec N A , where Vmolec is
the volume occupied by one molecule. �is volume is written in terms of the
radius r as 4πr 3 �3 so it follows that r = (3b�16πN A )1�3 .
E�C.�(a)
1�3
3b
3 × (0.0362 dm3 mol−1 )
r=�
� =�
�
16πN A
16π × (6.0221 × 1023 mol−1 )
1�3
= 1.53×10−9 dm = ��� pm
(i) In Section �C.�(b) on page �� it is explained that at the Boyle temperature
Z = 1 and dZ�dp = 0; this latter condition corresponds to the second
virial coe�cient, B or B′ , being zero. �e task is to �nd the relationship
between the van der Waals parameters and the virial coe�cients, and the
starting point for this is the expressions for the product pVm is each case
([�C.�b–��] and [�C.�b–��])
van der Waals: p =
RT
a
− 2
(Vm − b) Vm
hence
pVm =
RT Vm
a
−
(Vm − b) Vm
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
virial: pVm = RT �1 +
B
�
Vm
�e van der Waals expression for pVm is rewritten by dividing the denominator and numerator of the �rst fraction by Vm to give
pVm =
RT
a
−
(1 − b�Vm ) Vm
�e dimensionless parameter b�Vm is likely to be � 1, so the approximation (1 − x)−1 ≈ 1 + x is used to give
a
1
a
= RT �1 +
�b −
��
Vm
Vm
RT
pVm = RT(1 + b�Vm ) −
Comparison of this expression with the virial expansion shows that
B=b−
a
RT
It therefore follows that the Boyle temperature, when B = 0, is Tb = a�Rb.
For the van der Waals parameters from the Resource section
Tb =
a
6.260 atm dm6 mol−2
=
3
Rb (8.2057 × 10−2 dm atm K−1 mol−1 ) × (5.42 × 10−2 dm3 mol−1 )
= 1.41 × 103 K
(ii) In Section �C.�(a) on page �� it is argued that b = 4Vmolec N A , where
Vmolec is the volume occupied by one molecule. �is volume is written
in terms of the radius r as 4πr 3 �3 so it follows that r = (3b�16πN A )1�3 .
r=�
E�C.�(a)
1�3
3b
3 × (5.42 × 10−2 dm3 mol−1 )
� =�
�
16πN A
16π × (6.0221 × 1023 mol−1 )
1�3
= 1.75 × 10−9 dm = ��� pm
�e reduced variables are de�ned in terms of the critical constants,[�C.�–��]
Vr = Vm �Vc
p r = p�p c
Tr = T�Tc
If the reduced pressure is the same for two gases (�) and (�) it follows that
p(1)
=
(1)
pc
and similarly
p(2)
(2)
pc
hence
T (2) =
T (1)
(1)
Tc
p(2) =
× Tc(2)
p(1)
(1)
pc
× p(2)
c
�ese relationships are used to �nd the pressure and temperature of gas (�)
corresponding to a particular state of gas (�); it is necessary to know the critical
constants of both gases.
25
26
1 THE PROPERTIES OF GASES
(i) From the tables in the Resource section, for H� p c = 12.8 atm, Tc = 33.23 K,
and for NH� p c = 111.3 atm, Tc = 405.5 K. Taking gas (�) as H� and
gas (�) as NH� , the pressure and temperature of NH� corresponding to
p(H2 ) = 1.0 atm and T (H2 ) = 298.15 K is calculated as
p(NH3 ) =
T (NH3 ) =
p(H2 )
(H )
pc 2
T (H2 )
(H )
Tc 2
3)
× p(NH
=
c
× Tc(NH3 ) =
1.0 atm
× (111.3 atm) = �.� atm
12.8 atm
298.15 K
× (405.5 K) = 3.6 × 103 K
33.23 K
(ii) For Xe p c = 58.0 atm, Tc = 289.75 K.
p(Xe) =
T (Xe) =
p(H2 )
× p(Xe)
=
c
(H )
pc 2
T (H2 )
(H )
Tc 2
× Tc(Xe) =
1.0 atm
× (58.0 atm) = �.� atm
12.8 atm
298.15 K
× (289.75 K) = 2.6 × 103 K
33.23 K
(iii) For He p c = 2.26 atm, Tc = 5.2 K.
p(He) =
E�C.�(a)
p(H2 )
(H )
pc 2
T (He) =
× p(He)
=
c
T (H2 )
(H )
Tc 2
1.0 atm
× (2.26 atm) = �.�� atm
12.8 atm
× Tc(He) =
298.15 K
× (5.2 K) = �� K
33.23 K
�e van der Waals equation of state in terms of the molar volume is given by
[�C.�b–��], p = RT�(Vm − b) − a�Vm2 . �is relationship is rearranged to �nd b
RT
a
a
RT
−
hence p + 2 =
Vm − b Vm2
Vm Vm − b
pVm2 + a
RT
Vm2
Vm − b
hence
=
hence
=
2
2
Vm
Vm − b
pVm + a
RT
RT Vm2
hence b = Vm −
pVm2 + a
p=
With the data given
b = Vm −
−
RT Vm2
= (5.00 × 10−4 m3 mol−1 )
pVm2 + a
(8.3145 J K−1 mol−1 ) × (273 K) × (5.00 × 10−4 m3 mol−1 )2
(3.0 × 106 Pa) × (5.00 × 10−4 m3 mol−1 )2 + (0.50 m6 Pa mol−2 )
= 4.6 × 10−5 m3 mol−1
where 1 Pa = 1 kg m−1 s−2 and 1 J = 1 kg m2 s−2 have been used.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e compression factor Z is de�ned in [�C.�–��] as Z = Vm �Vm○ , where Vm○ is
the molar volume of a perfect gas under the same conditions. �is volume is
computed from the equation of state for a perfect gas, [�A.�–�], as Vm○ = RT�p,
hence Z = pVm �RT, [�C.�–��]. With the data given
E�C.�(a)
Z=
pVm (3.0 × 106 Pa) × (5.00 × 10−4 m3 mol−1 )
=
= �.��
RT
(8.3145 J K−1 mol−1 ) × (273 K)
�e van der Waals equation of state in terms of the volume is given by [�C.�a–
��], p = nRT�(V − b) − an 2 �V 2 . �e parameters a and b for ethane are
given in the Resource section as a = 5.507 atm dm6 mol−2 and b = 6.51 ×
10−2 dm3 mol−1 .
With these units it is convenient to use R = 8.2057 × 10−2 dm3 atm K−1 mol−1 .
(i) T = 273.15 K, V = 22.414 dm3 , n = 1.0 mol
nRT
an 2
− 2
V − nb V
(1.0 mol) × (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (273.15 K)
=
(22.414 dm3 ) − (1.0 mol) × (6.51 × 10−2 dm3 mol−1 )
(5.507 atm dm6 mol−2 ) × (1.0 mol)2
−
= 0.99 atm
(22.414 dm3 )2
p=
(ii) T = 1000 K, V = 100 cm3 = 0.100 dm3 , n = 1.0 mol
nRT
an 2
− 2
V − nb V
(1.0 mol) × (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (1000 K)
=
(0.100 dm3 ) − (1.0 mol) × (6.51 × 10−2 dm3 mol−1 )
(5.507 atm dm6 mol−2 ) × (1.0 mol)2
−
= 1.8 × 103 atm
(0.100 dm3 )2
p=
E�C.�(a)
Recall that 1 atm = 1.01325 × 105 Pa, 1 dm6 = 10−6 m6 , and 1 Pa = 1 kg m−1 s−2
a = (0.751 atm dm6 mol−2 ) ×
1.01325 × 105 Pa 10−6 m6
×
= 0.0761 Pa m6 mol−2
1 atm
1 dm6
= 0.0760 kg m−1 s−2 m6 mol−2 = 0.0761 kg m5 s−2 mol−2
b = (0.0226 dm3 mol−1 ) ×
E�C.�(a)
10−3 m3
= 2.26 × 10−5 m3 mol−1
1 dm3
�e compression factor Z is de�ned in [�C.�–��] as Z = Vm �Vm○ , where Vm○ is
the molar volume of a perfect gas under the same conditions. �is volume is
computed from the equation of state for a perfect gas, [�A.�–�], as Vm○ = RT�p,
hence Z = pVm �RT [�C.�–��].
27
28
1 THE PROPERTIES OF GASES
(i) If Vm is ��% smaller than the molar volume of a perfect gas, it follows that
Vm = Vm○ (1 − 0.12) = 0.88Vm○ . �e compression factor is then computed
directly as
Vm 0.88 × Vm○
Z= ○ =
= �.��
Vm
Vm○
(ii) From [�C.�–��] it follows that Vm = ZRT�p
Vm =
ZRT 0.88 × (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (250 K)
=
p
15 atm
= �.� dm3 mol−1
E�C.�(a)
Because Z < 1, implying that Vm < Vm○ , attractive forces are dominant.
�e van der Waals equation of state in terms of the volume is given by [�C.�a–
��], p = nRT�(V −b)−an 2 �V 2 . �e molar mass of N� is M = 2×14.01 g mol−1 =
28.02 g mol−1 , so it follows that the amount in moles is
n = m�M = (92.4 kg)�(0.02802 kg mol−1 ) = 3.29... × 103 mol
�e pressure is found by substituting the given parameters into [�C.�a–��],
noting that the volume needs to be expressed in dm3
nRT
an 2
− 2
V − nb V
(3.29... × 103 mol) × (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (500 K)
=
(1000 dm3 ) − (3.29... × 103 mol) × (0.0387 dm3 mol−1 )
(1.352 atm dm6 mol−2 ) × (3.29... × 103 mol)2
−
= 140 atm
(1000 dm3 )2
p=
E�C.�(a)
(i) �e pressure is computed from the equation of state for a perfect gas,
[�A.�–�], as p = nRT�V
nRT (10.0) × (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × ([27 + 273.15] K)
=
V
4.860 dm3
= ��.� atm
p=
(ii) �e van der Waals equation of state in terms of the volume is given by
[�C.�a–��], p = nRT�(V − b) − an 2 �V 2 . �is is used to calculate the
pressure
nRT
an 2
− 2
V − nb V
(10.0 mol) × (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × ([27 + 273.15] K)
=
(4.860 dm3 ) − (10.0 mol) × (0.0651 dm3 mol−1 )
(5.507 atm dm6 mol−2 ) × (10.0 mol)2
−
= 35.2... = 35.2 atm
(4.860 dm3 )2
p=
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e compression factor Z is given in terms of the molar volume and pressure by [�C.�–��], Z = pVm �RT. �e molar volume is V �n
Z=
=
pVm
pV
=
RT
nRT
(35.2... atm) × (4.860 dm3 )
= �.���
(10.0 mol) × (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (300.15 K)
Solutions to problems
P�C.�
�e virial equation is given by [�C.�b–��], pVm = RT(1 + B�Vm + . . .), and
from the Resource section the second virial coe�cient B for N� at ��� K is
−1
−10.5 cm3 mol . �e molar mass of N� is 2 × 14.01 = 28.02 g mol−1 , hence the
molar volume is
Vm =
V
V
2.25 dm3
=
=
= 13.8... dm3 mol−1
n m�M (4.56 g)�(28.02 g mol−1 )
�is is used to calculate the pressure using the virial equation. It is convenient
to use R = 8.2057 × 10−2 dm3 atm K−1 mol−1 and express all the volumes in
dm3
RT
B
�1 +
�
Vm
Vm
(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (273 K)
−1.05 × 10−2 dm3 mol−1
=
�1
+
�
13.8... dm3 mol−1
13.8... dm3 mol−1
= �.�� atm
p=
P�C.�
�e virial equation is [�C.�b–��], pVm = RT(1 + B�Vm + C�Vm2 + . . .). �e compression factor is de�ned in [�C.�–��] as Z = Vm �Vm○ , and the molar volume of
a perfect gas, Vm○ is given by Vm○ = RT�p.
It follows that
Vm = (RT�p)(1 + B�Vm + C�Vm2 ) = Vm○ (1 + B�Vm + C�Vm2 )
Vm
B
C
hence Z = ○ = 1 +
+
Vm
Vm Vm2
To evaluate this expression, the molar volume is approximated by the molar
volume of a perfect gas under the prevailing conditions
Vm○ =
RT (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (273 K)
=
= 0.224... dm3 mol−1
p
100 atm
�is value of the molar volume is then used to compute Z; note the conversion
of all the volume terms to dm3
Z =1+
=1+
B
C
+ 2
Vm Vm
−21.3 × 10−3 dm3 mol−1 1200 × 10−6 dm6 mol−2
+
= 0.928... = �.���
0.224... dm3 mol−1
(0.224... dm3 mol−1 )2
29
30
1 THE PROPERTIES OF GASES
�e molar volume is computed from the compression factor
Z=
hence Vm =
P�C.�
Vm
Vm
=
Vm○ RT�p
ZRT 0.928... × (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (273 K)
=
p
100 atm
= �.��� dm3 mol−1
In Section �C.�(b) on page �� it is explained that at the Boyle temperature
Z = 1 and dZ�dp = 0; this latter condition corresponds to the second virial
coe�cient, B or B′ , being zero. �e Boyle temperature is found by setting the
given expression for B(T) to zero and solving for T
0 = a + be−c�T hence − a�b = e−c�T
2
2
Taking logarithms gives ln(−a�b) = −c�T 2 hence
T =�
P�C.�
−c
�
ln(−a�b)
= ���.� K
1�2
=�
−1131 K2
�
ln[−(−0.1993 bar−1 )�(0.2002 bar−1 )]
1�2
(a) �e molar mass M of H� O is ��.�� g mol−1 . �e mass density ρ is related
to the molar density ρ m by ρ m = ρ�M, and the molar volume is simply
the reciprocal of the molar density Vm = 1�ρ m = M�ρ
Vm =
M 18.02 × 10−3 kg mol−1
=
= 1.352... × 10−4 m3 mol−1
ρ
133.2 kg m−3
�e molar volume is therefore �.���� dm3 mol−1
(b) �e compression factor Z is given by [�C.�–��], Z = pVm �RT
Z=
pVm
(327.6 atm) × (0.1352... dm3 mol−1 )
=
= �.����
RT
(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (776.4 K)
(c) �e virial equation (up to the second term) in terms of the molar volume
is given by [�C.�b–��]
Division of each side by p gives
B
�
Vm
Vm =
RT
B
�1 +
�
p
Vm
pVm = RT �1 +
�e quantity RT�p is recognised as the molar volume of a perfect gas, Vm○ ,
so it follows that
Vm = Vm○ �1 +
B
Vm
B
� hence ○ = Z = �1 +
�
Vm
Vm
Vm
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
In Problem P�C.� it is shown that B is related to the van der Waals constants by B = b − a�RT; using this, it is then possible to compute the
compression factor
a
= (0.03049 dm3 mol−1 )
RT
(5.464 atm dm6 mol−2 )
−
(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (776.4 K)
B=b−
= −0.552... dm3 mol−1
P�C.�
Z =1+
B
−0.552... dm3 mol−1
=1+
= �.����
Vm
0.1352... dm3 mol−1
According to Table �C.� on page ��, for the Dieterici equation of state the
critical constants are given by
pc =
a
4e2 b 2
Vc = 2b
Tc =
a
4bR
From the Resource section the values for Xe are Tc = 289.75 K, p c = 58.0 atm,
−1
Vc = 118.8 cm3 mol . �e coe�cient b is computed directly from Vc
b = Vc �2 = (118.8 × 10−3 dm3 mol−1 )�2 = �.���� dm3 mol−1
�e expressions for p c and Vc are combined to eliminate b
pc =
�is is then rearranged to �nd a
a
a
=
4e2 b 2 4e2 Vc2 �4
a = p c e2 Vc2 = (58.0 atm) × e2 × (118.8 × 10−3 dm3 mol−1 )2
= 6.049 atm dm6 mol−2
Alternatively, the expressions for Tc and Vc are combined to eliminate b
Tc =
�is is then rearranged to �nd a
a = 2Tc Vc R
a
a
=
4bR 4RVc �2
= 2 × (289.75 K) × (118.8 × 10−3 dm3 mol−1 )
× (8.2057 × 10−2 dm3 atm K−1 mol−1 ) = 5.649 atm dm6 mol−2
�e two values of a are not the same; their average is �.��� atm dm6 mol−2 .
From Table �C.� on page �� the expression for the pressure exerted by a Dieterici gas is
nRT exp(−a�[RT V �n])
p=
V − nb
31
32
1 THE PROPERTIES OF GASES
With the parameters given the exponential term evaluates to
−(5.849 atm dm6 mol−2 )
�
(8.2057 × 10−2 dm3 atm K−1 mol−1 )×(298.15 K)×(1.0 dm3 )�(1.0 mol)
= 0.787...
exp �
and hence the pressure evaluates to
(1.0 mol)×(8.2057 × 10−2 dm3 atm K−1 mol−1 )×(298.15 K)×(0.787...)
(1.0 dm3 ) − (1.0 mol)×(0.0594 dm3 mol−1 )
= ��.�� atm
p=
P�C.��
�e van der Waals equation in terms of the molar volume is given by [�C.�b–
��], p = RT�(Vm − b) − a�Vm2 . Multiplication of both sides by Vm gives
pVm =
RT Vm
a
−
(Vm − b) Vm
and then division of the numerator and denominator of the �rst fraction by Vm
gives
RT
a
pVm =
−
(1 − b�Vm ) Vm
�e approximation (1−x)−1 ≈ 1+x+x 2 is the used to approximate 1�(1−b�Vm )
to give
b
b2
a
pVm = RT �1 +
+ 2�−
Vm Vm
Vm
�e terms in 1�Vm and 1�Vm2 are gathered together to give
pVm = RT �1 +
1
a
b2
�b −
�+ 2�
Vm
RT
Vm
�is result is then compared with the virial equation in terms of the molar
volume, [�C.�b–��]
B
C
pVm = RT �1 +
+
�
Vm Vm2
�is comparison identi�es the virial coe�cients as
B=b−
a
RT
C = b2
√
From the given value C = 1200 cm6 mol−2 it follows that b = C = 34.64 cm3 mol−1 .
Expressed in the usual units this is b = �.����� dm3 mol−1 . �e value of a is
found by rearranging B = b − a�RT to
a = RT(b − B) = (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (273 K)×
[(0.03464 dm3 mol−1 ) − (−21.7 × 10−3 dm3 mol−1 )]
= �.��� atm dm6 mol−2
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P�C.��
In Section �C.�(b) on page �� it is explained that critical behaviour is associated
with oscillations in the isotherms predicted by a particular equation of state,
and that at the critical point there is a point of in�exion in the isotherm. At this
point it follows that
dp
d2 p
=0
=0
dVm
dVm2
�e procedure is �rst to �nd expressions for the �rst and second derivatives.
�en these are both set to zero give two simultaneous equations which can be
solved for the critical pressure and volume.
dp
RT 2B 3C
=− 2 + 3 − 4 =0
dVm
Vm Vm Vm
d2 p 2RT 6B 12C
= 3 − 4 + 5 =0
dVm2
Vm
Vm
Vm
�e �rst of these equations is multiplied through by Vm4 and the second by Vm5
to give
−RT Vm2 + 2BVm − 3C = 0
2RT Vm2 − 6BVm + 12C = 0
�e �rst equation is multiplied by � and added to the second, thus eliminating
the terms in Vm2 and giving
4BVm − 6C − 6BVm + 12C = 0
Vm = 3C�B
hence
�is expression for Vm is then substituted into −RT Vm2 + 2BVm − 3C = 0 to give
−RT
(3C)2
3C
+ 2B
− 3C = 0
B2
B
A term 3C is cancelled and the equation is multiplied through by B 2 to give
−RT(3C) + 2B 2 − B 2 = 0
hence
T = B 2 �3RC
Finally the pressure is found by substituting Vm = 3C�B and T = B 2 �3RC into
the equation of state
RT
B
C
−
+
Vm Vm2 Vm3
B2 R B
B3
CB 3
B3
B3
B3
B3
=
−
+
=
−
+
=
2
3
2
2
2
3RC 3C 9C
27C
9C
9C
27C
27C 2
p=
In summary, the critical constants are
P�C.��
Vm = 3C�B
T = B 2 �3CR
p = B 3 �27C 2
�e virial equation in terms of the pressure, [�C.�a–��], is (up to the second
term)
pVm = RT (1 + B′ p)
�e mass density ρ is given by m�V , and the mass m can be written as nM,
where n is the amount in moles and M is the molar mass. It follows that
33
1 THE PROPERTIES OF GASES
ρ = nM�V = M�Vm , where Vm is the molar volume. Rearranging gives Vm =
M�ρ: measurements of the mass density therefore lead to values for the molar
volume.
With this substitution for the molar volume the virial equation becomes
pM
= RT (1 + B′ p)
ρ
hence
p RT
=
(1 + B′ p)
ρ
M
�erefore a plot of p�ρ against p is expected to be a straight line whose slope is
related to B′ ; such a plot is shown in Fig. �.�.
p�kPa
12.22
25.20
36.97
60.37
85.23
101.30
ρ�(kg m−3 )
0.225
0.456
0.664
1.062
1.468
1.734
(p�ρ)�(kPa kg−1 m3 )
54.32
55.26
55.68
56.85
58.06
58.42
58
(p�ρ)�(kPa kg−1 m3 )
34
56
54
0
20
40
60
p�kPa
80
100
Figure 1.5
�e data fall on a reasonable straight line, the equation of which is
(p�ρ)�(kPa kg−1 m3 ) = 0.04610 × (p�kPa) + 53.96
�e slope is B′ RT�M
B′ RT
= 0.04610 kg−1 m3
M
For methoxymethane, CH� OCH� , M = 2 × 12.01 + 6 × 1.0079 + 16.00 =
46.0674 g mol−1 .
B′ =
(0.04610 kg−1 m3 ) × (46.0674 × 10−3 kg mol−1 )
= 8.57 × 10−7 m3 J−1
(8.3145 J K−1 mol−1 ) × (298.15 K)
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e units of the result can be simpli�ed by noting that 1 J = 1 kg m2 s−2 , so
1 m3 J−1 = 1 m kg−1 s2 . Recall that 1 Pa = 1 kg m−1 s−2 , so the units of the B′ are
Pa−1 , an inverse pressure, as expected: B′ = 8.57×10−7 Pa−1 or B′ = 0.0868 atm−1 .
�e virial coe�cient B is found using the result from Problem P�C.��, B = B′ RT
B = B′ RT
= (0.0868 atm−1 ) × (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (298.15 K)
P�C.��
P�C.��
= �.�� dm3 mol−1
A gas can only be lique�ed by the application of pressure if the temperature is
below the critical temperature, which for N� is ���.� K.
�e compression factor is given by [�C.�–��], Z = Vm �Vm○ = Vm p�RT. �e
given equation of state is rearranged to give an expression for Vm a�er putting
n=1
p(V − nb) = nRT
becomes
p(Vm − b) = RT
hence
It follows that the compression factor is given by
Z=
Vm =
RT
+b
p
Vm p (RT�p + b)p
bp
=
= 1+
RT
RT
RT
If Vm = 10b it follows from the previous equation that
Vm p 10bp
bp
=
=1+
RT
RT
RT
hence
b=
RT
9p
With this expression for b the compression factor is computed from Z = 1 +
bp�RT as
Z =1+
P�C.��
bp
RT p
1
=1+
= 1 + = 1.11
RT
9p RT
9
�e virial equation in terms of the molar volume, [�C.�b–��], is (up to the third
term)
pVm = RT �1 +
B
C
�
+
Vm Vm2
For part (a) only the �rst two terms are considered, and it then follows that a
plot of pVm against 1�Vm is expected to be a straight line with slope BRT; such
a plot is shown in Fig. �.�.
35
1 THE PROPERTIES OF GASES
p�MPa Vm �(dm3 mol−1 ) (pVm )�(MPa dm3 mol−1 ) (1�Vm )�(dm−3 mol)
0.400 0
6.220 8
2.488 3
0.160 75
0.500 0
4.973 6
2.486 8
0.201 06
0.600 0
4.142 3
2.485 4
0.241 41
0.800 0
3.103 1
2.482 5
0.322 26
1.000
2.479 5
2.479 5
0.403 31
1.500
1.648 3
2.472 5
0.606 69
2.000
1.232 8
2.465 6
0.811 16
2.500
0.983 57
2.458 9
1.016 7
3.000
0.817 46
2.452 4
1.223 3
4.000
0.609 98
2.439 9
1.639 4
linear
quadratic
(pVm )�(MPa dm3 mol−1 )
36
2.48
2.46
2.44
0.2
Figure 1.6
0.4
0.6
0.8
1.0
(1�Vm )�(dm
−3
1.2
1.4
1.6
mol)
�e data fall on a reasonable straight line, the equation of which is
(pVm )�(MPa dm3 mol−1 ) = −0.03302 × (1�Vm )�(dm−3 mol) + 2.4931
�e slope is BRT
BRT = (−0.03302 MPa dm6 mol−2 )
It is convenient to convert to atm giving BRT = (−0.3259 atm dm6 mol−2 )
hence
(−0.3259 atm dm6 mol−2 )
RT
(−0.3259 atm dm6 mol−2 )
=
(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (300 K)
B=
= −0.01324 dm3 mol−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
For part (b) the data points are �tted to polynomial of order � in 1�Vm using
mathematical so�ware; the data are a slightly better �t to such a function (see
the dashed line in the graph above) which is
(pVm )�(MPa dm3 mol−1 ) =
0.002652 × (1�Vm )2 �(dm−6 mol2 ) − 0.03748 × (1�Vm )�(dm−3 mol) + 2.494
�e coe�cient of the term in (1�Vm )2 is CRT
CRT = (0.002652 MPa dm9 mol−3 )
It is convenient to convert to atm giving CRT = (0.02617 atm dm9 mol−3 )
hence
(0.02617 atm dm9 mol−3 )
RT
(0.02617 atm dm9 mol−3 )
=
(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (300 K)
C=
P�C.��
= 1.063 × 10−3 dm6 mol−2
�e van der Waals equation of state in terms of the molar volume is given by
[�C.�b–��], p = RT�(Vm − b) − a�Vm2 . �is equation is a cubic in Vm , as is seen
by multiplying both sides by (Vm −b)Vm2 and then gathering the terms together
pVm3 − Vm2 (pb + RT) + aVm − ab = 0
From the Resource section the van der Waals parameters for Cl� are
a = 6.260 atm dm6 mol−2
b = 5.42 × 10−2 dm3 mol−1
It is convenient to convert the pressure to atm
p = (150 × 103 Pa) × (1 atm)�(1.01325 × 105 Pa) = 1.4804 atm
and to use R = 8.2057 × 10−2 dm3 atm K−1 mol−1 ; inserting all of these values
and the temperature gives the polynomial
1.4804Vm3 − 20.5946Vm2 + 6.260Vm − 0.3393 = 0
�e roots of this polynomial are found numerically using mathematical so�ware and of these roots only Vm = 13.6 dm3 mol−1 is a physically plausible
value for the molar volume.
�e molar volume of a perfect gas under corresponding conditions is
Vm =
RT (8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (250 K)
=
= 13.9 dm3 mol−1
p
1.48 atm
�e molar volume of the van der Waals gas is about �% smaller than that of
the perfect gas.
37
38
1 THE PROPERTIES OF GASES
Answers to integrated activities
I�.�
In Section �C.�(a) on page �� it is argued that b = 4Vmolec N A , where Vmolec is
the volume occupied by one molecule. �e collision cross-section σ is de�ned
in terms of a collision diameter d as σ = πd 2 , and in turn the diameter is
interpreted as twice the radius of the colliding spheres: d = 2r. It follows that
r = (σ�4π)1�2
b = 4Vmolec N A
4
16πN A σ 3�2
= 4 � πr 3 � N A =
� �
3
3
4π
=
I�.�
16π(6.0221 × 1023 mol−1 ) 0.46 × 10−18 m2
�
�
3
4π
3�2
= 7.1 × 10−5 m3 mol−1 = �.��� dm3 mol−1
According to the equipartition theorem (�e chemist’s toolkit � in Topic �A),
each quadratic contribution to the energy of a molecule contributes 12 kT to
the average energy per molecule. Translational kinetic energy is a quadratic
term, and because translation is possible in three dimensions, the quoted energy density of 0.15 J cm−3 is the result of three such contributions.
�e rotation of a molecule about an axis is also a quadratic contribution to the
energy, and in general three such contributions are expected corresponding to
rotation about three mutually perpendicular axes. However, linear molecules,
such as diatomics, do not show rotation about their long axes, so there are
only two contributions. �e contribution of rotation to the energy density will
therefore be 23 of that due to translation
��� � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
total energy denisty = (0.15 J cm−3 ) + 32 × (0.15 J cm−3 ) = 0.25 J cm−3
trans.
rot.
2
2A
Internal energy
Internal energy
Answers to discussion questions
D�A.�
Table �A.� on page �� lists four varieties of work: expansion, surface expansion, extension, and electrical. �ere is also work associated with processes in
magnetic and gravitational �elds which we will not describe in detail.
D�A.�
An isothermal expansion of a gas may be achieved by making sure that the gas
and its container are in thermal contact with a large ‘bath’ which is held at a
constant temperature – that is, a thermostat.
D�A.�
Work is done when a body is moves against an opposing force. For an in�nitesimal displacement in the x-direction, dx, against a force F along that direction
the work done by the body is F dx.
When the energy of a system changes as a result of a temperature di�erence
between the system and its surroundings, the resulting energy transfer from the
hotter to the cooler body is described as heat. In thermodynamic terms, both
heat and work cause the internal energy of an object to change: if heat ‘�ows in’
the internal energy of the body rises, if the body ‘does work’, its internal energy
decreases.
If the internal energy of an object increases, this is interpreted in molecular
terms as the molecules moving up to higher energy levels. If the molecules
drop down to lower levels the resulting energy is available as heat or work.
Solutions to exercises
E�A.�(a)
�e system is expanding against a constant external pressure, hence the expansion work is given by [�A.�–��], w = −pex ∆V . �e change in volume is the
cross-sectional area times the linear displacement
∆V = (50 cm2 ) × (15 cm) = 750 cm3 = 7.5 × 10−4 m3
�e external pressure is 1.0 atm = 1.01325 × 105 Pa, therefore the expansion
work is
w = −(1.01325 × 105 Pa) × (7.5 × 10−4 m3 ) = −76 J
Note that the volume is expressed in m3 . �e relationships 1 Pa = 1 kg m−1 s−2
and 1 J = 1 kg m2 s−2 are used to verify the units of the result.
40
2 INTERNAL ENERGY
E�A.�(a)
For all cases ∆U = 0, because the internal energy of a perfect gas depends on
the temperature alone.
(i) �e work of reversible isothermal expansion of a perfect gas is given by
[�A.�–��]
w = −nRT ln �
Vf
�
Vi
= −(1.00 mol) × (8.3145 J K−1 mol−1 ) × (293.15 K) × ln �
= −2.68 × 103 J = −�.�� kJ
30.0 dm3
�
10.0 dm3
Note that the temperature is expressed in K in the above equation. Using
the First Law of thermodynamics, [�A.�–��], gives
q = ∆U − w = 0 − (−2.68 kJ) = +�.�� kJ
(ii) �e �nal pressure of the expanding gas is found using the perfect gas law,
[�A.�–�]
pf =
nRT (1.00 mol) × (8.3145 J K−1 mol−1 ) × (293.15 K)
=
Vf
(30.0 × 10−3 m3 )
= 8.12... × 104 Pa
�is pressure equals the constant external pressure against which the gas
is expanding, therefore the work of expansion is
w = −pex × ∆V = (8.12... × 104 Pa) × (30.0 × 10−3 m3 − 10.0 × 10−3 m3 )
= −1.62 × 103 J = −1.62 kJ
and hence q = +1.62 kJ
E�A.�(a)
(iii) Free expansion is expansion against zero force, so w = 0 and therefore
q = 0 as well.
For a perfect gas at constant volume pi �Ti = pf �Tf therefore,
pf = pi ×
Tf
400 K
� = 1.33 atm
= (1.00 atm) × �
Ti
300 K
�e change in internal energy at constant volume is given by [�A.��b–��]
3
∆U = nC V ,m ∆T = (1.00 mol) × � × 8.3145 J K−1 mol−1 � × (400 K − 300 K)
2
= +1.25 × 103 J = +1.25 kJ
�e volume of the gas is constant, so the work of expansion is zero, w = 0 . �e
First Law of thermodynamics gives q = ∆U − w = +1.25 kJ − 0 = +1.25 kJ .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�A.�(a)
(i) �e work of expansion against constant external pressure is given by [�A.�–
��]
w = −pex ∆V
= −(200 Torr) × �
133.3 Pa
� × (3.3 × 10−3 m3 ) = −88 J
1 Torr
Note that the pressure is expressed in Pa and the change in volume in m3 ,
to give the work in J.
(ii) �e work done in a reversible isothermal expansion is given by [�A.�–��],
w = −nRT ln(Vf �Vi ). �e amount in moles of methane is
n=
m
(4.50 g)
=
= 0.280... mol
M (16.0416 g mol−1 )
w = −(0.280... mol) × (8.3145 J K−1 mol−1 ) × (310 K)
× ln �
[12.7 + 3.3] dm3
� = −1.7 × 102 J
12.7 dm3
Note that the modulus of the work done in a reversible expansion is greater
than the work for expansion against constant external pressure because
the latter is an irreversible process.
E�A.�(a)
�e chemist’s toolkit � in Topic �A gives an explanation of the equipartition
theorem. �e molar internal energy is given by
Um = 21 × (νt + νr + 2νv ) × RT
where νt is the number of translational degrees of freedom, νr is the number
of rotational degrees of freedom and νv is the number of vibrational degrees of
freedom. As each gas molecule can move independently along the x, y and z
axis, the number of translational degrees of freedom is three.
(i) Molecular iodine is a diatomic molecule, therefore it has two degrees of
rotational freedom. On account of its heavy atoms, molecular iodine is
likely to have one degree of vibrational freedom at room temperature.
�erefore, the molar internal energy of molecular iodine gas at room
temperature is
Um = 12 × (3 + 2 + 2) × RT = 72 × (8.3145 J K−1 mol−1 ) × (298.15 K)
= 8.7 kJ mol−1
(ii) and (iii) Both methane (tetrahedral) and benzene (planar) have three degrees of
rotational freedom. At room temperature it is unlikely that any of their
vibrational modes would be excited, therefore both are expected to have
approximately the same internal energy at room temperature:
Um = 12 × (3 + 3 + 0) × RT = 3 × (8.3145 J K−1 mol−1 ) × (298.15 K)
E�A.�(a)
= 7.4 kJ mol−1
A state function is a property with a value that depends only on the current state
of the system and is independent of how the state has been prepared. Pressure,
temperature and enthalpy are all state functions.
41
42
2 INTERNAL ENERGY
Solutions to problems
P�A.�
From the equipartition theorem (�e chemist’s toolkit � in Topic �A), the molar
internal energy is given by
Um = 12 × (νt + νr + 2νv ) × RT
where νt is the number of translational degrees of freedom, νr is the number
of rotational degrees of freedom and νv is the number of vibrational degrees
of freedom. Each gas molecule can move independently along the x, y and z
axis giving rise to three translational degrees of freedom. Carbon dioxide is
a linear molecule therefore it has two rotational degrees of freedom. None of
the vibrational modes of carbon dioxide are likely to be signi�cantly excited at
room temperature.
U = 12 × (3 + 2 + 0) × RT = 52 × (8.3145 J K−1 mol−1 ) × (298.15 K)
P�A.�
= 6.2 kJ mol−1
�e de�nition of work is given by [�A.�–��], dw = −�F�dz. Integrating both
sides gives:
� dw = �
w=�
l
0
0
l
F(x) dx
kf (x)x dx = �
l
0
�a − bx 2 � x dx
1
1
2 5 l
1
2 5
= � ax 2 − bx 2 � = al 2 − bl 2
2
5
2
5
0
Note that the second term arises from the non Hooke’s Law behaviour of the
elastomer, reducing the overall work done.
P�A.�
(a) �e natural logarithm can be expanded using the Taylor series as ln(1 +
ν) ≈ ν + ν 2 �2! + ν 3 �3! + ..., which, for ν << 1, can be approximated as
ln(1 + ν) ≈ ν, and similarly, ln(1 − ν) ≈ −ν. �erefore,
F=
kT
kT
νkT
[ln(1 + ν) − ln(1 − ν)] ≈
[ν − (−ν)] =
2l
2l
l
Because ν = n�N, it follows that
F=
νkT nkT
=
l
Nl
(b) Hooke’s law predicts F = const × x, that is the restoring force is directly
proportional to the displacement. Using n = x�l, the expression for the
force obtained in part (a) is rewritten as
nkT kTx
=
≡ const × x
Nl
Nl2
�erefore Hooke’s law applies and kT�N l 2 is the force constant.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P�A.�
�e van der Waals equation of state is given by [�C.�b–��],
nRT
n2 a
− 2
V − nb V
�e expansion work is given by [�A.�–��], dw = −pex dV . For a reversible
expansion pex is always equal to the pressure of the gas so
p=
dw = −pgas dV = − �
Integrating both sides give
nRT n 2 a
− 2 � dV
V
V
nRT
n2 a
− 2 dV
V
Vi V − nb
Vf
Vf dV
dV
= −nRT �
+ n2 a �
Vi V − nb
Vi V 2
Vf − nb
1
1
= −nRT ln �
� − n2 a � − �
Vi − nb
Vf Vi
w = −�
Vf
�e work done during the isothermal reversible expansion in the various cases
is calculated below and the results portrayed in Fig. �.�.
(a) A perfect gas.
Vf
� = −(1.0 mol) × (8.3145 J K−1 mol−1 ) × (298 K)
Vi
2.0 dm3
× ln �
� = −1.7 kJ
1.0 dm3
w = −nRT ln �
(b) A van der Waals gas in which repulsive forces dominate.
Using the general expression derived above with a = 0 and b = 5.11 ×
10−2 dm3 mol−1
w = −(1.0 mol) × (8.3145 J K−1 mol−1 ) × (298 K)
× ln �
(2.0 dm3 ) − (1.0 mol) × (5.11 × 10−2 dm3 mol−1 )
� = −1.8 kJ
(1.0 dm3 ) − (1.0 mol) × (5.11 × 10−2 dm3 mol−1 )
(c) A van der Waals gas in which attractive forces dominate.
−2
Using the general expression derived above with a = 4.2 dm6 atm mol
and b = 0. Constant a in SI units is
−2
a = (4.2 dm6 atm mol ) × �
= 0.425... m6 Pa mol
−2
1 m6
1.01325 × 105 Pa
�
6��
6
1 atm
10 dm
2.0 dm3
�
1.0 dm3
−2
1
1
− (1.0 mol)2 × (4.25... × 105 dm6 Pa mol ) × �
�
3 −
2.0 dm
1.0 dm3
= −1.71...kJ + 0.212...kJ = −1.5 kJ
w = −(1.0 mol) × (8.3145 J K−1 mol−1 ) × (298 K) × ln �
43
2 INTERNAL ENERGY
30
perfect
repulsion only
attraction only
25
p�atm
44
20
15
10
0.5
1.0
1.5
V �dm
Figure 2.1
P�A.�
2.0
2.5
3
(a) �e virial equation of state is given by [�C.�b–��]. �e �rst three terms
are
Using Vm = V �n gives
p = RT �
1
B
C
+
+
�
Vm Vm2 Vm3
p = nRT �
1 nB n 2 C
+
+ 3 �
V V2
V
�e work of expansion is calculated as
w = −�
Vf
Vi
pdV = − �
= −nRT ln
Vf
Vi
nRT �
1 nB n 2 C
+
+ 3 � dV
V V2
V
Vi
1
1
1
1
1
+ n 2 RTB � − � + n 3 RTC � 2 − 2 �
Vf
Vf Vi
2
Vf
Vi
−1
From Table �C.�, at ��� K, B = −21.7 cm3 mol
−2
C = 1.2 × 103 cm6 mol , therefore
and from the problem
nRT = (1.0 mol) × (8.3145 J K−1 mol−1 ) × (273 K) = 2.26... × 103 J
n 2 RTB = n × (nRT) × B = (1.0 mol) × (2.26... × 103 J)
−1
× (−21.7 cm3 mol )
= −4.92... × 104 J cm3
1 3
n RTC = 12 × n 2 × (nRT) × C
2
−2
= 12 × (1.0 mol)2 × (2.26... × 103 J) × (1.2 × 103 cm6 mol )
= 1.36... × 106 J cm6
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
1000 cm3
�
500 cm3
1
1
+ (−4.92... × 104 J cm3 ) × �
−
�
1000 cm3 500 cm3
1
1
+ (1.36... × 106 J cm6 ) × �
−
�
3
2
(1000 cm )
(500 cm3 )2
w = −(2.26... × 103 J) × ln �
= −1.5 kJ
(b) For a perfect gas, the work of reversible isothermal expansion is given by
[�A.�–��]
w = −nRT ln �
Vf
�
Vi
= −(1.0 mol) × (8.3145 J K−1 mol−1 ) × (273 K) × ln �
= −1.6 × 103 J = −1.6 kJ
1000 cm3
�
500 cm3
2B Enthalpy
Answers to discussion questions
D�B.�
If a substance is heated at constant volume all of the energy as heat is transformed into internal energy of the substance. If the same process is carried out
under conditions of constant pressure some of the energy as heat will be used
to expand the substance against the external pressure and so less of the energy
as heat is transformed into internal energy. �is e�ect is largest for gases whose
volumes change much more rapidly with temperature than do solids or liquids.
�e temperature of a substance is related to its internal energy, and therefore
as energy as heat is supplied the temperature increases. For a given amount
of energy as heat this increase in internal energy is smaller for the constantpressure case than for the constant-volume case. �erefore the rise in temperate is smaller for the constant-pressure process, and this implies that the heat
capacity is greater.
Solutions to exercises
E�B.�(a)
(i) �e heat capacity can be expressed as C p = a + bT where a = 20.17 J K−1
and b = 0.3665 J K−2 . Integrating the relationship dH = C p dT on both
45
46
2 INTERNAL ENERGY
sides gives
�
T2
T1
dH = �
T2
T1
C p dT = �
T2
(a + bT)dT = �aT + 12 bT 2 �T
T1
H(T2 )−H(T1 ) = na(T2 − T1 ) + 12 nb(T22 − T12 )
−1
T2
1
= (20.17 J K ) × (373.15 K − 298.15 K)
+ 12 × (0.3665 J K−2 ) × [(373.15 K)2 − (298.15 K)2 ]
= +10.7... kJ = +10.7 kJ
Under constant pressure conditions ∆H = qp = +10.7 kJ .
�e work of expansion against constant pressure pex is given by [�A.�–
��], w = −pex ∆V = −pex (Vf − Vi ). Assume that the gas is in mechanical
equilibrium with its surroundings, therefore pex is the same as the pressure of the gas, p. �e initial and �nal volumes are calculated from Tf and
Ti by Vf = nRTf �p and Vi = nRTi �p, therefore Vf − Vi = (Tf − Ti )nR�p.
Hence
nR
w = −p ×
(Tf − Ti ) = −nR∆T
p
= −(1.00 mol) × (8.3145 J K−1 mol−1 ) × (373.15 K − 298.15 K)
= −6.23... × 102 J = −624 J
∆U = q + w = (+10.7... kJ) + (−0.623... kJ) = +10.1 kJ
E�B.�(a)
(ii) �e energy and enthalpy of a perfect gas depends on the temperature
alone, hence ∆H and ∆U is the same as above, that is ∆H = +10.7 kJ and
∆U = +10.1 kJ . Under constant volume conditions there is no expansion
work, w = 0 , therefore the heat is equal to the change in internal energy,
qV = +10.1 kJ .
Under constant pressure qp = ∆H, therefore
qp = ∆H = nC p,m ∆T = (3.0 mol) × (29.4 J K mol−1 ) × (285 K − 260 K)
= 2.20... kJ = +2.2 kJ
�e de�nition of enthalpy is given by [�B.�–��], H = U + pV . For a change
at constant pressure, it follows that ∆H = ∆U + p∆V , where ∆V = Vf − Vi .
If the gas is assumed to be perfect, then Vf = nRTf �p and Vi = nRTi �p, so
∆V = (Tf − Ti )nR�p. Hence p∆V = nR(Tf − Ti ) = nR∆T.
∆U = ∆H − nR∆T
= (2.20... kJ) − (3.0 mol) × (8.3145 J K−1 mol−1 ) × (285 K − 260 K)
E�B.�(a)
= +1.6 kJ
�e heat transferred under constant pressure equals the change in enthalpy
of the system, [�B.�b–��], qp = ∆H. �e relationship between the change in
enthalpy, change in temperature and the heat capacity is given by [�B.�b–��]
C p,m =
∆H
(229 J)
=
= 29.9... J K mol−1 = 30 J K−1 mol−1
n∆T (3.0 mol) × (2.55 K)
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
For a perfect gas C p,m − C V ,m = R, [�B.�–��], therefore
C V ,m = C p,m −R = (29.9... J K−1 mol−1 )−(8.3145 J K−1 mol−1 ) = 22 J K−1 mol−1
E�B.�(a)
In the reaction � moles of gas are formed from � moles of gas. �e relationship
between ∆H and ∆U is given by [�B.�–��]
∆Hm −∆Um = ∆ng RT = (−2.0)×(8.3145 J K−1 mol−1 )×(298 K) = −5.0 kJ mol−1
Solutions to problems
P�B.�
�� g of benzene is n = m�M = (10 g)�(78.1074 g mol−1 ) = 0.128... mol. �e
heat needed to vaporize �� g of benzene under constant pressure is qp = n ×
∆ vap Hm = (0.128... mol) × (30.8 kJ mol−1 ) = 3.94... kJ. A current I �owing
through a potential ∆� corresponds to a power of I∆�. If the current �ows for
a time interval ∆t the energy is power × time , that is q = I∆�∆t. �erefore
∆t = q�(I∆�).
∆t =
P�B.�
qp
(3.94... × 103 J)
=
= 657 s = 11 min
I∆� (0.50 A) × (12 V)
�e relationships 1 A = 1 C s−1 and 1 V = 1 J C−1 are used in the last step.
Fitting the data set using a computer program to an expression in the form of
○
C −p,m
(T) = a+bT+cT −2 yields a = 48.0 J K−1 mol−1 , b = 6.49×10−3 J K−2 mol−1
and c = −9.33 × 105 J K mol−1 . �e change in enthalpy in response to a change
in temperature at constant pressure is given by [�B.�a–��], dHm = C p,m dT.
Substituting in the expression for C p,m and integrating both sides gives
�
(T2 )
(T1 )
dHm = (a + bT + cT −2 ) dT
dHm = �
T2
T1
(a + bT + cT −2 ) dT
By using Integral A.� in the Resource section for each term, it follows that
1
1
− �
T2 T1
= (48.0 J K−1 mol−1 ) × (1500 K − 298.15 K)
Hm (T2 ) − Hm (T1 ) = a(T2 − T1 ) + 12 b(T22 − T12 ) − c �
+ 12 × (6.49 × 10−3 J K−2 mol−1 )
× [(1500 K)2 − (298.15 K)2 ]
− (−9.33 × 105 J K mol−1 ) × �
= 62.2 kJ
1
1
−
�
1500 K 298.15 K
47
48
2 INTERNAL ENERGY
P�B.�
Since the volume is �xed no expansion work is done, hence w = 0 and
∆U = qV = +2.35 kJ . From H = U + pV it follows that ∆H = ∆U + V ∆p at
constant volume. Using the van der Waals equation of state, [�C.�b–��],
p=
RT
a
− 2
Vm − b Vm
it follows that
∆p =
R∆T
Vm − b
Note that the term in a cancels because the volume is constant. �erefore
RV ∆T
∆H = ∆U + v∆p = ∆U +
Vm − b
From the given data Vm = V �n = (15.0 dm3 )�(2.0 mol) = 7.5 dm3 mol−1 ,
∆T = 341 K − 300 K = 41 K and from Table �C.� b = 4.29 × 10−2 dm3 mol−1 .
�erefore
∆H = (+2.35 × 103 J) +
(8.3145 J K−1 mol−1 ) × (15.0 dm3 ) × (41 K)
(7.5 dm3 mol−1 ) − (4.29 × 10−2 dm3 mol−1 )
= 3.03... × 103 J = 3.0 kJ
2C Thermochemistry
Answers to discussion questions
D�C.�
When a system is subjected to constant pressure conditions, and only expansion work can occur, the energy supplied as heat is the change in enthalpy of the
system. �us enthalpy changes in the system can be determined by measuring
the amount of heat supplied under constant-pressure conditions.
A very simple example o�en encountered in elementary laboratory classes is a
thermally insulated vessel (for example, a foam plastic co�ee cup) le� open to
the atmosphere: the heat released in the reaction is determined by measuring
the change in temperature of the contents.
For a combustion reaction a constant-pressure �ame calorimeter (Section �B.�(b)
on page ��) may be used. In this apparatus a certain amount of substance
burns in a supply of oxygen and the rise in temperature is monitored. More
sophisticated methods include isothermal titration calorimetry and di�erential
scanning calorimetry, both described in Section �C.� on page ��.
D�C.�
�e main objection to the use of the term ‘heat’ to describe the energy change
associated with a physical or chemical process is that heat is not a state function.
�e value of the heat therefore depends on the path chosen.
If, in fact, the processes being described takes place at constant pressure, the
heat is equal to the enthalpy change. Because enthalpy is a state function, the
heat measured under these circumstance is a property of the physical or chemical change itself, and not a�ected by the path taken, and so is a meaningful and
useful quantity to discuss.
It is more appropriate to talk about the enthalpy change, the change in a state
function, rather to talk about the heat which, because of an unstated restriction,
just so happens to have the same value.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Solutions to exercises
E�C.�(a)
�e relationship between ∆ r H and ∆ r U is given by [�B.�–��], ∆ r H = ∆ r U +
∆ng RT where ∆ng is the change in the amount of gas molecules in the reaction.
For this reaction ∆ng = 2 mol + 3 mol − 3 mol = +2 mol
∆ r H −○ = ∆ r U −○ + ∆ng RT
= (−1373 × 103 J mol−1 ) + (+2.0 mol) × (8.3145 J K−1 mol−1 ) × (298 K)
= −1368 kJ mol−1
E�C.�(a)
(i) �e standard reaction enthalpy at 298 K is calculated from the standard
enthalpies of formation at 298 K using [�C.�a–��]
∆ r H −○ (298 K) =
−
○
� ν∆ f H −
products
�
reactants
ν∆ f H −○
= ∆ f H −○ (CO, g) + ∆ f H −○ (H2 , g) − ∆ f H −○ (H2 O, g) − ∆ f H −○ (graphite, s)
= (−110.53 kJ mol−1 ) + 0 − (−241.82 kJ mol−1 ) − 0
= +131.29 kJ mol−1
�e relationship between ∆ r H and ∆ r U is given by [�B.�–��], ∆ r H =
∆ r U + ∆ng RT where ∆ng is the change in the amount of gas molecules
in the reaction. For this reaction ∆ng = 1 mol + 1 mol − 1 mol = +1 mol
∆ r U −○ (298 K) = ∆ r H −○ (298 K) − ∆ng RT
= (+131.29 × 103 J mol−1 )
− (+1.0 mol) × (8.3145 J K−1 mol−1 ) × (298 K)
= +128.81 kJ mol−1
(ii) �e di�erence of the molar heat capacities of products and reactants is
calculated using [�C.�b–��] and the data from Table �C.�
∆ r C −p○ =
−
○
� νC p,m −
products
�
reactants
○
νC −p,m
○
○
○
○
= C −p,m
(CO, g) + C −p,m
(H2 , g) − C −p,m
(H2 O, g) − C −p,m
(graphite, s)
= (29.14 J K−1 mol−1 ) + (28.824 J K−1 mol−1 ) − (33.58 J K−1 mol−1 )
− (8.527 J K−1 mol−1 ) = +15.86 J K−1 mol−1
It is assumed that all heat capacities are constant over the temperature
range of interest, therefore the integrated form of Kirchho� ’s Law is ap-
49
50
2 INTERNAL ENERGY
plicable, [�C.�d–��]
∆ r H −○ (478 K) = ∆ r H −○ (298 K) + ∆T∆ r C −p○
= (+131.29 × 103 J mol−1 )
+ (478 K − 298 K) × (+15.86 J K−1 mol−1 )
= +134.1 kJ mol−1
∆ r U (478 K) = ∆ r H −○ (478 K) − ∆ng RT
−
○
= (+134.1 × 103 J mol−1 )
− (+1.0 mol) × (8.3145 J K−1 mol−1 ) × (478 K)
= +130 kJ mol−1
E�C.�(a)
�e reaction equation C(graphite, s) + O� (g) ��→ CO� (g) represents the formation of CO� (g) from its elements in their reference states, therefore ∆ r H −○ (298 K) =
∆ f H −○ (CO2 , g) = −393.51 kJ mol−1 . �e variation of standard reaction enthalpy with temperature is given by Kirchho� ’s Law, [�C.�a–��]
∆ r H −○ (T2 ) = ∆ r H −○ (T1 ) + �
T2
T1
∆ r C −p○ dT
�e di�erence of the molar heat capacities of products and reactants is given
by [�C.�b–��]
∆ r C −p○ =
−
○
� νC p,m −
products
�
reactants
○
νC −p,m
○
○
○
= C −p,m
(CO2 , g) − C −p,m
(O2 , g) − C −p,m
(graphite, s)
○
�e heat capacities in Table �B.� are expressed in the form of C −p,m
= a + bT +
2
−
○
2
c�T therefore ∆ r C p = ∆a + ∆bT + ∆c�T where ∆a = a(CO2 , g) − a(O2 , g) −
a(graphite, s) and likewise for ∆b and ∆c.
∆a = (44.22 J K−1 mol−1 ) − (29.96 J K−1 mol−1 ) − (16.86 J K−1 mol−1 )
= −2.60 J K−1 mol−1
∆b = (8.79 × 10−3 J K−2 mol−1 ) − (4.18 × 10−3 J K−2 mol−1 )
− (4.77 × 10−3 J K−2 mol−1 ) = −0.16 × 10−3 J K−2 mol−1
∆c = (−8.62 × 105 J K mol−1 ) − (−1.67 × 105 J K mol−1 )
− (−8.54 × 105 J K mol−1 ) = +1.59 × 105 J K mol−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Integrating Kirchho� ’s Law gives
∆ r H −○ (T2 ) = ∆ r H −○ (T1 ) + �
T2
T1
�∆a + ∆bT +
∆c
� dT
T2
1
1
− �
T2 T1
= (−393.51 × 103 J mol−1 ) + (−2.60 J K−1 mol−1 ) × (500 K − 298 K)
= ∆ r H (T1 ) + ∆a(T2 − T1 ) + 12 ∆b(T22 − T12 ) − ∆c �
−
○
+ 12 × (−0.16 × 10−3 J K−2 mol−1 ) × [(500 K)2 − (298 K)2 ]
1
1
− (1.59 × 105 J K mol−1 ) × �
−
�
500 K 298 K
E�C.�(a)
= −394 kJ mol−1
Tetrachloromethane is vaporized at constant pressure, therefore q = ∆H.
q = ∆H = n∆ vap H −○ = (0.75 mol) × (30.0 kJ mol−1 ) = +22.5 kJ
�e work of expansion under constant pressure is given by [�A.�–��], w =
−pex ∆V . Note that ∆V = Vf because the �nal state (gas) has a much larger
volume than the initial state (liquid). �e perfect gas law is used to calculate
Vf .
w = −pex ∆V = −pex (Vf − Vi ) ≈ −pex Vf = −pex ×
nRT
= −nRT
pex
= −(0.75 mol) × (8.3145 J K−1 mol−1 ) × (250 K) = −1.55... kJ = −1.6 kJ
�e First Law of thermodynamics [�A.�–��] gives
∆U = q + w = (+22.5 kJ) + (−1.55... kJ) = +21 kJ
E�C.�(a)
�e equation for combustion of ethylbenzene is C� H�� (l) + ��.� O� (g) ��→
� CO� (g) + � H� O(l). �e standard enthalpy of combustion is calculated using [�C.�a–��] and using values for the standard enthalpies of formation from
Table �C.�.
∆ c H −○ =
−
○
� ν∆ f H −
products
�
reactants
ν∆ f H −○
= 8∆ f H −○ (CO2 , g) + 5∆ f H −○ (H2 O, l) − 21
∆ H −○ (O2 , g) − ∆ f H −○ (C8 H10 , l)
2 f
= 8 × (−393.51 kJ mol−1 ) + 5 × (−285.83 kJ mol−1 ) − 0 − (−12.5 kJ mol−1 )
E�C.�(a)
= −4.57 × 103 kJ mol−1
�e standard enthalpy of formation of HCl(aq) is ∆ r H −○ for the reaction
1
H (g) + 12 Cl2 (g) ��→ HCl(aq)
2 2
51
52
2 INTERNAL ENERGY
Because HCl is a strong acid, the reaction is e�ectively
1
H (g) + 12 Cl2 (g) ��→ H+ (aq) + Cl− (aq)
2 2
By de�nition, ∆ f H −○ (H+ , aq) = 0, so ∆ r H −○ for this reaction is ∆ f H −○ (Cl− , aq).
E�C.�(a)
∆ r H −○ = ∆ f H −○ (Cl− , aq) = −167 kJ mol−1
�e equation for the combustion of naphthalene is C�� H� (s) + �� O� (g) ��→
�� CO� (g) + � H� O(l). ∆ c H −○ is computed using the thermochemical data from
Tables �C.� and �C.� and equation [�C.�a–��]
∆ c H −○ =
−
○
� ν∆ f H −
products
�
reactants
ν∆ f H −○
= 10∆ f H −○ (CO2 , g) + 4∆ f H −○ (H2 O, l)
− 12∆ f H −○ (O2 , g) − ∆ f H −○ (C10 H8 , s)
= 10 × (−393.51 kJ mol−1 ) + 4 × (−285.83 kJ mol−1 )
− 0 − (+78.53 kJ mol−1 ) = −5.16 × 103 kJ mol−1
In a bomb calorimeter the heat is at constant volume and is given by qv =
n∆ c U −○ . ∆ c U −○ is related to ∆ c H −○ by [�B.�–��], ∆ r H −○ = ∆ r U −○ + ∆ng RT, where
∆ng is the change in the amount of gas molecules in the reaction. In this case
∆ng = 10 mol − 12 mol = −2 mol, hence
∆ c U −○ = ∆ c H −○ − ∆ng RT
= (−5.15... × 106 J) − (−2.0 mol) × (8.3145 J K−1 mol−1 ) × (298 K)
= −5.15... × 103 kJ mol−1
�e heat released in the bomb calorimeter on combustion of 120 mg of naphthalene (M = 128.1632 g mol−1 ) is
qv = n∆ c U −○ = �
0.120 g
� × (−5.15... × 103 kJ mol−1 ) = −4.82 ...kJ
128.1632 g mol−1
�erefore the calorimeter constant is
C=
�qv � (4.82... kJ)
=
= 1.58... kJ K−1 = 1.58 kJ K−1
∆T
(3.05 K)
�e chemical equation for combustion of phenol is C� H� O(s) + � O� (g) ��→
� CO� (g)+� H� O(l). Following the same logic as above, ∆ c H −○ = −3054 kJ mol−1 ,
∆ng = −1 mol and ∆ c U −○ = −3.05... × 103 kJ mol−1 . �e heat released on
combustion of 150 mg of phenol (M = 94.10 g mol−1 ) is
qv = n∆ c U −○ = �
0.150 g
� × (−3.05... × 103 kJ mol−1 ) = −4.86... kJ
94.1074 g mol−1
�erefore ∆T = �qv ��C = (4.86... kJ)�(1.58...kJ K−1 ) = +3.07 K
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�C.�(a)
(i) Reaction(�) is reaction(�) − � × reaction(�), therefore
∆ r H −○ (3) = ∆ r H −○ (2) − 2∆ r H −○ (1)
= (−483.64 kJ mol−1 ) − 2 × (−184.62 kJ mol−1 )
= −114.40 kJ mol−1
�e relationship between ∆ r H and ∆ r U is given by [�B.�–��], ∆ r H =
∆ r U + ∆ng RT where ∆ng is the change in the amount of gas molecules
in the reaction. For this reaction ∆ng = 2 mol + 2 mol − 4 mol − 1 mol =
−1 mol
∆ r U −○ = ∆ r H −○ − ∆ng RT
= (−114.40 × 103 J mol−1 )
− (−1.0 mol) × (8.3145 J K−1 mol−1 ) × (298 K) = −112 kJ mol−1
(ii) Reaction(�) represents the formation of � moles of HCl(g) from its elements in their reference states, therefore the standard enthalpy of formation of HCl(g) is
∆ f H −○ (HCl, g) = 12 ∆ r H −○ (1) = 12 ×(−184.62 kJ mol−1 ) = −92.31 kJ mol−1
Reaction(�) represents the formation of � moles of H� O(g) from its elements in their reference states, therefore
∆ f H −○ (H2 O, g) = 12 ∆ r H −○ (2) = 12 ×(−483.64 kJ mol−1 ) = −241.82 kJ mol−1
Solutions to problems
P�C.�
At constant pressure the temperature rise is given by [�B.�–��], qp = C p ∆T.
○
�e heat capacity is approximated as nC −p,m
(H2 O, l) where n is the amount in
moles. �erefore
∆T =
qp
=
○
nC −p,m
�
(1.0 × 107 J)
65000 g
� × (75.29 J K−1 mol−1 )
18.0158 g mol−1
= 36.8... K = 37 K
Assuming that H� O (��� K, l) ��→ H� O (��� K, g) is the main process responsible for heat loss, and using data from Table �C.�, the amount of water to
evaporate is
n=
qp
(1.0 × 107 J)
=
= 2.27... × 102 mol
∆ vap H −○ (44.016 × 103 J mol−1 )
�e corresponding mass is
P�C.�
m = nM = (2.27... × 102 mol) × (18.0158 g mol−1 ) = 4.1 kg
(a) �e combustion equation for cyclopropane is C� H� (g) + �.� O� (g) ��→
� CO� (g) + � H� O(l). Using data from Table �C.� and applying [�C.�a–��]
for this case
∆ c H −○ = 3∆ f H −○ (CO2 , g) + 3∆ f H −○ (H2 O, l) − ∆ f H −○ (C3 H6 , g)
53
54
2 INTERNAL ENERGY
�erefore
∆ f H −○ (C3 H6 , g) = 3∆ f H −○ (CO2 , g) + 3∆ f H −○ (H2 O, l) − ∆ c H −○
= 3 × (−393.51 kJ mol−1 ) + 3 × (−285.83 kJ mol−1 ) − (2091 kJ mol−1 )
= +52.9... kJ mol−1 = +52.98 kJ mol−1
(b) �e reaction is cyclopropane ��→ propene
∆ r H −○ = ∆ f H −○ (propene, g) − ∆ f H −○ (cyclopropane, g)
P�C.�
= (+20.42 kJ mol−1 ) − (+52.9... kJ mol−1 ) = −32.56 kJ mol−1
�e combustion of 0.825 g of benzoic acid (M = 122.1174 g mol−1 ) is used to
determine the calorimeter constant. �e heat released in the calorimeter is
qv = n∆ c U −○ = �
0.825 g
� × (−3251 kJ mol−1 ) = −21.9... kJ
122.1174 g mol−1
�erefore the calorimeter constant is
C=
�qv � (21.9... kJ)
=
= 11.3... kJ K−1
∆T
(1.940 K)
It follows that the heat released during the combustion of ribose is �qv � = C∆T =
(11.3... kJ K−1 ) × (0.910 K) = 10.3... kJ, and so the standard internal energy of
combustion is
∆ c U −○ =
−�qv �
−(10.3... kJ)
=
= −2.12... × 103 kJ mol−1
n
(0.727 g)�(150.129 g mol−1 )
�e combustion reaction is C� H�� O� (s) + � O� (g) ��→ � CO� (g) + � H� O(l).
�ere is no change in the number of gaseous species when going from reactants
to products, therefore ∆ c U −○ = ∆ c H −○ = −2.12... × 103 kJ mol−1 . Applying
[�C.�a–��] for this reaction and using the values for the standard enthalpies
of formation from Table �C.� gives
∆ c H −○ = 5∆ f H −○ (CO2 , g) + 5∆ f H −○ (H2 O, l) − 5∆ f H −○ (O2 , g)
− ∆ f H −○ (C5 H10 O5 , s)
∆ f H −○ (C5 H10 O5 , s) = 5∆ f H −○ (CO2 , g) + 5∆ f H −○ (H2 O, l)
− 5∆ f H −○ (O2 , g) − ∆ c H −○
= 5 × (−393.51 kJ mol−1 ) + 5 × (−285.83 kJ mol−1 )
− 0 − (−2.12... × 103 kJ mol−1 )
P�C.�
= −1.27 × 103 kJ mol−1
�e chemical reaction for the combustion is C�� (s) + �� O� (g) ��→ �� CO� (g).
�e standard internal energy of combustion is
∆ c U −○ = (−36.0334 kJ g−1 ) × (60 × 12.01 g mol−1 ) = −2.59... × 104 kJ mol−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�ere is no change in the number of gaseous species when going from reactants to products, therefore ∆ c U −○ = ∆ c H −○ = −2.59... × 104 kJ mol−1 =
−25966 kJ mol−1 . Applying [�C.�a–��] for this case gives
∆ c H −○ = 60∆ f H −○ (CO2 , g) − 60∆ f H −○ (O2 , g) − ∆ f H −○ (C60 , s)
�erefore
∆ f H −○ (C60 , s) = 60∆ f H −○ (CO2 , g) − 60∆ f H −○ (O2 , g) − ∆ c H −○
= 60 × (−393.51 kJ mol−1 ) − 0 − (−2.59... × 104 kJ mol−1 )
P�C.�
= +2355.1 kJ mol−1
�e reaction equation for combustion of methane is CH� (g) + � O� (g) ��→
CO� (g) + � H� O(g). �e standard enthalpy of combustion can be calculated
using [�C.�a–��] and using the values for the standard enthalpies of formation
from Tables �C.� and �C.�
∆ c H −○ =
−
○
� ν∆ f H −
products
�
reactants
ν∆ f H −○
= ∆ f H −○ (CO2 , g) + 2∆ f H −○ (H2 O, g) − 2∆ f H −○ (O2 , g) − ∆ f H −○ (CH4 , g)
= (−393.51 kJ mol−1 ) + 2 × (−241.82 kJ mol−1 ) − 0 − (−74.81 kJ mol−1 )
= −802.3 kJ mol−1
�e variation of standard reaction enthalpy with temperature is given by Kirchho� ’s Law, [�C.�a–��]
∆ r H −○ (T2 ) = ∆ r H −○ (T1 ) + �
T2
T1
∆ r C −p○ dT
�e di�erence of the molar heat capacities of products and reactants is given
by [�C.�b–��]
∆ r C −p○ =
−
○
� νC p,m −
products
�
reactants
○
νC −p,m
○
○
○
○
= C −p,m
(CO2 , g) + 2C −p,m
(H2 O, g) − 2C −p,m
(O2 , g) − C −p,m
(CH4 , g)
○
�e heat capacities are expressed in the form of C −p,m
= α + βT + γT 2 therefore
−
○
2
∆ r C p = ∆α + ∆βT + ∆γT where ∆α = α(CO2 , g) + 2α(H2 O, g) − 2α(O2 , g) −
α(CH4 , g) and likewise for ∆β and ∆γ.
∆α = [(26.86) + 2 × (30.36) − 2 × (25.72) − (14.16)] J K−1 mol−1
= +21.98 J K−1 mol−1
∆β = [(6.97) + 2 × (9.61) − 2 × (12.98) − (75.5)] × 10−3 J K−2 mol−1
= −75.27 × 10−3 J K−2 mol−1
∆γ = [(−0.82) + 2 × (1.184) − 2 × (−3.862) − (−17.99)] × 10−6 J K−3 mol−1
= +27.262 × 10−6 J K−3 mol−1
55
56
2 INTERNAL ENERGY
Integrating Kirchho� ’s Law gives
∆ c H −○ (T2 ) = ∆ c H −○ (T1 ) + �
T2
T1
(∆α + ∆βT + ∆γT 2 ) dT
= ∆ c H −○ (T1 ) + ∆α(T2 − T1 ) + 12 ∆β(T22 − T12 ) + 13 ∆γ(T23 − T13 )
= (−802.3 kJ mol−1 ) + (+21.98 J K−1 mol−1 ) × (500 K − 298 K)
+ 12 × (−75.27 × 10−3 J K−2 mol−1 ) × [(500 K)2 − (298 K)2 ]
+ 13 × (27.262 × 10−6 J K−3 mol−1 ) × [(500 K)3 − (298 K)3 ]
P�C.��
= −803 kJ mol−1
(a) �e heat released in the calorimeter by the combustion of glucose is
�qv � = c∆T = (641 J K−1 ) × (7.793 K) = 4.99... kJ
�erefore the standard internal energy of combustion is
∆ c U −○ =
−�qv �
−(4.99... kJ)
=
= −2.80... × 103 kJ mol−1
n
(0.3212 g)�(180.1548 g mol−1 )
= −2.80 × 103 kJ mol−1
�e chemical equation for the combustion of glucose is C� H�� O� (s) +
� O� (g)
��→ � CO� (g) + � H� O(l). �ere is no change in the number of gaseous
species when going from reactants to products, therefore ∆ c U −○ = ∆ c H −○ =
−2.80... × 103 kJ mol−1 = −2.80 × 103 kJ mol−1 . Applying [�C.�a–��] for
this reaction and using the values for the standard enthalpies of formation
from Table �C.� gives
∆ c H −○ = 6∆ f H −○ (CO2 , g) + 6∆ f H −○ (H2 O, l) − 6∆ f H −○ (O2 , g)
− ∆ f H −○ (C6 H12 O6 , s)
∆ f H −○ (C6 H12 O6 , s) = 6∆ f H −○ (CO2 , g) + 6∆ f H −○ (H2 O, l)
− 6∆ f H −○ (O2 , g) − ∆ c H −○
= 6 × (−393.51 kJ mol−1 ) + 6 × (−285.83 kJ mol−1 )
− 0 − (−2.80... × 103 kJ mol−1 )
= −1.27 × 103 kJ mol−1
(b) �e reaction equation corresponding to anaerobic oxidation is C� H�� O� (s)
��→ � C� H� O� (s). �e standard enthalpy of reaction is
∆ r H −○ = 2∆ f H −○ (C3 H6 O3 (s)) − ∆ f H −○ (C6 H12 O6 (s))
= 2 × (−694.0 kJ mol−1 ) − (−1.27 × 103 kJ mol−1 )
= −1.13... × 102 kJ mol−1
�e amount of energy released in aerobic oxidation exceeds that of anaerobic glycolysis by
2.80... × 103 kJ mol−1 − 1.13... × 102 kJ mol−1 = 2.69 × 103 kJ mol−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
2D State functions and exact differentials
Answers to discussion questions
D�D.�
An inversion temperature is the temperature at which the Joule–�omson coe�cient µ changes sign from negative to positive or vice-versa. For a perfect
gas µ is always zero, thus it cannot have an inversion temperature. As explained
in detail in Section �D.� on page ��, the existence of the Joule–�omson e�ect
depends upon intermolecular attractions and repulsions. A perfect gas has by
de�nition no intermolecular attractions and repulsions, so it cannot exhibit the
Joule–�omson e�ect.
Solutions to exercises
E�D.�(a)
�e volume of the liquid can be written as
V = V ′ (a + bT + cT 2 )
where a = 0.75, b = 3.9 × 10−4 K−1 and c = 1.48 × 10−6 K−2 . �e expansion
coe�cient is de�ned in [�D.�–��] as α = (1�V )(∂V �∂T) p . �e derivative with
respect to T is
∂V
�
� = V ′ (b + 2cT)
∂T p
�erefore
α=
1
b + 2cT
× [V ′ (b + 2cT)] =
V ′ (a + bT + cT 2 )
a + bT + cT 2
Evaluating this expression at 320 K gives
α 320 =
E�D.�(a)
(3.9 × 10−4 K−1 ) + 2 × (1.48 × 10−6 K−2 ) × (320 K)
(0.75) + (3.9 × 10−4 K−1 ) × (320 K) + (1.48 × 10−6 K−2 ) × (320 K)2
= +1.3 × 10−3 K−1
�e isothermal compressibility is de�ned in [�D.�–��], κ T = −(1�V )(∂V �∂p)T ,
therefore at constant temperature dV �V = −κ T dp. �is question is concerned
with changes in density, so the next step is to rewrite the volume in terms of
the density, ρ. If the mass is m, V = m�ρ, and therefore dV = (−m�ρ 2 )dρ.
�erefore
dV
1
m
ρ
m
1
= �− 2 dρ� = − � � � 2 � dρ = − dρ
V
V
ρ
m
ρ
ρ
It therefore follows that
dρ
= κ T dp
ρ
and hence
dp =
1
κT ρ
dρ
�is expression gives the relationship between the change in pressure and the
change in density. Approximating dρ by δρ and dp by δ p for su�ciently small
changes gives
δp =
1
δρ
1
×
=�
� × (1.0 × 10−3 ) = +20 atm
κT
ρ
4.96 × 10−5 atm−1
57
58
2 INTERNAL ENERGY
E�D.�(a)
�e di�erence C p,m − C V ,m is given by [�D.��–��], C p,m − C V ,m = α 2 T Vm �κ T .
In this expression the molar volume is found from the mass density ρ and the
molar mass M by Vm = M�ρ. �e values of α and κ are available in the Resource
section, as is the mass density.
C p,m − C V ,m =
=
α 2 T Vm α 2 TM
=
κT
κT ρ
(12.4 × 10−4 K−1 )2 × (298 K) × (78.1074 g mol−1 )
[(92.1 × 10−6 bar−1 ) × (1 bar�105 Pa)] × (0.879 × 106 g m−3 )
= +44.2 J K−1 mol−1
E�D.�(a)
�e units are K−1 Pa m3 mol−1 = K−1 (N m−2 ) m3 mol−1 = K−1 N m mol−1
= J K−1 mol−1
�e molar volume of a perfect gas at 400 K is calculated as
Vm =
RT
(8.3145 J K−1 mol−1 ) × (400 K)
=
= 3.32... × 10−2 m3 mol−1
p
(1.00 bar) × [(105 Pa)�(1.00 bar)]
�e van der Waals parameter a of water vapour is found in Table �C.�, and
needs to be converted to SI units
−2
a = (5.464 dm6 atm mol ) × �
= 0.553... m6 Pa mol
−2
10−6 m6
1.01325 × 105 Pa
�
6 ��
1 atm
1 dm
�erefore the internal pressure is
E�D.�(a)
πT =
−2
a
(0.553... m6 Pa mol )
=
= 501 Pa
2
Vm (3.32... × 10−2 m3 mol−1 )2
�e internal energy of a closed system of constant composition is a function of
temperature and volume. For a change in V and T, dU is given by [�D.�–��],
dU = π T dV + C V dT. At constant temperature, this reduces to dU = π T dV .
Substituting in the given expression for π T for a van der Waals gas and using
molar quantities
a
dUm = 2 dVm
Vm
�is expression is integrated between Vm,i and Vm,f to give
hence
�
Vm,f
Vm,i
∆U m = −
dU m = �
Vm,f
Vm,f
Vm,i
a
dVm
Vm2
a
1
1
�
= −a �
−
�
Vm2 Vm,i
Vm,f Vm,i
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e van der Waals parameter a for nitrogen gas is found in Table �C.�, and
needs to be converted to SI units
−2
a = (1.352 dm6 atm mol ) × �
= 0.136... m6 Pa mol
−2
−2
∆Um = −(0.136... m6 Pa mol ) �
10−6 m6
1.01325 × 105 Pa
�
×
�
�
1 atm
1 dm6
1
1
−
�
20.00 × 10−3 m3 mol−1 1.00 × 10−3 m3 mol−1
= +1.30... × 102 J mol−1 = +130 J mol−1
�e work done by an expanding gas is given by [�A.�a–��], dw = −pex dV . For
a reversible expansion pex is the pressure of the gas, hence
w = − � pdVm
Substituting in the expression for the pressure of a van der Waals gas, [�C.�b–
��]
RT
a
RT
a
−
dVm = − �
dVm + �
dVm
Vm − b Vm2
Vm − b
Vm2
RT
= −�
dVm + ∆Um
Vm − b
w = −�
�e second term is identi�ed as ∆U m from the above. According to the First
Law, ∆U = q + w, the �rst term in the expression above must be −q, therefore
q=�
Vm,f
Vm,i
RT
Vm,f − b
V
dVm = RT ln(Vm − b)�Vm,f
= RT ln �
�
m,i
Vm − b
Vm,i − b
= (8.3145 J K−1 mol−1 ) × (298 K) × ln �
20.00 dm3 − 3.87 × 10−2 dm3
�
1.00 dm3 − 3.87 × 10−2 dm3
= +7.51... × 103 J mol−1 = +7.52 kJ mol−1
where the value for b is taken from the Resource section. From the First Law
the work done is w = −q + ∆Um , hence
w = −q + ∆Um = −7.51... ×103 J mol−1 +1.30... ×102 J mol−1 = −7.39 kJ mol−1
Solutions to problems
P�D.�
�e expansion coe�cient is de�ned in [�D.�–��] as α = (1�V )(∂V �∂T) p . For
su�ciently small changes the derivatives are approximated by �nite changes to
give
1 δV
α= �
�
V δT
59
60
2 INTERNAL ENERGY
Using the relationship given in the hint, δV = Aδr
α=
1 Aδr
�
�
V δT
therefore
δr =
αV
δT
A
From Table �D.� α = 2.1 × 10−4 K−1 . For a temperature rise of 1.0○ C
(2.1 × 10−4 K−1 ) × (1.37 × 109 km3 )
× (1.0 K) = 7.96... × 10−4 km
(361 × 106 km2 )
= 0.80 m
∆r =
P�D.�
As ∆r ∝ ∆T, for a temperature rise of 2.0 ○ C, ∆r = 1.6 m , and for a temperature rise of 3.5 ○ C, ∆r = 2.8 m is expected. �is calculation assumes that the
total surface area of the oceans remains constant and that α(H2 O) = α(ocean).
(a) If V = V (p, T) then
If p = p(V , T) then
dV = �
dp = �
∂V
∂V
� dp + �
� dT
∂p T
∂T p
∂p
∂p
� dV + � � dT
∂V T
∂T V
(b) Dividing the expression for dV by V gives
1
1 ∂V
1 ∂V
dV = �
� dp + �
� dT
V
V ∂p T
V ∂T p
Noting the de�nition of κ T , [�D.�–��], κ T = −(1�V )(∂V �∂p)T and that
of α, [�D.�–��], α = (1�V )(∂V �∂T) p , and rewriting (1�V )dV as d ln V
gives
d ln V = −κ T dp + αdT
Dividing the expression for dp by p gives
1
1 ∂p
1 ∂p
dp = �
� dV + � � dT
p
p ∂V T
p ∂T V
1 ∂p
1 ∂p
∂V
d ln p = �
� dV − �
� �
� dT
p ∂V T
p ∂V T ∂T p
=
1 ∂p
∂V
�
� �dV − �
� dT�
p ∂V T
∂T p
= �−
�
1 1�
1
∂V
� �dV − �
� �
� dT�
1
∂V
V p �− � � �
∂T p
V
∂p
T
1
dV 1 ∂V
=
�−
+ �
� dT�
pκ T
V
V ∂T p
1
dV
=
�−
+ αdT�
pκ T
V
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Going from line � to �, the identity (1�x)dx = d ln x and Euler’s chain
relationship are used. Going from line � to � the reciprocal relationship
is used. �e de�nition of κ T is used to go from line � to � and to go from
line � to � the de�nition of α is used.
P�D.�
�e isothermal compressibility is de�ned in [�D.��–��], κ T = −(1�V )(∂V �∂p)T .
Using the reciprocal identity this is equivalent to
κT = −
1
1
V (∂p�∂V )T
�e pressure of a van der Waals gas is p = nRT�(V − nb) − n 2 a�V 2 , therefore
�
∂p
nRT
2n 2 a −nRT V 3 + 2n 2 a(V − nb)2
� =−
+
=
2
∂V T
(V − nb)
V3
(V − nb)2 V 3
It follows that
and hence
�
∂V
(V − nb)2 V 3
� =
∂p T −nRT V 3 + 2n 2 a(V − nb)2
κT = −
1 ∂V
V 2 (V − nb)2
�
� =
V ∂p T nRT V 3 − 2n 2 a(V − nb)2
�e expansion coe�cient is de�ned in [�D.�–��] as α T = (1�V )(∂V �∂T) p ,
which is rewritten using the reciprocal identity as
α=
1
1
V (∂T�∂V ) p
Rearranging the van der Waals equation of state gives
T=
hence
p
na
(V − nb) +
(V − nb)
nR
RV 2
��� � � � � � � � � � ��� � � � � � � � � � � �
∂T
p
na 2na
T
2na
�
� =
+
−
(V − nb) =
−
(V − nb)
∂V p nR RV 2 RV 3
V − nb RV 3
A
=
RT V 3 − 2na(V − nb)2
(V − nb)RV 3
where the term A is identi�ed from the previous equation as being equal to
T�(V − nb).Using the reciprocal identity
therefore
�
α=
∂V
(V − nb)RV 3
� =
∂T p RT V 3 − 2na(V − nb)2
1 ∂V
(V − nb)RV 2
�
� =
V ∂T p RT V 3 − 2na(V − nb)2
61
62
2 INTERNAL ENERGY
It follows that
κT
V 2 (V − nb)2
(V − nb)RV 2
=
×
α
nRT V 3 − 2n 2 a(V − nb)2 RT V 3 − 2na(V − nb)2
(V − nb)
=
nR
P�D.�
For a molar quantity V → Vm and n → 1 to give κ T �α = (Vm − b)�R or
κ T R = α(Vm − b) .
�e equation of state of this gas is rearranged to give
From this it follows that
V=
�
nRT
+ nb
p
∂V
nR
� =
∂T p
p
Substituting this result in the expression for µ gives
µ=
=
1
∂V
�T �
� − V�
Cp
∂T p
nRT
nb
1
nR
�T � � − �
+ nb�� = −
Cp
p
p
Cp
�e heat capacity and the van der Waals parameter b are always positive, therefore the Joule–�omson coe�cient µ is negative for this gas. �is means that
the temperature of the gas will increase when expanding in an isenthalpic process.
P�D.�
From [�D.��–��], dH = −µC p dp+C p dT, it follows that at constant temperature
µ=−
1 ∂H
�
�
C p ∂p T
�e partial derivative (∂H�∂p)T is the slope of the plot of H against p measured
at constant temperature.
(a) �e plot for 300 K is shown in Fig. �.�.
�e slope of the plot is −17.93. �e Joule–�omson coe�cient is determined from the slope,
µ=−
(−17.93 kJ kg−1 MPa−1 )
= 23 K MPa−1
(0.7649 kJ K−1 kg−1 )
µ=−
(−14.46 kJ kg−1 MPa−1 )
= 14 K MPa−1
(1.0392 kJ K−1 kg−1 )
(b) �e plot for 350 K is shown in Fig. �.�.
�e slope of the plot is −14.46. �e Joule–�omson coe�cient is determined from the slope,
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
426.6
H�kJ kg−1
426.4
426.2
426.0
425.8
425.6
0.07
0.08
0.09
0.10 0.11
p�MPa
0.9
1.0
p�MPa
Figure 2.2
0.12
0.13
462
H�kJ kg
−1
460
458
456
0.8
Figure 2.3
1.1
1.2
2E Adiabatic changes
Answers to discussion questions
D�E.�
In an adiabatic expansion the system does work but as no energy as heat is
permitted to enter the system, the internal energy of the system falls and so
consequently does the temperature. From the First Law, ∆U = w because q =
0. However, the change in internal energy is also related to the temperature
change and the heat capacity: ∆U = C V ∆T. Equating these two expressions
for ∆U gives w = C V ∆T, or dw = C V dT for an in�nitesimal change.
For a reversible expansion, the work is dw = −pdV . Equating these two expressions for the work gives −pdV = C V dT. �is equation is the point from
which the relationships between pressure, volume and temperature for a reversible adiabatic expansion are found: the heat capacity comes into the �nal
expressions via this route.
In words, the key thing here is that in an adiabatic process there is a change
63
64
2 INTERNAL ENERGY
in temperature, so it is not surprising that the properties of such a process are
related to the heat capacity because this quantity relates the energy and the
temperature rise.
Solutions to exercises
E�E.�(a)
For a reversible adiabatic expansion the initial and �nal states are related by
γ
γ
[�E.�–��], p i Vi = p f Vf , where γ is the ratio of heat capacities, γ = C p,m �C V ,m .
�e initial volume of the sample is
Vi =
nRT (1.0 mol) × (8.3145 J K−1 mol−1 ) × (300 K)
=
= 5.79... × 10−3 m3
p
(4.25 atm) × [(1.01325 × 105 Pa)�(1 atm)]
= 5.79... dm3
For a perfect gas C p,m − C V ,m = R, hence
γ=
C p,m C V ,m + R (20.8 J K−1 mol−1 ) + (8.3145 J K−1 mol−1 )
=
=
= 1.39...
C V ,m
C V ,m
(20.8 J K−1 mol−1 )
�e �nal volume is given by Vf = (5.79... dm3 )×(4.25�2.50)1�1.39... = 8.46... dm3 .
�us Vf = 8.46 dm3 .
�e initial and �nal states are also related by (Tf �Ti ) = (Vi �Vf )1�c where c =
C V ,m �R. For this gas c = (20.8 J K−1 mol−1 )�(8.3145 J K−1 mol−1 ) = 2.50...
and hence
Vi 1�c
5.79... dm3
Tf = Ti � � = (300 K) × �
�
Vf
8.46 dm3
1�2.50...
= 2.57... × 102 K = 258 K
�e work done by a perfect gas during adiabatic expansion is given by [�E.�–��]
E�E.�(a)
w ad = C V ∆T = C V (Tf − Ti ) = (20.8 J K−1 )×(2.57... ×102 K−300 K) = −877 J
�e work done in a reversible adiabatic expansion is given by [�E.�–��], w ad =
C V ∆T. �e task is to �nd ∆T. �e initial and �nal states in a reversible adiabatic expansion are related by [�E.�b–��], Vi Tic = Vf Tfc , where c = C V ,m �R. If
the gas is assumed to be perfect, C p,m − C V ,m = R [�B.�–��], and so
c=
C V ,m C p,m − R (37.11 J K−1 mol−1 ) − (8.3145 J K−1 mol−1 )
=
=
= 3.46...
R
R
(8.3145 J K−1 mol−1 )
�e temperature change is given by
∆T = Tf − Ti = Ti �
Vi 1�c
Vi 1�c
� − Ti = Ti �� � − 1�
Vf
Vf
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�erefore the work done is
w ad = C V ∆T = nC V ,m ∆T = nC V ,m Ti ��
= n(C p,m − R)Ti ��
=�
E�E.�(a)
E�E.�(a)
Vi 1�c
� − 1�
Vf
Vi 1�c
� − 1�
Vf
2.45 g
� × [(37.11 J K−1 mol−1 )
44.01 g mol−1
− (8.3145 J K−1 mol−1 )] × (300.15 K)
�
1�3.46... �
�
�
500 cm3
�
�
�
− 1�
� 3.00 × 103 cm3
� = −194 J
�
�
�
�
�e initial and �nal states in a reversible adiabatic expansion are related by
γ
γ
[�E.�–��], p i Vi = p f Vf , therefore
Vi γ
0.50 dm3
p f = p i � � = (67.4 kPa) × �
�
Vf
2.00 dm3
1.4
= 9.7 kPa
Ammonia and methane are nonlinear polyatomic molecules which have three
degrees of translational and three degrees of rotational freedom. From the
equipartition theorem (�e chemist’s toolkit � in Topic �A)
C V ,m = 12 × (νt + νr + 2νv ) × R
where νt is the number of translational degrees of freedom, νr is the number
of rotational degrees of freedom and νv is the number of vibrational degrees of
freedom.
(i) Without any vibrational contribution the calculation is the same for both
molecules. C V ,m = 12 × (3 + 3 + 0) × R = 3R. For a perfect gas, [�B.�–��],
C p,m = C V ,m + R, therefore
γ=
C V ,m + R 3R + R 4
=
=
C V ,m
3R
3
(ii) If the vibrational contribution is included, the molecules have di�erent
values of γ. �e number of vibrational modes for a nonlinear polyatomic
molecule is νv = 3N − 6, where N is the number of atoms in the molecule.
�erefore, for ammonia νv = 3N − 6 = 6 and for methane νv = 3N − 6 =
9. �is gives C V ,m (NH3 ) = 12 × (3 + 3 + 2 × 6) × R = 9R, therefore
C p,m (NH3 ) = 10R and γ = 10�9. For methane C V ,m (CH4 ) = 12 × (3 + 3 +
2 × 9) × R = 12R, C p,m (CH4 ) = 13R and γ = 13�12.
�e experimental values of γ for ammonia and methane are the same, γ = 1.31,
which is closer to the value calculated without taking the vibrational degrees of
freedom into account.
65
66
2 INTERNAL ENERGY
E�E.�(a)
For a reversible adiabatic expansion the initial and �nal states are related by
[�E.�a–��], (Tf �Ti ) = (Vi �Vf )1�c , where c = C V ,m �R. For a monoatomic ideal
gas C V ,m = 32 R, so c = 32 . �erefore
2
1
Vi c
1.0 dm3 3
Tf = Ti � � = (273.15 K) × �
� = 1.3 × 102 K
Vf
3.0 dm3
Solutions to problems
P�E.�
�e work done in a reversible adiabatic expansion is given by [�E.�–��], w ad =
C V ∆T. �e initial and �nal states in a reversible adiabatic expansion are related
by [�E.�b–��], Vi Tic = Vf Tfc , where c = C V ,m �R. If the gas is assumed to be
perfect, C p,m − C V ,m = R [�B.�–��], and so
c=
C V ,m C p,m − R (35.06 J K−1 mol−1 ) − (8.3145 J K−1 mol−1 )
=
=
= 3.21...
R
R
(8.3145 J K−1 mol−1 )
�e �nal temperature is
Vi 1�c
0.50 dm3
Tf = Ti � � = (298 K) × �
�
Vf
2.00 dm3
1�3.21...
�erefore the work done is
= 1.93... × 102 K = 194 K
w ad = C V ∆T = nC V ,m ∆T = n(C p,m − R)(Tf − Ti )1�c
= (1.00 mol) × [(35.06 J K−1 mol−1 ) − (8.3145 J K−1 mol−1 )]
× [(1.93... × 102 K) − (298 K)]1�3.21... = −2.79 kJ
∆U = w ad = −2.79 kJ
Integrated activities
I�.�
�e change in reaction enthalpy with respect to temperature is described by
Kirchho� ’s Law, [�C.�a–��]. Assuming that the heat capacities are independent of temperature, the integrated form of Kircho� ’s Law is applicable and is
given by [�C.�d–��]
∆ r H −○ (T2 ) = ∆ r H −○ (T1 ) + (T2 − T1 )∆ r C −p○
If ∆ r C −p○ is negative, then the reaction enthalpy will decrease with increasing
temperature, whereas if ∆ r C −p○ is positive, then the reaction enthalpy will increase with increasing temperature.
(a)
∆ r C −p○ =
−
○
� νC p,m −
products
�
reactants
○
νC −p,m
○
○
○
= 2C −p,m
(H2 O, l) − C −p,m
(O2 , g) − 2C −p,m
(H2 , g)
= 2 × (9R) − � 72 R� − 2 × � 72 R� = +7 12 R
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
∆ r C −p○ is positive, therefore the standard reaction enthalpy of reaction will
increase with increasing temperature.
(b)
○
○
○
○
∆ r C −p○ = C −p,m
(CO2 , g) + 2C −p,m
(H2 O, l) − 2C −p,m
(O2 , g) − C −p,m
(CH4 , g)
= � 72 R� + 2 × (9R) − 2 × � 72 R� − (4R) = +10 12 R
∆ r C −p○ is positive, therefore the standard reaction enthalpy of reaction will
increase with increasing temperature.
I�.�
(a) From the equipartition principle (�e chemist’s toolkit � in Topic �A),
and assuming that only translational and rotational levels contribute, the
constant volume heat capacity of a gas of diatomic molecules is C V ,m =
5
R. For a perfect gas C p,m − C V ,m = R, [�B.�–��], therefore C p,m =
2
C V ,m + R = 72 R and so γ = C p,m �C V ,m = 75 . It follows that
cs = �
γRT 1�2
7RT 1�2
� =�
�
M
5M
cs = �
γRT 1�2
4RT 1�2
� =�
�
M
3M
(b) A linear triatomic has the same number of translational and rotational
modes of motion (three translational, two rotational) as a diatomic, so
the calculation is the same as in (a).
(c) A gas of non-linear triatomic molecules has one extra mode of rotational
motion when compared to a linear triatomic, so the constant volume heat
capacity is C V ,m = 3R. Because C p,m − C V ,m = R it follows that C p,m =
C V ,m + R = 4R and so γ = C p,m �C V ,m = 43 .
�e air is mostly composed of diatomic and linear triatomic molecules, N� , O�
and CO� , hence γ ≈ 75 . �e average molecular weight of air is approximately
29.0 g mol−1 , therefore
I�.�
cs = �
7RT 1�2
7 × (8.3145 J K−1 mol−1 ) × (298.15 K)
� =�
�
5M
5 × (29.0 × 10−3 kg mol−1 )
1�2
= 346 m s−1
A state function is a thermodynamic property, the value of which is independent of the history of the system. Examples of state functions are the properties
of pressure, temperature, internal energy, enthalpy as well as the properties of
entropy, Gibbs energy, and Helmholtz energy to be discussed fully in Focus �.
�e di�erentials of state functions are exact di�erentials. Hence, the mathematical properties of exact di�erentials can be used to draw far-reaching conclusions about the relations between physical properties and establish connections
that were unexpected but turn out to be very signi�cant.
One practical importance of these results is that the value of a property of interest can be obtained from the combination of measurements of other properties
without actually having to measure the required property itself, the measurement of which might be very di�cult.
67
68
2 INTERNAL ENERGY
I�.�
�e change in reaction enthalpy with respect to temperature is described by
Kirchho� ’s Law, [�C.�a–��]. Assuming that the heat capacities are independent of temperature, the integrated form of Kircho� ’s Law is applicable and is
given by [�C.�d–��]
∆ r H −○ (T2 ) = ∆ r H −○ (T1 ) + (T2 − T1 )∆ r C −p○
If ∆ r C −p○ is negative, then the reaction enthalpy will decrease with increasing
temperature, whereas if ∆ r C −p○ is positive, then the reaction enthalpy will increase with increasing temperature.
(a)
∆ r C −p○ =
−
○
� νC p,m −
products
�
reactants
○
νC −p,m
○
○
○
= 2C −p,m
(H2 O, g) − C −p,m
(O2 , g) − 2C −p,m
(H2 , g)
= 2 × (4R) − � 72 R� − 2 × � 72 R� = −2 12 R
∆ r C −p○ is negative, therefore the standard reaction enthalpy of reaction will
decrease with increasing temperature.
(b)
○
○
○
○
∆ r C −p○ = C −p,m
(CO2 , g) + 2C −p,m
(H2 O, g) − 2C −p,m
(O2 , g) − C −p,m
(CH4 , g)
= � 72 R� + 2 × (4R) − 2 × � 72 R� − (4R) = + 12 R
∆ r C −p○ is positive, therefore the standard reaction enthalpy of reaction will
increase with increasing temperature.
(c)
○
○
○
∆ r C −p○ = 2C −p,m
(NH3 , g) − C −p,m
(N2 , g) − 3C −p,m
(H2 , g)
= 2 × (4R) − � 72 R� − 3 × � 72 R� = −6R
∆ r C −p○ is negative, therefore the standard reaction enthalpy of reaction will
decrease with increasing temperature.
The second and third laws
3
3A
Entropy
Answers to discussion questions
D�A.�
Everyday experience indicates that the direction of spontaneous change in an
isolated system is accompanied by the dispersal of the total energy of the system. For example, for a gas expanding freely and spontaneously into a vacuum,
the process is accompanied by a dispersal of energy and matter. For a perfect
gas this entropy change of such an expansion is derived in Section �A.�(a) on
page �� as ∆S = nR ln(Vf �Vi ). �e entropy change is clearly positive if Vf is
greater than Vi .
�e molecular interpretation of this thermodynamic result is based on the identi�cation of entropy with molecular disorder. An increase in disorder results
from the chaotic dispersal of matter and energy and the only changes that can
take place within an isolated system (the universe) are those in which this kind
of dispersal occurs. �is interpretation of entropy in terms of dispersal and
disorder allows for a direct connection of the thermodynamic entropy to the
statistical entropy through the Boltzmann formula S = k ln W, where W is
the number of microstates, the number of ways in which the molecules of the
system can be arranged while keeping the total energy constant.
�e concept of the number of microstates makes quantitative the more illde�ned qualitative concepts of ‘disorder’ and ‘the dispersal of matter and energy’ used above to give a physical feel for the concept of entropy. A more
‘disorderly’ distribution of energy and matter corresponds to a greater number
of microstates associated with the same total energy.
D�A.�
�e Second Law of thermodynamics states only that the total entropy of both
the system (here, the molecules organizing themselves into cells) and the surroundings (here, the medium) must increase in a naturally occurring process.
It does not state that entropy must increase in a portion of the universe that
interacts with its surroundings. In this case, the cells grow by using chemical
energy from their surroundings (the medium) and in the process the increase
in the entropy of the medium outweighs the decrease in entropy of the system.
Hence, the Second Law is not violated.
Solutions to exercises
E�A.�(a)
�e e�ciency is de�ned in [�A.�–��], η = �w���q h �, and for a Carnot cycle
e�ciency is given by [�A.�–��], η = 1 − (Tc �Th ). �ese two are combined
70
3 THE SECOND AND THIRD LAWS
and rearranged into an expression for the temperature of the cold sink
�w���q h � = 1 − (Tc �Th )
hence
E�A.�(a)
Tc = �1 −
= �1 −
�w�
� × Th
�q h �
�3.00 kJ�
� × (273 K) = 191 K .
� − 10.00 kJ�
�e e�ciency of a Carnot cycle is given by [�A.�–��], η = 1 − (Tc �Th ). �us
η =1−
Tc
(273.15 K + 10 K)
=1−
= 0.241 = 24.1% .
Th
(273.15 K + 100 K)
Note that the temperatures must be in kelvins.
E�A.�(a)
E�A.�(a)
For the process to be spontaneous it must be irreversible and obey the Clausius
inequality [�A.��–��] implying that ∆S tot = ∆S + ∆S sur > 0. In this case,
∆S tot = 125 J K−1 + (−125 J K−1 ) = 0, thus the process is not spontaneous in
either direction and is at equilibrium.
�e thermodynamic de�nition of entropy is [�A.�a–��], dS = dq rev �T or for
a �nite change at constant temperature ∆S = q rev �T. �e transfer of heat is
speci�ed as being reversible, which can o�en be assumed for a large enough
metal block, therefore q rev = 100 kJ.
(i)
∆S =
(ii)
E�A.�(a)
∆S =
q rev
100 kJ
=
= 0.366 kJ = +366 J
T
273.15 K
q rev
100 kJ
=
= 0.309 kJ = +309 J
T
(273.15 K + 50 K)
As explained in Section �A.�(a) on page �� the change in entropy for an isothermal expansion of a gas is calculated using
Vf
m
Vf
� = R ln � �
Vi
M
Vi
15 g
3.0 dm3
−1
−1
=�
�
×
(8.3145
J
K
mol
)
×
ln
�
�
44.01 g mol−1
1.0 dm3
∆S = nR ln �
E�A.�(a)
= +3.1 J K−1 .
�e change in entropy for an isothermal expansion of a gas is ∆S = nR ln (Vf �Vi )
as explained in Section �A.�(a) on page ��. For a doubling of the volume
Vf �Vi = 2.
(i) Isothermal reversible expansion
∆S = �
14 g
−1
−1
−1
.
−1 � × (8.3145 J K mol ) × ln (2) = +2.9 J K
28.02 g mol
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Because the process is reversible ∆S tot = � .
Because ∆S tot = ∆S + ∆S sur
∆S sur = ∆S tot − ∆S = −2.9 J K−1 .
(ii) Isothermal irreversible expansion against p ex = 0 Because entropy is a
state function and the initial and �nal states of the system are the same as
in (a), ∆S is the same.
∆S = +2.9 J K−1 .
Expansion against an external pressure of � does no work, and for an
isothermal process of an ideal gas ∆U = 0. From the First Law if follows
that q = 0 and therefore ∆S sur = � .
∆S tot = ∆S + ∆S sur = +2.9 J K−1 .
(iii) Adiabatic reversible expansion For an adiabatic expansion there is no heat
�owing to or from the surroundings, thus ∆S sur = � . For a reversible
process ∆S tot = � , therefore it follows that ∆S = � as well.
Solutions to problems
P�A.�
(a) Isothermal reversible expansion
�e work of a reversible isothermal expansion of an ideal gas is given
by [�A.�–��], w = −nRT ln (Vf �Vi ). Because at �xed temperature p ∝
(1�V ) as given by Boyle’s law, an equivalent expression is
w = −nRT ln �
pi
�
pf
= −(1.00 mol) × (8.3145 J K−1 mol−1 )
3.00 atm
× (273.15 K + 27 K) × ln �
�
1.00 atm
= −2.74 × 103 ... J = −2.74 kJ .
For an isothermal process of a perfect gas ∆U = � and ∆H = � . �e
First Law is de�ned in [�A.�–��], ∆U = q + w, hence
q = ∆U − w = 0 − (−2.74... kJ) = +2.74 kJ .
�e heat transfer is reversible, therefore q rev = q.
q rev
2.74 × 103 ... J
=
T
273.15 K + 27 K
= +9.13... J K−1 = +9.13 J K−1 .
∆S =
�e process is reversible, therefore ∆S tot = � . Finally because ∆S tot =
∆S + ∆S sur
∆S sur = ∆S tot − ∆S = 0 − (+9.13... J K−1 ) = −9.13 J K−1 .
71
72
3 THE SECOND AND THIRD LAWS
(b) Isothermal expansion against p ex = 1.00 atm �e expansion work against
a constant external pressure is given by [�A.�–��], w = −p ex (Vf − Vi ).
�e volumes are written in terms of pressures by using the perfect gas law
[�A.�–�], pV = nRT.
w = −p ex (Vf − Vi )
= −p ex �
nRT nRT
p ex p ex
−
� = −nRT × �
−
�
pf
pi
pf
pi
= −(1.00 mol) × (8.3145 J K−1 mol−1 )
1.00 atm 1.00 atm
× (273.15 K + 27 K) × �
−
�
1.00 atm 3.00 atm
= −1.66... × 103 J = −1.66 kJ .
For an isothermal process in perfect gas ∆U = � and ∆H = � . Using the
First Law
q = ∆U − w = 0 − (−1.66 kJ) = +1.66 kJ .
Because entropy is a state function and the initial and �nal states of the
system are the same, the entropy change of the system is as in (a), ∆S =
+9.13 J K−1 .
�e entropy change of the surroundings in terms of the heat of the surroundings, q sur , is given by [�A.�b–��], ∆S sur = q sur �T. �is heat is
simply the opposite of the heat of the system: q sur = −q, therefore
q sur −q
=
T
T
−1.66... × 103 J
=
= −5.54... J K−1 = −5.54 J K−1 .
(273.15 K + 27 K)
∆S sur =
∆S tot = ∆S + ∆S sur
P�A.�
= (+9.13... J K−1 ) + (−5.53... J K−1 ) = +3.59 J K−1 .
(a) A�er Stage � the volume doubles, thus VB = 2 × VA = 2 × (1.00 dm3 ) =
2.00 dm3 . Assuming V T 3�2 = constant for the adiabatic stages, the
volume a�er Stage � is
VC = VB × �
Th 3�2
373 K 3�2
� = (2.00 dm3 ) × �
�
Tc
273 K
= 3.19... dm3 = 3.19 dm3 .
(b) Again assuming V T 3�2 = constant for the adiabatic stage, the volume
a�er Stage � can be related to the initial volume
VD = VA × �
Th 3�2
373 K 3�2
� = (1.00 dm3 ) × �
�
Tc
273 K
= 1.59... dm3 = 1.60 dm3 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(c) As shown in Section �A.�(a) on page �� the heat transferred reversibly
during an isothermal gas expansion is q rev = nRT ln (Vf �Vi ), thus the
heats for Stage � and Stage � are, respectively
VB
�
VA
= (0.100 mol) × (8.3145 J K−1 mol−1 ) × (373 K) × ln (2)
q 1 = q h = nRTh ln �
= +2.14... × 102 J = +215 J .
q 3 = q c = nRTc ln �
VD
�
VC
= (0.100 mol) × (8.3145 J K−1 mol−1 ) × (273 K) × ln �
= −1.57... × 102 J = −157 J .
1.59... dm3
�
3.19... dm3
Because there is no heat exchange during adiabatic processes, he heat
transfer for Stages � and � are q 2 = � and q 4 = � , respectively.
(d) At the beginning and end of the cycle the temperature is the same. Because the working substance is a perfect gas, ∆U = 0 over the cycle. �e
First Law [�A.�–��], ∆U = w + q, therefore implies that w = −q, that
is, the net heat over the cycle is converted to work. �is net heat is the
di�erence between that extracted from the hot source and deposited into
the cold sink.
(e) �e e�ciency is de�ned in [�A.�–��], η = �w���q h �. As has been explained,
�w� is the net heat.
�w� = �q h � − �q c � = � + 2.14... × 102 J� − � − 1.57... × 102 J�
hence
= +5.7... × 101 J = +58 J .
η=
�w�
� + 5.7... × 101 J�
=
= 0.268... = 27% .
�q h � � + 2.14... × 102 J�
(f) �e Carnot e�ciency is given by [�A.�–��],
η =1−
Tc
273 K
=1−
= 0.268 = 26.8% .
Th
373 K
the result is the same as the above (the di�erence is due to the use of fewer
signi�cant �gures in the previous calculation).
Using the values of the heat transfer calculated above in equation [�A.�–
��] gives
q c q c 214... J −157... J
+
=
+
Th Tc
373 K
273 K
= 0.0 J .
the result is zero, as expected from a Carnot cycle.
73
74
3 THE SECOND AND THIRD LAWS
P�A.�
(a) Consider a process in which heat dq c is extracted from the cold source at
temperature Tc , and heat dq h is discarded into the hot sink at temperature
Th . �e overall entropy change of such process is
dS =
dq c dq h
+
Tc
Th
Assume that dq c = −dq and dq h = +dq, where dq is a positive quantity.
It follows that
dS =
+dq −dq
1
1
+
= dq × � − �
Th
Tc
Th Tc
Because Th > Tc , the term in parentheses is negative, therefore dS is
negative. �e process is therefore not spontaneous and not allowed by
the Second Law. If work is done on the engine, �dq h � will become greater
than �dq c � and eventually dS will be greater than zero.
(b) Assuming q c = −�q� and q h = �q� + �w� the overall change in entropy is
∆S =
−�q� �q� + �w�
+
Tc
Th
For the process to be permissible by the Second Law the Clausius inequality de�ned in [�A.��–��], dS ≥ 0, must be satis�ed. �erefore
−�q� �q� + �w�
+
≥0
Tc
Th
which implies
P�A.�
�w� ≥ �q� × �
Th
Th
− 1� = �q� × � − 1� .
Tc
Tc
Suppose two adiabatic paths intersect at point A as shown in the �gure. Two
remote points corresponding to the same temperature on each adiabat, A and
B, are then connected by an isothermal path forming a cycle.
Consider energy changes for each Stage of the cycle. Stage � (A → B) is adiabatic
and, thus, no heat exchange takes place q 1 = 0. �erefore, the total change in
internal energy is ∆U 1 = w 1 + q 1 = w 1 . Stage � (B → C) is an isothermal change
and assuming that the system energy is a function of temperature only (e.g.
ideal gas): ∆U 2 = w 2 + q 2 = 0. Stage � (C → A) is again adiabatic, q 3 = 0, with
∆U 3 = w 3 + q 3 = w 3 . Because the system energy is a function of temperature
only, U B = U C and, thus
∆U 3 = U A − U C = U A − U B = −∆U 1
�is implies that w 1 = −w 3 .
Because internal energy is a state function and the cycle is closed:
U cycle = w cycle + q cycle = 0
= ∆U 1 + ∆U 2 + ∆U 3
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Pressure, p
A
Stage �
Stage �
C
Stage �
Figure 3.1
B
Volume, V
Finally, analyse the net work done, w cycle = w 1 + w 2 + w 3 = w 2 , and the net heat
absorbed, q cycle = q 1 + q 2 + q 3 = q 2 , over the cycle. It is apparent that the sole
result of the process is the absorption of heat q 2 and its convertion to work w 2 ,
which directly contradicts the statement of the Second Law by Kelvin, unless
the q 2 = w 2 = 0, i.e. points B and C are the same and correspond to the same
path. �erefore, no two such adiabatic paths exist.
3B Entropy changes accompanying specific processes
Answer to discussion question
D�B.�
�e explanation of Trouton’s rule is that a comparable change in volume is
expected whenever any unstructured liquid forms a vapour; accompanying this
will be a comparable change in the number of accessible microstates. Hence, all
unstructured liquids can be expected to have similar entropies of vaporization.
Liquids that show signi�cant deviations from Trouton’s rule do so on account of
strong molecular interactions that restrict molecular motion. As a result there
is a greater dispersal of matter and energy when such liquids vaporize.
Water is an example of a liquid with strong intermolecular interactions (hydrogen bonding) which tend to organize the molecules in the liquid, hence its
entropy of vaporization is expected to be greater than the value predicted by
Trouton’s rule. �e same is true for ethanol, which is also hydrogen bonded in
the liquid.
Mercury has quite strong interactions between the atoms, as evidenced by its
cohesiveness, and so its entropy of vaporization is expected to be greater than
that predicted by Trouton’s rule.
75
76
3 THE SECOND AND THIRD LAWS
Solutions to exercises
E�B.�(a)
Two identical blocks must come to their average temperature. �erefore the
�nal temperature is
Tf = 12 (T1 + T2 ) = 12 × (50 ○ C + 0 ○ C) = 25 ○ C = 298 K .
Although the above result may seem self-evident, the more detailed explaination is as follows. �e heat capacity at constant volume is de�ned in [�A.��–
��], C V = (∂U�∂T)V . As shown in Section �A.�(b) on page ��, if the heat
capacity is constant, the internal energy changes linearly with the change in
temperature. �at is ∆U = C V ∆T = C V (Tf − Ti ). For the two blocks at
the initial temperatures of T1 and T2 , the change in internal energy to reach
the �nal temperature Tf is ∆U 1 = C V ,1 (Tf − T1 ) and ∆U 2 = C V ,2 (Tf − T2 ),
respectively. �e blocks of metal are made of the same substance and are of
the same size, therefore C V ,1 = C V ,2 = C V . Because the system is isolated
the total change in internal energy is ∆U = ∆U 1 + ∆U 2 = 0. �is means that
∆U = C V ((Tf − T1 ) − (Tf − T2 )) = C V × (2Tf − (T1 + T2 )) = 0, which implies
that the �nal temperature is Tf = 12 (T1 + T2 ), as stated above.
�e temperature variation of the entropy at constant volume is given by [�B.�–
��], ∆S = C V ln (Tf �Ti ), with C p replaced by C V . Expressed with the speci�c
heat C V ,s = C V �m it becomes
∆S = mC V ,s ln �
Tf
�.
Ti
Note that for a solid the internal energy does not change signi�cantly with the
volume or pressure, thus it can be assumed that C V = C p = C. �e entropy
change for each block is found using this expression
∆S 1 = mC V ,s ln �
Tf
�
T1
= (1.00 × 103 g) × (0.385 J K−1 g−1 ) × ln �
298 K
�
50 K + 273.15 K
= (1.00 × 103 g) × (0.385 J K−1 g−1 ) × ln �
298 K
�
273.15 K
= −31.0... J K−1 = −31.0 J K−1 .
Tf
∆S 2 = mC V ,s ln � �
T2
= 33.7... J K−1 = +33.7 J K−1 .
�e total change in entropy is
∆S tot = ∆S 1 + ∆S 2 = (−31.0... J K−1 ) + (33.7... J K−1 )
= 27.2... J K−1 = +2.7 J K−1 .
Because ∆S tot > 0 the process is spontaneous, in accord with experience.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�B.�(a)
Because entropy is a state function, ∆S between the initial and �nal states is
the same irrespective of the path taken. �us the overall process can be broken
down into steps that are easier to evaluate. First consider heating the initial
system at constant pressure to the �nal temperature. �e variation of entropy
with temperature at constant pressure is given by [�B.�–��], S(Tf ) = S(Ti ) +
C p ln (Tf �Ti ). �us the change in entropy, ∆S = S(Tf ) − S(Ti ), of this step is
∆S 1 = C p ln �
Tf
Tf
� = nC p,m ln � �
Ti
Ti
Next consider an isothermal change in pressure. As explained in Section �A.�(a)
on page �� the change in entropy of an isothermal expansion of an ideal gas
is given by ∆S = nR ln (Vf �Vi ). Because for a �xed amount of gas at �xed
temperature p ∝ (1�V ) an equivalent expression for this entropy change is
∆S 2 = nR ln �
pi
�
pf
�erefore the overall entropy change for the system is
∆S = ∆S 1 + ∆S 2 = nC p,m ln �
Tf
pi
� + nR ln � �
Ti
pf
273.15 K + 125 K
�
273.15 K + 25 K
1.00 atm
+ (3.00 mol) × (8.3145 J K−1 mol−1 ) × ln �
�
5.00 atm
= (3.00 mol) × � 52 × 8.3145 J K−1 mol−1 � × ln �
E�B.�(a)
= (+18.0... J K−1 ) + (−40.1... J K−1 ) = −22.1 J K−1 .
Because entropy is a state function, ∆S between the initial and �nal states is
the same irrespective of the path taken. �us the overall process can be broken
down into steps that are easier to evaluate. First consider heating the ice at
constant pressure from the initial temperature to the melting point, Tm . �e
variation of entropy with temperature at constant pressure is given by [�B.�–
��], S(Tf ) = S(Ti ) + C p ln (Tf �Ti ). �us the change in entropy, ∆S = S(Tf ) −
S(Ti ), for this step is
∆S 1 = C p ln �
Tm
Tm
� = nC p,m (H2 O(s)) ln � �
Ti
Ti
Next consider the phase transition from solid to liquid at the melting temperature. �e entropy change of a phase transition is given by [�B.�–��], ∆ trs S =
∆ trs H�Ttrs , thus
−
○
∆ fus H m
∆S 2 = n
Tm
�en the liquid is heated to the boiling temperature, Tb . In analogy to the �rst
step
Tb
∆S 3 = nC p,m (H2 O(l)) ln � �
Tm
77
78
3 THE SECOND AND THIRD LAWS
�e next phase transition is from liquid to gas
∆S 4 = n
−
○
∆ vap H m
Tb
Finally, the vapour is heated from Tb to Tf
∆S 5 = nC p,m (H2 O(g)) ln �
Tf
�
Tb
�erefore the overall entropy change for the system is
∆S�n = ∆S 1 + ∆S 2 + ∆S 3 + ∆S 4 + ∆S 5
−
○
Tm
∆ fus H m
Tb
�+
+ C p,m (H2 O(l)) ln � �
Ti
Tm
Tm
−
○
∆ vap H m
Tf
+
+ C p,m (H2 O(g)) ln � �
Tb
Tb
273.15 K
= (37.6 J K−1 mol−1 ) × ln �
�
273.15 K − 10.0 K
6.01 × 103 J mol−1
+
273.15 K
273.15 K + 100.0 K
+ (75.3 J K−1 mol−1 ) × ln �
�
273.15 K
40.7 × 103 J mol−1
+
273.15 K + 100.0 K
273.15 K + 115.0 K
+ (33.6 J K−1 mol−1 ) × ln �
�
273.15 K + 100.0 K
= (+1.40... J K−1 mol−1 ) + (+22.0... J K−1 mol−1 )
= C p,m (H2 O(s)) ln �
+ (+23.4... J K−1 mol−1 ) + (+1.09... × 102 J K−1 mol−1 )
+ (+1.32... J K−1 mol−1 )
Hence
E�B.�(a)
= +1.57... × 102 J K−1 mol−1
∆S =
10.0 g
× (+1.57... × 102 J K−1 ) = +87.3 J K−1 .
18.02 g mol−1
�e entropy change of a phase transition is given by [�B.�–��], ∆ trs S = ∆ trs H�Ttrs .
As discussed in Section �B.� on page �� because there is no hydrogen bonding
in liquid benzene it is safe to apply Trouton’s rule. �at is ∆ vap S −○ = +85 J K−1 mol−1 .
It follows that
∆ vap H −○ = Tb × ∆ vap S −○
= (273.15 K + 80.1 K) × (+85 J K−1 mol−1 )
= 3.00... × 104 J mol−1 = +30 kJ mol−1 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�B.�(a)
(i) �e entropy change of a phase transition is given by [�B.�–��], ∆ trs S =
∆ trs H�Ttrs . For vaporisation this becomes
∆ vap S −○ =
∆ vap H −○
+ 29.4 × 103 J mol−1
=
Tb
334.88 K
= +87.8 J K−1 mol−1 .
E�B.�(a)
(ii) Because the system at the transition temperature is at equilibrium, ∆S tot =
0, thus
∆S sur = −∆ vap S −○ = −87.8 J K−1 mol−1 .
�e change in entropy with temperature is given by [�B.�–��],
∆S = S(Tf ) − S(Ti ) = �
Tf
Ti
Cp
dT
T
Assuming that C p is constant in the temperature range Ti to Tf , this becomes
∆S = C p ln (Tf �Ti ) as detailed in Section �B.� on page ��. �us, the increase
in the molar entropy of oxygen gas is
∆S m = S m (348 K) − S m (298 K) = (29.355 J K−1 mol−1 ) × ln �
E�B.�(a)
= +4.55 J K−1 mol−1 .
348 K
�
298 K
As explained in Section �B.� on page �� the temperature variation of the entropy at constant volume is given by
∆S = S(Tf ) − S(Ti ) = �
Tf
Ti
CV
dT
T
Assuming that C V = 32 R, the ideal gas limit, for the temperature range of
interest, the molar entropy at 500 K is given by
S m (500 K) = S m (298 K) + �
500 K
3 dT
R
2
T
500
K
= S m (298 K) + 32 R × ln �
�
298 K
= (146.22 J K−1 mol−1 )
298 K
+ ( 32 × 8.3145 J K−1 mol−1 ) × ln �
Solutions to problems
P�B.�
= 153 J K−1 mol−1 .
500 K
�
298 K
Because entropy is a state function, ∆S between the initial and �nal states is
the same irrespective of the path taken. �us the overall process can be broken
down into steps that are easier to evaluate.
79
80
3 THE SECOND AND THIRD LAWS
First consider heating the water at constant pressure from the initial temperature T to the melting point. �e variation of the entropy with temperature at
constant pressure is given by [�B.�–��], S(Tf ) = S(Ti ) + C p ln (Tf �Ti ). �us
the change in entropy for this step is
∆S 1 = C p ln �
Tm
Tm
� = nC p,m (H2 O(l)) ln � �
T
T
Next consider the phase transition from liquid to solid at the melting temper−
○
ature; note that freezing is just the opposite of fusion, thus ∆H 2 = n(−∆ fus H m
).
�e entropy change of a phase transition is given by [�B.�–��], ∆ trs S = ∆ trs H�Ttrs ,
thus
−
○
∆H 2
−∆ fus H m
∆S 2 =
=n
Tm
Tm
�e ice is then cooled to the �nal temperature, T. Similarly to the �rst step
∆S 3 = nC p,m (H2 O(s)) ln �
T
�
Tm
�erefore the overall entropy change for the system is
∆S = ∆S 1 + ∆S 2 + ∆S 3
○
−∆ fus H −m
Tm
T
�+n
+ nC p,m (H2 O(s)) ln � �
T
Tm
Tm
273.15 K
−1
−1
= (1.00 mol) × (75.3 J K mol ) × ln �
�
273.15 K − 5.00 K
−6.01 × 103 J mol−1
+ (1.00 mol) ×
273.15 K
273.15 K − 5.00 K
+ (1.00 mol) × (37.6 J K−1 mol−1 ) × ln �
�
273.15 K
= (+1.39... J K−1 ) + (−22.0... J K−1 ) + (−0.694... J K−1 )
= nC p,m (H2 O(l)) ln �
= −21.3... J K−1 = −21.3 J K−1 .
Consider enthalphy change for the same path. �e variation of enthalpy with
temperature at constant pressure is given by [�B.�b–��], ∆H = C p ∆T. �us for
the �rst and third steps, respectively
∆H 1 = nC p,m (H2 O(l))(Tm − T) and
∆H 3 = nC p,m (H2 O(s))(T − Tm )
�erefore the overall enthalpy change for the system is
∆H = ∆H 1 + ∆H 2 + ∆H 3
−
○
= nC p,m (H2 O(l))(Tm − T) + n(−∆ fus H m
) + nC p,m (H2 O(s))(T − Tm )
= (1.00 mol) × (75.3 J K−1 mol−1 ) × (+5.00 K)
+ (1.00 mol) × (−6.01 × 103 J mol−1 )
+ (1.00 mol) × (37.6 J K−1 mol−1 ) × (−5.00 K)
= (+3.76... × 102 J) + (−6.01... × 103 J) + (−1.88... × 102 J)
= −5.82... × 103 J
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
At constant pressure the heat released by the system is the enthalpy change of
the system, q = ∆H. Because q sur = −q, the entropy change of the surroundings
is
−q −(−5.82... × 103 J)
=
T
273.15 K − 5.00 K
= +21.7... J K−1 = +21.7 J K−1 .
∆S sur =
�erefore the total entropy change is
∆S tot = ∆S + ∆S sur = (−21.3... J K−1 ) + (+21.7... J K−1 )
= +0.403... J K−1 = +0.4 J K−1 .
Because the total entropy change is positive, the Second Law implies that the
process is spontaneous .
A similar method is used to �nd the entropy change when the liquid evaporates
at T2 . Consider heating the liquid to the boiling temperature Tb , then the phase
transition taking place, followed by cooling of the gas back to the temperature
T2 . �e entropy changes are calculated in an analagous way
∆S = ∆S 1 + ∆S 2 + ∆S 3
−
○
∆ vap H m
Tb
T2
�+n
+ nC p,m (H2 O(g)) ln � �
T2
Tb
Tb
273.15 K + 100 K
−1
−1
= (1.00 mol) × (75.3 J K mol ) ln �
�
273.15 K + 95.0 K
4.07 × 104 J mol−1
+ (1.00 mol) ×
273.15 K + 100 K
273.15 K + 95.0 K
+ (1.00 mol) × (33.6 J K−1 mol−1 ) ln �
�
273.15 K + 100 K
= (+1.01... J K−1 ) + (+1.09... × 102 J K−1 ) + (−0.453... J K−1 )
= nC p,m (H2 O(l)) ln �
= +1.09... × 102 J K−1 = +110 J K−1 .
−∆H
1
= − × (∆H 1 + ∆H 2 + ∆H 3 )
T2
T2
−
○
−∆ vap H m
Tb − T2
T2 − Tb
= − �nC p,m (H2 O(l))
+n
+ nC p,m (H2 O(g))
�
T2
T2
T2
5.00 K
= −(1.00 mol) × (75.3 J K−1 mol−1 )
273.15 K + 95.0 K
4.07 × 104 J mol−1
− (1.00 mol) ×
273.15 K + 95.0 K
−5.00 K
− (1.00 mol) × (33.6 J K−1 mol−1 ) ×
273.15 K + 95.0 K
= −(+1.02... J K−1 ) − (+1.10... × 102 J K−1 ) − (−0.456... J K−1 )
∆S sur =
= −1.11... × 102 J K−1 = −111 J K−1 .
81
82
3 THE SECOND AND THIRD LAWS
�erefore the total entropy change is
∆S tot = ∆S + ∆S sur = (+1.09... × 102 J K−1 ) + (−1.11... × 102 J K−1 )
= −1.48... J K−1 = −1.5 J K−1 .
Because the change in the entropy is negative, the Second Law implies that the
process is not spontaneous .
P�B.�
Consider heating trichloromethane at constant pressure from the initial to �nal temperatures. �e variation of the entropy with temperature is given by
T
[�B.�–��], S(Tf ) = S(Ti ) + ∫Ti f (C p �T)dT. �e contant-pressure molar heat
capacity is given as a function of temperature of a form C p,m = a + bT, with
a = +91.47 J K−1 mol−1 and b = +7.5 × 10−2 J K−2 mol−1 . �us the change in
molar entropy, ∆S m = S m (Tf ) − S m (Ti ), of this process is
∆S m = �
Tf
Ti
(C p,m �T)dT = �
= a�
Tf
Ti
= a × ln �
Ti
Tf a + bT
T
dT
Tf
1
dT + b × �
dT
T
Ti
Tf
� + b × (Tf − Ti )
Ti
300 K
�
273 K
+ (+7.5 × 10−2 J K−2 mol−1 ) × (300 K − 273 K)
= (+91.47 J K−1 mol−1 ) × ln �
= (+8.62... J K−1 mol−1 ) + (+2.02... J K−1 mol−1 )
P�B.�
= +10.7 J K−1 mol−1 .
Two identical blocks must come to their average temperature. �erefore the
�nal temperature is
T = 12 (Tc + Th )
Although the above result may seem self-evident, the more detailed explaination is as follows. �e heat capacity at constant volume is de�ned in [�A.��–
��], C V = (∂U�∂T)V . As shown in Section �A.�(b) on page ��, if the heat
capacity is constant, the internal energy changes linearly with the change in
temperature. �at is ∆U = C V ∆T = C V (Tf − Ti ). For the two blocks at
the initial temperatures of Tc and Th , the change in internal energy to reach
the �nal temperature T is ∆U c = C V ,c (T − Tc ) and ∆U h = C V ,h (T − Th ),
respectively. �e blocks of metal are made of the same substance and are of
the same size, therefore C V ,c = C V ,h = C V . Note that for a given solid the
internal energy does not change signi�cantly on the volume or pressure, thus
it can be assumed that C V = C p . Assuming the system is isolated the total
change in internal energy is ∆U = ∆U c + ∆U h = 0. �is means that ∆U =
C p ((T − Tc ) − (T − Th )) = C p × (2T − (T1 + T2 )) = 0, which implies that the
�nal temperature is T = 12 (Tc + Th ), as stated above.
At constant pressure the temperature dependence of the entropy is given by
[�B.�–��],
Tf
∆S = nC p,m ln � �
Ti
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�erefore for the two blocks
∆S c = nC p,m ln �
�e total change in entropy is
T
�
Tc
and
∆S h = nC p,m ln �
T
�
Th
∆S tot = ∆S c + ∆S h
T
T
� + nC p,m ln � �
Tc
Th
T2
= nC p,m × ln �
�
Tc × Th
= nC p,m ln �
� � 1 (Tc + Th )� �
m
= C p,m × ln 2
M
� Tc × Th
�
2
=
m
(Tc + Th )2
C p,m ln �
�.
M
4(Tc × Th )
where m is the mass of the block and M is the molar mass.
In the case given
∆S tot =
P�B.�
500 g
(250 K + 500 K)2
−1
−1
×
(24.4
J
K
mol
)
×
ln
�
�
4 × (250 K × 500 K)
63.55 g mol−1
= +22.6 J K−1 .
�e heat produced by the resistor over a time period ∆t is q = power × ∆t =
IV ∆t = I 2 R∆t, where the last expression was obtained using Ohm’s law, V =
IR. Note that care is needed handling the units. From the inside of the front
cover of the textbook use (1 A) ≡ (1 Cs−1 ) and (1 V) ≡ (1 JC−1 ), so that
(1 Ω) ≡ (1 JsC−2 ). �erefore the units of the �nal expression for the heat are
as expected
A2 × Ω × s ≡ (C2 s−2 ) × (JsC−2 ) × (s) ≡ J
Assuming that all the heat is absorbed by the large metal block at constant
pressure, this heat is the change of enthalpy of the system, ∆H = q. �e enthalpy
change on heating is given by [�B.�b–��], ∆H = C p ∆T. �is is rearranged to
give an expression for a temperature change
∆T =
∆H
q
I 2 R∆t
I 2 R∆t
=
=
=
Cp
Cp
Cp
(m�M)C p,m
where m is the mass, M the molar mass and C p,m the molar heat capacity. �us
the �nal temperature of the metal block is
Tf = Ti + ∆T = Ti +
= (293 K) +
I 2 R∆t
(m�M)C p,m
(1.00 A)2 × (1.00 × 103 Ω) × (15.0 s)
�(500 g)�(63.55 g mol−1 )� × (24.4 J K−1 mol−1 )
= (293 K) + 78.1... K = 3.71... × 102 K.
83
84
3 THE SECOND AND THIRD LAWS
�e variation of entropy with temperature at constant pressure is given by [�B.�–
��], S(Tf ) = S(Ti ) + C p ln (Tf �Ti ). �erefore the change in entropy is
Tf
m
Tf
� = � � C p,m ln � �
Ti
M
Ti
500 g
3.71... × 102 K
−1
−1
=�
�
−1 � × (24.4 J K mol ) × ln �
293 K
63.55 g mol
∆S = S(Tf ) − S(Ti ) = C p ln �
= +45.4 J K−1 .
For the second experiment, the initial and �nal states of the metal block is the
same, therefore ∆S = 0 . All the heat is released into surroundings, that is water
bath, which can be assumed to be large enough to retain constant temperature.
�us
q
I 2 R∆t
=
Tsur
Tsur
(1.00 A)2 × (1.00 × 103 Ω) × (15.0 s)
=
= +51.2 J K−1 .
293 K
∆S sur =
P�B.�
As suggested in the hint, �rst consider heating the folded protein at constant
pressure to from the initial temperature T to that of the transition, Ttrs . �e
variation of entropy with temperature at constant pressure is given by [�B.�–
��], S(Tf ) = S(Ti ) + C p ln (Tf �Ti ). �us the change in molar entropy, ∆S m =
S m (Tf ) − S m (Ti ), of this step is
∆S 1,m = C p,m (folded) ln �
Ttrs
�
T
Next consider the unfolding step. �e entropy change of such a transition is
given by [�B.�–��], ∆ trs S = ∆ trs H�Ttrs , thus
∆S 2,m =
○
∆ trs H −m
Ttrs
�e �nal step is cooling the unfolded protein to the initial temperature
∆S 3,m = C p,m (unfolded) ln �
T
Ttrs
� = −C p,m (unfolded) ln �
�.
Ttrs
T
�e overall entropy change is the sum of above steps
∆S m = ∆S 1,m + ∆S 2,m + ∆S 3,m
○
Ttrs
∆ trs H −m
Ttrs
�+
− C p,m (unfolded) ln �
�
T
Ttrs
T
○
∆ trs H −m
Ttrs
=
+ �C p,m (folded) − C p,m (unfolded)� × ln �
�
Ttrs
T
= C p,m (folded) ln �
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Given that C p,m (unfolded) − C p,m (folded) = 6.28 × 103 J K−1 mol−1 , the molar
entropy of unfolding at 25.0 ○ C is thus
∆S m =
5.09 × 105 J mol−1
273.15 K + 75.5 K
273.15 K + 75.5 K
�
273.15 K + 25.0 K
= (1.45... × 103 J K−1 mol−1 ) + (−9.82 × 102 J K−1 mol−1 )
P�B.��
+ (−6.28 × 103 J K−1 mol−1 ) × ln �
= +4.77... × 102 J K−1 mol−1 = +477 J K−1 mol−1 .
(a) Consider a process in which heat �dq� is extracted from the cold source at
temperature Tc , and heat q h = �dq� + �dw� is discarded into the hot sink at
temperature Th . �e overall entropy change of such process is
dS =
−�dq� �dq� + �dw�
+
Tc
Th
For the process to be permissible by the Second Law, the Clausius inequality de�ned in [�A.��–��], dS ≥ 0, must be satis�ed. �erefore
−�dq� �dq� + �dw�
+
≥0
Tc
Th
the equality implies the minimum amount of work for which the process
is permissible. Hence it follows that
�dq� �dq� + �dw�
=
.
Tc
Th
(b) �e expression in (a) is rearranged to �nd �dw� and the given relation,
dq = CdTc , is used to give
�dq�
− �dq�
Tc
dTc
�dw� = CTh �
� − C�dTc �
Tc
�dw� = Th
Integration of both sides between the appropriate limits gives
�
w
0
which evaluates to
�dw ′ � = CTh �
Tf
Ti
�w� = CTh �ln �
�
Tf
dTc
� − C � �dTc �
Tc
Ti
Tf
�� − C�Tf − Ti � .
Ti
85
86
3 THE SECOND AND THIRD LAWS
(c) Using C = (m�M)C p,m , the work needed is
�w� =
250 g
273 K
−1
−1
��
−1 × (75.3 J K mol ) × (293 K) × �ln �
293 K
18.02 g mol
250 g
−1
−1
−
−1 × (75.3 J K mol ) × �273 K − 293 K�
18.02 g mol
= � − 2.16... × 104 � J − � − 2.08... × 104 � J
= +7.47... × 102 J = +7.5 × 102 J .
(d) Assuming constant temperature, for �nite amounts of heat and work, the
expression derrived in (a) becomes
�q� �q� + �w�
=
Tc
Th
�is is rearranged to give the work as
�w� = �
Th
− 1� × �q�
Tc
�e heat transferred during freezing is equal to the enthalpy of the transition, which is the opposite of fusion, q = ∆ trs H = (m�M)(−∆ fus H −○ ).
�erefore the work needed is
�w� = �
293 K
250 g
− 1� × �
× (−6.01 × 103 J K−1 mol−1 )�
273 K
18.02 g mol−1
= 6.10... × 103 J = 6.11 × 103 J .
(e) �e total work is the sum of the two steps described in (c) and (d). �erefore
w tot = (+7.47... × 102 J) + (6.10... × 103 J)
= +6.85... × 103 J = +6.86 kJ .
(f) Assuming no energy losses, power is the total work divided by the time
interval over which the work is done, P = w tot �∆t, hence
∆t =
w tot 6.85... × 103 J
=
= 68.6 s .
P
100 W
3C The measurement of entropy
Answer to discussion question
D�C.�
Because solutions of cations cannot be prepared in the absence of anions, the
standard molar entropies of ions in solution are reported on a scale in which,
by convention, the standard entropy of H+ ions in water is taken as zero at all
temperatures: S −○ (H+ , aq) = 0.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Because the entropies of ions in water are values relative to the hydrogen ion in
water, they may be either positive or negative. A positive entropy means that
an ion has a higher molar entropy than H+ in water, and a negative entropy
means that the ion has a lower molar entropy than H+ in water. An ion with
zero entropy in fact has that same entropy as H+ .
Solutions to exercises
E�C.�(a)
Consider the chemical equation
1
N (g) + 32 H2 (g) �→ NH3 (g)
2 2
−
○
�e standard reaction entropy is given by [�C.�b–��], ∆ r S −○ = ∑J ν J S m
(J),
where ν J are singed stoichiometric coe�cients for a given reaction equation.
�erefore, using data from the Resource section
−
○
−
○
−
○
∆ r S −○ = nS m
(NH3 , (g)) − 32 nS m
(H2 , (g)) − 12 nS m
(N2 , (g))
= (1.00 mol) × (192.45 J K−1 mol−1 )
− � 32 × 1.00 mol� × (130.684 J K−1 mol−1 )
− � 12 × 1.00 mol� × (191.61 J K−1 mol−1 )
E�C.�(a)
= −99.38 J K−1 .
Assuming that the Debye extrapolation is valid, the constant-pressure molar
heat capacity is C p,m (T) = aT 3 . �e temperature dependence of the entropy
T
is given by [�C.�a–��], S(T2 ) = S(T1 ) = ∫T12 (C p,m �T)dT. For a given temperature T the change in molar entropy from zero temperature is therefore
S m (T) − S m (0) = �
Hence
p,m
dT ′ = �
T′
0
T aT ′ 3
dT ′
T′
T
aT 3 C p,m (T)
2
= a � T ′ dT ′ =
=
3
3
0
S m (4.2 K) − S m (0) =
E�C.�(a)
T C
0
C p,m (4.2 K) 0.0145 J K−1 mol−1
=
3
3
= 4.8 × 10−3 J K−1 mol−1 .
−
○
�e standard reaction entropy is given by [�C.�b–��], ∆ r S −○ = ∑J ν J S m
(J),
where ν J are the signed stoichiometric numbers.
(i)
−
○
−
○
−
○
∆ r S −○ = 2S m
(CH3 COOH, (l)) − 2S m
(CH3 CHO, (g)) − S m
(O2 , (g))
= 2 × (159.8 J K−1 mol−1 ) − 2 × (250.3 J K−1 mol−1 )
− (205.138 J K−1 mol−1 )
= −386.1 J K−1 mol−1 .
87
88
3 THE SECOND AND THIRD LAWS
(ii)
−
○
−
○
−
○
−
○
∆ r S −○ = 2S m
(AgBr, (s)) + S m
(Cl2 , (g)) − 2S m
(AgCl, (s)) − S m
(Br2 , (l))
= 2 × (107.1 J K−1 mol−1 ) + (223.07 J K−1 mol−1 )
− 2 × (96.2 J K−1 mol−1 ) − (152.23 J K−1 mol−1 )
(iii)
= +92.6 J K−1 mol−1 .
−
○
−
○
−
○
∆ r S −○ = S m
(HgCl2 , (s)) − S m
(Hg, (l)) − S m
(Cl2 , (g))
= (146.0 J K−1 mol−1 ) − (76.02 J K−1 mol−1 )
− (223.07 J K−1 mol−1 )
Solutions to problems
P�C.�
= −153.1 J K−1 mol−1 .
Consider the process of determining the calorimetric entropy from zero to
the temperature of interest. Assuming that the Debye extrapolation is valid,
the constant-pressure molar heat capacity at the lowest temperatures is of a
form C p,m (T) = aT 3 . �e temperature dependence of the entropy is given
T
by [�C.�a–��], S(T2 ) = S(T1 ) = ∫T12 (C p,m �T)dT. �us for a given (low)
temperature T the change in molar entropy from zero is
S m (T) − S m (0) = �
Hence
T C
0
= a�
T
0
p,m
dT ′ = �
T′
0
T ′ dT ′ =
2
T aT ′ 3
T′
dT ′
a 3 1
T = 3 C p,m (T)
3
−
○
−
○
Sm
(10 K) − S m
(0) = 13 × (4.64 J K−1 mol−1 ) = 1.54... J K−1 mol−1
−
○
�e increase in entropy on raising the temperature to the melting point is S m
(234.4 K)−
−1
−1
−
○
S m (10 K) = 57.74 J K mol . �e entropy change of a phase transition is
given by [�C.�b–��], ∆ trs S(Ttrs ) = ∆ trs H(Ttrs )�Ttrs . �us
−
○
∆ fus S m
(234.4 K) =
2322 J mol−1
= 9.90... J K−1 mol−1
234.4 K
Further raising the temperature to 298.0 K gives an increase in the entropy of
−
○
−
○
Sm
(298 K) − S m
(234.4 K) = 6.85 J K−1 mol−1 .
�e �ird-Law standard molar entropy at 298 K is the sum of the above con-
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
tributions.
−
○
−
○
−
○
−
○
−
○
−
○
Sm
(298 K) − S m
(0) = (S m
(10 K) − S m
(0)) + (S m
(234.4 K) − S m
(10 K))
−
○
−
○
−
○
+ ∆ fus S m
(234.4 K) + (S m
(298 K) − S m
(234.4 K))
= (1.54... J K−1 mol−1 ) + (57.74 J K−1 mol−1 )
+ (9.90... J K−1 mol−1 ) + (6.85 J K−1 mol−1 )
P�C.�
= 76.04 J K−1 mol−1 .
(a) Assuming that the Debye extrapolation is valid, the constant-pressure
molar heat capacity is of a form C p,m (T) = aT 3 . �e temperature depenT
dence of the entropy is given by [�C.�a–��], S(T2 ) = S(T1 ) = ∫T12 (C p,m �T)dT.
�us for a given (low) temperature T the change in the molar entropy
from zero is
S m (T) − S m (0) = �
T C
0
= a�
Hence
T
0
p,m
dT ′ = �
T′
0
T ′ dT ′ =
2
T aT ′ 3
T′
dT ′
a 3 1
T = 3 C p,m (T)
3
−
○
−
○
Sm
(10 K) − S m
(0) = 13 × (2.8 J K−1 mol−1 ) = 0.933... J K−1 mol−1
= �.�� J K−1 mol−1 .
(b) �e change in entropy is determined calorimetrically by measuring the
area under a plot of (C p,m �T) against T, as shown in Fig. �.�.
�e plot is rather irregular and is best �tted by two polynomials of order
�: one in the range 10 K to 30 K and the other in the range 30 K to 298 K.
De�ne y = (C p,m �T)�(J K−2 mol−1 ) and x = T�K, so that the �tted
function is expressed
y = c3 x 3 + c2 x 2 + c1 x + c0
where the best �tted coe�cients c i for the respective temperature ranges
are
ci
c3
c2
c1
c0
10 K to 30 K
+5.0222 × 10−5
−4.3010 × 10−3
+1.2025 × 10−1
−5.4187 × 10−1
30 K to 298 K
−5.2881 × 10−8
+3.5425 × 10−5
−8.1107 × 10−3
+7.5533 × 10−1
�e integral of the �tted functions over the range x i to x f is
I=�
=
xf
xi
c 3 x 3 + c 2 x 2 + c 1 x + c 0 dx
c3
c2
c1
�x f 4 − x i 4 � + �x f 3 − x i 3 � + �x f 2 − x i 2 � + c 0 (x f − x i )
4
3
2
89
3 THE SECOND AND THIRD LAWS
C p,m �(J K−1 mol−1 )
2.8
7.0
10.8
14.1
16.5
21.4
23.3
24.5
25.3
25.8
26.2
26.6
T�K
10
15
20
25
30
50
70
100
150
200
250
298
(C p,m �T)�(J K−2 mol−1 )
0.280 0
0.466 7
0.540 0
0.564 0
0.550 0
0.428 0
0.332 9
0.245 0
0.168 7
0.129 0
0.104 8
0.089 3
0.6
(C p,m �T)�(J K−2 mol−1 )
90
0.4
0.2
0.0
Figure 3.2
0
50
100
150
T�K
200
250
300
Using the appropriate coe�cients and limits the integrals are evaluated to
give the respective changes in entropy
−
○
−
○
Sm
(30 K) − S m
(10 K) = 10.0... J K−1 mol−1
−
○
−
○
Sm
(298 K) − S m
(30 K) = 53.8... J K−1 mol−1
�e total entropy change is the sum of the two integrals. �erefore
−
○
−
○
Sm
(298 K) − S m
(10 K) = (10.0... J K−1 mol−1 ) + (53.8... J K−1 mol−1 )
= 63.9... J K−1 mol−1 = ��.� J K−1 mol−1 .
(c) �e standard �ird-Law entropy at 298 K is the sum of the above calcu-
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
91
lated contributions. �us
−
○
−
○
−
○
−
○
Sm
(298 K) − S m
(0) = (S m
(298 K) − S m
(10 K))
−
○
−
○
+ (S m
(10 K) − S m
(0))
= (63.9... J K−1 mol−1 ) + (0.933... J K−1 mol−1 )
= ��.� J K−1 mol−1 .
For the standard �ird-Law entropy at 273 K, the second integral in part
(b) needs to be repeated with Tf = 273 K. �erefore
−
○
−
○
Sm
(273 K) − S m
(30 K) = 51.4... J K−1 mol−1
�e other contributions are the same, hence
−
○
−
○
Sm
(273 K) − S m
(0) = (10.0... J K−1 mol−1 + 51.4... J K−1 mol−1 )
+ (0.933... J K−1 mol−1 )
P�C.�
= ��.� J K−1 mol−1 .
−
○
�e standard reaction entropy is given by [�C.�b–��], ∆ r S −○ = ∑J ν J S m
(J),
where ν J are the signed stoichiometric numbers.
−
○
−
○
∆ r S −○ (298 K) = S m
(CO, (g)) + S m
(H2 O, (g))
−
○
−
○
− Sm
(CO2 , (g)) − S m
(H2 , (g))
= (197.67 J K−1 mol−1 ) + (188.83 J K−1 mol−1 )
− (213.74 J K−1 mol−1 ) − (130.684 J K−1 mol−1 )
= +42.07... J K−1 mol−1 = +42.08 J K−1 mol−1 .
Similarly, the standard reaction enthalpy is given by [�C.�b–��], ∆ r H −○ = ∑J ν J ∆ f H −○ (J).
∆ r H −○ (298 K) = ∆ f H −○ (CO, (g)) + ∆ f H −○ (H2 O, (g))
− ∆ f H −○ (CO2 , (g)) − ∆ f H −○ (H2 , (g))
= (−110.53 kJ mol−1 ) + (−241.82 kJ mol−1 )
− (−393.51 kJ mol−1 ) − 0
= +41.16 kJ mol−1 .
�e temperature dependence of the reaction entropy is given by [�C.�a–��],
T
∆ r S −○ (T2 ) = ∆ r S −○ (T1 ) + ∫T12 (∆ r C −p○ �T)dT. Similarly, the enthalpy dependence on temperature is given by Kirchho� ’s law [�C.�a–��], ∆ r H −○ (T2 ) =
T
∆ r H −○ (T1 ) + ∫T12 ∆ r C −p○ dT. �e quantity ∆ r C −p○ is de�ned in [�C.�b–��], ∆ r C −p○ =
○
(J). For the reaction at 298 K
∑J ν J C −p,m
○
○
∆ r C −p○ = C −p,m
(CO, (g)) + C −p,m
(H2 O, (g))
○
○
− C −p,m
(CO2 , (g)) − C −p,m
(H2 , (g))
= (29.14 J K−1 mol−1 ) + (33.58 J K−1 mol−1 )
− (37.11 J K−1 mol−1 ) − (28.824 J K−1 mol−1 )
= −3.21... J K−1 mol−1
92
3 THE SECOND AND THIRD LAWS
Assuming that ∆ r C −p○ is constant over the temperature range involved, the standard entropy and enthalpy changes of the reaction is given by, respectively,
[�C.�b–��], ∆ r S −○ (T2 ) = ∆ r S −○ (T1 )+∆ r C −p○ ln(T2 �T1 ), and [�C.�d–��], ∆ r H −○ (T2 ) =
∆ r H −○ (T1 ) + ∆ r C −p○ (T2 − T1 ).
∆ r S −○ (398 K) = ∆ r S −○ (298 K) + ∆ r C −p○ × ln �
= (+42.0... J K−1 mol−1 )
398 K
�
298 K
+ (−3.21... J K−1 mol−1 ) × ln �
= +41.15 J K−1 mol−1 .
398
�
298
∆ r H −○ (398 K) = ∆ r H −○ (298 K) + ∆ r C −p○ × (398 K − 298 K)
= (+41.1... × 103 J mol−1 )
+ (−3.21... J K−1 mol−1 ) × (100 K)
= +40.8... × 103 J mol−1 = +40.8 kJ mol−1 .
P�C.�
Assuming that the Debye extrapolation is valid, the constant-pressure molar
heat capacity is of a form C p,m (T) = aT 3 . �e temperature dependence of the
T
entropy is given by [�C.�a–��], S(T2 ) = S(T1 ) = ∫T12 (C p,m �T)dT. �us for a
given (low) temperature T the change in the molar entropy from zero is
S m (T) − S m (0) = �
T C
0
= a�
T
0
p,m
dT ′ = �
T′
0
T ′ dT ′ =
2
T aT ′ 3
T′
dT ′
a 3 1
T = 3 C p,m (T)
3
Hence
−
○
−
○
Sm
(14.14 K) − S m
(0) = 13 × (9.492 J K−1 mol−1 ) = 3.16... J K−1 mol−1
�e change in entropy is determined calorimetrically by measuring the area
under a plot of (C p,m �T) against T.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
T�K
14.14
16.33
20.03
31.15
44.08
64.81
100.90
140.86
183.59
225.10
262.99
298.06
C p,m �(J K−1 mol−1 )
9.492
12.70
18.18
32.54
46.86
66.36
95.05
121.3
144.4
163.7
180.2
196.4
(C p,m �T)�(J K−2 mol−1 )
0.671 29
0.777 71
0.907 64
1.044 62
1.063 07
1.023 92
0.942 02
0.861 14
0.786 54
0.727 23
0.685 20
0.658 93
(C p,m �T)�(J K−2 mol−1 )
1.2
1.0
0.8
0.6
0
50
100
150
T�K
200
250
300
�e plot is rather irregular and is best �tted by two polynomials of order �
and �, respectively, in the ranges 14.14 K to 44.08 K and 44.08 K to 298.06 K.
De�ne y = (C p,m �T)�(J K−2 mol−1 ) and x = T�K, so that the �tted function
is expressed
y = c3 x 3 + c2 x 2 + c1 x + c0
where the best �tted coe�cients c i for the respective temperature ranges are
ci
c3
c2
c1
c0
14.14 K to 44.08 K
�
−7.6119 × 10−4
+5.6367 × 10−2
+5.1090 × 10−2
44.08 K to 298.06 K
+7.1979 × 10−9
−3.0830 × 10−7
−2.2415 × 10−3
+1.1644 × 100
93
94
3 THE SECOND AND THIRD LAWS
�e integral of the �tted functions over the range x i to x f is
I=�
=
xf
xi
c 3 x 3 + c 2 x 2 + c 1 x + c 0 dx
c3
c2
c1
�x f 4 − x i 4 � + �x f 3 − x i 3 � + �x f 2 − x i 2 � + c 0 (x f − x i )
4
3
2
�e temperatures of interest are beyond 44.08 K. �us the contribution corresponding to the integral of the quadratic is included in the estimates of the
entropy. Using the appropriate coe�cients and limits the integral gives
−
○
−
○
Sm
(44.08 K) − S m
(14.14 K) = 29.6... J K−1 mol−1
Finally, the remaining contribution is found by estimating the integral of the
cubic polynomial to the temperature of interest. �erefore for T = 100 K, 200 K
and, a�er small extrapolation, 300 K
−
○
−
○
Sm
(100 K) − S m
(44.08 K) = 56.1... J K−1 mol−1
−
○
−
○
Sm
(200 K) − S m
(44.08 K) = 1.41... × 102 J K−1 mol−1
−
○
−
○
Sm
(300 K) − S m
(44.08 K) = 2.11... × 102 J K−1 mol−1
�e standard �ird-Law molar entropy at the temperatures of interests is the
sum of all the contributions up to that point. �us, the entropy at three temperatures are
−
○
−
○
−
○
−
○
Sm
(100 K) − S m
(0) = S m
(14.14 K) − S m
(0)
−
○
−
○
+ (S m
(44.08 K) − S m
(14.14 K))
−
○
−
○
+ (S m
(100 K) − S m
(44.08 K))
= (3.16... J K−1 mol−1 ) + (29.6... J K−1 mol−1 )
+ (56.1... J K−1 mol−1 )
= ��.� J K−1 mol−1 .
Similarly it is found for 200 K and 300 K, respectively
−
○
−
○
Sm
(200 K) − S m
(0) = (3.16... J K−1 mol−1 ) + (29.6... J K−1 mol−1 )
+ (1.41... × 102 J K−1 mol−1 )
= ���.� J K−1 mol−1 .
−
○
−
○
Sm
(300 K) − S m
(0) = (3.16... J K−1 mol−1 ) + (29.6... J K−1 mol−1 )
+ (2.11... × 102 J K−1 mol−1 )
P�C.�
= ���.� J K−1 mol−1 .
(a) Given the expression for the constant-pressure molar heat capacity, C p,m (T) =
aT 3 + bT, consider C p,m �T.
C p,m aT 3 + bT
=
= aT 2 + b
T
T
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�is expression is of the form of a straight line, y = (slope)×x+(intercept),
if y = C p,m (T) and x = T 2 . It follows that (slope) = a and (intercept) = b.
(b) �e data below are plotted in Fig. �.�.
T�K
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
C p,m �(J K−1 mol−1 )
0.437
0.560
0.693
0.838
0.996
1.170
1.361
1.572
T 2 �(K2 )
0.040
0.063
0.090
0.123
0.160
0.203
0.250
0.303
(C p,m �T)�(J K−2 mol−1 )
2.185 0
2.240 0
2.310 0
2.394 3
2.490 0
2.600 0
2.722 0
2.858 2
(C p,m �T)�(J K−2 mol−1 )
3.0
2.8
2.6
2.4
2.2
2.0
0.05
0.10
0.15
Figure 3.3
0.20
T�K
0.25
0.30
0.35
�e data lie on a good straight line, the equation of which is
(C p,m �T)�(J K−2 mol−1 ) = 2.569 × (T�K)2 + 2.080
�us a = 2.569 JK−4 mol−1 and b = 2.080 JK−2 mol−1 .
(c) �e dependence of the entropy on temperature is given by [�C.�a–��],
T
S(T2 ) = S(T1 ) + ∫T12 (C p,m �T)dT. �us for a given (low) temperature T
the molar entropy change from the zero temperature is
S m (T) = S m (0) + �
= S m (0) +
T C
0
p,m
dT ′ = S m (0) + �
T′
0
a 3
T + bT .
3
T
(aT ′ + b) dT ′
2
95
96
3 THE SECOND AND THIRD LAWS
(d) Assuming that the expression derived above can be extrapolated to 2.0 K
S m (2.0 K) − S m (0) =
(2.569 JK−4 mol−1 )
× (2.0 K)3
3
+ (2.080 JK−2 mol−1 ) × (2.0 K)
= 11.01 J K−1 mol−1 .
3D Concentrating on the system
Answers to discussion questions
D�D.�
As is discussed in detail in Topic �D the criteria for spontaneity at constant
volume and temperature is expressed in terms of the Helmholtz energy, dA ≤ 0,
and at constant pressure and temperature in terms of the Gibbs energy, dG ≤ 0.
Both the Helmholtz and Gibbs energies refer to properties of the system alone.
However, because of the way they are de�ned these quantities e�ectively allow
the entropy change of the system plus surroundings to be evaluated. For example, at constant volume and temperature the change in the Helmholtz energy
is expressed in terms of the internal energy change and the entropy change
of the system: dA = dU − TdS. If this expression is divided by −T to give
−dA�T = −dU�T + dS the two terms on the right can bothe be identi�ed as
entropy changes.
�e �rst term, −dU�T, is equal to the entropy change of the surroundings
because dq sur = −dq, and at constant volume dq = dU. �e second term is
the entropy change of the system. �us the sum of the two is the total entropy change, which the Second Law shows must be positive in a spontaneous
process. �erefore, the change in the Helmholtz energy is an indicator of the
total entropy change, even though the former refers only to the system. Similar
considerations can be applied to the Gibbs energy.
It is also possible to express the criterion for spontaneity in terms of the change
in H, U or S for the system. For example, as shown in Topic �D, dS U ,V ≥ 0.
However, the variables which are being held constant (here U and V ) do not
correspond to such easily realizable conditions such as constant temperature
and volume (or pressure) so such criteria are less applicable to chemical systems.
Solutions to exercises
E�D.�(a)
�e maximum non-expansion work is equal to the Gibbs free energy as explained in Section �D.�(e) on page ��. �e standard reaction Gibbs energy is
given by [�D.��b–��], ∆ r G −○ = ∑J ν J ∆ f G −○ (J), where ν J are the signed stoichio-
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
metric numbers. For the reaction CH4 (g) + 3O2 (g) �→ CO2 (g) + 2H2 O(l)
∆ r G −○ = ∆ f G −○ (CO2 ,(g)) + 2∆ f G −○ (H2 O ,(l)) − ∆ f H −○ (CH4 ,(g))
− 3∆ f H −○ (O2 ,(g))
= (−394.36 kJ mol−1 ) + 2 × (−237.13 kJ mol−1 )
− (−50.72 kJ mol−1 ) − 3 × 0
= −817.90 kJ mol−1 .
E�D.�(a)
�erefore, �w add,max � = �∆ r G −○ � = 817.90 kJ mol−1 .
�e standard reaction Gibbs energy is given by [�D.��b–��], ∆ r G −○ = ∑J ν J ∆ f G −○ (J),
where ν J are the signed stoichiometric numbers.
(i)
∆ r G −○ = 2∆ f G −○ (CH3 COOH ,(l)) − 2∆ f G −○ (CH3 CHO ,(g)) − ∆ f G −○ (O2 ,(g))
= 2 × (−389.9 kJ mol−1 ) − 2 × (−128.86 kJ mol−1 ) − 0
(ii)
= −522.1 kJ mol−1 .
∆ r G −○ = 2∆ f G −○ (AgBr ,(s)) + ∆ f G −○ (Cl2 ,(g)) − 2∆ f G −○ (AgCl ,(s))
− ∆ f G −○ (Br2 ,(l))
= 2 × (−96.90 kJ mol−1 ) + 0 − 2 × (−109.79 kJ mol−1 ) − 0
(iii)
E�D.�(a)
= +25.78 kJ mol−1 .
∆ r G −○ = ∆ f G −○ (HgCl2 ,(s)) − ∆ f G −○ (Hg ,(l)) − ∆ f G −○ (Cl2 ,(g))
= (−178.6 kJ mol−1 ) − 0 − 0 = −178.6 kJ mol−1 .
Consider the reaction
CH3 COOC2 H5 (l) + 5O2 (g) �→ 4CO2 (g) + 4H2 O(l)
�e standard reaction enthalpy is [�C.�b–��], ∆ r H −○ = ∑J ν J ∆ f H −○ (J), where ν J
are the signed stoichiometric numbers.
∆ r H −○ = 4∆ f H −○ (CO2 ,(g)) + 4∆ f H −○ (H2 O ,(l))
− ∆ f H −○ (CH3 COOC2 H5 ,(l)) − 5∆ f H −○ (O2 ,(g))
when rearranged this gives
∆ f H −○ (CH3 COOC2 H5 ,(l)) = 4∆ f H −○ (CO2 ,(g)) + 4∆ f H −○ (H2 O ,(l))
− 5∆ f H −○ (O2 ,(g)) − ∆ r H −○
= 4 × (−393.51 kJ) + 4 × (−285.83 kJ)
− 5 × 0 − (−2231 kJ mol−1 )
= −486.36 kJ mol−1
97
98
3 THE SECOND AND THIRD LAWS
−
○
�e standard reaction entropy is given by [�C.�b–��], ∆ r S −○ = ∑J ν J S m
(J).
�erefore, for the formation of the compound
−
○
−
○
∆ f S −○ (CH3 COOC2 H5 ,(l)) = S m
(CH3 COOC2 H5 ,(l)) − 4S m
(C ,(s))
−
○
−
○
− 4S m
(H2 ,(g)) − S m
(O2 ,(g))
= (259.4 J K−1 mol−1 ) − 4 × (5.740 J K−1 mol−1 )
− 4 × (130.684 J K−1 mol−1 ) − (205.138 J K−1 mol−1 )
= −4.91... × 102 J K−1 mol−1
�e standard reaction Gibbs energy is de�ned in [�D.�–��], ∆ r G −○ = ∆ r H −○ −
T∆ r S −○ , thus
∆ f G −○ (CH3 COOC2 H5 ,(l)) = (−486.36 kJ mol−1 )
− (298.15 K)(−0.491... kJ K−1 mol−1 )
E�D.�(a)
= −340 kJ mol−1 .
�e standard reaction Gibbs energy is given by [�D.�–��], ∆ r G −○ = ∆ r H −○ −
T∆ r S −○ . �e standard reaction enthalpy is given in terms of the enthalpies
of formation by [�C.�b–��], ∆ r H −○ = ∑J ν J ∆ f H −○ (J), where ν J are the signed
stoichiometric numbers.
(i)
∆ r H −○ = 2∆ f H −○ (CH3 COOH ,(l)) − 2∆ f H −○ (CH3 CHO ,(g))
− ∆ f H −○ (O2 ,(g))
= 2 × (−484.5 kJ mol−1 ) − 2 × (−166.19 kJ mol−1 ) − 0
= −636.62 kJ mol−1 = −636.6 kJ mol−1 .
Given the result for the previous execise, ∆ r S −○ = −386.1 J K−1 mol−1
∆ r G −○ = (−636.62 kJ mol−1 ) − (298.15 K) × (−0.3861 kJ K−1 mol−1 )
(ii)
= −521.5 kJ mol−1 .
∆ r H −○ = 2∆ f H −○ (AgBr ,(s)) + ∆ f H −○ (Cl2 ,(g)) − 2∆ f H −○ (AgCl ,(s))
− ∆ f H −○ (Br2 ,(l))
= 2 × (−100.37 kJ mol−1 ) + 0 − 2 × (−127.07 kJ mol−1 ) − 0
= +53.40 kJ mol−1 .
Given the result for the previous execise, ∆ r S −○ = +92.6 J K−1 mol−1
∆ r G −○ = (+53.40 kJ mol−1 ) − (298.15 K) × (+0.0926 kJ K−1 mol−1 )
= +25.8 kJ mol−1 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(iii)
∆ r H −○ = ∆ f H −○ (HgCl2 ,(s)) − ∆ f H −○ (Hg ,(l)) − ∆ f H −○ (Cl2 ,(g))
= (−224.3 kJ mol−1 ) − 0 − 0 = −224.3 kJ mol−1 .
Given the result for the previous execise, ∆ r S −○ = −153.1 J K−1 mol−1 .
∆ r G −○ = (−224.3 kJ mol−1 ) − (298.15 K) × (−0.1531 kJ K−1 mol−1 )
E�D.�(a)
= −178.7 kJ mol−1 .
−
○
�e standard reaction entropy is given by [�C.�b–��], ∆ r S −○ = ∑J ν J S m
(J),
where ν J are the signed stoichiometric numbers.
−
○
−
○
−
○
−
○
∆ r S −○ = 2S m
(I2 ,(s)) + 2S m
(H2 O ,(l)) − 4S m
(HI ,(g)) − S m
(O2 ,(g))
= 2 × (116.135 J K−1 mol−1 ) + 2 × (69.91 J K−1 mol−1 )
− 4 × (206.59 J K−1 mol−1 ) − (205.138 J K−1 mol−1 )
= −659.40... J K−1 mol−1
�e standard reaction enthalpy is given by [�C.�b–��], ∆ r H −○ = ∑J ν J ∆ f H −○ (J).
∆ r H −○ = 2∆ f H −○ (I2 ,(s)) + 2∆ f H −○ (H2 O ,(l)) − 4∆ f H −○ (HI ,(g))
− ∆ f H −○ (O2 ,(g))
= 2 × 0 + 2 × (−285.83 kJ mol−1 ) − 4 × (+26.48 kJ mol−1 ) − 0
= −677.58 kJ mol−1
�e standard reaction Gibbs energy is given by [�D.�–��], ∆ r G −○ = ∆ r H −○ −
T∆ r S −○ .
∆ r G −○ = (−677.58 kJ mol−1 ) − (298.15 K) × (−0.65940... kJ K−1 mol−1 )
= −480.98 kJ mol−1 .
Solutions to problems
P�D.�
(a) From the perfect gas law, pV = nRT, the �nal pressure is
p=
n B RTB (2.00 mol) × (8.3145 × 10−2 dm3 bar K−1 mol−1 ) × (300 K)
=
VB,f
1.00 dm3
= 49.8... bar = 49.9 bar .
�e �nal volume of Section A is VA,f = (VA + VB ) − VB,f = 3.00 dm3 .
�erefore the �nal temperature of Section A is
TA,f =
pVA,f
(49.8... bar) × (3.00 dm3 )
=
nR
(2.00 mol) × (8.3145 × 10−2 dm3 bar K−1 mol−1 )
= 900 K .
99
100
3 THE SECOND AND THIRD LAWS
(b) Taking the hint, �rst consider the heating at constant volume. �e entropy
dependence on temperature at constant volume is given by [�B.�–��],
∆S = nC V ,m ln (Tf �Ti ), with C p replaced by nC V ,m . �e volume is then
allowed to expand to the �nal. �e entropy change for the isothermal
expansion of a perfect gas is given by [�B.�–��], ∆S = nR ln(Vf �Vi ).
�erefore the total change in the entropy for the gas in Section A is
∆S A = nC V ,m ln �
TA,f
VA,f
� + nR ln �
�
TA
VA
900 K
�
300 K
3.00 dm3
+ (2.00 mol) × (8.3145 J K−1 mol−1 ) × ln �
�
2.00 dm3
= (2.00 mol) × (20.0 J K−1 mol−1 ) × ln �
= +50.6... J K−1 = +50.7 J K−1 .
(c) Section B is kept at the constant temperature throughout the process, thus
only the change in the volume needs to be considered
∆S B = nR ln �
VB,f
�
VB
= (2.00 mol) × (8.3145 J K−1 mol−1 ) × ln �
= −11.5... J K−1 = −11.5 J K−1 .
1.00 dm3
�
2.00 dm3
(d) �e change in internal energy as a result of a change in temperature assuming constant heat capacity is given by [�A.��b–��], ∆U = C V ∆T. Because the internal energy of a perfect gas depends only on the temperature
∆U A = (2.00 mol) × (20.0 J K−1 mol−1 ) × (900 K − 300 K)
= +2.40 × 104 J = +24.0 kJ .
∆U B = 0 .
(e) �e Helmholtz energy is de�ned in [�D.�a–��], A = U −T S. For the �nite
changes it becomes ∆A = ∆U − ∆(T S) = ∆U − T∆S − S∆T. Because
Section B is kept at constant temperature ∆T = 0 and so
∆A B = ∆U B − TB ∆S B
= 0 − (300 K) × (−11.5... J K−1 ) = +3.46 × 103 J .
Because ∆T is not zero and S is not given for the Section A, the equivalent
expression cannot be evaluated.
P�D.�
(f) Because the process is reversible, the total ∆A = ∆A A + ∆A B = 0 . �is
implies ∆A A = −∆A B .
Consider the thermodynamic cycle shown in Fig. �.�.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
H+ (g) + e – (g) + I (g)
∆ ec G −○ (I)
= −E a (I)
∆ ion G −○ (H) = I(H)
H+ (g) + I – (g)
H (g) + I (g)
∆ f G −○ (H (g))
+∆ f G −○ (I (g))
∆ solv G −○ (H+ )
+∆ solv G −○ (I− )
1
H (g) + 12 I� (g)
2 �
−∆ f G −○ (H+ (aq))
−∆ f G −○ (I− (aq))
H+ (aq) + I – (aq)
Figure 3.4
�e sum of the Gibbs energies for all the steps around a closed cycle is zero.
0 = − [∆ f G −○ (H+ (aq)) + ∆ f G −○ (I− (aq))] + ∆ f G −○ (H (g)) + ∆ f G −○ (I (g))
I(H) + (−E a (I)) + ∆ solv G −○ (H+ ) + ∆ solv G −○ (I− )
Because ∆ f G −○ (H+ (aq) = 0, by convention, the Gibbs energy of formation for
the I – (aq) is
∆ f G −○ (I− (aq)) = ∆ f G −○ (H (g)) + ∆ f G −○ (I (g)) + I(H)
− E a (I) + ∆ solv G −○ (H+ ) + ∆ solv G −○ (I− )
= (203.25 kJ mol−1 ) + (70.25 kJ mol−1 ) + (1312.0 kJ mol−1 )
− (295.3 kJ mol−1 ) + (−1090 kJ mol−1 ) + (−247 kJ mol−1 )
P�D.�
= -�� kJ mol−1 .
�e standard reaction Gibbs energy is given by [�D.�–��], ∆ r G −○ = ∆ r H −○ −
−
○
T∆ r S −○ . �e standard reaction entropy is [�C.�b–��], ∆ r S −○ = ∑J ν J S m
(J), where
ν J are the signed stoichiometric numbers. �erefore
−
○
−
○
−
○
∆ r S 1−○ = S m
(Li+ ,(g)) + S m
(F− ,(g)) − S m
(LiF ,(s))
= (133 J K−1 mol−1 ) + (145 J K−1 mol−1 ) − (35.6 J K−1 mol−1 )
= 242.4 J K−1 mol−1
101
102
3 THE SECOND AND THIRD LAWS
And so
∆ r G 1−○ = (1037 kJ mol−1 ) − (298 K) × (0.2424 kJ K−1 mol−1 )
= +9.64... × 102 kJ mol−1 = +965 kJ mol−1 .
For the second step ∆ r G 2−○ = ∆ solv G −○ (Li+ ) + ∆ solv G −○ (F− ). �e Gibbs energy of solvation in water is given by Born equation [�D.��b–��], ∆ solv G −○ =
−(z i 2 �[r i �pm]) × 6.86 × 104 kJ mol−1 , thus
(+1)2
(−1)2
+
� × 6.86 × 104 kJ mol−1
+
[r(Li )�pm] [r(F− )�pm]
1
1
= −�
+
� × 6.86 × 104 kJ mol−1 = −9.61... × 102 kJ mol−1
127 163
∆ r G 2−○ = − �
= −961 kJ mol−1
�erefore the total Gibbs energy change of the process
∆ r G −○ = ∆ r G 1−○ + ∆ r G 2−○ = (+9.64... × 102 kJ mol−1 ) + (−9.61... × 102 kJ mol−1 )
= +3.74... kJ mol−1 = +4 kJ mol−1 .
�e change in positive implying that the reverse process is spontaneous.
3E
Combining the First and Second Laws
Answer to discussion questions
D�E.�
�e relation (∂G�∂p)T = V , combined with the fact that the volume is always
positive, shows that the Gibbs function of a system increases as the pressure
increases (at constant temperature).
Solutions to exercises
E�E.�(a)
As explained in Section �E.�(c) on page ���, the change in Gibbs energy of a
p
phase transition varies with pressure as ∆ trs G m (p f ) = ∆ trs G m (p i ) ∫ p i f ∆ trs Vm dp.
Assuming that ∆ trs Vm changes little over the range of pressures considered
∆G m = ∆ trs G m (p f ) − ∆ trs G m (p i ) = (p f − p i )∆ trs Vm
= [(1000 × 105 Pa) − (1 × 105 Pa)] × (−1.6 × 10−6 m3 mol−1 )
E�E.�(a)
= −1.6 × 102 J mol−1 .
�e Gibbs energy dependence on pressure for a perfect gas is given by [�E.��–
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
���], G m (p f ) = G m (p i ) + RT ln(p f �p i ), thus
∆G m = RT ln �
pf
�
pi
= (8.3145 J K−1 mol−1 ) × (298 K) × ln �
E�E.�(a)
100.0 atm
�
1.0 atm
= +11.4... × 103 J mol−1 = +11 kJ mol−1 .
�e Gibbs energy dependence on temperature for a perfect gas is given by
[�E.��–���], G m (p f ) = G m (p i ) + RT ln(p f �p i ). From the perfect gas law p ∝
(1�V ). �is allows rewriting the previous equation for the change in Gibbs
energy due to isothermal gas expansion
∆G = nRT ln �
Vi
�
Vf
= (2.5 × 10−3 mol) × (8.3145 J K−1 mol−1 ) × (300 K) × ln �
E�E.�(a)
= −17 J .
�e variation of the Gibbs energy with pressure is given by [�E.�–���], (∂G�∂T) p =
−S. �e change in entropy is thus
∆S = S f − S i = − �
= −�
E�E.�(a)
∂G f
∂G i
∂(G f − G i )
� +�
� = −�
�
∂T p
∂T p
∂T
p
∂∆G
∂[(−85.40 J) + T × (36.5 J K−1 )]
� = −�
�
∂T p
∂T
p
= −(36.5 J K−1 ) = −36.5 J K−1 .
�e Gibbs-Helmholtz relation for the change in Gibbs energy is given by [�E.��–
���], (∂[∆G�T]�∂T) p = −∆H�T 2 . Expressing for the change in enthalpy gives
∆H = −T 2 �
E�E.�(a)
42 cm3
�
600 cm3
= T2 �
∂[∆G�T]
∂[(−85.40 J)�T + (36.5 J K−1 )]
� = −T 2 �
�
∂T
∂T
p
p
−85.40 J
� = −85.40 J .
T2
�e molar Gibbs energy dependence on pressure for an incompressible substance is given by [�E.��–���], G m (p f ) = G m (p i )+(p f − p i )Vm . Assuming that
the volume of liquid octane changes little over the range of pressures considered
∆G = n[G m (p f ) − G m (p i )] = (p f − p i )nVm = (p f − p i )V
1.01325 × 105 Pa
× (1.0 × 10−3 m3 )
1 atm
= +1.00... × 104 J = +10 kJ .
= (100 atm − 1.0 atm) ×
103
104
3 THE SECOND AND THIRD LAWS
For the molar Gibbs energy
∆G m =
=
Solutions to problems
P�E.�
∆G
∆G
M∆G
=
=
n
m�M
ρV
(114.23 g mol−1 ) × (+10.0... kJ)
= +1.6 kJ mol−1 .
(0.703 g cm−3 ) × (1.0 × 103 cm3 )
(a) �e Gibbs-Helmholtz relation for the change in Gibbs energy is given by
[�E.��–���], (∂[∆G�T]�∂T) p = −∆H�T 2 . Integrating the equation between the temperatures T1 and T2 and assuming that ∆H is temperature
independent gives
�
�erefore
T2
T1
�
T2
∂ ∆G(T)
1
� dT = ∆H �
− 2 dT
∂T T
T
T1
p
∆G(T2 ) ∆G(T1 )
1
1
−
= ∆H � − �
T2
T1
T2 T1
∆G(T2 ) ∆G(T1 )
1
1
=
+ ∆H � − �
T2
T1
T2 T1
(b) �e standard reaction entropy is given by [�C.�b–��], ∆ r G −○ = ∑J ν J ∆ f G −○ (J),
where ν J are the signed stoichiometric numbers.
∆ r G −○ (298 K) = 2∆ f G −○ (CO2 (g)) − 2∆ f G −○ (CO (g)) − ∆ f G −○ (O2 (g))
= 2 × (−394.36 kJ mol−1 ) − 2 × (−137.17 kJ mol−1 ) − 0
= −514.38 kJ mol−1 .
Similarly, the standard reaction enthalpy is given by [�C.�b–��], ∆ r H −○ =
∑J ν J ∆ f H −○ (J).
∆ r H −○ (298 K) = 2∆ f H −○ (CO2 (g)) − 2∆ f H −○ (CO (g)) − ∆ f H −○ (O2 (g))
= 2 × (−393.51 kJ mol−1 ) − 2 × (−110.53 kJ mol−1 ) − 0
= −565.96 kJ mol−1 .
(c) �e above derived expression is rearranged to give
Hence
∆G(T2 ) = ∆G(T1 )
T2
T2
+ ∆H �1 − �
T1
T1
∆G(375 K) = (−514.38 kJ mol−1 )
375 K
298 K
375 K
�
298 K
= (−6.47 × 102 kJ mol−1 ) + (+1.46 × 102 kJ mol−1 )
+ (−565.96 kJ mol−1 ) �1 −
= −501 kJ mol−1 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P�E.�
�e given expression for the reaction Gibbs energy dependence on temperature
is rearranged for ∆G(T2 ) and becomes
∆ r G −○ (T2 ) = ∆ r G −○ (T1 )
Hence at 37 ○ C = 310 K
T2
T2
+ ∆ r H −○ �1 − �
T1
T1
273.15 K + 37 K
298 K
273.15 K + 37 K
−1
+ (−5797 kJ mol ) �1 −
�
298 K
= −6355 kJ mol−1
∆ r G −○ (310 K) = (−6333 kJ mol−1 )
�e extra non-expansion work that is obtained by raising the temperature is
the di�erence
∆ r G −○ (310 K) − ∆ r G −○ (298 K) = (−6355 kJ mol−1 ) − (−6333 kJ mol−1 )
= −22 kJ mol−1
�erefore the result is an extra 22 kJ mol−1 of energy that is available for nonexpansion work.
P�E.�
Consider the exact di�erential of the Enthalpy, H = U + pV
dH = dU + d(pV ) = dU + V dp + pdV
�e exact di�erential of the internal energy is given by the fundamental equation [�E.�–���], dU = TdS − pdV , hence
dH = TdS − pdV + V dp + pdV = TdS + V dp
Because dH is the exact di�erential this implies
�
∂H
� =T
∂S p
�
∂ ∂H
∂ ∂H
�
� � =� �
� �
∂p ∂S p S
∂S ∂p S p
and
�
∂H
� =V
∂p S
�e mixed partial derivatives are equal irrespective of the order
�erefore
�
∂T
∂V
� =�
�
∂p S
∂S p
Similarly consider the exact di�erentials of the Helmholtz energy, A = U − T S,
and the Gibbs energy, G = H − T S. Starting with the Helmholtz energy
dA = dU − d(T S) = dU − TdS − SdT
= TdS − pdV − TdS − SdT = −pdV − SdT
105
106
3 THE SECOND AND THIRD LAWS
It follows that
�
∂p
∂S
� =�
�
∂T V
∂V T
For the Gibbs energy, the above derived result for dH is used
dG = dH − d(T S) = TdS + V dp − TdS − SdT
= V dp − SdT
It follows that
P�E.�
�
∂V
∂S
� = −� � .
∂T p
∂p T
(a) Assuming that a = 0 and b ≠ 0, the van der Waals equation becomes
p = RT�(Vm − b). �e molar volume is thus
Vm =
RT
+b
p
Consider the exact di�erential of the molar Gibbs energy at constant temperature, dG m = (∂G m �∂p)T dp. Integrating this gives
�
G m,f
G m,i
dG m = �
�erefore
pf
pi
�
pf
p f RT
∂G m
� dp = � Vm dp = � �
+ b� dp
∂p T
p
pi
pi
G m (p f ) = G m (p i ) + RT ln �
pf
� + b(p f − p i )
pi
�e change in Gibbs energy energy increases more rapidly with pressure
than the perfect gas due to the last term originating from the repulsion.
(b) Assuming that a ≠ 0 and b = 0, the van der Waals equation becomes
p = RT�Vm ) + a�Vm2 . �is is rearranged into a quadratic equation in Vm
�e solutions for Vm are
pVm2 − RT Vm + a = 0
�
(−RT)2 − 4pa
Vm =
2p
�
RT RT
4pa
=
±
1− 2 2
2p
2p
R T
−(−RT) ±
Because the van der Waals equation is a correction to the ideal gas, the
result should be approximately similar. Considering 4pa�(RT)2 � 1, it
is obvious that only a positive root reproduces the perfect gas and hence
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
is physically relevant. �is solution is used further to apply the suggested
approximate expansion
RT RT
Vm =
+
2p
2p
≈
�
1−
4pa
R2 T 2
RT RT
4pa
RT
a
+
�1 − 12 � 2 2 �� =
−
.
2p
2p
R T
p
pRT
Integrating this as before gives the Gibbs energy dependence on pressure
�erefore
�
G m,f
G m,i
dG m = �
pf
pi
Vm dp = �
G m (p f ) = G m (p i ) + RT ln �
p f RT
pi
p
−
a
dp
pRT
pf
a
pf
�−
ln � �
pi
RT
pi
�e change in Gibbs energy energy decreases more rapidly with pressure
than the perfect gas due to the last term originating from the attractive
interaction between the molecules term.
(c) Using the given data the change in molar Gibbs energy, ∆G m (p) = G m (p)−
G m (p−○ ), is plotted against (p�p−○ ) at 298 K using the requested units
(Fig. �.�). A zoomed version of the same plot is shown in Fig. �.�.
∆G m (p)�atm dm3 mol−1
150.0
100.0
50.0
0.0
perfect gas
vdW, repulsive
vdW, attractive
0
Figure 3.5
50
100
150
p�p−○
200
250
300
Answers to integrated activities
I�.�
(a) �e variation of entropy with volume at constant temperature is given by
one of the Maxwell relations from Table �E.� on page ���, (∂S�∂V )T =
(∂p�∂T)V . Working with molar quantities, the van der Waals equation
of state is p = RT�(Vm − b) − a�Vm2 , therefore (∂p�∂T)V = R�(Vm − b).
107
3 THE SECOND AND THIRD LAWS
100.0
∆G m (p)�atm dm3 mol−1
108
95.0
90.0
perfect gas
vdW, repulsive
vdW, attractive
85.0
80.0
25
30
35
p�p
Figure 3.6
−
○
40
45
50
�e integration is then straightforward
∆S m = �
Vm,f
Vm,i
R
Vm,f − b
dVm = R ln
Vm − b
Vm,i − b
= (8.3145 J K−1 mol−1 )
× ln
(10.0 dm3 mol−1 ) − (4.29 × 10−2 dm3 mol−1 )
(1.00 dm3 mol−1 ) − (4.29 × 10−2 dm3 mol−1 )
= +19.5 J K−1 mol−1
where the value of b is taken from the tables in the Resource section; note
the conversion of the molar volumes to dm3 mol−1 so as to match the
units of b.
(b) �e variation of entropy with temperature at constant volume and pressure are given by
∆S m = C V ,m ln(T2 �T1 ) and
∆S m = C p,m ln(T2 �T1 )
respectively; both relationships assume that the heat capacities do not
change in the temperature interval.
�e equipartition theorem, �e chemist’s toolkit � in Topic �A, is used to
estimate the value of C V ,m ; for a perfect gas C p,m = C V ,m + R. For atoms
there are just three translational degrees of freedom therefore C V ,m = 32 R
and C p,m = 52 R. For linear rotors there are in addition two rotational
degrees of freedom, therefore C V ,m = 52 R and C p,m = 72 R. For nonlinear rotors there are three rotational degrees of freedom, C V ,m = 3R
and C p,m = 4R.
Figures �.� and �.� show plots of ∆S m �R against ln(T2 �T1 ) for the constant volume and constant pressure cases, respectively.
(c) �e change in entropy as a function of temperature is given by [�B.�–��];
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
20
atoms
linear rotors
non-linear rotors
∆S m �R
15
10
5
0
0
1
Figure 3.7
2
3
ln(T2 �T1 )
4
5
20
atoms
linear rotors
non-linear rotors
∆S m �R
15
10
5
0
0
Figure 3.8
1
2
3
ln(T2 �T1 )
4
5
this is integrated for the particular form of the heat capacity suggested
a
c
� + b + 3 � dT
T
T
T
Ti
Tf
1
1
= a ln + b(Tf − Tf ) − 12 c � 2 − 2 �
Ti ��� � � � � � � � � �� � � � � � � � � � ��
Tf
Ti
��� � � � � � � �
term �
��
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
� � � � � � � � � � � ��
term �
∆S = �
Tf C
Ti
dT = �
Tf
term �
A convenient way of exploring this result is to choose a speci�c temperature range, say from ��� K to ��� K, and then plot the contribution of
each of the three terms as a function of the relevant parameter, a, b, or c.
Referring to the data in the Resource section it is seen that the ranges of
these parameters are: for a, between �� J K−1 mol−1 and �� J K−1 mol−1 ;
for b between � and 50 × 10−3 J K−2 mol−1 ; and for c between −10 ×
105 J K mol−1 and +2.0 × 105 J K mol−1 .
Figure �.� compares the contributions made by three terms over this tem-
109
3 THE SECOND AND THIRD LAWS
perature range; from the plots it is clear that the �rst term makes by far
the greatest contribution. Terms � and � both result in an increase in the
entropy with temperature, but term � will make a negative contribution
to the entropy change if c < 0, which is commonly the case.
(term �)�(J K−1 mol−1 )
(term �)�(J K−1 mol−1 )
40
40
20
0
20
20
40
a�(J K
−1
60
0
0.00
80
−1
mol )
0.02
b�(J K
(term �)�(J K−1 mol−1 )
110
0
−2
0.04
mol )
−1
−2
−4
Figure 3.9
−10
−5
0
c�(10 J K mol )
−1
5
(d) �e variation of G with p at constant T is given by [�E.�–���], (∂G�∂p)T =
V . �e physical signi�cance of the derivative is therefore that it is equal to
the volume of the system. For a perfect gas, V = nRT�p, which makes the
integration straightforward to give ∆G = nRT ln(p f �p i ) ([�E.��–���]).
Figure �.�� shows a plot of ∆G�nRT as a function of p f �p i . �e Gibbs
energy increases with pressure at constant temperature.
(e) �e fugacity coe�cient is given in terms of the compression factor Z by
ln � = �
0
p Z −1
p
Z=
dp
pVm
RT
�e pressure, volume and temperature can be expressed in terms of the
reduced variables p r , Vr , and Tr , given by
p r = p�p c
Vr = Vm �Vc
p c = a�27b 2
Vc = 3b
Tr = T�Tc
where the critical values of p, V , and T are given in terms of the van der
Waals parameters by
Tc = 8a�27bR
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
2.0
∆G�nRT
1.5
1.0
0.5
0.0
1
2
Figure 3.10
3
p f �p i
4
5
�e compression factor can therefore be written
Z=
pVm p c Vc p r Vr 1 a
27bR p r Vr 3 p r Vr
=
=
× 3b ×
×
=
RT
RTc Tr
R 27b 2
8a
Tr
8 Tr
�e aim is to write Z in terms of just Vr and Tr , therefore p r is substituted
by
8Tr
3
pr =
−
3Vr − 1 Vr2
to give
3 Vr
8Tr
3
3Vr
9
Z=
�
−
�=
−
8 Tr 3Vr − 1 Vr2
3Vr − 1 8Tr Vr
�e variable of integration is p, and it is desired to change this to Vr , for
which the following derivative is required
dp
dp r
d
8Tr
3
= pc
= pc
�
− 2�
dVr
dVr
dVr 3Vr − 1 Vr
= pc �
−24Tr
6
+
�
(3Vr − 1)2 Vr3
�e lower limit of the integral is p = 0, which corresponds to Vr = ∞. �e
integral therefore becomes
ln � = �
=�
0
p Z −1
Vr Z − 1
∞
pc pr
∞
pr
=�
=�
Vr Z − 1
Vr
∞
�
p
dp
pc �
�
−24Tr
6
+
� dVr
(3Vr − 1)2 Vr3
−24Tr
6
+
� dVr
(3Vr − 1)2 Vr3
−1
3Vr
9
8Tr
3
−
− 1� �
−
�
3Vr − 1 8Tr Vr
3Vr − 1 Vr2
�
−24Tr
6
+
� dVr
(3Vr − 1)2 Vr3
111
3 THE SECOND AND THIRD LAWS
Mathematical so�ware may be able to evaluate this integral analytically,
or failing that it will be necessary to resort to numerical methods. Some
representative results are shown in Fig. �.��.
1.2
Tr = 1
Tr = 2
Tr = 3
1.0
�
112
0.8
0.6
Figure 3.11
I�.�
1.0
1.5
2.0
Vr
2.5
3.0
�e statistical de�nition of entropy is given by [�A.�–��], S = k ln W, where W
is the number of microstates, the number of ways in which the molecules of a
system can be distributed over the energy states for a speci�ed total energy. As
explained in Section �A.�(a) on page �� the molecular interpretation of helps
to explain why, in the thermodynamic de�nition given by [�A.�a–��],dS =
q rev �T, the entropy change depends inversely on the temperature. In a system
at high temperature the molecules are spread out over a large number of energy
states. Increasing the energy of the system by the transfer of heat makes more
states accessible, but given that very many states are already occupied the proportionate change in W is small. In contrast, for a system at a low temperature
fewer states are occupied, and so the transfer of the same energy results in a
proportionately larger increase in the number of accessible states, and hence
a larger increase in W. �is argument suggests that the change in entropy for
a given transfer of energy as heat should be greater at low temperatures than
at high, as in the thermodynamic de�nition. As discussed in Section �C.�(a)
on page ��, the statistical de�nition of entropy also justi�es the �ird Law
of thermodynamics. �e law states that the entropy of all perfect crystalline
substances is zero at T = 0. At a molecular level the absence of thermal motion
in a perfectly localized crystalline solid is interpreted as there is only one way
to arrange the molecules like that. �us, W = 1 and from S = k ln W it follows
that S = 0 as stated by the law.
4
4A
Physical transformations
of pure substances
Phase diagrams of pure substances
Answers to discussion questions
D�A.�
For two phases to be in equilibrium, the chemical potentials of each component
must be equal in the two phases. In a one-component system, this means that
the chemical potential of that one component must be the same in all phases
that are in equilibrium. �e chemical potential is a function of two variables,
say p and T (and not of composition in a one-component system). �us, if there
are four phases α, β, γ, and δ in equilibrium the chemical potentials would need
to satisfy
µ α (p, T) = µ β (p, T) = µ γ (p, T) = µ δ (p, T)
�is is a set of three independent equations in only two variables (p and T),
which are not compatible.
D�A.�
Chemical potential is the single function that governs phase stability. �e phase
whose chemical potential is least under a set of given conditions is the most
stable. Conditions under which two or more phases have equal chemical potentials are conditions under which those phases are in equilibrium. Understanding how chemical potential varies with physical conditions such as temperature,
pressure, and composition makes it possible to compute chemical potentials for
various phases and to map out the conditions for stability of those phases and
for equilibrium between them.
Solutions to exercises
E�A.�(a)
E�A.�(a)
Use the phase rule [�A.�–���], F = C − P + 2, with C = 1 (one component).
Inserting P = 1 gives F = 1 − 1 + 2 = 2. �e condition P = 1 therefore
represents an area . An area has F = 2 because it is possible to vary pressure
and temperature independently (within limits) and stay within the area. P = 1
indicates that a single phase is present, so this result con�rms that a single phase
is represented by an area in a phase diagram.
(i) ��� K and �.� atm lies on the boundary between solid and gas phases.
Two phases , solid and gas, would therefore be present in equilibrium
under these conditions.
114
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
(ii) ��� K and � atm lies in the vapour region, so only one phase , vapour,
will be present.
(iii) ��� K is greater than the critical temperature, which means that there
is no distinction between gas and liquid. �erefore only one phase (a
supercritical �uid) will be present at all pressures.
E�A.�(a)
In a phase diagram, a single phase is represented by an area, while a line represents a phase boundary where two phases coexist in equilibrium. Point a lies
within an area and therefore only one phase is present. Points b and d each
lie on the boundary between two areas, and therefore in each case two phases
are present. Point c lies at the intersection of three phase boundaries, where
three phases are present in equilibrium.
E�A.�(a)
�e change in Gibbs energy when an in�nitesimal amount dn of substance is
moved from location � to location � is given by (Section �A.�(c) on page ���)
dG = (µ 2 − µ 1 )dn
Assuming that �.� mmol is a su�ciently small amount to be regarded as in�nitesimal, the Gibbs energy change in this case is
E�A.�(a)
∆G = (µ 2 − µ 1 )∆n = (7.1 × 103 J mol−1 ) × (0.1 × 10−3 mol) = �.�� J
Use the phase rule [�A.�–���], F = C − P + 2, with C = 2 (for two components).
Rearranging for the number of phases gives
P = C−F +2=2−F +2=4−F
�e number of variables that can be changed arbitrarily, F, cannot be smaller
than zero so the maximum number of phases in this case is � .
Solutions to problems
P�A.�
(a) ��� K and � atm lies in the solid region of the phase diagram, so initially
only solid carbon dioxide (dry ice) will be present. When the temperature reaches ���.� K, the sublimation point of CO2 at � atm, solid and
gas phases will be present in equilibrium. Above this temperature only
gaseous CO2 is present.
(b) ��� K and �� atm lies in the solid region of the phase diagram, so again
CO2 will initially be a solid. On heating, a point is reached at which the
solid melts; at this temperature solid and liquid phases are both present in
equilibrium. Above this temperature only a liquid phase is present until
the boiling temperature is reached, at which point liquid and gas will be in
equilibrium. Above this temperature, only the gas phase will be present.
P�A.�
A schematic phase diagram is shown in Fig �.�. Note that in reality the phase
boundaries may be curved rather than straight. �ere are two triple points
which are marked with dots.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
⇧
⇧
β
p
γ
⌅
s
⌅
⌅
δ
⇧
⇧ (((
s(
s Triple points
α
T
Figure 4.1
4B Thermodynamic aspects of phase transitions
Answers to discussion questions
D�B.�
D�B.�
Formally, the pressure derivative of the chemical potential is (∂µ�∂p)T = Vm .
Because the molar volume is always positive, the slope of the change in chemical
potential with respect to change in pressure is positive: that is, the chemical
potential increases with increasing pressure.
Formally, the temperature derivative of the chemical potential is (∂µ�∂T) p =
−S m . Because the molar entropy is always positive for all pure substances, the
slope of the change in chemical potential with respect to change in temperature
is negative: that is, the chemical potential decreases with increasing temperature.
Solutions to exercises
E�B.�(a)
�e relationship between pressure and temperature along the solid–liquid boundary is given by [�B.�–���], p = p∗ + (∆ fus H�T ∗ ∆ fus V )(T − T ∗ ). In this case
p∗ = 1 atm (corresponding to the normal melting point, T ∗ = 273.15 K) and
p = 1 bar (corresponding to the standard melting point). Rearranging for
(T − T ∗ ), the di�erence in melting points, gives
(T − T ∗ ) = (p − p∗ )
T ∗ ∆ fus V
∆ fus H
= �1 × 105 Pa − 1 atm ×
×
1.01325 × 105 Pa
�
1 atm
(273.15 K) × (−1.6 × 10−6 m3 mol−1 )
= 9.6 × 10−5 K
6.008 × 103 J mol−1
�is result shows that the standard melting point of ice is slightly higher than
the normal melting point, but the di�erence is negligibly small for most purposes.
115
116
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
E�B.�(a)
Since 1 W = 1 J s−1 , the rate at which energy is absorbed is (1.2 kW m−2 ) ×
(50 m2 ) = 60 kJ s−1 . �e rate of vaporization is then
rate of energy absorption
60 kJ s−1
=
= 1.36... mol s−1
∆ vap H
44 kJ mol−1
E�B.�(a)
Multiplication by the molar mass of water gives the rate of loss of water as
(1.36... mol s−1 ) × (18.0158 g mol−1 ) = �� g s−1 .
�e perfect gas equation [�A.�–�], pV = nRT, is used to calculate the amount
as n = pV �RT. V is the volume of the laboratory (75 m3 ) and p is the vapour
pressure. �e mass is found from m = nM, where M is the molar mass; hence
m = pV M�RT
Water:
Benzene:
m=
Mercury:
pV M (3.2 × 103 Pa) × (75 m3 ) × (18.0158 g mol−1 )
=
= �.� kg
RT
(8.3145 J K−1 mol−1 ) × ([25 + 273.15] K)
m=
m=
pV M (13.1 × 103 Pa)×(75 m3 )×(78.1074 g mol−1 )
=
= �� kg
RT
(8.3145 J K−1 mol−1 ) × ([25 + 273.15] K)
pV M
(0.23 Pa) × (75 m3 ) × (200.59 g mol−1 )
=
= �.� g
RT
(8.3145 J K−1 mol−1 ) × ([25 + 273.15] K)
1 Pa = 1 kg m−1 s−2 and 1 J = 1 kg m2 s−2 have been used. Note that an typically
sized bottle of benzene (containing less than �� kg of benzene) would evaporate
completely before saturating the air of the laboratory with benzene vapour.
E�B.�(a)
(i) �e integrated form of the Clausius–Clapeyron equation [�B.��–���] is
ln
∆ vap H 1
p
1
=−
� − ∗�
∗
p
R
T T
Rearranging for ∆ vap H and substituting in the numbers, taking p∗ , T ∗ at
��.� ○ C and p,T at ���.� ○ C, gives
∆ vap H = −R �
1
1 −1
p
− ∗ � ln ∗
T T
p
= −(8.3145 J K
× ln �
−1
−1
1
1
mol ) × �
−
�
[119.3 + 273.15] K [85.5 + 273.15] K
−1
5.3 kPa
� = 4.86... × 104 J mol−1 = �� kJ mol−1
1.3 kPa
(ii) �e integrated form of the Clausius–Clapeyron equation is now rearranged
for T. Substituting in p = 1 atm, or 1.01325 × 105 Pa, corresponding to
the normal boiling point, together with the value of ∆ vap H from above
and the same values for p∗ , T ∗ as before, gives
T =�
=�
−1
1
R
p
−
ln ∗ �
∗
T
∆ vap H p
−1
1
8.3145 J K−1 mol−1
1.01325 × 105 Pa
−
×
ln
�
[85.5 + 273.15] K
4.86... × 104
1.3 × 103 Pa
= 4.89... × 102 K = 4.9 × 102 K or 2.2 × 102 ○ C
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(iii) To �nd ∆ vap S at the boiling temperature, use [�B.�–��]:
E�B.�(a)
∆ vap S =
∆ vap H 4.86... × 104 J mol−1
=
= �� J K−1 mol−1
T
4.89... × 102 K
�e relationship between pressure and temperature along the solid–liquid boundary is given by [�B.�–���], p = p∗ + (∆ fus H�T ∗ ∆ fus V )(T − T ∗ ). �e molar
volume is Vm = M�ρ where M is the molar mass and ρ is the mass density
∆ fus V = Vm (l) − Vm (s) = M�ρ(l) − M�ρ(s)
�is expression is inserted into [�B.�–���], which is then rearranged for T. T ∗ ,
p∗ , and ∆ vap H are taken as the values corresponding to the normal melting
point of ice, that is, � ○ C (273.15 K) and � atm (101.325 kPa). It is assumed that
∆ vap H is constant over the temperature range of interest.
T = T ∗ + (p − p∗ )
T∗
M
M
�
−
�
∆ fus H ρ(l) ρ(s)
= (273.15 K) + ([50 × 105 − 1.01325 × 105 ] Pa) ×
E�B.�(a)
�
273.15 K
6.008 × 103 J mol−1
18.0158 g mol−1 18.0158 g mol−1
−
� = ��� K or −0.35 ○ C
1.00 × 106 g m−3 0.92 × 106 g m−3
�e variation of chemical potential with temperature is given by [�B.�a–���],
(∂µ�∂T) p = −S m . For a �nite change this gives ∆µ = −S m ∆T.
∆µ(liquid) = −(65 J K−1 mol−1 ) × (1 K) = −65 J mol−1
∆µ(solid) = −(43 J K−1 mol−1 ) × (1 K) = −43 J mol−1
�e chemical potentials of both solid and liquid are decreased at the higher
temperature, but the chemical potential of the liquid is decreased by a greater
amount. As they were at equilibrium before it follows that the liquid is the
more stable phase at the higher temperature, so melting will be spontaneous.
E�B.�(a)
E�B.�(a)
�e variation of chemical potential with temperature is given by [�B.�a–���],
(∂µ�∂T) p = −S m . For a �nite change this gives ∆µ = −S m ∆T, assuming that
S m is constant over the temperature range.
∆µ = −(69.9 J K−1 mol−1 ) × ([35 − 25] K) = −699 J mol−1
�e variation of chemical potential with pressure is given by [�B.�b–���], (∂µ�∂p)T =
Vm . For a �nite change, and assuming that Vm is constant over the pressure
range, this gives ∆µ = Vm ∆p. �e molar volume Vm is given by M�ρ where M
is the molar mass of copper and ρ is the mass density.
∆µ = Vm ∆p = (M�ρ)∆p
=
63.55 × 10−3 kg mol−1
× ([10 × 106 − 100 × 103 ] Pa) = +70 J mol−1
8960 kg m−3
Note that 1 Pa m3 = 1 J.
117
118
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
E�B.�(a)
�e variation of vapour pressure with applied pressure is given by [�B.�–���],
p = p∗ eVm (l)∆ p�RT .
p = (2.34 × 103 Pa)×exp �
= 2710 Pa = �.�� kPa
(18.1 × 10−6 m3 mol−1 )×([20 × 106 − 1 × 105 ] Pa)
�
(8.3145 J K−1 mol−1 ) × ([20 + 273.15] K)
E�B.��(a) �e relationship between pressure and temperature along the solid–liquid boundary is given by [�B.�–���], p = p∗ + (∆ fus H�T ∗ ∆ fus V )(T − T ∗ ), which is
rearranged to give ∆ fus H. In this case p∗ = 1.00 atm, T ∗ = 350.75 K, p =
100 atm and T = 351.26 K.
p − p∗ ∗
T ∆ fus V
T − T∗
([100 − 1] atm) × (1.01325 × 105 Pa�1 atm)
=
× (350.75 K)
([351.26 − 350.75] K)
∆ fus H =
× ([163.3 − 161.0] × 10−6 m3 mol−1 )
= 1.58... × 104 J mol−1 = ��.� kJ mol−1
�e entropy of transition is given by [�B.�–��], ∆ fus S = ∆ fus H�T, where T is
the transition temperature. At the melting temperature the entropy of fusion is
E�B.��(a)
∆ fus S =
1.58... × 104 J mol−1
= ��.� J K−1 mol−1
350.75 K
�e integrated version of the Clausius–Clapeyron equation [�B.��–���] is given
by ln(p�p∗ ) = −(∆ vap H�R)(1�T − 1�T ∗ ). Rearranging for T gives
T =�
=�
−1
1
R
p
−
ln �
T ∗ ∆ vap H p∗
−1
1
8.3145 J K−1 mol−1
70.0 kPa
−
× ln
�
[24.1 + 273.15] K 28.7 × 103 J mol−1
53.3 kPa
= ��� K or ��.� ○ C
E�B.��(a) �e Clausius–Clapeyron equation [�B.�–���] is d ln p�dT = ∆ vap H�RT 2 . �is
equation is rearranged for ∆ vap H, and the expression for ln p is di�erentiated.
It does not matter that the pressure is given in units of Torr because only the
slope of ln p is required.
∆ vap H = RT 2
E�B.��(a)
d ln p
d
2501.8 K
2501.8 K
= RT 2
�16.255 −
� = RT 2 �
�
dT
dT
T
T2
= (2501.8 K)R = (2501.8 K) × (8.3145 J K−1 mol−1 ) = ��.��� kJ mol−1
(i) �e Clausius–Clapeyron equation [�B.�–���] is d ln p�dT = ∆ vap H�RT 2 .
�is equation is rearranged for ∆ vap H and the expression for ln p is differentiated, noting from inside the front cover that ln x = (ln 10) log x. It
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
does not matter that the pressure is given in units of Torr because only
the slope of ln p is required.
d ln p
d log p
d
1780 K
= RT 2 ln 10
= RT 2 ln 10
�7.960 −
�
dT
dT
dT
T
1780 K
= RT 2 ln 10 �
� = (1780 K)R ln 10
T2
∆ vap H = RT 2
= (1780 K) × (8.3145 J K−1 mol−1 ) × ln 10 = ��.�� kJ mol−1
(ii) �e normal boiling point refers to the temperature at which the vapour
pressure is � atm which is ��� Torr. �e given expression, log(p�Torr) =
7.960 − (1780K)�T, is rearranged for T and a pressure of ��� Torr is
substituted into it to give
T=
1780 K
1780 K
=
= ���.� K or ��.�� ○ C
7.960 − log(p�Torr) 7.960 − log 760
Note that this temperature lies outside the range 10 ○ C to 30 ○ C for which
the expression for log(p�Torr) is known to be valid, and is therefore an
estimate.
E�B.��(a) �e relationship between pressure and temperature along the solid–liquid boundary is given by [�B.�–���], p = p∗ + (∆ fus H�T ∗ ∆ fus V )(T − T ∗ ). �e value of
∆ fus V is found by using Vm = M�ρ where M is the molar mass and ρ is the
mass density:
∆ fus V = Vm (l) − Vm (s) =
=
M
M
−
ρ(l) ρ(s)
78.1074 g mol−1
78.1074 g mol−1
−
= 1.19... × 10−6 m3 mol−1
0.879 × 106 g m−3 0.891 × 106 g m−3
Equation [�B.�–���] is then rearranged to �nd T:
T = T ∗ + (p − p∗ )
T ∗ ∆ fus V
∆ fus H
= ([5.5 + 273.15] K) + �([1000 − 1] atm) ×
×
1.01325 × 105 Pa
�
1 atm
([5.5 + 273.15] K) × (1.19... × 10−6 m3 )
= 2.8 × 102 K or �.� ○ C
10.59 × 103 J mol−1
Solutions to problems
P�B.�
�e work done in expanding against a constant external pressure is given by
equation [�A.�–��], w = −p ex ∆V . Because the molar volume of a gas is so
much greater the molar volume of a liquid, ∆ vap V ≈ Vm (g). In addition, if
the gas behaves perfectly, Vm = RT�p (from the perfect gas law, [�A.�–�]) with
p = p ex as the gas expands against constant external pressure. �e work of
119
120
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
expansion is therefore
RT
= −RT = −(8.3145 J K−1 mol−1 ) × ([100 + 273.15] K)
p ex
w = −p ex ×
= −3.10... × 103 J mol−1 = −3.10 kJ mol−1
�e negative sign indicates that the system has done work on the surroundings, so the internal energy of the system falls. �e fraction of the enthalpy of
vaporization spent on expanding the vapour is
P�B.�
3.10... kJ mol−1
× 100 % = �.�� %
40.7 kJ mol−1
�e variation of vapour pressure with temperature is given by [�B.��–���], p =
p∗ exp[(−∆ vap H�R)(1�T − 1�T ∗ )]. �e values of T ∗ and p∗ corresponding to
the normal boiling point are used
p = p∗ exp �−
= (1 atm)
∆ vap H 1
1
� − ∗ ��
R
T T
20.25 × 103 J mol−1
1
1
−
��
−1 × �
−1
(40 + 273.15) K (−29.2 + 273.15) K
8.3145 J K mol
= �.�� atm or ��� kPa
× exp �−
P�B.�
(a) From the variation of chemical potential with temperature (at constant
pressure) [�B.�a–���], (∂µ�∂T) p = −S m , the slope of the chemical potential against temperature is equal to the negative of the molar entropy.
�e di�erence in slope on either side of the normal freezing point of water
is therefore
�
∂µ(l)
∂µ(s)
� −�
� = −S m (l) − (−S m (s))
∂T p
∂T p
= −∆ fus S = −22.0 J K−1 mol−1
(b) In a similar way, the di�erence in slope on either side of the normal boiling point of water is
�
∂µ(g)
∂µ(l)
� −�
� = −S m (g) − (−S m (l))
∂T p
∂T p
(c) From part (a)
�
= −∆ vap S = −109.9 J K−1 mol−1
∂µ(l)
∂µ(s)
� −�
� = −∆ fus S
∂T p
∂T p
hence
�
∂[µ(l) − µ(s)]
� = −∆ fus S
∂T
p
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
For a �nite change ∆[µ(l) − µ(s)] = −∆ fus S × ∆T. For a 5 ○ C drop in
temperature:
∆[µ(l) − µ(s)] = −(22.0 J K−1 mol−1 ) × (−5 K) = +110 J mol−1
P�B.�
�erefore, since water and ice are in equilibrium (µ(l) − µ(s) = 0) at
0 ○ C it follows that the chemical potential of liquid water exceeds that of
ice by +110 J mol−1 at −5 ○ C. �e fact that µ(l) > µ(s) indicates that
supercooled water at −5, ○ C has a tendency to freeze to ice.
�e total pressure at the bottom of the column is
p = ρgd + 1 atm
= �13.6 g cm−3 ×
10−3 kg 106 cm3
×
� × (9.81 m s−2 ) × (10 m)
1g
1 m3
+ (1.01325 × 105 Pa) = 1.44... × 106 Pa
To �nd the freezing point, use [�B.�–���], p = p∗ +(∆ fus H�T ∗ ∆ fus V )(T −T ∗ ).
Rearranging for T gives
T ∗ ∆ fus V
(p − p∗ )
∆ fus H
(234.3 K) × (0.517 × 10−6 m3 mol−1 )
= (234.3 K) +
2.292 × 103 J mol−1
6
× ([1.44... × 10 − 1.01325 × 105 ] Pa) = ���.� K
T = T∗ +
Note that this is not a very large di�erence from the normal freezing point,
re�ecting the fact that the slope of the solid-liquid boundary is generally very
steep compared to the liquid-vapour boundary. Large changes in pressure are
therefore needed to bring about signi�cant changes in freezing point.
P�B.�
�e integrated form of the Clausius–Clapeyron equation [�B.��–���], ln(p�p∗ ) =
−(∆ vap H�R)(1�T − 1�T ∗ ), is rewritten
ln
∆ vap H 1 ∆ vap H
p
=−
+
∗
p
R T
RT ∗
�is implies that a plot of ln(p�p∗ ) against 1�T should be a straight line of slope
−∆ vap H�R and intercept ∆ vap H�RT ∗ ; such a plot is shown in Fig. �.�. If p∗ is
taken to be 1 atm, or 101.325 kPa, then T ∗ corresponds to the normal boiling
point which can then be obtained from the intercept.
121
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
θ�○ C p�kPa
0
1.92
20
6.38
40
17.70
50
27.70
70
62.30
80
89.30
90
124.90
100 170.90
T −1 �K−1
0.003 66
0.003 41
0.003 19
0.003 09
0.002 91
0.002 83
0.002 75
0.002 68
ln(p�p∗ )
−3.966
−2.765
−1.745
−1.297
−0.486
−0.126
0.209
0.523
0
ln(p�p∗ )
122
−2
−4
0.0025
Figure 4.2
0.0030
T −1 �K−1
0.0035
�e data fall on a good straight line, the equation of which is
ln(p�p∗ ) = (−4.570 × 103 ) × (T −1 �K−1 ) + 12.81
�e values of ∆ vap H and T ∗ are obtained from the slope and intercept respectively:
∆ vap H = −slope × R = −(−4.570 × 103 K) × (8.3145 J K−1 mol−1 )
= 3.79... × 104 J mol−1 = ��.� kJ mol−1
P�B.��
T∗ =
∆ vap H
3.79... × 104 J mol−1
=
= ��� K or �� ○ C
R × intercept (8.3145 J K−1 mol−1 ) × 12.81
(a) �e Clapeyron equation [�B.�a–���] is dp�dT = ∆ trs S�∆ trs V . For sublimation, and with ∆ trs S = ∆ trs H�T this becomes
dp
∆ sub H
=
dT T∆ sub V
Since the molar volume of a gas is much greater than that of a solid, ∆ sub V
can be approximated as ∆ sub V = Vm (g) − Vm (s) ≈ Vm (g), and if the gas
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
behaves perfectly, Vm = RT�p. Substituting these into the above equation
gives
dp
∆ sub H
p∆ sub H
=
=
dT T(RT�p)
RT 2
Using dx�x = d ln x this becomes written as d ln p�dT = ∆ sub H�RT 2
(b) Integration of the equation derived in (a) under the assumption that ∆ vap H
is independent of T gives
ln p = −
∆ sub H 1
+ constant
R T
�is implies that a plot of ln p against 1�T should be a straight line of
slope −∆ sub H�R; such a plot is shown in Fig. �.�.
T�K
145.94
147.96
149.93
151.94
153.97
154.94
p�Pa
13.07
18.49
25.99
36.76
50.86
59.56
T −1 �K−1
0.006 852
0.006 759
0.006 670
0.006 582
0.006 495
0.006 454
ln(p�Pa)
2.570
2.917
3.258
3.604
3.929
4.087
4.5
ln(p�Pa)
4.0
3.5
3.0
2.5
0.0064
Figure 4.3
0.0066
T �K
−1
−1
0.0068
�e data fall on a good straight line, the equation of which is
ln(p�Pa) = (−3.816 × 103 ) × (T −1 �K−1 ) + 28.71
�e slope is equal to −∆ vap H�R, so:
∆ vap H = −slope × R = −(−3.816 × 103 K) × (8.3145 J K−1 mol−1 )
= ��.� kJ mol−1
123
124
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
P�B.��
(a) If the mass of the liquid decreases by m, then the amount in moles of
vapour formed is n vap = m�M. �e amount in moles of the input gas
is given by n gas = PV �RT (from the perfect gas equation) so the mole
fraction of the vapour is
x vap =
n vap
n vap
m�M
mRT
=
=
=
n tot n vap + n gas PV �RT + m�M MPV + mRT
(b) If the total pressure remains at P, the partial pressure of the vapour is
p = x vap × P =
mRT
mRTP
×P =
MPV + mRT
MPV + mRT
(c) Dividing top and bottom of this expression by MPV gives
p=
AmP
1 + Am
where
A=
RT
MPV
(d) For geraniol, noting that P = 760 Torr = 1.01325 × 105 Pa,
A=
(8.3145 J K−1 mol−1 ) × ([110 + 273.15] K)
= 0.0407... g−1
(154.2 g mol−1 )×(1.01325 × 105 Pa)×(5.00 × 10−3 m3 )
hence
P�B.��
p=
AmP
(0.0407... g−1 ) × (0.32 g) × (1.01325 × 105 Pa)
=
= �.�� kPa
1 + Am
1 + (0.0407... g−1 ) × (0.32 g)
�e integrated form of the Clausius–Claypeyron equation [�B.��–���] is
ln
∆ vap H 1
p
1
=−
� − ∗�
∗
p
R
T T
Taking p∗ and T ∗ as corresponding to the pressure and boiling point at sea
level, p 0 and T0 , and inserting p�p 0 = e−a�H from the barometric formula gives
ln �e−a�H � = −
∆ vap H 1
1
� − �
R
T T0
hence
T =�
For water at 3 km, a = 3000 m, the boiling point is
T =�
−1
1
8.3145 J K−1 mol−1 3 km
+
×
�
373.15 K 40.7 × 103 J mol−1 8 km
−1
1
R a
+
�
T0 ∆ vap H H
= ��� K or ��.� ○ C
Solutions to integrated activities
I�.�
(a) �e expressions are plotted on the graph shown in Fig. �.�. Note that the
liquid-vapour line is only plotted for T3 ≤ T ≤ Tc because the liquid phase
does not exist below the triple point and there is no distinction between
liquid and vapour above the critical point. �e solid-liquid line is plotted
for T ≥ T3 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
50
p�bar
40
30
Liquid
Solid
20
10
Vapour
0
0
100
200
Figure 4.4
300 400
T�K
500
600
700
(b) �e standard melting point is the temperature corresponding to a pressure of 1 bar on the solid-liquid boundary. Setting p = 1 bar in the equation for the solid-liquid boundary and substituting in the value of p 3
gives:
1 = 0.4362 × 10−6 + 1000(5.60 + 11.727x)x
�is equation is rearranged to the standard quadratic form
11727x 2 + 5600x − 0.9999995638 = 0
which on solving for x gives x = 1.78... × 10−4 or x = −0.477.... �en,
since x = T�T3 − 1 where T3 = 178.15 K, it follows that
or
T = 178.15(1.78... × 10−4 + 1) = 178.18 K
T = 178.15(−0.477... + 1) = 93.11 K
�e 93.11 K solution is rejected since it lies below T3 where the liquid, and
therefore the solid-liquid boundary, does not exist. �e standard melting
point is therefore estimated to be 178.18 K .
(c) �e standard boiling point is the temperature at the point on the liquidvapour phase boundary corresponding to p = 1 bar. Substituting this
value of p into the equation for the liquid-vapour boundary and noting
that ln 1 = 0 gives
0 = −10.418�y+21.157−15.996y+14.015y 2 −5.0120y 3 +4.7334(1− y)1.70
Solving numerically gives
y = 0.645...
and so
T = y × Tc = 0.645... × 593.95 = 383.54 K
(d) Use the Clapeyron equation for the liquid-vapour boundary [�B.�–���]:
∆ vap H
dp
=
dT T∆ vap V
which rearranges to
∆ vap H = T∆ vap V ×
dp
dT
125
126
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
To �nd dp�dT, use d ln x = dx�x so that dp�dT = p × d ln p�dT. �e
expression for ln p is inserted and di�erentiated, and then evaluated at
the standard boiling point found above. For the evaluation of d ln p�dT it
does not matter that the expression has p in bar not Pa because the slope
of ln p is independent of the units of p due to the logarithm.
dp
d ln p
d ln p dy
=p
= p×
×
dT
dT
dy
dT
= p×
�en
d
10.418
�−
+ 21.157 − 15.996y + 14.015y 2 − 5.0120y 3
dy
y
+ 4.7334(1 − y)1.70 � ×
d T
� �
dT Tc
=
p
10.418
�
− 15.996 + 28.030y − 15.0360y 2
Tc
y2
=
105 Pa
10.418
�
593.95 K (0.645...)2
− 8.04678(1 − y)0.70 �
− 15.996 + 28.030(0.645...) − 15.0360(0.645...)2
− 8.04678(1 − 0.645...)0.70 � = 2.84... × 103 Pa K−1
dp
dT
= (383.54 K) × ([30.3 − 0.12] × 10−3 m3 mol−1 ) × (2.84... × 103 Pa K−1 )
∆ vap H = T∆ vap V ×
I�.�
= 33.0 kJ mol−1
(a) �e data are plotted in Fig. �.�.
�ese data �t well to the cubic
p�MPa = (4.989×10−6 K−3 )T 3 −(1.452×10−3 K−2 )T 2 +(0.1461 K−1 )T−5.058
�is equation is used to plot the line on the graph.
(b) �e standard boiling point corresponds to the temperature at which p =
1 bar or 0.1 MPa. �is value is substituted into the �tted function to give
0.1 = (4.989×10−6 K−3 )T 3 −(1.452×10−3 K−2 )T 2 +(0.1461 K−1 )T−5.058
which, on solving numerically using mathematical so�ware, yields
T = 1.11... × 10−2 K = ��� K
(c) �e Clapeyron equation for the liquid-vapour boundary is [�B.�–���]:
∆ vap H
dp
=
dT T∆ vap V
hence
∆ vap H = T∆ vap V ×
dp
dT
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
p�MPa
4.0
2.0
0.0
100
120
140
160
T�K
Figure 4.5
180
200
�e value of dp�dT is obtained by di�erentiating the �tted function and
substituting in the value of T found above.
d(p�MPa)
= (14.967 × 10−3 K−3 )T 2 − (2.904 × 10−3 K−2 )T + (0.1461 K−1 )
dT
= (14.967 × 10−3 K−3 ) × (1.11... × 102 K)2
− (2.904 × 10−3 K−2 ) × (1.11... × 102 K) + (0.1461 K−1 )
so that
= 8.57... × 10−3 K−1
�erefore
dp
= 8.57... × 10−3 MPa K−1 = 8.57... × 103 Pa K−1
dT
dp
dT
∆ vap H = T∆ vap V ×
= (1.11... × 102 K) × ([8.89 − 3.80 × 10−2 ] dm3 mol−1 ) ×
I�.�
× (8.57 × 103 Pa K−1 ) = 8.49 kJ mol−1
10−3 m3
1 dm3
�e relationship between p and T along the solid-liquid boundary is given by
equation [�B.�–���]:
p = p∗ +
∆ fus H
(T − T ∗ )
T ∗ ∆ fus V
Using Vm = M�ρ, ∆ fus V is calculated as
∆ fus V = Vm (l) − Vm (s) =
M
M
78.1074 g mol−1
78.1074 g mol−1
−
=
−
ρ(l) ρ(s) 0.879 × 106 g m−3 0.891 × 106 g m−3
= 1.19... × 10−6 m3 mol−1
127
128
4 PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
�is is used in equation [�B.�–���] together with the value of ∆ fus H quoted.
Taking p∗ and T ∗ as corresponding to the triple point, p∗ = 36 Torr = 4.80 kPa
and T ∗ = 5.50 ○ C = 278.65 K gives the equation of the solid-liquid boundary
as
p = (4.80 × 103 Pa) +
10.6 × 103 J mol−1
(T − 278.65 K)
(278.65 K) × (1.19... × 10−6 m3 mol−1 )
= (4.80 × 103 Pa) + (3.18 × 107 Pa K−1 ) × (T − 278.65 K)
so that
(p�kPa) = 4.80 + (3.18 × 104 ) × [(T�K) − 278.65]
�is takes the form of a steep straight line with a positive gradient extending
upwards from the triple point. �is is plotted in Fig. �.�. �e line is only drawn
for T ≥ T ∗ (p ≥ p∗ ) because the liquid does not exist below the triple point.
�e relationship between p and T along the liquid-vapour boundary is given by
equation [�B.��–���]; p∗ and T ∗ are again taken as corresponding to the triple
point.
∆ vap H 1
1
� − ∗ ��
R
T T
30.8 × 103 J mol−1 1
1
= (4.80 × 103 Pa) × exp �−
� −
��
8.3145 J K−1 mol−1 T 278.65 K
p = p∗ exp �−
or
(p�kPa) = 4.80 × exp �−3.70 × 103 �
1
1
−
��
T�K 278.65
�is equation is also plotted in Fig. �.�, again only for values of T in the range
T ≥ 278.65 K since the liquid does not exist below this temperature.
�e relationship between p and T along the solid-vapour boundary is given by
an equation that is analogous to the liquid-vapour one except that ∆ vap H is
replaced by ∆ sub H. ∆ sub H = ∆ fus H + ∆ vap H so the required equation is
∆ sub H 1
1
� − ∗ ��
R
T T
[10.6 + 30.8] × 103 J mol−1 1
1
= (4.80 × 103 Pa) × exp �−
� −
��
T 278.65 K
8.3145 J K−1 mol−1
p = p∗ exp �−
or
(p�kPa) = 4.80 × exp �−4.98 × 103 �
1
1
−
��
T�K 278.65
�is equation is plotted Fig. �.� for values in the range T ≤ 278.65 K since the
solid and vapour phases are only in equilibrium at the triple point and below.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
25
p�kPa
20
15
Liquid
Solid
10
Vapour
5
0
250
Figure 4.6
260
270
280 290
T�K
300
310
320
129
5
5A
Simple mixtures
The thermodynamic description of mixtures
Answers to discussion questions
D�A.�
Perfect gases spontaneously mix in all proportions. �ere are, however, conceivable circumstances under which two real gases might not mix spontaneously.
Consider allowing two gases initially at the same pressure p to mix (so that
mixing them would not change the pressure) under conditions of constant temperature. Mixing is spontaneous if ∆ mix G < 0, and this Gibbs energy change
has an entropic and an enthalpic contribution
∆ mix G = ∆ mix H − T∆ mix S
�e entropy change, ∆ mix S, is always positive, so mixing is always favoured
entropically. �e only circumstances under which mixing might not be spontaneous would be if ∆ mix H > T∆ mix S, that is if the change in enthalpy on mixing
was so unfavourable as to outweigh the entropic term.
For perfect gases, ∆ mix H = 0, so mixing always occurs. However, there are
liquids for which unfavourable interactions prevent mixing at least in some proportions and at some temperatures. If two such species were taken above their
critical temperatures and held at a pressure high enough to make their densities
more typical of liquids than gases, then it is possible to imagine that mixing
might not occur. Because the temperature is above the critical temperatures
the species are technically gases, although the term supercritical �uid might
be more appropriate. In conclusion, there might be examples of immiscibility
among supercritical �uids.
D�A.�
Raoult’s law, [�A.��–���] de�nes the behaviour of ideal solutions. Like perfect
gases, what makes the behaviour ideal can be expressed in terms of intermolecular interactions. Unlike perfect gases, however, the interactions in an ideal
solution cannot be neglected. Instead, ideal behaviour amounts to having the
same interactions between molecules of the di�erent components of the mixture as there are between molecules of the same type.
In short, ideal behaviour consists of A–B interactions being the same as A–A
and B–B interactions. If that is the case, then the cohesive forces that would
keep a molecule in the liquid phase would be the same in the solution as in a
pure liquid, and the vapour pressure of a component will di�er from that of a
pure liquid only in proportion to its abundance (mole fraction). �us, Raoult’s
132
5 SIMPLE MIXTURES
law is expected to be valid for mixtures of components that have very similar chemical structures. Similar structures imply both similar intermolecular
interactions and similar sizes.
In an ideal dilute solution, on the other hand, Raoult’s law holds for the solvent
in the limit as x A approaches �, not because A–B interactions are like A–A
interactions, but because there are so many more A–A interactions than A–
B interactions that A–A interactions dominate the behaviour of the solvent.
For the solute, on the other hand, there are many more A–B interactions than
B–B interactions in the limit as x B approaches zero. �us, only one kind of
interaction (A–B) is important in determining the a�nity of the solute for the
solution.
D�A.�
�e change in Gibbs energy at constant temperature is equal to the maximum
additional (non-expansion) work that the system can do dG = dw add,max , [�D.�–
��]. Changing the composition of a mixture gives rise to a change in Gibbs
energy, given by [�A.�–���], dG = µ A dn A + µ B dn B . . .. It therefore follows that
dw add,max = µ A dn A + µ B dn B . . .
and so non-expansion work can arise from the changing composition of a system.
Solutions to exercises
E�A.�(a)
�e partial pressure of gas A, p A above a liquid mixture is given by Raoult’s
Law, [�A.��–���], p A = x A p∗A , where x A is the mole fraction of A in the liquid
and p∗A is the vapour pressure over pure A. �e total pressure over a mixture
of A and B is p A + p B .
�e �rst step is to calculate the mole fractions. If the molar mass of A is M A
and the mass of A is m, then the amount in moles of A is m�M A , and likewise
because the mass of B is the same, the amount of B is m�M B . �e mole fraction
of A is therefore
xA =
m�M A
1�M A
1�M B
=
likewise x B =
m�M A + m�M B 1�M A + 1�M B
1�M A + 1�M B
�ese mole fractions are used with Raoult’s law to give the total pressure
p = x A p∗A + x B p∗B =
1�M A
1�M B
p∗ +
p∗
1�M A + 1�M B A 1�M A + 1�M B B
If A is benzene, M A = 6×12.01 g mol−1 +6×1.0079 g mol−1 = 78.1074 g mol−1 ,
and if B is methylbenzene M B = 7 × 12.01 g mol−1 + 8 × 1.0079 g mol−1 =
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
92.1332 g mol−1 .
p=
=
E�A.�(a)
1�M A
1�M B
p∗ +
p∗
1�M A + 1�M B A 1�M A + 1�M B B
1�(78.1074 g mol−1 )
× (10 kPa)
1�(78.1074 g mol−1 ) + 1�(92.1332 g mol−1 )
+
1�(92.1332 g mol−1 )
× (2.8 kPa)
1�(78.1074 g mol−1 ) + 1�(92.1332 g mol−1 )
= 5.41... kPa + 1.28... kPa = �.� kPa
�e total volume is calculated from the partial molar volumes of the two components using [�A.�–���], V = n A VA + n B VB . �e task is therefore to �nd the
amount in moles, n A and n B , of A and B in a given mass m of solution. If the
molar masses of A and B are M A and M B then it follows that
m = nA MA + nB MB
�e mole fraction of A is de�ned as x A = n A �(n A + n B ), hence n A = x A (n A +
n B ) and likewise for B. With these substitutions for n A and n B the previous
equation becomes
m = x A M A (n A + n B ) + x B M B (n A + n B ) hence (n A + n B ) =
m
xA MA + xB MB
�is latter expression for the total amount in moles, (n A + n B ), is used with
n A = x A (n A + n B ) to give
n A = x A (n A + n B ) =
and likewise
nB =
mx A
xA MA + xB MB
mx B
xA MA + xB MB
With these expressions for n A and n B the total volume is computed from the
partial molar volumes
V = n A VA + n B VB =
mx A VA
mx B VB
+
xA MA + xB MB xA MA + xB MB
m
[x A VA + x B VB ]
xA MA + xB MB
m
[x A VA + (1 − x A )VB ]
=
x A M A + (1 − x A )M B
=
where on the last line x B = (1 − x A ) is used.
Taking A as trichloromethane and B as propanone the molar masses are M A =
12.01+1.0079+3×35.45 = 119.3679 g mol−1 and M B = 3×12.01+6×1.0079+
16.00 = 58.0774 g mol−1 . With these values, the expression for the volume of
133
134
5 SIMPLE MIXTURES
�.��� kg evaluates as
V=
E�A.�(a)
1000 g
0.4693 × (119.3679 g mol ) + (1 − 0.4693) × (58.0774 g mol−1 )
−1
−1
−1
× �0.4693 × (80.235 cm3 mol ) + (1 − 0.4693) × (74.166 cm3 mol )�
= ���.� cm3
Consider a solution of A and B in which the fraction (by mass) of A is α (here
α = 12 ). �e total volume of a solution of A and B is calculated from the partial
molar volumes of the two components using [�A.�–���], V = n A VA + n B VB . In
this exercise V and VA are known, so the task is therefore to �nd the amount
in moles, n A and n B , of A and B in the solution of known mass density ρ.
�e mass of a volume V of the solution is ρV , so the mass of A is αρV . If
the molar mass of A is M A , then the amount in moles of A is n A = αρV �M A .
Similarly, n B = (1 − α)ρV �M B . �e volume is expressed using these quantities
as
αρV VA (1 − α)ρV VB
V = n A VA + n B VB =
+
MA
MB
�e term V cancels between the �rst and third terms to give
1=
αρVA (1 − α)ρVB
+
MA
MB
VA =
MA
(1 − α)VB ρ
�1 −
�
αρ
MB
�is equation is rearranged to give an expression for VA
In this exercise let B be H� O and A be ethanol, and as the mixture is ��%
by mass, α = 12 . �e molar mass of B (H� O) is M B = 16.00 + 2 × 1.0079 =
18.0158 g mol−1 and the molar mass of A (ethanol) is M A = 2 × 12.01 + 16.00 +
6 × 1.0079 = 46.0674 g mol−1 . �e above expression for VA evaluates as
VA =
=
MA
(1 − α)VB ρ
�1 −
�
αρ
MB
(46.0674 g mol−1 )
0.5 × (0.914 g cm−3 )
−1
�
(1 − 0.5) × (17.4 cm3 mol ) × (0.914 g cm−3 ) �
× 1−
�
�
18.0158 g mol−1
−1
E�A.�(a)
= ��.� cm3 mol
Henry’s law gives the partial vapour pressure of a solute B as p B = K B x B ,
[�A.��–���]. A test of this law is to make a plot of p B against x B which is
expected to be a straight line with slope K B ; such a plot is shown in Fig. �.�.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
p HCl �(kPa)
100
50
0
0.000
Figure 5.1
0.005
0.010
x HCl
0.015
0.020
�e data fall on a good straight line, the equation of which is
p HCl �(kPa) = 6.41 × 103 × (x HCl ) − 0.071
E�A.�(a)
If Henry’s law is obeyed the pressure should go to zero as x HCl goes to zero,
and the graph shows that this is almost achieved. Overall the conclusion is that
these data obey Henry’s law quite closely. �e Henry’s law constant K HCl is
computed from the slope as 6.4 × 103 kPa .
In Section �A.�(b) on page ��� it is explained that for practical applications
Henry’s law is o�en expressed as p B = K B b B , where b B is the molality of the
solute, usually expressed in mol kg−1 . �e molality is therefore calculated from
the partial pressure as b B = p B �K B .
Molality is the amount of solute per kg of solvent. �e mass m of a volume V
of solvent is given by m = ρV , where ρ is the mass density of the solvent. If the
amount of solute in volume V is n B , the molar concentration c B is related to
the molality by
�
nB
nB
nB
cB =
=
=ρ
= ρb B
V
m�ρ
m
Using Henry’s law the concentration is therefore given by
bB
�
pB
ρx B p
c B = ρb B = ρ
=
KB
KB
bB
where the partial pressure p B is expressed in terms of the mole fraction and the
total pressure p, p B = x B p.
�e mole fraction of N� in air is �.���, the Henry’s law constant for N� in
benzene is 1.87 × 104 kPa kg mol−1 and the density of benzene is �.��� g cm−3 .
If it is assumed that the total pressure is � atm then
c N2 =
ρx N2 p (0.879 × 103 kg m−3 )×(0.780)×(101.325 kPa)
=
= 3.71... mol m−3
K N2
1.87 × 104 kPa kg mol−1
135
136
5 SIMPLE MIXTURES
E�A.�(a)
�e molar concentration is therefore 3.7 × 10−3 mol dm−3 .
In Section �A.�(b) on page ��� it is explained that for practical applications
Henry’s law is o�en expressed as p B = K B b B , where b B is the molality of the
solute, usually expressed in mol kg−1 . �e molality is therefore calculated from
the partial pressure as b B = p B �K B . �e Henry’s law constant for CO� in water
is 3.01 × 103 kPa kg mol−1 .
For the case where the pressure of CO� is �.�� atm
b CO2 =
E�A.�(a)
p CO2 (0.10 atm) × (101.325 kPa�1 atm)
=
= 3.4 × 10−3 mol kg−1
−1
3
K CO2
3.01 × 10 kPa kg mol
When the pressure is ten times greater at �.�� atm the solubility is increased by
the same factor to 3.37 × 10−2 mol kg−1
As explained in Exercise E�A.�(a) the concentration of a solute is estimated
as c B = ρp B �K B where ρ is the mass density of the solvent. �e Henry’s law
constant for CO� in water is 3.01 × 103 kPa kg mol−1 and the density of water
is �.��� g cm−3 or ��� kg m−3 .
c CO2 =
ρp CO2 (997 kg m−3 ) × (5.0 atm) × (101.325 kPa�1 atm)
=
K CO2
3.01 × 103 kPa kg mol−1
= 1.67... × 102 mol m−3
E�A.�(a)
�e molar concentration is therefore �.�� mol dm−3 .
�e partial molar volume of B is de�ned from [�A.�–���] as
VB = �
∂V
�
∂n B p,T ,n ′
�e polynomial given relates υ to x, and so from this it is possible to compute
the derivative dυ�dx. �is required derivative is dV �dn B (where the partials
are dropped for simplicity), which is related to dυ�dx in the following way
�
dV
dV
dυ
dx
�=�
�� ��
�
dn B
dυ
dx dn B
Because x = n B �mol, dx�dn B = mol−1 , and because υ = V �cm3 , dυ�dV =
cm−3 and so dV �dυ = cm3 . Hence
�
dV
dV
dυ
dx
dυ
�=�
�� ��
� = � � cm3 mol−1
dn B
dυ
dx dn B
dx
�e required derivative is
hence
�
dυ
� = 35.677 4 − 0.918 46 x + 0.051 975 x 2
dx
VB = �35.677 4 − 0.918 46 x + 0.051 975 x 2 � cm3 mol−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�A.�(a)
�e partial molar volume of solute B (here NaCl) is de�ned from [�A.�–���] as
VB = �
∂V
�
∂n B p,T ,n ′
�e total volume is given as a function of the molality, but this volume is described as that arising from adding the solute to � kg of solvent. �e molality of
a solute is de�ned as (amount in moles of solute)/(mass of solvent in kg), therefore because in this case the mass of solvent is � kg, the molality is numerically
equal to the amount in moles, n B .
�e polynomial given relates υ to x, and so from this it is possible to compute
the derivative dυ�dx. �is required derivative is dV �dn B (where the partials
are dropped for simplicity), which is related to dυ�dx in the following way
�
dV
dV
dυ
dx
�=�
�� ��
�
dn B
dυ
dx dn B
�e quantity x is de�ned as b�b −○ , but it has already been argued that the molality can be expressed as n B �(1 kg), hence x = n B �(mol) and therefore dx�dn B =
mol−1 . Because υ = V �cm3 , dυ�dV = cm−3 and so dV �dυ = cm3 . Hence
�
dV
dV
dυ
dx
dυ
�=�
�� ��
� = � � cm3 mol−1
dn B
dυ
dx dn B
dx
�e required derivative is
dυ
� = 16.62 + 2.655 x 1�2 + 0.24 x
dx
Hence the expression for the partial molar volume of B (NaCl) is
�
VB = �16.62 + 2.655 x 1�2 + 0.24 x� cm3 mol−1
�e partial molar volume when b�b −○ = 0.1 is given by
VB �(cm3 mol−1 ) = (16.62 + 2.655 x 1�2 + 0.24 x)
= (16.62 + 2.655(0.100)1�2 + 0.24 × 0.100) = 17.4...
�erefore VB = 17.5 cm3 mol−1 .
�e total volume is calculated from the partial molar volumes of the two components, [�A.�–���], V = n A VA + n B VB . In this case V and VB are known,
so VA , the partial molar volume of the solvent water, can be found from VA =
(V − n B VB )�n A . �e total volume when b�b −○ = 0.1 is given by
V = 1003 + 16.62 × 0.100 + 1.77 × 0.1003�2 + 0.12 × 0.1002 = 1004.7... cm3
�e amount in moles of � kg of water is (1000 g)�[(16.00+2×1.0079) g mol−1 ] =
55.5... mol, hence
VA =
V − n B VB (1004.7... cm3 ) − (0.100 mol) × (17.4... cm3 mol−1 )
=
nA
55.5... mol
= 18.1 cm3 mol−1
where, as before, for this solution a molality of �.��� mol kg−1 corresponds to
n B = 0.100 mol.
137
138
5 SIMPLE MIXTURES
E�A.��(a) For a binary mixture the Gibbs–Duhem equation, [�A.��b–���], relates changes
in the chemical potentials of A and B
n A dµ A + n B dµ B = 0
If it is assumed that the di�erential can be replaced by the small change
(0.1 n B ) × (+12 J mol−1 ) + n B δµ B = 0
hence
E�A.��(a)
δµ B = −
(0.1 n B )
(+12 J mol−1 ) = −1.2 J mol−1
nB
Because the gases are assumed to be perfect and are at the same temperature
and pressure when they are separated, the pressure and temperature will not
change upon mixing. �erefore [�A.��–���], ∆ mix S = −nR(x A ln x A +x B ln x B ),
applies. �e amount in moles is computed from the total volume, pressure
and temperature using the perfect gas equation: n = pV �RT. Because the
separate volumes are equal, and at the same pressure and temperature, each
compartment contains the same amount of gas, so the mole fractions of each
gas in the mixture are equal at 0.5.
∆ mix S = −nR(x A ln x A + x B ln x B ) = −(pV �T)(x A ln x A + x B ln x B )
(1.01325 × 105 Pa) × (5.0 × 10−3 m3 )
(0.5 ln 0.5 + 0.5 ln 0.5)
298.15 K
= +1.2 J K−1
=−
Note that the pressure in expressed in Pa and the volume in m3 ; the units of the
result are therefore (N m−2 ) × (m3 ) × (K−1 ) = N m K−1 = J K−1 .
Under these conditions the Gibbs energy of mixing is given by [�A.��–���],
∆ mix G = nRT(x A ln x A + x B ln x B ); as before n = pV �RT.
∆ mix G = nRT(x A ln x A + x B ln x B ) = (pV )(x A ln x A + x B ln x B )
= [(1.01325 × 105 Pa) × (5.0 × 10−3 m3 )](0.5 ln 0.5 + 0.5 ln 0.5)
= −3.5 × 102 J
�e units of the result are (N m−2 )×(m3 ) = N m = J. As expected, the entropy
of mixing is positive and the Gibbs energy of mixing is negative.
Solutions to problems
P�A.�
�is problem is similar to the Example given in Section �A.�(d) on page ���.
�e Gibbs–Duhem equation [�A.��b–���], expressed in terms of partial molar
volumes is n A dVA + n B dVB = 0 which is rearranged to
dVA = −
nB
dVB
nA
If the variation of the solute partial molar volume VB with concentration is
described by a known function, then integration of this equation gives an expression for how the solvent partial molar volume VA varies.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e range of integration of VA is from pure A, at which the partial molar volume
is equal to the molar volume of the pure solvent VA∗ , up to some arbitrary
concentration. �e corresponding range for VB is from �, the molar volume
of B in the limit of of no B being present (that is pure A), up to some arbitrary
concentration.
VA
VB n
B
dVB
� ∗ dVA = − �
nA
VA
0
�e expression for VB is given as a function of the molality, which is the amount
in moles divided by the mass of the solvent in kg. In � kg of solvent the amount
in moles is n A = (1 kg)�M A , where M A is the molar mass of the solvent A.
With this expression the ratio n B �n A is rewritten
nB
nB
nB MA
=
=
n A (1 kg)�M A (1 kg)
�e quantity n B �(1 kg) is recognised as the molality b of solute B, hence n B �n A =
bM A . �e expression for VB is given in terms of x = b�b−○ , thus b = b −○ x and
hence n B �n A = M A b −○ x, With this, the integral to be evaluated becomes
VA
� ∗ dVA = − �
VA
VB
0
M A b −○ x dVB
�e partial molar volumes VJ are replaced throughout by the dimensionless
−1
quantities υ J = VJ �(cm3 mol ) to give
υA
� ∗ dυ A = − �
υA
υB
0
M A b −○ x dυ B
�e next step is to change the variable of integration on the right from υ B to x;
this is done by di�erentiating the relationship between these two quantities
υ B = 5.117 + 19.121 x 1�2
�e integral is then
υA
−
○
� ∗ dυ A = −M A b �
υA
x
0
hence
dυ B = 9.5605 x −1�2 dx
x(9.5605 x −1�2 ) dx = −M A b −○ �
x
0
9.5605 x 1�2 dx
Evaluating the integrals gives
υ A − υ∗A = −M A b −○ × 23 × 9.5605 x 3�2
�e molar mass of the solvent H� O is ��.���� g mol−1 ; for compatibility with
the units of molality this needs to be expressed as 1.80158 × 10−2 kg mol−1 . �e
value of υ∗A is given as ��.���; with these values the expression for υ A becomes
P�A.�
υ A = 18.079 − 0.11483 x 3�2
�e required molar masses are: N� ��.�� g mol−1 ; O� ��.�� g mol−1 ; Ar ��.�� g mol−1 ;
CO� ��.�� g mol−1 .
139
140
5 SIMPLE MIXTURES
Consider ��� g of the mixture. Of this ��.� g is N� so the amount in moles of this
gas is n N2 = (75.5 g)�(28.02 g mol−1 ) = 2.69... mol. Similar calculations are
made for the other cases to give the results shown below in the table. �e total
amount in moles n is found by summing these individual contributions and
this is then used to compute the mole fractions from x J = n J �n: the resulting
values are also shown in the table.
gas
mass %
n J �mol in ��� g
xJ
mass %
n J �mol in ��� g
xJ
N�
��.�
�.��...
�.���...
��.��
�.��...
�.���...
O�
��.�
�.���
�.���...
��.��
�.���...
�.���...
Ar
�.�
�.����...
9.42... × 10−3
�.��
�.����...
9.28... × 10−3
CO�
total
�.��...
�.���
1.04... × 10−3
3.02... × 10−4
�.��...
�e entropy of mixing (at constant pressure and temperature) is given by a
generalisation of [�A.��–���]
∆ mix S = −nR � x J ln x J
J
�e entropy of mixing per mole is (∆ mix S)�n is given by
(∆ mix S)�n = −R � x J ln x J
J
�is expression is used togther with the values given in the table to compute
the entropy of mixing for the �rst set of data as +4.70 J K−1 mol−1 and for the
second set of data as +4.711 J K−1 mol−1 . �e di�erence is of the order of
P�A.�
0.01 J K−1 mol−1 .
�e de�nition of the partial molar volume VB is
VB = �
∂V
�
∂n B n A
which is interpreted as the slope of a plot of V against n B , at constant n A . Let
B be the solute CuSO� and A be the solvent H� O.
�e task is therefore to calculate the volume of a solution with a �xed amount
of A as a function of the amount of B. �e data given refer to a particular mass
of the solution, whereas what is required is data for a particular mass of solvent,
so some manipulation is required. Imagine a solution created from a �xed mass
m A �(g) of solvent and which contains a mass m B �(g) of solute; the total mass is
therefore m A �(g) + m B �(g). From the data supplied ��� g of solution contains
m�(g) of CuSO� , so it follows that
��� � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � � � �
m A �(g) + m B �(g)
×m�(g) = m B �(g)
100
multiples of ��� g
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�is equation is rearranged to give an expression for m B �(g)
m B �(g) =
m A �(g) × m�(g)
100 − m�(g)
�e amount in moles of B is found using m B �M B , where M B is the molar mass
of B, which in this case is ���.�� g mol−1 .
�e volume of this solution is computed from the mass density as (m A +m B )�ρ.
�e following table of data is drawn up using m A = 1000 g as the �xed mass
of solvent, and using this a plot of V against n B is made, as shown in Fig. �.�.
Note that m tot �(g) = 1000 + m B �(g).
m(CuSO4 )�g
ρ�g cm−3
1.107
m B �g
111.1
n B �mol
m tot �g
V �cm3
10
15
1.167
176.5
1.106
1 176
1 008.1
20
1.230
250.0
1.566
1 250
1 016.3
5
1.051
52.6
0.330
0.696
1 053
1 111
1 001.6
1 003.7
1 015
V �cm3
1 010
1 005
1 000
0.4
0.6
Figure 5.2
0.8
1.0
n B �mol
1.2
1.4
1.6
�e data �t well to the polynomial (shown as the smooth curve on the plot)
V �(cm3 ) = 7.2249(n B �mol)2 − 1.8512(n B �mol) + 1001.4
�e partial molar volume is the slope of this curve which is the derivative with
respect to n B
−1
VB �(cm3 mol ) = 14.450(n B �mol) − 1.8512
�e following table gives values of VB for each of the data points. �ese are
plotted in Fig. �.�; the line is the function above.
141
5 SIMPLE MIXTURES
m(CuSO4 )�g
ρ�g cm−3
10
1.107
15
20
5
n B �mol
VB �cm3 mol
1.167
1.106
14.13
1.230
1.566
20.78
1.051
0.330
0.696
2.91
−1
8.21
20
−1
15
VB �cm3 mol
142
10
5
0.4
0.6
Figure 5.3
P�A.�
0.8
1.0
n B �mol
1.2
1.4
1.6
In Example �A.� on page ��� the partial molar volume of ethanol is found to be
given by
υ = 54.6664 − 0.72788 z + 0.084768 z 2
−1
where υ = VE �(cm3 mol ) and z = n E �mol. �e value of z at which υ is a
minimum or maximum is found by setting the derivative dυ�dz = 0
dυ
= −0.72788 + 0.169536 z = 0
dz
hence
z=
0.72788
= 4.2934
0.169536
�is value of z corresponds to 4.2934 mol in �.��� kg of solvent water (speci�ed
in the Example). �e molality is the amount in moles divided by the mass of the
solvent in kg, thus the corresponding molality is 4.2934 mol kg−1 . �e plot in
the text con�rms that this is indeed the position of the minimum in the partial
molar volume.
5B The properties of solutions
Answers to discussion question
D�B.�
�e boiling-point constant is given by [�B.�b–���], K = RT ∗2 �∆ vap H, where
T ∗ is the boiling point of the pure liquid and ∆ vap H is its enthalpy of vaporisation. However, by Trouton’s rule (Section �B.� on page ��), ∆ vap H�T ∗ is approximately constant, so the boiling-point constant is simply ∝ T ∗ . Di�erences
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
in boiling-point constants are therefore identi�ed as being due to di�erences in
the boiling points of the pure liquids. Water and benzene have di�erent boiling
points and so have di�erent boiling-point constants.
D�B.�
�e typical experimental arrangement for observing osmosis involves a pure
solvent being separated from a solution by a semipermeable membrane through
which only the solvent can pass. �e chemical potential of the solvent in the
solution is lower than that of the pure solvent, therefore there is a tendency for
the solvent to pass through the membrane from the side on which it is pure
into the solution because this results in a reduction in Gibbs energy.
At a molecular level the process involves an increase in ‘randomness’ as increasing the amount of solvent in the solution increases the number of possible
arrangements of solvent and solute molecules.
D�B.�
A regular solution has excess entropy S E of zero, but an excess enthalpy H E that
is non-zero. A regular solution of A and B can be thought of as one in which the
di�erent molecules of A and B are distributed randomly, as in an ideal solution,
but where the energy of A–A, B–B, and A–B interactions are di�erent.
In real solutions both S E and H E are non-zero, and in general both are likely
to vary with composition. �e non-zero value for S E is interpreted as arising
from the non-random distribution of molecules. �is is exempli�ed by ionic
solutions, in which ions of one charge are more likely to be surrounded by ions
of the opposite charge than of the same charge (Topic �F).
D�B.�
All of the colligative properties result from the lowering of the chemical potential of the solvent due to the presence of the solute. For an ideal solution, this
reduction is predicted by µ A = µ ∗A + RT ln x A . �e relationship shows that as
the amount of solute increases, the mole fraction of the solvent x A decreases
and hence the chemical potential the solvent A decreases.
If the chemical potential of the solvent is lowered, then the chemical potential
of the vapour in equilibrium with the solvent is also lowered because at equilibrium these two chemical potentials must be equal. �e chemical potential
of a perfect gas is given by µ A = µ −A○ + RT ln p A , so a lowering of the chemical
potential results in a reduction in the pressure.
�e overall result is that addition of a solute reduces the vapour pressure of
the solvent, and therefore the temperature at which the solvent boils is raised
because a greater increase in temperature is needed to make the vapour pressure equal to the external pressure. Similarly, the freezing point of the solvent
is decreased because the chemical potential of the solid will equal that of the
solvent at a lower temperature.
At a molecular level the decrease in vapour pressure can be thought of as being
due to the solute molecules getting in the way of the solvent molecules, thus
reducing their tendency to escape. Another way of looking at this is that the
presence of a solute increases the ‘randomness’, and hence the entropy, of the
solution, thus reducing the tendency for the formation of the (pure) vapour or
solid.
143
144
5 SIMPLE MIXTURES
Solutions to exercises
E�B.�(a)
In Exercise E�A.�(a) it is found that the vapour pressure obeys
p HCl �(kPa) = 6.41 × 103 × (x HCl ) − 0.071
(�.�)
�e task is to work out the mole fraction that corresponds to the given molality.
�e molality of HCl is de�ned as b HCl = n HCl �m GeCl4 , where n HCl is the amount
in moles of HCl and m GeCl4 is the mass in kg of solvent GeCl� . �e mole
fraction of HCl is n HCl �(n HCl + n GeCl4 ), where n GeCl4 is the amount in moles of
GeCl� , which is given by n GeCl4 = m GeCl4 �M GeCl4 , where M GeCl4 is the molar
mass of GeCl� . �ese relationships allow the mole fraction to be rewritten as
follows
n HCl
n HCl
x HCl =
=
n HCl + n GeCl4 n HCl + m GeCl4 �M GeCl4
�e amount in moles of HCl is written is n HCl = b HCl m GeCl4 ; using this the
above expression for the mole fraction becomes
x HCl =
n HCl
b HCl m GeCl4
b
=
=
n HCl + m GeCl4 �M GeCl4 b HCl m GeCl4 + m GeCl4 �M GeCl4 b + 1�M GeCl4
x HCl =
b
(0.10 mol kg−1 )
=
= 0.0209...
−1
b + 1�M (0.10 mol kg ) + 1�(214.44 × 10−3 kg mol−1 )
�e molar mass of GeCl� is ���.�� g mol−1 , therefore the mole fraction corresponding to b = 0.10 mol kg−1 is
�e pressure is found by inserting this value into eqn �.�
p HCl �(kPa) = 6.41 × 103 × (0.0209...) − 0.071 = 1.34... × 102
E�B.�(a)
�e vapour pressure of HCl is therefore 1.3 × 102 kPa .
Raoult’s law, [�A.��–���], p A = x A p∗A relates the vapour pressure to the mole
fraction of A, therefore from the given data is it possible to compute x A . �e
task is to relate the mole fraction of A to the masses of A (the solvent) and B
(the solute), and to do this the molar masses M A and M B are introduced. With
these n A = m A �M A , where m A is the mass of A, and similarly for n B . It follows
that
nA
m A �M A
MB mA
xA =
=
=
n A + n B m A �M A + m B �M B M B m A + M A m B
�e �nal form of this expression for x A is rearranged to given an expression for
M B , which is the desired quantity; then x A is replaced by p A �p∗A
MB =
(p A �p∗A )M A m B
xA MA mB
=
m A (1 − x A ) m A [1 − (p A �p∗A )]
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e molar mass of the solvent benzene, A, is 78.1074 g mol−1 , hence
MB =
=
E�B.�(a)
(p A �p∗A )M A m B
m A [1 − (p A �p∗A )]
[(51.5 kPa)�(53.3 kPa)] × (78.1074 g mol−1 ) × (19.0 g)
(500 g) × [1 − (51.5 kPa)�(53.3 kPa)]
= ��.� g mol−1
�e freezing point depression ∆Tf is related to the molality of the solute B, b B ,
by [�B.��–���], ∆Tf = K f b B , where K f is the freezing-point constant. From the
data and the known value of K f it is possible to calculate b B . �e task is then to
relate this to the given masses and the desired molar mass of the solute, M B .
�e molality of B is de�ned as b B = n B �m A , where m A is the mass of the solvent
A in kg. It follows that
nB
m B �M B
bB =
=
mA
mA
where m B is the mass of solute B. From the freezing point data b B = ∆Tf �K f ,
therefore
∆Tf m B �M B
mB Kf
=
hence M B =
Kf
mA
m A ∆Tf
With the data given and the value of the freezing-point constant from the Resource section
MB =
(100 g) × (30 K kg mol−1 )
= ��� g mol−1
(0.750 kg) × (10.5 K)
Note that because molality is de�ned as (amount in moles)/(mass of solvent in
kg), the mass of solvent m A is used as �.��� kg.
E�B.�(a)
�e freezing point depression ∆Tf is related to the molality of the solute B,
b B , by [�B.��–���], ∆Tf = K f b B , where K f is the freezing-point constant. �e
molality of the solute B is de�ned as b B = n B �m A , where n B is the amount in
moles of B and m A is the mass in kg of solvent A. �e amount is related to the
mass of B, m B , using the molar mass M B : n B = m B �M B . It therefore follows
that
Kf mB
∆Tf = K f b B =
MB mA
�e molar mass of sucrose C�� H�� O�� is ���.���� g mol−1 . A volume ��� cm3 of
water has mass ��� g to a good approximation. Using these values with the data
given and the value of the freezing-point constant from the Resource section
gives the freezing point depression as
∆Tf =
Kf mB
(1.86 K kg mol−1 ) × (2.5 g)
=
= 0.0679... K
M B m A (342.2938 g mol−1 ) × (0.200 kg)
Note that because molality is de�ned as (amount in moles)/(mass of solvent
in kg), the mass of solvent m A is used as �.��� kg. �e new freezing point is
therefore 273.15 K − 0.0679... K = ���.�� K
145
146
5 SIMPLE MIXTURES
E�B.�(a)
�e osmotic pressure Π is related to the molar concentration of solute B, [B],
by [�B.��–���], Π = [B]RT. �e freezing point depression ∆Tf is related to the
molality of B, b B , by [�B.��–���], ∆Tf = K f b B , where K f is the freezing-point
constant. �e task is to relate [B] to b B so that these two relationships can be
used together.
�e molar concentration [B] is given by [B] = n B �V , where n B is the amount
in moles of B and V is the volume of the solvent A. �is volume is related to
the mass of A, m A , using the mass density ρ: V = m A �ρ. It therefore follows
that
�
nB
nB
nB
[B] =
=
=
ρ = bB ρ
V
m A �ρ m A
bB
With this the osmotic pressure is related to the molality
[B] =
Π
RT
hence
bB ρ =
Π
RT
and so
bB =
Π
ρRT
�e freezing point depression for a solution exerting this osmotic pressure is
therefore
Kf Π
∆Tf = K f b B =
ρRT
Note that because molality is de�ned as (amount in moles)/(mass of solvent in
kg), the mass of solvent m A must be in kg and therefore the mass density must
be used in kg volume−1 .
With the data given, the value of the freezing-point constant from the Resource
section, and taking the mass density of water as 1 g cm−3 = 1000 kg m−3 gives
the freezing point depression as
∆Tf =
Kf Π
(1.86 K kg mol−1 ) × (120 × 103 Pa)
=
ρRT (1000 kg m−3 ) × (8.3145 J K−1 mol−1 ) × (300 K)
= 0.0894... K
In this expression all of the quantities are in SI units therefore the temperature
is expected to be in K, which is veri�ed as follows
(K kg mol−1 ) × (Pa)
Pa
=
−3
−1
−1
−3 × K−1
J
×
m
(kg m ) × (J K mol ) × (K)
=
E�B.�(a)
kg m−1 s−2
=K
(kg m2 s−2 ) × m−3 × K−1
�e freezing point is therefore 273.15 K − 0.0894... K = ���.�� K
�e Gibbs energy of mixing is given by [�B.�–���], ∆ mix G = nRT(x A ln x A +
x B ln x B ), the entropy of mixing by [�B.�–���], ∆ mix S = −nR(x A ln x A +x B ln x B ).
∆ mix H for an ideal solution is zero .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e total amount in moles is 0.50 mol+2.00 mol = 2.50 mol. With A as hexane
and B as heptane the thermodynamic quantities are calculated as
∆ mix G = nRT(x A ln x A + x B ln x B )
= (2.50 mol) × (8.3145 J K−1 mol−1 ) × (298 K) × �
= −3.10 × 103 J
∆ mix S = −nR(x A ln x A + x B ln x B )
= −(2.50 mol) × (8.3145 J K−1 mol−1 ) × �
E�B.�(a)
= +10.4 J K−1
0.50 0.50 2.00 2.00
ln
+
ln
�
2.50 2.50 2.50 2.50
0.50 0.50 2.00 2.00
ln
+
ln
�
2.50 2.50 2.50 2.50
�e entropy of mixing is given by [�B.�–���], ∆ mix S = −nR(x A ln x A +x B ln x B ),
and is a maximum when x A = x B = 12 . �is is evident from Fig. �B.� on page
���.
�e task is to relate the mole fraction of A (heptane) to the masses of A and
B (hexane), and to do this the molar masses M J are introduced. With these
n J = m J �M J , where m J is the mass of J. It follows that
xA =
nA
m A �M A
MB mA
=
=
n A + n B m A �M A + m B �M B M B m A + M A m B
�is is rearranged to give an expression for m B �m A
xA =
MB mA
MB
mB MB 1
=
hence
=
�
− 1�
M B m A + M A m B M B + M A (m B �m A )
mA MA xA
�e molar mass of A (heptane) is ���.���� g mol−1 , and that of B (hexane) is
��.���� g mol−1 . With these values and x A = 12
mB MB 1
86.1706 g mol−1
1
=
�
− 1� =
�
− 1� = �.����
mA MA xA
100.1964 g mol−1 1�2
E�B.�(a)
More simply, if equal amounts in moles of A and B are required, the ratio of
the corresponding masses of A and B must be equal to the ratio of their molar
masses: m B �m A = M B �M A .
�e ideal solubility of solute B at temperature T is given by [�B.��–���], ln x B =
(∆ fus H�R)(1�Tf − 1�T), where ∆ fus H is the enthalpy of fusion of the solute,
and Tf is the freezing point of the pure solute.
∆ fus H 1
1
� − �
R
Tf T
28.8 × 103 J mol−1
1
1
=
−
� = −4.55...
−1 �
−1
(217 + 273.15) K (25 + 273.15) K
8.3145 J K mol
ln x B =
hence x B = 0.0105....
147
148
5 SIMPLE MIXTURES
�e mole fraction is expressed in terms of the molality, b B = n B �m A , where m A
is the mass of the solvent in kg, in the following way
hence
nB
nB
n B �m A
bB
=
=
=
n A + n B m A �M A + n B 1�M A + n B �m A 1�M A + b B
xB
bB =
(1 − x B )M A
xB =
where M A is the molar mass of A, expressed in kg mol−1 . �e molar mass of
solvent benzene is ��.���� g mol−1 or 78.1074 × 10−3 kg mol−1 , therefore
bB =
E�B.�(a)
xB
0.0105...
=
= 0.136... mol kg−1
(1 − x B )M A (1 − 0.0105...) × (78.1074 × 10−3 kg mol−1 )
�e molality of the solution is therefore 0.137 mol kg−1 . �e molar mass of
anthracene (C�� H�� ) is ���.��� g mol−1 , so the mass of anthracene which is
dissolved in � kg of solvent is (0.136... mol kg−1 )×(1 kg)×(178.219 g mol−1 ) =
24.3 g .
Let the solvent CCl� be A and the solute Br� be B. �e vapour pressure of the
solute in an ideal dilute solution obeys Henry’s law, [�A.��–���], p B = K B x B ,
and the vapour pressure of the solvent obeys Raoult’s law, [�A.��–���], p A =
p∗A x A .
p B = K B x B = (122.36 Torr) × 0.050 = 6.11... Torr
p A = p∗A x A = (33.85 Torr) × (1 − 0.050) = 32.1... Torr
p tot = p A + p A = (6.11... Torr) + (32.1... Torr) = 38.2... Torr
�erefore the pressure are p B = 6.1 Torr , p A = 32 Torr , and p tot = 38 Torr .
�e partial pressure of the gas is given by p A = y A p tot , where y A is the mole
fraction in the vapour
32.1... Torr
pA
=
= �.��
p tot 38.2... Torr
pB
6.11... Torr
yB =
=
= �.��
p tot 38.2... Torr
yA =
E�B.��(a) Let methylbenzene be A and �,�-dimethylbenzene be B. If the solution is ideal
the vapour pressure obeys Raoult’s law, [�A.��–���], p J = p∗J x J . �e mixture
will boil when the sum of the partial vapour pressures of A and B equal the
external pressure, here �.�� atm.
p ext = p A + p B = x A p∗A + x B p∗B = x A p∗A + (1 − x A )p∗B
p ext − p∗A
p ext − p∗B
hence x A = ∗
and
by
analogy
x
=
B
p A − p∗B
p∗B − p∗A
(0.50 atm) × [(101.325 kPa)�(1 atm)] − (20.0 kPa)
= 0.920... = �.��
(53.3 kPa) − (20.0 kPa)
(0.50 atm) × [(101.325 kPa)�(1 atm)] − (53.3 kPa)
xB =
= 0.0792... = �.��
(20.0 kPa) − (53.3 kPa)
xA =
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e partial pressure of the gas is given by p J = y J p ext , where y J is the mole
fraction in the gas, and p J is given by p J = p∗J x J , hence y J = x J p∗J �p ext
x A p∗A
(0.920...) × (53.3 kPa)
=
= �.��
p ext
(0.50 atm) × [(101.325 kPa)�(1 atm)]
x B p∗B
(0.0792...) × (20.0 kPa)
yB =
= �.��
=
p ext
(0.50 atm) × [(101.325 kPa)�(1 atm)]
yA =
E�B.��(a)
�e vapour pressure of component J in the solution obeys Raoult’s law, [�A.��–
���], p J = p∗J x J , where x J is the mole fraction in the solution. In the gas the
partial pressure is p J = y J p tot , where y J is the mole fraction in the vapour.
�ese relationships give rise to four equations
p A = p∗A x A
p B = p∗B (1 − x A )
p A = p tot y A
p B = p tot (1 − y A )
where x A + x B = 1 is used and likewise for the gas. In these equations x A and
p tot are the unknowns to be found. �e expressions for p A are set equal, as are
those for p B , to give
p∗A x A = p tot y A hence p tot =
p∗A x A
yA
p∗B (1 − x A ) = p tot (1 − y A ) hence p tot =
p∗B (1 − x A )
1 − yA
�ese two expressions for p tot are set equal and the resulting equation rearranged to �nd x A
p∗A x A p∗B (1 − x A )
=
yA
1 − yA
With the data given
xA =
hence
xA =
p∗B y A
p∗A (1 − y A ) + p∗B y A
p∗B y A
(52.0 kPa) × (0.350)
=
∗
p A (1 − y A ) + p∗B y A (76.7 kPa)∗ (1 − 0.350) + (52.0 kPa) × (0.350)
= 0.267...
and
x B = 1 − 0.267... = 0.732...
�e composition of the liquid is therefore x A = 0.267 and x B = 0.733 .
�e total pressure is computed from p A = p tot y A and p A = p∗A x A to give p tot =
x A p∗A �y A
E�B.��(a)
p tot =
x A p∗A (0.267...) × (76.7 kPa)
=
= ��.� kPa
yA
0.350
If the solution is ideal, the vapour pressure of component J in the solution
obeys Raoult’s law, [�A.��–���], p J = p∗J x J , where x J is the mole fraction in
the solution. In the gas the partial pressure is p J = y J p tot , where y J is the mole
fraction in the vapour.
149
150
5 SIMPLE MIXTURES
Assuming ideality, the total pressure is computed as
p tot = p A + p B = p∗A x A + p∗B (1 − x A )
= (127.6 kPa) × (0.6589) + (50.60 kPa) × (1 − 0.6589) = 101 kPa
�e normal boiling point is when the total pressure is � atm, and this is exactly
the pressure found by assuming Raoult’s law applies. �e solution is therefore
ideal .
�e composition of the vapour is computed from p A = p tot y A and p A = p∗A x A
hence
p∗ x A (127.6 kPa) × (0.6589)
pA
yA =
= A
=
= 0.829...
p tot
p tot
101.325 kPa
It follows that y B = 1 − 0.829... = 0.170.... �e composition of the vapour is
therefore y A = 0.830 and y B = 0.170 .
Solutions to problems
P�B.�
�e freezing point depression in terms of mole fraction is predicted by [�B.��–
���]
RT ∗2
∆T = x B K ′
K′ =
∆ fus H
With the data given
K′ =
RT ∗2
(8.3145 J K−1 mol−1 ) × (290 K)2
=
= 61.3... K
∆ fus H
11.4 × 103 J mol−1
�e data are given in terms of molality, which is n B �m A , where n B is the amount
in moles of solute and m A is the mass of solvent in kg. �e mole fraction x B is
related to the molality by using the molar mass of the solvent, M A
xB =
nB
nB
n B �m A
bB
=
=
=
n A + n B m A �M A + n B 1�M A + n B �m A 1�M A + b B
�e molar mass of ethanoic acid CH� COOH is ��.���� g mol−1 . Because m A
must be in kg the molar mass must be expressed in kg volume−1 , M A = 60.0516×
10−3 kg mol−1 . For the data given 1�M A � b B therefore the expression for the
mole fraction is well approximated by x B = b B M A . With this, the freezing point
depression is given by
�
∆T = b B M A K ′
xB
hence
b B = ∆T�M A K ′
�e table below gives values of b B calculated from the given ∆T and this expression; to distinguish these values for the experimental values of b B , the calculated values are termed apparent molalities, b B,app
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
b B �(mol kg−1 )
∆T�K
b B,app �(mol kg−1 )
b B,app �b B
M B,app �(g mol−1 )
0.015
0.115
0.031
0.037
0.295
0.077
0.470
0.128
1.657
35.1
0.295
1.381
0.375
1.271
45.7
0.602
2.67
0.725
1.204
48.3
0.080
2.081
2.165
27.9
26.8
�e apparent molar mass of B, M B,app , is computed using
M B,app
bB
=
MB
b B,app
with M B = 58.01 g mol−1 , the molar mass of KF. �e argument that leads
to this is that the greater the apparent molality the smaller the molar mass:
M B,app ∝ 1�b B,app .
�e data in the table show that the molality predicted from the experimental
freezing point depression using the value of the freezing-point constant determined by [�B.��–���] is always greater than the molality know from the way
the solution was prepared. Presumably this latter molality is based on adding
a know mass of KF to a known mass of solvent, and assuming that the molar
mass of KF is ��.� g mol−1 . �e fact that the apparent molality is higher than
the molality of the prepared solution implies that the number of solute species
is greater than expected.
�e freezing point depression depends on the mole fraction of the solute, regardless of its identity. �erefore if the added KF were to dissociate completely
on dissolution in ethanoic acid the mole fraction of the solute would be twice as
large as expected on the basis of the amount of added KF, and in turn this would
mean that the apparent molality (based on the freezing-point depression) is
twice as large as expected.
P�B.�
�e data in the table can be interpreted as indicating that there is dissociation
of the KF, and that this dissociation is greater at lower molalities. However, this
only part of the picture as it does no explain why b B,app �b B is greater than � at
some molalities.
Let the two components of the mixture be labelled � (propionic acid) and �
(THP). �e de�nition of the partial molar volume of �, V1 , is
V1 = �
∂V
�
∂n 1 n 2
To use this de�nition an expression for V as a function of the n J is required.
�e excess volume V E is de�ned as V E = ∆V −∆V ideal , where ∆V is the volume
of mixing and ∆V ideal is the volume of mixing of the ideal solution, which is
zero. �erefore V E = ∆V .
151
152
5 SIMPLE MIXTURES
�e volume of mixing ∆V is written ∆V = V − Vseparated , which from the above
is also written V E = V − Vseparated . �e expression given in the problem for V E
is per mole, so for mixing n 1 moles of � is mixed with n 2 moles of � the excess
volume is in fact (n 1 + n 2 )V E .
�e volume of the separated components is computed from their mass densities: n 1 moles corresponds to a mass n 1 M 1 , where M 1 is the molar mass, which
has volume n 1 M 1 �ρ 1 , where ρ 1 is the mass density. It follows that
(n 1 + n 2 )V E = V −
n1 M1 n2 M2
n1 M1 n2 M2
−
hence V = (n 1 + n 2 )V E +
+
ρ1
ρ2
ρ1
ρ2
�e second equation above is the required expression for V as a function of n 1
and n 2 . �e expression given in the problem for V E is a function of the mole
fractions, which are easily written in terms of the amounts.
To compute the partial molar volume it is necessary to compute the derivative
of V with respect to n 1 , keeping in mind that V E is a function of n 1
hence V1 = �
V = (n 1 + n 2 )V E +
n1 M1 n2 M2
+
ρ1
ρ2
∂V
∂V E
M1
� = (n 1 + n 2 ) �
� + VE +
∂n 1 n 2
∂n 1 n 2
ρ1
(�.�)
To compute the derivative it has been recognised that (n 1 + n 2 )V E is a product
of two functions of n 1 .
�e �rst step is to compute (∂V E �∂n 1 )n 2 , and this requires rewriting the mole
fractions in terms of the n i
V E = x 1 x 2 a 0 + x 12 x 2 a 1 − x 1 x 22 a 1
�
=
n1 n2 a0
n 12 n 2 a 1
n 1 n 22 a 1
+
−
(n 1 + n 2 )2 (n 1 + n 2 )3 (n 1 + n 2 )3
∂V E
n2 a0
2n 1 n 2 a 0
2n 1 n 2 a 1
� =
−
+
2
3
∂n 1 n 2 (n 1 + n 2 )
(n 1 + n 2 )
(n 1 + n 2 )3
−
3n 12 n 2 a 1
n 22 a 1
3n 1 n 22 a 1
−
+
(n 1 + n 2 )4 (n 1 + n 2 )3 (n 1 + n 2 )4
�e quantity required is (n 1 +n 2 )(∂V E �∂n 1 )n 2 , so the above expression is multiplied by (n 1 + n 2 ). �is cancels a term (n 1 + n 2 ) in each of the denominators
and allows the expression to be rewritten in terms of the mole fractions
(n 1 + n 2 )(∂V E �∂n 1 )n 2 = x 2 a 0 −2x 1 x 2 a 0 +2x 1 x 2 a 1 −3x 12 x 2 a 1 − x 22 a 1 +3x 1 x 22 a 1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e parts of eqn �.� are now assembled
V1 = (n 1 + n 2 ) �
∂V E
M1
� + VE +
∂n 1 n 2
ρ1
= (x 2 a 0 − 2x 1 x 2 a 0 + 2x 1 x 2 a 1 − 3x 12 x 2 a 1 − x 22 a 1 + 3x 1 x 22 a 1 )
+ (x 1 x 2 a 0 + x 12 x 2 a 1 − x 1 x 22 a 1 ) + M 1 �ρ 1
= a 0 x 2 (1 − x 1 ) + a 1 x 2 (2x 1 − 2x 12 − x 2 + 2x 1 x 2 ) + M 1 �ρ 1
= a 0 x 2 (x 2 ) + a 1 x 2 (2x 1 [1 − x 1 ] − x 2 + 2x 1 x 2 ) + M 1 �ρ 1
= a 0 x 22 + a 1 x 2 (2x 1 x 2 − x 2 + 2x 1 x 2 ) + M 1 �ρ 1
= a 0 x 22 + a 1 x 2 (4x 1 x 2 − x 2 ) + M 1 �ρ 1
= a 0 x 22 + a 1 x 22 (4x 1 − 1) + M 1 �ρ 1
�e last four lines involve repeated use of x 1 + x 2 = 1 in order to simplify the
expression. A similar process is used to �nd an expression for V2 . In principle
all that is required is to swap the indices � and �, however when this is done for
the expression for V E the result is
V E = x 2 x 1 a 0 + x 22 x 1 a 1 − x 2 x 1 a 1
which, when compared with the original expression, shows that the sign of the
term in a 1 is reversed: this change needs to be carried through to the end. In
summary
V1 = a 0 x 22 + a 1 x 22 (4x 1 − 1) + M 1 �ρ 1
V2 = a 0 x 12 − a 1 x 12 (4x 2 − 1) + M 2 �ρ 2
�e molar mass of propionic acid CH� CH� COOH is M 1 = 74.0774 g mol−1
and that of THP C� H�� O is M 1 = 86.129 g mol−1 . For an equimolar mixture
x 1 = x 2 = 12 and therefore
V1 = a 0 x 22 + a 1 x 22 (4x 1 − 1) + M 1 �ρ 1 = 14 a 0 + 14 a 1 + M 1 �ρ 1
−1
−1
= 0.25 × (−2.4697 cm3 mol ) + 0.25 × (0.0608 cm3 mol )
−1
+ (74.0774 g mol−1 )�(0.97174 g cm−3 ) = ��.� cm3 mol
V2 = a 0 x 12 − a 1 x 12 (4x 2 − 1) + M 2 �ρ 2 = 14 a 0 − 14 a 1 + M 2 �ρ 2
−1
−1
= 0.25 × (−2.4697 cm3 mol ) − 0.25 × (0.0608 cm3 mol )
−1
+ (86.129 g mol−1 )�(0.86398 g cm−3 ) = ��.� cm3 mol
P�B.�
�e excess Gibbs energy G E is de�ned in [�B.�–���], G E = ∆ mix G − ∆ mix G ideal .
�e ideal Gibbs energy of mixing (per mole) is given by [�B.�–���], ∆ mix G ideal =
153
154
5 SIMPLE MIXTURES
RT(x A ln x A + x B ln x B ). Let A by MCH and B be THF. �e Gibbs energy of
mixing of n A moles of A with n B moles of B is therefore given by
∆ mix G = ∆ mix G ideal + G E
= (n A + n B )RT(x A ln x A + x B ln x B )
+ (n A + n B )RTx A (1 − x A )[0.4857 − 0.1077(2x A − 1)
+ 0.0191(2x A − 1)2 ]
With the values given (n A + n B ) = 4 mol, x A = 41 , and x B = 34
∆ mix G = (4.00 mol) × (8.3145 J K−1 mol−1 ) × (303.15 K) × ( 14 ln 14 + 34 ln 34 )
+ (4.00 mol) × (8.3145 J K−1 mol−1 ) × (303.15 K) × 14 × (1 − 14 )
× [0.4857 − 0.1077(2 × 14 − 1) + 0.0191(2 × 14 − 1)2 ]
P�B.�
= −4.64 kJ
�e osmotic pressure Π is related to the molar concentration [J] through a
virial-type equation, [�B.��–���], Π = [J]RT(1 + B[J]). �e data are given
in terms of the mass concentration, so the �rst task is to relate this to the molar
concentration. If the amount in moles of solute dissolved in volume V is n J and
the molar mass is M J it follows that
�
n J m J �M J m J 1
cJ
[J] =
=
=
=
V
V
V MJ MJ
cJ
where c J is the mass concentration. With this the virial equation is rewritten
Π = [J]RT(1 + B[J]) =
cJ
cJ
RT �1 + B �
MJ
MJ
Division of both sides by c J gives an equation of a straight line
Π RT BRT
=
+
cJ
cJ MJ
M J2
A plot of Π�c J against c J will have intercept RT�M J when c J = 0, and from this
it is possible to determine the molar mass.
�e pressure is given by Π = hρg; for the pressure to be in Pa the height needs
to be in m and ρ in kg m−3 ; for the present case ρ = 1 g cm−3 = 1000 kg m−3 .
�e data are plotted in Fig. �.�.
c�(mg cm−3 )
(Π�c)�(Pa mg−1 cm3 )
h�(cm)
Π�Pa
3.221
5.746
563.7
4.618
8.238
808.1
5.112
9.119
894.6
174.994 9
6.722
11.990
1 176.2
174.980 5
175.002 4
174.999 5
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(Π�c)�(Pa mg−1 cm3 )
175.02
175.00
174.98
0
1
2
Figure 5.4
3
4
c�(mg cm )
−3
5
6
7
�e data are a modest �t to a straight line, the equation of which is
(Π�c)�(Pa mg−1 cm3 ) = −6.628 × 10−3 × c�(mg cm−3 ) + 175.03
�e intercept is RT�M J ; before using the intercept in this expression it is best
to convert it from (Pa mg−1 cm3 ) to SI units, (Pa kg−1 m3 )
(175.03 Pa mg−1 cm3 ) ×
MJ =
P�B.�
1 m3
1 mg
× −6
= 175.03 Pa kg−1 m3
6
3
10 cm
10 kg
RT
(8.3145 J K−1 mol−1 ) × (293.15 K)
=
= 13.92... kg mol−1
intercept
175.03 Pa kg−1 m3
�e molar mass of the protein is therefore 1.39 × 104 g mol−1 .
�e osmotic pressure Π is related to the molar concentration [J] through a
virial-type equation, [�B.��–���], Π = [J]RT(1 + B[J]). As is shown in Problem
P�B.� this equation can be rewritten in terms of the mass concentration c J and
the molar mass of J, M J
Π RT BRT
=
+
cJ
cJ MJ
M J2
A plot of Π�c J against c J will have intercept RT�M J when c J = 0: from this it is
possible to determine the molar mass. �e second virial coe�cient is obtained
from the slope. �e plot is shown in Fig. �.�.
c�(mg cm−3 )
Π�Pa
(Π�c)�(Pa mg−1 cm3 )
1.33
30
22.6
2.10
51
4.52
132
29.2
7.18
246
34.3
9.87
390
39.5
24.3
155
5 SIMPLE MIXTURES
40
(Π�c)�(Pa mg−1 cm3 )
156
35
30
25
20
0
2
Figure 5.5
4
6
c�(mg cm )
−3
8
10
�e data are a good �t to a straight line, the equation of which is
(Π�c)�(Pa mg−1 cm3 ) = 1.975 × (c�(mg cm−3 )) + 20.09
�e intercept is RT�M J ; before using the intercept in this expression it is best
to convert it from (Pa mg−1 cm3 ) to SI units, (Pa kg−1 m3 )
MJ =
(20.09 Pa mg−1 cm3 ) ×
1 m3
1 mg
× −6
= 20.09 Pa kg−1 m3
6
3
10 cm
10 kg
RT
(8.3145 J K−1 mol−1 ) × (303.15 K)
=
= 1.25... × 102 kg mol−1
intercept
20.09 Pa kg−1 m3
�e molar mass of the polymer is therefore 1.25 × 105 g mol−1 .
�e second virial coe�cient is found by taking the ratio (slope)/(intercept)
slope
B
=
intercept M J
hence
B = MJ ×
slope
intercept
1.975 Pa mg−2 cm6
20.09 Pa mg−1 cm3
1.975 Pa mg−2 cm6
= (1.25... × 108 mg mol−1 ) ×
= 1.23... × 107 cm3 mol−1
20.09 Pa mg−1 cm3
B = (1.25... × 102 kg mol−1 ) ×
P�B.��
Given that [J] is usually in mol dm−3 it is convenient to quote the value of the
second virial coe�cient as B = 1.23 × 104 mol−1 dm3 .
�e excess enthalpy of mixing for this particular regular solution is given by
[�B.�–���], H E = nRT ξx A x B . �e plot in Fig. �.� shows H E �(nRT) as a function of x A for di�erent values of ξ; recall that x A +x B = 1, so x A x B = x A (1−x A ).
If ξ is �xed, the temperature dependence is explored by plotting H E �(nRξ) as
a function of x A : H E �(nRξ) = Tx A x B = Tx A (1− x A ). �is is shown in Fig. �.�.
Evidently the strongest temperature dependence is once more at x A = 12 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
ξ = −2
ξ = −1
ξ=0
ξ = +1
ξ = +2
0.4
H E �(nRT)
0.2
0.0
−0.2
−0.4
0.0
0.2
0.4
Figure 5.6
xA
0.6
0.8
1.0
200 K
250 K
300 K
350 K
(H E �(nRξ))�K
80
60
40
20
0
0.0
Figure 5.7
P�B.��
0.2
0.4
xA
0.6
0.8
1.0
�e osmotic pressures Π is related to the molar concentration [J] through a
virial-type equation, [�B.��–���], Π = [J]RT(1 + B[J]). As is shown in Problem
P�B.� this equation can be rewritten in terms of the mass concentration c J and
the molar mass of J, M J
Π RT BRT
=
+
cJ
cJ MJ
M J2
A plot of Π�c J against c J will have intercept RT�M J when c J = 0: from this it is
possible to determine the molar mass. �e second virial coe�cient is obtained
from the slope. Such a plot is shown in Fig. �.�.
157
5 SIMPLE MIXTURES
c�(g dm−3 )
(Π�c)�(Pa g−1 dm3 )
Π�Pa
1.00
27
27.0
2.00
70
4.00
197
49.3
7.00
500
71.4
9.00
785
87.2
35.0
80
(Π�c)�(Pa g−1 dm3 )
158
60
40
20
0
2
4
6
c�(g dm )
Figure 5.8
−3
8
10
�e data are a good �t to a straight line, the equation of which is
(Π�c)�(Pa g−1 dm3 ) = 7.466 × (c�(g dm−3 )) + 19.64
�e intercept is RT�M J ; before using the intercept in this expression it is best
to convert it from (Pa g−1 dm3 ) to SI units, (Pa kg−1 m3 )
MJ =
(19.64 Pa g−1 dm3 ) ×
1 m3
1g
×
= 19.64 Pa kg−1 m3
103 dm3 10−3 kg
RT
(8.3145 J K−1 mol−1 ) × (298 K)
=
= 1.26... × 102 kg mol−1
intercept
19.64 Pa kg−1 m3
�e molar mass of the polymer is therefore 1.26 × 105 g mol−1 . �e second
virial coe�cient is found by taking the ratio (slope)/(intercept)
slope
B
=
intercept M J
hence
B = MJ ×
slope
intercept
7.466 Pa g−2 dm6
19.64 Pa g−1 dm3
7.466 Pa g−2 dm6
= (1.26... × 105 g mol−1 ) ×
= 4.79... × 104 dm3 mol−1
19.64 Pa g−1 dm3
B = (1.26... × 102 kg mol−1 ) ×
�e second virial coe�cient is therefore B = 4.80 × 104 mol−1 dm3 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
5C Phase diagrams of binary systems: liquids
Answers to discussion questions
D�C.�
�e principal factor is the shape of the two-phase liquid-vapour region in the
phase diagram (usually a temperature-composition diagram). �e closer the
liquid and vapour lines are to each other, the more steps of the sort illustrated in
Fig. �C.� on page ��� are needed to move from a given mixed composition to an
acceptable enrichment in one of the components. However, the presence of an
azeotrope could prevent the desired degree of separation from being achieved.
D�C.�
A low-boiling azeotrope has a boiling temperature lower than that of either
component, so it is easier for the molecules to move into the vapour phase
than in a ‘normal’ (non-azeotropic) mixture. �erefore, the liquid phase has
less favorable intermolecular interactions than in a ‘normal’ mixture, a sign
that the components are less attracted to each other in the liquid phase than to
molecules of their own kind. �ese intermolecular interactions are determined
by factors such as dipole moment (polarity) and hydrogen bonding.
Conversely, a high-boiling azeotrope has a boiling temperature higher than that
of either component, so it is more di�cult for the molecules to move into the
vapour phase. �is re�ects the relatively unusual situation of components that
have more favorable intermolecular interactions with each other in the liquid
phase than with molecules of their own kind.
Solutions to exercises
E�C.�(a)
�e molar masses of phenol and water are ��.���� g mol−1 and ��.���� g mol−1 ,
respectively. �e mole fraction of phenol (P) is
xP =
(7.32 g)�(94.1074 g mol−1 )
= 0.149...
(7.32 g)�(94.1074 g mol−1 ) + (7.95 g)�(18.0158 g mol−1 )
Hence x P = 0.150 . Let the two phases be α (x P = 0.042) and β (x P = 0.161).
�e proportions of these two phases, n β �n α is given by the level rule, [�C.�–���]
n β l α 0.149... − 0.042
=
=
= �.��
n α l β 0.161 − 0.149...
�e phenol-rich phase is more abundant by a factor of almost ��.
E�C.�(a)
An approximate phase diagram is shown in Fig. �.�; the given data points are
shown with dots and the curve is a quadratic which is a modest �t to these
points. �e shape conforms to the expected phase diagram for such a system.
(i) A temperature of �� ○ C is above the highest temperature at which partial miscibility occurs, and therefore the expectation is that hexane and
per�uorohexane mix in all proportions to give a single phase.
(ii) At �� ○ C the possibility of phase separation exists; as the mole fraction
of per�uorohexane increases the phase diagram is traversed along the
159
5 SIMPLE MIXTURES
23.0
22.5
θ�○ C
160
22.0
21.5
21.0
0.0
0.2
0.4
x C6 F14
0.6
0.8
1.0
Figure 5.9
dashed line. When the mole fraction of per�uorohexane is low a single
phase forms, but as the mole fraction goes beyond �.�� phase separation occurs. Initially, according to the lever rule, the proportion of the
per�uorohexane-rich phase is very small, but as more and more per�uorohexane is added the proportion of this phase increases. When the mole
fraction is just under �.��, there is very little of the per�uorohexane-poor
phase present, and as the mole fraction increases further a one-phase zone
is reached in which there is complete miscibility.
E�C.�(a)
�e temperature–composition phase diagram is a plot of the boiling point against
(�) composition of the liquid, x M and (�) composition of the vapour, y M . �e
horizontal axis is labelled z M , which is interpreted as x M or y M according to
which set of data are being plotted. In addition to the data in the table, the
boiling points of the pure liquids are added. �e plot is shown in Fig �.��; in
this plot, the lines are best-�t polynomials of order �.
θ�○ C
xM
110.6
1
1
117.3
0.408
0.527
110.9
0.908
0.923
119.0
0.300
0.410
112.0
0.795
0.836
121.1
0.203
0.297
114.0
0.615
0.698
123.0
0.097
0.164
115.8
0.527
0.624
125.6
0
0
yM
θ�○ C
xM
yM
(i) �e vapour composition corresponding to a liquid composition of x M =
0.250 is found by taking the vertical line at this composition up to the
intersection with the liquid curve, and then moving across horizontally
to the intersection with the vapour curve; occurs at y M = 0.354 , which
gives the composition of the vapour. �e exact points of intersection can
be found either from the graph or by using the �tted functions.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
125
liquid
vapour
110
0.0
x M = 0.250
115
y M = 0.354
θ�○ C
120
0.2
Figure 5.10
E�C.�(a)
0.4
zM
0.6
0.8
1.0
(ii) A composition x O = 0.250 corresponds to x M = 0.750; from the graph
this corresponds to a vapour composition y M = 0.811 .
At the lowest temperature shown in the diagram the mixture is in the twophase region, and the two phases have composition of approximately x B =
0.88 and x B = 0.05. �e level rule shows that there is about � times more of
the B-rich than of the B-poor phase. As the temperature is raised the B-rich
phase becomes slightly less rich in B, and the other phase becomes richer in B.
�e lever rule implies that the proportion of the B-rich phase increases as the
temperature rises.
At temperature T1 the vertical line intersects the phase boundary. At this point
the B-poor phase disappears and only one phase, with x B = 0.8, is present.
Solutions to problems
P�C.�
If it is assumed that Raoult’s law applies, [�A.��–���], the partial vapour pressures of benzene (B) and methylbenzene (M) are
p B = x B p∗B
p M = x M p∗M
where x J are the mole fractions and p∗J are the vapour pressures over the pure
liquids. �e total pressure is taken to be p tot = p B + p M .
�e mole fraction in the vapour, y J , is related to the total pressure by p J = y J p tot ,
so it follows that
x J p∗J
pJ
yJ =
=
p tot
p tot
�erefore
x B p∗B
0.75 × (75 Torr)
=
= �.��
p tot
0.75 × (75 Torr) + 0.25 × (21 Torr)
x M p∗M
0.25 × (21 Torr)
yM =
=
= �.���
p tot
0.75 × (75 Torr) + 0.25 × (21 Torr)
yB =
161
5 SIMPLE MIXTURES
P�C.�
It is convenient to construct a pressure–composition phase diagram in order
to answer this question. If it is assumed that Raoult’s law applies, [�A.��–���],
the total pressure is computed from the sum of the partial vapour pressures of
benzene (B) and methylbenzene (M)
p tot = p B + p M = x B p∗B + x M p∗M
where x J are the mole fractions and p∗J are the vapour pressures over the pure
liquids. �is equation is used to construct the liquid line on the graph shown
in Fig. �.��, where z B is interpreted as x B .
�e mole fraction in the vapour, y J , is related to the total pressure by p J = y J p tot .
Using this it can be shown that the total pressure in terms of the mole fraction
in the vapour in given by [�C.�–���],
p tot =
p∗B p∗M
∗
p B + (p∗M − p∗B )y B
�is equation is used to construct the vapour line on the phase diagram, where
z B is interpreted as y B .
10.0
liquid
vapour
8.0
p tot �kPa
162
a1
6.0
4.0
a′′3
2.0
0.0
0.2
Figure 5.11
a ′′2
′
a2 a2
a ′1
a3
0.4
zB
0.6
0.8
1.0
(a) A mixture with equal amounts of B and M has mole fractions x B = x M =
1
. �e total pressure is therefore
2
p tot = x B p∗B + x M p∗M = 12 (9.9 kPa) + 12 (2.9 kPa) = �.� kPa
�is is the pressure at which boiling �rst occurs, point a 1 in the diagram.
(b) �e composition of the vapour is given by
yB =
1
pB
x B p∗B 2 (9.9 kPa)
=
=
= 0.773... = �.��
p tot
p tot
6.4 kPa
and therefore y M = 1 − 0.773... = �.�� . �is is point a ′1 on the diagram:
the lever-rule also indicates that the fraction of the phase with composition a′1 (the vapour) is very small.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(c) As the pressure is reduced further, say to point a 2 , the tie line indicates
that the liquid will have composition a ′′2 and the vapour will have composition a′2 , the latter being richer in the more volatile component B.
�e level rule indicates that the proportion of the vapour phase is now
signi�cant.
�e process continues until point a 3 is reached. At this pressure the composition of the liquid is given by point a ′′3 , and the level rule indicates that
the proportion of the liquid phase is very small. It is also evident from the
diagram that at point a 3 the vapour composition is y B = y M = 12 , therefore
p B = 12 p tot and p M = 12 p tot . Raoult’s law gives the partial vapour pressures
of B and M are p B = p∗B x B and p M = p∗M x M . It follows that
1
p = x B p∗B
2 tot
and
1
p = x M p∗M = (1 − x B )p∗M
2 tot
�ese two equations are combined to give
xB =
p∗M
(2.9 kPa)
=
= 0.226... = �.��
∗
p B + p∗M (9.9 kPa) + (2.9 kPa)
and x M = 1 − x B = 1 − 0.226... = �.�� . �is is point a ′′3 on the phase
diagram. �e vapour pressure of a mixture with this composition is
p tot = x B p∗B + x M p∗M = 0.226... × (9.9 kPa) + (1 − 0.226...) × (2.9 kPa)
P�C.�
= �.� kPa
�e annotated phase diagrams are shown in Fig. �.��. Given that the normal
boiling point of hexane is certainly lower than that of heptane the horizontal
scale should presumably be mole fraction of heptane; however, the solution
provided follows the labelling of the diagram in the text.
(a) �e phases present are indicated on the diagrams above.
(b) For an equimolar mixture x H = 0.5; the vertical line at this composition
intersects the liquid line at a, and reading across the pressure is ��� Torr .
(c) At this pressure the composition of the vapour is given by the intersection
of the horizontal line with the vapour line, which occurs at point d. �e
vapour is less rich in hexane than the liquid. As the solution continues
to evapourate the composition of the liquid moves along the liquid line
to point c. �is is at the pressure at which the composition of the vapour
matches the original composition of the liquid (x H = 0.5, point b). �e
composition of the liquid is read o� the scale as x H = 0.7 , and the pressure is ��� Torr .
(d) From part (b) the composition of the liquid is x H = 0.5 , and the composition of the vapour is read o� from where the horizontal line at ��� Torr
intersects the vapour curve, point d, which is at y H = 0.3 .
(e) From part (c) the composition of the vapour is y H = 0.5 and the composition of the liquid can be read o� from where the horizontal line at
��� Torr intersects the liquid curve, point c, which is at x H = 0.7 .
163
5 SIMPLE MIXTURES
Pressure, p/Torr
900
liquid line
70°C
liquid
700
a
d
625
500
c
b
300
vapour
0
0.2
500
vapour line
0.6 0.7 0.8
0.3 0.4 0.5
Mole fraction of hexane, z H
1
100
vapour line
vapour
Temperature, θ /°C
164
90
f
g
e
v
l
80
liquid
70
760 Torr
liquid line
60
0
Figure 5.12
0.2
0.4
0.44
0.6
Mole fraction of hexane,
0.8
1
0.74
zH
(f) Refer to the temperature–composition phase diagram; the stated composition is z heptane = 0.40 which corresponds to z H = 0.60. �e vertical line
at z H = 0.60 intersects the horizontal line at �� ○ C at point e. From the tie
line the composition of the vapour is read o� from point f, y H = 0.44; the
composition of the liquid is read o� from point g, y H = 0.74. From the
lever rule
n l v 0.60 − 0.44
= =
= 1.1
n v l 0.74 − 0.60
�e two phases are roughly equally abundant.
P�C.�
�e relationship between y A and x A is given in [�C.�–���]
yA =
P�C.�
x A p∗A
x A (p∗A �p∗B )
=
p∗B + (p∗A − p∗B )x A 1 + (p∗A �p∗B − 1)x A
�e form of the function on the right gives y A as a function of x A and the ratio
(p∗A �p∗B ) as required. �e plot if shown in Fig. �.��
If the excess enthalpy is modelled as H E = ξRTx A2 x B2 then, by anaolgy with
[�B.�–���], the expression for for Gibbs energy of mixing is
∆ mix G = nRT �x A ln x A + x B ln x B + ξx A2 x B2 �
�e minima and maxima in this function are located by setting the derivative
with respect to x A to zero; it is convenient to take the derivative of ∆ mix G�nRT
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
1.0
0.8
yA
0.6
0.4
0.2
0.0
0.0
Figure 5.13
0.2
0.4
xA
0.6
(p∗A �p∗B ) = 1
(p∗A �p∗B ) = 4
(p∗A �p∗B ) = 50
0.8
1.0
and before doing this x B is replaced by (1 − x A )
∆ mix G�nRT = x A ln x A + (1 − x A ) ln(1 − x A ) + ξx A2 (1 − x A )2
= x A ln x A + ln(1 − x A ) − x A ln(1 − x A ) + ξx A2 (1 − x A )2
d(∆ mix G�nRT)
1
xA
= 1 + ln x A −
+
− ln(1 − x A )
dx A
1 − xA 1 − xA
+ 2ξx A (1 − x A )2 − 2ξx A2 (1 − x A )
1 − xA − 1 + xA
xA
=
+ ln
+ 2ξx A (1 − x A )(1 − 2x A )
1 − xA
1 − xA
xA
= ln
+ 2ξx A (1 − x A )(1 − 2x A )
1 − xA
As before, the derivative is zero at x A = 0.5 for all values of ξ; this corresponds
either to a minimum when ξ is small, or to a maximum when ξ is su�ciently
large. Qualitatively the behaviour is similar to that shown in Fig. �B.�.
Apart from this solution at x A = 0.5, there are no analytical solutions for when
this derivative is zero. However, solutions can be found by graphically by looking for the intersection between ln (x A �[1 − x A ]) and −2ξx A (1− x A )(1−2x A ).
�is is done with the aid of Fig. �.�b. From the graph it is evident that for ξ = 1
there are no values of x A at which the curves intersect, and so no minima, but
at su�ciently high values of ξ (such as ξ = 6) such intersections do occur and
lead to two minima.
Overall, as is seen in Fig. �.��, the behaviour is qualitatively similar to that
for H E = ξRTx A x B . If ξ is below some particular positive value ∆ mix G is
always negative and shows a single minimum at x A = 0.5. Above some critical
value, ∆ mix G may become positive for some values of x A , and the plot shows
a maximum at x A = 0.5, �anked symmetically by two minima. �is indicates
that phase separation will occur.
165
5 SIMPLE MIXTURES
∆ mix G�nRT
ξ=0
ξ=3
ξ=6
ξ = 15
0.0
−0.5
0.0
0.2
0.4
Figure 5.14
xA
0.6
0.8
1.0
5D Phase diagrams of binary systems: solids
Answers to discussion questions
D�D.�
�e schematic phase diagram is shown in Fig. �.��. Incongruent melting means
that the compound AB� does not occur in the liquid phase.
liquid
A(s) + B(l)
+ A(l)
AB2(s),B(l)
A(l)
Temperature
166
B(l) + A(l)
+ B(s)
AB2(s),B(l)
A(l)
A(s) + AB2(s)
AB2(s) + B(s)
0
Figure 5.15
0.2
0.4
0.6
AB2
Mole fraction of B, xB
0.8
1
Solutions to exercises
E�D.�(a)
E�D.�(a)
�e feature that indicates incongruent melting is the intersection of the two
liquid curves at around x B = 0.6. �e incongruent melting point is marked as
T1 ≈ 350 ○ C. �e composition of the eutectic is x B ≈ 0.25 and its melting point
is labelled T2 ≈ 190 ○ C .
�e cooling curves are shown in Fig �.��; the break points are shown by the
short horizontal lines. For isopleth a the �rst break point is at ��� ○ C where
the isopleth crosses the liquid curve, there is a second break point where the
isopleth crosses the boundary at T1 ; there is then a eutectic halt at ��� ○ C. For
isopleth b the �rst break point is at ��� ○ C where the isopleth crosses the liquid
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
curve, there is a second break point where the isopleth crosses the boundary at
T1 , and then a eutectic halt at ��� ○ C.
500
T�K
400
300
a
b
200
time
Figure 5.16
Figure �.�� shows the relevant phase diagram to which dotted horizontal lines
have been added at the relevant temperatures.
1000
c1
800
Liquid
600
400
200
0
Figure 5.17
c2
Ag 3Sn
Temperature, θ/°C
E�D.�(a)
0
20
40
60
Mass percentage of Ag/%
c3
80
100
(i) �e solubility of silver in tin at ��� ○ C is determined by the point c 1 . (At
higher proportions of silver, the system separates into two phases, a liquid
and a solid phase rich in silver.) �e point c 1 corresponds to ��% silver
by mass.
(ii) �e compound Ag� Sn decomposes at this temperature. �ree phases are
in equilibrium here: a liquid containing atomic Ag and Sn about ��% Ag
by mass; a solid solution of Ag� Sn in Ag; and solid Ag� Sn. See point c 2 .
(iii) At point c 3 , two phases coexist: solid Ag� Sn and a solid solution of the
compound and metallic silver. Because this point is close to the Ag� Sn
composition, the solid solution is mainly Ag� Sn, at least when measured
in mass terms. �e composition of the solid solution is expressed as a ratio
of moles of compound (n c ) to moles of atomic silver (n a ). �ese quantities
are related to the silver mass fraction c Ag by employing the de�nition of
167
5 SIMPLE MIXTURES
mass fraction, namely the mass of silver (from the compound and from
atomic silver) over the total sample mass
c Ag =
m Ag
(3n c + n a )M Ag
=
m Ag + m Sn (3n c + n a )M Ag + n c M Sn
�is relationship is rearranged, collecting terms in n c on one side and n a
on the other
n c �3M Ag (c Ag − 1) + M Sn c Ag � = n a M Ag (1 − c Ag )
�e mole ratio of compound to atomic silver is given by
M Ag (1 − c Ag )
nc
=
n a 3M Ag (c Ag − 1) + M Sn c Ag
At ��� ○ C, c Ag = 0.78 (point c 3 on the coexistence curve), so
nc
(107.9 g mol−1 ) × (1 − 0.78)
=
= �.��
n a 3 × (107.9 g mol−1 ) × (0.78 − 1) + (118.7 g mol−1 ) × 0.78
At ��� ○ C, c Ag = 0.77 (point c 2 on the coexistence curve), so
E�D.�(a)
nc
(107.9 g mol−1 ) × (1 − 0.77)
=
= �.��
n a 3 × (107.9 g mol−1 ) × (0.77 − 1) + (118.7 g mol−1 ) × 0.77
�e schematic phase diagram is shown in Fig �.��. �e solid points are the data
given in the Exercise, and the lines are simply plausible connections between
these points; it is assumed that the compound in �:�. Note that to the right of
x B = 0.5 the solids are AB and B, whereas to the le� of this composition the
solids are AB and A.
130 A(s)
AB(s)
+AB(l)
+A(l)
+A(l)
+AB(l)
T�K
168
120
AB(s)+AB(l)
+B(l)
110
100
0.0
Figure 5.18
B(s)
+AB(l)
+B(l)
A(s)+AB(s)
B(s)+AB(s)
0.2
0.4
xB
0.6
0.8
1.0
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
94
(i)
(ii)
(iii)
liquid
(iv)
(v)
92
T�K
90
two-phase liquid
88
86
84
82
0.0
CF� (s)+CH� (s)
0.2
0.4
x CF4
0.6
0.8
1.0
Figure 5.19
E�D.�(a)
�e schematic phase diagram is shown in Fig �.��. �e solid points are the data
given in the Exercise, and lines are simply plausible connections between these
points. (�e dash-dotted lines are referred in to Exercise E�D.�(a).)
E�D.�(a)
�e compositions at which the cooling curves are plotted are indicated by the
vertical dash-dotted lines on the phase diagram for Exercise E�D.�(a), Fig. �.��.
�e cooling curves are shown in Fig �.��. �e break points, where solid phases
start to form are shown by the short horizontal lines, and the dotted lines indicate the temperatures of the two eutectics (�� K and �� K). �e horizontal
segments correspond to solidi�cation of a eutectic.
94
92
T�K
90
(i)
(ii)
(iii)
(iv)
(v)
88
86
84
82
Figure 5.20
time
Solutions to problems
P�D.�
�e schematic phase diagram is shown in Fig �.��. �e solid points are the
data given in the Exercise, and in the absence of any further information and
169
5 SIMPLE MIXTURES
because there are so few points, these have just been joined by straight lines.
solid
liquid
1,000
θ�○ C
170
900
800
0.0
0.2
0.4
x ZrF4
0.6
0.8
1.0
Figure 5.21
P�D.�
On cooling a liquid with composition x ZrF4 = 0.4 solid �rst starts to appear
when the isopleth intersects the liquid line (at about ��� ○ C). �e composition
of the small amount of solid that forms is given by the le�-hand open circle in
the diagram (about x = 0.29; containing less ZrF� than the liquid). As the
temperature drops further more solid is formed and its composition moves
along the solid line to the right becoming richer in ZrF� until it reaches the
point where the isopleth crosses the solid line. At this point what remains of
the liquid has composition given by the right-hand open circle (about x = 0.48).
A further drop in temperature results in complete solidi�cation.
�e phase diagram is shown in Fig. �.��, along with the relevant cooling curves.
�e fact that there is a phase boundary indicated by the vertical line at x B = 0.67
is taken to indicate the formation of compound AB� which has x B = 23 . By
analogy with the phase diagram shown in Fig. �D.� on page ���, the form of
the given phase diagram indicates that AB� does not exist in the liquid phase.
�e number of distinct chemical species (as opposed to components) and phases
present at the indicated points are, respectively
b(3, 2), d(2, 2), e(4, 3), f (4, 3), g(4, 3), k(2, 2)
Liquid A and solid A are considered to be distinct species.
P�D.�
(a) Note that, as indicated on the diagram, Ca� Si, CaSi, CaSi� appear at mole
fractions of Si 13 , 12 and 23 , as expected.
eutectics: x Si = 0.056 at approximately ��� ○ C, x Si = 0.402 at ���� ○ C,
x Si = 0.694 at ���� ○ C
congruent melting compounds: Ca� Si Tf = 1314 ○ C, CaSi Tf = 1324 ○ C
incongruent melting compound: CaSi� Tf = 1040 ○ C (melts into CaSi(s)
and Si-rich liquid with x Si around �.��)
(b) For an isopleth at x Si = 0.2 and at ���� ○ C the phases in equilibrium are
CaSi� and a Ca-rich liquid (x Si = 0.11). �e lever rule, [�C.�–���], gives
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Temperature, T
b
A(l)+B(l)+B(s)
f
A(l)+B(l)
+A(s)
g
A(l)+B(l)+AB2(s)
c
k
e
0.2
0.67
0.57
0.23
0.16
0
AB2(s)+B(s)
AB2(s)+A(s)
0.4
0.6
Mole fraction of B, xB
xB = 0.16
xB = 0.23
xB = 0.57
0.84
d
0.8
1
xB = 0.84
xB = 0.67
Figure 5.22
the relative amounts:
l liq
n Ca2 Si
0.2 − 0.11
=
=
= �.�
n liq
l Ca2 Si 0.333 − 0.2
(c) For an isopleth at x Si = 0.8 Si(s) begins to appear at about ���� ○ C. Further cooling causes more Si(s) to freeze out of the melt so that the melt
becomes more concentrated in Ca. �ere is a eutectic at x Si = 0.694 and
���� ○ C.
At a temperature just above the eutectic point the liquid has composition
x Si = 0.694 and the lever rule gives that the relative amounts of the Si(s)
and liquid phases as:
l liq 0.80 − 0.694
n Si
=
=
= �.��
n liq
l Si
1.0 − 0.80
At the eutectic temperature a third phase appears, CaSi� (s). As the melt
cools at this temperature, both Si(s) and CaSi� (s) freeze out of the melt
while the composition of the melt remains constant. At a temperature
slightly below the eutectic point all the melt will have frozen to Si(s) and
CaSi� (s) with the relative amounts:
P�D.�
n Si
l CaSi2 0.80 − 0.667
=
=
= �.��
n CaSi2
l Si
1.0 − 0.80
�e data are plotted as the phase diagram shown in Fig �.��; the �lled and open
circles are the data points and the solid/dashed line is a best-�t cubic function.
A mixture of �.��� mol of N,N-dimethylacetamide with �.��� mol of heptane
has mole fraction of the former of x 1 = 0.750�(0.750 + 0.250) = 0.750. �e
tie line at ���.� ○ C is shown on the diagram, and this intersects with the two
curves at x 1 = 0.167 and x 2 = 0.805 (determined from the best-�t polynomial -
171
5 SIMPLE MIXTURES
x1
x2
310
305
T�K
172
300
295
0.0
0.2
Figure 5.23
0.4
0.6
x 1 or x 2
0.8
1.0
an alternative would be to use the data points given for this temperature). �e
lever rule, [�C.�–���] gives the proportion of the two phases as
n x=0.805 0.750 − 0.167
=
= ��.�
n x=0.167 0.805 − 0.750
�e N,N-dimethylacetamide-rich phase is therefore more than ten times more
abundant than the other phase.
A mixture of this composition will become a single phase at the temperature at
which the x 1 = 0.750 isopleth intersects the right-hand phase boundary. Using
the �tted function, this intersection is at ���.� ○ C .
5E
Phase diagrams of ternary systems
Answers to discussion questions
D�E.�
D�E.�
�e composition represented by point c is approximately x Ni = 0.73, x Fe = 0.20,
x Cr = 0.07. �is is a three-phase region, with Fe, Ni and γFeNi present.
�e lever rule, [�C.�–���], applies in a ternary system, but with an important
caveat. For binary systems the tie lines to which the rule appplies are always
horizontal and so can be added to the phase diagram at will. In contrast, for
a ternary system the tie lines have no such simple orientation and have to be
determined experimentally. �us the lever rule applies, but in order to use
it additional information is needed about the tie lines at the composition of
interest.
Solutions to exercises
E�E.�(a)
�e points corresponding to the given compositions are marked with letters on
the phase diagram shown in Fig. �.��. Composition (i) is in a two-phase region,
(ii) is in a three-phase region, (iii) is in a region where there is only one phase.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Composition (iv) corresponds to the point at which the phase boundaries meet
and all of the phases are in equilibrium
1 H2O
0
S1
0.2
P = 1 0.8
S2
iii
P=2
P=2
i
0.4
0.6
0.8
NH4Cl 1
Figure 5.24
E�E.�(a)
iv
P=3
0.2
0
0.6
0.4
0.4
0.2
ii
0.6
0.8
1
0 (NH4)2SO4
(i) �e phase equilibrium between NH� Cl and H� O is indicated by the le�hand edge of the phase diagram shown in Fig. �.��. At the point S 1 the
system moves from two phases to one – in other words from a system of
solid NH� Cl in equilibrium with a solution of NH� Cl in H� O, to a system
in which there is just one phase, a solution of NH� Cl in H� O. �is point
therefore marks the solubility of NH� Cl, and from the diagram is occurs
at x NH4 Cl = 0.19.
�e task is to convert this mole fraction to a molar concentration. Imagine
a solution made from a mass m of H� O: the amount in moles of H� O is
m�M, where M is the molar mass of H� O. �e mole fraction of NH� Cl
is therefore
n NH4 Cl
n NH4 Cl
x NH4 Cl =
=
n NH4 Cl + n H2 O n NH4 Cl + m�M
�is rearranges to n NH4 Cl = x NH4 Cl (m�M)�(1 − x NH4 Cl ). If it is assumed
that the mass density of the solution is approximately the same as the mass
density of water, ρ, the volume of this solution is given by V = m�ρ. With
this, the molar concentration of NH� Cl is computed as
n NH4 Cl 1 x NH4 Cl (m�M) ρ x NH4 Cl (m�M)
=
=
V
V 1 − x NH4 Cl
m 1 − x NH4 Cl
x NH4 Cl ρ
=
M(1 − x NH4 Cl )
[NH4 Cl] =
With the data from the diagram and assuming ρ = 1000 g dm−3
[NH4 Cl] =
x NH4 Cl ρ
0.19 × (1000 g dm−3 )
=
M(1 − x NH4 Cl ) (18.0158 g mol−1 ) × (1 − 0.19)
= �� mol dm−3
�is high concentration rather casts doubt on assuming the density of the
solution is the same as that of water.
173
174
5 SIMPLE MIXTURES
(ii) �e solubility of (NH� )� SO� is indicated by point S 2 on the right-hand
edge, at x(NH4 )2 SO4 = 0.3. An analogous calculation to that in part (a)
E�E.�(a)
gives the concentration as [(NH4 )2 SO4 ] = �� mol dm−3 .
�e ternary phase diagram is shown in Fig �.��.
A
0.0
1.0
0.2
xC
0.8
0.4
0.6
0.6
0.8
C
1.0
0.0
(i)
(ii)
0.2
0.4
Figure 5.25
E�E.�(a)
0.4
(iii)
0.2
xA
0.6
xB
0.8
1.0
0.0
B
�e composition by mass needs to be converted to mole fractions, which requires the molar masses: M NaCl = 58.44 g mol−1 , M H2 O = 18.016 g mol−1 , and
M Na2 SO4 ⋅ 10 H2 O = 322.20 g mol−1 . Imagine that the solution contains �� g NaCl,
�� g Na� SO� ⋅ �� H� O and hence (100 − 25 − 25) = 50 g H2 O. �e mole fraction
of NaCl is
x NaCl =
m NaCl �M NaCl
m NaCl �M NaCl + m Na2 SO4 ⋅ 10 H2 O �M Na2 SO4 ⋅ 10 H2 O + m H2 O �M H2 O
(25 g)�(58.44 g mol−1 )
(25 g)�(58.44 g mol ) + (25 g)�(322.20 g mol−1 ) + (50 g)�(18.016 g mol−1 )
= 0.13
=
−1
Likewise, x Na2 SO4 ⋅ 10 H2 O = 0.024 and x H2 O = 0.85; this point is plotted in the
ternary phase diagram shown in Fig �.��.
�e line with varying amounts of water but the same relative amounts of the
two salts (in this case, equal by mass), passes through this point and the vertex
corresponding to x H2 O = 1. �is line intersects the NaCl axis at a mole fraction
corresponding to a ��:�� mixture (by mass) of the two salts
x NaCl =
(50 g)�(58.44 g mol−1 )
= 0.85
(50 g)�(58.44 g mol−1 ) + (50 g)�(322.20 g mol−1 )
�e line is shown on the diagram.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
NaCl
0.0
0.2
x H2 O
1.0
0.8
0.4
0.6
0.6
0.4
0.8
H� O
Figure 5.26
E�E.�(a)
0.2
1.0
0.0
x NaCl
0.2
0.4
0.6
x Na2 SO4 ⋅ 10 H2 O
0.8
1.0
0.0 Na� SO�
�e composition by mass needs to be converted to mole fractions, which requires the molar masses: M CHCl3 = 119.37 g mol−1 , M H2 O = 18.016 g mol−1 ,
and M CH3 COOH = 60.052 g mol−1 . �e mole fraction of CHCl� is
x CHCl3 =
=
m CHCl3 �M CHCl3
m CHCl3 �M CHCl3 + m CH3 COOH �M CH3 COOH + m H2 O �M H2 O
9.2 g
119.37 g mol−1
9.2 g
3.1 g
2.3 g
+ 60.052 g mol−1 + 18.016 g mol−1
119.37 g mol−1
= �.��
Likewise, x CH3 COOH = 0.20 and x H2 O = 0.50 . �is point in marked with the
open circle on the phase diagram shown in Fig. �.��; it falls clearly in the twophase region. �e point almost lies on the tie line a ′2 –a ′′2 , and using this as a
guide the lever rule indicates that the phase with composition a′2 , (x W = 0.57,
x T = 0.20, x E = 0.23), is approximately � times more abundant than the phase
with composition a′′2 , (x W = 0.06, x T = 0.82, x E = 0.12).
(i) When water is added to the mixture the composition moves along the
dashed line to the lower-le� corner. �e system will pass from the twophase to the one-phase region when the line crosses the phase boundary,
which is at approximately (x W = 0.75, x T = 0.14, x E = 0.10).
(ii) When ethanoic acid is added to the mixture the composition moves along
the dashed line to the vertex. �e system will pass from the two-phase to
the one-phase region when the line crosses the phase boundary, which is
at approximately (x W = 0.44, x T = 0.26, x E = 0.30).
Solutions to problems
P�E.�
�e given points are shown by �lled dots in the phase diagram shown in Fig. �.��,
and they are connected by a straight line, as indicated in the problem. Beneath
175
176
5 SIMPLE MIXTURES
CH� COOH
0.0
1.0
0.2
xW
0.8
0.4
0.6
0.4
a′2
0.8
H� O
xE
0.6
0.2
a ′′2
1.0
0.0
0.2
0.4
Figure 5.27
0.6
0.8
xT
1.0
0.0 CHCl�
this line lies a one-phase region because CO� and nitrobenzene are miscible
in all proportions. Addition of I� eventually causes phase separation into a
two-phase region for all compositions about the line.
0
1 I2
0.2
0.8
0.4
0.6
P=2
0.6
0.4
0.8
0.2
P=1
CO2 1
0 nitrobenzene
0
0.2
0.4
0.6
0.8
1
Figure 5.28
P�E.�
Consider the construction shown in Fig. �.��. �e line is interest is AW, where
as indicated the position of W is determined by the mole fractions of B and
C in the binary mixture; to avoid confusing these particular mole fractions
with others, they are denoted y B and y C . �e point P lies on this line and its
perpendicular distance from each of the edges of the triangle gives the mole
fractions of B and C, x B and x C .
Construct the line UV passing through P and parallel to the base of the triangle.
It follows that AWB and APV are similar triangles, therefore there exists the
following relationship between the ratios of the sides.
WB PV
=
AW AP
(�.�)
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
A
N
M
U
C
yB
xC
xB
V
P
B
W
yC
Figure 5.29
�e indicated angle is 60○ , so it follows that x C = PV sin 60○ .
Now consider the triangles AWC and APU, which are also similar and therefore
and as before x B = PU sin 60○ .
WC PU
=
,
AW AP
(�.�)
Dividing eqn �.� by eqn �.� gives
WB PV
=
WC PU
hence
y C PV
=
y B PU
Dividing x C = PV sin 60○ by x B = PU sin 60○ gives x C �x B = PV�PU. Equating
the two expressions for PV�PU gives the required result
yC xC
=
yB xB
Alternatively (and avoiding the use of the angle explicitly) PNV and PMU are
also recognised as similar triangles, so that it follows directly that x C �PV =
x B �PU.
5F
Activities
Answers to discussion questions
D�F.�
�e Debye–Hückel theory of electrolyte solutions formulates deviations from
ideal behaviour (essentially, deviations due to electrostatic interactions between
the ions) in terms of the work of charging the ions. �e assumption is that
the solute particles would behave ideally if they were not charged, and the
di�erence in chemical potential between real and ideal behaviour amounts to
the work of putting electrical charges onto the ions.
To �nd the work of charging, the distribution of ions must be found, and that is
done using the shielded Coulomb potential which takes into account the ionic
177
178
5 SIMPLE MIXTURES
strength of the solution and the dielectric constant of the solvent. �e Debye–
Hückel limiting law, [�F.��–���], relates the mean ionic activity coe�cient to
the charges of the ions involved, the ionic strength of the solution, and depends
on a constant that takes into account solvent properties and temperature.
D�F.�
If a solvent or solute has a certain chemical potential, then this is related to
the activity of the solvent or solute through [�F.�–���] and [�F.�–���]. At �rst
sight it seems odd to think of the chemical potential determining the activity,
but this approach is in fact logical as the chemical potential is the experimentally measurable quantity. For example, measurements of cell potentials or the
vapour pressure of liquids provide (slightly indirect, it must be admitted) ways
of measuring the chemical potential.
For ideal systems expressions are available which relate the chemical potential
to the concentration (in the form of its various measures, such as mole fraction
or molality). It therefore follows that the di�erence between the measured
chemical potential and that predicted from idealised models is attributable to
factors not taken into account in the ideal systems. Such factors are collectively
described as non-ideal interactions; the di�erence between activity and concentration is therefore ascribed to the presence of such interactions
Non-ideal interactions are no di�erent from the usual interactions between
molecular species. For example, they may include interactions between charged
or polar species, hydrogen bonding or more speci�c interactions.
D�F.�
�e main way of measuring activities described in this Topic is from measurements of partial vapour pressures, as given in [�F.�–���] for the solvent and in
[�F.��–���] for the solute activity. Other measurements from which the value
of the chemical potential can be inferred (for example, cell potentials) are used
to determine activities via the general relationship µ J = µ −J○ + RT ln a J .
Solutions to exercises
E�F.�(a)
Ionic strength is de�ned in [�F.��–���]
I = 12 � z 2i (b i �b −○ )
i
where the sum runs over all the ions in the solution, z i is the charge number
on ion i, and b i is its molality. For KCl the molality of K+ and Cl – are both
�.�� mol kg−1 ; z K+ = +1 and z Cl− = −1. For CuSO� the molality of Cu�+ and
SO� � – are both �.�� mol kg−1 ; z Cu2+ = +2 and z SO4 2− = −2. �e ionic strength
is therefore
I = 12 [1�(1 mol kg−1 )] �(+1)2 × (0.10 mol kg−1 ) + (−1)2 × (0.10 mol kg−1 )
E�F.�(a)
+(+2)2 × (0.20 mol kg−1 ) + (−2)2 × (0.20 mol kg−1 )� = �.�
Ionic strength is de�ned in [�F.��–���]
I = 12 � z 2i (b i �b −○ )
i
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
where the sum runs over all the ions in the solution, z i is the charge number
on ion i, and b i is its molality.
(i) �e aim here is to increase the ionic strength by 0.250−0.150 = 0.100; the
task is therefore to compute the mass m of Ca(NO� )� which, when added
to a mass m w of water, gives this increase in the ionic strength.
A solution of Ca(NO� )� of molality b contributes Ca�+ at molality b and
NO� – at molality 2b. �e contribution to the ionic strength is therefore
1
[(+2)2 × b + (−1)2 × 2b]�b −○ = 3b�b−○ .
2
�e molality arising from dissolving mass m of Ca(NO� )� in a mass m w
of solvent is (m�M)�m w , where M is the molar mass. It therefore follows
that to achieve the desired increase in ionic strength
3×
m
1
×
= 0.100
Mm w b−○
hence
m = 13 × 0.100 × Mm w b −○
�e molar mass of Ca(NO� )� is ���.�� g mol−1 ; using this, and recalling
that the molality is expressed in mol kg−1 , gives
m = 13 × 0.100 × (164.10 × 10−3 kg mol−1 ) × (0.500 kg) × (1 mol kg−1 )
= 2.73... × 10−3 kg
Hence the �.�� g of Ca(NO� )� needs to be added to achieve the desired
ionic strength.
(ii) �e argument is as in (a) except that the added solute is now NaCl which
contributes singly-charged ions at the same molality as the solute so the
contribution to the ionic strength is simply b�b−○ . It therefore follows that
to achieve the desired increase in ionic strength
m
1
×
= 0.100
Mm w b −○
hence
m = 0.100 × Mm w b −○
�e molar mass of NaCl is ��.�� g mol−1 , hence
m = 0.100 × (58.44 × 10−3 kg mol−1 ) × (0.500 kg) × (1 mol kg−1 )
= 2.92... × 10−3 kg
Hence the �.�� g of NaCl needs to be added to achieve the desired ionic
strength.
E�F.�(a)
�e Debye–Hückel limiting law, [�F.��–���], is used to estimate the mean activity coe�cient, γ± , at �� ○ C in water
log γ± = −0.509 �z+ z− � I 1�2
I = 12 � z 2i (b i �b −○ )
i
where z± are the charge numbers on the ions from the salt of interest and I is
the ionic strength, de�ned in [�F.��–���]. In the de�nition of I the sum runs
over all the ions in the solution, z i is the charge number on ion i, and b i is its
molality.
179
180
5 SIMPLE MIXTURES
A solution of CaCl� of molality b contributes Ca�+ at molality b and Cl – at
molality 2b. �e contribution to the ionic strength is therefore 12 [(+2)2 × b +
(−1)2 × 2b]�b −○ = 3b�b −○ . A solution of NaF of molality b ′ contributes Na+
at molality b′ and F – at molality b ′ . �e contribution to the ionic strength is
therefore 12 [(+1)2 × b ′ + (−1)2 × b ′ ]�b −○ = b ′ �b −○ . �e ionic strength of the
solution is therefore
(3b+b ′ )�b −○ = [3×(0.010 mol kg−1 )+1×(0.030 mol kg−1 )]�(1 mol kg−1 ) = 0.060
For solute CaCl� z+ = +2 and z− = −1 so the limiting law evaluates as
log γ± = −0.509 �z+ z− � I 1�2 = −0.509 �(+2) × (−1)� (0.060)1�2 = −0.249...
E�F.�(a)
�e mean activity coe�cient is therefore γ± = 10−0.249 ... = 0.563... = 0.56 .
�e Davies equation is given in [�F.��b–���]
log γ± =
−A �z+ z− � I 1�2
+ CI
1 + BI 1�2
Because the electrolyte is �:� with univalent ions, the ionic strength is simply
I = b HBr �b −○ . �ere is no obvious straight-line plot using which the data can
be tested against the Davies equation, therefore a non-linear �t is made using mathematical so�ware and assuming that A = 0.509; remember that the
molalities must be expressed in mol kg−1 . �e best-�t values are B = 1.96 and
C = 0.0183; With these values the predicted activity coe�cients are �.���, �.���
and �.���, which is very good agreement.
E�F.�(a)
�e activity in terms of the vapour pressure p is given by [�F.�–���], a = p�p∗ ,
where p∗ is the vapour pressure of the pure solvent. With the data given a =
p�p∗ = (1.381 kPa)�(2.3393 kPa) = �.���� .
E�F.�(a)
For the solute, Henry’s law is used as the basis and the activity is given by [�F.��–
���], a B = p B �K B , where K B is the Henry’s law constant expressed in terms of
mole fraction. In this case a B = p B �K B = (25 Torr�200 Torr) = �.��� .
E�F.�(a)
On the basis of Raoult’s law, the activity in terms of the vapour pressure p A
is given by [�F.�–���], a A = p A �p∗A , where p∗A is the vapour pressure of the
pure solvent. With the data given a A = p A �p∗A = (250 Torr)�(300 Torr) =
0.833... = �.��� . �e activity coe�cient is de�ned through [�F.�–���], a A =
γ A x A , therefore γ A = a A �x A = 0.833...�0.900 = �.��� .
On the basis of Raoult’s law, the activity in terms of the vapour pressure p J is
given by [�F.�–���], a J = p J �p∗J , where p∗J is the vapour pressure of the pure
solvent. �e partial vapour pressure of component J in the gas is given by p J =
y J p tot . In this case
aP =
p P y P p tot 0.516 × (1.00 atm) × [(101.325 kPa)�(1 atm)]
=
=
= 0.497...
p∗P
p∗P
105 kPa
�e activity of propanone is therefore a P = �.��� . �e activity coe�cient is
de�ned through [�F.�–���], a J = γ J x J , therefore γ P = a P �x P = 0.497...�0.400 =
�.�� .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
For the other component the mole fractions are y M = 1 − y P = 0.484 and
x M = 1 − x P = 0.600. �e rest of the calculation follows as before
aM =
p M y M p tot 0.484 × (1.00 atm) × [(101.325 kPa)�(1 atm)]
=
=
= 0.667...
p∗M
p∗M
73.5 kPa
�e activity of methanol is therefore a M = �.��� and its activity coe�cient is
given by γ M = a M �x M = 0.667...�0.600 = �.�� .
E�F.�(a)
For this model of non-ideal solutions the vapour pressures are given by [�F.��–
���], p A = p∗A x A exp(ξ[1 − x A ]2 ) and likewise for p B ; the total pressure is given
by p tot = p A + p B . �e vapour pressures are plotted in Fig. �.��.
20
pA
pB
p tot
p�kPa
15
10
5
0
0.0
Figure 5.30
0.2
0.4
xA
0.6
0.8
1.0
Solutions to problems
P�F.�
In the Raoult’s law basis the activity is given by [�F.�–���], a J = p J �p∗J , and the
activity coe�cient by [�F.�–���], a J = γ J x J . �e partial pressure of the vapour
is given by p J = y J p, where y J is the mole fraction of J in the vapour and p the
total pressure.
It follows that γ J = a J �x J = y J p�x J p∗J . In this binary system the activity coe�cient for �,�-epoxybutane (E) is found using the given data for trichloromethane
(T) by using and x E = 1 − x T , and likewise for y E . It follows that γ E = (1 −
y T )p�(1 − x T )p∗E . �e resulting activity coe�cients are shown in the table.
181
182
5 SIMPLE MIXTURES
P�F.�
p�kPa
xT
yT
γT
γE
23.40
0
0
21.75
0.129
0.065
0.417
0.998
20.25
0.228
0.145
0.490
0.958
18.75
0.353
0.285
0.576
0.885
18.15
0.511
0.535
0.723
0.738
20.25
0.700
0.805
0.885
0.563
22.50
0.810
0.915
0.966
0.430
26.30
1
1
1
1
�e Debye–Hückel limiting law, [�F.��–���], is
log γ± = −0.509 �z+ z− � I 1�2
I = 12 � z 2i (b i �b −○ )
b�(mmol kg−1 )
log γ±
i
where z± are the charge numbers on the ions from the salt of interest and I is
the ionic strength, de�ned in [�F.��–���]. For a �:� electrolyte of univalent ions
at molality b, I = b�b −○ (recall that b −○ = 1 mol kg−1 so b must also be in units
of mol kg−1 ). A test of this equation is to plot log γ± against I 1�2 , such a plot is
shown in Fig. �.��.
γ±
1.00
0.964 9
2.00
0.951 9
5.00
0.927 5
10.0
0.902 4
20.0
0.871 2
−0.015 52
−0.021 41
−0.032 69
−0.044 60
−0.059 88
I 1�2
0.031 6
0.044 7
0.070 7
0.100 0
0.141 4
�e data �t to quite a good straight line, the equation of which is
log γ± = −0.404 × I 1�2 − 3.395 × 10−3
�e Debye–Hückel limiting law is veri�ed to the extent that log γ± is linear
in I 1�2 , however the slope is −0.404 rather than the expected value of −0.509.
According to the limiting law the intercept at I 1�2 = 0 should be zero, which is
not the case. Overall, the conclusion is that the limiting law predicts the correct
functional dependence of γ± on ionic strength but fails to predict the correct
values.
�e Davies equation is given in [�F.��b–���]
log γ± =
−A �z+ z− � I 1�2
+ CI
1 + BI 1�2
�e data can be �tted to this equation using mathematical so�ware to implement a non-linear �t; the value of A is �xed as 0.509 for the �t. �e best-�t
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
0.00
log γ±
−0.02
−0.04
−0.06
0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14
I 1�2
Figure 5.31
parameters are B = 1.2975, C = −0.0470. Using these, a table is drawn up
comparing the experimental values of γ± with those predicted by the limiting
law and by the Davies equation.
b�(mmol kg−1 )
γ±
γ± (DH) error (%) γ± (Davies) error (%)
1.00
0.964 9
0.963 6
2.00
0.951 9
5.00
0.927 5
0.920 5
10.00
0.902 4
0.889 4
20.00
0.871 2
0.847 3
0.948 9
−0.13
−0.31
−0.76
−1.44
−2.75
0.965 1
0.02
0.951 9
0.00
0.927 4
0.902 4
0.871 2
−0.01
0.00
0.00
�e table shows that the values predicted by the limiting law are increasingly in
error as the ionic strength increases, whereas the Davies equation reproduces
most of the experimental data to within the stated precision. However, it must
be kept in mind that the B and C parameters in the Davies equation have been
adjusted speci�cally to �t these data.
Answers to integrated activities
I�.�
(a) To develop the expression for K into the form requested it is useful to
rewrite [MA] and [M]free in terms of the total concentration of macromolecule, [M]. �e total amount of A in the dialysis bag is [A]in = [A]free +
[A]bound , but the amount of A bound is equal to the amount of the macromolecule ligand complex, MA: [A]bound = [MA], therefore
[A]in = [A]free + [MA] hence [MA] = [A]in − [A]free
Recall that [A]free = [A]out and that, by de�nition ν = ([A]in −[A]out )�[M],
it therefore follows that
[MA] = [A]in − [A]out = ν[M]
183
184
5 SIMPLE MIXTURES
Now consider the macromolecule, the total concentration of which is
[M]. It follows that [M] = [MA] + [M]free . �e expression just derived
for [MA], [MA] = ν[M] is substituted in to give [M] = ν[M] + [M]free ,
from which it follows that [M]free = [M](1 − ν)
With these expressions for [M]free and [MA], the expression for K is
developed into the requested form
K=
[MA]c −○
ν[M]c −○
νc −○
=
=
[M]free [A]free [M](1 − ν)[A]out (1 − ν)[A]out
where [A]free = [A]out is also used.
(b) �e equilibrium constant K ′ describes the equilibrium between a macomolecule with a single binding site, S, and the bound complex, SA
K′ =
[SA]c −○
[S]free [A]free
In part (a) ν is de�ned as the average number of bound ligands per macromolecule, and is therefore given by ν = [A]bound �[M]. Whereas M has N
binding sites, S only has one site, so the average number of ligands bound
per S is ν�N. �is number is also expressed (by analogy with the earlier
discussion) as [A]bound �[S], so it follows that ν�N = [A]bound �[S]. �e
�nal step is to realise that the concentration of bound ligand is equal to
the concentration of the S–A complex, so [A]bound = [SA]. It therefore
follows that ν�N = [SA]�[S]. �is is rearranged to [SA] = [S]ν�N, which
is one of the terms needed in the expression for K ′ .
�e other term is [S]free which is related to [S] as follows. �e total concentration of S is given by [S] = [S]free + [SA], hence [S]free = [S] − [SA].
Substituting [SA] = [S]ν�N gives [S]free = [S] − [S]ν�N = (1 − ν�N)[S].
�e expression for the equilibrium constant is now developed as
K′ =
[SA]c −○
([S]ν�N)c −○
νc −○
=
=
[S]free [A]free (1 − ν�N)[S][A]free [A]free (N − ν)
where to go to the �nal expression the numerator and demoninator are
multiplied by N and [S] is cancelled. �e Scatchard equation follows by
taking the factor (N − ν) to the le�
K′ N − K′ν =
νc −○
[A]free
(c) �e straight line plot is ν�[A]free against ν. �e task is therefore to determine ν from the data. Note that the data given are the total concentrations
of EB in and outside the bag. It therefore follows that
[EB]total,in = [EB]bound + [EB]free = [EB]bound + [EB]out
It follows that [EB]bound = [EB]total,in − [EB]out . Recall that ν is de�ned
as ν = [EB]bound �[M], therefore
ν=
[EB]total,in − [EB]out
[M]
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e given data and the derived values of ν are shown in the following table
(1 µM = 1 µmol dm−3 ), and plotted in Fig. �.��.
[EB]out �µM
[EB]in �µM
0.042
ν
(ν�[EB]out )�(µM−1 )
0.292
0.250
0.590
0.50
0.204
1.204
1.00
4.90
0.526
2.531
2.01
3.81
1.150
4.150
3.00
2.61
0.092
5.95
5.41
(ν�[EB]out )�(µM−1 )
6.0
4.0
2.0
0.0
Figure 5.32
0.5
1.0
1.5
ν
2.0
2.5
3.0
3.5
�e data �t to quite a good straight line, the equation of which is
(ν�[EB]out )�(µM−1 ) = −1.17 × ν + 6.12
�e slope is −K ′ �c −○ where c −○ is the standard concentration, 1 mol dm−3 ,
but because [EB]out is used in µM, c −○ = 106 µM. It follows that the
(dimensionless) equilibrium constant is K ′ = 1.17 × 106 . �e intercept
gives K ′ N�c −○ , hence N = 5.23 : this is the average number of binding
sites per DNA molecule. �e graph is a good straight line, indicating that
the data �t the model quite well.
I�.�
�e dissolution of the protein according to the given equilibrium is described
by a solubility constant K s
⇀ Pv+ (aq) + v X− (aq)
PXv (s) ���
K s = a P a Xv
where the solubility constant is written in terms of the activities. Introducing
the activity coe�cients and molalities, b, gives
K s = γ±v+1 b P b Xv
At low to moderate ionic strengths the Debye–Hückel limiting law, [�F.��–���],
log γ± = −A�z− z+ �I 1�2 , is a reasonable approximation for the activity coe�cients.
185
186
5 SIMPLE MIXTURES
Addition of a salt, such as (NH� )� SO� , causes I to increase, log γ± to become
more negative, and hence γ± to decrease. However, K s is an equilibrium constant and remains unchanged. �erefore, the molality of Pv+ increases and the
protein solubility increases proportionately.
�is e�ect is also explicable in terms of Le Chatelier’s principle. As the ionic
strength increases by the addition of an inert electrolyte such as (NH� )� SO� ,
the ions of the protein that are in solution attract one another less strongly, so
that the equilibrium is shi�ed in the direction of increased solubility.
�e explanation of the salting out e�ect is somewhat more complicated and
can be related to the failure the Debye–Hückel limiting law at higher ionic
strengths. At high ionic strengths the Davies equation, [�F.��b–���], is a better
approximation
−A �z+ z− � I 1�2
+ CI
log γ± =
1 + BI 1�2
At low concentrations of inert salt, I 1�2 > I, the �rst term dominates, γ± decreases with increasing I, and salting in occurs; however, at high concentrations, I > I 1�2 , the second term dominates, γ± increases with increasing I, and
salting out occurs. �e Le Chatelier’s principle explanation is that the water
molecules are tied up by ion-dipole interactions and become unavailable for
solvating the protein, thereby leading to decreased solubility.
I�.�
In Section �B.�(e) on page ��� the derivation of the expression for the osmostic
pressure starts by equating the chemical potential of A as a pure liquid subject
to pressure p with that of A in a solution of mole fraction x A containing solute B
and subject to pressure p+Π: µ ∗A (p) = µ A (x A , p+Π). �e chemical potential of
A in the solution is then expressed as µ A (x A , p + Π) = µ∗A (p + Π) + RT ln x A ,
which assumes ideality. If the solution is not ideal, then the mole fraction is
replaced by the activity a A to give
µ ∗A (p) = µ ∗A (p + Π) + RT ln a A
�e derivation then proceeds as before yielding the intermediate result −RT ln a A =
Vm Π for the non-ideal solution.
�e osmotic coe�cient � is de�ned as � = −(x A �x B ) ln a A , hence ln a A =
−�x B �x A . �is expression for ln a A is substituted into −RT ln a A = Vm Π to
give RT�x B �x A = Vm Π. �e �nal steps assume that the solution is dilute so
that x A ≈ 1 and x B = n B �(n A + n B ) ≈ n B �n A
xB
= Vm Π
xA
nB
hence RT�
= Vm Π
nA
nB
RT�
=Π
n A Vm
RT�[B] = Π
RT�
To go to the last line V = n A Vm and [B] = n B �V are used.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
On the basis of Raoult’s law, the activity in terms of the vapour pressure p J
is given by [�F.�–���], a J = p J �p∗J , where p∗J is the vapour pressure of the
pure solvent. �e activity coe�cient is de�ned through [�F.�–���], a J = γ J x J ,
therefore γ J = p J �p∗J x J .
�e data as given do not include values for the vapour pressure over the pure
liquids, so the �rst task is to plot p J against x J and extrapolate to x J = 1 to
�nd p∗J . �e vapour pressures are plotted in this way Fig. �.��, and the linear extrapolations to �nd the vapour pressures of the pure substances are also
shown. �ese give the values p∗E = 7.45 kPa and p∗B = 35.41 kPa; using these
values the activity coe�cients are computed as shown in the table.
30
pE
pB
p J �kPa
I�.�
20
10
0
0.0
0.2
0.4
xJ
0.6
0.8
1.0
Figure 5.33
xE
0.016 0
p E �kPa
xB
p B �kPa
γE
γB
35.05
4.06
1.01
34.29
2.96
1.01
0.484
0.984 0
0.967
0.956 1
0.083 5
1.535
0.916 5
33.28
2.47
0.113 8
1.89
0.886 2
32.64
0.171 4
2.45
0.828 6
0.297 3
3.31
0.369 6
γ E (Henry) G E �kJ mol−1
1.70
0.08
1.24
0.16
1.03
1.03
0.26
2.23
1.04
0.93
0.34
30.90
1.92
1.05
0.80
0.42
0.702 7
28.16
1.49
1.13
0.63
0.55
3.83
0.630 4
26.08
1.39
1.17
0.58
0.59
0.583 4
4.84
0.416 6
20.42
1.11
1.38
0.47
0.53
0.660 4
5.36
0.339 6
18.01
1.09
1.50
0.46
0.52
0.843 7
6.76
0.156 3
10.00
1.07
1.81
0.45
0.993 1
7.29
0.006 9
0.47
0.98
1.92
0.41
0.043 9
0.41
−0.03
On the basis of Henry’s law, the activity in terms of the vapour pressure p J is
given by [�F.��–���], a J = p J �K J , where K J is the Henry’s law constant for J as
187
188
5 SIMPLE MIXTURES
a solute. �e activity coe�cient is de�ned as before, a J = γ J x J , and therefore
γ J = p J �K J x J .
To �nd the Henry’s law constant for E, the limiting slope of a plot of p E against
x E is taken. �e three data points given at the lowest values of x E do not extrapolate back to the origin, which is not in accord with Henry’s law. Arguably
there are several equally valid ways of proceeding here, but one is to force the
best-�t line to pass through the origin and then use the �rst three data points;
this leads to the slope is shown by the dashed line in Fig. �.��. �e limiting
slope, taken in this way is 17.77 and so K E = 17.77 kPa. �is value is used to
compute the activity coe�cients for E based on Henry’s law, and the results are
shown in column headed γ E (Henry) in the table above. �e outcome is not
satisfactory because the expecting limiting behaviour γ E → 1 as x E → 0 is not
evidenced.
�e excess Gibbs energy is de�ne in [�B.�–���] as G E = ∆ mix G − ∆ mix G ideal .
As explained in Section �F.� on page ���, the Gibbs energy of mixing is given
in terms of the activities as ∆ mix G = nRT (x A ln a A + x B ln a B ), whereas the
ideal Gibbs energy of mixing is ∆ mix G ideal = nRT (x A ln x A + x B ln x B ). �e
activities are written as a A = γ A x A and hence
G E = ∆ mix G − ∆ mix G ideal
= nRT (x A ln a A + x B ln a B ) − nRT (x A ln x A + x B ln x B )
= nRT (x A ln γ A x A + x B ln γ B x B ) − nRT (x A ln x A + x B ln x B )
= nRT (x A ln γ A + x B ln γ B )
Using the �nal expression G E �n is computed from the given data and using
the activity coe�cients (based on Raoult’s law) already derived. �e computed
values are given in the table.
I�.�
On the basis of Raoult’s law, the activity in terms of the vapour pressure p J is
given by [�F.�–���], a J = p J �p∗J , where p∗J is the vapour pressure of the pure
substance. �e activity coe�cient is de�ned through [�F.�–���], a J = γ J x J ,
therefore γ J = p J �p∗J x J . �e partial pressure in the gas phase is determined
from the mole fraction in the gas phase, y J , p J = y J p tot , so the �nal calculation
is γ J = y J p tot �p∗J x J .
�e total pressure is given in kPa, whereas the vapour pressure over pure oxygen is given in Torr. �e conversion is
(p kPa) = (p′ Torr) ×
(101.325 kPa)�(1 atm)
(760 Torr)�(1 atm)
�e temperature-composition phase diagram is shown in Fig. �.�� and the
computed values of the activity coe�cient are given in the table below. �e
fact that the activity coe�cient is close to � indicates near-ideal behaviour.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
95
T�K
90
vapour
liquid
85
80
75
0.0
0.2
0.4
0.6
x O2 or y O2
T�K
x O2
y O2
77.3
0
0
78
0.10
0.02
80
0.34
82
0.54
84
0.8
1.0
Figure 5.34
I�.��
p∗O2 �Torr
γ O2
154
171
0.88
0.11
225
1.08
0.22
294
1.04
0.70
0.35
377
0.99
86
0.82
0.52
479
0.99
88
0.92
0.73
601
0.99
90.2
1.00
1.00
760
0.99
In Section �B.�(a) on page ��� the derivation of the expression for the freezing
point depression starts by equating the chemical potential of A as a pure solid
with that of A in a solution of mole fraction x A containing solute B: µ ∗A (s) =
µ A (l, x A ). �e latter chemical potential is written, for an ideal solution, as
µ A (l, x A ) = µ∗A (l)+ RT ln x A . If the solution is not ideal, then the mole fraction
is replaced by the activity a A to give
µ∗A (s) = µ ∗A (l) + RT ln a A
�e derivation then proceeds as before. First, µ ∗A (s) − µ ∗A (l) is identi�ed as
−∆ fus G to give
µ ∗ (s) − µ∗A (l) −∆ fus G
ln a A = A
=
RT
RT
Next, both sides are di�erentiated with respect to T and the Gibbs–Helmholtz
equation, d(G�T)�dT = −H�T 2 , is applied to the right-hand side
d ln a A −1 d ∆ fus G
∆ fus H
=
�
�=
dT
R dT
T
RT 2
�e freezing point depression ∆T is de�ned as ∆T = T ∗ − T, where T is the
freezing point of the solution and T ∗ is the freezing point of the pure solvent.
189
190
5 SIMPLE MIXTURES
It follows that d∆T = −dT. Finally, because the freezing point depression is
small, T on the right-hand side of the previous equation can be replaced by T ∗
to give
d ln a A
∆ fus H
=−
(�.�)
d∆T
RT ∗ 2
�e empirical freezing-point constant K f is introduced in [�B.��–���], ∆T =
K f b B , where b B is the molality of the solvent. �is expression is developed
by writing b B = n B �m A , where m A is the mass of solvent A (in kg) and then
m A = n A M, where M is the molar mass of the solvent. It follows that ∆T =
K f n B �(n A M). For dilute solutions x B ≈ n B �n A so ∆T = K f x B �M.
�e expression for the freezing point depression given in [�B.��–���] is
∆T =
RT ∗ 2
xB
∆ fus H
Comparison of this with the expression just derived, ∆T = K f x B �M, gives
Kf
RT ∗ 2
=
M ∆ fus H
Using this expression for the right-hand side in eqn �.� gives the required form
d ln a A
M
=−
d∆T
Kf
(�.�)
Start with the Gibbs–Duhem equation, [�A.��b–���], n A dµ A + n B dµ B = 0 and
divide both sides by (n A + n B ) to give x A dµ A + x B dµ B = 0. Next introduce the
general dependence of the chemical potential on activity, µ A = µ −A○ + RT ln a A ,
from which it follows that dµ A = RTd ln a A . Introducing this into the Gibbs–
Duhem equation, and dividing both sides by RT gives
x A d ln a A + x B d ln a B = 0
It follows that
hence
d ln a A = −
xB
d ln a B
xA
(�.�)
d ln a A
x B d ln a B
=−
d∆T
x A d∆T
�is expression for (d ln a A )�d∆T is substituted into eqn �.�
−
x B d ln a B
M
=−
x A d∆T
Kf
�is expression is developed further by using the approximations x A ≈ 1 and
x B ≈ n B �n A which are appropriate for dilute solutions
d ln a B x A M
=
d∆T
xB Kf
1 M
1 nA M
1 mA
=
=
=
n B �n A K f n B K f
nB Kf
1
=
bB Kf
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
On the penultimate line m A = n A M, where m A is the mass of solvent A, is used,
and to go to the �nal line the molality of B, b B = n B �m A is introduced.
Recall the de�nition of the osmotic coe�cient � = −(x A �x B ) ln a A and the
result from eqn �.� that d ln a A = −(x B �x A )d ln a B . It follows that
xB
d ln a B
xA
� d ln a A = � −
hence
ln a A = � −
xB
d ln a B
xA
where the integration is from pure A to some arbitrary composition. �is result
is used with the de�nition of � to give
�=−
xA
xA
xB
ln a A = − � − d ln a B
xB
xB
xA
�e terms in x A and x B cannot be cancelled because those inside the integral
are functions of the variable of integration. For dilute dilute solutions x A ≈ 1
and x B ≈ n B �n A = n B �(m A �M) = M(n B �m A ) = Mb B , where M is the molar
mass of A, m A is the mass of the solvent, and b B is the molality of B. �e integral
for � therefore becomes
�=−
xA
1
ln a A =
� b B d ln a B
xB
bB
(�.�)
For a �:� univalent electrolyte the Debye–Hückel limiting law [�F.��–���], log γ± =
−A �z+ z− � I 1�2 , becomes log γ± = −A(b B �b −○ )1�2 , and changing to naural logarithms this becomes ln γ± = −A′ (b B �b −○ )1�2 , with A′ = 2.303 × A. Using this,
the activity and its derivative are developed as
ln a B = ln(b B �b−○ ) + ln γ = ln(b B �b −○ ) − A′ (b B �b−○ )1�2
hence d ln a B = �
1
1 1�2
− 12 A′ �
� � db B
bB
b B b −○
�is expression for d ln a B is used in eqn �.� and the integral is then evaluated
1
� b B d ln a B
bB
1
1
1 1�2
1 ′
=
b
�
−
A
�
� � db B
B
�
bB
bB 2
b B b −○
�
1
b B 1�2 �
�
�
=
1 − 12 A′ � −○ � � db B
� �
�
�
bB
b
�
�
�=
=
bB
1
1 1�2 3�2
�b B − 12 × 23 × A′ � −○ � b B �
bB
b
b B =0
b B 1�2
= 1 − 13 A′ � −○ �
b
191
6
6A
Chemical equilibrium
The equilibrium constant
Answers to discussion questions
D�A.�
�e terms appearing in the equilibrium constant are the activities of the species
involved in the equilibrium, and these terms arise because the chemical potential of each species depends on its activity. If a pure liquid or pure solid
is part of the equilibrium, its chemical potential contributes to the value of
∆ r G −○ . However, as the substance is in its pure form there is no composition
dependence of its chemical potential and hence no term in the equilibrium
constant. Put another way, such species have unit activity.
Solutions to exercises
E�A.�(a)
�e equilibrium constant is de�ned by [�A.��–���], K = �∏J a J J �equilibrium .
�e ‘equilibrium’ subscript indicates that the activities are those at equilibrium
rather than at an arbitrary stage in the reaction; however this subscript is not
usually written explicitly. In this case
ν
−6
4
K = a−1
P 4 (s) a H 2 (g) a PH 3 (g) =
4
a PH
3 (g)
6
a P4 (s) a H
2 (g)
�e activity of P4 (s) is 1, because it is a pure solid. Furthermore if the gases
are treated as perfect then their activities are replaced by a J = p J �p−○ . �e
equilibrium constant becomes
E�A.�(a)
K=
(p PH3 �p−○ )4 p4PH3 p−○
=
(p H2 �p−○ )6
p6H2
2
�e standard reaction Gibbs energy is given by [�A.��a–���]
∆ r G −○ =
−
○
� ν∆ f G −
Products
�
Reactants
ν∆ f G −○
�e relationship between ∆ r G −○ and K, [�A.��–���], ∆ r G −○ = −RT ln K, is then
used to calculate the equilibrium constant.
(i) For the oxidation of ethanal
∆ r G −○ = 2∆ f G −○ (CH3 COOH, l) − {2∆ f G −○ (CH3 CHO, g) + ∆ f G −○ (O2 , g)}
= 2∆ f G −○ (CH3 COOH, l) − 2∆ f G −○ (CH3 CHO, g)
= 2(−389.9 kJ mol−1 ) − 2(−128.86 kJ mol−1 ) = −5.22... × 105 J mol−1
194
6 CHEMICAL EQUILIBRIUM
�en
K = e−∆ r G �RT = exp �−
−
○
−5.22... × 105 J mol−1
� = 3.24 × 1091
(8.3145 J K−1 mol−1 ) × (298 K)
(ii) For the reaction of AgCl(s) with Br� (l)
∆ r G −○ = 2∆ f G −○ (AgBr, s) + ∆ f G −○ (Cl2 , g)
− {2∆ f G −○ (AgCl, s) + ∆ f G −○ (Br2 , l)}
= 2∆ f G −○ (AgBr, s) − 2∆ f G −○ (AgCl, s)
�en
= 2(−96.90 kJ mol−1 ) − 2(−109.79 kJ mol−1 ) = +25.7... kJ mol−1
K = e−∆ r G �RT = exp �−
−
○
E�A.�(a)
25.7... × 103 J mol−1
� = 3.03 × 10−5
(8.3145 J K−1 mol−1 ) × (298 K)
Of these two reactions, the �rst has K > 1 at 298 K.
�e relationship between ∆ r G −○ and the equilibrium constant is given by [�A.��–
���], ∆ r G −○ = −RT ln K. �e ratio of the equilibrium constants for the two
reactions is
K 1 e−∆ r G 1 �RT
∆ r G 1−○ − ∆ r G 2−○
= −∆ G −○ �RT = exp �−
�
K2 e r 2
RT
−
○
E�A.�(a)
= exp �−
(−320 × 103 J mol−1 ) − (−55 × 103 J mol−1 )
� = 1.4 × 1046
(8.3145 J K−1 mol−1 ) × (300 K)
�e reaction Gibbs energy at an arbitrary stage is given by [�A.��–���], ∆ r G =
∆ r G −○ + RT ln Q. In this case ∆ r G −○ = −32.9 kJ mol−1 . �e values of ∆ r G for at
value of Q are:
(i) At Q = 0.010
∆ r G = (−32.9 × 103 J mol−1 ) + (8.3145 J K−1 mol−1 )×(298 K)×ln(0.010)
= −4.43... × 104 J mol−1 = −44 kJ mol−1
(ii) At Q = 1.0
∆ r G = (−32.9 × 103 J mol−1 ) + (8.3145 J K−1 mol−1 ) × (298 K) × ln(1.0)
= −3.29... × 104 J mol−1 = −33 kJ mol−1
(iii) At Q = 10
(= ∆ r G −○ )
∆ r G = (−32.9 × 103 J mol−1 ) + (8.3145 J K−1 mol−1 ) × (298 K) × ln(10)
= −2.71... × 104 J mol−1 = −27 kJ mol−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(iv) At Q = 105
∆ r G = (−32.9 × 103 J mol−1 ) + (8.3145 J K−1 mol−1 ) × (298 K) × ln(105 )
= −4.37... × 103 J mol−1 = −4.4 kJ mol−1
(v) At Q = 106
∆ r G = (−32.9 × 103 J mol−1 ) + (8.3145 J K−1 mol−1 ) × (298 K) × ln(106 )
= +1.33... × 103 J mol−1 = +1.3 kJ mol−1
�e equilibrium constant K is the value of Q for which ∆ r G = 0. From the
above values, K will therefore be somewhere between 105 and 106 . To �nd
exactly where by linear interpolation, note that according to ∆ r G = ∆ r G −○ +
RT ln Q, a plot of ∆ r G against ln Q should be a straight line. Consider the two
points on either side of zero, that is, (iv) and (v). �e point ∆ r G = 0 occurs a
fraction (4.37...)�(1.33... + 4.37...) = 0.766... of the way between points (iv)
and (v), so is at
ln K = ln 105 + (0.766...) × (ln 106 − ln 105 ) = 13.2...
Hence K = e13.2 ... = 5.84 × 105
�e value is calculated directly by setting ∆ r G = 0 and Q = K in ∆ r G = ∆ r G −○ +
RT ln Q and rearranging for K
K = e−∆ r G �RT = exp �−
−
○
−32.9 × 103 J mol−1
� = 5.84 × 105
(8.3145 J K−1 mol−1 ) × (298 K)
which is the same result as obtained from the linear interpolation.
E�A.�(a)
For the reaction 2H2 O(g) � 2H2 (g) + O2 (g) the following table is drawn up
by supposing that there are n moles of H2 O initially and that at equilibrium a
fraction α has dissociated.
2H2 O
Initial amount
Change to reach equilibrium
Amount at equilibrium
Mole fraction, x J
Partial pressure, p J
n
−αn
(1 − α)n
1−α
1 + 12 α
(1 − α)p
1 + 12 α
�
2H2
�
+αn
αn
α
1 + 12 α
αp
1 + 12 α
+
O2
�
+ 12 αn
1
αn
2
1
α
2
1 + 12 α
1
αp
2
1 + 12 α
�e total amount in moles is n tot = (1 − α)n + αn + 12 αn = (1 + 12 α)n. �is
value is used to �nd the mole fractions. In the last line, p J = x J p has been
195
196
6 CHEMICAL EQUILIBRIUM
used. Treating all species as perfect gases so that a J = (p J �p−○ ), the equilibrium
constant is
2
2
aH
aO
(p H2 �p−○ )2 (p O2 �p−○ ) p2H2 p O2 � 1+ 12 α � � 1+ 12 α �
K = 22 2 =
=
=
(1−α)p 2 −
a H2 O
(p H2 O �p−○ )2
p2H2 O p−○
�
� p○
αp
=
2
1
αp
1+ 12 α
1 3 3
α p (1 + 12 α)2
α3
p
2
=
1
2
−
○
2
2
3
−
○
(1
−
α)
(2
+
α)
p
(1 − α) p (1 + 2 α) p
In this case α = 1.77% (= 0.0177) and p = 1.00 bar; recall that p−○ = 1 bar.
E�A.�(a)
K=
0.01773
1.00 bar
×
= 2.85 × 10−6
2
(1 − 0.0177) × (2 + 0.0177)
1 bar
�e relationship between K and K c is [�A.��b–���], K = K c × (c −○ RT�p−○ ) .
For the reaction H2 CO(g) � CO(g) + H2 (g)
∆ν = ν CO + ν H2 O − ν H2 CO = 1 + 1 − 1 = +1
E�A.�(a)
hence
∆ν
K = K c × (c −○ RT�p−○ )
p−○ �c −○ R evaluates to 12.03 K so the relationship can alternatively be written as
K = K c × (T�K)�(12.03).
�e following table is drawn up:
2A
Initial amount, n J,0 �mol
Change, ∆n J �mol
Equilibrium amount,
n J �mol
Mole fraction, x J
Partial pressure, p J
�.��
−0.60
+
B
�.��
�
3C
�
+
2D
�.��
�.��
−0.30
�.��
+0.90
�.��
+0.60
0.0869...
0.369...
0.195...
0.347...
�.��
(0.0869...)p (0.369...)p (0.195...)p (0.347...)p
To go to the second line, the fact that 0.90 mol of C has been produced is used to
deduce the changes in the other species given the stoichiometry of the reaction.
For example, � mol of A is consumed for every � mol of C produced so ∆ν A =
− 23 ∆ν C = 23 ×+0.90 mol = −0.60 mol. �e total amount in moles is (0.40 mol)+
(1.70 mol) + (0.90 mol) + (1.60 mol) = 4.6 mol. �is value has been used to
�nd the mole fractions. In the last line, p J = x J p has been used.
(i) �e mole fractions are given in the above table.
(ii) Treating all species as perfect gases so that a J = p J �p−○ the equilibrium
constant is
K=
=
2
a C3 a D
p3C p2D
x C3 x D2 p2
(p C �p−○ )3 (p D �p−○ )2
=
=
=
a A2 a B
(p A �p−○ )2 (p B �p−○ )
p2A p B p−○ 2 x A2 x B p−○ 2
(0.195...) (0.347...)
3
2
(0.0869...) (0.369...)
2
×
(1.00 bar)2
= 0.324... = �.��
(1 bar)2
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(iii) �e relationship between ∆ r G −○ and K [�A.��–���], ∆ r G −○ = −RT ln K, is
used to calculate ∆ r G −○ :
∆ r G −○ = −(8.3145 J K−1 mol−1 ) × ([25 + 273.15] K) × ln 0.324...
E�A.�(a)
= +2.8 kJ mol−1
�e reaction Gibbs energy for an arbitrary reaction quotient is given by [�A.��–
���], ∆ r G = ∆ r G −○ + RT ln Q. Treating borneol and isoborneol as perfect gases
so that a J = p J �p−○ , the reaction quotient Q is
Q=
a isoborneol p isoborneol �p−○ p isoborneol
=
=
a borneol
p borneol �p−○
p borneol
Because p J = x J p = (n J �n)p ∝ n J it follows that
Q=
Hence
n isoborneol 0.30 mol
=
= 2.0
n borneol
0.15 mol
∆ r G = ∆ r G −○ + RT ln Q
= (+9.4 × 103 J mol−1 ) + (8.3145 J K−1 mol−1 ) × (503 K) × ln 2.0
E�A.�(a)
= +12 kJ mol−1
�e reaction corresponding to the standard Gibbs energy change of formation
of NH3 is
1
N (g) + 32 H2 (g) � NH3 (g)
2 2
�is is the reaction in question. �e reaction Gibbs energy for an arbitrary
reaction quotient is given by [�A.��–���], ∆ r G = ∆ r G −○ + RT ln Q. All species
are treated as perfect gases so that a J = p J �p−○ . �erefore the reaction quotient
Q is
Q=
Hence
=
(p NH3 �p−○ )
p NH3 p−○
=
=
1�2 3�2
(p N2 �p−○ )1�2 × (p H2 �p−○ )3�2 p1�2 p3�2
a N2 a H2
N2 H2
a NH3
(4.0 bar) × (1 bar)
= 2.30...
(3.0 bar)1�2 × (1.0 bar)3�2
∆ r G = ∆ r G −○ + RT ln Q
= (−16.5 × 103 J mol−1 ) + (8.3145 J K−1 mol−1 ) × (298 K) × ln(2.30...)
= −14 kJ mol−1
Because ∆ r G < 0 the spontaneous direction of the reaction under these conditions is from le� to right.
197
198
6 CHEMICAL EQUILIBRIUM
E�A.��(a) �e standard Gibbs energy change for the reaction is given in terms of the
standard Gibbs energies of formation by [�A.��a–���]:
∆ r G −○ = ∆ f G −○ (CaF2 , aq) − ∆ f G −○ (CaF2 , s)
�is is rearranged for ∆ f G −○ (CaF2 , aq) and ∆ r G −○ is replaced by −RT ln K [�A.��–
���] to give
∆ f G −○ (CaF2 , aq) = ∆ r G −○ + ∆ f G −○ (CaF2 , s) = −RT ln K + ∆ f G −○ (CaF2 , s)
= −(8.3145 J K−1 mol−1 ) × ([25 + 273.15] K) × ln(3.9 × 10−11 )
E�A.��(a)
+ (−1167 × 103 J mol−1 ) = −1.1 × 103 kJ mol−1
In general if the extent of a reaction changes by an amount ∆ξ then the amount
of a component J changes by ν J ∆ξ where ν J is the stoichiometric number for
species J (positive for products, negative for reactants). In this case ν A = −1
and ν B = +2.
n A = n A,0 + ∆n A = n A,0 + ν A ∆ξ = (1.50 mol) + (−1) × (0.60 mol) = �.�� mol
n B = n B,0 + ∆n B = n B,0 + ν B ∆ξ = 0 + 2 × (0.60 mol) = �.�� mol
E�A.��(a) �e reaction Gibbs energy ∆ r G is de�ned by [�A.�–���], ∆ r G = (∂G�∂ξ) p,T .
Approximating the derivative by �nite changes gives
∆r G = �
∂G
∆G
−6.4 kJ
� ≈
=
= −64 kJ mol−1
∂ξ p,T ∆ξ +0.1 mol
E�A.��(a) A reaction is exergonic if ∆ r G < 0 and endergonic if ∆ r G > 0. From the Resource section the standard Gibbs energy change for the formation of methane
from its elements in their reference states at 298 K is ∆ f G −○ = −50.72 kJ mol−1 .
�is is negative so the reaction is exergonic .
E�A.��(a) �e reaction quotient is de�ned by [�A.��–���], Q = ∏J a J J . For the reaction
A + 2B → 3C, ν A = −1, ν B = −2, and ν C = +3. �e reaction quotient is then
ν
−2 3
Q = a−1
A aB aC =
a C3
a A a B2
Solutions to problems
P�A.�
(a) �e relationship between the equilibrium constant and ∆ r G −○ is [�A.��–
���], ∆ r G −○ = −RT ln K.
∆ r G −○ = −(8.3145 J K−1 mol−1 ) × (298.15 K) × ln 0.164 = +4.48 kJ mol−1
(b) �e following table is drawn up. Iodine is not included in the calculations
as it is a solid.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
I2 (s) +
Initial amount
Br2 (g)
n
—
Change to reach equilibrium
—
Amount at equilibrium
—
Mole fraction, x J
—
Partial pressure, p J
—
−αn
� 2IBr(g)
(1 − α)n
1−α
1+α
(1 − α)p
1+α
�
+2αn
2αn
2α
1+α
2α p
1+α
�e total amount in moles is n tot = (1 − α)n + 2αn = (1 + α)n. �is
value is used to �nd the mole fractions. Treating Br2 (g) and IBr(g) as
perfect gases, so that a J = p J �p−○ , and I2 as a pure solid, so that a I2 = 1,
the equilibrium constant is:
2α p
2
� 1+α
�
a IBr
(p IBr �p−○ )2
p2IBr
4α 2
p
K=
=
=
=
=
−
○
−
○
(1−α)p
−
○
a I2 a Br2 1 × (p Br2 �p ) p Br2 p
(1 − α)(1 + α) p−○
p
2
1+α
Note that (1 − α)(1 + α) = 1 − α . With this, an expression for α is found
by straightforward algebra.
2
2
K p−○
α=�
�
4p + K p−○
1
2
0.164 × (1 bar)
=�
�
4 × (0.164 atm) × (1.01325 bar)�(1 atm) + 0.164 × (1 bar)
1
= 0.444...
Hence
p IBr =
2α p
2 × 0.444... × (0.164 atm)
=
= 0.101 atm or 0.102 bar
1+α
1 + 0.444...
(c) �e issue here is that the reaction under discussion is that with I� (s). If
the partial pressure of I2 is not zero then p Br2 and p IBr no longer sum to
the total pressure p but rather to p − p I2 where p I2 is the partial pressure
of iodine. �e partial pressures in the last line of the above table therefore
become
p Br2 =
1−α
(p − p I2 ) and
1+α
p IBr =
2α
(p − p I2 )
1+α
�e equilibrium constant K for the I2 (s) + Br2 (g) � 2IBr(g) is still K =
p2IBr �p Br2 p−○ but now with the new partial pressures:
K=
2α
�� 1+α
� (p − p I2 )�
2
� 1−α
� (p − p I2 )p−○
1+α
=
4α 2
p − p I2
�
�
(1 + α)(1 − α)
p−○
Given the partial pressure of I2 this equation can be solved for α, and p IBr
calculated as before.
199
200
6 CHEMICAL EQUILIBRIUM
P�A.�
�e following table is drawn up for the reaction, assuming that to reach equilibrium the reaction proceeds by an amount z in the direction of the products.
Initial amount
Change to reach
equilibrium
Amount at
equilibrium
Mole fraction, x J
Partial pressure, p J
H2 (s)
n H2 ,0
−z
n H2 ,0 − z
n H2 ,0 − z
n tot
(n H2 ,0 − z)p
n tot
+
I2 (g)
n I2 ,0
�
2HI(g)
n HI,0
−z
n I2 ,0 − z
n I2 ,0 − z
n tot
(n I2 ,0 − z)p
n tot
+2z
n HI,0 + 2z
n HI,0 + 2z
n tot
(n HI,0 + 2z)p
n tot
where n tot = n H2 ,0 + n I2 ,0 + n HI,0 . Treating all species as perfect gases, so that
a J = p J �p−○ , the equilibrium constant is
K=
2
a HI
(p HI �p−○ )2
p2HI
(n HI,0 + 2z)2
=
=
=
a H2 a I2 (p H2 �p−○ )(p I2 �p−○ ) p H2 p I2 (n H2 ,0 − z)(n I2 ,0 − z)
Rearranging gives
Hence
K(n H2 ,0 − z)(n I2 ,0 − z) = (n HI,0 + 2z)2
2
(K − 4)z 2 − ([n H2 ,0 + n I2 ,0 ]K + 4n HI,0 )z + (n H2 ,0 n I2 ,0 K − n HI
)=0
Substituting in the values for n J and K, dividing through by mol2 and writing
x = z�mol yields the quadratic
866x 2 − 609.8x + 104.36 = 0
which has solutions x = 0.410... and x = 0.293... implying z = (0.410... mol)
or z = (0.293... mol). �e solution z = (0.410... mol) is rejected because z
cannot be larger than n H2 ,0 or n I2 ,0 . �e amounts of each substance present at
equilibrium are therefore
n H2 = n H2 ,0 − z = (0.300 mol) − (0.293... mol) = 6.67 × 10−3 mol
n I2 = n I2 ,0 − z = (0.400 mol) − (0.293... mol) = 0.107 mol
P�A.�
n HI = n HI,0 + 2z = (0.200 mol) + 2 × (0.293... mol) = 0.787 mol
If the extent of reaction at equilibrium is ξ, then from the stoichiometry of the
reaction the amounts of A and B that have reacted are ξ and 3ξ respectively and
the amount of C that has been formed is 2ξ. If the initial amounts of A, B and
C are n, 3n and 0, the following table is drawn up.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
+
A
n
Initial amount
−ξ
Change to reach equilibrium
n−ξ
Amount at equilibrium
n−ξ
4n − 2ξ
(n − ξ)p
4n − 2ξ
Mole fraction, x J
Partial pressure, p J
�
3B
3n
−3ξ
2C
�
+2ξ
3(n − ξ)
2ξ
3(n − ξ)
4n − 2ξ
3(n − ξ)p
4n − 2ξ
2ξ
4n − 2ξ
2ξp
4n − 2ξ
�e total amount in moles is n tot = (n − ξ) + 3(n − ξ) + 2ξ = 4n − 2ξ. �is value
is used to �nd the mole fractions. Treating all species as perfect gases, so that
a J = p J �p−○ , the equilibrium constant is
2
� 4n−2ξ � p−○
a2
p2C p−○
(p C �p−○ )2
K = C3 =
=
=
(n−ξ)p
3(n−ξ)p 3
a A a B (p A �p−○ )(p B �p−○ )3
p A p3B
�
��
�
2ξ p
=
16ξ (2n − ξ) p
� �
27(n − ξ)4
p
2
2
−
○
4n−2ξ
2
2
4n−2ξ
2
Rearranging and then taking the square root gives
ξ 2 (2n − ξ)2 27K p2
=
(n − ξ)4
16p−○ 2
ξ(2n − ξ) 1 √
= 4 27K(p�p−○ )
(n − ξ)2
hence
�e negative square root is rejected because 0 ≤ ξ ≤ n. �is requirement arises
because if ξ < 0 this would imply a negative amount of C, while if ξ > n this
would imply negative amounts of A and B. Because 0 ≤ ξ ≤ n the le� hand side
of the square rooted expression is always ≥ 0. Because p�p−○ cannot be negative
either it follows that the positive square root is required.
Rearranging further yields the quadratic
√
27K(p�p−○ )
√
(ξ�n) − 2(ξ�n) +
=0
1 + 14 27K(p�p−○ )
2
1
4
which is solved to give
�
�
1
√
(ξ�n) = 1 −
1
� 1 + 4 27K(p�p−○ ) �
1
2
�e positive square root is rejected in order to ensure that 0 ≤ (ξ�n) ≤ 1.
Inspection of this expression shows that ξ → 0 as p → 0, indicating that the
reactants are favoured at low pressures. On the other hand (ξ�n) → 1 as p → ∞
indicating that the products are favoured at high pressure. (ξ�n) is plotted
against p�p−○ in the graph shown in Fig. �.�, using three di�erent values of K.
201
6 CHEMICAL EQUILIBRIUM
1.0
K = 100
0.8
ξ�n
202
K=1
0.6
0.4
K = 0.01
0.2
0.0
0
10
20
p�p
Figure 6.1
−
○
30
40
50
6B The response of equilibria to the conditions
Answer to discussion question
D�B.�
�is is discussed in Section �B.�(a) on page ���.
D�B.�
�is is discussed in Section �B.� on page ��� and in Section �B.� on page ���.
Solutions to exercises
E�B.�(a)
�e relationship between ∆ r G −○ and K is given by [�A.��–���], ∆ r G −○ = −RT ln K.
Hence if K = 1, ∆ r G −○ = −RT ln 1 = 0. Furthermore ∆ r G −○ is related to ∆ r H −○
and ∆ r S −○ by [�D.�–��], ∆ r G −○ = ∆ r H −○ − T∆ r S −○ , so if K = 1
∆ r H −○ − T∆ r S −○ = 0
hence
T=
∆ r H −○
∆ r S −○
Values of ∆ r H −○ and ∆ r S −○ at ��� K are calculated using data from the Resource
section.
∆ r H −○ = ∆ f H −○ (CaO, s) + ∆ f H −○ (CO2 , g) − ∆ f H −○ (CaCO3 , s, calcite)
= (−635.09 kJ mol−1 ) + (−393.51 kJ mol−1 ) − (−1206.9 kJ mol−1 )
= +178.3 kJ mol−1
−
○
−
○
−
○
∆ r S −○ = S m
(CaO, s) + S m
(CO2 , g) − S m
(CaCO3 , s, calcite)
= (39.75 J K−1 mol−1 ) + (213.74 J K−1 mol−1 ) − (92.9 J K−1 mol−1 )
= 160.59 J K−1 mol−1
Substituting these values into the equation found above, assuming that ∆ r H −○
and ∆ r S −○ do not vary signi�cantly with temperature over the range of interest,
gives:
∆ r H −○ 178.13 × 103 J mol−1
T=
=
= 1109 K
∆ r S −○
160.59 J K−1 mol−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�B.�(a)
Treating the vapour as a perfect gas, so that a J = p J �p−○ , and noting that a A2 B = 1
because it is a pure solid, the equilibrium constant for the dissociation A2 B(s) �
A2 (g) + B(g) is
K=
a A2 ,g a B,g (p A2 �p−○ )(p B �p−○ ) p A2 p B2
=
=
a A2 B,s
1
p−○ 2
Furthermore, because A2 and B are formed in a 1 ∶ 1 ratio, they each have a
mole fraction of 1�2 and the partial pressure of each is half the total pressure:
p A2 = p B = 12 p. �e equilibrium constant is therefore
K=
( 12 p)( 12 p)
p2
4p−○ 2
p−○ 2
=
∆ r H −○ 1
1
� − �
R
T2 T1
hence
�e variation of K with temperature, assuming that ∆ r H −○ does not vary with
T over the temperature range of interest, is given by [�B.�–���]:
ln K 2 − ln K 1 = −
∆ r H −○ = −
R ln(K 2 �K 1 )
(1�T2 ) − (1�T1 )
Noting that the above expression for K implies that ln(K 2 �K 1 ) = ln(p22 �p21 ),
∆ r H −○ is calculated as
∆ r H −○ = −
(8.3145 J K−1 mol−1 ) × ln �(547 kPa)2 �(208 kPa)2 �
[1�(477 + 273.15) K] − [1�(367 + 273.15) K]
= 7.01... × 104 J mol−1 = 70.2 kJ mol−1
�e standard entropy of reaction, ∆ r S −○ is found by rearranging ∆ r G −○ = ∆ r H −○ −
T∆ r S −○ [�D.�–��] and replacing ∆ r G −○ by ∆ r G −○ = −RT ln K [�A.��–���]:
∆ r S −○ =
∆ r H −○ − ∆ r G −○ ∆ r H −○ + RT ln K ∆ r H −○
p2
=
=
+ R ln � −○ �
T
T
T
4p
Using the data for 367 ○ C (both temperatures give the same result) gives:
∆ r S −○ =
7.01... × 104 kJ mol−1
(208 kPa)2
+ (8.3145 J K−1 mol−1 ) × ln �
�
(367 + 273.15) K
4 × (100 kPa)2
= 1.10... × 102 J K−1 mol−1 = 110 J K−1 mol−1
An alternative (but equivalent) approach to �nding ∆ r H −○ and ∆ r S −○ is to �rst
calculate ∆ r G −○ at both temperatures and hence obtain two equations of the
form ∆ r G −○ = ∆ r H −○ − T∆ r S −○ . �ese can then be solved simultaneously to �nd
the two unknowns ∆ r H −○ and ∆ r S −○ , assuming them to be constant.
�e values of ∆ r H −○ and ∆ r S −○ are then used with ∆ r G −○ = ∆ r H −○ − T∆ r S −○
and ∆ r G −○ = −RT ln K to calculate ∆ r G −○ and K at the temperature of interest,
422 ○ C or 695.15 K. In making this calculation it is again assumed that ∆ r H −○
203
204
6 CHEMICAL EQUILIBRIUM
and ∆ r S −○ do not vary with temperature.
∆ r G −○ = ∆ r H −○ − T∆ r S −○
= (7.01... × 104 J mol−1 ) − (695.15 K) × (1.10... × 102 J K−1 mol−1 )
= −6.48... × 103 J mol−1 = −6.48 kJ mol−1
K = e−∆ r G �RT = exp �−
−
○
E�B.�(a)
−6.48... × 103
� = 3.07
(8.3145 J K−1 mol−1 ) × (695.15 K)
For the reaction N2 O4 (g) � 2NO2 (g) the following table is drawn up by
supposing that there are n moles of N2 O4 initially and that at equilibrium a
fraction α has dissociated.
� �NO2
N2 O4
n
Initial amount
Change to reach equilibrium
Amount at equilibrium
Mole fraction, x J
Partial pressure, p J
�
−αn
(1 − α)n
1−α
1+α
(1 − α)p
1+α
+2αn
2αn
2α
1+α
2α p
1+α
�e total amount in moles is n tot = (1 − α)n + 2αn = (1 + α)n. �is value
is used to �nd the mole fractions. In the last line, p J = x J p [�A.�–�] has been
used. Treating all species as perfect gases so that a J = (p J �p−○ ), the equilibrium
constant is
K=
2α p
2
� 1+α
�
a NO
p2NO2
(p NO2 �p−○ )2
4α 2
p
2
=
=
=
=
−
○
−
○
−
○
(1−α)p
−
○
a N2 O4 (p N2 O4 �p ) p N2 O4 p
(1
−
α)(1
+
α)
p
�
�p
2
1+α
In this case α = 0.1846 and p = 1.00 bar; recall that p−○ = 1 bar.
K=
4 × 0.18462
1.00 bar
×
= 0.141... = �.���
(1 − 0.1846) × (1 + 0.1846)
1 bar
�e temperature dependence of K is given by [�B.�–���],
ln K 2 − ln K 1 = −
∆ r H −○ 1
1
� − �
R
T2 T1
assuming that ∆ r H −○ is constant over the temperature range of interest. Taking
T1 = 25 ○ C (= 298.15 K) and T2 = 100 ○ C (= 373.15 K) gives
ln K 2 = ln(0.141...) −
56.2 × 103 J mol−1
1
1
−
� = 2.59...
−1 �
−1
373.15
K
298.15
K
8.3145 J K mol
�at is, K 2 = ��.� , a larger value than at 25 ○ C, as expected for this endothermic
reaction.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�B.�(a)
�e data in the Resource section is used to calculate ∆ r G −○ and ∆ r H −○ at ��� K
∆ r G −○ = ∆ f G −○ (CO2 , g) − ∆ f G −○ (PbO, s, red) − ∆ f G −○ (CO, g)
= (−394.36 kJ mol−1 ) − (−188.93 kJ mol−1 ) − (−137.17 kJ mol−1 )
= -��.�� kJ mol−1
∆ r H −○ = ∆ f H −○ (CO2 , g) − ∆ f H −○ (PbO, s, red) − ∆ f H −○ (CO, g)
= (−393.51 kJ mol−1 ) − (−218.99 kJ mol−1 ) − (−110.53 kJ mol−1 )
= −63.99 kJ mol−1
�e equilibrium constant at 298 K is calculated from ∆ r G −○ using [�A.��–���],
∆ r G −○ = −RT ln K
∆ r G −○
−68.26 × 103 J mol−1
=−
= 27.5...
RT
(8.3145 J K−1 mol−1 ) × (298 K)
hence K = e27.5 ... = 9.21... × 1011 = 9.22 × 1011
ln K = −
�e temperature dependence of K is given by [�B.�–���],
ln K 2 − ln K 1 = −
∆ r H −○ 1
1
� − �
R
T2 T1
assuming that ∆ r H −○ is constant over the temperature range of interest. �is is
used to calculate the equilibrium constant at 400 K
ln K 2 = ln(9.21... × 1011 ) −
E�B.�(a)
−63.99 × 103 J mol−1
1
1
�
−
� = 20.9...
8.3145 J K−1 mol−1 400 K 298 K
�at is, K 2 = 1.27 × 109 , a smaller value than at 298 K, as expected for this
exothermic reaction.
Assuming that ∆ r H −○ is constant over the temperature range of interest, the
temperature dependence of K is given by [�B.�–���],
ln K 2 − ln K 1 = −
∆ r H −○ 1
1
� − �
R
T2 T1
Using ∆ r G −○ = −RT ln K to substitute for K 1 and setting ln K 2 = ln 1 = 0 (the
crossover point) gives
∆ r G −○ (T1 )
∆ r H −○ 1
1
=−
� − �
RT1
R
T2 T1
Rearranging for T2 gives
T2 =
T1 ∆ r H −○
(1280 K) × (+224 kJ mol−1 )
=
= 1.5 × 103 K
∆ r H −○ − ∆ r G −○ (T1 ) (+224 kJ mol−1 ) − (+33 kJ mol−1 )
Note that this temperature is outside the range over which ∆ r H −○ is known to
be constant and is therefore an estimate.
205
206
6 CHEMICAL EQUILIBRIUM
E�B.�(a)
�e van ’t Ho� equation [�B.�–���], d ln K�dT = ∆ r H −○ �RT 2 , is rearranged to
obtain an expression for ∆ r H −○
d ln K
dT
B
C
B
2C
2C
2 d
= RT
�A + + 2 � = RT 2 �− 2 − 3 � = −R �B +
�
dT
T T
T
T
T
2 × (1.51 × 105 K2 )
= −(8.3145 J K−1 mol−1 ) × �(−1088 K) +
�
400 K
∆ r H −○ = RT 2
= +2.76... × 103 J mol−1 = +2.77 kJ mol−1
�e standard reaction entropy is obtained by �rst �nding an expression for
∆ r G −○ using [�A.��–���]
∆ r G −○ = −RT ln K = −RT �A +
B
C
C
+
� = −R �AT + B + �
T T2
T
�e equation ∆ r G −○ = ∆ r H −○ −T∆ r S −○ [�D.�–��] is then rearranged to �nd ∆ r S −○
�
�
�
∆ r H −○ − ∆ r G −○ 1 �
2C
C
C
∆r S =
= �
−R �B +
� + R �AT + B + ��
= R �A − 2 �
�
T
T�
T
T �
T
�
� � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � �� � −�○� � � � � � � � � � � � � � � � � � ���
���� � � � � � � � � � � � ∆� � � �
H −○
−∆ G
−
○
r
r
1.51 × 105 K2
= (8.3145 J K−1 mol−1 ) × �−1.04 −
� = −16.5 J K−1 mol−1
(400 K)2
An alternative approach to �nding ∆ r S −○ is to use the variation of G with T
which is given by [�E.�–���], (∂G�∂T) p = −S. �is implies that d∆ r G −○ �dT =
−∆ r S −○ where the derivative is complete (not partial) because ∆ r G −○ is independent of pressure. Using the expression for ∆ r G −○ from above it follows that
�
�
�
d∆ r G −○
d �
C
�−R �AT + B + �� = R �A − C �
∆r S = −
=−
�
dT
dT �
T �
T2
�
��
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
�
∆ G −○
−
○
r
which is the same expression obtained above.
E�B.�(a)
Treating all species as perfect gases so that a J = p J �p−○ , the equilibrium constant
for the reaction H2 CO(g) � CO(g) + H2 (g) is
a H2 a CO (p H2 �p−○ )(p CO �p−○ )
p H2 p CO
(x H2 p)(x CO p)
=
=
=
a H2 CO
(p H2 CO �p−○ )
p H2 CO p−○
(x H2 CO p)p−○
x H x CO p
p
= 2
= K x × −○
−
○
x H2 CO p
p
K=
where K x is the part of the equilibrium constant expression that contains the
equilibrium mole fractions of reactants and products. Because K is independent of pressure, if p doubles K x must halve in order to preserve the value of
K. In other words, K x is reduced by 50% .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�B.�(a)
�e following table is drawn up for the borneol � isoborneol reaction, denoting
the initial amounts of borneol and isoborneol by n b,0 and n iso,0 and supposing
that in order to reach equilibrium an amount z of borneol has converted to
isoborneol.
�
Borneol
n b,0
Initial amount
Change to reach equilibrium
Amount at equilibrium
Mole fraction, x J
Partial pressure, p J
−z
Isoborneol
n iso,0
+z
n b,0 − z
n b,0 − z
n b,0 + n iso,0
(n b,0 − z)p
n b,0 + n iso,0
n iso,0 + z
n iso,0 + z
n b,0 + n iso,0
(n iso,0 + z)p
n b,0 + n iso,0
�e total amount in moles is (n b,0 − z) + n iso,0 + z = n b,0 + n iso,0 . �is value is
used to �nd the mole fractions. Treating both species as perfect gases so that
a J = p J �p−○ the equilibrium constant is
K=
a borneol
p borneol
n iso,0 + z
=
=
a isoborneol p isoborneol
n b,0 − z
Rearranging for z gives z = (Kn b,0 − n iso,0 )�(1 + K). Noting that n = m�M
where M = 154.2422 g mol−1 is the molar mass of borneol and isoborneol,
gives
n b,0 = (7.50 g)�(154.2422 g mol−1 ) = 0.0486... mol
n iso,0 = (14.0 g)�(154.2422 g mol−1 ) = 0.0907... mol
z=
Kn b,0 − n iso,0 0.106 × (0.0486... mol) − (0.0907... mol)
=
= −0.0774... mol
1+K
1 + 0.106
�e negative value of z indicates that in order to reach equilibrium there is a net
conversion of isoborneol to borneol. Using this value of z, and the expressions
for x borneol and x isoborneol in the above table, the mole fractions at equilibrium
are calculated as
n b,0 − z
(0.0486... mol) − (−0.0774... mol)
=
= 0.904...
n b,0 + n iso,0
(0.0486... mol) + (0.0907... mol)
= 0.904
x borneol =
�en
E�B.�(a)
x isoborneol = 1 − x borneol = 1 − 0.904... = 0.096
�e temperature dependence of K is given by [�B.�–���]
ln K 2 − ln K 1 = −
∆ r H −○ 1
1
� − �
R
T2 T1
hence
∆ r H −○ = −
R ln(K 2 �K 1 )
(1�T2 ) − (1�T1 )
207
208
6 CHEMICAL EQUILIBRIUM
(i) If the equilibrium constant is doubled then K 2 �K 1 = 2
∆ r H −○ = −
(8.3145 J K−1 mol−1 ) × ln 2
= +52.9 kJ mol−1
[1�(308 K)] − (1�[298 K])
(ii) If the equilibrium constant is halved then K 2 �K 1 = 1�2
∆ r H −○ = −
Solutions to problems
P�B.�
Assuming ∆ r H −○ to be constant over the temperature range of interest, the temperature dependence of K is given by [�B.�–���]
ln K 2 − ln K 1 = −
�erefore
∆ r H −○ = −
P�B.�
(8.3145 J K−1 mol−1 ) × ln(1�2)
= −52.9 kJ mol−1
[1�(308 K)] − (1�[298 K])
∆ r H −○ 1
1
� − �
R
T2 T1
hence
∆ r H −○ = −
R ln(K 2 �K 1 )
(1�T2 ) − (1�T1 )
(8.3145 J K−1 mol−1 ) × ln �(1.75 × 105 )�(2.13 × 106 )�
= −92.2 kJ mol
[1�(308 K)] − [1�(288 K)]
−1
�e reaction for which ∆ r H −○ is the standard enthalpy of formation of UH3 is:
U(s) + 32 H2 (g) � UH3 (s)
Treating H2 as a perfect gas (so that a H2 = p H2 �p−○ ) and noting that pure solids
have a J = 1, the equilibrium constant for this reaction is written
1
p−○
K = 3�2
=
=
�
�
−
○ 3�2
p
a a U (p�p ) × 1
a UH3
3�2
H2
−3�2
p
= � −○ �
p
where p is the pressure of H2 . �e standard reaction enthalpy, which corresponds to ∆ f H −○ (UH3 , s), is obtained by rearranging the van ’t Ho� equation
[�B.�–���], d ln K�dT = ∆ r H −○ �RT 2 , for ∆ r H −○
−3�2
d ln K
d
p
d
p
= RT 2
ln � −○ �
= − 32 RT 2
ln � −○ �
dT
dT
p
dT
p
d
d
= − 32 RT 2
� ln p − ln p−○ � = − 32 RT 2
ln p
dT
dT
d
B
B
C
= − 32 RT 2
�A + + C ln T� = − 32 RT 2 �− 2 + �
dT
T
T
T
∆ f H −○ (UH3 , s) = RT 2
= − 32 R(CT − B)
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Heat capacity at constant pressure is de�ned by [�B.�–��], C p = (∂H�∂T) p ,
which implies that ∆ f C −p○ = d∆ f H −○ �dT where the derivative is complete (not
partial) because ∆ f H −○ does not depend on pressure. �erefore
∆ f C −p○ =
P�B.�
d
� − 32 R[−B + CT]� = − 32 RC
dT
= − 32 × (8.3145 J K−1 mol−1 ) × (−5.65) = +70.5 J K−1 mol−1
�e van ’t Ho� equation [�B.�–���] is:
d ln K ∆ r H −○
=
dT
RT 2
which can also be written
−
d ln K
∆ r H −○
=
d(1�T)
R
�e second form implies that a graph of − ln K against 1�T should be a straight
line of slope ∆ r H −○ �R. It is �rst necessary to relate K to α. To do this, the
following table is drawn up for the CO2 (g) � CO(g) + 12 O2 (g) equilibrium
CO2 (g)
n
Initial amount
Change to reach equilibrium
Amount at equilibrium
Mole fraction, x J
−αn
(1 − α)n
1−α
1 + 12 α
(1 − α)p
1 + 12 α
Partial pressure, p J
�
CO(g)
�
+αn
αn
α
1 + 12 α
α
1 + 12 α
+
1
O (g)
2 2
�
+ 12 αn
1
αn
2
1
α
2
1 + 12 α
1
α
2
1 + 12 α
�e total amount in moles is n tot = (1 − α)n + αn + 12 αn = (1 + 12 α)n. �is
value is used to �nd the mole fractions. Treating all species as perfect gases (so
that a J = p J �p−○ ) the equilibrium constant is
K=
=
1�2
a CO a O2
a CO2
1�2
1
αp
αp
2
�
�
�
�
1�2
1
1
p CO p O2
1+ α
1+ α
(p CO �p−○ )(p O2 �p−○ )1�2
2
2
=
=
=
(p CO2 �p−○ )
p CO2 p−○ 1�2
(1−α)p
� 1 � p−○
1+ α
2
1�2
1�2
3
� 12 α �
(1 − α)(1 + 12 α)1�2
p
� −○ �
p
Using this expression, with p = 1 bar, K is calculated at each temperature
and − ln K is plotted against 1�(T�K) as described above; the plot is shown
in Fig. �.�.
T�K
1 395
1 443
1 498
α
1.44 × 10−4
2.50 × 10−4
4.71 × 10−4
K
1.222 × 10−6
2.794 × 10−6
7.224 × 10−6
104 �(T�K) − ln K
7.168
13.62
6.930
12.79
6.676
11.84
209
6 CHEMICAL EQUILIBRIUM
14.0
13.0
− ln K
210
12.0
6.6
6.8
7.0
10 �(T�K)
7.2
4
Figure 6.2
�e data fall on a good straight line, the equation of which is
ln K = 3.607 × 104 �(T�K) − 12.23
∆ r H −○ �R is determined from the slope
∆ r H −○ = R × (slope × K) = (8.3145 J K−1 mol−1 ) × (3.607 × 104 K)
= +2.99... × 105 J mol−1 = +3.00 × 102 kJ mol−1
�e equilibrium constant K has already been calculated; from the table above
the value of K at 1443 K is 2.79 × 10−6 .
�e standard reaction Gibbs energy is then calculated using ∆ r G −○ = −RT ln K,
[�A.��–���], and the standard reaction entropy from ∆ r G −○ = ∆ r H −○ − T∆ r S −○
[�D.�–��].
∆ r G −○ = −RT ln K = −(8.3145 J K−1 mol−1 ) × (1443 K) × ln(2.79... × 10−6 )
= +1.53... × 105 J mol−1 = +153 kJ mol−1
∆ r S −○ =
∆ r H −○ − ∆ r G −○ (2.99... × 105 J mol−1 ) − (1.53... × 105 J mol−1 )
=
T
1443 K
= +102 J K−1 mol−1
P�B.�
�e equilibrium is 2CH3 COOH(g) � (CH3 COOH)2 (g), the dimer being
held together by hydrogen bonds. �e following table is drawn up, assuming
that the initial amount in moles of ethanoic acid is n and that at equilibrium a
fraction α of the ethanoic acid has dimerised.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
2CH3 COOH
n
Initial amount
Change to reach equilibrium
Amount at equilibrium
� (CH3 COOH)2
0
−αn
+ 12 αn
1
αn
2
1
α
2
1 − 12 α
(1 − α)n
1−α
1 − 12 α
Mole fraction, x J
(1 − α)p
1 − 12 α
Partial pressure, p J
1
αp
2
1 − 12 α
�e total amount in moles is n tot = (1 − α)n + 12 αn = (1 − 12 α)n. �is value is
used to �nd the mole fractions.
�e total amount in moles present at equilibrium is found from the pressure by
using the perfect gas law [�A.�–�], pV = n tot RT
pV = (1 − 12 α)nRT
hence
α =2−
2pV
2pV
=2−
nRT
(m�M)RT
where M = 60.0516 g mol−1 is the molar mass of ethanoic acid. �e value of
α is then used to calculate K. Assuming that both species present are perfect
gases (so that a J = p J �p−○ ) and using the expressions for p J from the above table,
the equilibrium constant is
K=
a(CH3 COOH)2 (p(CH3 COOH)2 �p ) p(CH3 COOH)2 p
=
=
=
2
a CH
(p CH3 COOH �p−○ )2
p2CH3 COOH
3 COOH
−
○
1
= 2
α(1 − 12 α)p−○
(1 − α)2 p
−
○
1
� 2 1 � p−○
αp
1− α
2
(1−α)p
� 1 �
1− α
2
2
�e values of α and K at the two temperatures are then calculated using these
formulae as
At 437 K
α =2−
2 × (101.9 × 103 Pa) × (21.45 × 10−6 m3 )
= 0.607...
(0.0519 g�60.0516 g mol−1 ) × (8.3145 J K−1 mol−1 ) × (437 K)
1
α(1 − 12 α)p−○
× (0.607...) × (1 − 12 × 0.607...) × (100 kPa)
2
=
= 1.35...
(1 − α)2 p
(1 − 0.607...)2 × (101.9 kPa)
= �.��
1
K= 2
211
212
6 CHEMICAL EQUILIBRIUM
At 471 K
α =2−
2 × (101.9 × 103 Pa) × (21.45 × 10−6 m3 )
= 0.235...
(0.0380 g�60.0516 g mol−1 ) × (8.3145 J K−1 mol−1 ) × (471 K)
1
α(1 − 12 α)p−○
× (0.235...) × (1 − 12 × 0.235...) × (100 kPa)
2
=
= 0.174...
(1 − α)2 p
(1 − 0.235...)2 × (101.9 kPa)
= �.���
1
K= 2
�e standard enthalpy of the dimerization reaction is found using the temperature dependence of K [�B.�–���]
ln K 2 − ln K 1 = −
∆ r H −○ 1
1
� − �
R
T2 T1
∆ r H −○ = −
hence
Taking T1 = 437 K and T2 = 471 K gives
∆ r H −○ = −
P�B.�
R ln(K 2 �K 1 )
(1�T2 ) − (1�T1 )
R ln(0.174...�1.35...)
= −103 kJ mol−1
[1�(471 K)] − [1�(437 K)]
�e relationship between K and ∆ r G −○ [�A.��–���], ∆ r G −○ = −RT ln K, implies
that
−
○
−
○
−
○
−
○
−
○
K = e−∆ r G �RT = e−(∆ r H −T ∆ r S )�RT = e−∆ r H �RT e∆ r S �R
�e ratio of the values of K that would be obtained using the lowest and highest
values of ∆ r H −○ is
∆ r H low − ∆ r H high
K lowH
e−∆ r H low �RT e∆ r S �R
= −∆ H −○ �RT
= exp �−
�
K highH e r high e∆ r S −○ �R
RT
−
○
−
○
−
○
−
○
For the given data, the value of this factor is
At ��� K:
K lowH
(243 − 289) × 103 J mol−1
= exp �−
� = 1.2 × 108
K highH
(8.3145 J K−1 mol−1 ) × (298 K)
At ��� K:
P�B.��
K lowH
(243 − 289) × 103 J mol−1
= exp �−
� = 2.7 × 103
K highH
(8.3145 J K−1 mol−1 ) × (700 K)
�e standard reaction Gibbs energy is related to the standard reaction enthalpy
and entropy according to [�D.�–��], ∆ r G −○ = ∆ r H −○ − T∆ r S −○ . However, ∆ r H −○
and ∆ r S −○ themselves vary with temperature according to Kirchho� ’s law [�C.�a–
��] for ∆ r H −○ and the analogous equation [�C.�a–��] for ∆ r S −○
∆ r H −○ (T2 ) = ∆ r H −○ (T1 ) + �
∆ r S −○ (T2 ) = ∆ r S −○ (T1 ) + �
T2
T1
∆ r C −p○ dT
○
T2 ∆ C −
r p
T1
T
dT
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
○
○
∆ r C −p○ is de�ned by [�C.�b–��], ∆ r C −p○ = ∑products νC −p,m
− ∑reactants νC −p,m
.
2
Because each species has C p,m given by C p,m = a + bT + c�T it is convenient
to write ∆ r C −p○ = A + BT + C�T 2 where A is de�ned by A = ∑products νa −
∑reactants νa and similarly for B and C. Expressions for ∆ r H −○ and ∆ r S −○ at
temperature T2 are then obtained by performing the integrations
∆ r H −○ (T2 ) = ∆ r H −○ (T1 ) + �
T2
T1
A + BT + C�T 2 dT
= ∆ r H −○ (T1 ) + A(T2 − T1 ) + 12 B �T22 − T12 � − C �
∆ r S −○ (T2 ) = ∆ r S −○ (T1 ) + �
= ∆ r S −○ (T1 ) + �
T2 A + BT + C�T 2
T1
T2 A
T
+B+
1
1
− �
T2 T1
dT
C
dT
T3
T
T
1
1
2
= ∆ r S −○ (T1 ) + A ln
+ B(T2 − T1 ) − 12 C � 2 − 2 �
T1
T2 T1
T1
�e standard reaction Gibbs energy at temperature T2 is then given by
∆ r G −○ (T2 ) = ∆ r H −○ (T2 ) − T2 ∆ r S −○ (T2 )
1
1
− ��
T2 T1
T
1
1
2
− T2 �∆ r S −○ (T1 ) + A ln
+ B(T2 − T1 ) − 12 C � 2 − 2 ��
T1
T2 T1
T
2
= ∆ r H −○ (T1 ) − T2 ∆ r S(T1 ) + A �T2 − T1 − T2 ln � ��
T1
+ B � 12 �T22 − T12 � − T2 (T2 − T1 )�
= �∆ r H −○ (T1 ) + A(T2 − T1 ) + 12 B �T22 − T12 � − C �
+C�
T2 1
1
1
1
� 2 − 2 � − � − ��
2 T2 T1
T2 T1
In order to obtain an expression that contains ∆ r G −○ (T1 ), it is necessary to
write the �rst part of the above expression as ∆ r H −○ (T1 ) − T1 ∆ r S −○ (T1 ) −(T2 −
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
∆ r G −○ (T1 )
T1 )∆ r S −○ (T1 ) so that:
∆ r G −○ (T2 ) = ∆ r G −○ (T1 ) − (T2 − T1 )∆ r S −○ (T1 ) + A �T2 − T1 − T2 ln �
+ B � 12 �T22 − T12 � − T2 (T2 − T1 )�
+C�
T2 1
1
1
1
�
−
� − � − ��
2 T22 T12
T2 T1
T2
��
T1
�e standard Gibbs energy for the formation of H2 O(l) is ∆ r H −○ for the reaction
H2 (g) + 12 O2 (g) → H2 O(l). From the data in the Resource section, at 298 K,
213
214
6 CHEMICAL EQUILIBRIUM
∆ f H −○ (H2 O, l) = −285.83 kJ mol−1 and ∆ f S −○ (H2 O, l) is calculated as
−
○
−
○
−
○
∆ f S −○ = S m
(H2 O, l) − S m
(H2 , g) − 12 S m
(O2 , g)
= (69.91 − 130.684 − 12 × 205.138) J K−1 mol−1 = −163.343 J K−1 mol−1
�e quantities A, B, and C are also calculated using the data from the Resource
section:
A = a H2 O,l − a H2 ,g − 12 a O2 ,g
= �75.29 − 27.28 − 12 × 29.96� J K−1 mol−1 = 33.03 J K−1 mol−1
B = b H2 O,l − b H2 ,g − 12 b O2 ,g
= �0 − 3.26 − 12 × 4.18� × 10−3 J K−2 mol−1 = −5.35 × 10−3 J K−2 mol−1
C = c H2 O,l − c H2 ,g − 12 c O2 ,g
= �0 − 0.50 − 12 × (−1.67)� × 105 J K mol−1 = 0.335 × 105 J K mol−1
Hence, using the expressions derived above with T1 = 298 K and T2 = 372 K:
∆ f H −○ (298 K)
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
∆ f H −○ (372 K) = (−285.83 × 103 J mol−1 )
+ (33.03 J K−1 mol−1 ) × ([372 − 298] K)
+ 12 (−5.35 × 10−3 J K−2 mol−1 ) × [(372 K)2 − (298 K)2 ]
1
1
− (0.335 × 105 J K mol−1 ) × �
−
�
372 K 298 K
= −283.49... kJ mol−1
∆ f S −○ (298 K)
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
372 K
∆ f S −○ (372 K) = (−163.343 J K−1 mol−1 ) +(33.03 J K−1 mol−1 ) × ln �
�
298 K
+ (−5.35 × 10−3 J K−2 mol−1 ) × [(372 K) − (298 K)]
− 12 (0.335 × 105 J K mol−1 ) × �
= −156.34... J K−1 mol−1
1
1
−
�
(372 K)2 (298 K)2
∆ f G −○ (372 K) = ∆ f H −○ (372 K) − (372 K) × ∆ f S −○ (372 K)
= (−283.49... × 103 J mol−1 )
− (372 K) × (−156.34... J K−1 mol−1 )
= −225.34 kJ mol−1
�is compares to −237.13 kJ mol−1 at 298 K (from the Resource section). Note
that ∆ f H −○ and ∆ r S −○ do not change very much between 298 K and 372 K in
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
this case. In fact, assuming that they are constant gives almost the same value
of ∆ f G −○ (372 K), as is seen by calculating ∆ f G −○ at ��� K using the values of
∆ f H −○ and ∆ f S −○ at ��� K:
∆ f G −○ (372 K) ≈ ∆ f H −○ (298 K) + (372 K) × ∆ f S −○ (298 K)
= (−285.83 × 103 J mol−1 ) − (372 K) × (−163.343 J K−1 mol−1 )
= −225.07 kJ mol−1
which di�ers from the value obtained above by less than 0.3 kJ mol−1 .
6C Electrochemical cells
Answers to discussion questions
D�C.�
�e role of a salt bridge is to minimise the liquid junction potential which
would otherwise occur as a result of the contact between the electrolytes in
the two half cells. For a cell to generate a potential these solutions must be
in electrical contact: the salt bridge achieves this without involving a physical
contact between the two solutions.
D�C.�
When a current is being drawn from an electrochemical cell, the cell potential
is altered by the formation of charge double layers at the surface of electrodes
and by the formation of solution chemical potential gradients (concentration
gradients). Resistive heating of the cell circuits may occur and junction potentials between dissimilar materials both external and external to the cell may
change.
D�C.�
A galvanic cell is an electrochemical cell that produces electricity as a result
of the spontaneous reaction occuring inside it. An electrolytic cell is an electrochemical cell in which a non-spontaneous reaction is driven by an external
source of current.
Solutions to exercises
E�C.�(a)
�e reduction half-reactions for the cell in question are:
R: Cd2+ (aq) + 2e− → Cd(s)
L: AgBr(s) + e− → Ag(s) + Br− (aq)
E −○ (R) = −0.40 V
E −○ (L) = +0.0713 V
�e cell reaction is obtained by subtracting the le�-hand half-reaction from the
right-hand half-reaction, a�er �rst multiplying the le�-hand half-reaction by
two so that both half-reactions involve the same number of electrons.
Cd2+ (aq) + 2Ag(s) + 2Br− (aq) → Cd(s) + 2AgBr(s)
�e cell potential is given by the Nernst equation [�C.�–���]
−
○
E cell = E cell
− (RT�νF) ln Q
215
216
6 CHEMICAL EQUILIBRIUM
In this case ν = 2 and the reaction quotient Q is
� ���2� � � �� � � � � � �
a Cd(s) a AgBr(s)
1
1
Q=
1
=
2
2
2
a Cd2+ (s) a Br
a Cd2+ (aq) a Ag(s)
a Br
− (aq)
− (aq)
�
1
where a J = 1 for pure solids has been used. For ions in solution the activity is
written as a = γ± b�b −○ , where γ± is the mean activity coe�cient, as established
in Section �F.� on page ���. �e Debye–Hückel limiting law [�F.��–���], which
applies at low molalities, is
log γ± = −A�z+ z− �I 1�2
where A = 0.502 for an aqueous solution at 298 K, z+ and z− are the charges
on the ions, and I is the dimensionless ionic strength of the solution which
for a solution containing two types of ion at molality b+ and b− is given by
[�F.��–���], I = 12 (b+ z+2 + b− z−2 )�b −○ .
For the cell in question, the right-hand electrode contains a solution of Cd(NO3 )2
of molality b R = 0.010 mol kg−1 . In this case z+ = +2 (for Cd2+ ), z− = −2 (for
NO−3 ), b+ = b Cd2+ = b R and b− = b NO−3 = 2b R . �e ionic strength is
I R = 12 �22 × b R + (−1)2 × (2b R )� �b −○ = 3b R �b −○
and the mean activity coe�cient for the right-hand electrode is therefore given
by
1
1
3b R 2
3b R 2
1�2
log γ±,R = −A�z+ z− �I R = −A�(+2)(−1)� � −○ � = −2A � −○ �
b
= −2 × 0.509 × �
b
1
2
3 × (0.010 mol kg−1 )
� = −0.176...
1 mol kg−1
Hence γ±,R = 10−0.176 ... = 0.666..., and so
a Cd2+ = γ±,R
b Cd2+
0.010 mol kg−1
=
(0.666...)
×
= 6.66... × 10−3
b −○
1 mol kg−1
In a similar way, the le�-hand electrode contains a solution of KBr of molality
b L = 0.050 mol kg−1 , so that z+ = +1 (for K+ ), z− = −1 (for Br− ), and b+ = b− =
b L . It follows that
I L = 12 (b+ z+2 + b− z−2 )�b −○ = 12 �12 × b L + (−1)2 × b L � �b−○ = b L �b−○
and therefore
1
1
bL 2
bL 2
1�2
log γ±,L = −A�z+ z− �I L = −A × �(+1) × (−1)� × � −○ � = −A � −○ �
= −0.509 × �
−1
1
2
b
0.050 mol kg
� = −0.113...
1 mol kg−1
b
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Hence γ±,L = 10−0.113 ... = 0.769..., and so
a Br− = γ±,L
b Br−
0.050 mol kg−1
=
(0.769...)
×
= 0.0384...
b −○
1 mol kg−1
Putting these activities into the Nernst equation for the cell with ν = 2 and the
expression for Q obtained above gives
−
○
E cell = E cell
−
RT
RT
1
ln Q = [E −○ (R) − E −○ (L)] −
ln �
�
νF
2F
a
a2
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
Cd 2+ Br−
−
○
E cell
= [(−0.40 V) − (+0.0713 V)] −
E�C.�(a)
× ln �
(8.3145 J K−1 mol−1 ) × (298 K)
2 × (96485 C mol−1 )
1
� = −0.619 V
(6.66... × 10−3 ) × (0.0384...)2
�e reduction half-reactions for the cell in question are
Cu2+ (aq) + 2e− → Cu(s)
Zn2+ (aq) + 2e− → Zn(s)
R:
L:
which reveal that ν = 2 for the given cell reaction. �e relationship between
−
○
−
○
∆ r G −○ and E cell
is given by [�C.�–���], ∆ r G −○ = −νFE cell
∆ r G −○ = −2 × (96485 C mol−1 ) × (+1.10 V) = −212 kJ mol−1
E�C.�(a)
where 1 C V = 1 J is used.
−
○
�e Nernst equation [�C.�–���] is E cell = E cell
− (RT�νF) ln Q. If Q changes
from Q 1 to Q 2 then the change in cell potential is given by
−
○
E cell,1 − E cell,2 = �E cell
−
RT
RT
RT
Q2
−
○
ln Q 2 � − �E cell
−
ln Q 1 � = −
ln � �
νF
νF
νF
Q1
For ν = 2 and Q 2 �Q 1 = 1�10 the change in cell potential is
E cell,1 − E cell,2 = −
E�C.�(a)
(8.3145 J K−1 mol−1 ) × (298 K)
1
× ln 10
= +0.030 V
2 × (96485 C mol−1 )
where 1 J C−1 = 1 V is used.
(i) �e reduction half-reactions for the cell Zn(s)|ZnSO4 (aq)||AgNO3 |Ag(s),
together with their standard electrode potentials from the Resource section, are
R: Ag+ (aq) + e− → Ag(s)
L: Zn2+ (aq) + 2e− → Zn(s)
E −○ (R) = +0.80 V
E −○ (L) = −0.76 V
217
218
6 CHEMICAL EQUILIBRIUM
�e cell reaction is obtained by subtracting the le�-hand reduction halfreaction from the right-hand reduction half-reaction, a�er �rst multiplying the right-hand half-reaction by two so that the numbers of electrons
in both half-reactions are the same
2Ag+ (aq) + Zn(s) → 2Ag(s) + Zn2+ (aq)
�e standard cell potential is calculated as the di�erence of the two stan−
○
dard electrode potentials, [�D.�–���], E cell
= E −○ (R) − E −○ (L)
−
○
E cell
= (+0.80 V) − (−0.76 V) = +1.56 V
(ii) Following the same approach as part (i), and noting that the Pt(s) is an
‘inert metal’ that is only present to act as a source or sink of electrons,
the half-reactions for the Cd(s)|CdCl2 (aq)||HNO3 (aq)|H2 (g)|Pt(s) cell
and their electrode potentials are
R: 2H+ (aq) + 2e− → H2 (g)
L: Cd2+ (aq) + 2e− → Cd(s)
�e cell reaction (R − L) is therefore
E −○ (R) = 0 (by de�nition)
E −○ (L) = −0.40 V
2H+ (aq) + Cd(s) → H2 (g) + Cd2+ (aq)
and the standard cell potential is
−
○
E cell
= 0 − (−0.40 V) = +0.40 V
(iii) For the Pt(s)|K3 [Fe(CN)6 ](aq),K4 [Fe(CN)6 ](aq)||CrCl3 (aq)|Cr(s) cell
the reduction half-reactions are:
R: Cr3+ (aq) + 3e− → Cr(s)
E −○ (R) = −0.74 V
3−
−
4−
L: [Fe(CN)6 ] (aq) + e → [Fe(CN)6 ] (aq) E −○ (L) = +0.36 V
�e cell reaction is obtained by subtracting the le�-hand half-reaction
from the right-hand half-reaction, a�er �rst multiplying the right-hand
half reaction by three so that both half-reactions involve the same number
of electrons.
Cr3+ (aq) + 3[Fe(CN)6 ]4− (aq) → Cr(s) + 3[Fe(CN)6 ]3− (aq)
�e standard cell potential is
E�C.�(a)
−
○
E cell
= (−0.74 V) − (+0.36 V) = −1.10 V
(i) �e required reduction half-reactions are
R: Cu2+ (aq) + 2e− → Cu(s)
L: Zn2+ (aq) + 2e− → Zn(s)
E −○ (R) = +0.34 V
E −○ (L) = −0.76 V
�e cell reaction (R − L) generated from these reduction half-reactions is
Zn(s)+Cu2+ (aq) → Zn2+ (aq)+Cu(s) which is equivalent to the required
reaction. �e cell required is
Zn(s)�ZnSO4 (aq)��CuSO4 (aq)�Cu(s)
and the standard cell potential is
−
○
E cell
= E −○ (R) − E −○ (L) = (+0.34 V) − (−0.76 V) = +1.10 V
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(ii) �e required reduction half-reactions are:
R: 2AgCl(s) + 2e− → 2Ag(s) + 2Cl− (aq)
L: 2H+ (aq) + 2e− → H2 (g)
E −○ (R) = +0.22 V
E −○ (L) = 0 (by de�nition)
�e cell reaction (R − L) generated from these reduction half-reactions is
2AgCl(s) + H2 (g) → 2Ag(s) + 2Cl− (aq) + 2H+ (aq) which is equivalent
to the required reaction. �e required cell is:
Pt(s)�H2 (g)�HCl(aq)�AgCl(s)�Ag(s)
�e Pt(s) electrode is an ‘inert metal’ that acts as an electron source or
sink. Note that there is no interface between the two half cells because
both electrodes have a common electrolyte (HCl). �e standard cell potential is
−
○
E cell
= E −○ (R) − E −○ (L) = +0.22 V
(iii) �e required reduction half reactions are
R: O2 (g) + 4H+ (aq) + 4e− → 2H2 O(l)
L: �H+ (aq) + 4e− → 2H2 (g)
E −○ (R) = +1.23 V
E −○ (L) = 0 (by de�nition)
�e cell reaction (R − L) generated from these reduction half-reactions is
the required reaction, O2 (g) + 2H2 (g) → 2H2 O(l). �e required cell is
Pt(s)�H2 (g)�HCl(aq)�O2 (g)�Pt(s)
As in part (ii) the platinum electrode is an ‘inert metal’ and there is no
interface between the half cells because they have a common electrolyte.
�e standard cell potential is
−
○
E cell
= E −○ (R) − E −○ (L) = +�.�� V
An alternative combination of reduction half-reactions is
R: O2 (g) + 2H2 O(l) + 4e− → 4OH− (aq)
L: �H2 O(l) + 4e− → 4OH− (aq) + 2H2 (g)
E −○ (R) = +0.40 V
E −○ (L) = −0.83 V
which uses alkaline instead of acidic conditions. �e cell required is
Pt(s)�H2 (g)�NaOH(aq)�O2 (g)�Pt(s)
�e overall cell reaction (R − L) is the same and so, therefore, is the standard cell potential:
Solutions to problems
P�C.�
−
○
E cell
= (+0.40 V) − (−0.83V) = +1.23 V
(a) �e reaction of hydrogen and oxygen, 2H2 (g) + O2 (g) → 2H2 O(l), can
be broken down into the reduction half-reactions
R: O2 (g) + 4H+ (aq) + 4e− → 2H2 O(l)
L: 4H+ (aq) + 4e− → 2H2 (g)
E −○ (R) = +1.23 V
E −○ (L) = 0 (by de�nition)
219
220
6 CHEMICAL EQUILIBRIUM
�e standard cell potential is given by
−
○
E cell
= E −○ (R) − E −○ (L) = +1.23 V
(b) �e balanced chemical equation for the combustion of butane, C4 H10 (g)+
13
O2 (g) → 4CO2 (g) + 5H2 O(g) can be broken down into the reduction
2
half-reactions
R: 13
O2 (g) + 26H+ (aq) + 26e− → 13H2 O(l)
2
L: 4CO2 (g) + 26H+ (aq) + 26e− → C4 H10 (g) + 8H2 O(l)
�e standard electrode potential for the le�-hand reduction half-reaction
−
○
is not in the Resource section, so E cell
cannot be calculated from E −○ (R) −
−
○
−
○
E (L) as in part (a). Instead ∆ r G is calculated for the cell reaction by
−
○
�rst using standard Gibbs energies of formation and then using E cell
=
−
○
−
○
−∆ r G �νF [�C.�–���] to calculate E cell . Note from the above half-reactions
that ν = 26.
∆ r G −○ = 4∆ f G −○ (CO2 , g) + 5∆ f G −○ (H2 O, l) − ∆ f G −○ (C4 H10 , g)
= 4 × (−394.36 kJ mol−1 ) + 5 × (−237.13 kJ mol−1 )
Hence
P�C.�
− (−17.03 kJ mol−1 ) = −2746.06 kJ mol−1
−
○
E cell
=−
∆ r G −○
−2746.06 × 103 J mol−1
=−
= +1.09 V
νF
26 × (96485 C mol−1 )
�e reduction half-reactions for the cell are
R: Q(aq) + 2H+ (aq) + 2e− → QH2 (aq)
L: Hg2 Cl2 (s) + 2e− → 2Hg(l) + 2Cl− (aq)
E −○ (R) = +0.6994 V
E −○ (L) = +0.27 V
for which ν = 2. �e value of E −○ (L) is taken from the Resource section. �e cell
reaction (R − L) is
Q(aq) + 2H+ (aq) + 2Hg(l) + 2Cl− (aq) → QH2 (aq) + Hg2 Cl2 (s)
−
○
E cell
= E −○ (R) − E −○ (L) = (+0.6994 V) − (+0.27 V) = +0.4294 V
Noting that ν = 2 and that a J = 1 for pure solids and liquids, the Nernst
equation is
a QH2
RT
−
○
E cell = E cell
−
ln �
�
2 a2
2F
aQ aH
+ Cl−
Taking a QH2 = a Q , because Q and QH2 are present at the same concentration,
and a H+ = a Cl− gives:
RT
1
2RT
2RT
−
○
−
○
ln � 4 � = E cell
+
ln a H+ = E cell
+
ln 10 × log a H+
2F
a H+
F
F
2RT ln 10
−
○
= E cell
−
× pH
F
−
○
E cell = E cell
−
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
where pH = − log a H+ and ln x = ln 10 × log x (from inside the front cover)
have been used. Rearranging for pH gives:
F
�E −○ − E cell �
2RT ln 10 cell
96485 C mol−1
=
2 × (8.3145 J K−1 mol−1 ) × (298 K) × ln 10
× [(+0.4294 V) − (+0.190 V)] = 2.0
pH =
6D Electrode potentials
Answer to discussion questions
D�D.�
�is is discussed in Impact ��.
Solutions to exercises
E�D.�(a)
�e reduction half-reactions for the given cell are
R:
L:
Ag+ (aq) + e− → Ag(s)
AgI(s) + e− → Ag(s) + I− (aq)
�e cell reaction (R − L) is Ag+ (aq) + I− (aq) → AgI(s), with ν = 1. �e equi−
○
librium constant for this reaction is calculated from E cell
using [�C.�–���],
−
○
E cell = (RT�νF) ln K. Rearranging gives
ln K =
νF −○
1 × (96485 C mol−1 )
E cell =
× (0.9509 V) = 37.0...
RT
(8.3145 J K−1 mol−1 ) × (298.15 K)
where 1 V = 1 J C−1 is used. Hence K = 1.18... × 1016 .
�e dissolution reaction, AgI(s) → Ag+ (aq) + I− (aq), corresponds to the reverse of the cell reaction as written above. �e required equilibrium constant
is therefore the reciprocal of the one just calculated
E�D.�(a)
K diss =
1
= 8.445 × 10−17
1.18... × 1016
(i) �e reduction half-reactions for the speci�ed cell and their corresponding
electrode potentials from the Resource section are
R: Cu2+ (aq) + 2e− → Cu(s)
L: 2Ag+ (aq) + 2e− → 2Ag(s)
�e overall cell reaction is
E −○ (R) = +0.34 V
E −○ (L) = +0.80 V
Cu2+ (aq) + 2Ag(s) → Cu(s) + 2Ag+ (aq)
�e standard cell potential is
ν=2
−
○
E cell
= E −○ (R) − E −○ (L) = (+0.34 V) − (+0.80 V) = -�.�� V
221
222
6 CHEMICAL EQUILIBRIUM
�e reaction Gibbs energy is related to the cell potential according to
−
○
[�C.�–���], ∆ r G −○ = −νFE cell
. �erefore
−
○
∆ r G −○ = −νFE cell
= −2 × (96485 C mol−1 ) × (−0.46 V) = 88.7... kJ mol−1
= +89 kJ mol−1
�e standard reaction enthalpy is calculated using standard enthalpies of
formation from the Resource section, noting that elements in their reference states have ∆ f H −○ = 0.
∆ r H −○ = 2∆ f H −○ (Ag+ , aq) − ∆ f H −○ (Cu2+ , aq)
= 2 × (+105.58 kJ mol−1 ) − (+64.77 kJ mol−1 ) = +146.39 kJ mol−1
(ii) �e standard entropy change of reaction is obtained from ∆ r G −○ and ∆ r H −○
using [�D.�–��], ∆ r G −○ = ∆ r H −○ − T∆ r S −○ . Rearranging for ∆ r S −○ gives
∆ r H −○ − ∆ r G −○ (146.39 × 103 J mol−1 ) − (88.7... × 103 J mol−1 )
=
T
298 K
−1
2
−1
= +1.93... × 10 J K mol
∆ r S −○ =
�e value of ∆ r G −○ at ��� K is then calculated using [�D.�–��], ∆ r G −○ =
∆ r H −○ − T∆ r S −○ , assuming that ∆ r H −○ and ∆ r S −○ do not vary signi�cantly
with temperature over this range
∆ r G −○ = (+146.39 × 103 J mol−1 ) − (308 K) × (+1.93... × 102 J K−1 mol−1 )
E�D.�(a)
= +87 kJ mol−1
Assuming that the mercury forms Hg2 SO4 (s) in the reaction, the required
reduction half-equations and the corresponding standard electrode potentials
are
R: Zn2+ (aq) + 2e− → Zn(s)
L: Hg2 SO4 (s) + 2e− → 2Hg(l) + SO2−
4 (aq)
E −○ (R) = −0.76 V
E −○ (L) = +0.62 V
�e cell reaction is Zn2+ (aq) + SO2−
4 (aq) + 2Hg(l) → Zn(s) + Hg2 SO4 (s), and
the standard cell potential is
−
○
E cell
= E −○ (R) − E −○ (L) = (−0.76 V) − (+0.62 V) = −1.38 V
−
○
�e negative value of E cell
indicates that the cell reaction as written will not be
spontaneous. �is means that no , mercury cannot produce zinc metal from
aqueous zinc sulfate under standard conditions.
E�D.�(a)
(i) �e following electrodes are combined
R: Sn4+ (aq) + 2e− → Sn2+ (aq)
L: Sn2+ (aq) + 2e− → Sn(s)
E −○ (R) = +0.15 V
E −○ (L) = −0.14 V
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e overall cell reaction (R−L) is therefore Sn4+ (aq)+Sn(s) → 2Sn2+ (aq),
which is the required reaction, and has ν = 2. �e standard cell potential
−
○
is given by [�D.�–���], E cell
= E −○ (R) − E −○ (L)
−
○
E cell
= (+0.15 V) − (−0.14 V) = +0.29V
�e relationship between the equilibrium constant and the standard cell
−
○
potential is given by [�C.�–���], E cell
= (RT�νF) ln K. Rearranging gives
ln K =
νF −○
2 × (96485 C mol−1 )
E cell =
× (+0.29 V) = 22.5...
RT
(8.3145 J K−1 mol−1 ) × (298 K)
where 1 V = 1 J C−1 is used. Hence K = 6.4 × 109 .
(ii) �e following electrodes are combined
R: 2AgCl(s) + 2e− → 2Ag(s) + 2Cl− (aq)
L: Sn2+ (aq) + 2e− → Sn(s)
E −○ (R) = +0.22 V
E −○ (L) = −0.14 V
�e cell reaction is 2AgCl(s) + Sn(s) → 2Ag(s) + 2Cl− (aq) + Sn2+ (aq)
which is equivalent to the required reaction, and has ν = 2. �erefore,
using the same equations as in part (i)
−
○
E cell
= E −○ (R) − E −○ (L) = (+0.22 V) − (−0.14 V) = +0.36 V
ln K =
νF −○
2 × (96485 C mol−1 )
E cell =
× (+0.36 V) = 28.0...
RT
(8.3145 J K−1 mol−1 ) × (298 K)
Hence K = 1.5 × 1012
Solutions to problems
P�D.�
�e given reaction can be broken down into the following reduction half equations
R:
L:
2Fe3+ (aq) + 2e− → 2Fe2+ (aq)
Ag2 CrO4 (s) + 2e− → 2Ag(s) + CrO2−
4 (s)
where the K+ and Cl− spectator ions have been ignored. �ese half-equations
show that ν = 2 for the given reaction.
(a) �e standard potential is calculated from the standard reaction Gibbs
−
○
energy using [�C.�–���], E cell
= −∆ r G −○ �νF.
−
○
E cell
=−
−∆ r G −○
−62.5 × 103 J mol−1
=−
= 0.323... V = +0.324 V
νF
2 × (96485 C mol−1 )
−
○
(b) �e standard potential of the Ag2 CrO4 /Ag,CrO2−
4 couple, equal to E (L)
−
○
of the cell considered above, is calculated from E cell and the known value
−
○
of E −○ (R) using [�D.�–���], E cell
= E −○ (R) − E −○ (L). �e value of E −○ (R),
−
○
3+
2+
in this case E (Fe �Fe ), is taken from the Resource section.
−
○
E −○ (L) = E cell
− E −○ (R) = (+0.77 V) − (+0.323... V) = +0.45 V
223
224
6 CHEMICAL EQUILIBRIUM
P�D.�
+
(a) �e equilibrium HCO−3 (aq) � CO2−
3 (aq) + H (aq) is broken down into
the following reduction half-reactions
R: HCO−3 (aq) + e− → 12 H2 (g) + CO2−
3 (aq)
L: H+ (aq) + e− → 12 H2 (g)
�e standard cell potential for this cell is given by
−
○
−
○
+
E cell
= E −○ (R) − E −○ (L) = E −○ (HCO−3 �CO2−
3 , H2 ) − E (H �H2 )
�e standard electrode potential of the H+ /H2 electrode is zero by de�−
○
−
○
nition, so it follows that E −○ (HCO−3 �CO2−
3 , H2 ) = E cell . �e value of E cell
−
○
−
○
is calculated using [�C.�–���], E cell = −∆ r G �νF, noting that ν = 1. �e
value of ∆ r G −○ is calculated using the data in the question, noting from
Section �D.�(a) on page �� that ∆ f G −○ (H+ , aq) = 0.
−
○
−
∆ r G −○ = ∆ f G −○ (CO2−
3 , aq) − ∆ f G (HCO3 , aq)
Hence
= (−527.81 kJ mol−1 ) − (−586.77 kJ mol−1 ) = +58.96 kJ mol−1
−
○
E cell
=−
∆ r G −○
58.96 × 103 J mol−1
=−
= −0.6111 V
νF
1 × (96485 C mol−1 )
As shown above, this is equal to E −○ (HCO−3 �CO2−
3 , H2 ).
(b) �e reaction Na2 CO3 (aq) + H2 O(l) → NaHCO3 (aq) + NaOH(aq) is broken down into the following reduction half-equations, in which the Na+
counterions are ignored because they play no part in the reaction. �e
value of E −○ (L) is as calculated in part (a), and E −○ (R) is taken from the
Resource section.
R: H2 O(l) + e− → 12 H2 (g) + OH− (aq)
L: HCO−3 (aq) + e− → 12 H2 (g) + CO2−
3 (aq)
�e standard cell potential is given by
E −○ (R) = −0.83 V
E −○ (L) = −0.611... V
−
○
E cell
= E −○ (R)−E −○ (L) = (−0.83 V)−(−0.611... V) = −0.218... V = −0.22 V
(c) �e cell reaction for the cell considered in part (b) is
−
−
CO2−
3 (aq) + H2 O(l) → HCO3 (aq) + OH (aq)
ν=1
It is assumed that a H2 O = 1 because solvent water is close to being in its
standard state. �erefore the Nernst equation is
−
○
E cell = E cell
−
RT � a HCO−3 a OH− �
ln
F
� a CO2−
�
3
(d) �e standard cell potential corresponds to all species involved in the cell
reaction, which includes OH− , being present at unit activity. �is means
that the pH will need to be approximately ��, in order to give a OH− = 1. At
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
pH �.�, the concentration of OH− will be lower than at pH ��, which will
mean that the cell reaction as written above will have a greater tendency
to move in the forward direction. As a result E cell is predicted to be larger
at pH �.� than when a OH− = 1.
Assuming that the activities of all other species remain the same, the
change in cell potential on going from a OH− = 1 to pH � is
�
� �
�
− × 1 ��
a
� −○
�
RT � a HCO−3 a OH− ��
� − �E −○ − RT ln � HCO3
�
∆E cell = �
E
−
ln
� cell
� � cell
�
2−
F
F
a
� a CO2−
�
�
�
�
�
�
�
CO
3
3
�
� �
�
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
E cell at pH=�
E cell at a OH− =�
RT
RT ln 10
=−
ln a OH− = −
log a OH−
F
F
where ln x = ln 10 × log x from inside the front cover is used. To relate
a OH− to the pH, use the relation K w = a H+ a OH− so that
a OH− =
P�D.�
Kw
a H+
hence
log a OH− = log K w − log a H+ = −pK w + pH
where pH = − log a H+ and pK w = − log K w . Taking K w = 1.00 × 10−14 , or
pK w = 14.0, the change in cell potential when the pH is changed to � is
therefore
RT ln 10
∆E cell = −
(pH − pK w )
F
(8.3145 J K−1 mol−1 ) × (298 K) × ln 10
=−
× (−14.0 + 7.0)
(96485 C mol−1 )
= +0.4139 V
�erefore the cell potential has increased on going from a OH− = 1 to pH �.
−
○
−
○
�e relationship between ∆ r S −○ and E cell
is given by [�C.�–���], dE cell
�dT =
−
○
−
○
∆ r S �νF. If it is assumed that ∆ r S is independent of temperature over the
−
○
range of interest, integration of dE cell
= (∆ r S −○ �νF)dT between T1 and T2 gives
∆ r S −○
(T2 − T1 )
νF
−
○
where E cell
(T) is the potential at temperature T. �is equation is conveniently
−
○
written as ∆ r S −○ = νF∆E cell
�∆T.
−
○
−
○
E cell
(T2 ) − E cell
(T1 ) = −
∆ r S −○ = νF ×
−
○
∆E cell
∆T
= 4×(96485 C mol−1 )×
(+1.2251 V)−(+1.2335 V)
= −324 J K−1 mol−1
(303 K)−(293 K)
�e standard reaction enthalpy is then calculated from [�D.�–��], ∆ r G −○ =
−
○
∆ r H −○ − T∆ r S −○ , with ∆ r G −○ being given by [�C.�–���], ∆ r G −○ = −νFE cell
.
−
○
∆ r H −○ = ∆ r G −○ + T∆ r S −○ = −νFE cell
+ T∆ r S −○
= −4 × (96485 C mol−1 ) × (+1.2335 V)
+ (293 K) × (−3.24... × 102 J K−1 mol−1 ) = −571 kJ mol−1
225
226
6 CHEMICAL EQUILIBRIUM
−
○
In calculating ∆ r H −○ , the value of E cell
at 293 K has been used. However, because
−
○
∆ r S has been assumed to be constant over the temperature range, the data at
��� K will give the same value for ∆ r H −○ .
Solutions to integrated activities
I�.�
�e reduction half-reactions, and the overall cell reaction, for the speci�ed cell
(R − L) are:
R:
L:
R − L:
AgCl(s) + e− → Ag(s) + Cl− (aq)
H+ (aq) + e− → 12 H2 (g)
AgCl(s) + 12 H2 (g) → Ag(s) + Cl− (aq) + H+ (aq)
ν=1
Noting that a J = 1 for pure solids and that in this cell a H2 = 1 because the
hydrogen is at standard pressure, the Nernst equation is
−
○
E cell = E cell
−
RT
ln (a Cl− a H+ )
F
In addition, the base B and its conjugate acid are in equilibrium:
BH+ (aq) � B(aq) + H+ (aq)
Ka =
a B a H+
a BH+
�e expression for K a is rearranged to give a H+ = K a a BH+ �a B and this is substituted into the Nernst equation to give
−
○
E cell = E cell
−
RT
RT
a Cl− a BH+ K a
−
○
ln (a Cl− a H+ ) = E cell
−
ln �
�
F
F
aB
Replacing activities by a J = γ J (b J �b −○ ) [�F.��–���] gives
−
○
E cell = E cell
−
RT
(γ Cl− b Cl− �b −○ )(γ BH+ b BH+ �b −○ )K a
ln �
�
F
(γ B b B �b −○ )
In this case b Cl− = b BH+ = b B so the Nernst equation simpli�es to
−
○
E cell = E cell
−
RT
γ Cl− γ BH+ bK a
RT
γ 2 bK a
−
○
ln �
× −○ � = E cell
−
ln � ± −○ �
F
γB
b
F
b
where the mean activity coe�cient of the BH+ and Cl – ions is given by [�F.��–
���], γ± = (γ Cl− γ BH+ )1�2 and the neutral base B is assumed to be an ideal solute
so that γ B = 1. Noting from inside the front cover that ln x = ln 10 log x, the
Nernst equation becomes
RT ln 10
γ 2 bK a
log � ± −○ �
F
b
RT
ln
10
b
−
○
= E cell
−
�2 log γ± + log � −○ � − pK a �
F
b
−
○
E cell = E cell
−
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
where pK a = − log K a has been used. Next the Davies equation [�F.��b–���],
log γ± = −A�z+ z− �I 1�2 �(1+BI 1�2 )+CI, is used to substitute for log γ± . �e ionic
strength I is given by [�F.��–���], I = 12 ∑ i z 2i (b i �b −○ ), where z i is the charge
on ion species i and the sum extends over all the ions present in the solution.
In this case, b BH+ = b Cl− = b, and b H+ is neglected because it will be much
smaller on account of the equilibrium involving the base B. �erefore the ionic
strength is
2
2
−
○
I = 12 �z BH
= 12 �12 × b + (−1)2 × b� �b −○ = b�b−○
+ b + z Cl− b� �b
and therefore
log γ± = −
A�z BH+ × z Cl− �I 1�2
A(b�b −○ )1�2
b
+
CI
=
−
+ C � −○ �
1�2
−
○
1�2
b
1 + BI
1 + B(b�b )
Substitution of this expression into the Nernst equation derived above gives
−
○
E cell = E cell
−
RT ln 10
A(b�b−○ )1�2
b
b
�2 �−
+ C � −○ �� + log � −○ � − pK a �
−
○
1�2
F
b
b
1 + B(b�b )
which rearranges to
−
○
F(E cell − E cell
)
2A(b�b−○ )1�2
b
b
=
− 2C � −○ � − log � −○ � + pK a
RT ln 10
b
b
1 + B(b�b −○ )1�2
��� � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � �
y
De�ning (b�b −○ )1�2 as x and the le�-hand side as y, and introducing A = 0.5091
gives
1.0182x
− 2Cx 2 − 2 log x + pK a
y=
1 + Bx
which is �tted to the data using mathematical so�ware to give the following values for the parameters: B = 2.54 , C = −0.204 , and pK a = 6.74 . �ese values
have been used to draw the line on the graph shown in Fig. �.�.
b�mmol kg−1
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
E cell �V (b�b−○ )1�2
0.744 52
0.100
0.728 53
0.141
0.719 28
0.173
0.713 14
0.200
0.708 09
0.224
0.703 80
0.245
0.700 59
0.265
0.697 90
0.283
0.695 71
0.300
0.693 38
0.316
y
8.823 73
8.553 44
8.397 08
8.293 30
8.207 94
8.135 42
8.081 16
8.035 69
7.998 67
7.959 29
227
6 CHEMICAL EQUILIBRIUM
8.8
−
○
F(E cell − E cell
)�RT ln 10
228
8.6
8.4
8.2
8.0
0.05
0.10
0.15
Figure 6.3
I�.�
0.20
(b�b )
−
○ 1�2
0.25
0.30
0.35
From Impact � the reaction for the hydrolysis of ATP to ADP and inorganic
phosphate P−i is
ATP(aq) + H2 O(l) → ADP(aq) + P−i (aq) + H3 O+ (aq)
Under biological standard conditions, that is, pH = 7, the standard reaction
Gibbs energy at �� ○ C is given in Impact � as ∆ r G ⊕ = −31 kJ mol−1 .
In an environment in which pH = 7.0 and the ATP, ADP and P−i concentrations
are all �.� mmol dm−3 , the reaction Gibbs energy is given by [�A.��–���],
∆ r G = ∆ r G ⊕ + RT ln Q ⊕
where Q ⊕ is the reaction quotient calculated relative to the biological standard
state. Because pH is de�ned by pH = − log a H3 O+ , pH � corresponds to a H3 O+ =
10−7 so that when computing Q ⊕ the activity of H� O+ is measured relative to
an activity of 10−7 rather than an activity of � as is usually the case. In practice
this means that a H3 O+ is replaced by (a H3 O+ �10−7 ) in the expression for Q ⊕ .
For the ATP hydrolysis reaction this gives
∆ r G = ∆ r G ⊕ + RT ln �
a ADP × a P−i × (a H3 O+ �10−7 )
a ATP × a H2 O
�
Water is a pure liquid so a H2 O = 1, and for the environment speci�ed in the
question, pH = 7 so a H3 O+ = 10−7 . For the other species activities are approxi-
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
mated by concentrations according to a J = [J]�c −○ where c −○ = 1 mol dm−3 .
∆ r G = ∆ r G ⊕ + RT ln �
= ∆ r G ⊕ + RT ln �
([ADP]�c −○ )([P−i ]�c −○ )(10−7 �10−7 )
�
([ATP]�c −○ )
[ADP][P−i ]
�
[ATP]c −○
= (−31 × 103 J mol−1 ) + (8.3145 J K−1 mol−1 ) × ([37 + 273.15] K)
× ln �
(1.0 × 10−3 mol dm−3 ) × (1.0 × 10−3 mol dm−3 )
�
(1.0 × 10−3 mol dm−3 ) × (1 mol dm−3 )
= −49 kJ mol−1
�is is to be compared with the value under standard biological conditions,
which is −31 kJ mol−1 , and also with the value under the usual standard conditions. �e di�erence between ∆ r G −○ and ∆ r G ⊕ is that the former has a H+ = 1
and the latter has a H+ = 10−7 . Given that
∆ r G = ∆ r G −○ + RT ln �
a ADP a Pi a H3 O+
�
a ATP a H2 O
setting all the activities to � except for that for H� O+ which is set to 10−7 gives
∆r G ⊕
∆ r G ⊕ = ∆ r G −○ + RT ln 10−7
hence
∆ r G −○ = ∆ r G ⊕ − RT ln 10−7
= (−31 × 103 J mol−1 )−(8.3145 J K−1 mol−1 )×([37 + 273.15] K)×ln 10−7
I�.�
= +11 kJ mol−1
A bacterium could potentially oxidise ethanol to ethanal, ethanoic acid, or
CO2 (g), while nitrate, NO−3 (aq), could potentially be reduced to a number
of possible species including NO2 (g), NO−2 (aq), NO(g), N2 (g), or NH+4 (aq).
Assuming complete oxidation of ethanol to CO2 and complete reduction of
NO−3 to NH+4 the reduction half-reactions are
R:
L:
NO−3 (aq) + 10H+ + 8e− → NH+4 (aq) + 3H2 O(l)
2CO2 (g) + 12H+ + 12e− → CH3 CH2 OH(aq) + 3H2 O(l)
�e right-hand half reaction is multiplied by three and the le�-hand half reaction by two in order that both involve the same number of electrons. �e
overall reaction is
2CH3 CH2 OH(aq) + 3NO−3 (aq) + 6H+ (aq) → 4CO2 (g) + 3NH+4 (aq) + 3H2 O(l)
229
230
6 CHEMICAL EQUILIBRIUM
�e data in the Resource section is used to calculate ∆ r G −○ for this reaction:
∆ r G −○ = 4∆ f G −○ (CO2 , g) + 3∆ f G −○ (NH+4 , aq) + 3∆ f G −○ (H2 O, l)
− 2∆ f G −○ (CH3 CH2 OH, aq) − 3∆ f G −○ (NO−3 , aq) − 6∆ f G −○ (H+ , aq)
= 4 × (−394.36 J mol−1 ) + 3 × (−79.31 J mol−1 ) + 3 × (−237.13 J mol−1 )
− 2 × (−174.78 J mol−1 ) − 3 × (−108.74 J mol−1 ) = −1851 kJ mol−1
�e negative value of ∆ r G −○ indicates that the reaction is exergonic, so yes , a
bacterium could evolve to use this reaction to drive endergonic processes such
as the formation of ATP for use in cellular processes.
�e calculation is valid under standard conditions, which includes a H+ = 1
(pH = 0). As explained in Impact � on the website of this text, pH = 0 is
not normally appropriate for biological conditions so it is common to adopt
the biological standard state in which pH = 7.0. �e reaction Gibbs energy
for the oxidation of ethanol by nitrate under standard biological conditions is
calculated by using the appropriate value of a H+ in [�A.��–���], ∆ r G = ∆ r G −○ +
RT ln Q, leaving all other species with a J = 1. �e reaction consumes six
6
moles of H+ so under standard biological conditions Q = 1�a H
+ . Noting from
inside the front cover that ln x = ln 10 log x, and also that pH = − log a H+ , and
assuming T = 298 K, gives
1
∆ r G = ∆ r G −○ + RT ln Q = ∆ r G −○ + RT ln � 6 �
a H+
= ∆ r G −○ − 6RT ln 10 log(a H+ ) = ∆ r G −○ + 6RT ln 10×pH
= (−1.85... × 106 J mol−1 ) + 6×(8.3145 J K−1 mol−1 )×(298 K)×ln 10×7
= −1611 kJ mol−1
�us the reaction remains exergonic under standard biological conditions.
I�.�
(a) �e standard reaction enthalpy is found using the van ’t Ho� equation
[�B.�–���]:
d ln K ∆ r H −○
=
dT
RT 2
−
d ln K
∆ r H −○
=
d(1�T)
R
− ln K =
∆ r H −○ 1 ∆ r S −○
−
R T
R
which can also be written
�e second form implies that a graph of − ln K against 1�T should be a
straight line of slope ∆ r H −○ �R, from which ∆ r H −○ can be determined. �e
value of ∆ r S −○ is found by combining ∆ r G −○ = ∆ r H −○ − T∆ r S −○ [�D.�–��]
and ∆ r G −○ = −RT ln K [�A.��–���]. Equating these expressions for ∆ r G −○
gives
−RT ln K = ∆ r H −○ − T∆ r S −○
hence
Assuming that ∆ r H −○ and ∆ r S −○ do not very signi�cantly over the temperature range of interest this equation implies that a plot of − ln K against
1�T should be a straight line of intercept −∆ r S −○ �R, from which ∆ r S −○ can
be determined; such a plot is shown in Fig. �.�. �e plot will have a slope
of ∆ r H −○ �R, as already deduced above.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
T�K
233
248
258
268
273
280
288
295
303
K
4.13 × 108
5.00 × 107
1.45 × 107
5.37 × 106
3.20 × 106
9.62 × 105
4.28 × 105
1.67 × 105
6.02 × 104
1�(T�K) − ln K
0.004 29 −19.8
0.004 03 −17.7
0.003 88 −16.5
0.003 73 −15.5
0.003 66 −15.0
0.003 57 −13.8
0.003 47 −13.0
0.003 39 −12.0
0.003 30 −11.0
−10
− ln K
−15
−20
0.0032
Figure 6.4
0.0036
0.0040
1�(T�K)
0.0044
�e data fall on a reasonable straight line, the equation of which is
− ln K = −8 787 × 1�(T�K) + 17.62
∆ r H −○ �R is determined from the slope
∆ r H −○ = R × (slope × K) = (8.3145 J K−1 mol−1 ) × (−8787 K)
= −73.0... kJ mol−1 = −73.1 kJ mol−1
−∆ r S −○ �R is determined from the intercept
∆ r S −○ = −R × intercept = −(8.3145 J K−1 mol−1 ) × (+17.62)
= −1.46... × 102 J K−1 mol−1 = −147 J K−1 mol−1
(b) �e standard reaction enthalpy for the reaction 2ClO(g) → (ClO)2 (g) is
expressed in terms of standard enthalpies of formation
∆ r H −○ = ∆ f H −○ [(ClO)2 , g] − 2∆ f H −○ (ClO, g)
231
232
6 CHEMICAL EQUILIBRIUM
Hence
∆ f H −○ [(ClO)2 , g] = ∆ r H −○ + 2∆ f H −○ (ClO, g)
= (−73.0... kJ mol−1 ) + 2 × (+101.8 kJ mol−1 )
= +131 kJ mol−1
Similarly
−
○
−
○
∆ r S −○ = S m
[(ClO)2 , g] − 2S m
(ClO, g)
Hence
−
○
−
○
Sm
[(ClO)2 , g] = ∆ r S −○ + 2S m
(ClO, g)
= (−1.46... × 102 J K−1 mol−1 ) + 2 × (226.6 J K−1 mol−1 )
I�.�
= 307 J K−1 mol−1
(a) �e ionic strength is given by [�F.��–���], I = 12 �b+ z+2 + b− z−2 � �b −○ , where
z+ and z− are the charges on the ions. For the CuSO4 compartment,
z+ = 2, z− = −2, and b+ = b− = b CuSO4 :
I = 12 �b+ z+2 + b− z−2 � �b−○ = 12 �b CuSO4 × (+2)2 + b CuSO4 × (−2)2 � �b −○
= 4(b CuSO4 �b−○ ) = 4 ×
1.00 × 10−3 mol kg−1
= 4.00 × 10−3
1 mol kg−1
Because the charges are the same for ZnSO� it follows that I = 4(b ZnSO4 �b−○ )
= 1.20 × 10−2 .
(b) According to the Debye–Hückel limiting law (Section �F.�(b) on page
���), the mean activity coe�cient is given by [�F.��–���], log γ± = −A�z+ z− �I 1�2 ,
where A = 0.509 for aqueous solutions at �� ○ C. For the CuSO4 solution
log γ±,CuSO4 = −(0.509) × �(2) × (−2)� × (4.00 × 10−3 )1�2 = −0.128...
Hence γ±,CuSO4 = 10−0.128 ... = 0.743... = 0.743 . For the ZnSO4 solution
log γ±,ZnSO4 = −(0.509) × �(2) × (−2)� × (1.20 × 10−2 )1�2 = −0.223...
Hence γ±,ZnSO4 = 10−0.223 ... = 0.598... = 0.598 .
(c) Noting that pure solids have a J = 1 and writing the activities of ions in
solution as a = γ± (b�b−○ ), the reaction quotient for the reaction
Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s)
is given by
Q=
=
a Zn2+ γ±,ZnSO4 (b Zn2+ �b −○ ) γ±,ZnSO4 b Zn2+
=
=
×
a Cu2+ γ±,CuSO4 (b Cu2+ �b −○ ) γ±,CuSO4 b Cu2+
0.598... 3.00 × 10−3 mol kg−1
×
= 2.41... = �.��
0.743... 1.00 × 10−3 mol kg−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(d) �e reaction is thought of as being composed of the reduction half-reactions
R:
L:
Cu2+ (aq) + 2e− → Cu(s)
Zn2+ (aq) + 2e− → Zn(s)
which show that ν = 2 for this reaction. �e standard cell potential is
−
○
calculated from ∆ r G −○ using [�C.�–���], E cell
= −∆ r G −○ �νF:
−
○
E cell
=−
∆ r G −○
−212.7 × 103 J mol−1
=−
= +1.102 V
νF
2 × (96485 C mol−1 )
Note that 1 J C−1 = 1 V.
(e) �e cell potential is given by the Nernst equation [�C.�–���]:
RT
ln Q
νF
(8.3145 J K−1 mol−1 ) × ([25 + 273.15] K)
= (+1.102 V) −
× ln(2.41...)
2 × (96485 C mol−1 )
= +1.09 V
−
○
E cell = E cell
−
I�.��
�e reaction for the autoprotolysis of liquid water is
H2 O(l) → H+ (aq) + OH− (aq)
�is reaction is split into the reduction half-reactions
R:
L:
H2 O(l) + e− → 12 H2 (g) + OH− (aq)
H+ (aq) + e− → 12 H2 (g)
Because the standard electrode potential for the le�-hand half-reaction is zero
by de�nition, the standard cell potential for this cell is equal to the standard
electrode potential of the H2 O/H2 ,OH− electrode. �e equilibrium constant
−
○
K w for the cell reaction is then given by [�C.�–���], E cell
= (RT�νF) ln K.
−
○
Rearranging for ln K and using ν = 1, K = K w , and E cell = E −○ (H2 O�H2 , OH− )
gives
F
ln K w =
× E −○ (H2 O�H2 , OH− )
RT
Noting from inside the front cover that ln x = ln 10 × log x, and also that pK w =
− log K w allows the above equation to be rewritten as
pK w = − log K w = −
ln K w
F
=−
× E −○ (H2 O�H2 , OH− )
ln 10
RT ln 10
�e task is therefore to �nd E −○ (H2 O�H2 , OH− ) from the given data. To do
this, the speci�ed cell is written in terms of its reduction half-reactions
R:
L:
AgCl(s) + e− → Ag(s) + Cl− (aq)
H2 O(l) + e− → 12 H2 (g) + OH− (aq)
233
234
6 CHEMICAL EQUILIBRIUM
�e cell reaction, which has ν = 1, is
AgCl(s) + 12 H2 (g) + OH− (aq) → Ag(s) + Cl− (aq)
Noting that a J = 1 for pure solids, and that in this cell a H2 = 1 because the
hydrogen is at standard pressure, the Nernst equation for the cell is
−
○
E cell = E cell
−
RT
RT
a Cl−
−
○
ln Q = E cell
−
ln �
�
νF
F
a OH−
Writing the activities as a = γ± (b�b −○ ), the Nernst equation becomes
−
○
E cell = E cell
−
RT
γ± (b Cl− �b −○ )
RT
b Cl−
−
○
ln �
� = E cell
−
ln �
�
−
○
F
γ± (b OH− �b )
F
b OH−
�e standard cell potential is split into contributions from the two electrodes
−
○
using [�D.�–���], E cell
= E −○ (R) − E −○ (L)
E cell = E −○ (AgCl�Ag, Cl− ) − E −○ (H2 O�H2 , OH− ) −
RT
b Cl−
ln �
�
F
b OH−
E −○ (H2 O�H2 , OH− ) = E −○ (AgCl�Ag, Cl− ) − E cell −
RT
b Cl−
ln �
�
F
b OH−
Hence
�is equation is used with b Cl− = 0.01125 mol kg−1 , b OH− = 0.0100 mol kg−1 ,
and the values of E cell and E −○ (AgCl�Ag, Cl− ) to calculate E −○ (H2 O�H2 , OH− )
at each temperature. �e relation derived earlier
pK w = −(F�RT ln 10)E −○ (H2 O�H2 , OH− )
is then used to calculate pK w . �e results are given in the following table.
θ�○ C T�K E cell �V E −○ (AgCl�Ag, Cl− )�V E −○ (H2 O�H2 , OH− )�V pK w
20.0 293.15 1.047 74
0.225 02
−0.825 70
14.20
25.0 298.15 1.048 64
0.222 30
−0.829 37
14.02
30.0 303.15 1.049 42
0.219 59
−0.832 91
13.85
−
○
To �nd ∆ r S −○ for the autoprotolysis, the relationship between E cell
and temper−
○
−
○
−
○
ature [�C.�–���], dE cell �dT = ∆ r S �νF is used. If ∆ r S is constant over the
−
○
temperature range this equation implies that a plot of E cell
against T should
−
○
be a straight line of slope ∆ r S −○ �νF. In this case E cell
for the autoprotolysis
reaction is equal to E −○ (H2 O�H2 , OH− ) as explained earlier. �e plot is shown
in Fig. �.�.
�e data fall on a good straight line, the equation of which is
E −○ (H2 O�H2 , OH− )�V = −7.229 × 10−4 × (T�K) − 0.6137
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E −○ (H2 O�H2 , OH− )�V
−0.825
−0.830
−0.835
292
294
296
298
T�K
300
302
304
Figure 6.5
∆ r S −○ �νF is determined from the slope
∆ r S −○ = νF × slope × V K−1
= 1 × (96485 C mol−1 ) × (−7.229 × 10−4 V K−1 )
= −68.5... J K−1 mol−1 = −68.6 J K−1 mol−1
�e standard enthalpy change for the autoprotolysis is calculated from [�D.�–
��], ∆ r G −○ = ∆ r H −○ − T∆ r S −○ , with ∆ r G −○ being given by [�C.�–���], ∆ r G −○ =
−
○
−
○
−νFE cell
. In this case E cell
= E −○ (H2 O�H2 , OH− ) and ν = 1. Using the value
○
for 25.0 C gives
∆ r H −○ = ∆ r G −○ + T∆ r S −○ = −νFE −○ (H2 O�H2 , OH− ) + T∆ r S −○
= −1 × (96485 C mol−1 ) × (−0.829...V) + (298.15 K) × (−68.5... J K−1 mol−1 )
= +59.6 kJ mol−1
235
7
7A
Quantum theory
The origins of quantum mechanics
Answers to discussion question
D�A.�
By wave-particle duality it is meant that in some experiments an entity behaves as a wave while in other experiments the same entity behaves as a particle. Electromagnetic radiation behaves as a wave in di�raction experiments
but it behaves as particulate photons in absorption and emission spectroscopy.
Electrons behave as waves in di�raction experiments, but as particles in the
photoelectric e�ect.
�e development of quantum theory is much concerned with the need to embrace this wave-particle duality and, as is explained in the following Topics,
this is exempli�ed by the introduction of the wavefunction to describe the
properties of a particles and the notion of ‘complementary variables’ such as
position and momentum.
D�A.�
�e ultimately unsuccessful classical approach to the description of black-body
radiation involved assuming that the radiation resulted from oscillating electric
charges in the walls of the body, and that each oscillator has the same average
energy as predicted by the equipartition principle. �is view results in the
ultra-violet catastrophe, in which the radiation increases without limit as the
wavelength becomes shorter.
Planck assumed two things: �rst, that the oscillators could only have energies
given by E = nhν, where ν is the frequency and n is 0, 1, 2, . . .; second, that the
probability of an individual oscillator having a particular energy is described
by the Boltzmann distribution. As the frequency of the oscillator or the value
of n increases, so does its energy and the Boltzmann distribution predicts that
such a state is less likely. In addition, the highest frequency oscillations may
not be excited at all, that is have n = 0, on the grounds that the resulting state
has too high an energy to be populated. Planck’s theory therefore avoids the
ultraviolet catastrophe by having no excitation of highest frequency oscillators.
Solutions to exercises
E�A.�(a)
If the power, P, is constant, the total energy emitted in time ∆t is P∆t. �e
energy of each emitted photon is E photon = hν = hc�λ. �e total number of
photons emitted in this time period is therefore the total energy emitted divided
238
7 QUANTUM THEORY
by the energy per photon
N = P∆t�E photon = P∆tλ�hc
�e conservation of linear momentum requires that the loss of a photon must
impart an equivalent momentum in the opposite direction to the glow-worm,
hence the total momentum p imparted to the glow-worm in time ∆t is
p = N p photon = N h�c = (P∆tλ�hc) × (h�λ) = P∆t�c
Because p = (mυ)glow-worm , the �nal speed of the glow-worm is
υ = P∆t�cmglow-worm
=
(0.10 W) × (10 y) × (3.1536 × 107 s y−1 )
= �� m s−1
(2.9979 × 108 m s−1 ) × (0.0050 kg)
Noting that the number of seconds in one year is
E�A.�(a)
365 × 24 × 60 × 60 = 3.1536 × 107
�e de Broglie relation is [�A.��–���], λ = h�p = h�(mυ). �erefore,
υ=
h
6.6261 × 10−34 J s
=
= 7.27 × 106 m s−1
m e λ (9.1094 × 10−31 kg) × (100 × 10−12 m)
�e kinetic energy acquired by an electron accelerated through a potential E is
eE: E k = 12 m e υ 2 = eE. Solving for the potential gives
E�A.�(a)
E�A.�(a)
E�A.�(a)
E=
m e υ 2 (9.1094 × 10−31 kg) × (7.27 × 106 m s−1 )2
=
= ��� V
2e
2 × (1.6022 × 10−19 C)
�e de Broglie relation is [�A.��–���] λ = h�p = h�(mυ).
υ=
h
6.6261 × 10−34 J s
=
= 2.4 × 10−2 m s−1
m e λ (9.1094 × 10−31 kg) × (3 × 10−2 m)
�e de Broglie relation is [�A.��–���] λ = h�p = h�(mυ). �erefore
λ=
h
α −1 h
137 × (6.6261 × 10−34 J s)
=
=
= ��� pm
m e αc
me c
(9.1094 × 10−31 kg) × (2.9979 × 108 m s−1 )
�e de Broglie wavelength is [�A.��–���], λ = h�p = h�(mυ)
(i)
(ii)
λ=
λ=
6.6261 × 10−34 J s
(1.0 × 10−3 kg) × (1.0 × 10−2 m s−1 )
= 6.6 × 10−29 m
6.6261 × 10−34 J s
= 6.6 × 10−36 m
(1.0 × 10−3 kg) × (100 × 103 m s−1 )
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(iii)
6.6261 × 10−34 J s
((4.00 × 10−3 kg mol−1 )�(6.0221 × 1023 mol−1 )) × (1.00 × 103 m s−1 )
= ��.� pm
λ=
E�A.�(a)
E�A.�(a)
Wien’s law [�A.�–���], λ max T = 2.9 × 10−3 m K, is rearranged to give the
wavelength at which intensity is maximised
λ max = (2.9 × 10−3 m K)�T = (2.9 × 10−3 m K)�(298 K) = 9.7 × 10−6 m
Assuming that the object is a black body is equivalent to assuming that Wien’s
law [�A.�–���], λ max T = 2.9 × 10−3 m K, holds. Using ν̃ = λ−1 , Wien’s law is
expressed in terms of the wavenumber of maximum intensity (ν̃ max )
T�ν̃ max = 2.9 × 10−3 m K
�is is rearranged to give the temperature
T = (2.9 × 10−3 m K) × ν̃ max
E�A.�(a)
= (2.9 × 10−3 m K) × (2000 × 102 m−1 ) = ��� K
Molar heat capacities of monatomic non-metallic solids obey the Einstein relation [�A.�a–���],
C V ,m (T) = 3R f E (T),
f E (T) = �
θ E 2 eθ E �2T
� � θ �T
�
T
e E −1
2
where the solid is at temperature T and is characterized by an Einstein temperature θ E . �us, for a solid at 298 K with an Einstein temperature of 2000 K
f E (298 K) = �
E�A.�(a)
2000 K 2 e(2000 K)�2(298 K)
� � (2000 K)�298 K
� = 5.49... × 10−2
298 K
e
−1
2
Hence, C V ,m (298 K) = (5.49 × 10−2 ) × 3R
�e energy of the quantum is given by the Bohr frequency condition [�A.�–
���], ∆E = hν, and the frequency is ν = 1�T. �e energy per mole is ∆E m =
N A ∆E.
(i) For T = 1.0 fs
∆E = (6.6261 × 10−34 J s)�(1.0 × 10−15 s) = 6.6 × 10−19 J
∆E m = (6.6... × 10−19 J) × (6.0221 × 1023 mol−1 ) = 4.0 × 102 kJ mol−1
(ii) For T = 10 fs
∆E = (6.6261 × 10−34 J s)�(10 × 10−15 s) = 6.6 × 10−20 J
∆E m = (6.6... × 10−20 J) × (6.0221 × 1023 mol−1 ) = �� kJ mol−1
239
240
7 QUANTUM THEORY
(iii) For T = 1.0 s
∆E = (6.6261 × 10−34 J s)�(1.0 s) = 6.6 × 10−34 J
∆E m = (6.6... × 10−34 J) × (6.0221 × 1023 mol−1 ) = 4.0 × 10−13 kJ mol−1
E�A.��(a) �e energy of a photon with wavelength λ is given by
E = hν = hc�λ = (6.6261 × 10−34 J s) × (2.9979 × 108 m s−1 )�λ
= (1.9825 × 10−25 J)�(λ�m)
�e energy per mole is then given by
E m = N A E = (0.11939 J mol−1 )�(λ�m)
Hence, the following table is drawn up
E�A.��(a)
λ�nm
E�zJ
(a)
600
330
(b)
550
360
(c)
400
496
E m �kJ mol−1
199
217
298
When a photon is absorbed by a free hydrogen atom, the law of conservation of
energy requires that the kinetic energy acquired by the atom is E k , the energy
of the absorbed photon. Assuming relativistic corrections are negligible the
kinetic energy is E k = E photon = 12 m H υ 2 . �e atom is accelerated to the speed
υ=�
2E photon 1�2
2N A E photon 1�2
� =�
�
mH
MH
2 × (6.0221 × 1023 mol−1 ) × E photon
=�
�
(1.0079 × 10−3 kg mol−1 )
= (3.45... × 1013 m s−1 ) × �E photon �J�
1�2
1�2
�e photon energies have been calculated in Exercise E�A.��(a), and thus the
following table is drawn up
υ�km s−1
λ�nm
E�zJ
(a)
600
330
(b)
550
360
20.8
(c)
400
496
24.4
19.9
E�A.��(a) �e energy emitted from a lamp at (constant) power P in a time interval ∆t is
P∆t. �e energy of a single photon of wavelength λ is E = hc�λ. Hence, the
total number of photons emitted in this time interval is the total energy emitted
divided by the energy per photon, N = P∆t�E photon = P∆tλ�hc. �us, for a
time interval of 1 s and a wavelength of 550 nm
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(i) P = 1 W
N=
(1 W) × (1 s)(550 × 10−9 m)
= 2.77 × 1018
(6.6261 × 10−34 J s) × (2.9979 × 108 m s−1 )
N=
(100 W) × (1 s)(550 × 10−9 m)
= 2.77 × 1020
(6.6261 × 10−34 J s) × (2.9979 × 108 m s−1 )
(ii) P = 100 W
E�A.��(a) As described in Section �A.� on page ���, photoejection can only occur if the
energy of the incident photon is greater than or equal to the work function of
the metal �. If this condition is ful�lled, the energy of the emitted photon is
given by [�A.��–���], E k = hν − Φ = hc�λ − Φ. To convert the work function to
Joules, multiply through by the elementary charge, as described in Section �A.�
on page ���,
Φ = (2.14 eV) × e = (2.14 eV) × (1.602 × 10−19 J eV−1 ) = 3.42... × 10−19 J
�
and since E k = 1�2m e υ 2 , υ = 2E k �m e
(i) For λ = 700 nm
hc (6.6261 × 10−34 J s) × (2.9979 × 108 m s−1 )
=
λ
700 × 10−9 m
−19
= 2.84... × 10 J
E photon =
�is is less than the threshold energy, hence no electron ejection occurs.
(ii) For λ = 300 nm
hc (6.6261 × 10−34 J s) × (2.9979 × 108 m s−1 )
=
λ
300 × 10−9 m
−19
= 6.62... × 10 J
E photon =
�is is greater than the the threshold frequency, and so photoejection can
occur. �e kinetic energy of the electron is,
E k = hc�λ − Φ = 6.62... × 10−19 J − 3.42... × 10−19 J = 3.19 × 10−19 J
�
2E photon �m e
�
= 2 × (3.19 × 10−19 J)�(9.109 × 10−31 kg) = 837 km s−1
υ=
Solutions to problems
P�A.�
A cavity approximates an ideal black body, hence the Planck distribution [�A.�a–
���], applies
8πhc
ρ(λ, T) =
λ 5 �e hc�λk T − 1�
241
242
7 QUANTUM THEORY
Because the wavelength range is small (5 nm), the energy density is approximated by
∆E(T) = ρ(λ, T)∆λ
Taking λ = 652.2 nm gives
and
hc (6.6261 × 10−34 J s) × (2.9979 × 108 m s−1 )
=
= 2.20... × 104 K
λk (652.5 × 10−9 m) × (1.3806 × 10−23 J K−1 )
8πhc 8π × (6.6261 × 10−34 J s) × (2.9979 × 108 m s−1 )
=
= 4.22... × 107 J m−4
λ5
(652.5 × 10−9 m)5
It follows that
∆E(T) = (4.22... × 107 J m−4 ) ×
(a)
(b)
P�A.�
=
0.211... J m−3
e(2.20 ...×104 K)�T − 1
∆E(298 K) =
∆E(3273 K) =
1
e(2.20 ...×104 K)�T − 1
× (5 × 10−9 m)
0.211... J m−3
= 1.54 × 10−33 J m−3
e(2.20 ...×104 K)�(298 K) − 1
0.211... J m−3
= 2.51 × 10−4 J m−3
e(2.20 ...×104 K)�(3273 K) − 1
As λ increases, hc�λkT decreases, and at very long wavelengths hc�λkT � 1.
Hence, the exponential can be expanded in a power series. Let x = hc�λkT,
then ex = 1 + x + 2!1 x 2 + 3!1 x 3 ..., and the Planck distribution becomes
ρ=
8πhc
λ 5 �1 + x + 2!1 x 2 + 3!1 x 3 ... − 1�
When x << 1 the second and higher order terms in x become negligibly small
compared to x. Consequently
lim ρ =
λ→∞
8πhc 8πhc
1
8πkT
= 5 �
�=
λ5 x
λ
hc�λkT
λ4
�is is the Rayleigh–Jeans law
P�A.�
Each data point is used to �nd a value for λ max T, and then the mean of these is
used with λ max T = hc�(4.965k) to �nd a value of h, h = (4.965k�c) (λ max T)mean .
�e following table is drawn up
θ� ○ C
T� K
1000
1500
2000
2500
3000
3500
1273
1773
2273
2773
3273
3773
2181
1600
1240
1035
878
763
m K) 2.776
2.837
2.819
2.870
2.874
2.879
λ max � nm
−3
λ max T�(10
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e mean is 2.84... × 10−3 m K with a standard deviation of 4.01... × 10−5 m K.
�erefore
P�A.�
h=
4.965 × (1.3806 × 10−23 J K−1 ) × (2.84... × 10−3 m K)
= 6.54 × 10−34 J s
2.9979 × 108 m s−1
�e total energy density is given by [�A.�–���],
E(T) = �
0
∞
ρ(λ, T) dλ = 8πhc �
0
∞
1
1
dλ
5
hc�λk
T −1
λ e
Let x = hc�λkT, or λ = hc�xkT. �en, dλ = −hc�x 2 kT dx
E(T) = 8πhc �
0
∞
∞ x3
1
1
hc
(kT)4
dx
=
8π
dx
�
(hc�xkT)5 ex − 1 x 2 kT
(hc)3 0 ex − 1
= 8π(kT)4 �(hc)3 × π 4 �15 = 8π 5 k 4 T 4 �15(hc)5
P�A.�
�is is the Stefan–Boltzmannn law, that the total energy density is proportional
to T 4 , with a constant of proportionality of 8π 5 k 4 �15(hc)5 .
Assuming the Sun approximates an ideal black body, Wien’s law, λ max T = 2.9 ×
10−3 m K, applies. Hence, λ max = (2.9×10−3 m K)�(5800 K) = ��� nm which
corresponds to blue-green .
7B Wavefunctions
Answers to discussion questions
D�B.�
�is is discussed in Section �B.�(b) on page ���.
D�B.�
�ese terms are best illustrated by referring to a one-dimensional system. �e
probability density P(x) is de�ned so that P(x) dx is the probability of �nding
the system between x and x + dx. �e value of the wavefunction itself, ψ(x),
is the probability amplitude. �e probability density is given in terms of this
amplitude as P(x) = ψ ∗ (x)ψ(x).
Solutions to exercises
E�B.�(a)
(i) �e function ψ = x 2 cannot be normalized as the area under ψ ∗ ψ = x 4
is in�nite when the integral is evaluated over all space. However, if the
function were restricted to a �nite region of space, it could be normalized.
(ii) �e function ψ = 1�x cannot be normalized as it goes to ∞ as x goes to
0 so the integral of ψ ∗ ψ over all space is in�nite. However, if the function
were restricted to a �nite region of space which did not include x = 0 it
could be normalized.
(iii) �e function ψ = exp(−x 2 ) can be normalized as ψ ∗ ψ = exp(−2x 2 )
goes to 0 at x = ±∞ so that the integral of ψ ∗ ψ over all space is �nite.
243
244
7 QUANTUM THEORY
E�B.�(a)
E�B.�(a)
A function is an acceptable wavefunction if it: (�) is not in�nite over a �nite
region; (�) is single-valued; (�) is continuous; (�) has a continuous �rst derivative. �e functions ψ = x 2 and ψ = exp(−x 2 ) satisfy all these conditions. �e
function ψ = 1�x satis�es these conditions provided that the wavefunction is
restricted to a region which excludes x = 0.
�e normalized wavefunction is ψ(x) = (2�L)1�2 sin(2πx�L), and so the probability density is P(x) = �ψ(x)�2 = (2�L) sin2 (2πx�L). �is is maximized when
sin2 (2πx�L) = 1, and so when sin(2πx�L) = ±1. �ese values occur when
2πx�L = π�2, 3π�2 and hence x = L�4, 3L�4 .
Nodes occur when the wavefunction goes through zero: sin(2πx�L) = 0. �e
sine function is zero when 2πx�L = π, hence x = L�2 . �e wavefunction goes
to zero at x = 0 and x = L, but these do not count as nodes as the wavefunction
does not pass through zero.
�e task is to �nd N such that ψ = N sin(2πx�L) satis�es the normalization
condition [�B.�c–���], ∫ ψ ∗ ψ dτ = 1. In this case the integration is over x and
the range is 0 to L; the function is real, so ψ = ψ ∗ .
N2 �
E�B.�(a)
L
0
=sin 4π=0
��� � � � � � � � � � � �� � � � � � � � � � � � ��
sin2 (2πx�L)dx = N 2 [L�2 − (L�8π) sin(4πL�L)] = N 2 (L�2)
�e integral is of the form of Integral T.� with a = L and k = 2π�L. For
the wavefunction to be normalized, the integral must be 1 and therefore the
normalizing factor is N = (2�L)1�2 .
�e task is to �nd N such that ψ = N exp(−ax 2 ) satis�es the normalization
condition [�B.�c–���], ∫ ψ ∗ ψ dτ = 1. In this case the integration is over x
and the range is −∞ to ∞; the function is real, so ψ = ψ ∗ and the integral is
∞
therefore N 2 ∫−∞ exp(−2ax 2 ) dx. �e integrand is even, meaning that it has
the same value at +x and −x, so the integral between −∞ and +∞ is simply
twice that between 0 and +∞. �e integral is then of the form of Integral G.�
with k = 2a
2N 2 �
E�B.�(a)
0
∞
exp(−2ax 2 )dx = 2N 2 [ 12 (π�2a)1�2 ]
Setting this equal to 1 gives N = (2a�π)1�4 .
(i) �e function ψ = exp(−ax 2 ) can be normalized as ψ ∗ ψ = exp(−2ax 2 )
goes to 0 at x = ±∞ so that the integral over all space of ψ ∗ ψ is �nite.
(ii) �e function ψ = exp(−ax) cannot be normalized as it goes to ∞ as x
goes to −∞, therefore the integral over all space of ψ ∗ ψ is in�nite. If the
wavefunction were limited to a �nite region of space it could, however, be
normalized.
A function is an acceptable wavefunction if it: (�) is not in�nite over a �nite region; (�) is single-valued; (�) is continuous; (�) has a continuous �rst derivative.
Both functions satisfy all of these conditions and so are acceptable wavefunctions.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�B.�(a)
�e probability of �nding an electron in an in�nitesimal region dx around x
is P(x)dx = ψ ∗ (x)ψ(x)dx, provided that ψ(x) is a normalized wavefunction.
�e wavefunction is real so that ψ ∗ = ψ, hence the probability is given by
P(x)dx = �(2�L)1�2 sin(2πx�L)� dx = (2�L) sin2 (2πx�L)dx
2
E�B.�(a)
Hence, at x = L�2, P(L�2)dx = (2�L) sin(π)dx = 0 .
�e normalized wavefunction is ψ(x) = (2�L)1�2 sin(2πx�L). �e probability
of �nding the electron between x = L�4 and x = L�2 is, using Integral T.�
�
L�2
L�4
(2�L) sin2 (2πx�L) dx = (2�L) [x�2 − (L�8π) sin(4πx�L)�L�4 �
L�2
= (2�L)([L�4 − (L�8π) sin(2π)] − [L�8 − (L�8π) sin(π)])
E�B.�(a)
= (2�L)[L�4 − L�8] = �/�
In two dimensions �ψ(x, y)�2 dxdy is the probability of �nding the particle in
the in�nitesimal area dxdy at the position (x, y). �e probability is a dimensionless quantity, and the area dxdy has units of (length)2 . Hence, �ψ(x, y)�2
must have units of (length)−2 , implying that ψ has units of length−1 .
Solutions to problems
P�B.�
(a) �e task is to �nd N such that ψ = Nei� satis�es the normalization condition [�B.�c–���], ∫ ψ ∗ ψ dτ = 1. In this case the integration is over � and
the range is 0 to 2π. �e function is complex so ψ ∗ = N exp(−i�); note
that the normalization factor N is always real. �e integrand is therefore
ψ ∗ ψ = e−i� ei� = e−i�+i� = e0 = 1
�e integral to evaluate is therefore
N2 �
2π
0
1 d� = N 2 ��0 = 2πN 2
2π
Setting this equal to 1 gives N = (2π)−1�2 .
(b) For the function eim l � exactly the same argument applies and so the normalization factor is the same.
P�B.�
(a) �e task is to �nd N such that ψ = N sin(πx�L x ) sin(πy�L y ) satis�es the
normalization condition [�B.�c–���], ∫ ψ ∗ ψ dτ = 1. �e integration is
over x from 0 to L x , and y from 0 to L y . �e wavefunction is real, and so
the integral to evaluate is
N2 �
Lx
0
�
Ly
0
sin2 (πx�L x ) sin2 (πx�L y )dxdy
�is separates into the product of two integrals
N 2 ��
Lx
0
sin2 (πx�L x )dx� ��
Ly
0
sin2 (πy�L y )dy�
245
246
7 QUANTUM THEORY
Each of these is evaluated using Integral T.�
�
Lx
0
sin2 (πx�L x )dx = L x �2 − (L x �4π) sin(2πL�L) = L x �2
and similarly the integral over y is equal to L y �2. Hence the integral is
�
evaluated as N 2 (L x �2)(L y �2). Setting this equal to 1 gives N = 2� L x L y .
P�B.�
(b) In the case when L x = L y = L this reduces to N = 2�L .
�e task is to �nd N such that ψ(x) = Ne−ax , satis�es the normalization
condition [�B.�c–���], ∫ ψ ∗ ψ dτ = 1. �e integration is over x ranging from 0
to ∞.
N2 �
0
∞
∞
e−2ax dx = −(N 2 �2a) e−2ax �0 = −(N 2 �2a)(e−∞ − e0 ) = N 2 �2a
where e−∞ = 0 and e0 = 1 are used. Setting this equal to 1 gives N = (2a)1�2 .
Hence, the total probability of �nding the particle at a distance x ≥ x 0 is
�
P�B.�
∞
x0
�ψ(x)�2 dx = 2a �
∞
x0
∞
e−2ax dx = −(2a�2a) e−2ax �x = e−2ax 0
0
With a = 2 m−1 and x 0 = 1 m, the probability is exp(−2 × (2 m−1 ) × (1 m))
= exp(−4) = �.����
�e probability of �nding the particle in the range x = a to x = b is
P(a → b) = �
b
a
�ψ(x)�2 dx = �
b
a
�is is evaluated using Integral T.� to to give
(2�L) sin2 (πx�L) dx
P(a → b) = (2�L) � x�2 − (L�4π) sin(2πx�L)� a �
b
Hence,
=
b−a
1
2πb
2πa
−
�sin �
� − sin �
��
L
2π
L
L
(a) L = 10.00 nm, a = 4.95 nm, b = 5.05 nm, P(a → b) = 2.00 × 10−2
(b) L = 10.00 nm, a = 1.95 nm, b = 2.05 nm, P(a → b) = 6.91 × 10−3
(c) L = 10.00 nm, a = 9.90 nm, b = 10.00 nm, P(a → b) = 6.58 × 10−6
P�B.�
(d) L = 10.00 nm, a = 5.00 nm, b = 10.00 nm, P(a → b) = �.� . �is is
expected as for this wavefunction the probability density is symmetric
around x = 5.00 nm.
�e probability of being in a small volume δV at distance r from the nucleus
is ψ(r)2 δV . �e probability of the electron being in a small sphere of radius r ′
centred at r from the nucleus is
P(r) = �(πa 03 )−1�2 e−r�a 0 � × �(4�3)π(r ′ )3 � = �4(r ′ )3 �3a 03 � e−2r�a 0
2
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
where a 0 = 53 pm is the Bohr radius. If the sphere has radius 1.0 pm then the
expression evaluates to
4 × (1.0 pm)3
× e−2r�a 0 = 8.95... × 10−6 × e−2r�a 0
3 × (53 pm)3
(a) If the sphere is centered at the nucleus r = 0, P(0) = 8.95 × 10−6
P�B.��
(b) If the sphere is centered at r = a 0 , P(a 0 ) = 8.95...×10−6 ×e−2 = 1.21 × 10−6
�e probability �nding the atom between x and x +dx, where dx is an in�nitesimal displacement, is given by
P(x)dx = ψ(x)2 dx = N 2 x 2 e−x �a dx
2
2
�e task is to �nd the position of the maximum in P(x), which is done by
setting dP(x)�dx = 0. To compute the derivative it is necessary to use the
product rule.
� d x2
2
2
2
2
dP(x)
d
d e−x �a �
=
�N 2 x 2 e−x �a � = N 2
× e−x �a + x 2 ×
dx
dx
dx �
� dx
2
= N 2 �2xe−x �a − x 2 ×
2
2
2
2x −x 2 �a 2
x 2
2 −x 2 �a 2
�
=
2xN
e
�1
−
�
e
� �
a2
a
�e derivative goes to zero at x = 0, ±a, ±∞. Inspection of the form of P(x)
shows that x = 0 is a minimum (P(x) = 0), and at x = ±∞, P(x) also goes
asymptotically to zero. �e maxima are therefore at x = ±a .
7C Operators and observables
Answers to discussion questions
D�C.�
D�C.�
In quantum mechanics an observable quantity (such as energy, position or
momentum) is represented by a particular operator Ω̂. If the wavefunction
is ψ the average value of the quantity represented by the operator Ω̂ is given by
�Ω� = ∫ ψ ∗ Ω̂ψ dτ, called the expectation value. For the special case that ψ is
an eigenfunction of Ω̂, the expectation value is the eigenvalue corresponding
to this eigenfunction.
�e operator which represents kinetic energy is T̂ = (−ħ 2 �2m)d2 �dx 2 . �e
expectation value of the kinetic energy is therefore related to the average value
of d2 ψ�dx 2 , that is the average of the second derivative or curvature of the
wavefunction. Sharply curved regions of the wavefunction will make a larger
contribution to the kinetic energy than less sharply curved regions. �e expectation value of the kinetic energy will have contributions from all parts of the
wavefunction.
247
248
7 QUANTUM THEORY
Solutions to exercises
E�C.�(a)
Two wavefunctions ψ i and ψ j are orthogonal if ∫ ψ ∗i ψ j dτ = 0, [�C.�–���].
In this case the integration is from � = 0 to � = 2π. Let ψ i = exp(i�), the
wavefunction with m l = +1, and ψ j = exp(2i�), the wavefunction with m l =
+2. Note that the functions are complex, so ψ ∗i = exp(−i�). �e integrand is
therefore
ψ ∗i ψ j = exp(−i�) exp(2i�) = exp(i�)
and the integral evaluates as
�
2π
0
=1
=1
��� � � � � � � � � � � �� � � � � � � � � � � �� ��� � � � � � � � ��� � � � � � � � � �
2π
exp(i�) d� = (1�i) exp(i�)�0 = (1�i)[exp(i × 2π) − exp(i × 0)] = 0
�e identity exp(ix) = cos x + i sin x (�e chemist’s toolkit �� in Topic �C on
page ���) is used to evaluate exp(i2π) = cos(2π) + i sin(2π) = 1 + 0 = 1. �e
integral is zero, so the functions are indeed orthogonal.
E�C.�(a)
�e normalized wavefunction is ψ(x) = (2�L)1�2 sin(2πx�L). �e operator for
position is x̂ = x, therefore the expectation value of the position of the electron
is [�C.��–���]
�x� = � ψ ∗ x̂ψ dτ = (2�L) �
L
0
x sin2 (2πx�L) dx
�is integral is of the form of Integral T.�� with k = 2π�L and a = L
=sin 4π=0
=cos 4π=1
��� � � � � � � � � �� � � � � � � � � � ��
��� � � � � � � � � � � � � � � � � � � � � � ��
� 2
�
�
�
2�
L
L
4πL
1
4πL
�
��
�
= � −
sin �
�−
� cos �
� −1��
�
2
�
�
L � 4 4 × 2π�L
L
8 × (2π�L) �
L
�
�
��
�
�
2 L2
= ×
= L�2
L
4
E�C.�(a)
Because the probability density �ψ(x)�2 is symmetric about x = L�2, the expected result is �x� = L�2.
�e normalized wavefunction is ψ(x) = (2�L)1�2 sin(2πx�L). �e expectation
value of the momentum is ∫ ψ ∗ p̂ x ψ dx, and the momentum operator is p̂ x =
(ħ�i)d�dx, therefore
�p x � = (2�L) �
L
0
= (2ħ�iL) �
sin(2πx�L) p̂ x sin(2πx�L) dx
L
0
sin(2πx�L)(d�dx) sin(2πx�L) dx
Using (d�dx) sin(2πx�L) = (2π�L) cos(2πx�L) gives
�p x � = (4πħ�iL 2 ) �
L
0
sin(2πx�L) cos(2πx�L) dx
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e integral is of the form of Integral T.� with a = L, k = 2π�L
�p x � =
4πħ
1
2πL
×
sin2 �
�= 0
iL 2 2 × 2π�L
L
�is result is interpreted as meaning that there are equal probabilities of having
momentum in the positive and negative x directions.
E�C.�(a)
For the case when m l = +1 the normalized wavefunction is ψ+1 (�) = (2π)−1�2 ei� .
∗
�is is complex, and so ψ+1
= (2π)−1�2 e−i� . �e expectation value of the
position, speci�ed by the operator �, is
��� = �
2π
0
∗
ψ+1
�ψ+1 d� = (1�2π) �
= (1�2π) �
where e−i� ei� = 1 is used.
2π
0
2π
0
e−i� � ei� d�
� d� = (1�2π) × (� 2 �2)�0 = π
2π
For the general case the integral is
(1�2π) �
E�C.�(a)
2π
0
e−im l � �eim l � d�
which evaluates to give the same result, ���m l = π. �is result is interpreted
by noting that the probability density around the ring is (1�2π)e−im l � eim l � =
1�2π, which is constant. �erefore the average position is half way round the
ring at � = π.
�e uncertainty in the momentum is given by ∆p = m∆υ, where m is the mass
and ∆υ is the uncertainty in the velocity. �e uncertainties in position (∆q)
and momentum (∆p) must obey the Heisenberg uncertainty principle [�C.��a–
���], ∆p q ∆q ≥ (ħ�2), which in this case is expressed as m∆υ∆q ≥ (ħ�2). �is
is rearranged to give the uncertainty in the velocity, ∆υ ≥ ħ�(2m∆q). �e
minimum uncertainty in the speed is therefore ∆υ min = ħ�(2m∆q), which is
evaluated as
1.0546 × 10−34 J s
= 1.05 × 10−28 m s−1
2 × (500 × 10−3 kg) × (1.0 × 10−6 m)
�e uncertainty principle can be rearranged for the uncertainty in the position
∆q ≥ ħ�2m∆υ, and so the minimum uncertainty is ħ�(2m∆υ). �e uncertainty in the position of the bullet is 1 × 10−5 m s−1 , and hence
E�C.�(a)
∆q min =
1.0546 × 10−34 J s
= 1.05 × 10−27 m
2 × (5.0 × 10−3 kg) × (1 × 10−5 m s−1 )
�e desired uncertainty in the momentum is
∆p = 0.0100 × 10−2 p = 1.00 × 10−4 m p υ
= 1.00 × 10−4 × (1.6726 × 10−27 kg) × (0.45 × 106 m s−1 )
= 7.52... × 10−26 kg m s−1
249
250
7 QUANTUM THEORY
�e Heisenberg uncertainty principle, [�C.��a–���] is rearranged to give the
uncertainty in the position as ∆q ≥ ħ�(2∆p), which gives a minimum uncertainty of ∆q min = ħ�(2∆p). �is is evaluated as
E�C.�(a)
E�C.�(a)
1.0546 × 10−34 J s
= 7.01 × 10−10 m
2 × (7.52... × 10−26 kg m s−1 )
To construct the potential energy operator, replace the position x in the classical expression by the operator for position x̂ to give V̂ = 12 k f x̂ 2 . However,
because x̂ = x× the potential energy operator is V̂ = 12 k f x 2 .
A function ψ is an eigenfunction of an operator Ω̂ if Ω̂ψ = ωψ where ω is a
constant called the eigenvalue.
(i) (d�dx) cos(kx) = −k sin(kx). Hence cos kx is not an eigenfunction of
the operator d�dx.
(ii) (d�dx)eik x = ikeik x . Hence eik x is an eigenfunction of the operator
d�dx, with eigenvalue ik .
(iii) (d�dx)kx = k. Hence kx is not an eigenfunction of the operator d�dx.
(iv) (d�dx)e−ax = −2axe−ax . Hence e−ax is not an eigenfunction of the
operator d�dx.
2
E�C.�(a)
2
2
Wavefunctions ψ 1 and ψ 2 are orthogonal if ∫ ψ ∗1 ψ 2 dτ = 0, [�C.�–���]. Here
ψ 1 (x) = sin(πx�L), ψ 2 (x) = sin(2πx�L), and the region is 0 ≤ x ≤ L. �e
integral is evaluated using Integral T.�
∗
� ψ 1 ψ 2 dτ = �
L
0
sin(πx�L) sin(2πx�L) dx
= [(−L�2π) sin(−π) − (L�6π) sin(3π)] = 0
where sin(nπ) = 0 for integer n is used. �us, the two wavefunctions are
orthogonal.
E�C.��(a) Wavefunctions ψ 1 and ψ 2 are orthogonal if ∫ ψ ∗1 ψ 2 dτ = 0, [�C.�–���]. Here
ψ 1 (x) = cos(πx�L), ψ 2 (x) = cos(3πx�L), and the region is −L�2 ≤ x ≤ L�2.
�e integral is evaluated using Integral T.�
∗
� ψ 1 ψ 2 dτ = �
L�2
−L�2
cos(πx�L) cos(3πx�L) dx
= (−L�4π) sin(−2πx�L) + (L�8π) sin(4πx�L)�−L�2
= [(−L�4π) sin(−π) + (L�8π) sin(2π)]
− [(−L�4π) sin(π) + (L�8π) sin(−2π)] = 0
L�2
where sin(nπ) = 0 for integer n is used. �us, the two wavefunctions are
orthogonal.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Solutions to problems
P�C.�
ˆ the inversion operator, which has the e�ect
Operate on each function f with i,
of making the replacement x → −x. If the result of the operation is f multiplied
by a constant, f is an eigenfunction of iˆ and the constant is the eigenvalue.
ˆ 3 − kx) = (−x)3 − k(−x) = −(x 3 − kx). Yes , f is an eigenfunction,
(a) i(x
eigenvalue −1 .
ˆ
(b) i(cos(kx))
= cos(k(−x)) = cos(kx). Yes , f is an eigenfunction, eigenvalue +1 .
ˆ 2 + 3x − 1) = (−x)2 + 3(−x) − 1 = x 2 − 3x − 1. No , f is not an
(c) i(x
eigenfunction.
P�C.�
∗
An operator Ω̂ is hermitian if ∫ ψ ∗i Ω̂ψ j dτ = �∫ ψ ∗j Ω̂ψ i dτ� , [�C.�–���]. For
the kinetic energy operator, it is necessary to integrate by parts (�e chemist’s
toolkit �� in Topic �C on page ���) twice
∗
� ψ i �−
2
∞
ħ 2 d2
ħ2
∗ d ψj
�
ψ
dx
=
−
ψ
dx
j
�
i
2m dx 2
2m −∞
dx 2
�
�
∞
� ħ2
∞ dψ ∗ dψ
∞ dψ ∗ dψ
dψ
ħ2 �
j
j
� ∗ j
�
i
i
� ψi
=−
� −�
dx � =
dx
� 2m �−∞ dx dx
2m �
dx
dx
dx
−∞
�
�
−∞
���� � � � � � � � � � ��� � � � � � � � � � � �
�
A
Term A evaluates to zero because, as stated in the problem, the wavefunction
and its derivative can be assumed to go to zero at x = ±∞. �en integrate the
remaining term by parts once again,
�
�
∞
∗
2 �
∞ dψ ∗ dψ
∞ d2 ψ ∗
�
dψ
ħ2
ħ
j
i
i
� i ψ j� − �
dx =
ψ j dx �
�
�
2
2m −∞ dx dx
2m �
dx
dx
−∞
�
�
−∞
���� � � � � � � � � � �� � � � � � � � � � � ��
�
B
As before, term B evaluates to zero, so the �nal result is
∗
� ψ i �−
∞
ħ 2 d2
ħ 2 d2
�
ψ
dx
=
ψ
�−
� ψ ∗i dx
j
j
�
2m dx 2
2m dx 2
−∞
�e term on the right can be rewritten as a complex conjugate to give
∗
� ψ i �−
∗
ħ 2 d2
ħ 2 d2
∗
�
ψ
dx
=
�
ψ
�−
� ψ i dx�
j
�
j
2m dx 2
2m dx 2
Note that because the complex conjugate of the whole term is taken, to compensate for this the complex conjugate of the terms inside the bracket need to
be taken too ψ = [ψ ∗ ]∗ . �is �nal equation is consistent with [�C.�–���] and
so demonstrates that the kinetic energy operator is hermitian.
251
252
7 QUANTUM THEORY
P�C.�
(a) Consider the sum of two arbitrary hermitian operators  and B̂,
∗
∗
∗
� ψ i (Â + B̂)ψ j dτ = � ψ i Âψ j dτ + � ψ i B̂ψ j dτ
∗
As  is hermitian, ∫ ψ ∗i Âψ j dτ = �∫ ψ ∗j Âψ i dτ� and similarly for B̂.
∗
∗
∗
∗
∗
� ψ i ( + B̂)ψ j dτ = �� ψ j Âψ i dτ� + �� ψ j B̂ψ i dτ�
∗
= �� ψ ∗j ( + B̂)ψ i dτ�
which demonstrates that the sum of two hermitian operators is also hermitian.
(b) As Ω̂ψ j = ψ k is also a function, the integral can be rewritten
∗
∗
� ψ i Ω̂ Ω̂ψ j dτ = � ψ i Ω̂ψ k dτ
As Ω̂ is a hermitian operator it follows that
∗
∗
∗
� ψ i Ω̂ψ k dτ = �� ψ k Ω̂ψ i dτ�
Next realise that Ω̂ψ i is a function which will be called ψ l : Ω̂ψ i = ψ l .
With this substitution
∗
∗
∗
∗
� ψ i Ω̂ψ k dτ = �� ψ k ψ l dτ� = � ψ k ψ l dτ
= � ψ ∗l ψ k dτ
= � ψ ∗l Ω̂ψ j dτ
To go to the second line the two functions have been re-ordered, which
is always permitted, and to go to the third line the de�nition Ω̂ψ j = ψ k
has been used. Because Ω̂ is hermitian the �nal term can be rewritten as
[∫ ψ ∗j Ω̂ψ l dτ]∗ and so
∗
∗
∗
� ψ i Ω̂ψ k dτ = �� ψ j Ω̂ψ l dτ�
In the term on the right the de�nition Ω̂ψ i = ψ l is used, and in the term
on the le� Ω̂ψ j = ψ k is used to give
∗
∗
∗
� ψ i Ω̂ Ω̂ψ j dτ = �� ψ j Ω̂ Ω̂ψ i dτ�
�is is exactly the property which proves that Ω̂ Ω̂ is hermetian.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P�C.�
�e operator for position is multiplication by x, so it follows that the expectation value of the position operator is
�x� = � ψ ∗ x̂ψdτ = a �
P�C.�
0
∞
xe−ax dx = a ×
1!
= 1�a
a2
�e integral is of the form of Integral E.� with n = 1 and k = a.
(a) �e expectation value is given by [�C.��–���], �Ω� = ∫ ψ ∗ Ω̂ψ dτ, where
ψ is normalized. A hermitian operator has the property ∫ ψ ∗i Ω̂ψ j dτ =
[∫ ψ ∗j Ω̂ψ i dτ]∗ , which for the case ψ i = ψ j = ψ becomes
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � � � � � � � �∗
�
�
∗
∗
�
� ψ Ω̂ψ dτ = �
� � ψ Ω̂ψ dτ �
�
�
�
�
A
B
�e quantities A and B and the same, so it follows that A = B∗ = A∗ ,
which is only true if A is real. �e quantity A is the expectation value, so
it follows that the expectation value of a hermitian operator is real.
(b) �e expectation value of the square of the hermitian operator Ω̂ Ω̂ is
�ΩΩ� = � ψ ∗ Ω̂ Ω̂ψ dτ
Recognise that Ω̂ψ is a function, which will be written �: Ω̂ψ = �; with
this, and using the fact that Ω̂ is hermitian it follows that
∗
∗
∗
∗
� ψ Ω̂ Ω̂ψ dτ = � ψ Ω̂� dτ = �� � Ω̂ψ dτ�
Once again use Ω̂ψ = � in the term on the right to give
∗
∗
∗
� ψ Ω̂ Ω̂ψ dτ = �� � � dτ�
�e integral ∫ �∗ � dτ is the sum over all space of the square magnitude
of a wavefunction. Using the Born interpretation, this is proportional to
the total probability and therefore must be a real positive number. It is
therefore shown that the expectation value of the square of a hermitian
operator is real and positive.
P�C.��
(a) �e normalized wavefunction is ψ(x) = (2a�π)1�4 e−ax , leading to a
2
probability density of �ψ(x)�2 = (2a�π)1�2 e−2ax . �e expectation value
of the position is therefore
2
�x� = (2a�π)1�2 �
∞
−∞
xe−2ax dx = 0
2
as the integrand is odd and the integration is over a symmetric range. For
�x 2 � the required integral is
�x 2 � = (2a�π)1�2 �
∞
−∞
x 2 e−2ax dx = 2(2a�π)1�2 �
2
0
∞
x 2 e−2ax dx
2
253
254
7 QUANTUM THEORY
�e integrand is an even function, and so its integral between −∞ and ∞
is twice that from 0 to ∞. �is integral is of the form of Integral G.� with
k = 2a
�x 2 � = 2(2a�π)1�2 × 14 (π�(2a)3 )1�2 = 1�4a
�e e�ect of the momentum operator, (ħ�i)(d�dx), on the wavefunction
2
2
is (ħ�i)(d�dx)(2a�π)1�4 e−ax = (ħ�i)(2a�π)1�4 × (−2axe−ax ). Hence
the expectation value of the momentum is
�p x � = � ψ ∗ p̂ x ψ dτ =
2
ħ 2a 1�2 ∞
� � � −2axe−2ax dx = 0
i π
−∞
as the integrand is odd.
To evaluate the e�ect of the operator p̂2x = −ħ 2 (d2 �dx 2 ) on the wavefunction the product rule is used
p̂2x ψ = p̂ x ( p̂ x )ψ =
�is is equal to
which gives
2
ħ d
�(ħ�i)(2a�π)1�4 × [−2axe−ax ]�
i dx
−ħ 2 (2a�π)
1�4
(4a 2 x 2 e−ax − 2ae−ax )
�p2x � = � ψ ∗ p̂2x ψ dτ
= −ħ 2 (2a�π)
2
+∞
2
(4a 2 x 2 e−2ax − 2ae−2ax )
�
−∞
�
�
1�2
2a 1�2 � 2
π
π 1�2 �
�
= −ħ 2 � � × �
2a
�
�
−
2a
�
�
�
�
π
(2a)3
2a
�
�
�
�
2
= ħ a
1�2
2
2
where Integrals G.� and G.� are used.
(b) �e uncertainty in the position is given by
∆x = ��x 2 � − �x�2 �
1�2
= (1�4a − 02 )1�2 = (4a)−1�2
and the uncertainty in the momentum is given by
∆p x = ��p2x � − �p x �2 �
1�2
�e product of the uncertainties is
√
= (ħ 2 a − 02 )1�2 = ħ a
√
∆x∆p x = (4a)−1�2 × ħ a = ħ�2
�is uncertainty is ≥ ħ�2 and so is consistent with the uncertainty principle; in fact it is the smallest possible uncertainty.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P�C.��
To evaluate the commutator of the operators consider the e�ect of the commutator on an arbitrary function ψ; use the product rule to evaluate the derivatives
of products as necessary.
(a)
�
(b)
d 1
d ψ
1 dψ 1 dψ
dx −1 1 dψ
, �ψ =
� �−
=
+ψ
−
dx x
dx x
x dx x dx
dx
x dx
1
=− 2 ψ
x
�e commutator is therefore identi�ed as the term multiplying ψ, −1�x 2 .
�
d 2
d(x 2 ψ)
dψ
dψ
dx 2
dψ
, x �ψ =
− x2
= x2
+ψ
− x2
dx
dx
dx
dx
dx
dx
= 2x ψ
�e commutator is therefore identi�ed as the term multiplying ψ, 2x .
P�C.��
To complete this problem it is useful to establish some general properties of
commutators. First, an operator  commutes with powers of itself: for example,
[Â, Â2 ] = 0. �is is proved by simply multiplying out the commutator:
[Â, Â2 ] = Â(ÂÂ) − (ÂÂ)Â = 0
Note that multiplying either of the terms by a constant does not alter this property. Second, any operator commutes with a constant: [Â, c] = 0, which follows
trivially on multiplying out the commutator. Finally, the useful property that
[Â + B̂, Ĉ] = [Â, Ĉ] + [B̂, Ĉ]. Again, this follows by multiplying out the commutator
[Â + B̂, Ĉ] = (Â + B̂)Ĉ − Ĉ(Â + B̂) = ÂĈ − Ĉ Â + B̂Ĉ − Ĉ B̂ = [Â, Ĉ] + [B̂, Ĉ]
(a) For the case V̂ = V0 :
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � �
A
B
[Ĥ, p̂ x ] = [ p̂2x �2m + V0 , p̂ x ] = [ p̂2x �2m, p̂ x ] + [V0 , p̂ x ] = 0
Term A is zero because it is a commutator of an operator with a power of
itself, and term B is zero because it is a commutator with a constant.
For the case V̂ = 12 k f x 2 :
[Ĥ, p̂ x ] = [ p̂2x �2m + 12 k f x 2 , p̂ x ] = [ p̂2x �2m, p̂ x ] + [ 12 k f x 2 , p̂ x ]
�e �rst commutator is zero. To evaluate the second commutator consider its e�ect on an arbitrary wavefunction ψ; to simplify the calculation
set aside the constant multiplying term 12 k f and re-introduce it at the end
[x 2 , p̂ x ]ψ = (x 2 p̂ x − p̂ x x 2 )ψ
=
ħ 2 dψ d(x 2 ψ)
ħ
dψ
dψ
�x
−
� = �x 2
− x2
− ψ × 2x�
i
dx
dx
i
dx
dx
= −2x(ħ�i)ψ
Reintroducing the constants gives the result [Ĥ, p̂ x ] = −(k f ħ�i)x .
255
256
7 QUANTUM THEORY
(b) For the case V̂ = V0 :
[Ĥ, x̂] = [ p̂2x �2m + V0 , x̂] = [ p̂2x �2m, x̂] + [V0 , x̂]
�e second term is zero because it is a commutator with a constant. �e
�rst term is evaluated in the usual way, �rst setting aside the constants
[ p̂2x , x̂]ψ = ( p̂2x x − x p̂2x )ψ
= −ħ 2 �
= −ħ 2 �
d2 (xψ)
d2 ψ
d
dψ
d2 ψ
2
−
x
�
=
−ħ
�
(ψ
+
x
)
−
x
�
dx 2
dx 2
dx
dx
dx 2
dψ
d2 ψ dψ
d2 ψ
dψ
+x 2 +
− x 2 � = −2ħ 2
dx
dx
dx
dx
dx
Reintroducing the constants gives the result [Ĥ, x̂] = −(ħ 2 �m)(d�dx)
For the case V̂ = 12 k f x 2 :
[Ĥ, x̂] = [ p̂2x �2m + 12 k f x 2 , x] = [ p̂2x �2m, x] + [ 12 k f x 2 , x]
�e second commutator is zero. �e �rst has already been evaluated in
the preceding calculation hence [Ĥ, x̂] = −(ħ 2 �m)(d�dx)
7D Translational motion
Answers to discussion questions
D�D.�
�e physical origin of tunnelling is related to the probability density of the
particle, which according to the Born interpretation is the square of the wavefunction that represents the particle. �is interpretation requires that the wavefunction of the system be everywhere continuous, even at barriers. �erefore,
if the wavefunction is non zero on one side of a barrier it must be non zero on
the other side of the barrier and this implies that the particle has tunnelled into
the barrier. �e transmission probability depends upon the mass of the particle: the greater the mass the smaller the probability of tunnelling. Electrons
and protons have small masses, molecular groups large masses; therefore, tunnelling e�ects are more observable in process involving electrons and protons.
D�D.�
�e hamiltonian for a particle in a two- or three-dimensional box is separable into a sum of terms each of which depends on just one of the variables
x, y or z. �erefore the solutions to the Schrödinger equation are products
of separate functions of each of these variables, Ψ(x, y, z) = ψ(x)ψ(y)ψ(z).
In the hamiltonian the terms in the three variables are all of the same form,
so the wavefunctions are likewise of the same form. Similarly, the boundary
conditions along each direction are the same, so the quantization which this
imposes is the same. As a result the wavefunctions are simply a product of the
wavefunctions that would be found by solving the Schrödinger equation along
each direction separately.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e energy levels of two- and three-dimensional boxes can be degenerate, that
is there are di�erent wavefunctions which have the same energy. �is is usually
a consequence of the box having a certain kind of symmetry, for example in a
two-dimensional box in which the two sides are the same length the levels with
quantum numbers n 1 = l, n 2 = m and n 1 = m, n 2 = l are degenerate.
Solutions to exercises
E�D.�(a)
E�D.�(a)
�e energy levels of a particle in a box with length L are given by [�D.�–���],
E n = h 2 n 2 �8mL 2 and when the length is increased to 1.1 × L the energies
become E n′ = h 2 n 2 �8m(1.1 × L)2 = h 2 n 2 �8m(1.21 × L 2 ) giving a fractional
change of
E n′ − E n [h 2 n 2 �8m(1.21 × L 2 )] − [h 2 n 2 �8mL 2 ]
1
=
=
− 1 = −0.174
2
2
2
En
h n �8mL
1.21
�e energy levels of a particle in a box of length L are given by [�D.�–���],
E n = h 2 n 2 �8mL 2 . Hence the spacing between adjacent levels (n and n + 1) is
given by
h 2 (n + 1)2 h 2 n 2
h2
�(n + 1)2 − n 2 �
−
=
8mL 2
8mL 2 8mL 2
h2
h2
=
(n 2 + 2n + 1 − n 2 ) =
(2n + 1)
2
8mL
8mL 2
∆E(n) = E n+1 − E n =
Setting ∆E(n) to the average thermal energy gives h 2 (2n + 1)�8mL 2 = kT�2,
and so (2n + 1) = 4mkTL 2 �h 2 , leading to
n=
2mkTL 2 1
−2
h2
For a helium atom, mass 4.00 m u in a box for length 1 cm,
n = �2(4.00 × 1.6605 × 10−27 kg) × (1.3806 × 10−23 J K−1 ) × (298 K)
E�D.�(a)
×
(1 × 10−2 m)2
� − 12 = 1.24 × 1016
(6.6261 × 10−34 J s)2
�e wavefunction of a particle in a square box of side length L with quantum numbers n 1 = 2, n 2 = 2 is ψ 2,2 (x, y) = (2�L) sin (2πx�L) sin (2πy�L).
2
�e probability density is P2,2 (x, y) = (2�L) sin2 (2πx�L) sin2 (2πy�L). �e
2
probability density is maximized when sin (2πx�L) × sin2 (2πy�L) = 1 which
occurs only when each sin term is equal to ±1,
sin (2πx�L) = ±1
2πx�L = π�2, 3π�2
hence
x = L�4, 3L�4
and similarly for y. Hence the maxima in probability density occurs at (x, y) =
(L�4, L�4), (L�4, 3L�4), (3L�4, L�4), (3L�4, 3L�4)
257
258
7 QUANTUM THEORY
E�D.�(a)
Nodes occur when the wavefunction passes through zero, which is when either
of the sin terms are zero, excluding the boundaries at x = 0, L y = 0, L because
at these points the wavefunction does not pass through zero. �ere is thus a
node when sin (2πx�L) = 0, corresponding to x = L�2 and any value of y. �is
node is therefore a line at x = L�2 and parallel to the y axis . Similarly there is
another nodal line at y = L�2 and parallel to the x axis .
�e energy levels for a �D rectangular box, side lengths L 1 , L 2 are
E n 1 ,n 2 =
h 2 n 12 n 22
� + �
8m L 21 L 22
where n 1 and n 2 are integers greater than or equal to 1.
For the speci�c case where L 1 = L, L 2 = 2L,
E n 1 ,n 2 =
h 2 n 12
n 22
h2
n 22
2
� 2+
�
=
�n
+
�
1
8m L
(2L)2
8mL 2
4
�e energy of the state with n 1 = n 2 = 2 is then
E 2,2 =
h2
22
h2
�22 + � =
(5)
2
8mL
4
8mL 2
E(1,4) =
h2
42
h2
2
�1
+
�
=
(5)
8mL 2
4
8mL 2
�is is degenerate with the state with n 1 = 1, n 2 = 4
E�D.�(a)
�e question notes that degeneracy frequently accompanies symmetry, and
suggests that one might be surprised to �nd degeneracy in a box with unequal
lengths. Symmetry is a matter of degree. �is box is less symmetric than a
square box, but it is more symmetric than boxes whose sides have a non-integer
or irrational ratio. Every state of a square box except those with n 1 = n 2 is
degenerate (with the state that has n 1 and n 2 reversed). Only a few states in this
rectangular box are degenerate. In this system, a state (n 1 , n 2 ) is degenerate
with a state (n 2 �2, 2n 1 ) as long as the latter state (a) exists (that is, n 2 �2 must
be an integer) and
√ (b) is distinct from (n 1 , n 2 ). A box with incommensurable
sides, say, L and 2L, would have no degenerate levels.
�e energy levels of a cubic box are given by [�D.��b–���]
h2
(n 2 + n 22 + n 32 )
8mL 2 1
where n 1 , n 2 , n 3 are integers greater than or equal to 1. Hence the lowest energy
state is that with n 1 = n 2 = n 3 = 1, with energy
E n 1 ,n 2 ,n 3 =
h2
h2
2
2
2
(1
+
1
+
1
)
=
(3)
8mL 2
8mL 2
and so the energy of the level with energy three times that of the lowest is
9h 2 �8mL 2 , which will be produced by states for which n 12 + n 22 + n 32 = 9. �ere
are three states with this energy, (n 1 , n 2 , n 3 ) = (1, 2, 2), (2, 1, 2), (2, 2, 1), and
so the degeneracy is � .
E(1,1,1) =
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�D.�(a)
�e transmission probability [�D.��a–���] depends on the energy of the tunnelling particle relative to the barrier height (ε = E�V0 = (1.5 eV)�(2.0 eV) =
0.75), the width of the barrier, (L = 100 pm), and the decay parameter of the
wavefunction inside the barrier (κ)
(2m(E − V0 ))1�2
ħ
(2 × (9.1094 × 10−31 kg) × [(2.0 − 1.5) eV × 1.6022 × 10−19 J eV−1 ])1�2
=
1.0546 × 10−34 J s
9
−1
= 3.68... × 10 m
κ=
�e product κL = (3.68... × 109 m−1 ) × (100 × 10−12 m) = 0.368..., and so the
transmission probability is given by
E�D.�(a)
T = �1 +
−1
(eκ L − e−κL )2
�
16ε(1 − ε)
= �1 +
−1
(e0.368 ... − e−0.368 ... )2
� = �.��
16 × 0.75 × (1 − 0.75)
�e linear momentum of a free electron is given by
p = ħk = (1.0546 × 10−34 J s) × (3 × 109 m−1 ) = 3 × 10−25 kg m s−1
where 1 J = 1 kg m2 s−2 is used; note that 1 nm−1 = 1 × 109 m−1 . �e kinetic
energy is given by [�D.�–���]
E�D.�(a)
Ek =
(ħk)2 ((1.0546 × 10−34 J s) × (3 × 109 m−1 ))2
=
= 5 × 10−20 J
2m
2 × (9.1094 × 10−31 kg)
�e electron is travelling in the negative x direction hence the momentum,
and the value of k, are both negative. �e kinetic energy is related to k through
[�D.�–���], E k = k 2 ħ 2 �2m, which is rearranged to give k
k = −�
2mE k 1�2
2 × (2.0 × 10−3 kg) × (20 J)
�
=
−
�
�
ħ2
(1.0546 × 10−34 J s)2
1�2
�e wavefunction is then, with x is measured in metres,
E�D.�(a)
ψ(x) = eik x = e−i(2.7×10
33
= −2.7 × 1033 m−1
m−1 )x
�e energy levels of a particle in a box are given by [�D.�–���], E n = n 2 h 2 �8mL 2 ,
where n is the quantum number. With the mass equal to that of the electron
and the length as �.� nm, the energies are
En =
n2 h2
(6.6261 × 10−34 J s)2
2
=
n
×
8mL 2
8 × (9.1094 × 10−31 kg) × (1.0 × 10−9 m)2
= n 2 × (6.02... × 10−20 J)
To convert to kJ mol−1 , multiply through by Avogadro’s constant and divide by
1000.
= n 2 × (6.02... × 10−20 J) ×
6.0221 × 1023 mol−1
= n 2 × (36.2... kJ mol−1 )
1000
259
260
7 QUANTUM THEORY
To convert to electronvolts divide through by the elementary charge
= n 2 × (6.02... × 10−20 J) ×
1
= n 2 × (0.376... eV)
1.6022 × 10−19 C
To convert to reciprocal centimetres, divide by hc, with c in cm s−1
= n2 ×
(6.02... × 10−20 J)
= n 2 × (3.03... × 103 cm−1 )
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
�e energy separation between two levels with quantum numbers n 1 and n 2 is
∆E(n 1 , n 2 ) = E n 2 − E n 1 = (n 22 − n 12 ) × (6.02... × 10−20 J)
�e values in the other units are found by using the appropriate value of the
constant, computed above.
(i) ∆E(1, 2) = 1.8 × 10−19 J , 1.1 × 102 kJ mol−1 , �.� eV , or 9.1 × 103 cm−1
(ii) ∆E(5, 6) = 6.6 × 10−19 J , 4.0 × 102 kJ mol−1 , �.� eV , or 3.3 × 104 cm−1
E�D.��(a) �e wavefunctions are ψ 1 (x) = (2�L)1�2 sin(πx�L) and ψ 2 (x) = (2�L)1�2 ×
sin(2πx�L). �ey are orthogonal if ∫ ψ ∗1 ψ 2 dτ = 0. In this case the integral is
taken from x = 0 to x = L as outside this range the wavefunctions are zero. �e
required integral is of the form of Integral T.�, with A = π�L, B = 2π�L, and
a=L
L
2
πx
2πx
2 sin(−πL�L) sin(3πL�L)
� dx = �
−
�
� sin � � sin �
L 0
L
L
L
−2π�L
6π�L
=
E�D.��(a)
2 sin(−π) sin(3π)
�
−
�=0
L −2π�L
6π�L
where sin nπ = 0 for integer n is used. �e two wavefunctions are orthogonal.
�e particle in a box wavefunction with quantum number n is given by ψ n (x) =
(2�L)1�2 sin(nπx�L) �e probability of �nding the electron in a small region of
space δx centred on position x is approximated as ψ 2 (x)δx. For this Exercise
x = 0.50L, δx = 0.02L.
For the case where n = 1
�
2
ψ 1 (0.50L)2 × 0.02L = � 2�L sin (π(0.50L)�L)� × 0.02L
For the case where n = 2
= (2�L) sin2 (0.50π) × 0.02L = �.��
�
2
ψ 2 (0.50L)2 × 0.02L = � 2�L sin (2 × π(0.50L)�L)� × 0.02L
= (2�L) sin2 (1.00π) × 0.02L = �
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
ψ1
ψ 12
0.2
0.4
0.6
0.8
x�L
1.0
Figure 7.1
E�D.��(a) �e wavefunction
� of the lowest energy state for a particle in a box has n = 1
and is ψ 1 (x) = 2�L sin(πx�L), which leads to a probability density P1 (x) =
�ψ 1 (x)�2 = (2�L) sin2 (πx�L). Graphs of these functions are shown in Fig �.�.
�e probability density is symmetric about x = L�2. �erefore, there is an equal
probability of observing the particle at an arbitrary position x ′ and at L − x ′ , so
it follows that the average position of the particle must be at L�2.
�
E�D.��(a) �e wavefunction for the state with n = 1 is ψ 1 (x) = 2�L sin(πx�L), which
leads to a probability density P1 (x) = �ψ 1 (x)�2 = (2�L) sin2 (πx�L); these are
plotted in Fig �.�.
�e plotted probability density gives the probability of the particle being found
in an interval at a particular value of x – that is, it is the probability density of
x. It is not the probability density of x 2 , so there is no simple way of inferring
from the plot the value of �x 2 �. �e fact that the probability density of x is
symmetric about L�2 does not imply that the probability density of x 2 has the
same property. �ere is therefore no reason to assume that �x 2 � = (L�2)2 .
E�D.��(a) For an electron in a square well of side L the energy of the state characterized
by quantum numbers n 1 and n 2 is given by [�D.��b–���], E n x ,n y = h 2 (n 2x +
n 2y )�8m e L 2 , where n x , n y are integers greater than or equal to one. Hence, the
minimum or zero-point energy has n x = n y = 1,
E 1,1 = h 2 (12 + 12 )�8m e L 2 = h 2 �4m e L 2
Setting this equal to the rest energy m e c 2 , and rearranging for the length, L
h 2 �4m e L 2 = m e c 2
L = h�2m e c = λ C �2
where λ C = h�(m e c) is the Compton wavelength.
E�D.��(a) �e wavefunction for a particle
�in a one-dimensional box with length L and in
the state with n = 3 is ψ(x) = 2�L sin(3πx�L) giving a probability density of
P(x) = �ψ(x)�2 = (2�L) sin2 (3πx�L).
261
262
7 QUANTUM THEORY
As this is a trigonometric function, the function is maximized when sin2 (3πx�L) =
1, which is when sin(3πx�L) = ±1; these values occur when the argument of
the sine function is an odd multiple of π�2
3πx�L = π�2, 3π�2, 5π�2...
hence
x = L�6, L�2, 5L�6
�e probability density is zero when the wavefunction is zero, which is when
the argument of the sine function is a multiple of π
Solutions to problems
P�D.�
3πx�L = 0, π, 2π, ...
hence
x = 0, L�3, 2L�3, L
�e energy levels of a particle in a one dimensional box are E n = h 2 n 2 �8mL 2 .
�e mass of an O� molecule is 2 × 16.00 amu × 1.6605 × 10−27 kg = 5.3136 ×
10−26 kg. �e separation in energy of the lowest two levels, being n = 1 and
n = 2, is given by
h 2 22
h 2 12
h2
−
=
(4 − 1)
8mL 2 8mL 2 8mL 2
(6.6261 × 10−34 J s)2
=
×3
8 × (5.3136 × 10−26 kg) × (5.0 × 10−2 m)
∆E = E 2 − E 1 =
= 6.2 × 10−41 J
To �nd the value of n that makes the energy equal to the average thermal energy,
set E n = 12 kT and solve for n
√
2L mkT
n=
h
=
2×(5.0 × 10−2 m)×�(5.3136 × 10−26 kg)×(300 K)×(1.3806 × 10−23 J K−1 )�
1�2
6.6261 × 10−34 J s
= 2.23... × 109 = 2.2 × 109
�e separation of level n from the one immediately below it, n − 1, is E n − E n−1
h 2 n 2 h 2 (n − 1)2 h 2 (2n − 1)
−
=
8mL 2
8mL 2
8mL 2
−34
2
(6.6261 × 10 J s) × (2.23... × 109 − 1)
=
8 × (5.3136 × 10−26 kg) × (5.0 × 10−2 m)2
E n − E n−1 =
P�D.�
= 1.8 × 10−30 J
�e normalized wavefunction with n = 1 is ψ 1 (x) = (2�L)1�2 sin(πx�L), and
this is used in [�C.��–���], �Ω� = ∫ ψ ∗ Ω̂ψ dτ, to compute the expectation value
for the cases Ω̂ = x and Ω̂ = x 2 .
�x� = �
L
0
ψ ∗1 xψ 1 dx = (2�L) �
L
0
x sin2 (πx�L) dx
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e integral is of the form of Integral T.�� with a = L and k = π�L
=sin 2π=0
=cos(2π)=1
��� � � � � � � � � � � �� � � � � � � � � � � � ��
��� � � � � � � � � � � � � � � � � � � � � � � � � �
= (2�L)�L �4 − (L �4π) sin(2πL�L) −(L 2 �8π 2 ){cos(2πL�L) −1}�
2
= L�2
�x 2 � = �
L
0
2
ψ ∗1 x 2 ψ 1 dx = (2�L) �
L
0
x 2 sin2 (πx�L) dx
�e integral is of the form of Integral T.�� with a = L and k = π�L
= (2�L) �L 3 �6 − (L 3 �4π − L 3 �8π 3 ) sin(2πL�L) − (L 3 �4π 2 ) cos(2πL�L)�
P�D.�
= L 2 �3 − 1�2π 2
(a) Assuming that the system can be modelled as a �D particle in a box, the
energy levels of the system are given by [�D.�–���], E n = h 2 n 2 �8mL 2 .
Hence, the separation between the levels n and n + 1 is
∆E = E n+1 − E n = h 2 [(n + 1)2 − n 2 ]�8mL 2 = h 2 (2n + 1)�8mL 2
�e conjugated system consists of �� atoms, and so there are �� bonds,
therefore the length of the box is �� times the average internuclear distance
L = 11×140 pm = 1.54 nm. Hence the separation between the levels with
n = 6 and n = 7 is
∆E =
(6.6261 × 10−34 J s)2 × (2 × 6 + 1)
= 3.30 × 10−19 J
8 × (9.1094 × 10−31 kg) × (1.54 × 10−9 m)2
(b) �e energy and frequency of a transition are related by [�A.�–���], ∆E =
hν, rearranging for the frequency gives
ν = ∆E�h = (3.30... × 10−19 J)�(6.6261 × 10−34 J s) = 4.98 × 1014 Hz
(c) �e terms in the energy expression that change with the number of conjugated atoms N are the quantum number of the highest energy occupied
state, n, and the length of the box, L. �e energy, and hence frequency, of
the transition is directly proportional to 2n +1 and inversely proportional
to L 2 . Because the there are N electrons which occupy the states in pairs,
the quantum number of the highest occupied state is n = N�2 (rounded
up if N is odd). �e length L is proportional to the number of bonds
in the molecule, which is N − 1. Hence, the frequency of the transition is
proportional to (2(N�2)+1)�(N −1)2 = (N +1)�(N −1)2 ≈ N −1 , and so
the absorption spectrum of a linear polyene shi�s to a lower frequency
as the number of conjugated atoms increases .
P�D.�
�e expectation values are �x� = L�2, �x 2 � = L 2 (1�3−1�2n 2 π 2 ), �p x � = 0, �p2x � =
n 2 h 2 �4L 2 .
263
264
7 QUANTUM THEORY
(a) �e uncertainty in the position is
∆x = [�x 2 � − �x�2 ]1�2 = [L 2 (1�3 − 1�2n 2 π 2 ) − (L�2)2 ]1�2
= L(1�12 − 1�2n 2 π 2 )1�2
and in the momentum
∆p x = [�p2x � − �p x �2 ]1�2 = [n 2 h 2 �4L 2 − 02 ]1�2 = nh�2L
(b) �e product of these is then
∆x∆p x = L(1�12 − 1�2n 2 π 2 )1�2 × nh�2L
= (nh�2)(1�12 − 1�2n 2 π 2 )1�2
(c) �e Heisenberg uncertainty principle states that ∆x∆p x ≥ ħ�2. Rewriting
the expression in (b) in terms of ħ gives ∆x∆p x = (ħ�2) × 2πn(1�12 −
1�2n 2 π 2 )1�2 . For n = 1
∆x∆p x = (ħ�2) × 2π(1�12 − 1�2π 2 )1�2 = 1.13... × (ħ�2)
and for n = 2,
∆x∆p x = (ħ�2) × 4π(1�12 − 1�8π 2 )1�2 = 3.34... × (ħ�2)
In both cases ∆x∆p x ≥ (ħ�2), which is consistent with the uncertainty
principle. It is evident that ∆x∆p x is an increasing function of n and
therefore it can be inferred that the uncertainly principle is obeyed for all
n.
P�D.�
(a) In �D the Schrödinger equation with a potential energy function V (x, y, z)
is given by
−
ħ2 ∂2
∂2
∂2
� 2 + 2 + 2 � ψ + V (x, y, z)ψ = Eψ
2m ∂x
∂y
∂z
In this case V (x, y, z) = 0 inside the box and in�nite elsewhere. It follows that the wavefunction is zero outside the box. Inside the box the
Schrödinger equation is
−
ħ2 ∂2
∂2
∂2
� 2 + 2 + 2 � ψ = Eψ
2m ∂x
∂y
∂z
Assume that the solution is a product of three functions of a single variable, that is let ψ = X(x)Y(y)Z(z). Substituting this into the Schrödinger
equation gives
−
ħ2
d2 X
d2 Y
d2 Z
�Y Z 2 + XZ 2 + Y Z 2 � = EXY Z
2m
dx
dy
dz
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Dividing through by the function XY Z gives
−
ħ 2 1 d2 X 1 d2 Y 1 d2 Z
�
+
+
�=E
2m X dx 2 Y dy 2 Z dz 2
Now rearrange such that the terms that depend on x only are on the le�
hand side
ħ 2 d2 X
ħ 2 1 d2 Y 1 d2 Z
−
=E+
�
+
�
2
2mX dx
2m Y dy 2 Z dz 2
�e le� hand side depends only on x, whilst the right side depends only
on y and z. �e only way that this can be true for all x, y, z is if both sides
are equal to a constant, arbitrarily called E x , such that
−
ħ 2 d2 X
= Ex
2mX dx 2
Ex = E +
ħ 2 1 d2 Y 1 d2 Z
�
+
�
2m Y dy 2 Z dz 2
�e same procedure is followed with the functions of y and z to give
−
ħ 2 d2 Y
= Ey
2mY dy 2
−
ħ 2 d2 Z
= Ez
2mZ dz 2
where E = E x + E y + E z . �e assumption that the wavefunction can be
written as a product of single-variable functions is a valid one, as ordinary
di�erential equations can be found for the assumed factors. �at is what
it means for a partial di�erential equation to be separable.
(b) �e di�erential equation involving the function X is recognized as being
the same as the Schrödinger equation for a particle in a one-dimensional
box, whose solutions are already known: X(x) = (2�L)1�2 sin(nπx�L)
with energies E n = n 2 h 2 �8mL 2 . �e product X(x)Y(y)Z(z) which is
the solution to this three-dimensional problem will therefore be a product
with three such one-dimensional functions along x, y and z, each with the
appropriate length L i and quantum number n i
ψ(x, y, z) = (2�L 1 )1�2 sin(n x πx�L 1 ) × (2�L 2 )1�2 sin(n y πy�L 2 )
× (2�L 3 )1�2 sin(n z πz�L 3 )
Likewise, the total energy will be the sum of energies along each direction
2
h 2 n 2x n y n 2z
E = Ex + E y + Ez =
� +
+ �
8m L 21 L 22 L 23
(c) For a cubic box, L 1 = L 2 = L 3 = L and so
E = h 2 (n 2x + n 2y + n 2z )�8mL 2
Generally states with all three quantum numbers the same are non degenerate, if two quantum numbers are the same they are triply degenerate,
and if all three quantum numbers are the di�erent they are six-fold degenerate. In addition, there may be ‘accidental’ degeneracies, such as that
between (n x , n y , n z ) = (3, 3, 3) and (5, 1, 1). �e energy level diagram is
shown in Fig. �.�; in the diagram the states are labelled (n x , n y , n z ).
265
7 QUANTUM THEORY
��
(�,�,�)
(�,�,�),(�,�,�)
(�,�,�)
(�,�,�)
��
E n x ,n y ,n z �(h 2 �8ml 2 )
266
��
(�,�,�)
(�,�,�)
(�,�,�)
(�,�,�)
(�,�,�)
(�,�,�)
(�,�,�)
(�,�,�)
(�,�,�)
(�,�,�)
(�,�,�)
Figure 7.2
P�D.��
�
(d) Comparing the �D and �D energy diagrams, the one dimensional case is
more sparse than the three dimensional case, as the ��eenth level is not
reached until E n �(h 2 �8ml 2 ) = 225. In addition, none of the levels in the
one-dimensional case are degenerate.
�e rate of tunnelling is proportional to the transmission probability, given by
[�D.��b–���], T = 16ε(1 − ε)e−2κL . �erefore the ratio of tunnelling rates
is equal to the corresponding ratio of transmission probabilities. �e desired
factor is T1 �T2 where the subscripts denote di�erent tunnelling distances.
T1 16ε(1 − ε)e−2κ L 1
=
= e−2κ(L 1 −L 2 )
T2 16ε(1 − ε)e−2κ L 2
With κ = 7 nm−1 , L 1 = 1 nm and L 1 = 2 nm
P�D.��
−1
T1
= e−2κ(L 1 −L 2 ) = e−2(7 nm )[(1 nm)−(2 nm)] = 1.20 × 106
T2
(a) �e wavefunctions in each region are ψ 1 (x) = eik 1 x + B 1 ei k 1 x , ψ 2 (x) =
A 2 e k 2 x + B 2 e−k 2 x , ψ 3 (x) = A 3 eik 3 x . With this choice of A 1 = 1, the
transmission probability is simply T = �A 3 �3 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e wavefunction coe�cients are determined by the criteria that both the
wavefunction and its �rst derivative with respect to x are continuous at
the potential boundaries, ψ 1 (0) = ψ 2 (0), dψ 1 (0)�dx = dψ 2 (0)�dx, ψ 2 (W) =
ψ 3 (W), dψ 3 (W)�dx = dψ 2 (W)�dx. �ese establish the algebraic relationships
(a) 1 + B 1 = A 2 + B 2
(b) ik 1 − ik 1 B 1 = A 2 k 2 − B 2 k 2
(c) A 2 e k 2 W + B 2 e−k 2 W = A 3 eik 3 W
(d) A 2 k 2 e k 2 W − B 2 k 2 e−k 2 W = ik 3 A 3 eik 3 W
�ese equations need to be solved for A 3 and hence B 1 , A 2 , B 2 need to be
eliminated. A 3 appears in (c) and (d) only. Solving these equations for
A 3 eik 3 W and setting the results equal to each other yields
A 2 e k 2 W + B 2 e−k 2 W = (k 2 �ik 3 )(A 2 k 2 e k 2 W − B 2 e−k 2 W )
�is equation can be solved for B 2 in terms of A 2
B2 =
A 2 e2k 2 W (k 2 �ik 3 − 1) A 2 e2k 2 W (k 2 − ik 3 )
=
k 2 �ik 3 + 1
k 2 + ik 3
Note that B 1 appears only in (a) and (b). Solving these equations for B 1
and setting the results equal to each other yields
B 1 = A 2 + B 2 − 1 = 1 − (1�ik 1 )(A 2 k 2 − B 2 k 2 )
Substituting the expression for B 2 into this equation yields
A 2 �1 +
e2k 2 W (k 2 − ik 3 )
e2k 2 W (k 2 − ik 3 )
� = 2ik 1 − A 2 k 2 �1 −
�
k 2 + ik 3
k 2 + ik 3
�is can be rearranged to give
−1
1 e2k 2 W (k 2 − ik 3 )
k 2 (k 2 − ik 3 )
k 2 e2k 2 W (k 2 − ik 3 )
A2 = � +
+
−
�
2
2(k 2 + ik 3 )
2ik 1 (k 2 + ik 3 )
2ik 1 (k 2 + ik 3 )
Hence,
B2 =
�is gives
−1
A 2 e2k 2 W (k 2 − ik 3 )
k 2 + ik 3
= A 2 � 2k W
�
2
k 2 + ik 3
e
(k 2 − ik 3 )
=�
−1
k 2 + ik 3
1
k2
k2
+ +
−
�
2k
W
2k
W
2
2
2e
(k 2 − ik 3 ) 2 2ik 1 e
2ik 1
A 3 = e−ik 3 W (A 2 e k 2 W + B 2 e−k 2 W ) =
4k 1 k 2 eik 2 W
(ia + b)e k 2 W − (ia − b)e−k 2 W
Since sinh z = 12 (ez − e−z ), substitute e k 2 L = 2 sinh k 2 L + e k 2 L
A3 =
2k 1 k 2 eik 2 L
(ia + b) sinh(k 2 W) + be−k 2 W
267
268
7 QUANTUM THEORY
�e transmission coe�cient is then
4k 12 k 22
(a 2 + b 2 ) sinh2 (k 2 L) + b 2
T = �A 3 �2 =
where a 2 + b 2 = (k 12 + k 22 )(k 22 + k 32 ), and b 2 = k 22 (k 1 + k 3 )2 .
(b) In the special case, for which V1 = V3 = 0, k 1 = k 3 . Additionally, k 1 �k 2 =
E�(V2 − E) = ε�(1 − ε), where ε = E�V2 . Also
a 2 + b 2 = (k 12 + k 22 )2 = k 24 [1 + (k 1 �k 2 )2 ]2
b 2 = 4k 12 k 22
Hence
T=
−1
b2
a2 + b2
= �1 +
sinh2 k 2 W�
2
b2
b 2 + (a 2 + b 2 ) sinh (k 2 W)
−1
sinh2 (k 2 W)
= �1 +
�
4ε(1 − ε)
−1
[ 1 (e k 2 W − e−k 2 W )]2
= �1 + 2
�
4ε(1 − ε)
−1
(e k 2 W − e−k 2 W )2
= �1 +
�
16ε(1 − ε)
(�.�)
Which is eqn �D.��a as expected. In the case of a high, wide barrier,
k 2 W � 1. �is implies that e−k 2 W is negligibly small in comparison to
e k 2 W , and that 1 is negligibly small compared to e2k 2 W �16ε(1 − ε), such
that
−1
T ≈ �e2k 2 W �16ε(1 − ε)� = 16ε(1 − ε)e−2k 2 W
(c) �e function plotted in Fig. �.� is eqn �.� with the parameters given and
P�D.��
k 2 = [2m(V2 − E)]1�2 �ħ
ε = E�V2
�
(a) �e particle in a box wavefunctions are given by ψ n (x) = 2�L sin(nπx�L),
giving a probability density of Pn (x) = �ψ n (x)�2 = (2�L) sin2 (nπx�L).
�is probability density is plotted in Fig. �.� for n = 1 . . . 5, and for n = 50
in Fig. �.�.
�e correspondence principle is that as the quantum number becomes
large the quantum result must converge with the classical result. For a
particle in a box, the classical result is an even probability. As n increases
the oscillations become more rapid leading to a more uniform distribution especially if it is averaged over short distance which is a small fraction
of the width of the box.
(b) �e transmission probability of a particle of mass m passing through a
barrier of height V0 , length W is given by, [�D.��a–���]
−1
(eκW − e−κW )2
T = �1 +
�
16ε(1 − ε)
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
T
0.2
0.1
0.0
10
12
14
16
V2 �kJ mol
Figure 7.3
18
−1
20
n=1
n=2
n=3
n=4
n=5
ψ 2n �(2�L)
1.0
0.5
0.0
0.0
Figure 7.4
0.2
0.4
0.6
0.8
1.0
x�L
�
�
where ε = E�V0 and κ = 2m(V0 − E)�ħ = 2mV0 (1 − ε)�ħ. A plot of
T versus ε is shown in Fig. �.� for the passage by a H� molecule, a proton,
and an electron.
�e curves in the �gure di�er in the value of W(mV0 )1�2 �ħ , a measure of
the obstruction that the barrier represents taking into account its height,
width and the type of particle. �e curves can be thought of as having the
same value of W and V0 , but di�ering only in m. �e values of W and V0
were chosen such that the proton and hydrogen molecule exhibit ‘typical’
tunnelling behaviour: if the incident energy is small enough, there is
practically no transmission, and if the incident energy is high enough,
transmission is virtually complete. A barrier through which a proton and
hydrogen molecule exhibit tunnelling behaviour is practically no barrier
for an electron on account of the much smaller mass of the latter. On this
plot, T for the electron is essentially �.
(c) �e wavefunction of a �D particle in a box with quantum numbers n 1 and
269
7 QUANTUM THEORY
2
ψ 50
�(2�L)
1.0
0.5
0.0
0.0
0.2
0.4
0.6
0.8
1.0
x�L
Figure 7.5
H�
p+
e–
1.0
T
270
0.5
0.0
0.0
1.0
Figure 7.6
2.0
ε
3.0
4.0
n 2 is given by [�D.��a–���]
ψ n 1 ,n 2 (x, y) = (2�L x )1�2 (2�L y )1�2 sin(n 1 πx�L x ) sin(n 2 πy�L y )
�is is plotted for n 1 = 1, n 2 = 1, and for n 1 = 1, n 2 = 2 in Fig. �.�; the
wavefunctions with n 1 = 2, n 2 = 1, and with n 1 = 2, n 2 = 2 are plotted in
Fig. �.�.
As can be seen from the plots, the function sin(n 1 πx�L x ) has n 1 −1 nodes
where the function passes through zero; the boundaries do not count
as nodes because the wavefunction does not pass through zero at these
points. Similarly the function sin(n 2 πy�L y ) has n 2 − 1 nodes. In two
dimensions a node in the wavefunction along x or y leads to a nodal line,
and the total number of these is (n 1 − 1) + (n 2 − 1) = n 1 + n 2 − 2 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
n 1 = 1, n 2 = 1
ψ�(2�L x )1�2 (2�L y )1�2
1
0.5
0
0
1
0.2
0.4
0.6
0.5
0.8
x�L x
1 0
y�L y
psi�(2�L x )1�2 (2�L y )1�2
n 1 = 1, n 2 = 2
1
0
−1
0
Figure 7.7
1
0.2
0.4
x�L x
0.6
0.5
0.8
1 0
y�L y
7E Vibrational motion
Answers to discussion questions
D�E.�
If the harmonic oscillator were to have zero energy the particle would be at
rest and located at the bottom of the potential energy well (x = 0). Such
an arrangement has no uncertainty in either the position or the momentum,
which is not in accord with the uncertainty principle. By giving the oscillator
zero-point energy, there is some uncertainty in the position as the particle
oscillates back and forth, and then it is possible for there to be the appropriate
uncertainty in the momentum such that the uncertainty principle is satis�ed.
�e existence of zero-point energy is thus traced to the need to satisfy the
uncertainty principle.
D�E.�
For the harmonic oscillator the spacing of the energy levels is constant. �erefore, relative to the energy of the oscillator, the spacing becomes progressively
smaller as the quantum number increases. In the limit of very high quantum
numbers this spacing becomes negligible compared to the total energy, and
e�ectively the energy can take any value, as in the classical case.
271
7 QUANTUM THEORY
psi�(2�L x )1�2 (2�L y )1�2
n 1 = 2, n 2 = 1
1
0
−1
0
1
0.2
0.4
x�L x
psi�(2�L x )1�2 (2�L y )1�2
272
0.6
0.5
0.8
1 0
y�L y
n 1 = 2, n 2 = 2
1
0
−1
0
Figure 7.8
1
0.2
0.4
x�L x
0.6
0.5
0.8
1 0
y�L y
As is shown in Fig. �E.� on page ���, as the quantum number becomes large the
probability density clusters more and more around the classical turning points
of the classical harmonic oscillator (that is, the points at which the kinetic
energy is zero). Because the classical oscillator is moving most slowly near
these points, they are the displacements at which it is most probable that the
oscillator will be found. Again, the quantum and classical results converge at
high quantum numbers
Solutions to exercises
E�E.�(a)
�e wavefunctions are depicted in Fig. �E.� on page ���. �e general form of
2
the harmonic oscillator wavefunctions is ψ υ = N υ H υ (y)e−y �2 , where N υ is
a normalization constant and H υ (y) is the Hermite polynomial of order υ in
the reduced position variable y. Nodes occur when the wavefunction passes
through zero. �e wavefunction asymptotically approaches zero at y = ±∞,
but as the function does not pass through zero these limits do not count as
nodes. �e nodes in the wavefunction therefore correspond to the solutions of
H υ (y) = 0. H υ (y) is a polynomial of order υ, meaning that the highest power
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�E.�(a)
E�E.�(a)
of y that occurs is y υ ; such polynomials in general have υ solutions and hence
there are υ nodes. �erefore (a) the wavefunction with υ = 3 has 3 nodes; (b)
the wavefunction with υ = 4 has 4 nodes.
�e wavefunction with υ = 2 is ψ 2 (y) = N 2 (y 2 − 1)e−y �2 . Nodes occur when
the wavefunction passes through zero; the wavefunction approaches zero at
y = ±∞, but these do not count as nodes as the wavefunction does not pass
through zero. It is evident that the nodes occur when y 2 − 1 = 0, which solves
to give nodes at y = −1, +1 .
2
�e wavefunction with υ = 1 is ψ 1 (y) = N 1 ye−y �2 , which gives a probability
2
density of P(y) = �ψ 2 (y)�2 = N 12 y 2 e−y . �e extrema are located by di�erentiating the wavefunction, setting the result to 0 and solving for y. �e di�erential
is evaluated using the product rule
2
� dy 2 −y 2
2
2
dψ 1 (y)
de−y �
= N 12
e + y2
= N 12 [2ye−y + y 2 × (−2ye−y )]
dy
dy �
� dy
2
= N 12 2y(1 − y 2 )e−y
2
�is function is equal to zero when y = 0, or 1 − y 2 = 0 or when e−y = 0. �ese
correspond to extrema at y = 0, ±1, ±∞.
2
To identify the maxima, consider the form of the probability density. �is goes
to zero at y = 0, ±∞, and so these must be minima, implying that y = ±1
correspond to maxima.
E�E.�(a)
�e zero-point energy of a harmonic oscillator is given by [�E.�–���], E 0 =
1
ħω, where the frequency ω is given by [�E.�–���], ω = (k f �m)1�2 . For this
2
system,
E 0 = 12 × (1.0546 × 10−34 J s) × �(155 N m−1 )�(2.33 × 10−26 kg)�
E�E.�(a)
1�2
= 4.30 × 10−21 J
�e separation between adjacent energy levels of a harmonic oscillator is [�E.�–
���], ∆E = ħω, where the frequency, ω is given by [�E.�–���], ω = (k f �m)1�2 .
�is is rearranged for the force constant as k f = m(∆E�ħ)2 . Evaluating this
gives
k f = (1.33 × 10−25 kg) × �(4.82 × 10−21 J)�(1.0546 × 10−34 J s)� = 278 N m−1
2
E�E.�(a)
�e separation between adjacent energy levels of a harmonic oscillator is [�E.�–
���], ∆E = ħω, where the frequency, ω is given by [�E.�–���], ω = (k f �m)1�2 .
�e Bohr frequency condition [�A.�–���], ∆E = hν, can be rewritten in terms
of the wavelength as ∆E = hc�λ. �e wavelength of the photon corresponding
to a transition between adjacent energy levels is therefore given by ħω = hc�λ,
or ħ(k f �m)1�2 = hc�λ. Solving for λ gives λ = 2πc�(k f �m)1�2 ; with the data
273
274
7 QUANTUM THEORY
given
λ=
E�E.�(a)
E�E.�(a)
2π × (2.9979 × 108 m s−1 )
�(855 N m−1 )�(1.0078 × 1.6605 × 10−27 kg)�
1�2
= 2.64 × 10−6 m
�e wavefunctions are depicted in Fig. �E.� on page ���; they are real. Two
wavefunctions are orthogonal if ∫ ψ ∗i ψ j dτ = 0. In this case the wavefunctions
2
2
are ψ 0 (y) = N 0 e−y �2 and ψ 1 (y) = N 1 ye−y �2 , and the integration is from
2
y = −∞ to +∞. �e integrand ψ 0 ψ 1 is N 0 N 1 ye−y , which is an odd function,
meaning that its value at −y is the negative of its value at +y. �e integral of
an odd function over a symmetric range is zero, hence these wavefunctions are
orthogonal.
�e zero-point energy of a harmonic oscillator is given by [�E.�–���], E 0 =
1
ħω, where the frequency ω is given by [�E.�–���], ω = (k f �µ)1�2 . �e e�ec2
tive mass µ of a diatomic AB is given by [�E.�–���], µ = (m A m B )�(m A + m B ).
In the case of a homonuclear diatomic A� this reduces to µ = m A �2. With the
data given
E 0 = 12 × (1.0546 × 10−34 J s)
× �(329 N m−1 )�( 12 × 34.9688 × 1.6605 × 10−27 kg)�
E�E.�(a)
1�2
= 5.61 × 10−21 J
For the state with υ = 0 the energy of a harmonic oscillator is given by [�E.�–
���], E 0 = 12 ħω, where the frequency ω is given by [�E.�–���], ω = (k f �m)1�2 .
(i) For the system with k f = 1000 N m−1 the energy of the state with υ = 0 is
E 0 = 12 × (1.0546 × 10−34 J s) × [(1000 N m−1 )�(1 × 1.6605 × 10−27 kg)]1�2
= 4.09... × 10−20 J = 4.09 × 10−20 J
2
�e classical turning points of this state occur when E 0 = 12 k f x tp
. Solving
�
this for x tp leads to x tp = ± 2E 0 �k f , giving a separation of
�
�
2 2E 0 �k f = 2 2 × (4.09... × 10−20 J)�(1000 N m−1 ) = 18.1 pm
(ii) For the system with k f = 100 N m−1 the energy of the state with υ = 0 is
E 0 = 12 × (1.0546 × 10−34 J s) × [(100 N m−1 )�(1 × 1.6605 × 10−27 kg)]1�2
= 1.29 × 10−20 J
2
�e classical turning points of this state occur when E 0 = 12 k f x tp
. Solving
�
this for x tp leads to x tp = ± 2E 0 �k f , giving a separation of
�
�
2 2E 0 �k f = 2 2 × (1.29... × 10−20 J)�(100 N m−1 ) = 32.2 pm
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Solutions to problems
P�E.�
�e intermolecular potential is electrostatic in origin and arises from the interaction between the protons and electrons, the number and distribution of
which remain the same on isotopic substitution and so therefore does the potential. �e force constant, which re�ects the shape of the intermolecular potential, is therefore una�ected by such a change.
(a) �e force constants for the molecules are the same therefore ω A′ B �ω AB =
(k f �µ A′ B )1�2 �(k f �µ AB )1�2 = (µ AB �µ A′ B )1�2 ; multiplying through by ω AB ,
gives ω A′ B = ω AB (µ AB �µ A′ B )1�2 . �e vibrational frequency in Hz, ν, is
related to ω by ν = ω�2π, hence ν A′ B = ν AB (µ AB �µ A′ B )1�2 .
(b) �e e�ective mass of �H ��Cl is µ 1 = (1 × 35)�(35 + 1) = 0.972... m u , that
of �H ��Cl is µ 2 = (2 × 35)�(35 + 2) = 1.89... m u , and that of �H ��Cl is
µ 3 = (1 × 37)�(1 + 37) = 0.973... m u .
�e vibrational frequency of �H ��Cl is
ν 2 = ν 1 (µ 1 �µ 2 )1�2 = (5.63 × 1014 Hz)[(0.972... m u )�(1.89... m u )]1�2
= 4.04 × 1014 Hz
and for �H ��Cl
ν 3 = ν 1 (µ 1 �µ 3 )1�2 = (5.63 × 1014 Hz)[(0.972... m u )�(0.973... m u )]1�2
P�E.�
P�E.�
= 5.63 × 1014 Hz
Assuming that the force constant is the same
then the
� for all
� the molecules,
�
ratio of frequencies is ω 2 �ω 1 = ν 2 �ν 1 = k f �µ 2 � k f �µ 1 = µ 1 �µ 2 . For a
homonuclear diatomic molecule A� the e�ective mass is µ A2 = m A2 �2m A =
m A �2. Hence, µ 1H2 = 1�2 = 0.5 m u , µ 2H2 = 2�2 = 1 m u , µ 3H2 = 3�2 = 1.5 m u .
�
�
ν 2H2 = ν 1H2 µ 1H2 �µ 2H2 = 131.9 THz × 0.5 m u �1 m u = 93.27 THz
�
�
ν 3H2 = ν 1H2 µ 1H2 �µ 3H2 = 131.9 THz × 0.5 m u �1.5 m u = 76.15 THz
(a) �e wavenumber, ν̃ is given by ν̃ = ν�c, where c is the speed of light (see
�e chemist’s toolkit ��). �e frequency in Hz is related to the angular
frequency by ν = ω�2π, hence ν̃ = ω�2πc .
(b) �e vibrational frequency of �H ��Cl is 5.63×1014 s−1 , which gives a wavenumber of
ν̃ = (5.63 × 1014 s−1 )�[2π × (2.9979 × 1010 cm s−1 )] = 2.99 × 103 cm−1
Note that in order to express the wavenumber in cm−1 , the speed of light
is used in cm s−1 .
(c) �e frequency ω is given by [�E.�–���], ω = (k f �µ) , and so the vibra1�2
tional frequency expressed as a wavenumber is ν̃ = (1�2πc) (k f �µ) .
�is rearranges to give k f = µ(2πν̃c)2 .
1�2
275
276
7 QUANTUM THEORY
(d) �e e�ective mass of ��C ��O is µ = (12 × 16)�(12 + 16) = 6.85... m u .
Using the previous result, and converting the mass to kg, gives the force
constant
k f = [6.85... × (1.6605 × 10−27 kg)]
× [2π × (2170 cm−1 ) × (2.9979 × 1010 cm s−1 )]2 = 1902 N m−1
P�E.�
P�E.�
For an isotopic substitution the vibrational�frequencies are related by the
e�ective masses (Problem P�E.�) ν̃ 2 �ν̃ 1 = µ 1 �µ 2 . �e e�ective mass of
�� ��
C O is µ 14C 16O = (14 × 16)�(14 + 16) = 7.46... m u . �is gives the
wavenumber for ��C ��O as
�
ν̃ 14C 16O = ν̃ 12C 16O µ 12C 16O �µ 14C 16O
�
= (2170 cm−1 ) × 6.85... m u �7.46 m u = 2080 cm−1
If the C atom is immobilized, only the O will be moving, and the force constant
will be the same as that in the CO molecule. �e wavenumber of the CO
vibration is 2170 cm−1 , as given in Problem P�E.�, and as the force constant is
the�same this is linked to the wavenumber of the vibration when bound as ν̃ 2 =
ν̃ 1 µ 1 �µ 2 , where 1 refers to the free molecule and 2 that bound to the haem
group. �e e�ective mass of ��C ��O is µ 1 = (12×16)�(12+16) = 6.85... m u , and
in the bound case the e�ective mass is simply the mass of the O atom, 16 m u ,
as this is the only atom which is moving. Hence
�
ν̃ 2 = (2170 cm−1 ) × (6.85... m u )�(16 m u ) = 1420 cm−1
Expressed as a frequency this is ν = 4.259 × 1013 Hz.
(a) �e second derivative of the function ψ = e−gx is evaluated using the
chain rule
2
� dx −gx 2
d2 ψ
d2
d
de−gx �
−gx 2
−gx 2
=
�e
�
=
�−2gxe
�
=
−2g
e
+
x
dx 2 dx 2
dx
dx �
� dx
2
= −2g �e−gx + x × −2gxe−gx � = 2g(2gx 2 − 1)e−gx
2
2
2
(b) �e Hamiltonian for the harmonic oscillator is Ĥ = −(ħ 2 �2m)d2 �dx 2 +
2
1
k x 2 . To �nd the condition for e−gx to be an eigenfunction of Ĥ, �rst
2 f
allow the operator to act on the function
Ĥe−gx = �−(ħ 2 �2m)d2 �dx 2 + 12 k f x 2 � e−gx
2
2
= −(ħ 2 �2m)2g(2gx 2 − 1)e−gx + 21 k f x 2 e−gx
2
2
(�.�)
Evidently e−gx is not an eigenfunction of Ĥ because when this operator
2
acts on e−gx the result is not a constant times that function. However, this
condition would be satis�ed if the terms in x 2 were to cancel one another,
that is
−(ħ 2 �2m)2g(2gx 2 ) + 12 k f x 2 = 0
2
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Solving this for g gives g = (mk f )1�2 �2ħ ; this is the value of g needed
for e−gx to be an eigenfunction of the hamiltonian. �e resulting wave1�2 2
function is ψ = e−(mk f ) x �2ħ , which is in agreement with that quoted in
[�E.�b–���].
(c) Returning to eqn �.�, if the terms in x 2 cancel the remaining terms are
2
−(ħ 2 �2m)2g(−1)e−gx
2
which is of the form of a constant times the original function. �e constant is the eigenvalue, which in this case is the energy
E = (ħ 2 �2m)2g = (ħ 2 �2m) × 2 × (mk f )1�2 �2ħ = 12 ħ(k f �m)1�2
�e energy is indeed the same as that of the lowest level of the harmonic
oscillator as described in the text.
P�E.��
(a) �e variable y is de�ned as y = x�α where α = (ħ 2 �mk f )1�4 , and the kinetic energy operator in terms of x is T̂ = −(ħ 2 �2m)d2 �dx 2 . Substituting
in y leads to
T̂ = −
=−
ħ 2 d2
ħ 2 d2
=− 2
2
2m d(α y)
2α m dy 2
ħ 2 mk f 1�2 d2
k f 1�2 d2
d2
1
1
� 2 �
=
−
ħ
�
�
=
−
ħω
2
2
2m ħ
dy 2
m
dy 2
dy 2
where for the last line the de�nition of the frequency ω is used, ω =
(k f �m)1�2 .
(b) �e expectation value of the kinetic energy of the state with quantum
number υ is given by
�T�υ = − 12 ħωN υ2 �
∞
−∞
H υ e−y �2
2
2
d2
H υ e−y �2 dy
2
dy
To evaluate the derivative, the product rule is employed,
2
d2
d
d e−y �2
d
−y 2 �2
′ −y 2 �2
H
e
=
(H
e
+
H
)=
(H ′ − yH υ )e−y �2
υ
υ
υ
dy 2
dy
dy
dy υ
2
�e product rule is applied once more
2
d(H ′υ − yH υ ) −y 2 �2
d2
d e−y �2
H υ e−y �2 =
e
+ (H ′υ − yH υ )
2
dy
dy
dy
2
= (H ′′υ − yH ′υ − H υ )e−y �2 − y(H ′υ − yH υ )e−y �2
2
= [H ′′υ − 2yH ′υ + (y 2 − 1)H υ ]e−y �2
2
2
From Table �E.� on page ��� one of the properties of the Hermite polynomials is H ′′υ − 2yH ′υ + 2υH υ = 0, so it follows that H ′′υ − 2yH ′υ = −2υH υ .
Using this in the last line above gives
2
2
2
d2
H υ e−y �2 = −(2υ + 1)H υ e−y �2 + y 2 H υ e−y �2
2
dy
277
278
7 QUANTUM THEORY
(c) A further property of the Hermite polynomials is H υ+1 −2yH υ +2υH υ−1 =
0, which rearranges to yH υ = υH υ−1 + 12 H υ+1 . Using this, the term y 2 H υ
is rewritten
y 2 H υ = y(yH υ ) = y(υH υ−1 + 12 H υ+1 ) = yυH υ−1 + 12 yH υ+1
�e same relationship is used to substitute for the two terms on the right,
but rewritten �rstly by making the substitution υ → υ − 1 to give yH υ−1 =
(υ − 1)H υ−2 + 12 H υ , and secondly with with the substitution υ → υ + 1 to
give yH υ+1 = (υ + 1)H υ + 12 H υ+2 . With these substitutions
yυH υ−1 + 12 yH υ+1 = υ[(υ − 1)H υ−2 + 12 H υ ] + 12 [(υ + 1)H υ + 12 H υ+2 ]
= υ(υ − 1)H υ−2 + (υ + 12 )H υ + 14 H υ+2
Hence, the second derivative is given by
2
2
d2
H υ e−y �2 = �−(2υ + 1)H υ + y 2 H υ � e−y �2
dy 2
= �−(2υ + 1)H υ + υ(υ − 1)H υ−2 + (υ + 12 )H υ + 14 H υ+2 � e−y �2
2
= �υ(υ − 1)H υ−2 − (υ + 12 )H υ + 14 H υ+2 � e−y �2
2
(d) �e expectation value of the kinetic energy is therefore given by
�T�υ = − 12 ħωN υ2 �
∞
−∞
H υ e−y �2 [υ(υ − 1)H υ−2
2
= − 12 ħωN υ2 �υ(υ − 1) �
− (υ + 12 )H υ + 14 H υ+2 ]e−y �2 dy
∞
−∞
2
H υ H υ−2 e−y dy + 14 �
2
∞
H υ H υ+2 e−y dy
2
−∞
∞
−y 2
1
− (υ + 2 ) �
H υ H υ e dy�
−∞
�e �rst two integrals are zero due to orthogonality of the Hermite polynomials, leaving
�T�υ = 12 (υ + 12 )ħωN υ2 �
∞
−∞
H υ2 e−y dy
2
(�.�)
�e next step to to evaluate the normalization constant N υ which is found
from
∞
2
N υ2 �
H υ2 e−y dy = 1
−∞
A property of the Hermite polynomials is (Table �E.� on page ���)
�
∞
−∞
H υ2 e−y dy = π 1�2 2υ υ!
2
If follows that N υ2 = 1�(π 1�2 2υ υ!). However, by the same property of
the Hermite polynomials the integral in eqn �.� is also equal to π 1�2 2υ υ!,
therefore this term cancels with N υ2 to leave
�T�υ = 12 (υ + 12 )ħω = 12 E υ
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P�E.��
As is shown in Example �E.� on page ���, in terms of the dimensionless variable
y the classical turning points are at y tp = ±(2υ + 1)1�2 , where υ is the quantum
√
number of the state. For the �rst excited state υ = 1, and so y tp = ± 3.
�e wavefunction is ψ 1 = N 1 ye−y �2 , which is normalized by solving
2
+∞
N 12 �
−∞
y 2 e−y dy = 2N 12 �
2
0
+∞
y 2 e−y dy = 1
2
�e integral is of the form of Integral G.� with k = 1 and evaluates to 14 π 1�2 ,
thus N 12 = 2π−1�2 . �e probability of �nding the particle outside the range of
the turning points is then
∞
P = 2 × 2π−1�2 �√ y 2 e−y dy =
2
3
P�E.��
∞
4
�√
3
π 1�2
y 2 e−y dy
2
Evaluating this integral numerically gives P = 0.112 .
∞
�e integral to be evaluated is I = ∫−∞ ψ ∗υ′ xψ υ dx. �is can be written in terms
of the dimensionless variable y, de�ned as x = α y where α = (ħ 2 �mk f )1�4 ,
∞
∞
as I = ∫−∞ ψ ∗υ′ (α y)ψ υ d(α y) = α 2 ∫−∞ ψ ∗υ′ yψ υ dy. �e wavefunctions are
2
ψ υ = N υ H υ e−y �2 , where N υ is a normalization constant, and H υ is a Hermite
polynomial of order υ. With these substitutions I is
I = α 2�
∞
−∞
(N υ′ H υ′ e−y �2 )y(N υ H υ e−y �2 ) dy
2
∞
= α N υ N υ′ �
2
−∞
2
H υ′ (yH υ )e−y dy
2
Using the properties of the Hermite polynomials given in Table �E.� on page
���, the term yH υ can be rewritten as yH υ = 12 H υ+1 +υH υ−1 , and so the integral
can be rewritten as
I = α 2 N υ N υ′ � 12 �
∞
−∞
H υ′ H υ+1 e−y dy + υ �
2
∞
−∞
H υ′ H υ−1 e−y dy�
2
∞
(�.�)
Due to orthogonality of the Hermite polynomials, the integral ∫−∞ H υ′ H υ e−y dy
is only non-zero when υ′ = υ. Hence, the integral I is only non-zero if υ′ = υ+1
or υ′ = υ − 1. Hence, the only transitions with intensities not equal to zero are
those with υ′ = υ ± 1 .
2
�e normalization constant N υ is found from
N υ2 �
∞
−∞
H υ2 e−y dy = 1
2
∞
It is a property of the Hermite polynomials that ∫−∞ H υ2 e−y dy = π 1�2 2υ υ! so
it follows that N υ = (π 1�2 2υ υ!)−1�2 .
2
For υ′ = υ+1 only the �rst integral in eqn �.� is non-zero. From the properties of
2
∞
2
Hermite polynomials it follows that ∫−∞ H υ+1
e−y dy = [π 1�2 2υ+1 (υ + 1)!]1�2 .
279
280
7 QUANTUM THEORY
Using this, and the normalization factors already quoted, gives
I = α 2 N υ N υ+1 12 �
∞
−∞
2
H υ+1
e−y dy
2
1
1
α2
1�2 υ+1
= 12 α 2 1�2 υ 1�2 1�2 υ+1
(π
2
(υ
+
1)!)
=
(υ + 1)1�2
(π 2 υ!) [π 2 (υ + 1)!]1�2
21�2
In the case that υ′ = υ − 1 only the second integral in eqn �.� is non-zero. Using
2
∞
2
the same procedure as before with ∫−∞ H υ−1
e−y dy = [π 1�2 2υ−1 (υ − 1)!]1�2 ,
gives
I = α 2 N υ N υ−1 υ �
∞
−∞
P�E.��
2
H υ−1
e−y dy
2
1
1
α2 √
1�2 υ−1
= υα 2 1�2 υ 1�2 1�2 υ−1
[π
2
(υ
−
1)!]
=
υ
[π 2 υ!] (π 2 (υ − 1)!)1�2
21�2
(a) �e harmonic oscillator potential energy is V (x) = 12 k f x 2 which is symmetric about x = 0. It therefore follows that the probability density must
also be symmetric about x = 0, for all wavefunctions. Hence the probability of �nding the particle at x = x ′ is the same as �nding it at x = −x ′
and so, for all υ, �x�υ = 0 .
(b) As the potential is symmetric about x = 0 the probability of �nding the
particle moving to the right with a particular positive momentum is equal
to that of �nding the particle moving to the le� with same momentum,
but with opposite sign. It therefore follows that �p x �υ = 0 .
(c) From the virial theorem, �E k �υ = 12 E υ = 12 (υ + 12 )ħω. As the kinetic
energy is E k = p2x �2m, where m is the mass of the particle, it follows that
�p2x �υ = 2m�E k �υ = 2m × 12 (υ + 12 )ħω = m(υ + 12 )ħω .
(d) �e value of �x 2 �υ is given in [�E.��b–���] as �x 2 �υ = (υ + 12 )ħ�(mk f )1�2 .
∆x υ = [�x 2 �υ − �x�2υ ]1�2 = [(υ + 12 )ħ�(mk f )1�2 − 02 ]1�2
= (υ + 12 )1�2 (ħ 2 �mk f )1�4
∆p υ = [�p2x �υ − �p x �2υ ]1�2 = [m(υ + 12 )ħω − 02 ]1�2
= (υ + 12 )1�2 (ħmω)1�2
(e) �e product of the uncertainties is
∆x υ ∆p υ = (υ + 12 )(ħ 2 �mk f )1�4 × (ħmω)1�2 = (υ + 12 )(ħ 4 mω 2 �k f )1�4
�is can be simpli�ed by noting that ω = (k f �m)1�2 ; using this in the
expression above gives ∆x υ ∆p υ = (υ + 12 )ħ . �is function increases
linearly with υ and its minimum value is when υ = 0: ∆x 0 ∆p 0 = ħ�2.
�is value satis�es the uncertainty principle, ∆x∆p ≥ ħ�2, as do wavefunctions with higher values of υ.
(f) �e minimum product of the uncertainties is for υ = 0 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
7F
Rotational motion
Answers to discussion questions
D�F.�
�e vector model of angular momentum is described in Section �F.�(c) on page
���. �e model captures the key idea that the magnitude and the z-component
of the angular momentum can be known simultaneously and precisely. However, if this is so there is no knowledge of the other components of the angular
momentum other than that the sum of their squares must be equal to a particular value.
D�F.�
Rotational motion on a ring and on a sphere share the following features: (a)
quantization arising as a result of the need to satisfy a cyclic boundary condition; (b) energy levels which go inversely with the moment of inertia; (c)
the lack of zero-point energy; (d) degeneracy; (e) quantization of the angular
momentum about one axis.
Solutions to exercises
E�F.�(a)
�e diagrams shown in Fig. �.� are drawn by forming a vector of length [l(l +
1)]1�2 and√with a projection m l on the z-axis. For l = 1, m l = +1 the vector is
of length √2 and has projection +1 on the z-axis; for l = 2, m l = 0 the vector is
of length 6 and has projection 0 on the z-axis. Each vector may lie anywhere
on a cone described by rotating the vector about the z-axis.
z
+1
0
Figure 7.9
E�F.�(a)
l = 1, m l = 1
l = 2, m l = 0
�e spherical harmonic is
Y3,0 = 14 (7�π)1�2 (5 cos3 θ − 3 cos θ) = 14 × (7�π)1�2 cos θ(5 cos2 θ − 3)
E�F.�(a)
�e nodes occur when Y3,0 passes through zero, which
happens when either
�
cos θ = 0 or 5 cos2 θ − 3 = 0, that is when cos θ = ± 3�5; recall that θ is in the
range 0 to π. Hence, the nodes are θ = π�2, 0.684, 2.46 . �ere are 3 angular
nodes.
�
�e real part of the spherical harmonic Y1,+1 is − 12 3�π sin θ cos �. When �
varies, an angular node therefore occurs when cos � = 0, i.e. at � = π�2, 3π�2 .
�is plane corresponds to the plane x = 0, i.e. the yz plane.
�
�e imaginary part of the same spherical harmonic is − 12 3�π sin θ sin �. When
� varies, an angular node therefore occurs when sin � = 0, i.e. at � = 0, π. �is
plane corresponds to the plane y = 0, i.e. the xz plane.
281
282
7 QUANTUM THEORY
E�F.�(a)
E�F.�(a)
�e rotational energy depends only on the quantum number l [�F.��–���], but
there are distinct states for every allowed value of m l , which can range from −l
to +l in integer steps. �ere are 2l + 1 such states, as there are l of these with
m l > 0, l of these with m l < 0 and m l = 0. Hence l = 3 has a degeneracy of � .
E�F.�(a)
�e projection of the angular momentum onto the z-axis is given by [�F.�–���],
ħm l , where m l is a quantum number that takes values between −l and +l in
integer steps, m l = −l , −l + 1, . . . + l. Hence the possible projections onto the
z-axis are −ħ, 0, ħ .
�e magnitude of the angular momentum associated with a wavefunction with
angular momentum quantum number l is given by [�F.��–���], magnitude =
ħ[l(l + 1)]1�2 . Hence for l = 1 the magnitude is ħ[1(1 + 1)]1�2 = 21�2 ħ .
�e wavefunction of a particle on a ring, with quantum number m l is ψ m l =
eim l � = cos(m l �) + i sin(m l �) in the range 0 ≤ � ≤ 2π. �e real and imaginary
parts of the wavefunction are therefore cos(m l �) and sin(m l �) respectively.
Nodes occur when the function passes through zero, which for trigonometric
functions are the same points at which the function is zero. Hence in the real
part, nodes occur when cos(m l �) = 0, and so when m l � = (2n + 1)π�2 for
integer n, which gives � = (2n + 1)π�2m l . In the imaginary part, nodes occur
when sin(m l �) = 0 and so when m l � = nπ for an integer n, which gives � =
nπ�m l .
(i) With m l = 0 the real part is a constant and has no nodes ; the imaginary
part is zero everywhere.
(ii) With m l = 3, nodes in the real part occur at π�6, π�2, 5π�6, 7π�6, 3π�2 ,
11π�6 . In the imaginary part nodes occur at 0, π�3, 2π�3, π, 4π�3, 5π�3 .
�ere are � nodes in each of the parts.
E�F.�(a)
�e normalization condition is ∫ ψ ∗m l ψ m l dτ = 1. In this case the integral is
over � in the range 0 ≤ � ≤ 2π, and the wavefunction is ψ m l = Neim l � . Hence,
noting that the wavefunction is complex
N 2 � ψ ∗m l ψ m l dτ = N 2 �
E�F.�(a)
2π
0
e−im l � eim l � d� = N 2 �
2π
0
d� = 2πN 2
where e−iθ e+iθ = e0 = 1 is used. Setting this integral to � gives N = (2π)−1�2 .
�e integral to evaluate is ∫0 ψ ∗m ′ ψ m l d�, with ψ m l = eim l � . �is wavefunction
2π
l
is complex and so ψ ∗m ′ = e−im l � . Note that both m l and m′l are integers and
l
therefore m l ± m′l is also an integer. �e integral is then
′
�
2π
0
e−im l � eim l � d� = �
′
2π
0
′
ei(m l −m l )� d�
= (i[m l − m′l ])−1 (e2πi(m l −m l ) − e0 )
′
= (i[m l − m′l ])−1 (1 − 1) = 0
where e2πin = 1 for any integer n is used. Hence, wavefunctions for a particle
on a ring with di�erent quantum numbers are orthogonal.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�F.�(a)
E�F.��(a)
E�F.��(a)
E�F.��(a)
E�F.��(a)
�e energy levels of a particle on a ring are [�F.�–���], E m l = m 2l ħ 2 �2I where I
is the momentum of inertia of the system, I = mr 2 , see �e chemist’s toolkit ��
in Topic �F on page ���. �e minimum excitation is therefore ∆E = E 1 − E 0 =
(ħ 2 �2I)(12 − 02 ) = ħ 2 �2I Evaluating this gives
∆E =
(1.0546 × 10−34 J s)2
= 3.32 × 10−22 J
2 × (1.6726 × 10−27 kg) × (100 × 10−12 m)2
�e energy levels are [�F.��–���], E l = ħ 2 l(l + 1)�2I, where I is the moment
of inertia. �e minimum energy to start it rotating is the minimum excitation
energy, the energy to take it from the motionless l = 0 to the rotating l = 1
state, ∆E = E 1 − E 0 = (ħ 2 �2I)(1(1 + 1) − 0(0 + 1)] = ħ 2 �I. Evaluating this gives
∆E = (1.0546 × 10−34 J s)2 �(5.27 × 10−47 kg m2 ) = 2.11 × 10−22 J
�e energy levels are [�F.��–���], E l = ħ 2 l(l + 1)�2I, where I is the moment
of inertia. So that the excitation energy is ∆E = E 2 − E 1 = (ħ 2 �2I)[2(2 + 1) −
1(1 + 1)] = 2ħ 2 �I. Evaluating this gives
∆E = 2(1.0546 × 10−34 J s)2 �(5.27 × 10−47 kg m2 ) = 4.22 × 10−22 J
�e energy levels are [�F.��–���], E l = ħ 2 l(l + 1)�2I, where I is the
�moment
of inertia. �e corresponding angular momentum is �l 2 �1�2 = ħ l(l + 1).
Hence, the minimum energy allowed is 0, through this corresponds to zero
angular momentum, and so rest and not motion. So the minimum energy of
rotation occurs for the state that has l = 1. �e angular momentum of that state
�
√
√
1�2
is �l 2 �1 = ħ 1(1 + 1 = 2ħ = 2 × (1.0546 × 10−34 J s) = 1.49 × 10−34 J s .
�e diagrams shown in Fig. �.�� are drawn by forming a vector of length [l(l +
1�2
1)]
and with a projection m l on the z-axis. For l = 1 the vector is of length
√
2
and
has projection −1, 0, +1 on the z-axis. For l = 2 the vector is of length
√
6 and has projection −2, . . . + 2 in integer steps on the z-axis. Each vector
may lie anywhere on a cone described by rotating the vector about the z-axis.
z
+2
z
+1
+1
−1
−1
0
Figure 7.10
+1
+2
0
0
−1
−2
−2
+1
−1
0
283
284
7 QUANTUM THEORY
E�F.��(a)
�e angle in question is that between the z-axis and the vector representing the
angular momentum. �e projection
of the vector onto the z-axis is m l ħ, and
�
the length of the vector is ħ l(l + 1). �erefore the angle θ that the vector
�
makes to the z-axis is given by cos θ = m l � l(l + 1).
�
√
When m l = l, cos θ = l� l(l + 1), which for l = 1 gives cos θ = 1� 2, and so
√
θ = π�4 , and for l = 5 gives cos θ = 5� 30, and so θ = 0.420 .
Solutions to problems
P�F.�
�e angular momentum states have quantum numbers m l = 0, ±1, ±2.... �e
energy levels for a particle on a ring are given by [�F.�–���], E m l = m 2l ħ 2 �2I,
and have angular momentum [�F.�–���], J z = m l ħ. �e moment of inertia for
an electron on this ring is I = mr 2 = (9.1094 × 10−31 kg) × (440 × 10−12 m)2 =
1.76... × 10−49 kg m2
(a) If there are 22 electrons in the system the highest occupied state will be
the degenerate levels m l = ±5. �ese states have an energy of E±5 =
(±5)2 (1.0546 × 10−34 J s)2 �2(1.76... × 10−49 kg m2 ) = 7.88 × 10−19 J , and
angular momenta of
J z = ±5ħ = ±5 × (1.0546 × 10−34 J s) = 5.273 × 10−34 J s
(b) �e lowest unoccupied levels are those with m l = ±6, and so the difference in energy between the highest occupied and lowest unoccupied
levels is
∆E = E±6 − E±5 = (ħ 2 �2I)(62 − 52 )
= 11 × 1.0546 × 10−34 J s)2 �2(1.76... × 10−49 kg m2 ) = 3.46... × 10−19 J
�e Bohr frequency condition, [�A.�–���] states that the frequency of
radiation that will excite such a transition is
P�F.�
ν = ∆E�h =
3.46... × 10−19 J
= 5.23 × 1014 Hz
6.6261 × 10−34 J s
In Cartesian coordinates, the equation de�ning the ellipse is x 2 �a 2 + y 2 �b 2 = 1.
An appropriate change of variables can transform this ellipse into a circle. �at
change of variable is most conveniently described in terms of new Cartesian
coordinates (X, Y) where X = x and Y = ay�b. In these coordinates, the
equation for the ellipse can be rewritten as X 2 + Y 2 = a 2 , which is the equation
of a circle radius a centered on the origin. �e text found the eigenfunctions
and eigenvalues for a particle on a circular ring by transforming from Cartesian
coordinates to plane polar coordinates. A similar transformation can be made
by de�ning coordinates (R, Φ) such that X = R cos Φ, Y = R sin Φ.
In these coordinates, this is simply a particle on a ring, as described in the text,
for which the Schrödinger equation is separable .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P�F.�
�e Schrödinger equation for a particle on a sphere is −(ħ 2 �2I)Λ̂ 2 ψ = Eψ,
where
1 ∂2
1 ∂
∂
Λ̂ 2 =
+
sin θ
∂θ
sin2 θ ∂� 2 sin θ ∂θ
(a) Y0,0 = (1�2)π−1�2 , which is a constant, and so its derivatives with respect
to all θ and � are zero, so Λ̂ 2 Y0,0 = 0 implying that E = 0 and lˆz Y0,0 = 0,
so that J z = 0 .
(b) Y2,−1 = N sin θ cos θe−i� , thus
∂Y2,−1 �∂θ = Ne−i� (cos2 θ−sin2 θ)
∂Y2,−1 �∂� = −iN sin θ cos θe−i� = −iY2,−1
In addition, ∂ 2 Y2,−1 �∂� 2 = N sin θ cos θe−i�
1 ∂ 2 Y2,−1
1 ∂
∂Y2,−1
+
sin θ
2
2
sin θ ∂θ
∂θ
sin θ ∂�
N cos θe−i� Ne−i� ∂
=
+
sin θ(cos2 θ − sin2 θ)
sin θ
sin θ ∂θ
Λ̂ 2 Y2,−1 =
�e derivative is evaluated using the product rule
=
N cos θe−i�
sin θ
Ne−i�
�sin θ(−4 cos θ sin θ) + cos θ(cos2 θ − sin2 θ)�
+
sin θ
as cos3 θ = cos θ cos2 θ = cos θ(1 − sin2 θ)
N cos θe−i�
cos θ
+ Ne−i� �−6 sin θ cos θ +
�
sin θ
sin θ
= −6Ne−i� sin θ cos θ
=
�is has an eigenvalue of −6, giving an energy eigenvalue of 6ħ 2 �2I . For
angular momentum, lˆz Y2,−1 = (ħ�i)×−iY2,−1 = ħY2,−1 , giving an angular
momentum eigenvalue of J z = −ħ .
(c) Y3,+3 = Ne3i� sin3 θ, thus
∂Y3,+3 �∂θ = 3Ne3i� sin2 θ cos θ
∂Y3,+3 �∂� = 3iNe3i� sin3 θ = 3iY3,+3
In addition, ∂ 2 Y3,+3 �∂� 2 = −9Ne3i� sin3 θ. Hence,
1 ∂ 2 Y3,+3
1 ∂
∂Y3,+3
+
sin θ
sin θ ∂θ
∂θ
sin2 θ ∂� 2
3
3i�
−9Ne sin θ
1 ∂
=
+
3Ne3i� sin3 θ cos θ
sin θ ∂�
sin2 θ
Λ̂ 2 Y3,+3 =
285
286
7 QUANTUM THEORY
�e derivative is evaluated using the product rule
3Ne3i�
(3 sin2 θ cos2 θ − sin4 θ)
sin θ
= −9Ne3i� sin θ + 3Ne3i� (3 sin θ cos2 θ − sin3 θ)
= −9Ne3i� sin θ +
= −12Ne3i� sin3 θ = −12Y3,+3
�is has an eigenvalue of −12, giving an energy eigenvalue of 12ħ 2 �2I .
For angular momentum, lˆz Y2,−1 = (ħ�i) × 3iY2,−1 = 3ħY2,−1 , giving an
angular momentum eigenvalue of J z = 3ħ .
P�F.�
�e energies are given by [�F.��–���], E l = ħ 2 l(l + 1)�2I, and therefore E 0 =
0, E 2 = 2(3)ħ 2 �2I = 6ħ 2 �2I and E 3 = 4(3)ħ 2 �2I = 12ħ 2 �2I, all of which
are consistent with the calculated eigenvalues.
�
�
�e function Y1,+1 = − 12 3�2π sin θei� and Y1,0 = 12 3�π cos θ. �e integral
to evaluate is
�
π
θ=0
�
2π
�=0
∗
Y1,0
Y1,+1 sin θ dθ d�
Y1,0 is real and so the integrand is
�
�
√
− 12 3�2π sin θei� × 12 3�π cos θ × sin θ = −(3�4π 2) sin2 θ cos θei�
�is gives the integral as
I=�
π
θ=0
�
2π
�=0
∗
Y1,0
Y1,+1 sin θ dθ d� = −
π
2π
3
√ � �
sin2 θ cos θei� dθ d�
θ=0
�=0
4π 2
�e integrand is the product of separate functions of θ and �, and so the integral
can be separated
I=−
2π
π
3
√ �
ei� d� � sin2 θ cos θ dθ
0
4π 2 0
Evaluating the �rst integral gives
Hence
�
2π
0
ei� d� = (1�i) ei� �0 = (1�i)[e2πi − e0 ] = (1�i)(1 − 1) = 0
2π
�
π
θ=0
�
2π
�=0
∗
Y1,0
Y1,+1 sin θ dθ d� = 0
so the two functions are orthogonal .
P�F.�
(a) Multiplying out the brackets, noting that the derivatives in the le� brackets act on the whole term in the right brackets, e.g. x∂ f �∂x
∂
∂
∂f
∂f
lˆx lˆy f = −ħ 2 �y − z � �z
−x �
∂z
∂y
∂x
∂z
= −ħ 2 �y
∂
∂f
∂
∂f
∂
∂f
∂
∂f
�z � − y �x � − z �z � + z �x ��
∂z ∂x
∂z
∂z
∂y ∂x
∂y
∂z
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
To evaluate the �rst term in this, the product rule is used
= −ħ 2 �y
= −ħ 2 �y
(b) Similarly,
∂z ∂ f
∂2 f
∂2 f
∂2 f
∂2 f
+ yz
− x y 2 − z2
+ zx
�
∂z ∂x
∂z∂x
∂z
∂y∂x
∂y∂z
∂f
∂2 f
∂2 f
∂2 f
∂2 f
+ yz
− x y 2 − z2
+ zx
�
∂x
∂z∂x
∂z
∂y∂x
∂y∂z
∂
∂f
∂f
∂
lˆy lˆx f = −ħ 2 �z
− x � �y
−z �
∂x
∂z
∂z
∂y
= −ħ 2 �z
∂
∂f
∂
∂f
∂
∂f
∂
∂f
�y � − z
�z � − x �y � + x �z ��
∂x
∂z
∂x
∂y
∂z
∂z
∂z ∂y
To evaluate the �nal term in this, the product rule must be used
= −ħ 2 �yz
= −ħ 2 �yz
∂2 f
∂2 f
∂2 f
∂z ∂ f
∂2 f
− z2
− xy 2 + x
+ xz
�
∂x∂z
∂x∂y
∂z
∂z ∂y
∂z∂y
∂2 f
∂2 f
∂2 f
∂f
∂2 f
− z2
− xy 2 + x
+ xz
�
∂x∂z
∂x∂y
∂z
∂y
∂z∂y
(c) Due to the symmetry of mixed partial derivatives, the only terms that are
not repeated in both of these terms are the �rst derivatives. Hence,
∂f
∂f
ħ
∂f
∂f
lˆx lˆy f − lˆy lˆx f = ħ 2 �x
− y � = iħ × �x
− y � = iħ lˆz f
∂y
∂x
i
∂y
∂x
where the de�nition of the lˆz operator given in [�F.��–���] is used. It
follow that [ lˆx , lˆy ] = lˆz .
(d) Applying cyclic permutation to lˆx gives (ħ�i)(z∂�∂x − x∂�∂z) = lˆy . Likewise lˆy gives (ħ�i)(x∂�∂y − y∂�∂x) = lˆz , and lˆz gives (ħ�i)(y∂�∂z −
z∂�∂y) = lˆx .
P�F.��
(e) Applying this permutation to the expression [ lˆx , lˆy ] = iħ lˆz gives [ lˆy , lˆz ] =
iħ lˆx . Permuting this expression gives [ lˆz , lˆx ] = iħ lˆy , and permuting that
expression gives [ lˆx , lˆy ] = iħ lˆz .
�e Cartesian coordinates expressed in terms of the spherical polar coordinates
are x = r sin θ cos �, y = r sin θ sin �, z = r cos θ, see �e chemist’s toolkit �� in
Topic �F on page ���. �e chain rule is therefore used to express ∂�∂� in terms
of derivatives of x, y and z.
∂
∂x ∂
∂y ∂
∂z ∂
=
+
+
∂� ∂� ∂x ∂� ∂y ∂� ∂z
Evaluating the derivatives gives
= −r sin θ sin �
∂
∂
∂
∂
∂
+ r sin θ cos �
+ 0 = −y
+x
∂x
∂y
∂z
∂x
∂y
where it has been noted that the factors multiplying the derivatives are Cartesian coordinates. Hence, lˆz = (ħ�i)(x∂�∂y − y∂�∂x) = (ħ�i)∂�∂�
287
288
7 QUANTUM THEORY
Answers to integrated activities
I�.�
(a) �e �rst step is to compute the total energy of the system of N A particles,
which is identi�ed as the internal energy U. �e energy levels for a particle in a cubic box of side L are given by [�D.��b–���], E n = h 2 n 2 �8mL 2 ,
where n 2 = n 12 + n 22 + n 32 . If there are N A particles, all occupying the
level corresponding to a particular value of n, the internal energy of the
system is U = N A E n = N A h 2 n 2 �8mL 2 . Using V = L 3 the length is
written in terms of the volume as L = V 1�3 , hence L 2 = V 2�3 and therefore
U = N A h 2 n 2 �8mV 2�3 .
If the expansion is adiabatic (that is, not heat �ows into or out of the
system) then from the First Law, dU = dq + dw, it follows that dU = dw.
�e work done on expansion is therefore computed by �nding how U
changes with volume, speci�cally by �nding ∂U�∂V .
NA h2 n2
∂U
∂ NA h2 n2
−2 N A h 2 n 2
=
�
�
=
×
=
−
∂V ∂V 8mV 2�3 adia
3
8mV 5�3
12mV 5�3
�e change in internal energy on expansion through dV will therefore be
��� � � � � � � ��� � � � � � � � �
∂U
NA h2 n2
dU = �
� dV = −
dV
∂V adia
12mV 5�3
A
(�.�)
�e work is equal to this change in internal energy. For a �nite change
the expression is integrated with respect to V between limits V1 and V2 ,
with ∆V = V2 − V1 .
(b) It is evident from eqn �.� that dU, and hence the work, goes as n 2 .
(c) �e work of expansion against an external pressure p ex is given by [�A.�a–
��], dw = −p ex dV . In eqn �.� the term A which multiplies dV refers to the
sample itself, and so must presumably in some way re�ect the pressure
of the sample, not the external pressure. However, if the expansion is
reversible, the external pressure is equal to the internal pressure and the
term A can then be identi�ed as the pressure. �erefore, if it is assumed
that the expansion is both adiabatic and reversible
p=
NA h2 n2
12mV 5�3
�e expression can be rewritten in terms of the average energy of each
particle which, because they all occupy the same level, is simply E av =
n 2 h 2 �8mL 2 = n 2 h 2 �8mV 2�3 , hence
��� � � � � �� � � � � � ��
N A h 2 n 2 8N A n 2 h 2
2N A E av
p=
=
=
5�3
2�3
12V 8mV
3V
12mV
E av
�is expression is reminiscent of the form of the pressure derived using
the kinetic theory of gases (Topic �B): pV = 13 nMυ 2rms , where n is the
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
amount in moles, M is the molar mass, and υ rms is the root-mean-square
speed. Because M = mN A , where m is the mass of a molecule, the expression can be rewritten
��� � � � � �� � � � � �
Ek
pV = 13 nmN A υ 2rms = n 23 N A 12 mυ 2rms
�e term 12 mυ 2rms is identi�ed as the average kinetic energy of one molecule
and, because in the kinetic theory the only energy a molecule possesses is
kinetic, E k can further be identi�ed as the average energy, E av . �us, for
one mole (n = 1)
pV = 23 N A E av
hence
p=
2N A E av
3V
�e two expressions for the pressure are therefore directly comparable
within the restrictions imposed.
(d) For an isothermal expansion heat would have to enter the system in order
to maintain its temperature, and this would involve promoting particles
to higher energy levels. As the volume increases the energy levels move
closer together, so the promotion of particles to higher levels needs to
o�set this e�ect as well.
I�.�
(a) In Problem P�D.� and Problem P�D.� it is shown that for a particle in a
box in a state with quantum number n
hence
∆x = L(1�12 − 1�2n 2 π 2 )1�2
and
∆p x = nh�2L
∆x∆p x = L(1�12 − 1�2n 2 π 2 )1�2 × nh�2L = (nh�2)(1�12 − 1�2n 2 π 2 )1�2
For n = 1
∆x∆p x = L(1�12 − 1�2π 2 )1�2 × h�2L = (h�2)(1�12 − 1�2π 2 )1�2 ≈ 0.57ħ
and for n = 2 ∆x∆p x ≈ 1.7ħ. �e Heisenberg uncertainty principle
is satis�ed in both cases, and it is evident that ∆x∆p x is an increasing
function of n. �e principle is therefore satis�ed for all n > 1.
(b) In Problem P�E.�� it is shown that for a harmonic oscillator in a state with
quantum number υ
∆x υ ∆p υ = (υ + 12 )ħ
I�.�
�erefore, for the ground state with υ = 0, ∆x∆p = 12 ħ: the Heisenberg
uncertainty principle is satis�ed with the smallest possible uncertainty. It
follow that for υ > 0 the principle is also satis�ed because ∆x υ ∆p υ is an
increasing function of υ.
Macroscopic synthesis and material development always contains elements of
randomness at the molecular level. Crystal structures are never perfect. A
289
290
7 QUANTUM THEORY
product of organic synthesis is never absolutely free of impurities, although
impurities may be at a level that is lower than measurement techniques make
possible. Alloys are grainy and slightly non-homogeneous within any particular grain. Furthermore, the random distribution of atomic/molecular positions
and orientations within, and between, macroscopic objects causes the conversion of energy to non-useful heat during manufacturing processes. Production
e�ciencies are di�cult to improve. Nanometre technology on the 1 nm to
100 nm scale may resolve many of these problems. Self-organization and production processes by nanoparticles and nanomachines may be able to exclude
impurities and greatly improve homogeneity by e�ective examination and selection of each atom/molecule during nanosynthesis and nanoproduction processes. Higher e�ciencies of energy usage may be achievable as nanomachines
produce idealized materials at the smaller sizes and pass their products to larger
nanomachines for production of larger scale materials.
�e directed, non-random, use of atoms and molecules by nanotechniques
holds the promise for the production of smaller transistors and wires for the
electronics and computer industries. Unusual material strengths, optical properties, magnetic properties, and catalytic properties may be achievable. Higher
e�ciencies of photo-electronic conversion would have a great impact.
8
8A
Atomic structure and
spectra
Hydrogenic Atoms
Answers to discussion questions
D�A.�
(i) A boundary surface for a hydrogenic orbital is drawn so as to contain
most (say ��%) of the probability density of an electron in that orbital. Its
shape varies from orbital to orbital because the electron density distribution is di�erent for di�erent orbitals.
(ii) �e radial distribution function P(r) gives the probability density that the
electron will be found at a distance r from the nucleus. It is de�ned such
that P(r) dr is the probability of �nding the electron in a shell of radius r
and thickness dr. Because the radial distribution function gives the total
density, summed over all angles, it has no angular dependence and, as a
result, perhaps gives a clearer indication of how the electron density varies
with distance from the nucleus.
D�A.�
(i) �e principal quantum number n determines the energy of a hydrogenic
atomic orbital through [�A.�–���].
(ii) �e azimuthal quantum number l determines the magnitude of the orbital angular momentum, given by [l(l + 1)]1�2 ħ.
(iii) �e magnetic quantum number m l determines the z-component of the
orbital angular momentum, given by m l ħ.
(iv) �e spin quantum number s determines the magnitude of the spin angular momentum, given by [s(s + 1)]1�2 ħ; for hydrogenic atomic orbitals s
can only be 12 .
(v) �e quantum number m s determines the z-component of the spin angular momentum, given by m s ħ; for hydrogenic atomic orbitals m s can only
be ± 12 .
Solutions to exercises
E�A.�(a)
�e radial wavefunction of a �s orbital is given in Table �A.� on page ��� as
R 3,0 = N(6 − 6ρ + ρ 2 )e−ρ�2 , where ρ = 2Zr�3a 0 . Radial nodes occur when the
wavefunction passes through 0, which is when 6 − 6ρ + ρ 2 = 0. �e roots of this
292
8 ATOMIC STRUCTURE AND SPECTRA
quadratic equation are at ρ = 3 ±
√
3 and hence the nodes are at
r = (3 ±
√
3)(3a 0 �2Z)
�e wavefunction goes to zero as ρ → ∞, but this does not count as a node as
the wavefunction does not pass through zero.
E�A.�(a)
Angular nodes occur when cos θ sin θ cos � = 0, which occurs when any of
cos θ, sin θ, or cos � is equal to zero; recall that the range of θ is 0 → π and of
� is 0 → 2π.
Although the function is zero for θ = 0 this does not describe a plane, and so
is discounted. �e function is zero for θ = π�2 with any value of �: this is the
x y plane. �e function is also zero for � = π�2 with any value of θ: this is the
yz plane. �ere are two nodal planes, as expected for a d orbital.
E�A.�(a)
E�A.�(a)
�e radial distribution function is de�ned in [�A.��b–���], P(r) = r 2 R(r)2 .
For the �s orbital R(r) is given in Table �A.� on page ��� as R 2,0 = N(2 −
ρ)e−ρ�2 where ρ = 2Zr�na 0 , which for n = 2 is ρ = Zr�a 0 . With the substitution r 2 = ρ 2 (a 0 �Z)2 , the radial distribution function is therefore P(ρ) =
N 2 (a 0 �Z)2 ρ 2 (2 − ρ)2 e−ρ .
Mathematical so�ware is used √
to �nd the values of ρ for which dP(ρ)�dρ = 0,
giving the results ρ = 0, 2, 3 ± 5. �e simplest way to identify which of these
is a maximum is to plot
√ P(ρ) against ρ, from which it is evident that ρ = 2 is a
minimum√
and ρ = 3± 5 are both maxima, with the principal maximum being
at ρ = 3 + 5. �e maximum in the radial distribution function is therefore at
√
r = (3 + 5)(a 0 �Z) .
�e radius at which the electron is most likely to be found is that at which the
radial distribution function is a maximum. �e radial distribution function is
de�ned in [�A.��b–���], P(r) = r 2 R(r)2 . For the �p orbital R(r) is given in
Table �A.� on page ��� as R 2,1 = N ρe−ρ�2 where ρ = 2Zr�na 0 , which for n = 2
is ρ = Zr�a 0 . With the substitution r 2 = ρ 2 (a 0 �Z)2 , the radial distribution
function is therefore P(ρ) = N 2 (a 0 �Z)2 ρ 4 e−ρ .
To �nd the maximum in this function the derivative is set to zero; the multiplying constants can be discarded for the purposes of this calculation
d 4 −ρ
ρ e = 4ρ 3 e−ρ − ρ 4 e−ρ
dr
E�A.�(a)
Setting this derivative to zero gives the solutions ρ = 0 and ρ = 4. P(ρ) is zero
for ρ = 0 and as ρ → ∞, therefore ρ = 4 must be a maximum. �is occurs at
r = 4a 0 �Z .
�e M shell has n = 3. �e possible values of l (subshells) are �, corresponding
to the s orbital, l = 1 corresponding to the p orbitals, and l = 2 corresponding
to the d orbital; there are therefore � subshells . As there is one s orbital, � p
orbitals and � d orbitals, there are � orbitals in total.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�A.�(a)
�e magnitude
� of the orbital angular momentum of an orbital with quantum
number l is l(l + 1)ħ. �e total number of nodes for an orbital with quantum
number n is n − 1, l of these are angular and so the number of radial nodes is
n − l − 1.
orbital
�s
�s
�d
n
�
�
�
l
�
�
�
ang. mom.
�
�
√
6ħ
angular nodes
�
�
�
radial nodes
�
�
�
E�A.�(a)
All the �p orbitals have the same value of n and l, and hence have the same
radial function, which is given in Table �A.� on page ��� as R 2,1 = N ρe−ρ�2
where ρ = 2Zr�na 0 , which for n = 2 is ρ = Zr�a 0 . Radial nodes occur when
the wavefunction passes through zero. �e function goes to zero at ρ = 0 and as
ρ → ∞, but it does not pass through zero at these points so they are not nodes.
�e number of radial nodes is therefore � .
E�A.�(a)
�e energy of the level of the H atom with quantum number n is given by
[�A.�–���], E n = −hc R̃ H �n 2 . As described in Section �A.�(d) on page ���,
the degeneracy of a state with quantum number n is n 2 .
E�A.�(a)
�e state with E = −hc R̃ H has n = 1 and degeneracy (1)2 = � ; that with
E = −hc R̃ H �9 has n = 3 and degeneracy (3)2 = � ; and that with E = −hc R̃ H �25
has n = 5 and degeneracy (5)2 = �� .
�e task is to �nd the value of N such that the integral ∫ ψ ∗ ψ dτ = 1, where
ψ = Ne−r�a 0 . �e integration is over the range r = 0 to ∞, θ = 0 to π, and
� = 0 to 2π; the volume element is r 2 sin θ dr dθ d�. �e required integral is
therefore
N2 �
0
∞
�
π
0
�
2π
0
r 2 e−2r�a 0 sin θ dr dθ d�
�e integrand is a product of functions of each of the variables, and so the
integral separates into three
N2 �
0
2
∞
r 2 e−2r�a 0 dr �
π
0
sin θ dθ �
2π
0
d�
= N [2!�(2�a 0 )3 ] × (− cos θ)�0 × ��0
π
2π
= N 2 [2!�(2�a 0 )3 ] × 2 × 2π = N 2 a 03 π
�e integral over r is evaluated using Integral E.� with n = 2 and k = 2�a 0 .
Setting the full integral equal to � gives N = (a 03 π)−1�2 .
E�A.��(a) �e wavefunction is given by [�A.��–���], ψ n,l ,m l = Yl ,m l (θ, �)R n,l (r); for the
state with n = 2, l = 0, m l = 0 this is
ψ 2,0,0 = Y0,0 (θ, �)R 1,0 (r) = (4π)−1�2 (Z�2a 0 )3�2 (2 − ρ)e−ρ�2
293
294
8 ATOMIC STRUCTURE AND SPECTRA
where the radial wavefunction is taken from Table �A.� on page ���, the angular
wavefunction (the spherical harmonic) is taken from Table �F.� on page ���,
and ρ = 2Zr�na 0 . �e probability density is therefore
P2,0,0 = �ψ 2,0,0 �2 = (4π)−1 (Z�2a 0 )3 (2 − ρ)2 e−ρ
E�A.��(a)
�e probability density at the nucleus, ρ = 0, is then (1�4π)Z 3 �(8a 03 )(2 −
0)2 e0 = Z 3 �(8πa 03 ) .
�e radial wavefunction of a �s orbital is taken from Table �A.� on page ���,
R 2,0 (r) = N(2−ρ)e−ρ�2 , where ρ = 2Zr�na 0 ; for n = 2, ρ = Zr�a 0 . �e extrema
are located by �nding the values of ρ for which dR 2,0 �dρ = 0; the product rule
is required
dR 2,0
d(2 − ρ) −ρ�2
de−ρ�2
=N
e
+ N(2 − ρ)
dρ
dρ
dρ
= N(−1)e−ρ�2 + N(2 − ρ)(− 12 e−ρ�2 ) = N(ρ�2 − 2)e−ρ�2
�e derivative is zero when ρ = 4, which corresponds to r = 4a 0 �Z . �e wavefunction is positive at ρ = 0, negative at ρ = 4, and asymptotically approaches
zero as ρ → ∞; ρ = 4 must therefore correspond to a minimum.
E�A.��(a) Assuming that the electron is in the ground state, the wavefunction is ψ =
Ne−r�a 0 , and so the probability density is P(r) = ψ 2 = N 2 e−2r�a 0 . P(r) is a
maximum at r = 0 and then simply falls o� as r increases; it falls to 12 its initial
value when P(r ′ )�P(0) = 12
P(r ′ )�P(0) = e−2r �a 0 = 12
′
Hence r ′ = − 12 ln 12 a 0 = 0.347a 0 .
Solutions to problems
P�A.�
�e �p orbitals only di�er in the axes along which they are directed. �erefore,
the distance from the origin to the position of maximum probability density
will be the same for each.
�e radial function for the �p orbitals is R 2,1 = N ρe−ρ�2 , where ρ = Zr�a 0 . �e
2
probability density is the square of the radial function, R 2,1
= N 2 ρ 2 e−ρ , and the
2
maximum in this is found by setting dR �dρ = 0
2
dR 2,1
= 2N 2 ρe−ρ − N 2 ρ 2 e−ρ = 0
dρ
Turning points occur at ρ = 0, 2, and it is evident from a plot of R(ρ) that ρ = 2
is the maximum. �is corresponds to r = 2a 0 �Z. For the 2pz orbital, for which
the angular part goes as cos θ, the maximum will be at θ = 0, which corresponds
to x = y = 0. �e position of maximum probability density is therefore at
x = 0, y = 0, z = 2a 0 �Z . �e corresponding positions for the other �p orbitals
are found by permuting the x, y and z coordinates.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P�A.�
�e energy levels of a hydrogenic atom with atomic number Z are given by
[�A.��–���], E n = −hcZ 2 R̃ N �n 2 , where the Rydberg constant for the atom is
given by [�A.��–���], R̃ N = (µ�m e )R̃∞ ; µ is the reduced mass of the atom.
For the D atom, with nuclear mass m D , the reduced mass is
mD me
µ=
mD + me
and therefore the Rydberg constant for D is
µ
mD me
mD
R̃ D =
R̃∞ =
R̃∞ =
R̃∞
me
m e (m D + m e )
mD + me
2.01355 × (1.660 539 × 10−27 kg)
=
[2.01355 × (1.660 539 × 10−27 kg)]+(9.109 382 × 10−31 kg)
× (109 737 cm−1 ) = 109 707 cm−1
where the constants have been used to su�cient precision to match the data.
�e energy of the ground state is
E 1 = −hc R̃ D = −(6.626 070 × 10−34 J s) × (2.997 925 × 1010 cm s−1 )
× (109 707 cm−1 ) = −2.179 27 × 10−18 J
P�A.�
Expressed as a molar quantity this is −1312.39 kJ mol−1 .
(a) By analogy with [�A.��–���], the wavefunction for a �px orbital is ψ 3p x =
R 3,1 (r) × [Y1,+1 (θ, �) − Y1,−1 (θ, �)](2)−1�2 . �e required integral to
verify normalization is
�
0
∞
2
R 3,1
r 2 dr �
π
0
�
2π
0
2−1 �Y1,+1 (θ, �) − Y1,−1 (θ, �)� sin θ dθ d�
2
Consider �rst the integral over r. From Table �A.� on page ��� R 3,1 (ρ) =
(486)−1�2 (Z�a 0 )3�2 (4 − ρ)ρe−ρ�2 , where ρ = 2Zr�na 0 which is this case
is 2Zr�3a 0 . It is convenient to calculate the integral over ρ, noting that
r = ρ(3a 0 �2Z) so that r 2 = ρ 2 (3a 0 �2Z)2 and dr = dρ (3a 0 �2Z). �e
integral becomes
∞
1 Z3
(4 − ρ)2 ρ 2 (3a 0 �2Z)2 ρ 2 e−ρ (3a 0 �2Z) dρ
3 �
486 a 0 0
∞
∞
1 33
1
2 4 −ρ
4
5
6 −ρ
� (4 − ρ) ρ e dρ =
� (16ρ − 8ρ + ρ )e dρ
3
486 2 0
144 0
1
�16(4!�15 ) − 8(5!�16 ) + (6!�17 )� = 1
=
144
where Integral E.� is used, with k = 1 and the appropriate value of n. �e
radial part of the function is therefore normalized.
�e angular function is found using the explicit form of the spherical
harmonics listed in Table �F.� on page ���
=
3 1�2 1
�sin θ ei� + sin θ e−i� �
�
8π
2
21�2
3 1�2 1
3 1�2
= −� �
×
2
sin
θ
cos
�
=
−
�
� sin θ cos �
8π
4π
21�2
Yx =
1
(Y1,+1 − Y1,−1 ) = − �
1�2
295
296
8 ATOMIC STRUCTURE AND SPECTRA
�e normalization integral over the angles becomes
�
π
2π
3
�� �
sin3 θ cos2 � dθ d�
4π 0
0
Using Integral T.� with a = π and k = 1 gives the integral over θ as 4�3.
2π
�e term cos2 � is written as 1 − sin2 �. �e integral ∫0 1 d� = 2π and,
2π
using Integral T.� with a = 2π and k = 1, gives ∫0 sin2 � d� = π. Hence
the integral over � is 2π − π = π. �e overall integral over the angles is
therefore (3�4π) × (4�3) × (π) = 1; the angular part is also normalized,
and as a result the complete wavefunction is normalized.
�e next task is to show that ψ 3p x and ψ 3d x y are mutually orthogonal;
taking the hint from question, attention is focused on the angular parts,
because if these are orthogonal the overall wavefunctions will also be
orthogonal. Setting aside all of the normalization factors, which will not
be relevant to orthogonality, the angular part of the wavefunction for px
is (Y1,+1 − Y1,−1 ). In a similar way, it is expected that the angular parts of
a d orbital can be constructed from spherical harmonics with l = 2.
To �nd the combination that represents dx y recall that this function is of
the form x y f (r) and that in spherical polar coordinates x = r sin θ cos �
and y = r sin θ sin �, therefore
x y f (r) = (r sin θ cos �)(r sin θ sin �) f (r) = r 2 f (r) sin2 θ cos � sin �
= 12 r 2 f (r) sin2 θ sin 2�
�e spherical harmonics Y2,±2 have the form (again, omitting the normalization factors) sin2 θ e±2i� . �e angular part of dx y is therefore obtained
by the combination
Y2,+2 − Y2,−2 = sin2 θ �e2i� − e−2i� �
= sin2 θ [cos 2� + i sin 2� − cos 2� + i sin 2�] = 2i sin2 θ sin 2�
It therefore follows that, to within some numerical factors, the angular
part of dx y is given by Y2,+2 − Y2,−2 . �e orthogonality of the angular
parts of dx y and px therefore involves the following integral
�
π
0
�
2π
0
(Y2,+2 − Y2,−2 )∗ (Y1,+1 − Y1,−1 ) sin θ dθ d�
(�.�)
Concentrating on the integral over �, this will involve terms such as
�
2π
0
∗
Y2,+2
Y1,+1 d� = �
2π
0
sin2 θ e−2i� sin θ ei� d� = �
2π
0
sin3 θ e−i� d�
�is integral is zero because ∫0 eni� d� is zero for integer n. All of the
terms in eqn �.� follow this pattern and therefore the overall integral is
zero; the orbitals are therefore orthogonal.
2π
(b) �e radial nodes for the �s, �p and �d orbitals are found by examining
the radial wavefunctions, which are listed in Table �A.� on page ���, expressed as functions of ρ = 2Zr�3a 0 . �ese functions all go to zero as
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
ρ → ∞ and, in some cases they are also zero at ρ = 0; these do not
count as radial nodes as the wavefunction does not pass through zero
at these points. �e nodes are located by �nding the values of ρ at which
the polynomial part of the radial function is zero.
�e positions of the nodes in the �s√orbital are given by the solutions to
6 − 6ρ + ρ 2 = 0, which are at ρ = 3 ± 3. In terms of r these nodes occur at
√
r = (3a 0 �2Z)(3 ± 3) . �e positions of the nodes in the �p orbital are
given by the solutions to (4 − ρ)ρ = 0; there is just one node t ρ = 4 which
corresponds to r = 6a 0 �Z . For the �d orbital the polynomial is simply
ρ 2 , which does not lead to any nodes.
�e �s orbital has no angular nodes , as it has no angular variation. �e
3px orbital has an angular node when x = 0, that is the yz plane. �e
3dx y orbital has angular nodes when x = 0 or y = 0, corresponding to the
yz and xz planes.
(c) �e mean radius is calculated as
�r� = � ψ ∗3s rψ 3s dτ = �
0
∞
�
π
0
�
2π
0
ψ ∗3s rψ 3s r 2 sin θ dr dθ d�
�e wavefunction is written in terms of its radial and angular parts: ψ 3s =
R 3,0 (r)Y0,0 (θ, �). �e angular part, the spherical harmonic Y0,0 (θ, �), is
normalized with respect to integration over the angles
�
π
0
�
2π
0
Y0,0 (θ, �)∗ Y0,0 (θ, �) sin θ dθ d� = 1
All that remains is to compute the integral over r
�r� = �
0
∞
R 3,0 (r)2 r 3 dr
�e form of R(ρ) is given in Table �A.� on page ���, where ρ = 2Zr�3a 0 .
It is convenient to compute the integral over ρ using r 3 = ρ 3 (3a 0 �2Z)3
and dr = (3a 0 �2Z)dρ
�r� = �
0
∞
R 3,0 (r)2 r 3 dr = (3a 0 �2Z)4 �
0
∞
R 3,0 (ρ)2 ρ 3 dρ
1 3a 0 4 Z 3 ∞
�
� � � � (6 − 6ρ + ρ 2 )2 ρ 3 e−ρ dρ
243 2Z
a0
0
4
∞
1 3 a0
2 2 3 −ρ
=
� (6 − 6ρ + ρ ) ρ e dρ
243 24 Z 0
∞
1 a0
7
6
5
4
3
−ρ
=
� (ρ − 12ρ + 48ρ − 72ρ + 36ρ ) e dρ
48 Z 0
1 a0
1 a0
=
[7! − 12(6!) + 48(5!) − 72(4!) + 36(3!)] =
× 648
48 Z
48 Z
= (27a 0 )�(2Z)
=
where the integrals are evaluated using Integral E.� with k = 1 and the
appropriate value of n.
297
8 ATOMIC STRUCTURE AND SPECTRA
(d) �e radial distribution function is de�ned in [�A.��b–���], P(r) = r 2 R(r)2 .
It is convenient to express this in terms of ρ = 2Zr�3a 0 using R(ρ) from
Table �A.� on page ���, and with r 2 = ρ 2 (3a 0 �2Z)2
1
Z 3 3a 0 2
� � �
� (6 − 6ρ + ρ 2 )2 ρ 2 e−ρ
243 a 0
2Z
1 Z 6
�ρ − 12ρ 5 + 48ρ 4 − 72ρ 3 + 36ρ 2 � e−ρ
=
108 a 0
1
Z 3 3a 0 2
P3p =
� � �
� (4 − ρ)2 ρ 2 ρ 2 e−ρ
486 a 0
2Z
1 Z 6
�ρ − 8ρ 5 + 16ρ 4 � e−ρ
=
216 a 0
1
Z 3 3a 0 2 2 2 2 −ρ
P3d =
� � �
� (ρ ) ρ e
2430 a 0
2Z
1 Z 6 −ρ
=
ρ e
1080 a 0
P3s =
Plots of these three functions are shown in Fig. �.�
�s
�p
�d
0.10
P�(Z�a 0 )
298
0.05
0.00
0
5
10
ρ
Figure 8.1
15
20
�e radial distribution function for the �s orbital has two subsidiary maxima which lie close in to the nucleus, and that for �p has one such maximum. In multi-electron atoms this density close to the nucleus results
in the energies of the �s, �p and �d orbitals no longer being equal: see
Section �B.� on page ���.
P�A.�
�e probability of �nding an electron within a sphere of radius σ is found by
integrating the probability density over all angles and from r = 0 to r = σ
P(σ) = �
σ
0
�
π
0
�
2π
0
�ψ(r, θ, �)�2 r 2 sin θ dr dθ d�
�e ground state of the H atom is the �s orbital for which the wavefunction is
ψ 1s = (πa 03 )−1�2 e−r�a 0 therefore P(r) = (πa 03 )−1 e−2r�a 0 . Because P(r) does not
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
depend on the angles, the integral over the angles can be evaluated separately
to give 4π. �e expression for P(σ) therefore becomes
P(σ) = (4�a 03 ) �
σ
0
r 2 e−2r�a 0 dr
�e integral is evaluated using integration by parts
�
σ
0
r 2 e−2r�a 0 dr = −(a 0 �2) r 2 e−2r�a 0 �0 + a 0 �
σ
σ
= −(a 0 σ 2 �2)e−2σ�a 0
0
re−2r�a 0 dr
+ a 0 �−(a 0 �2) re−2r�a 0 �0 + (a 0 �2) �
σ
σ
0
e−2r�a 0 dr�
= −(a 0 σ 2 �2)e−2σ�a 0 + a 0 �−(a 0 σ�2)e−2σ�a 0 − (a 02 �4) e−2r�a 0 �0 �
σ
= a 03 �4 − e−2σ�a 0 �(a 0 σ 2 �2) + (a 02 σ�2) + a 03 �4�
hence
P(σ) = 1 − e−2σ�a 0 [2(σ�a 0 )2 + 2(σ�a 0 ) + 1]
To �nd the radius at which P(σ) = 0.9 needs the solution to the equation
0.9 = 1 − e−2σ�a 0 [2(σ�a 0 )2 + 2(σ�a 0 ) + 1], which is found numerically to be
σ = 2.66a 0 . Figure �.� is a plot of P(σ) as a function of σ; it is a sigmoid curve
and shows, as expected, that the the radius of the sphere increases as the total
enclose probability increases.
1.0
P(σ)
0.8
0.6
0.4
0.2
0.0
0.0
0.5
1.0
1.5 2.0
σ�a 0
2.5
3.0
3.5
Figure 8.2
P�A.�
�e repulsive centrifugal force of an electron travelling with an angular momentum J in a circle radius r is J 2 �m e r 3 . Bohr postulated that J = nħ where
n can only take integer values, making the centrifugal force (nħ)2 �m e r 3 . �e
atom is in a stationary state when repulsive force is balanced by the attractive
299
300
8 ATOMIC STRUCTURE AND SPECTRA
Coulombic force Ze 2 �4πε 0 r 2 , that is Ze 2 �4πε 0 r 2 = (nħ)2 �m e r 3 . �is relationship is rearranged to give an expression for the radius r n of the orbit for an
electron in state n, r n = 4π(nħ)2 ε 0 �Ze 2 m e .
�e total energy of the state with an electron orbiting at radius r n is the sum
of the kinetic and potential energies. �e kinetic energy is written in terms of
the angular momentum as J 2 �2I = J 2 �2m e r n2 , with I the moment of inertia and
J = nħ. �e potential energy depends only on r n
J2
Ze 2
−
2m e r n2 4πε 0 r n
(nħ)2
Ze 2
=
−
2
2
2
2m e [4π(nħ) ε 0 �Ze m e ]
4πε 0 [4π(nħ)2 ε 0 �Ze 2 m e ]
E n = Ek + V =
P�A.��
= −
Z 2 e 4 me
1
× 2
2
2
2
32π ε 0 ħ
n
�e Bohr radius a 0 is given by [�A.�–���] and the Hartree is de�ned as E h =
2hc R̃∞ , where the Rydberg constant is given by [�A.��–���]
a 0,H =
4πε 0 ħ 2
me e 2
E h,H =
me e 4
4ε 20 h 2
�ese constants are based on the approximation that the nucleus is in�nitely
heavy. If this is not the case, then the mass of the electron m e must be replaced
by the reduced mass of the atom, µ = m e m N �(m e + m N ), where m N is the mass
of the nucleus.
In the case of positronium the ‘nucleus’ has the same mass as the electron, so
µ = m e �2 and hence
a 0,pos =
4πε 0 ħ 2
= 2a 0,H
e 2 (m e �2)
E h,pos =
(m e �2)e 4
= 12 E h,H
4ε 20 h 2
8B Many-electron atoms
Answers to discussion questions
D�B.�
See Section �B.� on page ���.
D�B.�
�is is covered in any introductory or general chemistry text.
Solutions to exercises
E�B.�(a)
All con�gurations have the [Ar] core.
Sc
2
4s 3d
Ti
1
Fe
2
4s 3d
2
4s 3d
V
2
Co
6
2
4s 3d
2
4s 3d
Cr
4s 3d
8
1
Ni
7
2
4s 3d
1
Mn
3
5
4s2 3d5
10
2
Cu
4s 3d
Zn
4s 3d10
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�B.�(a)
[Ar] �d8
E�B.�(a)
Across the period the energy of the orbitals generally decreases as a result of
the increasing nuclear charge. �erefore Li is expected to have the lowest
ionization energy as its outer electron has the highest orbital energy.
E�B.�(a)
Hydrogenic orbitals are written in the form [�A.��–���], R n, l (r)Yl ,m l (θ, �),
where the appropriate radial function R n, l is selected from Table �A.� on page
��� and the appropriate angular function Yl ,m l is selected from Table �F.� on
page ���. Using Z = 2 for the �s and Z = 1 for the �s gives
ψ 1s (r) = R 1,0 Y0,0 = 2(2�a 0 )3�2 e−2r�a 0 × (4π)−1�2
ψ 2s (r) = R 2,0 Y0,0 = (8)−1�2 (1�a 0 )3�2 [2 − (r�a 0 )]e−r�2a 0 × (4π)−1�2
�e overall wavefunction is simply the product of the orbital wavefunctions
E�B.�(a)
Ψ(r 1 , r 2 ) = ψ 1s (r 1 )ψ 2s (r 2 )
For a subshell with angular momentum quantum number l there are 2l + 1
values of m l , each of which corresponds to a separate orbital. Each orbital can
accommodate two electrons, therefore the total number of electrons is 2×(2l +
1). �e subshell with l = 3 can therefore accommodate 2(6+1) = �� electrons.
Solutions to problems
P�B.�
�e radial distribution function for a �s orbital is given by [�A.��–���], P(r) =
(4Z 3 �a 03 )r 2 e−2Zr�a 0 . �is gives the probability density of �nding the electron
in a shell of radius r. �e most probable radius is found by �nding the maximum in P(r), when dP(r)�dr = 0. In �nding this maximum the multiplying
constants are not relevant and can be discarded
d 2 −2Zr�a 0
r e
= [2r − (2Z�a 0 )r 2 ] e−2Zr�a 0 = 0
dr
P�B.�
It follows that r max = a 0 �Z; that this is a maximum is most easily seen by
plotting P(r). For Z = 126 the most probable radius will be a 0 �126 .
Toward the middle of the �rst transition series (Cr, Mn, and Fe) elements exhibit the widest ranges of oxidation states. �is is due to the large number
of electrons in the �d and �s subshells that have similar energies, and as the
�d electrons that are generally removed provide very little shielding to the �s
orbitals, the e�ective nuclear charge does not increase signi�cantly between
adjacent oxidation states, meaning that the ionization energies of these levels
are close, and as these are the outermost electrons the ionization levels are
relatively small, meaning that large numbers of reactions will release enough
energy to lose many electrons.
However, it should be noted that the higher oxidation states of the middle
transition metals do not exist as cations, but only in compounds or compound
ions where there is a signi�cant stabilization of this ion by electron rich atoms,
301
8 ATOMIC STRUCTURE AND SPECTRA
for example the MnVII state only exists in the MnO� – ion where there is large
electron donation from bonding with four O� – ions, as here the e�ective nuclear charge has increased a lot over the neutral atom.
�is phenomenon is related to the availability of both electrons and orbitals
favourable for bonding. Elements to the le� (Sc and Ti) of the series have few
electrons and relatively low e�ective nuclear charge leaves d orbitals at high
energies that are relatively unsuitable for bonding. To the far right (Cu and
Zn) e�ective nuclear charge may be higher but there are few, if any, orbitals
available for bonding. Consequently, it is more di�cult to produce a range of
compounds that promote a wide range of oxidation states for elements at either
end of the series. At the middle and right of the series the +� oxidation state is
very commonly observed because normal reactions can provide the requisite
ionization energies for the removal of �s electrons.
P�B.�
�e �rst, second and third ionization energies for the group �� elements are
plotted in Fig. �.�
40
Ionization energy/eV
302
�rst IE
second IE
third IE
30
20
10
0
0
Figure 8.3
20
40
60
Atomic number, Z
80
�e following trends are identi�ed.
(a) In all cases, I 1 < I 2 < I 3 because of decreased nuclear shielding as each
successive electron is removed.
(b) �e ionization energies of boron are much larger than those of the remaining group elements because the valence shell of boron is very small
and compact with little nuclear shielding. �e boron atom is much smaller
than the aluminum atom.
(c) �e ionization energies of Al, Ga, In, and Tl are comparable even though
successive valence shells are further from the nucleus because the ionization energy decrease expected from large atomic radii is balanced by an
increase in e�ective nuclear charge.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
8C Atomic spectra
Answers to discussion questions
D�C.�
�is is discussed in Section �C.�(b) on page ���.
D�C.�
�e selection rules are given in [�C.�–���]. In part these can be rationalised
by noting that a photon has one unit of (spin) angular momentum and that
in the spectroscopic transition this angular momentum must be conserved.
�e selection rule for l, ∆l = ±1, can be understood as a single electron in
the atom changing angular momentum by one unit in order to accommodate
the angular momentum from the photon. �is selection is derived in How
is that done? �C.� on page ��� by considering the relevant transition dipole
moment. �e selection rule for the total spin, ∆S = 0, stems from the fact that
the electromagnetic radiation does not a�ect the spin directly.
�e selection rules for multi-electron atoms are harder to rationalise not least
because the change in the overall angular momentum (L and J) is a�ected both
by changes in the angular momenta of individual electrons and by the way in
which these couple together.
Solutions to exercises
E�C.�(a)
E�C.�(a)
E�C.�(a)
E�C.�(a)
E�C.�(a)
For a d electron l = 2 and s = 12 . Using the Clebsh–Gordon series, [�C.�–���],
the possible values of j are l + s, l + s − 1, . . . �l − s�, which in this case are
j = 52 , 32 .
For an f electron l = 3 and s = 12 hence j = 72 , 52 .
�e Clebsch–Gordan series [�C.�–���], in the form j = l + s, l + s − 1, . . . �l − s�,
with s = 12 implies that there are two possible values of j, j = l ± 12 . Hence, given
that j = 32 , 12 it follows that l = 1 .
�e symbol D implies that the total orbital angular momentum L = 2 , the superscript � implies that the multiplicity 2S + 1 = 1, so that the total spin angular
momentum S = 0 . �e subscript � implies that the total angular momentum
J =2.
�e Clebsch–Gordan series, [�C.�–���], is used to combine two spin angular
momenta s 1 and s 2 to give S = s 1 + s 2 , s 1 + s 2 − 1 ..., �s 1 − s 2 �. (i) For two
electrons, each with s = 12 , S = 1, 0 with multiplicities, 2S = 1, of 3, 1 . (ii)
�ree electrons are treated by �rst combining the angular momenta of two of
them to give S ′ = 0, 1 and then combining each value of S ′ with s 3 = 12 for
the third spin. �erefore, for S ′ = 1, S = 1 + 12 , �1 − 12 � = 32 , 12 . Combining
S ′ = 0 with s = 12 simply results in S = 12 . �e overall result is S = 32 , 12 with
corresponding multiplicities �, � .
�e valence electron con�guration of the Ni�+ is [Ar] �d8 . In principle the
same process could be adopted as in Exercise E�C.�(a), in which the spin angular momenta of all eight electrons are coupled together in successive steps to
303
304
8 ATOMIC STRUCTURE AND SPECTRA
�nd the overall spin angular momentum. Such an approach would be rather
tedious and would also run the risk of generating values of S which come from
arrangements of electrons which violate the Pauli principle. A quicker method,
and one which ensures that the Pauli principle is not violated, is to consider
combinations of the quantum number m s which gives the z-component of the
spin angular momentum and which takes values ± 12 . �e total z-component of
the spin angular momentum is found by simply adding together the m s values:
M S = m s 1 + m s 2 + . . ..
E�C.�(a)
With � electrons in the � d orbitals, � of these electrons must doubly occupy
three of the orbitals, and the Pauli principle requires that the two electrons in
each orbital are spin paired: one has m s = + 12 and one has m s = − 12 . �ese six
electrons therefore make no net contribution to M S , in the sense that the sum
of the individual m s values is �. �e remaining two electrons can either occupy
the same orbital with spins paired, giving M S = + 12 − 12 = 0, or they can occupy
di�erent orbitals with either their spins paired, giving M S = 0 once more, or
with their spins parallel, giving M S = + 12 + 12 = +1 or M S = − 12 − 12 = −1. Recall
that a total spin S gives M S values of S, (S − 1) . . . − S. �erefore the �rst
arrangement with just M S = 0 is interpreted as arising from S = 0 , and the
second arrangement with M S = 0, ±1 is interpreted as arising from S = 1 .
�ese electrons are not equivalent, as they are in di�erent subshells, hence all
the terms that arise from the vector model and the Clebsch–Gordan series are
allowed. �e orbital angular momentum of the s and d electrons are l 1 = 0 and
l 2 = 2 respectively, and these are combined using L = l 1 +l 2 , l 1 +l 2 −1, ... �l 1 −l 2 �
which in this case gives L = 2 only. �e spin angular momenta of each electron
is s 1 = s 2 = 12 , and these combine in the same way to give S = 1, 0; these values
of S have spin multiplicities of 2S +1 = 3, 1. �e terms which arise are therefore
3
D and 1 D.
�e possible values of J are given by J = L+S, L+S −1, ..., �L−S�, and hence for
S = 1, L = 2 the values of J are �, �, and �. For S = 0, L = 2 only J = 2 is possible.
�e term symbols are therefore 3 D3 , 3 D2 , 3 D1 , and 1 D2 . From Hund’s rules,
described in Section �C.�(d) on page ���, the lowest energy state is the one with
the greatest spin and then, because the shell is less than half full, the smallest J.
�is is 3 D1 .
E�C.�(a)
E�C.�(a)
(i) 1 S has L = 0, S = 0 and so J = 0 only; there are 2J + 1 values of M J , which
for J = 0 is just � state. (ii) 2 P has L = 1, S = 12 , and so J = 32 , 12 ; the former has
� states and the latter has � states. (iii) 3 P has L = 1, S = 1, and so J = 2, 1, 0 ,
with 5, 3, 1 states, respectively.
Closed shells have total spin and orbital angular momenta of zero, and so do
not contribute to the overall values of S and L. (i) For the con�guration �s1
there is just one electron to consider with l = 0 and s = 12 , so L = 0, S = 12 , and
J = 12 . �e term symbol is 2 S1�2 . (ii) For the con�guration �p1 there is just one
electron to consider with l = 1 and s = 12 , so L = 1, S = 12 , and J = 32 , 12 . �e
term symbols are therefore 2 P3�2 and 2 P1�2 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�C.�(a)
�e two terms arising from a d1 con�guration are 2 D3�2 , 2 D5�2 , which have
S = 12 , L = 2 and J = 32 , 52 . �e energy shi� due to spin-orbit coupling is given
by [�C.�–���], E L,S , J = 12 hc Ã[J(J + 1) − L(L + 1) − S(S + 1)], where à is the
spin-orbit coupling constant. Hence, E 2,1�2,3�2 = −(3�2)hc à , and E 2,1�2,5�2 =
+hc à .
E�C.��(a) �e selection rules for a many-electron atom are given in [�C.�–���].
(i) 3 D2 (S = 1, L = 2, J = 2) → 3 P1 (S = 1, L = 1, J = 1) has ∆S = 0, ∆L = −1,
∆J = −1 and so is allowed .
(ii) 3 P2 (S = 1, L = 1, J = 2) → 1 S0 (S = 0, L = 0, J = 0) has ∆S = −1, ∆L = −1,
∆J = −2 and so is forbidden by the S and J selection rules.
E�C.��(a)
(iii) 3 F4 (S = 1, L = 3, J = 4) → 3 D3 (S = 1, L = 2, J = 3) has ∆S = 0, ∆L = −1,
∆J = −1 and so is allowed .
�e spectral lines of a hydrogen atom are given by [�A.�–���], ν̃ = R̃ H (n−2
1 −
n−2
2 ), where R̃ H is the Rydberg constant and ν̃ is the wavenumber of the transition.
�e Lyman series corresponds to n 1 = 1. �e lowest energy transition, which
would involve a photon with the longest wavelength, is to the next highest
energy level which has n 2 = 2 . Transitions to higher energy levels involve more
an more energy, and the limit of this is the transition to n 2 = ∞ which involves
the greatest possible energy change and hence the shortest wavelength.
E�C.��(a) �e energy levels of a hydrogenic atom are E n = −hcZ 2 R̃ N n−2 , where Z is the
atomic number; for all but the most precise work it is su�cient to approximate
R̃ N by R̃∞ . �e wavenumber of the transition between states with quantum
numbers n 1 and n 2 in the He+ ion is given by a modi�ed version of [�A.�–���],
−2
ν̃ = Z 2 R̃∞ (n−2
1 − n 2 ). For the 2 → 1 transition and with Z = 2
ν̃ = 22 × (1.0974 × 105 cm−1 ) × (1−2 − 2−2 ) = 3.29 × 105 cm−1
λ = ν̃ −1 = 1�[22 × (1.0974 × 105 cm−1 ) × (1−2 − 2−2 )]
= 3.03... × 10−6 cm = ��.� nm
ν = c�λ = (2.9979 × 108 m s−1 )�(3.03... × 10−8 m) = �.�� PHz
E�C.��(a) �e selection rules for a many-electron atom are given in [�C.�–���]. For a
single electron these reduce to ∆l = ±1; there is no restriction on changes in n.
(i) �s (n = 2, l = 0) → �s (n = 1, l = 0) has ∆l = 0, and so is forbidden .
(ii) �p (n = 2, l = 1) → �s (n = 1, l = 0) has ∆l = −1, and so is allowed .
(iii) �d (n = 3, l = 2) → �p (n = 2, l = 1) has ∆l = −1, and so is allowed .
E�C.��(a) �e single electron in a p orbital has l = 1 and hence L = 1, and s = 12 hence
S = 12 . �e spin multiplicity is 2S + 1 = 2. Using the Clebsh–Gordon series,
[�C.�–���], the possible values of J are J = L + S, L + S − 1, . . . �L − S� = 32 , 12 .
Hence, the term symbols for the levels are 2 P1�2 , 2 P3�2 .
305
306
8 ATOMIC STRUCTURE AND SPECTRA
Solutions to problems
P�C.�
�e wavenumbers of the spectral lines of the H atom for the n 2 → n 1 transition
−2
are given by [�A.�–���], ν̃ = R̃ H (n−2
1 − n 2 ), where R̃ H is the Rydberg constant
−1
for Hydrogen, R̃ H = 109677 cm . Hence, the wavelength of this transition is
−2
−2 −1
λ = ν̃−1 = R̃−1
H (n 1 − n 2 ) .
�e lowest energy, and therefore the longest wavelength transition (the one at
λ max = 12368 nm = 1.2368 × 10−3 cm) corresponds to the transition from
n 1 + 1 → n 1 , therefore
1
λ max R̃ H
=
1
1
(n 1 + 1)2 − n 12
2n 1 + 1
−
=
= 2
2
2
2
2
n 1 (n 1 + 1)
n 1 (n 1 + 1)
n 1 (n 1 + 1)2
From the given data (λ max R̃ H )−1 = [(1.2368 × 10−3 cm) × (109677 cm−1 )]−1 =
(135.6...)−1 . �e value of n 1 is found by seeking an integer value of n 1 for
which n 12 (n 1 + 1)2 �(2n 1 + 1) = 135.6.... For n 1 = 6 the fraction on the le� is
62 × 72 �13 = 135.6.... �erefore, the Humphreys series is that with n 1 = 6 .
P�C.�
�e wavelengths of the transitions in the Humphreys series are therefore given
−1
by λ = (109677 cm−1 )−1 × (6−2 − n−2
for n 2 = 7, 8, .... �e next few lines,
2 )
with n 2 = 8, 9, and 10 are at ����.� nm, ����.� nm, ����.� nm, respectively.
�e convergence limit, corresponding to n 2 = ∞ is ����.� nm, as given in the
data.
�e wavenumbers of transitions between energy levels in hydrogenic atoms are
given by a modi�ed version of [�A.�–���]
−2
ν̃ = Z 2 R̃ N �n−2
1 − n2 �
(�.�)
where Z is the nuclear charge and R̃ N is the Rydberg constant for the nucleus
in question. In turn this is given by [�A.��–���]
R̃ N =
µ
R̃∞
me
µ=
me mN
me + mN
where m N is the mass of the nucleus. �e spectra of 4 He+ and 3 He+ di�er
because R̃ N is di�erent for the two atoms. However, this di�erence is very
small because the value of the reduced mass µ is dominated by the mass of
the electron (m e � m N ). It is therefore necessary to work at high precision.
�e �rst step is to compute R̃ N for each nucleus, using m 4 He = 4.002 602m u
and m3 He = 3.016 029m u .
µ
mN
R̃∞ =
R̃∞
me
me + mN
4.002 602 × (1.660 539 × 10−27 kg)
=
(9.109 383 × 10−31 kg) + 4.002 602 × (1.660 539 × 10−27 kg)
R̃4 He =
× (1.097 373 × 105 cm−1 ) = 1.097 223 × 105 cm−1
A similar calculation gives R̃ 3 He = 1.097 173 × 105 cm−1 . With these values of
the Rydberg constant the wavenumber of the relevant transitions is computed
using eqn �.�; the results are given in the table.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
ν̃ 3→2 /cm−1
ν̃ 2→1 /cm−1
He+
�� ���.�
��� ���
He
�� ���.�
��� ���
di�erence
�.�
��
4
3
+
If the spectrometer has su�cient resolution these di�erences are detectable; the
greatest di�erence is for the higher wavenumber transition.
�e three transitions originate from the same level, the 2 P, with energy E 2 P ,
and if it is assumed that the nd 2 D states are hydrogenic their energies may be
written E n = −A�n 2 , where A is some constant. It follows that the wavenumber
of the transitions can be written
ν̃ n = E 2 P �hc − (A�hc)�n 2
A plot of ν̃ n against 1�n 2 is therefore expected to a straight line with y-intercept
(at 1�n 2 = 0) E 2 P �hc. �e data are tabulated below and the graph is given in
Fig. �.�.
n
3
4
5
30 000
1�n 2
0.111
0.063
0.040
λ�nm
610.36
460.29
413.23
ν̃ n �cm−1
16 384
21 725
24 200
ν̃∞
25 000
ν̃ n �cm−1
P�C.�
20 000
15 000
0.00
0.02
0.04
0.06
1�n
0.08
0.10
0.12
2
Figure 8.4
�e data fall on a good straight line which has y-intercept ν̃∞ = E 2 P �hc =
28 595 cm−1 . �e transition from the 2 S ground state to the 2 P state is at a
wavelength of ���.�� nm, which corresponds to a wavenumber of �� ��� cm−1 .
�erefore the transition from 2 S to the ionization limit of the 2 P–2 D series will
be at wavenumber
14 908 cm−1 + ν̃∞ = 14 908 cm−1 + 28 595 cm−1 = 43 503 cm−1
307
308
8 ATOMIC STRUCTURE AND SPECTRA
�is corresponds to the ionization energy of the ground state, which can be
expressed in eV as �.�� eV . Although the data are given to high precision,
quite a long extrapolation is needed to �nd the energy of the 2 P state and it also
has been assumed that the constant A is independent of n, which may not be
the case. As a result, the ionization energy is quoted to more modest precision.
P�C.�
�e outer electron in K can occupy an s, p or d orbital and such con�gurations gives rise to 2 S1�2 , 2 P3�2,1�2 , and 2 D5�2,3�2 states, respectively. Taking into
account the selection rules and the e�ect of spin-orbit coupling, two closely
spaced lines are expected as a result of the transitions 2 S1�2 → 2 P3�2 and 2 S1�2 →
2
P1�2 . �e separation of the two lines will re�ect the separation of the 2 P3�2 and
2
P1�2 levels, which is computed using [�C.�–���]; the terms in L and S cancel
as they take the same value for the two states
∆E = E 1,1�2,3�2 − E 1,1�2,1�2
= 12 hc Ã[ 32 ( 32 + 1) − L(L + 1) − S(S + 1)]
− 12 hc Ã[ 12 ( 12 + 1) − L(L + 1) − S(S + 1)] = 32 hc Ã
�e wavenumber of the separation between the two lines is therefore 32 Ã, hence
à = 23 �(766.70 × 10−7 cm)−1 − (770.11 × 10−7 cm)−1 �
= 23 (57.7... cm−1 ) = ��.� cm−1
P�C.�
�e Rydberg constant for positronium is [�A.��–���] R̃ Ps = R̃∞ × (µ Ps �m e ),
where the reduced mass of the positron–electron system is µ Ps = m e2 �(2m e ) =
m e �2, as the mass of the nucleus is equal to that of the electron. Hence, R̃ Ps =
R̃∞ �2 = (109737 cm−1 )�2 = 54868.5 cm−1 . �e spectral lines of the positron−2
ium atom are given by ν̃ = R̃ Ps (n−2
1 − n 2 ).
�e Balmer series are those lines with n 1 = 2, and so the wavenumbers of these
are
−1
−2
ν̃ = R̃ Ps (2−2 − n−2
− n−2
2 ) = (54868.5 cm ) × (2
2 )
P�C.��
n 2 = 3, 4 . . .
�e �rst three lines have n 2 = 3, 4, 5 and are at 7 621 cm−1 , 10 288 cm−1 , and
11 522 cm−1 , respectively. �e ionization energy is simply the binding energy
of the ground state, which is hc R̃ Ps . Hence I = hc R̃ Ps �e = �.��� eV
�e derivation follows the method used in How is that done? �C.� on page
���. For a transition to be allowed the transition dipole moment µfi must
be non-zero. It is convenient to explore this condition by examining the x-,
y-, and z-components of the moment: if any of these are non-zero, the overall moment will also be non-zero. �e x- and y-components are given by
µ x ,fi = −e ∫ ψ ∗f xψ i dτ and µ y,fi = −e ∫ ψ ∗f yψ i dτ, respectively. �e limits of
integration are r = 0 to ∞, θ = 0 to π, and � = 0 to 2π, and the volume element
is dτ = r 2 sin θ dr dθ d�.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e �rst step is to express the Cartesian co-ordinates in spherical polar coordinates and then in terms of the spherical harmonics. From �e chemist’s toolkit
�� in Topic �F on page ��� it is seen that x = r sin θ cos �, and y = r sin θ sin �.
From Table �F.� on page ��� the spherical harmonics Y1,±1 are ∓N sin θe±i� ,
where N is the normalization constant. Using the identity e±i� = cos � ± i sin �
it follows that
Y1,+1 + Y1,−1 = −N sin θ(cos � + i sin �) + N sin θ(cos � − i sin �)
Similarly
= −2iN sin θ sin �
Y1,+1 − Y1,−1 = −N sin θ(cos � + i sin �) − N sin θ(cos � − i sin �)
= −2N sin θ cos �
With these relationships x and y are expressed as
x = r sin θ cos � = −r(Y1,+1 − Y1,−1 )�2N
y = r sin θ sin � = −r(Y1,+1 + Y1,−1 )�2iN
�e wavefunctions of the atomic orbitals are expressed in terms of a radial and
an angular part: ψ n, l ,m l = R n,l (r)Yl ,m l (θ, �). As is seen in the calculation in
the text, the selection rule is derived by considering only the integral over the
angles. Focusing just on this, and setting aside all the normalization and other
factors, the integral to consider for the x-component of the transition moment
is
µ x ,fi ∝ �
π
0
�
2π
0
Yl∗f ,m l ,f (Y1,+1 − Y1,−1 )Yl i ,m l ,i sin θ dθ d�
It is a property of spherical harmonics that the ‘triple integral’
�
π
0
�
2π
0
Yl∗f ,m l ,f Yl ,m Yl i ,m l ,i sin θ dθ d�
vanishes unless l f = l i ± l and m l ,f = m l ,i ± m. In this case the integrals of
interest have l = 1 and m = ±1, therefore they are non-zero only if l f = l i ± 1
and m l ,f = m l ,i ± 1. It follows that the integral on which the x-component
of the transition depends is only non-zero if these conditions are satis�ed. A
similar argument applies to the y-component. �e selection rules are therefore
∆l = ±1, ∆m l = ±1 .
In the text it is seen that the selection rule deriving from the z-component of
the transition moment is ∆l = ±1, ∆m l = 0; when the x- and y- components
are considered as well, transitions with ∆m l = ±1 are also allowed.
Answers to integrated activities
I�.�
(a) �e ground state of the He+ ion is �s1 with S = 21 , L = 0 and hence J = 12 .
�e term symbol is therefore 2 S1�2 . �e excited state con�guration is �p1
which has S = 12 , L = 1 and hence J = 32 or 12 ; the term symbols are
309
310
8 ATOMIC STRUCTURE AND SPECTRA
P3�2 and 2 P1�2 , the lowest of which is that with J = 12 . According to the
selection rules the transitions from 2 S1�2 to both 2 P states are allowed:
hence, the transitions are 2 S1�2 → 2 P1�2 and 2 S1�2 → 2 P3�2 .
2
(b) �e wavenumber the spectral line corresponding to the n 1 → n 2 transition is given by a modi�ed version of [�A.�–���] which takes into account
−2
the nuclear charge Z: ν̃ = Z 2 R̃ H (n−2
1 − n 2 ), where R̃ H is the Rydberg
−1
constant for hydrogen, 109 677 cm ; Z = 2 for He+ . In principle the
Rydberg constant is di�erent for He, but the change is so small that it can
safely be ignored. Hence for a transition for n = 1 → 4, the wavenumber
is ν̃ = 4 × (109 677 cm−1 ) × (1−2 − 4−2 ) = 411 289 cm−1 .
�is corresponds to a wavelength of λ = ν̃−1 = (411 289 cm−1 )−1 =
2.43... × 10−6 cm = ��.��� � nm . �e corresponding frequency is
ν = cλ−1 = c ν̃ = (2.997 925 × 1010 cm s−1 ) × (411 289 cm−1 )
= 1.233 01 × 1016 Hz
(c) �e mean radius of a hydrogenic orbital, characterized by quantum numbers, n, l , m l is given by
�r�n, l ,m l =
n2 a0
l(l + 1)
�1 + 12 �1 −
��
Z
n2
�r�1,0,0 =
(12 )a 0
0(0 + 1)
3a 0
�1 + 12 �1 −
�� =
2
2
1
4
�r�4,1,0 =
(4)2 a 0
1(1 + 1)
23a 0
�1 + 12 �1 −
�� =
2
42
2
For the ground state orbital, with Z = 2, n = 1 and l = 0 in He+
For the upper state with n = 4 and l = 1
I�.�
Hence, the mean radius of the atom increases by 23a 0 �2−3a 0 �4 = 43a 0 �4 .
Because the beam splits into two, with de�ections ±(µ B L 2 �4E k )dB�dz, a splitting between the two beams of ∆x is achieved by satisfying the condition ∆x =
(µ B L 2 �2E k )dB�dz, which is rearranged to give an expression for the �eld gradient dB�dz = 2E k ∆x�µ B L 2 . A reasonable estimate for the mean kinetic energy
is to take the equipartition value of 32 kT.
dB 2E k ∆x 3kT∆x
=
=
dz
µB L2
µB L2
=
3 × (1.3806 × 10−23 J K−1 ) × (1000 K) × (1.00 × 10−3 m)
(9.2740 × 10−24 J T−1 ) × (50 × 10−2 m)2
= 17.9 T m−1
9
9A
Molecular Structure
Valence-bond theory
Answers to discussion questions
D�A.�
See Section �A.�(b) on page ��� for details on hybridization applied to simple
carbon compounds. �e carbon atoms in alkanes are sp3 hybridized. �is
explains the nearly tetrahedral bond angles about the carbon atoms in such
molecules.
�e double-bonded carbon atoms in alkenes are sp2 hybridized. �is explains
the bond angles of approximately ���○ about these atoms. �e simultaneous
overlap of sp2 hybridized orbitals and unhybridized p orbitals in C=C double
bonds explains the resistance of such bonds to torsion and the co-planarity of
the atoms attached to those atoms.
�e triple-bonded carbon atoms in alkynes are sp hybridized, which explains
the ���○ bond angles about these atoms. �e central carbon atom in allene is
also sp hybridized. Each of its C=C double bonds involves one of its sp hybrids
and one unhybridized p orbital. �e two resulting π orbitals are oriented perpendicular to one another, which is why the two CH� groups are rotated by
��○ relative to one another. �is arrangement of orbitals also accounts for the
resistance to the two CH� groups being rotated relative to one another about
the long axis.
D�A.�
Resonance refers to the superposition of the wave functions representing different electron distributions in the same nuclear framework. �e wavefunction
resulting from the superposition is called a resonance hybrid.
Resonance allows for a more re�ned description of the electron distribution,
and hence bonding, than is given by a single valence bond wavefunction. Different valence bond structures are allowed to contribute to di�erent extents,
meaning that the overall wavefunction is built up from contributions from
di�erent valence-bond wavefunctions.
�is approach makes it possible to describe polar bonds as a combination of a
purely covalent and a purely ionic structure, and delocalized bonding in terms
of combinations of valence-bond structures in which, for example, a double
bond is located in di�erent parts of a molecule. Resonance is a device for
calculating an improved wavefunction: it does not imply that wavefunction
�ickers between those for the di�erent structures.
D�A.�
Promotion and hybridization are two modi�cations to the simplest version of
valence-bond (VB) theory, adopted to overcome obvious mismatches between
312
9 MOLECULAR STRUCTURE
predictions of that theory and observations. In its simplest form VB theory
assumes that the functions ψ A and ψ B that appear in a VB wavefunction, [�A.�–
���], are orbitals in free atoms occupied by unpaired electrons. For example, such a theory would predict that carbon, with the electronic con�guration
2s2 2p2 , would form two bonds on account if it having two unpaired electrons.
�is prediction is at odds with the characteristic valency of four shown by
carbon.
To account for the tetravalence of carbon it is supposed that one of the �s
electrons is excited (‘promoted’) to the empty �p orbital, giving a con�guration
of 2s1 2p3 . �ere are now four unpaired electrons (in the �s and �p orbitals)
available for forming four valence bonds.
Hybrid orbitals are invoked to account for the fact that valence bonds formed
from atomic orbitals would have di�erent orientations in space than are commonly observed. For instance, the four bonds in CH� are observed to be equivalent and directed toward the corners of a regular tetrahedron. By contrast,
bonds made from the three distinct �p orbitals in carbon would be expected to
be oriented at ��○ angles from each other, and those three bonds would not be
equivalent to the bond made from a �s orbital. Hybrid atomic orbitals, in this
case sp3 hybrids, are formed by combining the atomic orbitals in such a way
that the hybrid orbitals have the required directional properties.
Solutions to exercises
E�A.�(a)
�e ammonium ion is iso-electronic with methane, therefore the two species
are expected to have the same description of bonding. Four sp3 hybrid atomic
orbitals are formed from the �s and the three �p orbitals of the nitrogen atom;
each hybrid then forms a σ bond by overlapping with a hydrogen �s orbital.
E�A.�(a)
All the carbon atoms in �,�-butadiene are sp2 hybridized. �e σ framework of
the molecule consists of C–H and C–C σ bonds. Each C–H σ bond is formed
by the overlap of an sp2 hybrid atomic orbital on a carbon atom with a �s atomic
orbital on a neighbouring hydrogen atom. Similarly, C–C σ bonds are formed
by the overlap of sp2 hybrid atomic orbitals on neighbouring carbon atoms.
�e two π bonds are formed by the side-by-side overlap of unhybridized �p
orbitals on carbon atoms C� and C�, and likewise between C� and C�.
E�A.�(a)
�e carbon and nitrogen atoms in methylamine are sp3 hybridized. �e C–N
bond is formed by the overlap of an sp3 orbital on carbon with an sp3 orbital
on nitrogen. �e C–H bonds are formed by the overlap of a carbon sp3 hybrid
atomic orbital with a hydrogen �s atomic orbital. Similarly, the N–H bonds are
formed by the overlap of a nitrogen sp3 hybrid atomic orbital with a hydrogen
�s atomic orbital. �e lone pair on nitrogen resides on an sp3 hybrid atomic
orbital.
E�A.�(a)
�e condition of orthogonality is given by [�C.�–���], ∫ Ψi∗ Ψ j dτ = 0 for i ≠ j.
�e atomic orbitals are all real, therefore Ψi∗ = Ψi . �e orthogonality condition
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
becomes
∗
� h 1 h 2 dτ = � (s + px + p y + pz )(s − px − p y + pz ) dτ
��� � � � � ��� � � � � � ��� � � � � � � � � � � � � � � � � � ��� � � � � � � � ��� � � � � � � � � ��� � � � � � � ��� � � � � � � � �
= � s2 dτ − � spx dτ − � sp y dτ + � spz dτ
1
0
0
0
��� � � � � � �� � � � � � ��
��� � � � � � �� � � � � � ��
��� � � � � � � � � � � � � �
2
2
+ ... − � px dτ +.... − � p y dτ +... + � p2z dτ
1
1
1
=1−1−1+1=0
All the integrals of the form ∫ sp i dτ are zero because the s and p orbitals are
orthogonal, and all the integrals of the form ∫ s2 dτ and ∫ p2i dτ are � because
the orbitals are normalized. �e condition for the orthogonality of h 1 and h 2
is satis�ed.
E�A.�(a)
A normalized wavefunction satis�es [�B.�c–���], ∫ Ψ ∗ Ψ dτ = 1. �e wavefunction is normalized by �nding the value of N for which h = N(s + 21�2 p)
satis�es this condition. �e orbital wavefunctions s and p are real as is N,
therefore
∗
2
1�2 2
� h h dτ = N � (s + 2 p) dτ
1
0
� 1
�
���� � � � � ��� � � � � � ��� � � � � � � � � � � � � �
��� � � � � � ��� � � � � � ��
�
�
�
�
= N 2 �� s2 dτ +2 � p2 dτ +23�2� sp dτ � = 3N 2
�
�
�
�
�
�
�
�
E�A.�(a)
E�A.�(a)
�e integral ∫ sp dτ is zero because the s and p orbitals are orthogonal, and the
integrals ∫ s2 dτ and ∫ p2 dτ are � because the orbitals are normalized. From
the normalization condition it follows that 3N 2 = 1 and hence N = 1�31�2 .
Using [�A.�–���] and assuming that the valence-bond in HF is formed between the H�s and F�pz atomic orbitals, the spatial part of the valence-bond
wavefunction is written as Ψ(1, 2) = ψF�pz (1)ψH�s (2) + ψF�pz (2)ψH�s (1). �e
overall wavefunction must be antisymmetric to satisfy the Pauli principle, therefore the symmetric spatial part has to be combined with the antisymmetric twoelectron spin wavefunction given by [�B.�–���], σ− (1, 2). �e (unnormalized)
complete two-electron wavefunction is therefore
Ψ(1, 2) = [ψF�pz (1)ψH�s (2) + ψF�pz (2)ψH�s (1)] × [α(1)β(2) − β(1)α(2)]
�e resonance hybrid wavefunction constructed from one two-electron wavefunction corresponding to the purely covalent form of the bond and one twoelectron wavefunction corresponding to the ionic form of the bond is given
in [�A.�–���] as Ψ = Ψcovalent + λΨionic . �erefore the (unnormalized) resonance hybrid wavefunction of HF with two ionic structures is written as ΨHF =
ΨH–F + λΨH+ F− + κΨH− F+ . ΨH–F is written as in Exercise E�A.�(a), ΨH–F =
313
314
9 MOLECULAR STRUCTURE
E�A.�(a)
[ψF�pz (1)ψH�s (2) + ψF�pz (2)ψH�s (1)] × σ− (1, 2). �e wavefunction ΨH+ F− describes the electron distribution when both electrons reside on the F2pz orbital. �e spatial part of this wavefunction is given by ψF2pz (1) ψF2pz (2) , which
is symmetric, therefore it has to be combined with the antisymmetric spin
wavefunction resulting in ΨH+ F− = [ψF2pz (1) ψF2pz (2) ] × σ− (1, 2). Similarly, the
other ionic structure has ΨH− F+ = [ψH�s(1) ψH�s(2) ] × σ− (1, 2).
Both phosphorus and nitrogen are in Group ��, therefore the valence bond
description of the bonding in P2 is similar to that of N2 . �ere is a triple bond
between the two sp hybridized phosphorus atoms. A σ bond is formed by the
overlap of two sp hybrid atomic orbitals projecting towards each other along
the internuclear axis. �e two π bonds are the result of the side-by-side overlap
of 3px with 3px and 3p y with 3p y orbitals. �ere is one lone pair on each
phosphorus atom, contained in the sp orbital projecting outwards along the
internuclear axis.
⇀ 2 P2 . In the tetrahedral P4 there are six σ
Consider the equilibrium P4 ���
bonds, whereas in two molecules of P2 there are two σ and four π bonds overall.
π bonds are generally weaker than σ bonds, therefore the equilibrium favors P4 .
Solutions to problems
P�A.�
�e wavefunction in terms of the polar coordinates of each electron is given in
Brief illustration �A.� on page ��� as
Ψ(1, 2) =
1
�e−(rA� +rB� )�a 0 + e−(rA� +rB� )�a 0 �
πa 03
Given that the internuclear separation along the z-axis is R, in Cartesian coordinates rAi and rBi becomes
rAi = (x i2 + y 2i + z 2i )1�2
and
�erefore the wavefunction is
Ψ(1, 2) =
1
πa 03
× �e−[(x 1 +y 1 +z 1 )
2
P�A.�
2
2 1�2
+(x 22 +y 22 +(z 2 −R)2 )1�2 ]�a 0
rBi = (x i2 + y 2i + (z i − R)2 )1�2
+ e−[(x 2 +y 2 +z 2 )
2
2
2 1�2
+(x 12 +y 12 +(z 1 −R)2 )1�2 ]�a 0
�
For the purposes of this problem, the px and p y orbitals are represented by
unit vectors along the x-and y-axes, respectively. �e given hybrid atomic
orbitals are created by the linear combination of the s, px and p y orbitals. �e s
orbital is spherically symmetric about the origin, therefore it does not modify
the directions in which the hybrids point. �e vector representations of the
hybrid atomic orbitals are
h1 =
√
2j
h2 =
�
�
3�2 i − 1�2 j
�
�
h3 = − 3�2 i − 1�2 j
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
y
α
h1
�
α
3
2
h3
�x
1
2
h2
√
From the diagram it is evident that α = tan−1 �1� 3� = 30○ . It follows that the
angle between adjacent hybrids is ���○ .
9B Molecular orbital theory: the hydrogen molecule-ion
Answer to discussion questions
D�B.�
As described in Section �B.�(b) on page ���, the reason why the bonding molecular orbital is lower in energy than the atomic orbitals is not entirely clear.
However, it is clear that bonding character correlates strongly with molecular
orbitals that have an accumulation of electron density between nuclei due to
overlap and constructive interference of their component atomic orbitals. A
simple and plausible explanation of this correlation is that enhanced electron
probability between nuclei lowers the potential energy by putting electrons in
a position where they can be attracted to two nuclei at the same time; however,
the source of the reduced energy may be more complicated.
D�B.�
�e Born–Oppenheimer approximation treats the nuclei of the multi-particle
system of electrons and nuclei as if they were �xed. �e dependence of energy on nuclear positions is then obtained by solving the Schrödinger equation at many di�erent (�xed) nuclear geometries. Molecular potential energy
curves and surfaces are plots of molecular energy (computed under the Born–
Oppenheimer approximation) as a function of nuclear coordinates.
Solutions to exercises
E�B.�(a)
�e energy of the σ bonding orbital in H� + is given by [�B.�–���], E σ = EH�s +
j 0 �R − ( j + k)�(1 + S). Molecular potential energy curves are usually plotted
with respect to the energy of the separated atoms, therefore the energies to be
plotted are E σ − EH�s = j 0 �R − ( j + k)�(1 + S).
Using [�B.�d–���], j 0 �a 0 = 27.21 eV = 1 Eh the energy for R�a 0 = 1 is computed as
E σ − EH�s =
(1 E h ) (0.729 E h ) + (0.736 E h )
−
= +0.211 E h
1
(1 + 0.858)
Similar calculations give the following energies
315
9 MOLECULAR STRUCTURE
R�a 0
(E σ − EH�s )�Eh
�
+0.211
�
−5.32 × 10
−2
�
�
−5.88 × 10
−3.76 × 10−2
−2
0.2
(E σ − EH�s )�E h
316
0.1
0.0
Figure 9.1
R�a 0
1.0
1.5
2.0
2.5
3.0
3.5
4.0
−0.1
�ese data are plotted in Fig. �.�; with so few data points it is di�cult to locate
the minimum. �e data are �tted well by the following cubic
(E σ − EH�s )�E h = −0.0386(R�a 0 )3 + 0.3611(R�a 0 )2 − 1.0771(R�a 0 ) + 0.9656
E�B.�(a)
E�B.�(a)
Note that this cubic equation has no physical meaning, it is only used to draw
the line on the plot above and to locate the minimum by setting the derivative
to zero; in particular the maximum close to the �nal data point has no physical
basis. �is minimum is found to be at R = 2.5 a 0 , which corresponds to the
predicted equilibrium bond length. �e depth of the potential energy well at
this distance is about −0.073 E h which is 2.0 eV .
A sketch of the bonding and the antibonding molecular orbitals resulting from
the side-by-side overlap of two p orbitals is shown in Fig. �C.� on page ���. �e
bonding molecular orbital is antisymmetric with respect to inversion, therefore
it is denoted as a πu orbital. �e antibonding molecular orbital is symmetric
with respect to inversion, therefore it is a πg orbital.
�e normalization condition is given by [�B.�c–���], ∫ ψ ∗ ψ dτ = 1. �e wavefunction is normalized by �nding N such that ψ = N(ψA + λψB ) satis�es this
condition. �e wavefunctions ψA and ψB are real, as is N, therefore
∗
2
2
� ψ ψ dτ = N � (ψA + λψB ) dτ
1
1
S
�
�
���� � � � � � � � � � � � � � � �
��� � � � � � � �� � � � � � � �
��� � � � � � � � � � � � ��� � � � � � � � � � � � ��
�
�
�
�
= N 2 �� ψA2 dτ +λ 2 � ψB2 dτ +2λ� ψA ψB dτ �
�
�
�
�
�
�
�
�
2
2
= N (1 + λ + 2λS)
�e integrals ∫ ψA2 dτ and ∫ ψB2 dτ are � because the wavefunctions ψA and ψB
are normalized. It follows that N = 1�(1 + λ 2 + 2λS)1�2 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�B.�(a)
�e condition of orthogonality is given by [�C.�–���], ∫ ψ ∗i ψ j dτ = 0 for i ≠ j.
�e given molecular orbital, ψ i = 0.145A + 0.844B is real, therefore ψ ∗i = ψ i .
�e new linear combination for A and B, which is orthogonal to ψ i must have
the form of ψ j = A + βB, where the coe�cient of wavefunction A is chosen
to be � for simplicity. Substitution of these wavefunctions in the condition of
orthogonality gives
∗
� ψ i ψ j dτ = � (0.145A + 0.844B) × (A + βB) dτ
��� � � � � � � � � � � � � �
��� � � � � � ��� � � � � � �
��� � � � � � � � � � � � � � � �
2
2
= 0.145� A dτ +0.844β � B dτ +(0.145β + 0.844)� AB dτ
1
S
1
= 0.145 + 0.844β + (0.145β + 0.844)S
Using S = 0.250 the value of the integral becomes 0.356 + 0.88025β. �is value
must be zero for the two wavefunctions to be orthogonal, therefore β = −0.404
and hence ψ j = A − 0.404B.
Normalization of ψ i follows the same logic as in Exercise E�B.�(a). First the
wavefunction is written as ψ i = N(0.145A + 0.844B) and then the normalization constant N is found such that ∫ ψ ∗ ψ dτ = 1.
∗
2
� ψ i ψ i dτ = � [N(0.145A + 0.844B)] dτ
��� � � � � � � � � � � � � �
��� � � � � � ��� � � � � � �
��� � � � � � � � � � � � � � � �
2
2
2
2
2
= N �0.145 � A dτ +0.844 β� B dτ +(2 × 0.145 × 0.844)� AB dτ �
1
1
S
= N 2 (0.733 + 0.245S)
√
Using S = 0.250 gives a value of 0.794N 2 for the integral, therefore N = 1� 0.794 =
1.12. �erefore the normalized wavefunction is
ψ i = 1.12 × (0.145A + 0.844B) = 0.163A + 0.947B
Normalization of ψ j follows a similar procedure as for ψ i , giving N = 1.02 and
therefore ψ j = 1.02A − 0.412B .
Solutions to problems
P�B.�
Inspection of [�B.�–���] reveals that the repulsion energy between two hydrogen nuclei is given by e 2 �4πε 0 R, where R is the internuclear separation. In
molar quantities, the repulsion energy is N A e 2 �4πε 0 R, which, at an equilibrium
separation of R = 74.1 pm becomes
(6.0221 × 1023 mol−1 ) × (1.6022 × 10−19 C)2
= 1.87 × 106 J mol−1
4π × (8.8542 × 10−12 J−1 C2 m−1 ) × (74.1 × 10−12 m)
317
318
9 MOLECULAR STRUCTURE
�e molar gravitational potential energy between two hydrogen nuclei is
N A Gmp2
R
(6.0221 × 1023 mol−1 ) × (6.6738 × 10−11 N m2 kg−2 ) × (1.6726 × 10−27 kg)2
=
(74.1 × 10−12 m)
= 1.52 × 10−30 J mol−1
�erefore the gravitational attraction is entirely negligible compared to the electrostatic repulsion between the two nuclei.
P�B.�
Refer to the data presented in Exercise E�B.�(a). �e energy of the σ bonding
orbital in H� + is given by [�B.�–���], E σ = EH�s + j 0 �R − ( j + k)�(1 + S). �is
energy in usually measured with respect to the energy of the separated atoms,
therefore the energy is E σ − EH�s = j 0 �R − ( j + k)�(1 + S). Likewise for the
σ∗ antibonding orbital the energy is given by [�B.�–���], E σ∗ = EH�s + j 0 �R −
( j − k)�(1 − S). Relative to the separated atoms, the energy is E σ∗ − EH�s =
j 0 �R − ( j − k)�(1 − S). With the data given, and using j 0 �a 0 = 27.21 eV = 1 Eh ,
the energies of these molecular orbitals are
R�a 0
(E σ − EH�s )�Eh
(E σ∗ − EH�s )�Eh
�
+0.211
+1.05
�
−5.32 × 10
+0.340
−2
�
−5.88 × 10
+0.132
−2
�
−3.76 × 10−2
+5.52 × 10−2
It is evident that at each distance the antibonding molecular orbital is raised in
energy by more than the bonding molecular orbital is lowered. �is appears to
be generally true for any reasonable internuclear separation.
P�B.�
�e bonding and antibonding MOl wavefunctions are ψ± = N± (ψ A ± ψ B ),
where N± is the normalizing factor, given by (Example �B.� on page ���)
N± =
1
[(2(1 ± S)]1�2
where for two �s AOs separated by a distance R the overlap integral is given
by [�B.�a–���], S = �1 + R�a 0 + 13 (R�a 0 )2 � e−R�a 0 . �e form of ψ A and ψ B are
given in Brief illustration �B.� on page ���
ψ A = (1�πa 03 )1�2 e−r A1 �a 0
ψ B = (1�πa 03 )1�2 e−r B1 �a 0
Without loss of generality, it is assumed that atom A is located at z A1 = 0 and
atom B at z B1 = R, the internuclear separation. �e requirement is to plot the
wavefunction along the z-axis, so x A1 = y A1 = 0, and likewise for orbital B.
With all of these conditions imposed the function to plotted is
ψ± =
1
1
�e−�z��a 0 ± e−�(z−R)��a 0 �
[(2(1 ± S)]1�2 (πa 03 )1�2
�e modulus signs are needed because the argument of the exponential is the
distrance from the nucleus, which is always positive.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Figure �.� shows plots of (a) the bonding and antibonding wavefunctions, and
(b) the squares of these functions (the probability density) for the case R = 2a 0 .
�e quantity plotted in (a) is (a 0 )3�2 ψ± , and in (b) it is (a 0 )3 ψ±2 ; the same scale
is used for each orbital.
�e antibonding orbital has a node at the mid-point of the bond, whereas the
bonding orbital has signi�cant electron density at this point. From the diagrams it appears that the antibonding orbital has greater overall probability, but
this is not in fact the case – both orbitals are normalized. It is just that when
the functions are plotted along the z-axis there appears to be such a di�erence
due to the di�erent distribution of electron density elsewhere in the orbitals.
�e di�erence density is the di�erence in electron density between the molecular orbital and two non-interacting �s orbitals, one on each atom. It is a measure
of the way in which the electron density is changed when the molecular orbitals
are compared to non-interacting atomic orbitals. �e di�erence density is given
by
ψ±2 − 12 �ψ A2 + ψ B2 �
Also shown in Fig. �.� is this di�erence density for (c) the bonding and (d)
the antibonding molecular orbital; the same scale is used for each plot. If
the di�erence density is positive the implication is that the electron density
is greater than for two non-interacting atomic orbitals, whereas if the di�erence density is negative the implication is that there is a reduction in electron
density. It is evident from the plots that in the bonding molecular orbital there
is an increase in the electron density in the internuclear region, whereas for
the antibonding orbital the density in this region is reduced, but the density is
increased further away. �ese observations account (partially, at least) for the
fact that occupying the bonding molecular orbital promotes bond formation.
�e apparent di�erence in size between the di�erence densities between (c) and
(d) is a result of simply plotting the function along the z-axis.
9C Molecular orbital theory: homonuclear diatomic molecules
Answer to discussion questions
D�C.�
�e building-up principle for homonuclear diatomic molecules is essentially
the same as for atoms, but the diatomic molecular orbitals used in the former
are di�erent in name and in nature than the atomic orbitals used in the latter.
A diagram of energy levels (orbitals) and degeneracies is needed. For diatomic
molecules, these energy levels are either nondegenerate (for σ bonds) or doubly
degenerate (for all others). �e orbitals are populated with electrons, placing
each successive electron in the lowest-energy orbital available, no more than
two electrons per orbital. Hund’s rule indicates that di�erent degenerate orbitals should be populated �rst, with electrons that have parallel spins, before
pairing two electrons in the same degenerate orbital.
D�C.�
�e bond strength is related to the extent to which the occupied bonding molecular orbitals are lowered in energy compared to the constituent atomic orbitals.
319
320
9 MOLECULAR STRUCTURE
(a)
(b)
antibonding
bonding
–4
–2
0
2
4
6
z/a0
–4
(c)
0
2
4
6
z/a0
0
2
4
6
z/a0
(d)
antibonding
bonding
–4
–2
–2
0
2
4
6
z/a0
–4
–2
Figure 9.2
As described in Topic �B for the case of H� + , this lowering in energy depends
on the size of the term k, [�B.�c–���], which is a measure of the interaction
between a nucleus and the excess electron density in the internuclear region
arising from overlap.
�e overlap integral, S, is a di�erent quantity than k, but its behaviour with
(for example) internuclear distance is quite similar. �us the overlap integral
is o�en taken as a proxy for k, not least as it much easier to imagine how
the overlap varies when the orbital or the internuclear distance is varied. It is
therefore common to speak of a bond being strong when ‘there is good overlap’.
�e fact that there is a correlation between overlap and bond strength may,
however, simply be fortuitous as the theory does not indicate such a connection.
Solutions to exercises
E�C.�(a)
�e molecular orbital energy level diagram for Li2 , Be2 , B2 , C2 and N2 is shown
in Fig. �C.�� on page ���, and for O2 , F2 and Ne2 in Fig. �C.�� on page ���.
Following the same logic as in Exercise E�C.�(a) and Exercise E�C.�(a) gives
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Li2
Be2
B2
C2
N2
O2
F2
Ne2
E�C.�(a)
1 + 1 = 2 VE
2 + 2 = 4 VE
3 + 3 = 6 VE
4 + 4 = 8 VE
5 + 5 = 10 VE
6 + 6 = 12 VE
7 + 7 = 14 VE
8 + 8 = 16 VE
1σ 2g
1σ 2g 1σ∗2
u
2
1σ 2g 1σ∗2
u 1π u
4
1σ 2g 1σ∗2
u 1π u
4
2
1σ 2g 1σ∗2
u 1π u 2σ g
2
4
∗2
1σ 2g 1σ∗2
u 2σ g 1π u 1π g
2
4
∗4
1σ 2g 1σ∗2
u 2σ g 1π u 1π g
2
4
∗4
∗2
1σ 2g 1σ∗2
u 2σ g 1π u 1π g 2σ u
b = 12 (2 − 0) = 1
b = 12 (2 − 2) = 0
b = 12 (4 − 2) = 1
b = 12 (6 − 2) = 2
b = 12 (8 − 2) = 3
b = 12 (8 − 4) = 2
b = 12 (8 − 6) = 1
b = 12 (8 − 8) = 0
�e molecular orbital energy level diagram for Li2 , Be2 , B2 , C2 , N2 and their
ions is shown in Fig. �C.�� on page ���, and for O2 , F2 , Ne2 and their ions in
Fig. �C.�� on page ���. �e highest occupied molecular orbital (HOMO) is the
molecular orbital which is the highest in energy and is at least singly occupied.
�e HOMO of each of the listed ions is indicated by a box around it.
Li2 +
Be2 +
B2 +
C2 +
N2 +
O2 +
F2 +
Ne2 +
Li2 −
Be2 −
B2 −
C2 −
N2 −
O2 −
F2 −
Ne2 −
1 + 1 − 1 = 1 VE
1σ 1g
2 + 2 − 1 = 3 VE
1σ 2g 1σ∗1
u
5 + 5 − 1 = 9 VE
4
1
1σ 2g 1σ∗2
u 1π u 2σ g
3 + 3 − 1 = 5 VE
4 + 4 − 1 = 7 VE
6 + 6 − 1 = 11 VE
7 + 7 − 1 = 13 VE
8 + 8 − 1 = 15 VE
1
1σ 2g 1σ∗2
u 1π u
3
1σ 2g 1σ∗2
u 1π u
2
4
∗1
1σ 2g 1σ∗2
u 2σ g 1π u 1π g
2
4
∗3
1σ 2g 1σ∗2
u 2σ g 1π u 1π g
2
4
∗4
∗1
1σ 2g 1σ∗2
u 2σ g 1π u 1π g 2σ u
1 + 1 + 1 = 3 VE
1σ 2g 1σ∗1
u
4 + 4 + 1 = 9 VE
4
1
1σ 2g 1σ∗2
u 1π u 2σ g
2 + 2 + 1 = 5 VE
3 + 3 + 1 = 7 VE
5 + 5 + 1 = 11 VE
6 + 6 + 1 = 13 VE
7 + 7 + 1 = 15 VE
8 + 8 + 1 = 17 VE
1
1σ 2g 1σ∗2
u 1π u
3
1σ 2g 1σ∗2
u 1π u
4
2
∗1
1σ 2g 1σ∗2
u 1π u 2σ g 1π g
2
4
∗3
1σ 2g 1σ∗2
u 2σ g 1π u 1π g
2
4
∗4
∗1
1σ 2g 1σ∗2
u 2σ g 1π u 1π g 2σ u
2
4
∗4
∗2
1
1σ 2g 1σ∗2
u 2σ g 1π u 1π g 2σ u 3σ g
Note that the extra electron in Ne� is accommodated on a bonding molecular
orbital resulting from the overlap of the �s atomic orbitals.
321
322
9 MOLECULAR STRUCTURE
E�C.�(a)
�e energy of the incident photon must equal the sum of the ionization energy
of the orbital and the kinetic energy of the ejected photoelectron, [�C.�–���],
hν = I + 12 me υ 2 . �e energy of the incident photon is given by hν = hc�λ =
(6.6261 × 10−34 J s) × (2.9979 × 108 m s−1 )�(100 × 10−9 m) = 1.98... × 10−18 J.
Rearranging the equation to give the speed of the ejected electron gives
1�2
2 hc
� − I��
me λ
2
=�
× [(1.98... × 10−18 J)
(9.1094 × 10−31 kg)
υ=�
E�C.�(a)
− (12.0 eV) × (1.6022 × 10−19 J eV−1 )]�
1�2
= 3.70 × 105 m s−1
�e molecular orbital diagram for the homonuclear diatomic molecules Li2 ,
Be2 , and C2 is shown in Fig. �C.�� on page ���. According to the Pauli principle,
up to two valence electrons can be placed in each of the molecular orbitals. First
the lowest energy orbital is �lled up, then the next lowest and so on, until all
the valence electrons are used up.
(i) Li2 has 1+1 = 2 valence electrons (VE) overall, therefore the ground-state
electron con�guration is 1σ 2g . �e bond order is de�ned in [�C.�–���] as
b = 12 (N − N ∗ ), therefore b = 12 (2 − 0) = 1 .
1
(ii) Be2 : 2 + 2 = 4 VE; 1σ 2g 1σ∗2
u ; b = 2 (2 − 2) = 0 .
E�C.�(a)
1
4
(iii) C2 : 4 + 4 = 8 VE; 1σ 2g 1σ∗2
u 1π u ; b = 2 (6 − 2) = 2 .
�e molecule with the greater bond order is expected to have the larger dissociation energy. Qualitatively B2 and C2 share the same molecular orbital
energy level diagram, shown in Fig. �C.�� on page ���. B2 has 3 + 3 = 6
valence electrons overall, therefore its ground-state electron con�guration is
1
2
∗
1σ 2g 1σ∗2
u 1π u . �e bond order is de�ned in [�C.�–���] as b = 2 (N − N ),
therefore b = 12 (4 − 2) = 1.
4
C2 has 4 + 4 = 8 valence electrons, its con�guration is 1σ 2g 1σ∗2
u 1π u , and the
bond order is b = 12 (6 − 2) = 2. C2 has greater bond order than B2 , therefore
C2 is expected to have the larger bond dissociation energy.
E�C.�(a)
�e molecule with the greater bond order is expected to have the larger dissociation energy. �e molecular orbital energy level diagram of F2 and F2 + is shown
in Fig. �C.�� on page ���. F2 has 7 + 7 = 14 valence electrons overall, therefore
2
4
∗4
the ground-state electron con�guration is 1σ 2g 1σ∗2
u 2σ g 1π u 1π g . �e bond
1
1
∗
order is de�ned in [�C.�–���] as b = 2 (N − N ), therefore b = 2 (8 − 6) = 1.
Removing one electron from F� gives F� + , which has one fewer electron in the
antibonding π∗g orbital, therefore the bond order is b = 12 (8 − 5) = 32 . F2 + has
greater bond order than F2 , therefore F2 + is expected to have the larger bond
dissociation energy.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
323
Solutions to problems
P�C.�
(a) Figure �.� shows a plot of the overlap integral (Z = 1 is assumed)
S(2s, 2s) = �1 +
1 R
1 R 2
1
R 4
� �+
� � +
� � � e−R�2a 0
2 a0
12 a 0
240 a 0
1.0
S(2s,2s)
S(2p,2p)
0.8
S
0.6
0.4
0.2
0.0
0
Figure 9.3
5
10
R�a 0
15
20
(b) �e overlap integral S(2s, 2s) reaches a value of �.�� at R�a 0 = 8.03 ; this
value can be read o� a graph or found by using mathematical so�ware.
(c) Figure �.� shows a plot of the overlap integral (Z = 1 is assumed)
S(2p, 2p) = �1 +
1 R
1 R 2
1
R 3
� �+
� � +
� � � e−R�2a 0
2 a0
10 a 0
120 a 0
(d) �e value of the overlap integral at R�a 0 = 8.03 is
P�C.�
1
1
S(2p, 2p) = �1 + 12 ×8.03 + 10
×(8.03)2 + 120
×(8.03)3 �×e−8.03�2 = �.��
Figure �.� shows contour plots of the bonding and antibonding �pσ and �pπ
molecular orbitals for a representative internuclear distance of R = 6a 0 ; negative amplitude is indicated by dashed contours, and the locations of the nuclei
are shown by the black dots. Density plots, in which the intensity of the shading
is proportional to the square of the wavefucntion, are shown in Fig. �.�.
�ese plots are useful as they identify aspects of the symmetry of the wavefunctions and the positions of nodal planes. In addition, they illustrate that for the
bonding orbitals electron density is accumulating in the internuclear region.
9D Molecular orbital theory: heteronuclear diatomic molecules
Answer to discussion questions
D�D.�
�e Coulomb integral is essentially the energy of an electron when it occupies
an atomic orbital in the molecule.
9 MOLECULAR STRUCTURE
2pσg
2pσu
4
0
0
−4
−4
x / a0
4
−12 −8
−4
0
4
z / a0
8
12
−12 −8
−4
0
4
z / a0
2pπu
x / a0
324
8
4
4
0
0
−4
−4
−8
−8
Figure 9.4
−4
0
4
z / a0
8
12
2pπg
8
−12 −8
8
12
−12 −8
−4
0
4
z / a0
8
12
�e resonance integral is a contribution to the energy of a molecule that can be
associated with an electron interacting with more than one nucleus at once.
D�D.�
In forming a bond an atom must, to some extent, give up electron density to
be shared with other atoms in the molecule. �e energy needed to do this is
connected with the value of the ionization energy of the orbital. Equally, the
atom will to some extent acquire additional electron density to interact with,
and the energy gained from acquiring this density is connected in some way to
the electron a�nity. �us both electron gain and electron loss, in the loosest
sense, are involved in the process of bonding. It is for this reason that ionization
energy and electron a�nity are involved in the estimation of atomic orbital
energies for participation in bonding. See Section �D.�(a) on page ���.
Solutions to exercises
E�D.�(a)
A suitable MO diagram in shown in the solution to Exercise E�D.�(a). �e ion
with the greater bond order is expected to have the shorter bond length. NO+
has 5 + 6 − 1 = 10 valence electrons, just enough to completely �ll up all the
bonding molecular orbitals, leading to a ground state electron con�guration
of 1σ 2 2σ 2 3σ 2 1π 4 . NO− has two more electrons, both accommodated in the
antibonding 2π orbital. It follows that NO+ has a greater bond order than NO− ,
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
2pσg
4
0
0
−4
−4
x / a0
4
−12 −8
−4
0
4
z / a0
8
12
2pπu
8
x / a0
2pσu
−12 −8
4
0
0
−4
−4
−8
−8
−4
0
4
z / a0
8
0
4
z / a0
12
8
12
2pπg
8
4
−12 −8
−4
−12 −8
−4
0
4
z / a0
8
12
Figure 9.5
therefore NO+ is expected to have the shorter bond length.
E�D.�(a)
�e relationship between the Pauling and Mulliken electronegativities is given
1�2
by [�D.�–���], χPauling = 1.35χMulliken − 1.37. A plot of the Pauling electronegativities of Period � atoms against the square root of their Mulliken electronegativities is shown in Fig. �.�.
�e equation of the best �t line is χPauling = 3.18χMulliken − 2.57, which is very
far from the expected relationship.
1�2
E�D.�(a)
�e orbital energy of an atomic orbital in a given atom is estimated using the
procedure outlined in Brief illustration �D.� on page ���, and using data from
the Resource section. �e orbital energy of hydrogen is
αH = − 12 [I + Eea ]
= − 12 ×[(1312.0 kJ mol−1 )+(72.8 kJ mol−1 )]×
(1 eV)
= −7.18 eV
(96.485 kJ mol−1 )
�e conversion factor between kJ mol−1 and eV is taken from inside the front
325
9 MOLECULAR STRUCTURE
4.0
3.0
χPauling
326
2.0
1.0
1.0
1.2
1.4
Figure 9.6
1.6
1.8
2.0
2.2
1�2
χMulliken
cover . Similarly for chlorine
αCl = − 12 [I + Eea ]
E�D.�(a)
= − 12 ×[(1251.1 kJ mol−1 )+(348.7 kJ mol−1 )]×
�e orbital energies of hydrogen (αH = −7.18 eV) and chlorine (αCl = −8.29 eV)
are calculated in Exercise E�D.�(a). Taking β = −1.0 eV as a typical value and
setting S = 0 for simplicity, substitution into [�D.�c–���] gives
E± = 12 (αH + αCl ) ± 12 (αH − αCl ) �1 + �
= 12 [(−7.18 eV) + (−8.29 eV)]
�
�
2
2β
� �
αH − αCl
± 12 [(−7.18 eV) − (−8.29 eV)] �
�1 + �
= (−7.73... eV) ± (1.14... eV)
E�D.�(a)
(1 eV)
= −8.29 eV
(96.485 kJ mol−1 )
�
�
1�2
2 �1�2
�
(−2.0 eV)
� �
(−7.18 eV) − (−8.29 eV) �
�
�
�erefore the energy of the bonding molecular orbital is E− = (−7.73... eV) −
(1.14... eV) = −8.88 eV , and the antibonding orbital is at an energy level of
E+ = (−7.73... eV) + (1.14... eV) = −6.59 eV .
�e orbital energies of hydrogen (αH = −7.18 eV) and chlorine (αCl = −8.29 eV)
are calculated in Exercise E�D.�(a). Taking β = −1.0 eV as a typical value, and
setting S = 0.2, substitution into [�D.�a–���] gives
αH + αCl − 2βS ± [(2βS − (αH + αCl ))2 − 4(1 − S 2 )(αH αCl − β 2 )]1�2
2(1 − S 2 )
(−15.0... eV) ± (1.54... eV)
=
= (−7.84... eV) ± (0.803... eV)
(1.92)
E± =
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�D.�(a)
�erefore the energy of the bonding molecular orbital is E− = (−7.48... eV) −
(0.803... eV) = −8.65 eV , and the antibonding orbital is at an energy level of
E+ = (−7.48... eV) + (0.803... eV) = −7.05 eV .
�e molecular orbital energy level diagram for a heteronuclear diatomic AB
is similar to that for a homonuclear diatomic A2 (Fig. �C.�� on page ��� or
Fig. �C.�� on page ���) except that the atomic orbitals on A and B are no longer
at the same energies. As a result the molecular orbitals no longer have equal
contributions from the orbitals on A and B; furthermore, it is more likely that
the �s and �p orbitals will mix. From simple considerations it it not possible to
predict the exact ordering of the resulting molecular orbitals, so the diagram
shown in Fig. �.� is simply one possibility. Note that because the heteronuclear
diatomic no longer has a centre of symmetry the g/u labels are not applicable.
�e electronic con�gurations are: (i) CO (�� valence electrons) 1σ 2 2σ 2 3σ 2 1π 4 ;
(ii) NO (�� valence electrons) 1σ 2 2σ 2 3σ 2 1π 4 2π 1 ; (iii) CN – is isoelectronic
with CO and therefore has the same con�guration.
A
Molecule
B
4σ
2p
2π
2p
1π
3σ
2σ
2s
2s
1σ
Figure 9.7
E�D.�(a)
�e molecular orbital energy level diagram of the heteronuclear diatomic molecule
XeF is similar to the one shown in the solution to Exercise E�D.�(a) except that
the orbitals on atom A are �s and �p. It is not possible to predict the precise
energy ordering of the orbitals from simple considerations, so this diagram is
simply a plausible suggestion.
XeF has 8 + 7 = 15 valence electrons, therefore the ground state electron con�guration is 1σ 2 2σ 2 3σ 2 1π 4 2π 4 4σ 1 . �e con�guration of XeF+ is the same
except that, as there is one fewer electrons, the antibonding 4σ orbital is not
occupied. �is means that the bond order in XeF+ (b = 1) is greater than the
bond order in XeF (b = 12 ), therefore XeF+ is likely to have a shorter bond
length than XeF.
327
9 MOLECULAR STRUCTURE
Solutions to problems
P�D.�
(a) A normalized wavefunction satis�es the condition given by [�B.�c–���],
∗
∗
∫ ψψ dτ = 1. �e given wavefunction is real, therefore ψ = ψ .
� ψψ dτ = � (ψA cos θ + ψB sin θ) dτ
∗
2
��� � � � � � ��� � � � � � �
��� � � � � � � � � � � � � �
��� � � � � � � � � � � �� � � � � � � � � � � ��
2
2
2
2
= cos θ � ψA dτ + sin θ � ψB dτ +2 cos θ sin θ � ψA ψB dτ
1
1
0
= cos2 θ + sin2 θ = 1
�e values of the integrals come from the fact that ψA and ψB are orthonormal.
(b) �e wavefunction which describes the bonding molecular orbital is formed
by the in-phase interference of the atomic orbitals ψA and ψB , therefore
the coe�cients of ψA and ψB must have the same sign. Similarly, the
antibonding orbital is the result of the out-of-phase interference of the
basis atomic orbitals, therefore the corresponding coe�cients must have
opposite signs. A plot of the coe�cients, cos θ and sin θ as a function of
θ is shown in Fig. �.�.
1.0
value of function
328
0.5
0.0
−0.5
−1.0
0.0
Figure 9.8
P�D.�
cos θ
sin θ
0.2
0.4
0.6
0.8
1.0
θ�π
�erefore ψ describes a bonding molecular orbital for 0 < θ < π�2, and
an antibonding molecular orbital for π�2 < θ < π.
�e energy of ψA and ψC are kept constant in the following, but the energy of
ψB is progressively lowered.
(a) Taking the energy of the orbital ψB to be −12.0 eV, the secular determinant becomes
��� (−7.2 eV) − E
(−1.0 eV)
(−0.8 eV) ����
���
���
(−12.0 eV) − E
0
��� (−1.0 eV)
��
��� (−0.8 eV)
0
(−8.4 eV) − E ����
�
= −E 3 − (27.6 eV)E 2 − (246.04 eV2 )E − (709.68 eV3 )
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Setting the polynominal to zero and solving the cubic gives the following
energies: E 1 = −12.2 eV , E 2 = −8.75 eV and E 3 = −6.65 eV . �e matrix which diagonalizes the hamiltonian matrix is
0.394
� 0.202
� 0.978 −0.121
� 0.0425 0.911
0.897 �
−0.168 �
−0.409 �
�e entries in each column of the matrix above give the coe�cients of the
atomic orbitals for the corresponding molecular orbital. Note that E 1 is
close to the energy of ψB , and that in the �rst column the orbital with the
largest coe�cient by far is ψB .
(b) If the energy of ψB is lowered further to −15.0 eV, the secular determinant
becomes
��� (−7.2 eV) − E
(−1.0 eV)
(−0.8 eV) ����
���
���
(−15.0 eV) − E
0
��� (−1.0 eV)
��
��� (−0.8 eV)
0
(−8.4 eV) − E ����
�
= −E 3 − (30.6 eV)E 2 − (292.84 eV2 )E − (889.2 eV3 )
�e energies are E 1 = −15.1 eV , E 2 = −8.77 eV and E 3 = −6.70 eV , and
the matrix which diagonalizes the hamiltonian is
� 0.127
� 0.992
� 0.0151
0.419
−0.0672
0.906
0.899 �
−0.108 �
−0.424 �
As before E 1 is close in energy to the energy of ψB , and in the �rst column the coe�cient of that atomic orbital is close to �. Furthermore, in
columns � and � the other molecular orbitals are seen to have only small
contributions from ψB .
�e interpretation is that as the energy of ψB becomes more and more
separate from the other orbitals, one of the molecular orbitals becomes
very much like ψB and would be classed as non-bonding, and the other
two molecular orbitals have little contribution from ψB .
9E Molecular orbital theory: polyatomic molecules
Answer to discussion questions
D�E.�
See Section �E.� on page ���.
D�E.�
In ab initio methods an attempt is made to evaluate all integrals that appear
in the secular determinant. Approximations are still employed, but these are
mainly associated with the construction of the wavefunctions involved in the
integrals. In semi-empirical methods, many of the integrals are expressed in
terms of spectroscopic data or physical properties. Semi-empirical methods
exist at several levels. At some levels, in order to simplify the calculations, many
of the integrals are set equal to zero.
329
330
9 MOLECULAR STRUCTURE
Density functional theory (DFT) is di�erent from the Hartree-Fock (HF) selfconsistent �eld (HF-SCF) methods, the ab initio methods, in that DFT focuses
on the electron density while HF-SCF methods focus on the wavefunction.
However, they both attempt to evaluate integrals from �rst principles, so DFT
methods are in that sense ab initio methods: both are iterative self-consistent
methods in that the calculations are repeated until the energy and wavefunctions (HF-SCF) or energy and electron density (DFT) are unchanged to within
some acceptable tolerance.
D�E.�
�ese are all terms originally associated with the Hückel approximation used
in the treatment of conjugated π electron molecules, in which the π electrons
are considered independent of the σ electrons. �e π electron binding energy is
the sum of the energies of each π electron in the molecule. �e delocalization
energy is the di�erence in energy of the π electrons between the conjugated
molecule with n π bonds and the energy of n ethene molecules, each of which
has one π bond. �e π bond formation energy is the energy released when a
π bond is formed. It is obtained from the total π electron binding energy by
subtracting the contribution from the Coulomb integrals, α.
Solutions to exercises
E�E.�(a)
(i) Following the same logic as in Exercise E�E.�(a) and applying the Hückel
approximations as explained there the secular determinant for anthracene
is written as (the numbers in bold refer to the numbering of the carbon
atoms in the molecule)
�
�
�
�
�
�
�
�
�
�� �� �� �� ��
� α−E β
0
0
0
0
0
0
0
0
0
0
0
β
� β α−E β
0
0
0
0
0
0
0
0
0
0
0
� 0
β α−E β
0
0
0
0
0
0
0
0
0
0
� 0
0
β α−E β
0
0
0
0
0
0
0
0
0
� 0
0
0
β α−E β
0
0
0
0
0
0
0
β
� 0
0
0
0
β α−E β
0
0
0
0
0
0
0
� 0
0
0
0
0
β α−E β
0
0
0
β
0
0
� 0
0
0
0
0
0
β α−E β
0
0
0
0
0
� 0
0
0
0
0
0
0
β α−E β
0
0
0
0
�� 0
0
0
0
0
0
0
0
β α−E β
0
0
0
�� 0
0
0
0
0
0
0
0
0
β α−E β
0
0
�� 0
0
0
0
0
0
β
0
0
0
β α −E β
0
�� 0
0
0
0
0
0
0
0
0
0
0
β α−E β
�� β
0
0
0
β
0
0
0
0
0
0
0
β α−E
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(ii) Similarly for phenanthrene
�
�
�
�
�
�
�
�
�
�� �� �� �� ��
� α−E β
0
0
0
0
0
0
0
0
0
0
0
β
� β α−E β
0
0
0
0
0
0
0
0
0
0
0
� 0
β α−E β
0
0
0
0
0
0
0
0
0
0
� 0
0
β α−E β
0
0
0
0
0
0
0
0
0
� 0
0
0
β α−E β
0
0
0
0
0
0
0
β
� 0
0
0
0
β α−E β
0
0
0
0
0
0
0
� 0
0
0
0
0
β α−E β
0
0
0
0
0
0
� 0
0
0
0
0
0
β α−E β
0
0
0
β
0
� 0
0
0
0
0
0
0
β α−E β
0
0
0
0
�� 0
0
0
0
0
0
0
0
β α−E β
0
0
0
�� 0
0
0
0
0
0
0
0
0
β α−E β
0
0
�� 0
0
0
0
0
0
0
0
0
0
β α−E β
0
�� 0
0
0
0
0
0
0
β
0
0
0
β α−E β
�� β
0
0
0
β
0
0
0
0
0
0
0
β α−E
E�E.�(a)
To calculate the π-electron binding energy of the given systems, it is necessary
to calculate the energies of the occupied molecular orbitals. �is is done by diagonalising the hamiltonian matrix: the diagonal elements of the resulting matrix are the energies of the molecular orbitals. �e hamiltonian matrix has the
same form as the secular matrix except that the diagonal elements are α instead
of α − E. Alternatively, the energies can be found my �nding the eigenvalues of
the hamiltonian matrix, or by multiplying out the secular determinant, setting
the resulting polynomial in E to zero and then �nding the roots. Mathematical
so�ware is needed for all of these approaches.
�e secular determinants are derived in Exercise E�E.�(a), and from these the
form of the hamiltonain matrix is easily found.
(i) �e orbital energies for anthracene are E = α + 2.41β, α + 2β, α + 1.41β
(doubly degenerate), α + β (doubly degenerate), α + 0.414β, α − 0.414β,
α − β (doubly degenerate), α − 1.41β (doubly degenerate), α − 2β, α −
2.41β. �e π system of anthracene accommodates �� electrons, therefore
the � lowest energy π molecular orbitals are �lled. �e π-electron binding
energy is therefore E π = 2(α + 2.41β) + 2(α + 2β) + 4(α + 1.41β) + 4(α +
β) + 2(α + 0.414β) = 14α + 19.3β .
E�E.�(a)
(ii) �e orbital energies for phenanthrene are E = α + 2.43β, α + 1.95β, α +
1.52β, α +1.31β, α +1.14β, α +0.769β, α +0.605β, α −0.605β, α −0.769β,
α − 1.14β, α − 1.31β, α − 1.52β, α − 1.95β, α − 2.43β. �e π system of
anthracene accommodates �� electrons, therefore the � lowest energy π
molecular orbitals are �lled. �e π-electron binding energy is therefore
E π = 2(α + 2.43β) + 2(α + 1.95β) + 2(α + 1.52β) + 2(α + 1.31β) + 2(α +
1.14β) + 2(α + 0.769β) + 2(α + 0.605β) = 14α + 19.5β
�e hamiltonian for a single electron in H2 + is given by [�B.�–���]. It has a
kinetic energy term, T̂ = −(ħ 2 �2me )∇21 , and a potential energy term, V̂ . �e
331
332
9 MOLECULAR STRUCTURE
species HeH+ has two electrons, therefore the kinetic energy term is written as
T̂ = −
ħ2 2
ħ2 2
∇1 −
∇
2me
2me 2
�e energy of interaction between an electron and a nucleus with charge number Z at distance r is given by −Ze 2 �4πε 0 r. �e potential energy operator
consists of terms for each electron interacting with the H nuclues (Z = 1) and
the He nucleus (Z = 2)
V̂ = −
E�E.�(a)
e2
1
1
2
2
1
�
+
+
+
−
�
4πε 0 r 1H r 2H r 1He r 2He r 12
�e �rst term represent the interaction between electron � and the H nucleus,
and the second is for electron � with the same nucleus. �e third and fourth
terms represent the interactions of the two electrons with the He nucleus. �e
last term accounts for the repulsion between the two electrons. �e complete
electronic hamiltonian is Ĥ elec = T̂ + V̂ . Because only the electronic hamiltonian is required, the repulsion between the two nuclei is not included.
(i) Without making the Hückel approximations, the secular determinant of
the H3 molecule is written as
��� α 1 − E
���
��� β 21 − S 21 E
��� β − S E
31
� 31
β 12 − S 12 E
α2 − E
β 32 − S 32 E
β 13 − S 13 E ����
β 23 − S 23 E ����
�
α 3 − E ����
where α n is the Coulomb integral of the orbital on atom n, β nm is the
resonance integral accounting for the interaction between the orbitals on
atoms n and m, E is the energy of the molecular orbital and S nm is the
overlap integral between the orbtials on atoms n and m.
Within the Hückel approximations the energy of the basis atomic orbitals
is taken to be independent of the position of the corresponding atoms in
the molecule, therefore all Coulomb integrals are set equal to α (given
that there is only one type of basis atomic orbital and only one type of
atom is involved in the problem). Interaction between orbitals on nonneighbouring atoms is neglected, that is β nm = 0 if atoms n and m are
not neighbouring. All other resonance integrals are set equal to β. �e
overlap between atomic orbitals is also neglected, therefore all overlap
integrals S nm with n ≠ m are set to zero. Hence the secular determinant
for linear H� is
��� α − E
β
0 ����
���
α−E
β ����
��� β
�
��� 0
β
α − E ����
�
(ii) In this case hydrogen atoms � and � are neighbours, therefore β 13 = β,
and the secular determinant is
��� α − E
���
��� β
��� β
�
β
α−E
β
β ����
β ����
�
α − E ����
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E�E.�(a)
E�E.�(a)
�e energies of the molecular orbitals in benzene are given by [�E.��–���] as
E = α ± 2β, α ± β, α ± β. Note that α and β are negative quantities, therefore the
a�u molecular orbital is the lowest in energy with energy of α + 2β, as shown in
Fig. �E.� on page ���.
(i) �e benzene anion has 6 + 1 = 7 electrons in its π system, its electronic
2 4 1
con�guration is a�u
e�g e�u and the total π-electron binding energy is E π =
2(α + 2β) + 4(α + β) + (α − β) = 7α + 7β .
(ii) �e benzene cation has 6 − 1 = 5 electrons in its π system, its electronic
2 3
con�guration is a�u
e�g and the total π-electron binding energy is E π =
2(α + 2β) + 3(α + β) = 5α + 7β .
�e delocalization energy is the energy di�erence between the π-electron binding energy E π in the given species and the hypothetical π-electron binding
energy if the given species had isolated π bonds. �erefore the delocalization
energy is given by Edeloc = E π − N π (α + β), where N π is the number of π
electrons. �e π-bond formation energy is de�ned in [�E.��–���] as Ebf =
E π − N π α.
(i) �e benzene anion has � π electrons and its π-electron binding energy
is calculated in Exercise E�E.�(a) as E π = 7α + 7β. �erefore Edeloc =
(7α + 7β) − 7(α + β) = 0 and Ebf = (7α + 7β) − 7α = 7β .
(ii) �e benzene cation has � π electrons and its π-electron binding energy
is calculated in Exercise E�E.�(a) as E π = 5α + 7β. �erefore Edeloc =
(5α + 7β) − 5(α + β) = 2β and Ebf = (5α + 7β) − 5α = 7β .
Solutions to problems
P�E.�
Number the oxygen atoms from � to � and the carbon atom as number �. Taking
the Hückel approximations the secular determinant of the carbonate ion is
written as
���� αO −E 0
0
β ����
���
�
β ����
��� 0 αO −E 0
= (αO − E)2 [(αO − E)(αC − E) − 3β 2 ]
��� 0
0 αO −E β ����
���
�
β
β αC −E ����
�� β
Hence the energies of the molecular orbitals of the carbonate ion are the solutions of the equation (αO − E)2 [(αO − E)(αC − E) − 3β 2 ] = 0. �e solution
E = αO occurs twice and represents a degenerate pair of orbitals. �e other two
roots are the solutions of the equation (αO − E)(αC − E) − 3β 2 = 0, which gives
the energies
�
E± = 12 [αO + αC ± (αO − αC )2 + 12β 2 ]
where E− < E+ . Because oxygen is the more electronegative element it is likely
that the three lowest energy molecular orbitlas will be those with E = αO and
E = E− . �e carbonate ion has � π electrons which will occupy these orbitals
and hence give a π-electron binding energy of
�
�
E π = (αO +αC − (αO − αC )2 + 12β 2 )+4αO = 5αO +αC − (αO − αC )2 + 12β 2
333
334
9 MOLECULAR STRUCTURE
In a localized description of the bonding two electrons occupy a π-bonding
molecular orbital between C and O, and four electrons will be localized on oxygen atoms in orbitals with energy α O . Using the orbital energies from [�D.�c–
���], the total
� energy of the two electrons in the bonding molecular orbital is
αO + αC + (αO − αC )2 + 4β 2 . Hence, in a localized structure the π-electron
binding energy is
�
E loc = 5αO + αC + (αO − αC )2 + 4β 2
�e delocalization energy is given by E π − E loc
Edeloc = E π − E loc
P�E.�
�
�
= [5αO + αC + (αO − αC )2 + 12β 2 ] − [5αO + αC + (αO − αC )2 + 4β 2 ]
�
�
= (αO − αC )2 + 12β 2 − (αO − αC )2 + 4β 2
�e secular equations are
(αA − E)cA + (βAB − SAB E)cB + (βAC − SAC E)cC = 0
(βBA − SBA E)cA + (αB − E)cB + (βBC − SBC E)cC = 0
(βCA − SCA E)cA + (βCB − SCB E)cB + (αC − E)cC = 0
In this case, orbitals B and C are on the same atom. It follows that the resonance
integral βBC and the overlap integral SBC are zero, as the atomic orbitals on one
atom are orthogonal to each other. �erefore the secular equations simplify to
(αA − E)cA + (βAB − SAB E)cB + (βAC − SAC E)cC = 0
(βBA − SBA E)cA + (αB − E)cB = 0
(βCA − SCA E)cA + (αC − E)cC = 0
and hence the secular determinant is
P�E.�
���
αA − E
���
��� βBA − SBA E
��� β − S E
CA
� CA
βAB − SAB E
αB − E
0
βAC − SAC E ����
���
0
���
���
αC − E
Within the Hückel approximations, the secular determinant of cyclobutadiene
is
��� α − E
β
0
β ����
����
�
β
0 ����
��� β α − E
��� 0
β α−E
β ����
���
�
0
β α − E ����
�� β
�e hamiltonian matrix is of the same form, but with diagonal elements α
� α
� β
H=�
� 0
� β
β
α
β
0
0
β
α
β
β �
0 �
�
β �
α �
� 0
� 1
H = α� + β �
� 0
� 1
1
0
1
0
0
1
0
1
1 �
0 �
�
1 �
0 �
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
As explained in the text, in order to �nd the energies it is su�cient to diagonalize the matrix on the right, and this is convenient because most mathematical
so�ware packages are only able to diagonalize numerical matrices. �e diagonal elements in the resulting diagonalized matrix are +2, 0, 0, −2 giving the
energies as α + 2β, α (doubly degenerate), and α − 2β.
Similarly, the secular determinant and hamiltonian matrix for benzene is
��� α − E β
0
0
0
β ���
���
�
0
0
0 ����
���� β α − E β
�
��� 0
β α−E β
0
0 ����
���
�
0
β α−E β
0 ����
��� 0
�
��� 0
0
0
β α − E β ����
���
0
0
0
β α − E ����
�� β
�0 1 0 0 0 1�
�1 0 1 0 0 0�
�
�
�0 1 0 1 0 0�
�
H = α� + β �
�0 0 1 0 1 0�
�
�
�0 0 0 1 0 1�
�
�
�1 0 0 0 1 0�
�e diagonal elements in the resulting diagonalized matrix are +2, +1, +1, −1,
−1, and −2, giving the energies α +2β, α + β (doubly degenerate), α − β (doubly
degenerate), and α − 2β.
�e secular determinant and hamiltonian matrix for cyclooctatetraene is
��� α − E β
0
0
0
0
0
β ����
���
0
0
0
0
0 ����
���� β α − E β
�
��� 0
β α−E β
0
0
0
0 ����
���
�
0
β α−E β
0
0
0 ����
���� 0
�
��� 0
0
0
β α−E β
0
0 ����
���
�
0
0
0
β α−E β
0 ����
��� 0
��� 0
0
0
0
0
β α − E β ����
���
�
0
0
0
0
0
β α − E ����
�� β
�0 1 0 0 0 0 0 1�
�1 0 1 0 0 0 0 0�
�
�
�0 1 0 1 0 0 0 0�
�
�
�0 0 1 0 1 0 0 0�
�
H = α�+β �
�0 0 0 1 0 1 0 0�
�
�
�0 0 0 0 1 0 1 0�
�
�
�
�
�0 0 0 0 0 1 0 1�
�1 0 0 0 0 0 1 0�
�e diagonal elements in the resulting diagonalized matrix are +2.00, +1.41,
+1.41, 0, 0, −1.41, −1.41, and −2, giving energies α + 2β, α + 1.41β (doubly
degenerate), α (doubly degenerate), α − 1.41β (doubly degenerate), and α − 2β.
For all three molecules the lowest and highest energy molecular orbital is not
degenerate, and all the other orbitals occur as degenerate pairs.
P�E.�
(a) Within the Hückel approximations, the secular determinant of the triangular species H� is
��� α − E
β
β ����
���
β ����
��� β α − E
�
��� β
β α − E ����
�
�e hamiltonian matrix is of the same form, but with diagonal elements
α
� α β β �
� 0 1 1 �
H=� β α β �
H = α� + β � 1 0 1 �
� β β α �
� 1 1 0 �
As explained in the text, in order to �nd the energies it is su�cient to
diagonalize the matrix on the right, and this is convenient because most
mathematical so�ware packages are only able to diagonalize numerical
335
9 MOLECULAR STRUCTURE
matrices. �e diagonal elements in the resulting diagonalized matrix are
+2, −1, and −1 giving the energies as α+2β, and α−β (doubly degenerate).
�e molecular orbital energy level diagram is shown below.
E =α−β
energy
336
E = α + 2β
�e number of valence electrons (VE) and electron binding energies of
each species are
H3 +
H3
H3 −
2 VE
3 VE
4 VE
Etot = 2α + 4β
Etot = 3α + 3β
Etot = 4α + 2β
(b) Consider the following set of equations
H3 + (g) ��→ 2 H(g) + H+ (g)
H2 (g) ��→ 2 H(g)
H+ (g) + H2 (g) ��→ H3 + (g)
∆ r U −○ (1) = +849 kJ mol−1
∆ r U −○ (2) = +432.1 kJ mol−1
∆ r U −○ (3)
Reaction(�) is reaction(�) − reaction(�), therefore
∆ r U −○ (3) = ∆ r U −○ (2) − ∆ r U −○ (1) = (+432.1 kJ mol−1 ) − (+849 kJ mol−1 )
= −4.16... × 102 kJ mol−1 = −417 kJ mol−1
(c) �e change in the total electron binding energy directly gives the change
in the internal energy in the reaction H+ (g)+H2 (g) ��→ H3 + (g). �erefore ∆ r U −○ is expressed in terms of the resonance and Coulomb integrals
as
∆ r U −○ = Etot (products) − Etot (reactants)
H3 +
H+
��� � � � � � � � �� � � � � � � � � �� � ��� � � � � � �� � � � � � � �
= (2α + 4β) − 0 − 2(α + β) = 2β
P�E.�
H2
�erefore β = (−4.16... × 102 kJ mol−1 )�2 = −208 kJ mol−1 . �e electron
binding energies are 2α − 834 kJ mol−1 for H3 + , 3α − 625 kJ mol−1 for H3
and 4α − 416 kJ mol−1 for H3 − .
For H� the �-��G* basis set is equivalent to the �-��G basis set because the star
indicates that the basis set adds d-type polarization functions for each atom
other than hydrogen. Consequently, the basis sets (a) �-��G* and (b) �-���+G**
were chosen. Since the calculated energy is with respect to the energy of widely
separated stationary electrons and nuclei, the experimental ground electronic
energy of dihydrogen is calculated as D e + 2I.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
H�
�-��G*
�-���+G**
R�pm
��.�
��.�
ground-state energy, E 0 �eV −30.6626 −30.8167
F�
R�pm
���.�
���.�
ground-state energy, E 0 �eV −5406.30 −5407.92
experimental
��.�
−32.06
���.�
Both computational basis sets give satisfactory bond length agreement with the
experimental value for H� . However the �-��G* basis set is not as accurate as the
larger basis set as illustrated by consideration of both its higher ground-state
energy and the variation principle that the energy of a trial wavefunction is
never less than the true energy. �at is, the energy provided by the �-���+G**
basis set is closer to the true energy.
P�E.��
(a) Linear conjugated polyenes do not have degenerate energy levels, so each
value of k corresponds to a molecular orbital which can be occupied by
up to two electrons. �erefore for ethene, with � electrons, the HOMO
has k = 1, and the LUMO has k = 2. �e HOMO–LUMO energy gap is
∆E = �α + 2β cos
2π
π
� − �α + 2β cos � = (α − β) − (α + β) = −2β
3
3
�is energy gap corresponds to 61500 cm−1 , therefore β = −3.07... ×
104 cm−1 .
Butadiene has � electrons, the HOMO has k = 2 and the LUMO has k = 3;
the HOMO–LUMO energy gap is given by
∆E = �α + 2β cos
3π
2π
� − �α + 2β cos � = −1.23β
5
5
∆E = �α + 2β cos
4π
3π
� − �α + 2β cos � = −0.890β
7
7
∆E = �α + 2β cos
5π
4π
� − �α + 2β cos � = −0.695β
9
9
∆E corresponds to 46080 cm−1 , therefore β = −3.72... × 104 cm−1 .
Hexatriene has � electrons, the HOMO has k = 3 and the LUMO has
k = 4; the HOMO–LUMO energy gap is given by
∆E corresponds to 39750 cm−1 , therefore β = −4.46... × 104 cm−1 .
Octatetraene has � electrons, the HOMO has k = 4 and the LUMO has
k = 5; the HOMO–LUMO energy gap is given by
∆E corresponds to 32900 cm−1 , therefore β = −4.73... × 104 cm−1 .
�e average value of β is −4.00... × 104 cm−1 , which in electronvolts is
(−4.00... × 104 cm−1 )×(6.6261 × 10−34 J s)×(2.9979 × 1010 cm s−1 )
(1.6022 × 10−19 J eV−1 )
= −4.96 eV
β=
337
338
9 MOLECULAR STRUCTURE
(b) Octatetraene has � π electrons, therefore the orbitals with quantum numbers k = 1, �, � and � are all fully occupied. Hence the π-electron binding
energy is
π
2π
3π
E π = 2 �α + 2β cos � + 2 �α + 2β cos � + 2 �α + 2β cos �
9
9
9
4π
+ 2 �α + 2β cos � = 8α + 9.52β
9
�erefore the delocalization energy is Edeloc = 8α + 9.52β − 8(α + β) =
1.52β .
(c) �e energies and the orbital coe�cients are calculated according to the
given formulae and presented in the following tabe; k is the index for the
molecular orbital.
k
�
�
�
�
�
�
Ek
α + 1.80β
α + 1.25β
α + 0.445β
α − 0.445β
α − 1.25β
α − 1.80β
c k,1
�.���
�.���
�.���
�.���
�.���
�.���
c k,2
�.���
�.���
�.���
−0.232
−0.521
−0.418
c k,3
�.���
�.���
−0.418
−0.418
�.���
�.���
c k ,4
�.���
−0.232
−0.418
�.���
�.���
−0.521
c k,5
�.���
−0.521
�.���
�.���
−0.521
�.���
c k,6
�.���
−0.418
�.���
−0.521
�.���
−0.232
For the lowest energy molecular orbital (k = 1) all the coe�cients of the
atomic orbitals are positive, therefore this molecular orbital is bonding
between all pairs of carbon atoms. As the energy of the molecular orbitals
increase, the number of nodes increases as indicated by the number of
sign changes of the coe�cients of neighbouring atomic orbitals. In the
highest energy molecular orbital (k = 6) the sign of the neighbouring
coe�cients alternates, hence it is a fully antibonding molecular orbital.
Integrated activities
I�.�
(a) �e calculated and the measured values of the standard enthalpy of formation (∆ f H −○ �kJ mol−1 ) of ethene, butadiene, hexatriene and octatetraene
are shown in the table below, together with the relative error in the calculated values.
molecule
C� H�
C� H�
C� H�
C� H��
computed
��.��
���.�
���.�
���.�
experimental
��.�����
108.8 ± 0.79
168 ± 3
���.�
% error
��.�
��.�
��.�
��.�
�e experimental values are taken from webbook.nist.gov/chemistry/ and
book �ermodynamic Data of Organic Compounds by Pedley, Naylor and
Kirby.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(b) �e % errors shown in the table above are calculated using the expression
% error =
�∆ f H −○ (calc) − ∆ f H −○ (expt)�
∆ f H −○ (expt)
(c) For all of the molecules, the computed enthalpies of formation deviate
from the experimental values by much more than the uncertainty in the
experimental value. It is clear that molecular modeling so�ware is not a
substitute for experimentation when it comes to quantitative measures.
(a) �e energies of the LUMOs of the given molecules are calculated using the
DF/B�LYP/�-��G* method. �e results along with the standard reduction
potentials are listed in the table below.
molecule
A
B
C
D
E
ELUMO �eV
−3.54
−3.39
−3.24
−3.11
−3.01
E −○ �V
�.���
�.���
−0.067
−0.165
−0.260
�e plot of ELUMO against E −○ is shown in Fig. �.�
−3.0
−3.2
ELUMO �eV
I�.�
−3.4
Figure 9.9
−3.6
−0.3
−0.2
−0.1
E −○ �V
0.0
0.1
�e data points are a moderate �t to a straight line, the equation of which
is
ELUMO �eV = (−1.53) × E −○ �V − 3.38
(b) �e energy of the LUMO of this molecule is calculated using the same
method as above as −2.99 eV. Hence the predicted reduction potential is
E −○ �V =
ELUMO �eV + (3.38...) (−2.99) + (3.38...)
=
= −0.25
(−1.53...)
(−1.53...)
339
340
9 MOLECULAR STRUCTURE
(c) �e energy of the LUMO of the given molecule is calculated as −3.11 eV.
Hence the predicted reduction potential is
E −○ �V =
ELUMO �eV + (3.38...)
(−3.11) + (3.38...)
V=
= −0.18
(−1.53...)
(−1.53...)
Plastoquinone has less negative reduction potential, therefore it is the
better oxidizing agent.
I�.�
∗
�e energy of a normalized trial wavefunction Ψtrial is E = ∫ Ψtrial
ĤΨtrial dτ.
�e hamiltonian operator for the hydrogen atom can be inferred from [�A.�–
���] as
ħ2
e2
Ĥ = − ∇2 −
2µ
4πε 0 r
�e Laplacian operator ∇2 is given in Section �F.�(a) on page ��� as
∇2 =
1 ∂2
1
r + 2 Λ2
r ∂r 2
r
but an entirely equivalent form, which is more convenient here, is
∇2 =
∂2
2 ∂
1
+
+ 2 Λ2
2
∂r
r ∂r r
�e legendrian operator Λ 2 contains derivatives with respect to angles only. As
the given trial wavefunction is independent of angles, Λ 2 Ψ = 0 and therefore
the laplacian operating on the trial wavefunction gives
∇2 Ψtrial =
2
2
2
2
∂2
2 ∂
�Ne−αr � +
�Ne−αr � + 0 = 4N α 2 r 2 e−αr − 6N αe−αr
2
∂r
r ∂r
∗
�e wavefunction Ψtrial is real, therefore Ψtrial
= Ψtrial . �erefore the hamiltonian operating on the trial wavefunction gives
ĤΨtrial = −
2
2
N αħ 2
Ne 2 −αr 2
�2αr 2 e−αr − 3e−αr � −
e
µ
4πε 0 r
Hence the energy of the trial wavefunction is given by
∗
E = � Ψtrial
ĤΨtrial dτ
=�
0
∞
−
2
2
N 2 αħ 2
N 2 e 2 −2αr 2
�2αr 2 e−2αr − 3e−2αr � −
e
dτ
µ
4πε 0 r
�e volume element dτ in polar coordinates is r 2 sin θ dθ d� dr. �ere is no
angular dependence in the integrand, hence integrating over all angles gives
4π. �us the integral becomes
E = 4πN 2 �
0
∞
−
2
2
αħ 2
e 2 r −2αr 2
�2αr 4 e−2αr − 3r 2 e−2αr � −
e
dr
µ
4πε 0
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e integral is best evaluated term by term. To evaluate the �rst term, Integral
G.� is used from the Resource section to give
�
0
∞
−
2α 2 ħ 2 4 −2αr 2
3ħ 2 π 1�2
r e
dr = −
� �
µ
16µ 2α
Using Integral G.�, the second term gives
�
∞ 3αħ 2
µ
0
r 2 e−2αr dr =
2
�e last term is evaluated using Integral G.�
�
0
∞
−
3ħ 2 π 1�2
� �
8µ 2α
e 2 r −2αr 2
e2
e
dr = −
4πε 0
16πεα
�erefore the energy is given by
E = N2 �
3ħ 2 π π 1�2
e2
� � −
�
4µ 2α
4ε 0 α
�e value of N 2 is found using the normalization condition given by [�B.�c–
∗
���], ∫ Ψtrial
Ψtrial dτ = 1. Again, note that in polar coordinates the volume
element dτ is given by r 2 sin θ dθ d� dr, and because the wavefunction is spherically symmetric, integration over all angles gives 4π. �erefore the normalization condition becomes
4πN 2 �
0
∞
r 2 e−2αr dr = 1
2
�e integral is evaluated using Integral G.� to give
1 π 1�2
4πN 2 � 3 � = 1
4 8α
and therefore
N2 =
2α 2α 1�2
� �
π π
Hence the energy corresponding to the trial wavefunction is
E=
3ħ 2 α
e2
− 1�2 3�2 α 1�2
2µ
2 π ε0
According to the variation principle the minimum energy is obtained by taking
the derivative of the trial energy with respect to the adjustable parameter, which
is in this case α. Setting the derivative equal to zero and then solving the
equation for α gives the value of α which minimises the energy of the trial
wavefunction.
dE 3ħ 2
e2
1
=
− 3�2 3�2
=0
dα
2µ 2 π ε 0 α 1�2
Hence α is given by
µ2 e 4
α=
18ħ 4 π 3 ε 20
Substituting this back into the energy expression yields the minimum energy
for this trial wavefunction.
E=
µe 4
µe 4
µe 4
−
=
−
12π 3 ħ 2 ε 20 6π 3 ħ 2 ε 20
12π 3 ħ 2 ε 20
341
342
9 MOLECULAR STRUCTURE
I�.�
(a) All the coe�cients of the atomic orbitals for the molecular orbital ψ 1 are
positive, therefore it is a fully bonding molecular orbital. �e molecular
orbital ψ 3 has alternating positive, negative and then positive coe�cients,
therefore it is a fully antibonding molecular orbital. �e molecular orbital
ψ 2 is non-bonding, with no interaction between atomic orbitals on adjacent atoms.
(b) �e oxygen, carbon and nitrogen atoms all contribute one p orbital (the
�pz atomic orbital) to the π system, therefore they must be sp2 hybridized.
All the σ bonds to an sp2 hybridized atom must lie in the same plane,
therefore all the atoms connected to the oxygen, carbon and nitrogen
atoms must lie in the same plane together with the oxygen, carbon and
the nitrogen atoms.
(c) �e molecular orbital energy level diagram is shown in part (a). �ere
are four electrons in the π system of the peptide link. Two electrons come
from the formal π bond between the oxygen and the carbon atoms, and
two from the formal lone pair on the nitrogen atom. Hence the two lowest energy molecular orbitals, that is the bonding and the non-bonding
molecular orbitals, are �lled.
(d) �is picture of bonding in the peptide group does not invoke delocalization. �e lowest energy, bonding molecular orbital, ψ 4 , represents a
localized π bond between the oxygen and the carbon atoms. According
to this representation, the next highest energy molecular orbital, ψ 5 , is
the lone pair on nitrogen, a non-bonding molecular orbital. �e highest energy molecular orbital, ψ 6 is the antibonding π∗ molecular orbital
between the oxygen and the carbon atoms. �e four electrons in the π
system occupy the two lowest energy molecular orbitals.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E
O
O
O
C
N
ψ6
C
N
ψ5
C
N
ψ4
(e) Assume that the oxygen, carbon and nitrogen atoms are sp2 hybridized.
�e p orbital on nitrogen is in the plane in which the oxygen, carbon and
their neighbouring atoms are, therefore the two other atoms connected
to nitrogen must lie in a plane perpendicular to this one. �erefore the
oxygen, carbon, nitrogen and the neighbouring atoms cannot all lie in the
same plane.
(f) �e bonding molecular orbital ψ 1 must be lower in energy than ψ 4 , because ψ 1 extends over three atoms and is bonding between two pairs of
atoms, whereas ψ 4 extends over two atoms only and is bonding between
one pair of atoms only. Following the same logic, ψ 3 must be higher in
energy than ψ 6 . Disregarding the small di�erence in the Coulomb integral of nitrogen and oxygen, the nonbonding orbitals have approximately
the same energies.
(g) Only the two lowest energy molecular orbitals are occupied in either case.
Occupation of the non-bonding orbital does not result in any di�erence
in the π-electron binding energies. �e bonding molecular orbital in the
case of the planar peptide link is lower in energy, hence occupation of
this orbital results in greater π-electron binding energy when compared
with the occupation of the bonding molecular orbital in the non-planar
peptide link. �erefore planar geometry is the lower energy conformation
of the peptide link.
343
Molecular symmetry
10
10A
Shape and symmetry
Answers to discussion questions
D��A.�
A molecule may be chiral (optically active) only if it does not possess an axis of
improper rotation, S n .
D��A.�
symmetry operation
symmetry element
identity, E
the entire object
n-fold rotation
n-fold axis of symmetry, C n
re�ection
mirror plane, σ
inversion
centre of symmetry, i
n-fold improper rotation
n-fold improper rotation axis, S n
Solutions to exercises
E��A.�(a)
�e molecules to be assigned are shown in Fig. ��.�. For clarity not all symmetry
elements are shown.
C3,S3
σv
C2,σv
σv
N
O
O
NO� C 2v
C2
F
σh
F
P
PF� D 3h
H
C2',σv
F
F
F
C3
σv
σh
Cl
Cl
Cl
CHCl� C 3v
F
C2,i
F
C2',σv
�,�-di�uorobenzene D 2h
Figure 10.1
In each case the point group is identi�ed using the �ow diagram in Fig. ��A.� on
page ��� having identi�ed the symmetry elements, or by drawing an analogy
with one of the shapes in the summary table in Fig. ��A.� on page ���.
346
10 MOLECULAR SYMMETRY
(i) Nitrogen dioxide, NO� , is a V-shaped molecule shape with a bond angle
of approximately 134○ . It possesses
• the identity, E
• a C 2 axis that lies in the plane of the molecule and passes through
the nitrogen atom.
• two di�erent vertical mirror planes σv , both containing the C 2 axis
but with one lying in the plane of the molecule and the other perpendicular to it.
�e point group is C 2v .
(ii) PF� has a trigonal bipyramidal shape with two axial and three equatorial
�uorine atoms. It possesses
• the identity, E
• a C 3 axis passing through the phosphorus and the two axial chlorine
atoms
• a horizontal mirror plane σh perpendicular to the C 3 axis and containing the phosphorus and the three equatorial �uorine atoms
• three C 2′ axes perpendicular to the C 3 axis, each passing through
the phosphorus and one of the equatorial �uorine atoms. Only one
of the C 2′ axes is shown in Fig. ��.�.
• a S 3 axis coincident with the C 3 axis
• three vertical mirror planes σv , each containing the phosphorus, the
two axial �uorine atoms, and one of the three equatorial �uorine
atoms. Only one of the σv mirror planes is shown in Fig. ��.�.
�e point group is D 3h
(iii) CHCl� , chloroform, possesses
• the identity E
• a C 3 axis passing through the hydrogen and the carbon atom
• three vertical mirror planes σv , each containing the hydrogen, carbon, and one of the chlorine atoms; only one of these is shown in
Fig. ��.�.
�e point group is C 3v
(iv) �,�-di�uorobenzene possesses
• the identity, E
• a C 2 axis perpendicular to the plane of the molecule and passing
through the centre of the benzene ring
• two di�erent C 2′ axes, both perpendicular to the C 2 axis and lying
in the plane of the molecule. One C 2′ axis passes through the two
�uorine atoms, while the other perpendicular to this.
• a horizontal mirror plane σh lying in the plane of the molecule
• two di�erent vertical mirror planes σv , both containing the C 2 axis
and perpendicular to the plane of the molecule. One contains the two
�uorine atoms while the other is perpendicular to the line joining the
two �uorine atoms.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
• a centre of inversion i , which lies at the centre of the benzene ring
�e point group is D 2h
E��A.�(a) �e cis and trans isomers are shown in Fig. ��.�.
C2,σv
σh
σv
Cl
Cl
Cl
cis isomer C 2v
Cl
C2
trans isomer C 2h
Figure 10.2
(i) cis-dichloroethene possesses a C 2 axis lying in the plane of the molecule
and bisecting the C=C double bond. It also has two σv mirror planes
containing this axis, one lying in the plane of the molecule and the other
perpendicular to it. �e �ow diagram in Fig. ��A.� on page ��� is then
used to establish that the point group is C 2v .
(ii) trans-dichloroethene possesses a C 2 axis that is perpendicular to the plane
of the molecule and passes through the centre of the C=C double bond. It
also has a horizontal mirror plane σh perpendicular to this axis and lying
in the plane of the molecule, and also has a centre of inversion i. �e �ow
diagram in Fig. ��A.� on page ��� is then used to establish that the point
group is C 2h .
E��A.�(a) �e molecules are shown in Fig. ��.� along with their point groups. For clarity
not all symmetry elements are shown.
C2
C2,σv
σv
σ
N
H
O
C2
N
O
Pyridine, C 2v Nitroethane, C s
H
i
H
C∞
BeH� , D∞h (Be not
shown for clarity)
H
H
C2
B
i B
H
H
H
B� H� , D 2h
Figure 10.3
As explained in Section ��A.�(a) on page ���, only molecules belonging to the
groups C n , C nv , and C s may be polar.
(i) Pyridine has point group C 2v , so it may be polar . �e dipole must lie
along the C 2 axis, which passes through the nitrogen and the carbon atom
opposite it in the ring.
347
348
10 MOLECULAR SYMMETRY
(ii) Nitroethane, CH� CH� NO� , belongs to point group C s , so it may be polar .
(iii) Linear BeH� belongs to point group D∞h , so it may not be polar.
(iv) Diborane, B� H� , belongs to point group D 2h , so it may not be polar.
E��A.�(a) �ere are �� distinct isomers of dichloronaphthalene, shown in Fig. ��.� together with their point groups. All isomers have a mirror plane in the plane of
the paper; additional symmetry elements are marked.
Cl
Cl
Cl 2 1
8
7
3
Cl
Cl
C2
C2, σv
6
4
5
Cl
Cl
�,� isomer C s
�,� isomer C s
Cl
�,� isomer C 2v
Cl
Cl
�,� isomer C 2h
Cl
Cl
Cl
Cl
C2, σv
Cl
�,� isomer C s
�,� isomer C s
Cl
�,� isomer C 2v
C2, σv
Cl
�,� isomer C 2v
Cl
Cl
C2
Cl
�,� isomer C 2h
C2, σv
�,� isomer C 2v
Figure 10.4
E��A.�(a) As explained in Section ��A.�(b) on page ���, to be chiral a molecule must not
possess an axis of improper rotation, S n . �is includes mirror planes, which
are the same as S 1 , and centres of inversion, which are the same as S 2 . �e
table in Section ��A.�(c) on page ��� shows that a molecule belonging to D 2h
possesses three mirror planes, so such a molecule may not be chiral . Similarly
the table in Section ��A.�(b) on page ��� shows that a molecule belonging to
C 3h possesses a σh mirror plane, so such a molecule may not be chiral .
E��A.�(a) From the table in Section ��A.�(b) on page ��� a molecule belonging to the
point group C 3v , such as chloromethane, possesses
• the identity E
• a C 3 axis passing through the chlorine and carbon atoms
• three vertical mirror planes σv , each containing the chlorine, carbon and
one of the hydrogen atoms
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
C3
σv
Cl
H
H
H
Figure 10.5
�e C 3 axis and one of the σv mirror planes is shown in Fig. ��.�
E��A.�(a) �e point group of the naphthalene molecule is identi�ed by following the �ow
diagram in Fig. ��A.� on page ���. Firstly, naphthalene is not linear, which
leads to the question “Two or more C n , n > 2?” Naphthalene has three two-fold
axes of rotation but no higher order axes, so the answer to this question is ‘No’.
However the fact that it does have three C 2 axes means that the answer to the
next question “C n ?” is ‘Yes’.
�is leads to the question “Select C n with the highest n; then, are there nC 2 perpendicular to C n ?” As already identi�ed there are three mutually perpendicular
C 2 axes and therefore no one axis with highest n, that is, no principal axis. As
explained in Section ��A.� on page ��� in the case of a planar molecule such as
naphthalene with more than one axis competing for the title of principal axis,
it is common to choose the one perpendicular to the plane of the molecule.
However, because all three axes are mutually perpendicular, whichever of the
three is selected it still has two other C 2 axes perpendicular to it, so the answer
to this question is necessarily ‘Yes’.
�is leads to the question “σh ?” to which the answer is ‘Yes’ because, whichever
C 2 axis is selected, there is a mirror plane perpendicular to it. In the case of the
C 2 axis perpendicular to the plane of the molecule, the σh mirror plane lies in
the plane of the molecule. Answering ‘Yes’ to this question leads to the result
D nh , and because the C n axis with highest n in this molecule is C 2 , it follows
that the point group is D 2h .
Alternatively, the point group may be identi�ed from the table in Fig. ��A.� on
page ��� by drawing an analogy between the planar naphthalene molecule and
the rectangle which belongs to D 2h .
From the table in Section ��A.�(c) on page ��� a molecule belonging to the
point group D 2h possesses
• the identity E
• a C 2 axis , which in the case of a planar molecule such as naphthalene is
commonly taken to be the axis perpendicular to the plane of the molecule.
�is axis passes through the mid-point of the central bond joining the two
rings.
• two C 2′ axes , perpendicular to the C 2 axis and lying in the plane of the
molecule. One of these passes along the line of the central bond joining
349
350
10 MOLECULAR SYMMETRY
the two rings, while the other bisects this bond.
• a horizontal mirror plane σh in the plane of the molecule
In addition
• �e presence of the C 2 axis and the two C 2′ axes jointly imply the presence
of two vertical mirror planes σv , both containing the principal axis. One
of these σv planes also contains the central bond joining the two rings,
while the other bisects this bond.
• �e C 2 and σh elements jointly imply a centre of inversion i , which lies
at the midpoint of the central bond.
�ese symmetry elements are shown in Fig. ��.�.
C2',σv
σh
C2,i
C2',σv
Figure 10.6
E��A.�(a) �e objects to be assigned are shown in Fig. ��.�. For clarity not all symmetry
elements are shown.
C2,σv
C2,σv
C3,S3
σv
C2,σv
C2,σv
i
C∞
Figure 10.7
(i) As explained in Section ��A.�(f) on page ��� a sphere possesses an in�nite
number of rotation axes with all possible values of n, and belongs to the
full rotation group R 3 .
(ii) �e point group of an isosceles triangle is identi�ed using the �ow diagram in Fig. ��A.� on page ���. It has a C 2 axis lying in the plane of the
paper on which the shape is drawn which bisects the non-equal side and
passes through the vertex opposite this side. �ere are no other C n axes,
and there is not a σh mirror plane perpendicular to the C 2 axis. �ere
are, however, two σv mirror planes which contain the C 2 axis, one in the
plane of the paper and the other perpendicular to it. �ese considerations
establish the point group as C 2v .
(iii) �e point group of an equilateral triangle is readily identi�ed as D 3h by
reference to the table of shapes in Fig. ��A.� on page ���.
351
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(iv) Modelling the unsharpened cylindrical pencil as a cylinder (Fig. ��.�) and
assuming no lettering or other pattern on it, its point group is determined
using the �ow diagram in Fig. ��A.� on page ���. It is linear, in the sense
that rotation by any angle around the long axis of the pencil is a symmetry
operation so that the pencil possesses a C∞ axis. Because the pencil has
not been sharpened, both ends are the same and this means that the pencil
possesses a centre of inversion i. Using the �ow diagram this establishes
the point group as D∞h .
Solutions to problems
P��A.�
�e molecules concerned are shown in Fig. ��.�. For clarity not every symmetry
element is shown in each diagram.
C2
(a)
H
C3,S6
H
H
H
H
i
σd H
H
σd
C3
H
H
H
H
H
C2
H
C3 H
H
H
H
H
σd
σd
C2
(b)
C3
i
C2
σd
C2
C3
C2
σd
σv
σd
σd
σv
C2
C2
C3,S6
H2
N
H
H
H
C2
B
C2
i B
H
C2
(d)
C2
(c)
H
H
H2N
Co
N
H2 H N
NH2
NH2
C3
C2
2
C2
(e)
C2
C2
C2
S
σd
S
S
S
S
S
S
S
σd
C2
C4
σd
σd
C4
σd
C4,S8
Figure 10.8
(a) �e staggered conformation of ethane possesses the following symmetry
352
10 MOLECULAR SYMMETRY
elements
• �e identity E
• A C 3 axis along the the C–C bond
• �ree C 2 axes perpendicular to the C 3 axis. �ese are best seen in
the Newman projection also shown in Fig. ��.�(a), that is, the view
along the C 3 axis
• �ree dihedral mirror planes σd , each containing the C 3 principal
axis and two hydrogen atoms. �ese mirror planes are ‘dihedral’
rather than ‘vertical’ because they bisect the angles between the C 2
axes. �e three σd mirror planes are seen most easily in the Newman
projection; for clarity only one σd plane is shown
• An S 6 axis coincident with the C 3 axis
• A centre of inversion i at the midpoint of the C–C bond.
Using the �ow diagram in Fig. ��A.� on page ��� the point group is D 3d .
(b) �e chair conformation of cyclohexane possesses the following symmetry
elements
• �e identity E
• A C 3 axis perpendicular to the approximate plane of the molecule
• �ree C 2 axis perpendicular to the C 3 axis, each passing through
the midpoint of two opposite C–C bonds
• �ree dihedral mirror planes σd , each containing the C 3 axis as well
as two opposite carbon atoms
• An S 6 axis coincident with the C 3 axis
• A centre of inversion i at the intersection of the C 3 and C 2 axes.
Using the �ow diagram in Fig. ��A.� on page ��� the point group is D 3d .
�e boat conformation of cyclohexane possesses
• �e identity E
• A C 2 axis passing vertically through the centre of the boat where the
mast would be
• Two vertical mirror planes σv , both containing the C 2 axis but one
also containing the bow and stern of the boat and the other perpendicular to this
Using the �ow diagram in Fig. ��A.� on page ��� the point group is C 2v .
(c) Diborane, B� H� , possesses
• �e identity E
• �ree di�erent C 2 axes , all perpendicular to each other. One passes
through the two bridging hydrogen atoms, one passes through the
two boron atoms, and the third is perpendicular to the plane de�ned
by the boron and bridging hydrogen atoms.
• �ree di�erent mirror planes σ , one perpendicular to each of the
three C 2 axes. One mirror plane contains the boron atoms and the
four terminal hydrogen atoms. �e second contains the two bridging
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
hydrogen atoms and forms the perpendicular bisector of a line joining the two boron atoms. �e third contains the boron atoms and the
two bridging hydrogen atoms. For clarity these mirror planes are not
shown in Fig. ��.�(c).
• A centre of inversion i at the intersection of the C 2 axes.
Using the �ow diagram in Fig. ��A.� on page ��� the point group is D 2h .
(d) Ignoring the detailed structure of the en ligands, [Co(en)3 ]3+ can be
drawn in a way that resembles a propeller. It possesses the following
symmetry elements
• �e identity E
• A C 3 axis corresponding to the axis of the propeller
• �ree C 2 axes , each passing through the cobalt and the centre of one
of the en ligands
Using the �ow diagram in Fig. ��A.� on page ��� the point group is D 3 .
(e) Crown-shaped S� possesses
• �e identity E
• A C 4 axis perpendicular to the approximate plane of the molecule
• Four C 2 axes perpendicular to the C 4 axis, analogous to the C 2 axes
in the chair conformation of cyclohexane
• Four dihedral mirror planes σd , again analogous to the σd planes in
the chair conformation of cyclohexane
• An S 8 axis coincident with the C 4 axis
Using the �ow diagram in Fig. ��A.� on page ��� the point group is D 4d .
(i) As explained in Section ��A.�(a) on page ���, only molecules belonging to
the groups C n , C nv , or C s may have a permanent electric dipole moment.
�erefore of the molecules considered only boat cyclohexane (C 2v ) can
be polar.
P��A.�
(ii) As explained in Section ��A.�(b) on page ���, a molecule may be chiral only if it does not possess an axis of improper rotation S n . �is includes mirror planes, which are equivalent to S 1 , and a centre of inversion,
which is equivalent to S 2 . �erefore of the molecules considered only
[Co(en)3 ]3+ is chiral.
�e molecules are shown in Fig. ��.� along with some of their key symmetry
elements.
(a) In addition to the identity element, ethene possesses three C 2 axes, three
mirror planes, and a centre of inversion i. As explained in Section ��A.�
on page ���, in the case of a planar molecule with several C n axes competing for the title of principal axis it is common to choose the axis perpendicular to plane of the molecule to be the principal axis. �is axis is
labelled C 2 in Fig. ��.�, while the other C 2 axes are labelled C 2′ . �e mirror
plane that lies in the plane of the molecule is denoted σh while the other
353
354
10 MOLECULAR SYMMETRY
(a) Ethene
C2',σv
σh
H
Allene
C2,i H
H
σd
σd
C2',σv
H
H
C2,S4 H
C
C
C
H
H
C2'
H
H
C2
H σ
d
H
C2'
(b) (i) 0o
(ii) 90o
C2',σv
σd
C2'
σd
σh
C2,i
C2
C2',σv
C2,S4
σd
C2'
(iii) 45o
(iv) 60o
C2'
C2
45o
C2'
C2'
C2
60o
C2'
Figure 10.9
two are denoted σv . Using the �ow diagram in Fig. ��A.� on page ��� the
point group is established as D 2h .
In the case of allene there is a C 2 axis, coincident with an S 4 axis, that
passes through all three carbon atoms. �ere are also two other C 2 axes,
denoted C 2′ , that are perpendicular to C 2 and which pass through the
central carbon; these are most clearly seen in the Newman projection
shown in Fig. ��.�(a). Finally there are two dihedral mirror planes σd
which bisect the angle between the two C 2′ axes and which each pass
through all three carbon atoms and two of the hydrogen atoms. Using the
�ow diagram in Fig. ��A.� on page ��� these symmetry elements establish
the point group as D 2d .
(b) (i) �e biphenyl molecule with a dihedral angle of 0○ , that is, with both
phenyl groups in the same plane, has a shape analogous to that of the
ethene molecule from part (a). Like ethene it belongs to the point
group D 2h and possesses the symmetry elements E, C 2 , 2C 2′ , 2σv ,
σh and i.
(ii) �e biphenyl molecule with a dihedral angle of 90○ , that is, with the
two rings perpendicular, has a shape analogous to that of allene. Like
allene it belongs to the point group D 2d and possesses the symmetry
elements E, C 2 , 2C 2′ , 2σd , and S 4 .
(iii) If the dihedral angle is changed from 90○ to 45○ the C 2 and C 2′ axes
are retained but the σd mirror planes and the S 4 axis are no longer
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
present. Using the �ow diagram in Fig. ��A.� on page ��� the symmetry elements establish the point group as D 2 .
(iv) �e biphenyl molecule with a dihedral angle of 60○ possesses the
same set of symmetry elements as when the dihedral angle is 45○
and therefore belongs to the same point group of D 2 .
P��A.�
�e ion is shown in Fig. ��.��. In each diagram only the mirror plane referred
to in the question is shown; any other mirror planes are omitted for clarity.
C2
C2
σ
σh
F3C
i
CN
CF3
C2
σv
F
F
i
CN
F
F
F
C2
F
CN
NC
C2
Structureless CF� D 2h
F
F
F
Staggered CF� C 2h
F
F
F
Eclipsed CF� C 2v
Figure 10.10
(a) If the CF� groups are treated as structureless ligands then in addition to
the identity E the ion possesses three C 2 axes, a centre of inversion, and
three mirror planes, only one of which is shown in Fig. ��.��. �e point
group is D 2h .
(b) Fig. ��.�� shows the staggered and eclipsed conformations of the CF� groups.
In the staggered arrangement the centre of inversion i is retained, as is the
C 2 axis that lies along the CN–Ag–CN bond and the mirror plane perpendicular to this, but the other C 2 axes and mirror planes are lost. �e point
group is C 2h . In the eclipsed arrangement, the C 2 axis perpendicular to
the plane of the Ag and the four ligands is retained, as are the two mirror
planes containing this axis, only one of which is shown in Fig. ��.��. �e
other C 2 axes, the other mirror plane, and the centre of inversion are lost.
�e point group is C 2v .
10B Group theory
Answer to discussion questions
D��B.�
�e top row of the table lists the operations of the group; operations in the
same class are grouped together. Subsequent rows list the characters of each
of the irreducible representations (symmetry species), the symbols for which
are shown if the far le� column. Operations in the same class have the same
character, which is why they can conveniently be grouped together.
One-dimensional irreducible representations are labelled A or B, and have character � under the operation E. Two- and three-dimensional irreducible repre-
355
356
10 MOLECULAR SYMMETRY
sentations are labelled E and T, respectively, and have characters � and � under
the operation E.
On the right of the table various cartesian functions are listed on the row corresponding to the irreducible representation as which they transform. Where
a pair (or more) of functions transform as a two (or higher) dimensional representation, the functions are bracketed together.
�e order of the group, h, is equal to the number of operations. Alternatively,
as the operations are grouped into classes, the order is the sum of the number
of operations belonging to each class.
D��B.�
�e number of irreducible representations equals the number of classes. Also,
apart from the groups C n with n > 2, the sum of the squares of the dimensions
of each irreducible representation is equal to the order.
�e letters and subscripts of a symmetry species provide information about
the symmetry behaviour of the species. An A or a B is used to denote a onedimensional representation; A is used if the character under the principal rotation is +1 (symmetric behaviour), and B is used if the character is −1 (antisymmetric behaviour). Subscripts are used to distinguish the irreducible representations if there is more than one of the same type: A1 is reserved for the
representation with the character � under all operations.
E denotes a two-dimensional irreducible representation, and T a three dimensional irreducible representation. For groups with an inversion centre, a subscript g (gerade) indicates symmetric behaviour under the inversion operation;
a subscript u (ungerade) indicates antisymmetric behaviour. If a horizontal
mirror plane is present, ′ or ′′ superscripts are added if the behaviour is symmetric or antisymmetric, respectively, under the σ h operation.
D��B.�
A representative is a mathematical operator, usually taking the form of a matrix,
that represents the e�ect of a particular symmetry operation. It is constructed
by considering the e�ect of the operation on a particular set of basis functions.
�e set of all these mathematical operators corresponding to all the operations
of the group is called a representation.
Solutions to exercises
E��B.�(a)
BF� belongs to the point group D 3h . �e σh operation corresponds to a re�ection in the plane of the molecule, while the C 3 operation corresponds to
rotation by 120○ around the C 3 axis which passes through the boron atom and
is perpendicular to the plane of the molecule (Fig. ��.��).
Using the orbital numbering shown in Fig. ��.��, the matrix representatives
for the σh and C 3 operations were found in Exercise E��B.�(a) and Exercise
E��B.�(b) to be
� −1 0 0 0 �
� 0 −1 0 0 �
�
D(σh ) = �
� 0 0 −1 0 �
� 0 0 0 −1 �
and
�1 0 0 0�
�0 0 0 1�
�
D(C 3 ) = �
�0 1 0 0�
�0 0 1 0�
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
z
C3
p1
p3
p2
σh
p4
Figure 10.11
�e matrix representative of the operation σh C 3 is found by multiplying the
matrix representatives of σh and C 3 . Basic information about how to handle
matrices is given in �e chemist’s toolkit �� in Topic �E on page ���.
� −1 0 0 0 � � 1 0 0 0 � � −1 0 0 0 �
� 0 −1 0 0 � � 0 0 0 1 � � 0 0 0 −1 �
��
�= �
�
D(σh )D(C 3 ) = �
� 0 0 −1 0 � � 0 1 0 0 � � 0 −1 0 0 �
� 0 0 0 −1 � � 0 0 1 0 � � 0 0 −1 0 �
�e operation corresponding to this representative is found by considering its
e�ect on the starting basis
� −1 0 0 0 �
� 0 0 0 −1 �
� = ( −p1 −p3 −p4 −p2 )
( p1 p2 p3 p4 ) �
� 0 −1 0 0 �
� 0 0 −1 0 �
�e operation σh C 3 therefore changes the sign of p1 , and converts p2 , p3 and
p4 into −p3 , −p4 , and −p2 respectively. �is is precisely the same outcome as
achieved by the S 3 operation , that is, a ���○ rotation around the C 3 axis followed by a re�ection in the σh plane. �us, D(σh )D(C 3 ) = D(S 3 ), as expected
from the fact that by de�nition the S 3 operation corresponds to a C 3 rotation
followed by a re�ection in the plane perpendicular to the C 3 axis.
E��B.�(a) Fig. ��.�� shows BF� , an example of a molecule with D 3h symmetry. �e three
C 2 axes are labelled C 2 , C 2′ and C 2′′ , and likewise for the σv mirror planes.
C2',σv'
C2',σv'
C2',σv'
4
1
F
2
F
C2'',σv''
F
3
F
3
4
3
F
C2,σv
σv′′−1 C 2 σv′′ = C 2′
Figure 10.12
F
2
1
C2'',σv''
F
F
4
C2,σv
σv−1 C 2′ σv = C 2′′
C2'',σv''
F
2
1
C2,σv
σv′−1 C 2′′ σv′ = C 2
�e criteria for two operations R and R ′ to be in the same class is given by
[��B.�–���], R′ = S −1 RS where S is another operation in the group. �e task is
357
358
10 MOLECULAR SYMMETRY
therefore to �nd an operation S in D 3h such that this equation is satis�ed when
R and R ′ are two of the C 2 axes.
Referring to the �rst diagram in Fig. ��.��, to show that C 2 and C 2′ are in the
same class consider the operation σv′′−1 C 2 σv′′ . Start at the arbitrary point �, and
recall that the operations are applied starting from the right. �e operation
σv′′ moves the point to �, and then C 2 moves the point to �. �e inverse of a
re�ection is itself, σv′′−1 = σv′′ , so the e�ect of σv′′−1 is to move the point to �.
From the diagram it can be seen that � can be reached by applying C 2′ to point
�, thus demonstrating that σv′′−1 C 2 σv′′ = C 2′ and hence that C 2 and C 2′ belong to
the same class.
In a similar way the second diagram in Fig. ��.�� shows that σv−1 C 2′ σv = C 2′′ and
hence that C 2′ and C 2′′ belong to the same class, while the third diagram shows
that C 2′′ and C 2 belong to the same class.
E��B.�(a) �e orthonormality of irreducible representations is de�ned by [��B.�–���],
1
0 for i ≠ j
Γ (i)
Γ ( j)
� N(C)χ (C)χ (C) = �
1 for i = j
h C
where the sum is over all classes of the group, N(C) is the number of operations
(i)
in class C, and h is the number of operations in the group (its order). χ Γ (C)
(i)
is the character of class C in irreducible representation Γ , and similarly for
( j)
χ Γ (C).
�e character table for the point group C 2h is available in the Online resource
centre. �e operations of the group are {E, C 2 , i, σh }, so h = 4. �ere are four
classes and in this group each class has just one member, so all N(C) = 1. �e
irreducible representations have the following characters
Ag {�, �, �, �}
Au {�, �, −1, −1}
Bg {�, −1, �, −1}
Bu {�, −1, −1, �}
To prove orthogonality, [��B.�–���] is used with each pair of irreducible representations; the result is zero in each case which shows that each pair are
orthogonal
Ag and Bg
Ag and Au
Ag and Bu
Bg and Au
Bg and Bu
Au and Bu
1
{1×1×1 + 1×1×(−1) + 1×1×1 + 1×1×(−1)} = 0
4
1
{1×1×1 + 1×1×1 + 1×1×(−1) + 1×1×(−1)} = 0
4
1
{1×1×1 + 1×1×(−1) + 1×1×(−1) + 1×1×1} = 0
4
1
{1×1×1 + 1×(−1)×1 + 1×1×(−1) + 1×(−1)×(−1)} = 0
4
1
{1×1×1 + 1×(−1)×(−1) + 1×1×(−1) + 1×(−1)×1} = 0
4
1
{1×1×1 + 1×1×(−1) + 1×(−1)×(−1) + 1×(−1)×1} = 0
4
To prove that each irreducible representation is normalised, [��B.�–���] is used
with i = j for each irreducible representation Γ(i) ; the result is � in each case.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Ag and Ag
Bg and Bg
Au and Au
Bu and Bu
1
{1×1×1 + 1×1×1 + 1×1×1 + 1×1×1} = 1
4
1
{1×1×1 + 1×(−1)×(−1) + 1×1×1 + 1×(−1)×(−1)} = 1
4
1
{1×1×1 + 1×1×1 + 1×(−1)×(−1) + 1×(−1)×(−1)} = 1
4
1
{1×1×1 + 1×(−1)×(−1) + 1×(−1)×(−1) + 1×1×1} = 1
4
E��B.�(a) �e D 3h character table is given in the Resource section. As explained in Section ��B.�(a) on page ���, the symmetry species of s, p and d orbitals on a
central atom are indicated at the right hand side of the character table. �e
position of z in the D 3h character table shows that pz , which is proportional
to z f (r), has symmetry species A′′2 . Similarly the positions of x and y in
the character table shows that px and p y , which are proportional to x f (r)
and y f (r) respectively, jointly span the irreducible representation of symmetry
species E′ .
In the same way the positions of z 2 , x 2 − y 2 , x y, yz and zx in the character table
show that dz 2 has symmetry species A′1 , dx 2 −y 2 and dx y jointly span E′ , and
d yz and dzx jointly span E′ .
E��B.�(a) As explained in Section ��B.�(c) on page ���, the highest dimensionality of
irreducible representations in a group is equal to the maximum degree of degeneracy in the group. �e highest dimensionality is found by noting the maximum value of χ(E) in the character table. An octahedral hole in a crystal
has O h symmetry, and from the O h character table in the Resource section the
maximum value of χ(E) is three, corresponding to the T irreducible representations. �erefore the maximum degeneracy of a particle in an octahedral hole
is three .
E��B.�(a) As explained in Section ��B.�(c) on page ���, the highest dimensionality of
any irreducible representation in a group is equal to the maximum degree of
degeneracy in the group. �e highest dimensionality is found by noting the
maximum value of χ(E) in the character table. Benzene has D 6h symmetry, the
character table for which is available in the Online resource centre. �e maximum value of χ(E) is two, corresponding to the E irreducible representations.
�erefore the maximum degeneracy of the orbitals in benzene is two .
E��B.�(a) BF� belongs to the point group D 3h . �e σh operation corresponds to re�ection
in the plane of the molecule.
z
z
C3
C3
p1
p3
p2
σh
σh
σh
p4
Figure 10.13
As shown in Fig. ��.�� the e�ect of this operation is to change the sign of all
four p orbitals. �is e�ect is written as ( −p1 −p2 −p3 −p4 ) ← ( p1 p2 p3 p4 )
359
360
10 MOLECULAR SYMMETRY
which is expressed using matrix multiplication as
D(σ h )
��� � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � �
� −1 0 0 0 �
� 0 −1 0 0 �
� = ( p1 p2 p3 p4 )D(σh )
( −p1 −p2 −p3 −p4 ) = ( p1 p2 p3 p4 ) �
� 0 0 −1 0 �
� 0 0 0 −1 �
�e representative of this operation is therefore
� −1 0 0 0 �
� 0 −1 0 0 �
�
D(σh ) = �
� 0 0 −1 0 �
� 0 0 0 −1 �
Solutions to problems
P��B.�
Fig. ��.�� shows trans-di�uoroethene, an example of a molecule with C 2h symmetry.
C2
1
2
σh
i
H
F
F
H
3
Figure 10.14
�e group multiplication table is constructed by considering the e�ect of successive transformations. For example, Fig. ��.�� shows the e�ect on an arbitrary
point of a C 2 rotation followed by a σh re�ection. �e C 2 operation moves
the point from � to �, and the σh re�ection moves the point to �. �e overall
e�ect, 1 → 3, is equivalent to carrying out the i operation. �us, σh C 2 = i.
Considering all other pairs of operations in the same way gives the following
table.
R ↓ R′ →
E
C2
σh
i
P��B.�
E
E
C2
σh
i
C2
C2
E
i
σh
σh
σh
i
E
C2
i
i
σh
C2
E
Fig. ��.�� shows that a water molecule together with the speci�ed axis system
and the six basis orbitals, which are labelled sA , sB , sO , px , p y , and pz . �e σv
plane is the xz plane, and the σv′ plane is the yz plane.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
z
z
C2
O
x
HA
HB
z
sO
y
y
x
sA
x
px
pz
py y
sB
Figure 10.15
�e matrix representatives are obtained using the approach described in Section ��B.�(a) on page ���. Beginning with C 2 , this operation exchanges sA and
sB , leaves sO and pz unchanged, and changes the sign of px and of p y . Its e�ect
is written ( sB sA sO −px −p y pz ) ← ( sA sB sO px p y pz ) which is expressed
using matrix multiplication as
D(C 2 )
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
0 0 �
� 0 1 0 0
� 1 0 0 0
0 0 �
�
�
� 0 0 1 0
0 0 �
�
( sB sA sO −px −p y pz ) = ( sA sB sO px p y pz )�
� 0 0 0 −1 0 0 �
�
�
� 0 0 0 0 −1 0 �
�
�
� 0 0 0 0
0 1 �
Similarly, the σv operation exchanges sA and sB , leaves sO , px and pz unchanged,
and changes the sign of p y : ( sB sA sO px −p y pz ) ← ( sA sB sO px p y pz ) �is
is expressed using matrix multiplication as
D(σ v )
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��
� 0 1 0 0 0 0 �
� 1 0 0 0 0 0 �
�
�
� 0 0 1 0 0 0 �
�
( sB sA sO px −p y pz ) = ( sA sB sO px p y pz )�
� 0 0 0 1 0 0 �
�
�
� 0 0 0 0 −1 0 �
�
�
� 0 0 0 0 0 1 �
�e σv′ operation leaves all orbitals unchanged except px , which changes sign.
D(σ v′ )
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��
� 1 0 0 0 0 0 �
� 0 1 0 0 0 0 �
�
�
� 0 0 1 0 0 0 �
�
( sA sB sO −px p y pz ) = ( sA sB sO px p y pz )�
� 0 0 0 −1 0 0 �
�
�
� 0 0 0 0 1 0 �
�
�
� 0 0 0 0 0 1 �
361
362
10 MOLECULAR SYMMETRY
Finally, the E operation, ‘do nothing’, leaves all orbitals unchanged.
D(E)
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
� 1 0 0 0 0 0 �
� 0 1 0 0 0 0 �
�
�
� 0 0 1 0 0 0 �
�
( sA sB sO px p y pz ) = ( sA sB sO px p y pz )�
� 0 0 0 1 0 0 �
�
�
� 0 0 0 0 1 0 �
�
�
� 0 0 0 0 0 1 �
(a) �e representative of the operation C 2 σv is found by multiplying together
the matricies D(C 2 ) and D(σv ) found above. Basic information about
how to handle matrices is given in �e chemist’s toolkit �� in Topic �E on
page ���.
D(C 2 )
D(σ v′ )
D(σ v )
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � �� ��� � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � ��
� 0 1 0 0 0 0 �� 0 1 0 0 0 0 � � 1 0 0 0 0 0 �
� 1 0 0 0 0 0 �� 1 0 0 0 0 0 � � 0 1 0 0 0 0 �
�
��
� �
�
� 0 0 1 0 0 0 �� 0 0 1 0 0 0 � � 0 0 1 0 0 0 �
��
� �
�
D(C 2 )D(σv ) = �
� 0 0 0 −1 0 0 � � 0 0 0 1 0 0 � = � 0 0 0 −1 0 0 �
�
��
� �
�
� 0 0 0 0 −1 0 � � 0 0 0 0 −1 0 � � 0 0 0 0 1 0 �
�
��
� �
�
� 0 0 0 0 0 1 �� 0 0 0 0 0 1 � � 0 0 0 0 0 1 �
= D(σv′ )
�is establishes that C 2 σv = σv′ . Similarly, the representative of the operation σv σv′ is found by representative matricies of σv and σv ’; the result is
the representative of C 2 , establishing that σv σv′ = C 2 .
D(σ v′ )
D(σ v )
D(C 2 )
��� � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � �� ��� � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � �� ��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
� 0 1 0 0 0 0 �� 1 0 0 0 0 0 � � 0 1 0 0 0 0 �
� 1 0 0 0 0 0 �� 0 1 0 0 0 0 � � 1 0 0 0 0 0 �
�
��
� �
�
� 0 0 1 0 0 0 �� 0 0 1 0 0 0 � � 0 0 1 0 0 0 �
′
��
�=�
�
D(σv )D(σv ) = �
� 0 0 0 1 0 0 � � 0 0 0 −1 0 0 � � 0 0 0 −1 0 0 �
�
��
� �
�
� 0 0 0 0 −1 0 � � 0 0 0 0 1 0 � � 0 0 0 0 −1 0 �
�
��
� �
�
� 0 0 0 0 0 1 �� 0 0 0 0 0 1 � � 0 0 0 0 0 1 �
= D(C 2 )
(b) �e representation is reduced using the method in Section ��B.�(c) on
page ���. Inspection of the matrix representatives reveales that they are
all of block-diagonal format
�
�
�
� 0
D=�
� 0
�
� 0
�
� 0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 �
0 �
�
0 �
�
0 �
�
0 �
�
�
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�is implies that the symmetry operations of C 2v never mix the sO , px ,
p y and pz orbitals together, nor do they mix these orbitals with sA and sB ,
but sA and sB are mixed together by the operations of the group. Consequently the basis can be cut into �ve parts: four one-dimensional bases
for the individual orbitals on the oxygen and a two-dimensional basis for
the two hydrogen s orbitals. �e representations corresponding to the
four one-dimensional bases are
For sO :
For px :
For p y :
For pz :
D(E) = 1
D(E) = 1
D(E) = 1
D(E) = 1
D(C 2 ) = 1
D(C 2 ) = −1
D(C 2 ) = −1
D(C 2 ) = 1
D(σv ) = 1
D(σv ) = 1
D(σv ) = −1
D(σv ) = 1
D(σv′ ) = 1
D(σv′ ) = −1
D(σv′ ) = 1
D(σv′ ) = 1
�e characters of one-dimensional representatives are just the representatives themselves. �erefore, inspection of the C 2v character table from
the Resource section indicates that the one-dimensional representations
for these orbitals correspond to A1 , B1 , B2 , and A1 respectively. �e
oxygen-based orbitals therefore span 2A1 + B1 + B2 .
�e remaining two orbitals, sA and sB , are a basis for a two-dimensional
representation
D(E) =
�1 0�
�0 1�
�0 1�
�1 0�
D(C 2 ) =
D(σv ) =
D(σv′ ) =
�0 1�
�1 0�
�1 0�
�0 1�
D(E) =
�1 0�
�1 0 �
�1 0 �
�1 0�
D(C 2 ) =
D(σv ) =
D(σv′ ) =
�0 1�
� 0 −1 �
� 0 −1 �
�0 1�
where the matrices correspond to the top le�-hand corners of the 6 ×
6 matricies found above. �at this two-dimensional representation is
reducible is demonstrated by considering the linear combinations s1 =
sA + sB and s2 = sA − sB . �e C 2 and σv operations both exchange sA and
sB : ( sB sA ) ← ( sA sB ) which means that (sB + sA ) ← (sA + sB ), corresponding to (s1 ) ← (s1 ). Similarly, (sB − sA ) ← (sA − sB ), corresponding
to (−s2 ) ← (s2 ). On the other hand, the E and σv′ operations leave sA and
sB unchanged, which leads to the results (s1 ) ← (s1 ) and (s2 ) ← (s2 ).
�e representation in the basis ( s1 s2 ) is therefore
�e new representatives are all in block diagonal format, in this case in
0
the form �
�, which means that the two combinations are not mixed
0
with each other by any operation of the group. �e two dimensional basis
( s1 s2 ) can therefore be cut into two one-dimensional bases corresponding to s1 and s2 . �e representations for these bases are
For s1 : D(E) = 1 D(C 2 ) = 1 D(σv ) = 1 D(σv′ ) = 1
For s2 : D(E) = 1 D(C 2 ) = −1 D(σv ) = −1 D(σv′ ) = 1
Because the characters of one-dimensional representatives are just the
representatives themselves, inspection of the C 2v character table immediately indicates that these one-dimensional representations correspond
363
364
10 MOLECULAR SYMMETRY
P��B.�
to A1 and B2 respectively. Combining these results with those from the
oxygen orbitals found earlier, which span 2A1 + B1 + B2 , the original sixdimensional representation therefore spans 3A1 + B1 + 2B2 .
�e ethene molecule and axis system are shown in Fig. ��.��.
y
y
C2,σyz
σxy
sB H
sC
H
C2z
H s
A
H
sD
x
x
C2,σzx
Figure 10.16
�e C 2x operation exchanges sA with sD , and sB with sC :
( sD sC sB sA ) ← ( sA sB sC sD )
�is is written using matrix multiplication as
D(C 2x )
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 0 0 0 1 �
� 0 0 1 0 �
�
( sD sC sB sA ) = ( sA sB sC sD )�
� 0 1 0 0 �
� 1 0 0 0 �
�e matrix D(C 2x ) is the representative of the C 2x operation in this basis. �e
representatives of the other seven operations are obtained in the same way.
D(C 2 )
y
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 0 1 0 0 �
� 1 0 0 0 �
y
�
For C 2 : ( sB sA sD sC ) = ( sA sB sC sD )�
� 0 0 0 1 �
� 0 0 1 0 �
D(C 2z )
For C 2z :
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 0 0 1 0 �
� 0 0 0 1 �
�
( sC sD sA sB ) = ( sA sB sC sD )�
� 1 0 0 0 �
� 0 1 0 0 �
D(i)
For i:
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 0 0 1 0 �
� 0 0 0 1 �
�
( sC sD sA sB ) = ( sA sB sC sD )�
� 1 0 0 0 �
� 0 1 0 0 �
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
D(σ x y )
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 1 0 0 0 �
� 0 1 0 0 �
�
For σ x y : ( sA sB sC sD ) = ( sA sB sC sD )�
� 0 0 1 0 �
� 0 0 0 1 �
D(σ yz )
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 0 1 0 0 �
� 1 0 0 0 �
�
For σ yz : ( sB sA sD sC ) = ( sA sB sC sD )�
� 0 0 0 1 �
� 0 0 1 0 �
D(σ zx )
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 0 0 0 1 �
� 0 0 1 0 �
�
For σ zx : ( sD sC sB sA ) = ( sA sB sC sD )�
� 0 1 0 0 �
� 1 0 0 0 �
D(E)
For E:
P��B.�
P��B.�
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 0 0 1 0 �
� 0 0 0 1 �
�
( sC sD sA sB ) = ( sA sB sC sD )�
� 1 0 0 0 �
� 0 1 0 0 �
For one-dimensional representatives, the characters as just the representatives
themselves. �us for the �rst representation, D(C 3 ) = 1 and D(C 2 ) = 1, the
characters are χ(C 3 ) = 1 and χ(C 2 ) = 1. Inspection of the C 6v character table
in the Resource section shows that this combination of characters corresponds
to either A1 or A2 . Both of these irreducible representations have χ(C 6 ) = +1,
as expected from the matrix multiplication D(C 6 ) = D(C 3 )D(C 2 ) = 1 × 1 = 1
which implies that χ(C 6 ) = +1. �e character table gives the characters of
σv and σd as both +1 for A1 or both −1 for A2 , implying that D(σv ) = +1 ,
D(σd ) = +1 in the case of A1 , or D(σv ) = −1, D(σd ) = −1 in the case of A2 .
For the second representation, D(C 3 ) = 1 and D(C 2 ) = −1, the characters are
χ(C 3 ) = 1 and χ(C 2 ) = −1. Inspection of the character table shows that this
corresponds to either B1 or B2 ; as expected these both have χ(C 6 ) = −1. �e
characters of σv and σd are +1 and −1 in the case of B1 , or −1 and +1 in the
case of B2 . Hence D(σv ) = +1, D(σd ) = −1 in the case of B1 , or D(σv ) = −1 ,
D(σd ) = +1 in the case of B2 .
(a) �e C 2v character table is given in the Resource section. As explained in
the question, r 2 is invariant to all operations of the group, and furthermore the C 2v character table shows that both z and z 2 belong to the totally
symmetric symmetric representation A1 which means that they are also
invariant to all operations of the group. Because all parts of the function z(5z 2 − 3r 2 ) are invariant to all operations, it follows that the entire
function and therefore the f orbital that it represents is invariant to all
365
366
10 MOLECULAR SYMMETRY
operations. Consequently this f orbital belongs to the totally symmetric
representation A1 .
(b) �e function y(5y 2 − 3r 2 ) is considered as a product of the functions
y and (5y 2 − 3r 2 ). �e C 2v character table shows that the function y 2
belongs to the totally symmetric representation and is therefore invariant
to all operations of the group; because this is also true of r 2 it follows that
the factor (5y 2 −3r 2 ) is similarly invariant to all operations. �e character
table also shows that the function y belongs to the B2 representation, for
which the characters are +1 under E and σv′ and −1 under C 2 and σv . �is
indicates that the function y changes sign under C 2 and σv and, therefore,
because (5y 2 −3r 2 ) is invariant to all operations, the product y(5y 2 −3r 2 )
behaves in the same way as y. �e f orbital therefore belongs to the B2
representation.
(c) In the same way, the function x(5x 2 − 3r 2 ) behaves in the same way as
x, because (5x 2 − 3r 2 ) is invariant to all operations. �e character table
shows that x, and therefore this f orbital, belongs to the B1 representation.
(d) �e function z(x 2 − y 2 ) belongs to the totally symmetric A1 representation, because as shown in the character table each of z, x 2 and y 2 belongs
to A1 and are therefore invariant to all operations of the group. It follows
that the product z(x 2 − y 2 ) is also invariant to all operations and hence
belongs to A1 .
(e) �e function y(x 2 − z 2 ) behaves in the same way as y, because the factor
(x 2 − z 2 ) is invariant to all operations of the group. Hence this f orbital
belongs to B2 , the same as y.
(f) Similarly the function x(z 2 − y 2 ) behaves the same way as x and therefore
belongs to B1 .
(g) �e function x yz is considered as the product x y × z. �e character
table shows that z belongs to the totally symmetric representation A1 , so
the function x yz behaves in the same way as x y which as shown in the
character table belongs to A2 .
�e seven f orbitals therefore span 2A1 + A2 + 2B1 + 2B2 .
10C Applications of symmetry
Answers to discussion questions
D��C.�
�e key point is that only orbitals (or combinations of orbitals) which transform
as the same symmetry species (irreducible representation) can overlap to form
molecular orbitals. �e �rst step is therefore to classify the valence orbitals
according to symmetry. Usually, it is possible to identify sets of such orbitals
which are interconverted by the operations of the group and so can be considered separately from other sets. Having classi�ed a set of orbitals according to
symmetry the projection operator is then used to construct symmetry-adapted
linear combinations (SALCs) each of which transforms as a single symmetry
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
species (irreducible representation). �e molecular orbitals are formed from
the overlap of these SALCs.
Solutions to exercises
E��C.�(a)
�e D 2h character table shows that x y spans B1g in D 2h . To show this explicitly,
the direct product is formed between the irreducible representations spanned
by x and y individually. �e character table shows that x spans B3u and y spans
B2u ; the direct product B3u × B2u is calculated using the method described in
Section ��C.�(a) on page ���
y
E C 2x C 2 C 2z
i σ x y σ yz σ zx
B3u
� −1 −1
� −1
� −1
�
B2u
� −1
� −1 −1
�
� −1
product �
� −1 −1
�
� −1 −1
�e characters in the product row are those of symmetry species B1g , thus
con�rming that the function x y has symmetry species B1g in D 2h .
E��C.�(a) As explained in Section ��C.�(a) on page ���, only orbitals of the same symmetry species may have a nonzero overlap. Inspection of the C 2v character table
shows that the oxygen �s and �pz orbitals both have A1 symmetry, so they
can interact with the A1 combination of �uorine orbitals. �e oxygen �p y
orbital has B2 symmetry, so it can interact with the B2 combination of �uorine
orbitals. �e oxygen �px orbital has B1 symmetry, so cannot interact with either
combination of �uorine orbitals and therefore remains nonbonding.
In SF� the same interactions with the sulfur s and p orbitals are possible but
there is now the possibility of additional interactions involving the d orbitals.
�e C 2v character table shows that dz 2 and dx 2 −y 2 have A1 symmetry, so they
can interact with the A1 combination of �uorine orbitals. �e d yz orbital
has B2 symmetry so can interact with the B2 combination. �e dx y and dzx
orbitals have A2 and B1 symmetry respectively, so they cannot interact with
either combination of �uorine orbitals and therefore remain nonbonding.
E��C.�(a) As explained in Section ��C.�(a) on page ���, only orbitals of the same symmetry species may have a nonzero overlap. Inspection of the C 2v character table
shows that the nitrogen �s, px , p y , and pz orbitals span A1 , B1 , B2 , and A1
respectively. Because none of these orbitals have A2 symmetry, none of them
can interact with the A2 combination of oxygen orbitals.
�e character table also shows that the dz 2 , dx 2 −y 2 , dx y , d yz , and dzx orbitals
of the sulfur in SO� transform as A1 , A1 , A2 , B2 and B1 respectively. �erefore
the dx y orbital has the correct symmetry to interact with the A2 combination
of oxygen p orbitals.
E��C.�(a) As explained in Section ��C.� on page ���, a transition from a state with symmetry Γ(i) to one with symmetry Γ(f) is only allowed if the direct product
367
368
10 MOLECULAR SYMMETRY
Γ(f) × Γ(q) × Γ(i) contains the totally symmetric irreducible representation,
which in C 2v is A1 . If the ground state is A1 , then the direct product becomes
Γ(f) × Γ(q) × A1 . �is is simply Γ(f) × Γ(q) because, from the �rst simplifying
feature of direct products listed in Section ��C.�(a) on page ���, the direct
product of the totally symmetric irreducible representation A1 with any other
representation is the latter representation itself.
If Γ(f) × Γ(q) is to be A1 , then Γ(f) must equal Γ(q) because, from the second
simplifying feature of direct products listed in Section ��C.�(a) on page ���,
the direct product of two irreducible representations only contains the totally
symmetric irreducible representation if the two irreducible representations are
identical. �e C 2v character table shows that Γ(q) = B1 for x polarized light (q =
x), B2 for y polarised light, and A1 for z polarised light. It follows that x, y and
z polarised light can excite the molecule to B1 , B2 , and A1 states respectively.
E��C.�(a) �e number of times n(Γ) that a given irreducible representation Γ occurs in a
representation is given by [��C.�a–���], n(Γ) = (1�h) ∑C N(C)χ(Γ) (C)χ(C),
where h is the order of the group, N(C) is the number of operations in class C,
χ(Γ) is the character of class C in the irreducible representation Γ, and χ(C) is
the character of class C in the representation being reduced.
n(A1 ) = 18 �1× χ(A1 ) (E)× χ(E) + 1× χ(A1 ) (C 2 )× χ(C 2 ) + 2× χ(A1 ) (C 4 )× χ(C 4 )
+ 2× χ(A1 ) (σv )× χ(σv ) + 2× χ(A1 ) (σd )× χ(σd )�
= 18 �1×1×5 + 1×1×1 + 2×1×1 + 2×1×3 + 2×1×1� = 2
Similarly
n(A2 ) = 18 �1×1×5 + 1×1×1 + 2×1×1 + 2×(−1)×3 + 2×(−1)×1� = 0
n(B1 ) = 18 �1×1×5 + 1×1×1 + 2×(−1)×1 + 2×1×3 + 2×(−1)×1� = 1
n(B2 ) = 18 �1×1×5 + 1×1×1 + 2×(−1)×1 + 2×(−1)×3 + 2×1×1� = 0
n(E) = 18 �1×2×5 + 1×(−2)×1 + 2×0×1 + 2×0×3 + 2×0×1� = 1
�e representation therefore spans 2A1 + B1 + E .
E��C.�(a) �e number of times n(Γ) that a given irreducible representation Γ occurs in a
representation is given by [��C.�a–���], n(Γ) = (1�h) ∑C N(C)χ(Γ) (C)χ(C),
where h is the order of the group, N(C) is the number of operations in class C,
χ(Γ) is the character of class C in the irreducible representation Γ, and χ(C) is
the character of class C in the representation being reduced. In the case of D 4h ,
h = 16.
Because the representation being reduced has characters of zero for all classes
except E, C 2′ , σh , and σv , only these latter four classes make a non-zero contribution to the sum and therefore only these classes need be considered. �e
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
number of times that the irreducible representation A1g occurs is therefore
1
�N(E)× χ(A1g ) (E)× χ(E) + N(C 2′ )× χ(A1g ) (C 2′ )× χ(C 2′ )
n(A1 ) = 16
+ N(σh )× χ(A1g ) (σh )× χ(σh ) + N(σv )× χ(A1g ) (σv )× χ(σv )�
Similarly
1
= 16
(1×1×4 + 2×1×2 + 1×1×4 + 2×1×2) = 1
1
n(A2g ) = 16
(1×1×4 + 2×(−1)×2 + 1×1×4 + 2×(−1)×2) = 0
1
n(B1g ) = 16
(1×1×4 + 2×1×2 + 1×1×4 + 2×1×2) = 1
1
n(B2g ) = 16
(1×1×4 + 2×(−1)×2 + 1×1×4 + 2×(−1)×2) = 0
1
n(Eg ) = 16
(1×2×4 + 2×0×2 + 1×(−2)×4 + 2×0×2) = 0
1
n(A1u ) = 16
(1×1×4 + 2×1×2 + 1×(−1)×4 + 2×(−1)×2) = 0
1
n(A2u ) = 16
(1×1×4 + 2×(−1)×2 + 1×(−1)×4 + 2×1×2) = 0
1
n(B1u ) = 16
(1×1×4 + 2×1×2 + 1×(−1)×4 + 2×(−1)×2) = 0
1
n(B2u ) = 16
(1×1×4 + 2×(−1)×2 + 1×(−1)×4 + 2×1×2) = 0
1
n(Eu ) = 16
(1×2×4 + 2×0×2 + 1×0×4 + 2×1×2) = 1
�e representation therefore spans A1g + B1g + Eu .
E��C.�(a) As explained in Section ��C.� on page ���, a transition from a state with symmetry Γ(i) to one with symmetry Γ(f) is only allowed if the direct product Γ(f) ×
Γ(q) × Γ(i) contains the totally symmetric irreducible representation, which
for both molecules is A1g . �e ground state is totally symmetric, implying
that it transforms as A1g . �erefore the direct product becomes Γ(f) × Γ(q) ×
A1g . �is is simply Γ(f) × Γ(q) because, from the �rst simplifying feature of
direct products listed in Section ��C.�(a) on page ���, the direct product of the
totally symmetric representation A1g with any other representation is the latter
representation itself.
If Γ(f) × Γ(q) is to be A1g , then Γ(f) must equal Γ(q) because, from the second
simplifying feature of direct products listed in Section ��C.�(a) on page ���,
the direct product of two irreducible representations only contains the totally
symmetric irreducible representation if the two irreducible representations are
identical.
(i) Benzene belongs to point group D 6h . �e D 6h character table in the Online resource centre shows that shows that z transforms as A2u , and x and
y together transform as E1u . �erefore light polarized along z can excite
benzene to an A2u state, and x or y polarised light can excite it to an E1u
state.
(ii) Naphthalene belongs to point group D 2h . �e D 2h character table in the
Resource section shows that x transforms as B3u , y transforms as B2u , and z
transforms as B1u . �erefore naphthalene can be excited to B3u , B2u , or B1u
states by x, y, and z polarised light respectively.
369
370
10 MOLECULAR SYMMETRY
E��C.�(a) As explained in Section ��C.� on page ��� an integral can only be non-zero if
the integrand spans the totally symmetric irreducible representation, which in
C 2v is A1 . From Section ��C.�(a) on page ��� the symmetry species spanned by
the integrand px zpz is found by the forming the direct product of the symmetry
species spanned by px , z, and pz separately. �ese are read o� the C 2v character
table by looking for the appropriate Cartesian functions listed on the right of
the table: x and hence px spans B1 , while z and pz both span A1 . �e direct
product required is therefore B1 × A1 × A1 .
�e order does not matter, so this is equal to A1 × A1 × B1 , which is equal to B1
because, from the �rst simplifying feature described in Section ��C.�(a) on page
���, the direct product of the totally symmetric representation with any other
representation is the latter representation itself: A1 × Γ(i) = Γ(i) . �e integrand
therefore spans B1 . �is is not the totally symmetric irreducible representation,
therefore the integral is zero .
E��C.�(a) As explained in Section ��C.� on page ���, an electric dipole transition is forbidden if the electric transition dipole moment µ q,fi is zero. �e transition
dipole moment is given by [��C.�–���], µ q,fi = −e ∫ ψ ∗f qψ i dτ where q is x,
y, or z. �e integral is only non-zero if the integrand contains the totally symmetric representation, which from the C 3v character table is A1 .
For a transition A1 → A2 , the symmetry species of the integrand is given by
the direct product A2 × Γ(q) × A1 . �e order does not matter so this is equal
to A1 × A2 × Γ(q) , which is simply equal to A2 × Γ(q) because, from the �rst
simplifying feature listed in Section ��C.�(a) on page ���, the direct product of
the totally symmetric irreducible representation with any other representation
is the latter representation itself. �erefore A1 × A2 = A2 .
�e direct product A2 × Γ(q) contains the totally symmetric irreducible representation only if Γ(q) spans A2 because, according to the second simplifying
feature listed in Section ��C.�(a) on page ���, the direct product of two irreducible representations contains the totally symmetric irreducible representation only if the two irreducible representations are identical. �e C 3v character
table shows that none of x, y or z span A2 , so it follows that the integrand does
not contain A1 and hence the transition is forbidden .
Solutions to problems
P��C.�
Methane belongs to point group Td . �e methane molecule and its H�s orbitals
are shown in Fig. ��.��, along with one operation of each class. It is su�cient to
consider just one operation in each class because, by de�nition, all operations
the same class have the same character.
�e C 3 operation shown in Fig. ��.�� leaves sA unchanged but converts sB into
sC , sC into sD , and sD into sB : ( sA sC sD sB ) ← ( sA sB sC sD ). �is is written
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
C2,S4
C3
σd
H
sA
sB
H
H
H
sD
sC
Figure 10.17
using matrix multiplication as
D(C 3 )
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 1 0 0 0 �
� 0 0 0 1 �
�
( sA sC sD sB ) = ( sA sB sC sD )�
� 0 1 0 0 �
� 0 0 1 0 �
�e matrix D(C 3 ) is the representative of the C 3 operation in this basis. Similarly, the C 2 operation shown in Fig. ��.�� exchanges sA and sB , and also exchanges sC and sD . �is is written using matrix multiplication as
D(C 2 )
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 0 1 0 0 �
� 1 0 0 0 �
�
( sB sA sD sC ) = ( sA sB sC sD )�
� 0 0 0 1 �
� 0 0 1 0 �
�e σd operation shown in Fig. ��.�� leaves sA and sB unchanged and exchanges
sC and sD ; this gives
D(σ d )
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 1 0 0 0 �
� 0 1 0 0 �
�
( sA sB sD sc ) = ( sA sB sC sD )�
� 0 0 0 1 �
� 0 0 1 0 �
�e S 4 operation shown in Fig. ��.�� converts sA to sD , sB to sC , sC to sA , and
sD to sB , giving
D(S 4 )
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 0 0 1 0 �
� 0 0 0 1 �
�
( sD sC sA sB ) = ( sA sB sC sD )�
� 0 1 0 0 �
� 1 0 0 0 �
371
372
10 MOLECULAR SYMMETRY
Finally, the E operation leaves all orbitals unchanged, meaning that its representative is simply the identity matrix
D(E)
��� � � � � � � � � � � � � � � � � � � � � � ��� � � � � � � � � � � � � � � � � � � � � � �
� 1 0 0 0 �
� 0 1 0 0 �
�
( sA sB sC sD ) = ( sA sB sC sD )�
� 0 0 1 0 �
� 0 0 0 1 �
�e representatives found above have the following characters
χ(E) = 4
χ(C 3 ) = 1
χ(C 2 ) = 0
χ(σd ) = 2
χ(S 4 ) = 0
�is result can be arrived at much more quickly by noting that: (�) only the
diagonal elements of the representative matrix contribute to the trace; (�) orbitals which are unmoved by an operation will result in a � on the diagonal; (�)
orbitals which are moved to other positions by an operation will result in a � on
the diagonal. �e character is found simply by counting the number of orbitals
which do not move. In the present case � are unmoved by E, � is unmoved by
C 3 , none are unmoved by C 2 , � are unmoved by σd , and none are unmoved by
S 4 . �e characters are thus {4, 1, 0, 2, 0}.
�is representation is decomposed using the method described in Section ��C.�(b)
on page ���. �e number of times n(Γ) that a given irreducible representation
Γ occurs in a representation is given by [��C.�a–���],
n(Γ) =
1
(Γ)
� N(C)χ (C)χ(C)
h C
where h is the order of the group, N(C) is the number of operations in class C,
χ(Γ) is the character of class C in the irreducible representation Γ, and χ(C) is
the character of class C in the representation being reduced. In the case of the
group Td , h = 24. �e number of times that the irreducible representation A1
occurs is
1
�N(E)× χ(A1 ) (E)× χ(E) + N(C 3 )× χ(A1 ) (C 3 )× χ(C 3 )
n(A1 ) = 24
+ N(C 2 )× χ(A1 ) (C 2 )× χ(C 2 ) + N(σd )× χ(A1 ) (σd )× χ(σd )
+ N(S 4 )× χ(A1 ) (S 4 )× χ(S 4 )�
Similarly
1
= 24
(1×1×4 + 8×1×1 + 3×1×0 + 6×1×2 + 6×1×0) = 1
1
n(A2 ) = 24
(1×1×4 + 8×1×1 + 3×1×0 + 6×(−1)×2 + 6×(−1)×0) = 0
1
n(E) = 24
(1×2×4 + 8×(−1)×1 + 3×2×0 + 6×0×2 + 6×0×0) = 0
1
n(T1 ) = 24
(1×3×4 + 8×0×1 + 3×(−1)×0 + 6×(−1)×2 + 6×1×0) = 0
1
n(T2 ) = 24
(1×3×4 + 8×0×1 + 3×(−1)×0 + 6×1×2 + 6×(−1)×0) = 1
�e four H�s orbitals in methane therefore span A1 + T2 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
As explained in Section ��C.�(a) on page ���, only orbitals of the same symmetry species may have a nonzero overlap. �e carbon �s orbital spans the
totally symmetric irreducible representation A1 as it is unchanged under all
symmetry operations in the group. It can therefore form molecular orbitals
with the A1 combination of hydrogen orbitals. Inspection of the Td character
table shows that the carbon px , p y , and pz orbitals jointly span T2 , so they can
form molecular orbitals with the T2 combinations of hydrogen orbitals.
�e character table also shows that the dz 2 and dx 2 −y 2 orbitals on the silicon in
SiH� jointly span E; note that the dz 2 orbital appears as (3z 2 − r 2 ) in the character table. �ese d orbitals therefore cannot form molecular orbitals with the
A1 or T2 combinations of hydrogen orbitals. However, the dx y , d yz , and dzx
orbitals on the silicon span T2 and can therefore form molecular orbitals with
the T2 combinations of hydrogen orbitals.
P��C.�
As explained in Section ��C.� on page ���, an integral over a region will necessarily vanish unless the integrand, or part thereof, spans the totally symmetric
irreducible representation of the point group of the region. In the case of the
function 3x 2 − 1, the −1 part will always span the totally symmetric representation because it is invariant to all symmetry operations. �erefore the integral
will not necessarily vanish in any of the ranges.
P��C.�
�e p 1 symmetry adapted linear combination is shown in Fig. ��.��, along with
some of the symmetry elements.
y
C2',σv
σh
C2'',σd
A
B
x
D
C4,C2,i,S4
C
Figure 10.18
�is SALC is una�ected by any of the operations E, C 2 , C 2′′ , S 4 , and σv , but
changes sign under C 4 , C 2′ , i, σh , and σd . �e representatives, which are the
same as the characters because the representatives are one-dimensional, are
therefore
class C E C 4 C 2 C 2′ C 2′′ i S 4 σh σv σd
D(C) � −1 � −1 � −1 � −1 � −1
�e D 4h character table shows that this representation corresponds to the B2u
irreducible representation. �e s orbital on the xenon spans the totally symmetric representation A1g , and the D 4h character table indicates that px and p y
jointly span Eu ; pz spans A2u . It also shows that d yz and dzx jointly span Eg .
373
374
10 MOLECULAR SYMMETRY
�e orbitals dz 2 , dx 2 −y 2 and dx y span respectively A1g , B1g and B2g . Because
only orbitals of the same symmetry may have a non-zero overlap it follows that
none of the s, p and d orbitals on the xenon can form molecular orbitals with
the B2u orbital p 1 .
P��C.�
�e ethene molecule is shown in Fig. ��.��.
y
y
C2,σyz
σxy
sB H
sC
H
C2z
H s
A
H
x
x
C2,σzx
sD
Figure 10.19
�e SALCs are generated using the method described in Section ��C.�(b) on
page ���, applying each operation to sA . �e results are given in the following
table.
E
sA
C 2z
sC
C2
sB
y
C 2x
sD
i
sC
σxy
sA
σ yz
sB
σ zx
sD
characters for Ag
�
product of rows � and � sA
characters for B2u
�
product of rows � and � sA
characters for B3u
�
product of rows � and � sA
characters for B1g
�
product of rows � and � sA
characters for B1u
�
product of rows � and �� sA
�
sC
−1
−sC
−1
−sC
�
sC
�
sC
�
sB
�
sB
−1
−sB
−1
−sB
−1
−sB
�
sD
−1
−sD
�
sD
−1
−sD
−1
−sD
�
sC
−1
−sC
−1
−sC
�
sC
−1
−sC
�
sA
�
sA
�
sA
�
sA
−1
−sA
�
sB
�
sB
−1
−sB
−1
−sB
�
sB
�
sD
−1
−sD
�
sD
−1
−sD
�
sD
Row
� e�ect on sA
�
�
�
�
�
�
�
�
��
��
�e SALCs are formed by summing rows �, �, �, � and �� and dividing each by
the order of the group (h = 8).
Row �: ψ (A1g ) = 18 (sA + sC + sB + sD + sC + sA + sB + sD ) = 14 (sA + sB + sC + sD )
Row �: ψ (B2u ) = 18 (sA − sC + sB − sD − sC + sA + sB − sD ) = 14 (sA + sB − sC − sD )
Row �: ψ (B3u ) = 18 (sA − sC − sB + sD − sC + sA − sB + sD ) = 14 (sA − sB − sC + sD )
Row �: ψ (B1g ) = 18 (sA + sC − sB − sD + sC + sA − sB − sD ) = 14 (sA − sB + sC − sD )
Row ��: ψ (B1u ) = 18 (sA + sC − sB − sD − sC − sA + sB + sD ) = 0
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e results from row �� show that attempting to project out a SALC with symmetry B1u gives zero . �is is because the four hydrogen �s orbitals do not span
B1u .
375
11
11A
Molecular Spectroscopy
General features of molecular spectroscopy
Answers to discussion questions
D��A.�
�is is discussed in Section ��A.� on page ���.
D��A.�
Doppler broadening. �is contribution to the linewidth is due to the Doppler
e�ect which shi�s the frequency of the radiation emitted or absorbed when the
molecules involved are moving towards or away from the detecting device. In
a gas, molecules have a wide range of speeds in all directions and the detected
spectral line is the absorption or emission pro�le arising from the resulting
Doppler shi�s. �e shape of a Doppler-broadened spectral line re�ects the
Maxwell distribution of speeds in the sample.
Lifetime broadening. �is kind of broadening is a quantum mechanical e�ect
which predicts that for a state with a lifetime τ there is an energy uncertainty
δE given by δE ≈ ħ�τ. �is uncertainty in the energy translates to absorption
(or emission) over a range of frequencies and hence a linewidth. �e lifetime
of a state may be limited by the rate of spontaneous emission from the state, in
which case the resulting broadening is called natural line broadening.
Collisions between molecules are e�cient at changing their rotational and vibrational energies, and therefore the lifetimes of such states are limited by the
the collision rate. �e resulting line broadening is called collisional or pressure
line broadening.
�e rate of spontaneous emission cannot be changed; hence its contribution is
the same regardless of phase. Doppler broadening is expected to contribute in a
similar way for both gases and liquids. �e higher density of liquids compared
to gases implies that collisions will be more frequent and hence the collisional
line broadening will be greater for a liquid.
Solutions to exercises
E��A.�(a)
If a light source of frequency ν 0 is approached at a speed s, the Doppler shi�ed
frequency ν a is [��A.��a–���],
νa = ν0 �
1 + s�c
�
1 − s�c
1�2
378
11 MOLECULAR SPECTROSCOPY
Writing the frequencies in terms of the wavelength as ν = c�λ and then inverting before sides gives
λa = λ0 �
1 − s�c
�
1 + s�c
1�2
At nonrelativistic speeds, s � c, this simpli�es to λ a = λ 0 (1 − s�c)
1�2
. Hence
λ a = (680 nm) × [1 − (60 km h−1 ) × (1 h�3600 s)
× (1000 m�1 km)�(2.9979 × 108 m s−1 )]1�2 = ��� nm
Within the precision of the data given, the Doppler shi� is insigni�cant.
E��A.�(a)
�e uncertainty in the energy of a state with lifetime τ is δE ≈ ħ�τ. �erefore a
spectroscopic transition involving this state has an uncertainty in its frequency,
and hence a linewidth, of the order of δν = δE�h ≈ (2πτ)−1 . �is expression
is rearranged to give the lifetime as τ = (2πδν)−1 ; expressing the linewidth as
a wavenumber gives τ = (2πδ ν̃c)−1 .
(i) For δν̃ = 0.20 cm−1
τ = [2π×(0.20 cm−1 )×(2.9979×1010 cm s−1 )]−1 = 2.65...×10−11 s = �� ps
(ii) For δ ν̃ = 2.0 cm−1
E��A.�(a)
τ = [2π×(2.0 cm−1 )×(2.9979×1010 cm s−1 )]−1 = 2.65...×10−12 s = �.� ps
�e uncertainty in the energy of a state with lifetime τ is δE ≈ ħ�τ. �erefore a
spectroscopic transition involving this state has an uncertainty in its frequency,
and hence a linewidth, of the order of δν = δE�h ≈ (2πτ)−1 . If the linewidth is
expressed as a wavenumber the expression becomes δν̃ = δE�hc ≈ (2πτc)−1 .
If each collision deactivates the molecule, the lifetime is �/(collision frequency),
but if only � in N of the collisions deactivates the molecule, the lifetime is
N/(collision frequency). �us τ = N�z, where z is the collision frequency. �e
linewidth is therefore δ ν̃ = (2πcN�z)−1 .
(i) If each collision is e�ective at deactivation, N = 1 and with the data given
δ ν̃ = [2π × (2.9979 × 1010 cm s−1 ) × 1�(1.0 × 1013 s−1 )]−1 = �� cm−1
(ii) If only � in ��� collisions are e�ective at deactivation, N = 100
E��A.�(a)
δ ν̃ = [2π × (2.9979 × 1010 cm s−1 ) × 100�(1.0 × 1013 s−1 )]−1 = �.�� cm−1
�e ratio A�B is given by [��A.�a–���], A�B = 8πhν 3 �c 3 ; the frequency ν is
related to the wavelength though ν = c�λ, and to the wavenumber through
ν = ν̃c.
(i) For X-rays with λ = 70.8 pm
A 8πh(c�λ)3 8πh 8π × (6.6261 × 10−34 J s)
=
= 3 =
= 0.0469 J s m−3
B
c3
λ
(70.8 × 10−12 m)3
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(ii) For visible light with λ = 500 nm
A 8πh 8π × (6.6261 × 10−34 J s)
= 3 =
= 1.33 × 10−13 J s m−3
B
λ
(500 × 10−9 m)3
(iii) For infrared radiation with ν̃ = 3000 cm−1
A 8πh(c ν̃)3
=
= 8πh ν̃ 3
B
c3
= 8π × (6.6261 × 10−34 J s) × (3000 × 102 m−1 )3 = 4.50 × 10−16 J s m−3
E��A.�(a)
Note the conversion of the wavenumber from cm−1 to m−1 .
�e Beer–Lambert law [��A.�–���], I = I 0 10−ε[J]L relates the intensity of the
transmitted light I to that of the incident light I 0 .
I�I 0 = 10−ε[J]L = 10−(723 dm mol
3
= 0.171...
E��A.�(a)
−1
cm−1 )×(4.25×10−3 mol dm−3 )×(0.250 cm)
Using this, the percentage reduction in intensity is calculated as 100(I 0 −I)�I 0 =
100(1 − I�I 0 ) = 100(1 − 0.171...) = ��.�% . Note the conversion of L to cm and
[J] to mol dm−3 in order to match the units of ε.
�e Beer–Lambert law [��A.�–���], I = I 0 10−ε[J]L relates the intensity of the
transmitted light I to that of the incident light I 0 . If a fraction T of the incident
light passes through the sample, I = TI 0 and hence I�I 0 = T; T is the transmittance. It follows that log T = −ε[J]L hence ε = −(log T)�[J]L. If ��.�% of the
light is transmitted, T = 0.181
ε = −(log T)�[J]L = −[log(0.181)]�(0.139 × 10−3 mol dm−3 ) × (1.00 cm)
= 5.34 × 103 dm3 mol−1 cm−1
Note the use of L in cm and the conversion of [J] to mol dm−3 in order to give
the usual units of ε.
E��A.�(a)
�e Beer–Lambert law [��A.�–���], I = I 0 10−ε[J]L relates the intensity of the
transmitted light I to that of the incident light I 0 . If a fraction α is absorbed,
then a fraction T = 1−α of the incident light passes through the sample, I = TI 0
and hence I�I 0 = T; T is the transmittance. It follows that log T = −ε[J]L
hence [J] = −(log T)�εL. If ��.�% of the light is absorbed, α = 0.385 and
T = 1 − 0.385 = 0.615
[J] = −(log T)�εL = −[log(0.615)]�(386 dm3 mol−1 cm−1 ) × (0.500 cm)
= 1.09... × 10−3 mol dm−3 = 1.09 mM
Note the use of L in cm.
E��A.�(a)
�e transmittance T is the ratio I�I 0 , hence the Beer–Lambert law [��A.�–���]
can be written T = I�I 0 = 10−ε[J]L . It follows that log T = −ε[J]L and hence
ε = −(log T)�[J]L. With this, the following table is drawn up, with L = 0.20 cm
379
380
11 MOLECULAR SPECTROSCOPY
[dye]/(mol dm−3 )
0.0010
0.0050
0.0100
T�%
81.4
35.6
12.7
T
0.814
0.356
0.127
447
449
448
ε�(dm3 mol−1 cm−1 )
E��A.�(a)
0.0500
3.0 × 10−3
3.0 × 10−5
452
�e average value of ε from these measurements is 449 dm3 mol−1 cm−1 .
�e transmittance T is the ratio I�I 0 , hence the Beer–Lambert law [��A.�–���]
can be written T = I�I 0 = 10−ε[J]L . It follows that log T = −ε[J]L and hence
ε = −(log T)�[J]L. With the given data, T = 0.48 and L = 0.20 cm, the molar
absorption coe�cient is calculated as
ε = −(log 0.48)�[(0.010 mol dm−3 )×(0.20 cm)] = 1.59...×102 dm3 mol−1 cm−1
�e molar absorption coe�cient is therefore ε = 1.6 × 102 dm3 mol−1 cm−1 .
For a path length of �.�� cm the transmittance is
T = 10−(1.59 ...×10 dm mol
2
3
Hence T = 23% .
−1
cm−1 )×(0.40 cm)×(0.01 mol dm−3 )
= 0.230
E��A.��(a) �e ratio of the incident to the transmitted intensities of light a�er passing
through a sample of length L, molar concentration [H2 O], and molar absorption coe�cient ε is given by [��A.�–���], T = I�I 0 = 10−ε[H2 O]L . It follows that
log T = −ε[H2 O]L, which rearranges to give L = −(log T)�ε[H2 O].
�e molar concentration of H� O is calculated by noting that its mass density
is ρ = 1000 kg m−3 and its molar mass is M = 18.016 g mol−1 . �e concentration is therefore ρ�M = (1000 kg m−3 )�(18.016 × 10−3 kg mol−1 ) =
55.5... × 103 mol m−3 = 55.5... mol dm−3 .
�e light intensity is half that at the surface when T = 0.5, hence the depth is
calculated as
L = −(log 0.5)�[(6.2 × 10−5 dm3 mol−1 cm−1 ) × (55.5... mol dm−3 )]
= 87.4... cm = 0.875 m
�e light intensity reaches a tenth of at the surface when T = 0.1
L = −(log 0.1)�[(6.2 × 10−5 dm3 mol−1 cm−1 ) × (55.5... mol dm−3 )]
= 290... cm = 2.90 m
E��A.��(a) �e integrated absorption coe�cient is given by [��A.��–���], A = ∫band ε(ν̃) dν̃,
where the integration is over the band and ν̃ = λ−1 is the wavenumber. �e initial, peak, and �nal wavenumbers of the lineshape are given by (220×10−7 cm)−1 =
4.54...×104 cm−1 , (270×10−7 cm)−1 = 3.70...×104 cm−1 and (300×10−7 cm)−1 =
3.33... × 104 cm−1 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Assuming that the lineshape is triangular the area under it is 12 × base × height
A = 21 × [(4.54... − 3.33...) × 104 cm−1 ] × (2.21 × 104 dm3 mol−1 cm−1 )
= 1.33... × 108 dm3 mol−1 cm−2
�e integrated absorption coe�cient is therefore 1.34 × 108 dm3 mol−1 cm−2 .
E��A.��(a) �e Doppler linewidth is given by [��A.��a–���], δν obs = (2ν 0 �c)(2kT ln 2�m)1�2 .
Because λ = c�ν this may be rewritten δν obs = (2�λ 0 )(2kT ln 2�m)1�2 . Taking
the mass of a hydrogen atom as 1 m u gives the linewidth as
δν obs = (2�λ 0 )(2kT ln 2�m)1�2
= [2�(821 × 10−9 m)] × �
= 4.53... × 109 Hz
2 × (1.3806 × 10−23 J K−1 ) × (300 K) × ln 2
�
1.6605 × 10−27 kg
1�2
where 1 J = 1 kg m2 s−2 is used. Expressed as a wavenumber the linewidth is
(4.53... × 109 Hz)�(2.9979 × 1010 cm s−1 ) = �.��� cm−1 .
Solutions to problems
P��A.�
P��A.�
�e fraction of the incident photons that reach the retina is (1 − 0.30) × (1 −
0.25)×(1−0.09)×(1−0.43) = 0.272.... Hence the number of photons reaching
the retina in �.� s is
(4.0 × 103 mm−2 s−1 ) × (40 mm2 ) × (0.1 s) × 0.272... = 4.4 × 103
�e absorbance at λ 1 and λ 2 are A 1 and A 2 , respectively
A 1 = ε A1 [A]L + ε B1 [B]L
A 2 = ε A2 [A]L + ε B2 [B]L
(��.�)
(��.�)
At each wavelength the absorbance depends on the concentration of each species
and the relevant molar absorption coe�cient. Equation ��.� is multiplied by ε A2
and eqn ��.� is multiplied by ε A1 to give
ε A2 A 1 = ε A2 ε A1 [A]L + ε A2 ε B1 [B]L
ε A1 A 2 = ε A1 ε A2 [A]L + ε A1 ε B2 [B]L
Subtracting the two equations eliminates [A], and rearrangement gives the required expression for [B]
ε A2 A 1 − ε A1 A 2 = ε A2 ε B1 [B]L − ε A1 ε B2 [B]L
[B] =
ε A2 A 1 − ε A1 A 2
(ε A2 ε B1 − ε A1 ε B2 )L
Simply exchanging the labels A and B gives the corresponding expression for
[A]
ε B2 A 1 − ε B1 A 2
[A] =
(ε B2 ε A1 − ε B1 ε A2 )L
381
11 MOLECULAR SPECTROSCOPY
P��A.�
Following the hint, a plot is made of ln ε against ν̃; the data are shown in the
following table and the plot is shown in Fig. ��.�.
ε�(dm3 mol−1 cm−1 ) ln[ε�(dm3 mol−1 cm−1 )]
1 512
7.32
865
6.76
477
6.17
257
5.55
135.9
4.91
69.5
4.24
34.5
3.54
λ�nm
292.0
296.3
300.8
305.4
310.1
315.0
320.0
ν̃�(104 cm−1 )
3.425
3.375
3.324
3.274
3.225
3.175
3.125
7
ln[ε�(dm3 mol−1 cm−1 )]
382
6
5
4
3.10 3.15 3.20 3.25 3.30 3.35 3.40 3.45
ν̃�(104 cm−1 )
Figure 11.1
�e data are quite a good �t to the line
ln(ε�dm3 mol−1 cm−1 ) = 12.609 × [ν̃�(104 cm−1 )] − 35.793
�is can be expressed as ln(ε�dm3 mol−1 cm−1 ) = a(ν̃�cm−1 ) + b with a =
1.2609 × 10−3 and b = −35.793. It follows that ε = e a ν̃ eb , where the units have
been omitted for clarity. With this expression for ε, the integrated absorption
coe�cient is found by evaluating the integral
A=�
ν̃ max
ν̃ min
b
ε dν̃ = �
= e (1�a) �e
ν̃ max
ν̃ min
e a ν̃ eb dν̃ = eb (1�a)e a ν̃ �ν̃
ν̃ max
min
�
−e
−3
4
−3
4
1
= e−35.793
�e(1.2609×10 )×(3.425×10 ) − e(1.2609×10 )×(3.125×10 ) �
−3
1.2609 × 10
= 1.26 × 106 dm3 mol−1 cm−2
a ν̃ max
a ν̃ min
Again, units have been omitted for clarity.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P��A.�
(a) �e area of a triangle is 12 × base × height, so the integrated absorption
coe�cient is
A = 12 × [(34483 − 31250) cm−1 ] × (150 dm3 mol−1 cm−1 )
= 2.42 × 105 dm3 mol−1 cm−2
⇀ M� , where M is CH� I.
(b) Assume that the equilibrium involved is � M ���
�e total pressure is known, and from this it is possible to compute the
total concentration of M and M� together; the fraction of the total present
as M� is also known. Using these data it is possible to �nd the concentration of M, and hence the absorbance.
Suppose that initially there are n 0 moles of M which then come to equilibrium by forming n moles of M� : the amount in moles of M is then
n M = n 0 − 2n, and the total amount in moles of all species is n tot = n 0 − n.
Let the fraction that is present as dimer be α, α = n�n tot .
�e aim is to express n M = n 0 − 2n in terms of the known quantities n tot
and α
=n tot +n
=αn tot
�
�
n M = n 0 −2n = n tot − n = n tot (1 − α)
Assuming that the perfect gas law applies
c tot =
n tot
p
=
V
RT
where c tot is the total concentration of both M and M� . It follows that the
concentration of M is
[M] =
n M n tot (1 − α)
p
=
= c tot (1 − α) =
(1 − α)
V
V
RT
With the data given
[M] =
(2.4 Torr)×[(1 atm)�(760 Torr)]×[(1.01325 × 105 Pa)�(1 atm)]
(8.3145 J K−1 mol−1 ) × (373 K)
× (1 − 0.01) = 0.102... mol m−3 = 1.02... × 10−4 mol dm−3
�e absorbance at the mid-point is
A = ε[M]L
= (150 dm3 mol−1 cm−1 ) × (1.02... × 10−4 mol dm−3 ) × (12.0 cm)
= �.��
(c) With the data at ��� Torr and ��% dimers, the concentration of the monomer
is
[M] =
(100 Torr)×[(1 atm)�(760 Torr)]×[(1.01325 × 105 Pa)�(1 atm)]
(8.3145 J K−1 mol−1 ) × (373 K)
× (1 − 0.18) = 3.52... mol m−3 = 3.52... × 10−3 mol dm−3
383
384
11 MOLECULAR SPECTROSCOPY
�e absorbance at the mid-point is
A = ε[M]L
= (150 dm3 mol−1 cm−1 ) × (3.52... × 10−3 mol dm−3 ) × (12.0 cm)
= 6.34...
�e absorbance at the mid-point is therefore A = 6.35 . From this value
the molar absorption coe�cient would be inferred as ε = A�c tot L and c tot
is computed as before from the pressure
c tot =
(100 Torr)×[(1 atm)�(760 Torr)]×[(1.01325 × 105 Pa)�(1 atm)]
(8.3145 J K−1 mol−1 ) × (373 K)
= 4.29... mol m−3 = 4.29... × 10−3 mol dm−3
Hence
ε = A�c tot L = (6.34...)�[(4.29... × 10−3 mol dm−3 ) × (12.0 cm)]
P��A.�
= 123 dm3 mol−1 cm−1
�e line from the star is at longer wavelength, and hence lower frequency, than
for the Earth-bound observation, therefore the object is receding. �e Doppler
shi� is given by [��A.��a–���]
f = ν r �ν 0 = �
1 − (s�c)
�
1 + (s�c)
1�2
�e ratio f is equal to λ 0 �λ r because the frequency is inversely proportional to
the wavelength. Writing x = s�c gives
f =�
1 − x 1�2
�
1+x
hence
f 2 (1 + x) = (1 − x) hence
It follows that s = c[1 − (λ 0 �λ r )2 ]�[1 + (λ 0 �λ r )2 ].
s = (2.9979×108 m s−1 )×
x=
1− f2
1+ f2
1 − [(654.2 nm)�(706.5 nm)]2
= 2.301 × 106 m s−1
1 + [(654.2 nm)�(706.5 nm)]2
�e Doppler linewidth is given by [��A.��a–���], δν�ν 0 = (2�c)(2kT ln 2�m)1�2 .
Provided that the linewidth is small compared to the absolute frequency of the
line (which is the case here), δν�ν 0 is well approximated by δλ�λ 0
δλ 2 2kT ln 2 1�2
= �
�
λ0 c
m
With the data given
T =�
hence
T =�
δλ 2 c 2 m
�
λ 0 8k ln 2
0.0618 nm 2 (2.9979 × 108 m s−1 )2 × 47.95 × (1.6605 × 10−27 kg)
�
706.5 nm
8 × (1.3806 × 10−23 J K−1 ) × ln 2
= 7.15 × 105 K
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P��A.��
If each collision is e�ective at changing the energy of a state, the lifetime is
simply the inverse of the collision rate: τ = 1�z .
�e uncertainty in the energy of a state with lifetime τ is δE ≈ ħ�τ. �erefore a
spectroscopic transition involving this state has an uncertainty in its frequency,
and hence a linewidth, of the order of δν = δE�h ≈ (2πτ)−1 . Using τ = 1�z
and the given expression for z gives the linewidth as
δν = 1�2πτ = z�2π =
4σ kT 1�2 p
4σ 2
�
�
=� 3
�
2π πm
kT
π mkT
1�2
p
With the given data and taking m = 36 m u
4 × (0.30 × 10−18 m2 )2
δν = � 3
�
π ×(36)×(1.6605 × 10−27 kg)×(1.3806 × 10−23 J K−1 ) × (298 K)
1�2
× (1.01325 × 105 Pa) = �.�� GHz
�e Doppler linewidth is given by [��A.��a–���], δν�ν 0 = (2�c)(2kT ln 2�m)1�2 ;
with the data given
δν =
2ν 0 2kT ln 2 1�2 2ν̃ 0 c 2kT ln 2 1�2
2kT ln 2 1�2
�
� =
�
� = 2ν̃ 0 �
�
c
m
c
m
m
2 × (1.3806 × 10−23 J K−1 ) × (298 K) × ln 2
= 2×(6356 m )×�
�
(36)×(1.6605 × 10−27 kg)
= 3.93 MHz
P��A.��
−1
1�2
Note that ν̃ 0 is used in m−1 . For the collisional broadening to be equal to
the Doppler broadening the former must be reduced by a factor 700�3.93 =
178; because the linewidth is proportional to the pressure, this means that the
pressure must be reduced by this factor to (1.01325 × 105 Pa)�178 = ��� Pa or
�.�� Torr .
�e best way to approach this is to generate the interferogram in a numerical
form, that is as a table of data points. As is seen in the previous Problem it is necessary to have at least two data points per cycle in order to represent the wavenumber correctly, which implies that the distance by which the mirror must
be moved in one step is δ = 1�2ν̃ max , where ν̃ max is the highest wavenumber
which will be represented correctly. �e i th data point in the interferogram is
constructed using the expression
��� � � � �� � � � ��
2
I i = � �a j [1 + cos(2πν̃ j iδ)]� e−α(i δ)
apodization
j
where a j and ν̃ j are the intensity and wavenumber, respectively, of the jth peak
in the spectrum; i runs from 0 to N, the number of data points. �e value of
N is a matter of choice, but a sensible starting value might be ���; the reason
385
11 MOLECULAR SPECTROSCOPY
for this apparently odd choice is that some numerical implementations of the
Fourier transform require that the number of points be a power of � (256 = 28 ).
�e apodization term is there in order to force the interferogram to go smoothly
to zero (or at least near to zero) for the largest value of the pathlength di�erence
N δ. If this is not done, the peaks in the spectrum will have ‘wiggles’ around
their bases, as seen in Fig. ��A.� on page ���. In a practical spectrometer this
term might not be required because with radiation passing through the interferometer covers a wide range of frequencies and interference between these
will naturally drive the interferogram to zero. In this simulation, with only a
few frequencies present, apodization is required. �e parameter α is adjusted
to achieve the desired smoothing of the envelope.
Figure ��.� shows an interferogram computed using the following parameters;
the data points have been joined up by a continuous line
N = 256
ν̃ max = 100 cm−1
a 1 = 0.25 ν̃ 1 = 5.0 cm−1
δ = 1�(2 × 100 cm−1 ) = 0.005 cm
a 2 = 1.00 ν̃ 2 = 15 cm−1
α = 2.5 cm−2
a 3 = 0.75 ν̃ 3 = 50 cm−1
4
3
I
386
2
1
0
0.0
0.2
0.4
0.6
0.8
p�cm
1.0
1.2
Figure 11.2
To �nd the spectrum it is necessary to compute the Fourier transform of the interferogram. �ere are many variants of the way this transform is implemented
as a numerical procedure, and the one needed here is usually referred to as the
discrete cosine Fourier transform. As can be seen from Fig. ��.�, the interferogram is always positive and decays to zero; this will give a large peak in the
spectrum at a wavenumber of zero, in addition to the peaks corresponding to
the wavenumbers of the oscillating terms that have been introduced. Figure ��.�
shows the spectrum obtained by Fourier transformation of the interferogram;
the peak at zero wavenumber has been truncated.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
0
20
40
60
80
100
ν̃�cm
Figure 11.3
11B Rotational spectroscopy
Answers to discussion questions
D��B.�
�is is discussed in Section ��B.� on page ���.
D��B.�
��
C has spin zero and so is a boson; ��C and �H have spin half and so are
fermions; �H has spin � and so is a boson. All the molecules are linear so the
same considerations as described in Section ��B.� on page ��� apply.
For �H ��C ≡ ��C �H the ��C have no e�ect as they are spin �, so the rotational
levels behave in just the same way as 1 H2 : the (odd J)/(even J) statistical weight
ratio is therefore �/�. Similarly, the rotational levels of �H ��C ≡ ��C �H behave in
just the same way as 2 H2 : the (odd J)/(even J) statistical weight ratio is therefore
I�(I + 1) = 1�2.
For �H ��C ≡ ��C �H there are four nuclear spin wavefunctions arising from the
�
H nuclei, three symmetric and one antisymmetric with respect to exchange of
the nuclei. In addition there are four nuclear spin wavefunctions arising from
the ��C nuclei, three symmetric and one antisymmetric. Overall, there are ��
nuclear spin wavefunctions. Of these, � arise from combining a symmetric
wavefunction for �H� and a symmetric wavefunction for ��C� , giving overall
symmetric wavefunctions. In addition there is one more overall symmetric
wavefunction obtained by combining the antisymmetric wavefunction for ��C�
with that for �H� . �e total number of symmetric wavefunctions is therefore
��, and the remaining � are therefore antisymmetric.
�e ratio of symmetric to antisymmetric nuclear spin functions is therefore
��/�, therefore the (odd J)/(even J) statistical weight ratio is ��/�.
D��B.�
Symmetric rotor: �e energy depends on J and K 2 , hence each level except the
K = 0 level is doubly degenerate. In addition, states of a given J have (2J + 1)
values of the component of their angular momentum along an external axis,
characterized by the quantum number M J . �e energy is not a�ected by M J ,
387
388
11 MOLECULAR SPECTROSCOPY
so there is a degeneracy of 2J + 1 for each J. It follows that a symmetric rotor
level is 2(2J + 1)-fold degenerate for K ≠ 0, and 2J + 1 degenerate for K = 0.
Linear rotor: A linear rotor has K �xed at �, but there are still 2J + 1 values of
M J , so the degeneracy is 2J + 1.
Spherical rotor: A spherical rotor can be regarded as a version of a symmetric
rotor in which A = B; consequently the energy is independent of the 2J + 1
values that K can assume. Hence, there is a degeneracy of 2J + 1 associated
with both K and M J , resulting in a total degeneracy of (2J + 1)2 .
If a decrease in rigidity a�ects the symmetry of the molecule, the rotational
degeneracy could be a�ected also.
D��B.�
A molecule has three principal moments of inertia about perpendicular axes:
these moments are labelled I a , I b , and I c , with I c ≥ I b ≥ I a . A prolate symmetric
rotor has I a ≠ I b = I c ; examples include a thin rod, any linear molecule, CH� F
and CH� CN. An oblate symmetric rotor has I a = I b ≠ I c ; examples include a
�at disc, benzene and BF� . In terms of I�� and I� , prolate rotors have I�� < I� and
oblate tops have I�� > I� .
Solutions to exercises
E��B.�(a)
�e wavenumbers of the lines in the rotational spectrum are given by [��B.��a–
���], ν̃(J) = 2B̃(J + 1); the J = 3 ← 2 transition is therefore at ν̃(2) = 2B̃(2 +
1) = 6B̃. �e rotational constant is given by [��B.�–���], B̃ = ħ�4πcI, and the
moment of inertia is given by m eff R 2 , where m eff = m 1 m 2 �(m 1 + m 2 ).
I=
(14.0031 × 15.9949)m u2 1.6605 × 10−27 kg
×
× (115 × 10−12 m)2
(14.0031 + 15.9949)m u
1 mu
= 1.63... × 10−46 kg m2
B̃ =
ħ
1.0546 × 10−34 J s
=
4πcI 4π × (2.9979 × 1010 cm s−1 ) × (1.63... × 10−46 kg m2 )
= 1.70... cm−1
�e transition occurs at 6B̃ = 6 × (1.70... cm−1 ) = ��.� cm−1 . Expressed in
frequency units this is 6c B̃ = 6 × (2.9979 × 1010 cm s−1 ) × (1.70... cm−1 ) =
3.07... × 1011 Hz = ��� GHz .
Centrifugal distortion will lower the frequency.
E��B.�(a)
�e wavenumbers of the lines in the rotational spectrum are given by [��B.��a–
���], ν̃(J) = 2B̃(J + 1). �e J = 3 ← 2 transition is therefore at ν̃(2) = 2B̃(2 +
1) = 6B̃, hence B̃ = (63.56�6) cm−1 . �e rotational constant is given by [��B.�–
���], B̃ = ħ�4πcI, and the moment of inertia is given by m eff R 2 , where m eff =
m 1 m 2 �(m 1 + m 2 ). It follows that R = (ħ�4πcm eff B̃)1�2 .
m eff =
(1.0078 × 34.9688)m u2 1.6605 × 10−27 kg
×
= 1.62 . . . × 10−27 kg
(1.0078 + 34.9688)m u
1 mu
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
R=�
E��B.�(a)
1.0546 × 10−34 J s
�
4π×(2.9979 × 1010 cm s−1 )×(1.62... × 10−27 kg)×[(65.36�6)cm−1 ]
1�2
= 125.7 pm
�e wavenumbers of the lines in the rotational spectrum are given by [��B.��a–
���], ν̃(J) = 2B̃(J + 1); the lines are therefore spaced by 2B̃, it therefore follows
that B̃ = (12.604�2) cm−1 . �e rotational constant is given by [��B.�–���],
B̃ = ħ�4πcI, and the moment of inertia is given by m eff R 2 , where m eff =
m 1 m 2 �(m 1 + m 2 ). It follows that I = ħ�4πc B̃ and R = (I�m eff )1�2 .
I = ħ�4πc B̃
=
1.0546 × 10−34 J s
= 4.4420 × 10−47 kg m2
4π × (2.9979 × 1010 cm s−1 ) × [(12.604�2] cm−1 )
m eff =
(1.0078 × 26.9815)m u2 1.6605 × 10−27 kg
×
= 1.61... × 10−27 kg
(1.0078 + 26.9815)m u
1 mu
R = (I�m eff )1�2 = [(4.44... × 10−47 kg m2 )�(1.61... × 10−27 kg)]1�2
E��B.�(a)
= 165.9 pm
�e most occupied J state is given by [��B.��–���], J max = (kT�2hc B̃)1�2 − 12 .
(i) At 25 ○ C, ��� K, this gives
J max =
�
(1.3806 × 10−23 J K−1 )×(298 K)
� − 12
2×(6.6261 × 10−34 J s)×(2.9979 × 1010 cm s−1 )×(0.244 cm−1 )
1�2
= ��
(ii) At 100 ○ C, ��� K, this gives
J max =
�
E��B.�(a)
E��B.�(a)
(1.3806 × 10−23 J K−1 )×(373 K)
� − 12
2×(6.6261 × 10−34 J s)×(2.9979 × 1010 cm s−1 )×(0.244 cm−1 )
1�2
= ��
For a molecule to show a pure rotational Raman spectrum it must have an
anisotropic polarizability. With the exception of spherical rotors, all molecules
satisfy this requirement. �erefore H� , HCl, CH� Cl all give rotational Raman
spectra.
�e Stokes lines appear at wavenumbers given by [��B.��a–���], ν̃(J + 2 ← J) =
ν̃ i − 2B̃(2J + 3), where the wavenumber of the incident radiation is ν̃ i , and J is
the quantum number of the initial state. With the given data
ν̃(2 ← 0) = 20 487 cm−1 − 2 × (1.9987 cm−1 )(2 × 0 + 3) = 20 475 cm−1
389
390
11 MOLECULAR SPECTROSCOPY
E��B.�(a)
�e Stokes lines appear at wavenumbers given by [��B.��a–���], ν̃(J + 2 ← J) =
ν̃ i − 2B̃(2J + 3), where the wavenumber of the incident radiation is ν̃ i , and J is
the quantum number of the initial state. It therefore follows that the separation
between adjacent lines is 4B̃, hence B̃ = (0.9752�4) cm−1 .
�e rotational constant is given by [��B.�–���], B̃ = ħ�4πcI, and the moment
of inertia is given by m eff R 2 , where m eff = m 1 m 2 �(m 1 + m 2 ). It follows that
I = ħ�4πc B̃ and R = (I�m eff )1�2 .
I = ħ�4πc B̃
=
1.0546 × 10−34 J s
= 1.14... × 10−45 kg m2
4π × (2.9979 × 1010 cm s−1 ) × [(0.9752�4) cm−1 ]
For a homonuclear diatomic the e�ective mass is simply m eff = 12 m
1.14... × 10−45 kg m2
�
× 34.9688 × (1.6605 × 10−27 kg)
2
R = (I�m eff )1�2 = � 1
E��B.�(a)
E��B.�(a)
= 198.9 pm
1�2
�e ratio of the weights for (odd J)/(even J) is given by [��B.��–���]. For ��Cl,
I = 32 and the nucleus is therefore a fermion. �e ratio is (odd J)/(even J)
= (I + 1)�I = ( 32 + 1)�( 32 ) = 53 .
�e moment of inertia I of a molecule about a speci�ed axis is given by [��B.�–
���], I = ∑ i m i r 2i where the sum is over all the atoms, m i is the mass of atom
i and r i is its perpendicular distance to the axis. For the calculation of the
moment of inertia about the bisector, the central atom makes no contribution.
R θ�2
Each of the other atoms is at a perpendicular distance R sin(θ�2), where θ is
the bond angle and R the bond length. �e moment of inertia is therefore
I = 2 × m O R 2 sin2 (θ�2)
= 2 × (15.9949) × (1.6605 × 10−27 kg)
× [(128 × 10−12 m) × sin(117○ �2)]2
= 6.32... × 10−46 kg m2 = 6.33 × 10−46 kg m2
�e corresponding rotational constant is given by [��B.�–���],
B̃ =
ħ
1.0546 × 10−34 J s
=
4πcI 4π × (2.9979 × 1010 cm s−1 ) × (6.32... × 10−46 kg m2 )
= 0.442 cm−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E��B.��(a) �e required expressions are the �rst listed under symmetric rotors in Table ��B.�
on page ���
I� = m A f 1 (θ)R 2 +
m A (m B + m A )
f 2 (θ)R 2
m
mC
�(3m A + m B )R ′ + 6m A R[ 13 f 2 (θ)]1�2 � R′
m
I�� = 2m A f 1 (θ)R 2
+
Note that the molecule described by these relationships is BA� C, which becomes BA� by letting C=A; the question refers to a molecule AB� , but for consistency with the main text the exercise will be continued with BA� . Let m C =
m A and R ′ = R to give
I� = m A f 1 (θ)R 2 +
+
m A (m B + m A )
f 2 (θ)R 2
m
mA
�(3m A + m B ) + 6m A [ 13 f 2 (θ)]1�2 � R 2
m
with m = m B + 4m A . To simplify the expression somewhat let m B = αm A . �is
gives m = αm A + 4m A = (4 + α)m A
I� = m A f 1 (θ)R 2 +
m A2 (1 + α)
f 2 (θ)R 2
(4 + α)m A
mA
�m A (3 + α) + 6m A [ 13 f 2 (θ)]1�2 � R 2
(4 + α)m A
(1 + α)
1
�(3 + α) + 6[ 13 f 2 (θ)]1�2 �
I� �(m A R 2 ) = f 1 (θ) +
f 2 (θ) +
(4 + α)
(4 + α)
+
I�� �(m A R 2 ) = 2 f 1 (θ)
�e variation of the moments of inertia with θ are shown in Fig. ��.�; I� is shown
for three representative values of α. Not surprisingly, I� and I�� converge onto
the same value when θ is the tetrahedral angle (shown by the vertical dotted
line). �is is because in this limit the molecule becomes tetrahedral and is then
a spherical rotor, for which all the moments of inertia are the same.
At the tetrahedral angle cos θ tet = − 13 ; hence f 1 (θ tet ) = 43 and f 2 (θ tet ) = 13
(1 + α) 1
1
�(3 + α) + 6[ 13 × 13 ]1�2 �
×3+
(4 + α)
(4 + α)
(1 + α) 1
1
= 43 +
×3+
{(3 + α) + 2}
(4 + α)
(4 + α)
4(4 + α) + (1 + α) + 3(5 + α) 32 + 8α
=
=
=8
3(4 + α)
3(4 + α) 3
I� �(m A R 2 ) = 43 +
�e moment of inertia for a tetrahedral molecule is, from the table, I�(m A R 2 ) =
8
, in agreement with the result just derived. In this limit the moment of inertia
3
does not depend on the mass of B (the central atom), as the axes pass through
this atom.
391
11 MOLECULAR SPECTROSCOPY
3.0
I��
I� α = 1
I� α = .2
I� α = 5
2.8
I�m A R 2
392
2.6
2.4
2.2
2.0
90.0
95.0
Figure 11.4
100.0
θ�○
105.0
110.0
E��B.��(a) To be a symmetric rotor a molecule most possess an n-fold axis with n > 2.
(i) O� is bent (like H� O), it has a two-fold axis and so is an asymmetric rotor.
(ii) CH� CH� has a three-fold axis and so is a symmetric rotor. (iii) XeO� is
tetrahedral, and so is a spherical rotor. (iv) Ferrocene has a �ve-fold axis and
so is a symmetric rotor.
E��B.��(a) In order to determine two unknowns, data from two independent experiments
are needed. In this exercise two values of B for two isotopologues of HCN are
given; these are used to �nd two moments of inertia. �e moment of inertia of
a linear triatomic is given in Table ��B.� on page ���, and if it is assumed that
the bond lengths are una�ected by isotopic substitution, the expressions for
the moment of inertia of the two isotopologues can be solved simultaneously
to obtain the two bond lengths.
�e rotational constant in wavenumber is given by [��B.�–���], B̃ = ħ�4πcI;
multiplication by the speed of light gives the rotational constant in frequency
units B = ħ�4πI, which rearranges to I = ħ�4πB
I HCN = (1.0546 × 10−34 J s)�[4π × (44.316 × 109 Hz)] = 1.89... × 10−46 kg m2
I DCN = (1.0546 × 10−34 J s)�[4π × (36.208 × 109 Hz)] = 2.31... × 10−46 kg m2
It is somewhat more convenient for the subsequent manipulations to express
the moments of inertia in units of the atomic mass constant m u and nm.
I HCN = (1.89... × 10−46 kg m2 ) × �
= 0.114... m u nm2
I DCN = (2.31... × 10−46 kg m2 ) × �
= 0.139... m u nm2
2
109 nm
1 mu
� ×
1m
1.6605 × 10−27 kg
2
109 nm
1 mu
� ×
1m
1.6605 × 10−27 kg
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Using the expressions from Table ��B.� on page ���, the moments of inertia
are expressed in terms of the masses and bond lengths, where the former are
expressed as multiples on m u . In this case A = �H or �H, B = ��C and C = ��N.
(m H R − m N R ′ )2
mH + mC + mN
(1.0078R − 14.0031R ′ )2
= 1.0078R 2 + 14.0031R′2 −
1.0078 + 12.0000 + 14.0031
(1.0078R − 14.0031R′ )2
= 1.0078R 2 + 14.0031R′2 −
27.0109
(m D R − m N R′ )2
2
′2
I DCN = m D R + m N R −
mD + mC + mN
(2.0141R − 14.0031R ′ )2
= 2.0141R 2 + 14.0031R′2 −
2.0141 + 12.0000 + 14.0031
(2.0141R − 14.0031R′ )2
= 2.0141R 2 + 14.0031R′2 −
28.0172
I HCN = m H R 2 + m N R′2 −
�ese two equations need to be solved simultaneously for R and R ′ , but because they are quadratics this is a very laborious process by hand: it is best
achieved using mathematical so�ware. �is gives the resulting bond lengths as
R = R CH = 0.1062 nm and R′ = R CN = 0.1157 nm .
E��B.��(a) �e centrifugal distortion constant is given by [��B.��–���], D̃ J = 4B̃ 3 �ν̃ 2 . With
the given data D̃ J = 4(6.511 cm−1 )3 �(2308 cm−1 )2 = 2.073 × 10−4 cm−1 .
�e rotational constant is inversely proportional to the moment of inertia of
the molecule, I = m eff R 2 where R is the bond length and m eff is the e�ective
mass. Assuming that isotopic substitution does not a�ect the bond length, it
follows that B̃ ∝ m−1
eff . Assuming that isotopic substitution does not a�ect
−1�2
the force constant, the vibrational frequency is proportional to m eff . �us
−1�2
3
2
−2
D̃ ∝ (m−1
eff ) �(m eff ) = m eff . For this estimation it is su�cient to use integer
masses, and because a ratio is involved these can be expressed as multiples of
mu .
D̃ 2HI �D̃ 1HI = (m eff 1HI �m eff 2HI ) = �
2
1 × 127 2 + 127 2
×
� = �.��
1 + 127 2 × 127
E��B.��(a) For a molecule to show a pure rotational (microwave) absorption spectrum is
must have a permanent dipole moment. Of the molecules given, the only ones
to satisfy this requirement are HCl, CH� Cl and CH� Cl� .
Solutions to problems
P��B.�
Suppose that the bond length is R and that the centre of mass is at a distance x
from mass m 1 and therefore (R − x) from mass m 2 . Balancing moments gives
m 1 x = m 2 (R − x), hence x = m 2 R�(m 1 + m 2 ). Using this result it follows that
(R − x) = R − m 2 R�(m 1 + m 2 ) = m 1 R�(m 1 + m 2 ). �e moment of inertia is
393
394
11 MOLECULAR SPECTROSCOPY
therefore
m 1 m 22 R 2
m 2 m 12 R 2
+
(m 1 + m 2 )2 (m 1 + m 2 )2
2
m 1 m 2 (m 2 + m 1 )R
m1 m2 R2
=
=
= m eff R 2
(m 1 + m 2 )2
(m 1 + m 2 )
I = m 1 x 2 + m 2 (R − x)2 =
P��B.�
�e rotational terms for a symmetric rotor are given by [��B.��a–���], F̃(J, K) =
B̃J(J+1)+(Ã− B̃)K 2 . �e selection rules are ∆J = ±1 and ∆K = 0, and therefore
the term in K does not a�ect the wavenumber of the lines in the spectrum; the
result is that the lines are at exactly the same wavenumbers as for a linear rotor,
[��B.��a–���], ν̃(J) = 2B̃(J + 1). �e separation of the lines is 2B̃.
In frequency units the spacing is 2B = 2 × (298 GHz) = 596 GHz . Expressed
as a wavenumber this spacing is (596 × 109 Hz)�(2.9979 × 1010 cm s−1 ) =
��.� cm−1 .
�e rotational constant is given by [��B.�–���], B̃ = ħ�4πcI. Expressed in
frequency units this is B = ħ�4πI. It follows that I = ħ�4πB
I = ħ�4πB =
1.0546 × 10−34 J s
= 2.82 × 10−47 kg m2
4π × (298 × 109 Hz)
Expressions for the moment of inertia are given in Table ��B.� on page ���;
NH� is a symmetric rotor and the second entry under symmetric rotors is the
required one. �e moment of inertia corresponding to the rotational constant
B is I� . With the data given
I� = m H (1 − cos θ)R 2 +
mH mN
(1 + 2 cos θ)R 2
m N + 3m H
It is convenient to work with the masses as multiples of m u and R in nm
I� = (0.1014 nm)2 × �(1.0078) × (1 − cos 106.78○ )
1.0078 × 14.0031
(1 + 2 cos 106.78○ )� × m u
14.0031 + 3 × 1.0078
= 0.0169... m u nm2
+
Converting to the usual units gives
I = (0.0169... m u nm2 ) ×
10−18 m2 1.6605 × 10−27 kg
×
= 2.815 × 10−47 kg m2
1 nm2
1 mu
�is value is consistent with the moment of inertia determined from the given
rotational constant.
P��B.�
Bonding is essentially the result of electrostatic interactions so to a very good
approximation it is expected that adding an uncharged neutron will have no
e�ect on the bond length.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e wavenumbers of the lines expected for a diatomic are given by [��B.��a–
���], ν̃(J) = 2B̃(J + 1); the separation of the lines is 2B̃. �e rotational constant
is inversely proportional to the e�ective mass, therefore if the bond length is
una�ected by isotopic substitution the ratio of the rotational constants should
be equal to the inverse ratio of the e�ective masses. With the data given
B̃ 1H 35Cl �B̃ 2H 35Cl = (20.8784 cm−1 )(10.7840 cm−1 ) = 1.93605
m eff , 2H 35Cl
m 1H m 35Cl
m 2 + m 35Cl
=
× H
m eff , 1H 35Cl m 1H + m 35Cl
m 2H m 35Cl
1.007 825 + 34.968 85 2.0140 × 34.968 85
=
×
1.007 825 × 34.968 85 2.0140 + 34.968 85
= 1.93440
�ese two quantities di�er by less than �.�% so the hypothesis that the bond
length is invariant to isotopic substitution is con�rmed to quite a high level of
precision; with the accuracy of the data given there is, however, some perceptible change.
P��B.�
Note: there is an error in the problem; for the 34 S isotopologue the line at
��.��� �� GHz is for J = 3.
�e wavenumbers of the lines expected for a linear rotor are given by [��B.��a–
���], ν̃(J) = 2B̃(J + 1); the separation of the lines is 2B̃. For OC32 S the average
spacing of the lines is ��.����� GHz, so the best estimate for the rotational
constant is B OCS = 12 × (12.16272 GHz) = 6.08136 GHz. For OC34 S there
are just two lines, one for J = 1 and one for J = 3; these are separated by
��.����� GHz, which is 4B. �e best estimate for the rotational constant is
B OCS′ = 14 × (23.73007 GHz) = 5.93252 GHz.
�e rotational constant in wavenumber is given by [��B.�–���], B̃ = ħ�4πcI;
multiplication by the speed of light gives the rotational constant in frequency
units B = ħ�4πI, hence I = ħ�4πB
I OCS = (1.0546 × 10−34 J s)�[4π × (6.08136 × 109 Hz)] = 1.37... × 10−45 kg m2
I OCS′ = (1.0546 × 10−34 J s)�[4π × (5.93252 × 109 Hz)] = 1.41... × 10−45 kg m2
where for short S implies ��S and S′ implies ��S. It is somewhat more convenient
for the subsequent manipulations to express the moments of inertia in units of
the atomic mass constant m u and nm.
I OCS = (1.37... × 10−45 kg m2 ) × �
= 0.831... m u nm2
I OCS′ = (1.41... × 10−45 kg m2 ) × �
= 0.851... m u nm2
2
109 nm
1 mu
� ×
1m
1.6605 × 10−27 kg
2
109 nm
1 mu
� ×
1m
1.6605 × 10−27 kg
395
396
11 MOLECULAR SPECTROSCOPY
Using the expressions from Table ��B.� on page ���, the moments of inertia
are expressed in terms of the masses and bond lengths, where the former are
expressed as multiples on m u . In this case A = ��O, B = ��C and C = ��S or ��S.
(m O R − m S R′ )2
mO + mC + mS
(15.9949R − 31.9721R′ )2
= 15.9949R 2 + 31.9721R ′2 −
15.9949 + 12.0000 + 31.9721
(15.9949R − 31.9721R ′ )2
= 15.9949R 2 + 31.9721R ′2 −
59.967
′ 2
′R )
(m
R
−
m
O
S
I OCS′ = m O R 2 + m S′ R ′2 −
m O + m C + m S′
(15.9949R − 33.9679R′ )2
= 15.9949R 2 + 33.9679R ′2 −
15.9949 + 12.0000 + 33.9679
(15.9949R − 33.9679R ′ )2
= 15.9949R 2 + 33.9679R ′2 −
61.9628
I OCS = m O R 2 + m S R′2 −
P��B.�
�ese two equations need to be solved simultaneously for R and R ′ , but because they are quadratics this is a very laborious process by hand: it is best
achieved using mathematical so�ware. �is gives the resulting bond lengths as
R = R OC = 0.1167 nm and R′ = R CS = 0.1565 nm .
�e wavenumbers of the lines expected for a linear rotor are given by [��B.��a–
���], ν̃(J) = 2B̃(J +1); the separation of the lines is 2B̃. However, the separation
between adjacent lines in the given data is not constant, but increases along the
series. To account for this, the e�ects of centrifugal distortion are included,
and in this case the frequencies of the lines are given by [��B.��b–���], ν(J) =
2B(J+1)−4D J (J+1)3 (written with the constants in frequency units). Division
of both side of this expression by 2(J +1) indicates that a plot of [ν(J)]�2(J +1)
against (J + 1)2 should be a straight line with slope −2D J and intercept B. �e
data are tabulated below; δ is the di�erence between successive lines. �e plot
is shown in Fig. ��.�.
J
24
25
26
27
28
29
ν(J)�MHz
214 777.7
223 379.0
231 981.2
240 584.4
249 188.5
257 793.5
δ�MHz
8 601.3
8 602.2
8 603.2
8 604.1
8 605.0
[ν(J)�2(J + 1)]�MHz
4 295.6
4 295.8
4 296.0
4 296.2
4 296.4
4 296.6
(J + 1)2
625
676
729
784
841
900
�e data are a good �t to the line
{ν(J)]�2(J + 1)}�MHz = 3.652 × 10−3 × (J + 1)2 + 4293.28
�e value of the rotational constant is found from the intercept: (B�MHz) =
intercept. Some elementary statistics on the best-�t line indicates an error of
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
[ν(J)�2(J + 1)]�MHz
4 296.5
4 296.0
4 295.5
600
Figure 11.5
650
700
750
(J + 1)
800
850
900
2
about 0.03 MHz in the intercept, so the best estimate for the rotational constant
is B = 4293.28 ± 0.03 MHz or, expressed as a wavenumber, B̃ = 0.1432 cm−1 .
It is somewhat unusual that the centrifugal distortion constant appears to be
negative.
�e most occupied J state is given by [��B.��–���], J max = (kT�2hc B̃)1�2 − 12 .
At ��� K
J max =
�
(1.3806 × 10−23 J K−1 )×(298 K)
�
−34
2×(6.6261 × 10 J s)×(2.9979 × 1010 cm s−1 )×(0.1432 cm−1 )
= ��
P��B.��
1�2
− 12
A similar calculation at ��� K gives J max = 15 .
�e population of level J, N J , is given by N J ∝ g J e−E J �k T . In this expression g J
is the degeneracy of level J, g J = (2J + 1), and E J is the energy of that level, E J =
hc B̃J(J+1). To �nd the level with the greatest population the derivative dN J �dJ
is computed and then set to zero; it is not necessary to know the constant of
proportion, which will be written A. To compute the derivative requires the
product rule and the chain rule
d
A(2J + 1)e−hc B̃ J(J+1)�k T
dJ
= A×2×e−hc B̃ J(J+1)�k T − A(2J + 1)×(2J + 1)×(hc B̃�kT)e−hc B̃ J(J+1)�k T
setting the derivative to zero and gathering terms gives
0 = Ae−hc B̃ J(J+1)�k T �2 − (2J + 1)2 (hc B̃�kT)�
�e exponential term goes to zero as J → ∞, but this is not a maximum; rather,
397
398
11 MOLECULAR SPECTROSCOPY
the maximum is when the term in square brackets is zero
0 = �2 − (2J max + 1)2 (hc B̃�kT)�
(2J max + 1)2 = 2kT�hc B̃
hence
(2J max + 1) = (2kT�hc B̃)1�2
J max = 12 × (2kT�hc B̃)1�2 − 12
�e level with the greatest population is therefore J max = (kT�2hc B̃)1�2 − 12 .
With the given data
(1.3806 × 10−23 J K−1 )×(298 K)
J max = �
� − 12
2(6.6261 × 10−34 J s)×(2.9979 × 1010 cm s−1 )×(0.1142 cm−1 )
1�2
= ��
For a spherical rotor the degeneracy of each level is (2J + 1)2 . Finding the most
populated level proceeds as before
d
A(2J + 1)2 e−hc B̃ J(J+1)�k T
dJ
= A × 4(2J + 1) × e−hc B̃ J(J+1)�k T
− A(2J + 1)2 × (2J + 1) × (hc B̃�kT)e−hc B̃ J(J+1)�k T
setting the derivative to zero and gathering terms gives
0 = Ae−hc B̃ J(J+1)�k T (2J + 1) �4 − (2J + 1)2 (hc B̃�kT)�
As before the maximum occurs when the term in square brackets is zero
0 = �4 − (2J + 1)2 (hc B̃�kT)�
(2J max + 1)2 = 4kT�hc B̃
hence
(2J max + 1) = (4kT�hc B̃)1�2
J max = 12 × (4kT�hc B̃)1�2 − 12
�e level with the greatest population is therefore J max = (kT�hc B̃)1�2 − 12 .
With the given data
(1.3806 × 10−23 J K−1 )×(298 K)
J max = �
� − 12
(6.6261 × 10−34 J s)×(2.9979 × 1010 cm s−1 )×(5.24 cm−1 )
1�2
= �
P��B.��
In such calculations it may be helpful to use kT�hc = 207.225 cm−1 at ��� K
(from inside the front cover).
Temperature e�ects. At extremely low temperatures (�� K) only the lowest rotational states are populated. No emission spectrum is expected for the CO in the
cloud and star-light microwave absorptions by the CO in the cloud are from the
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
lowest rotational states. At higher temperatures additional high-energy lines
appear because higher energy rotational states are populated. Circumstellar
clouds may exhibit infrared absorptions due to vibrational excitation as well as
electronic transitions in the ultraviolet. Ultraviolet absorptions may indicate
the photodissocation of carbon monoxide. High temperature clouds exhibit
emissions.
Density e�ects. �e density of an interstellar cloud may range from one particle
to a billion particles per cm3 . �is is still very much a vacuum compared
to the laboratory high vacuum of a trillion particles per cm3 . Under such
extreme vacuum conditions the half-life of any quantum state is expected to
be extremely long and absorption lines should be very narrow. At the higher
densities the vast size of nebulae obscures distant stars. High densities and high
temperatures may create conditions in which emissions stimulate emissions of
the same wavelength by molecules. A cascade of stimulated emissions greatly
ampli�es normally weak lines – the maser phenomena of microwave ampli�cation by stimulated emission of radiation.
Particle velocity e�ects. Particle velocity can cause Doppler broadening of spectral lines. �e e�ect is extremely small for interstellar clouds at �� K but is appreciable for clouds near high temperature stars. Out�ows of gas from pulsing
stars exhibit a red Doppler shi� when moving away at high speed and a blue
shi� when moving toward us.
�ere will be many more transitions observable in circumstellar gas than in
interstellar gas, because many more rotational states will be accessible at the
higher temperatures. Higher velocity and density of particles in circumstellar
material can be expected to broaden spectral lines compared to those of interstellar material by shortening collisional lifetimes. (Doppler broadening is not
likely to be signi�cantly di�erent between circumstellar and interstellar material in the same astronomical neighbourhood. �e relativistic speeds involved
are due to large-scale motions of the expanding universe, compared to which
local thermal variations are insigni�cant.)
A temperature of ���� K is not high enough to signi�cantly populate electronically excited states of CO; such states would have di�erent bond lengths,
thereby producing transitions with di�erent rotational constants. Excited vibrational states would be accessible, though, and ro-vibrational transitions with
P and R branches as detailed in this following Topic would be observable in
circumstellar but not interstellar material. �e rotational constant for CO is
�.��� cm−1 . �e �rst excited rotational energy level, J = 1, with energy 2hc B̃, is
thermally accessible at about � K (based on the rough equation of the rotational
energy to thermal energy kT). In interstellar space, only two or three rotational
lines would be observable; in circumstellar space (at about ���� K) the number
of transitions would be more like ��.
399
400
11 MOLECULAR SPECTROSCOPY
11C Vibrational spectroscopy of diatomic molecules
Answers to discussion questions
D��C.�
�e rotational constant depends inversely on the moment of inertia, which in
turn depends on the square of the bond length. However, because the molecule
is vibrating, the bond length is constantly changing. Vibration is much faster
than rotation, so for the purposes of calculating the moment of inertia it is
generally a good approximation to take an average over the vibrational motion
and use �R 2 � in place of R 2 . It follows that B ∝ 1��R 2 �.
If the vibration is assumed to be harmonic �R 2 � increases with increasing vibrational energy. However, for a typical anharmonic vibration there is a much
greater e�ect on �R 2 � arising from the asymmetry of the potential. Put simply,
instead of the molecule oscillating symmetrically about the equilibrium position, the bond stretches more than it is compressed, resulting in the average
bond length increasing. As the vibrational energy increases the potential curve
becomes shallower for bond extension and the average bond length increases
further.
D��C.�
�e value of 1��R 2 � therefore decreases as the vibrational quantum number υ
increases, and as a result the rotational constant B is a decreasing function of
the υ. For typical molecules, this e�ect of the anharmonicity is dominant, and
it is not unusual for the rotational constant to decrease by �–� per cent when
going from the υ = 0 to the υ = 1 vibrational level.
Because bonding is principally a matter resulting from electrostatic interactions, the addition of a neutral particle to the nucleus is not expected to alter the
geometry of a molecule (bond lengths, bond angles), nor is it expected to alter
the force constants which describe the stretching of bonds. However, rotational
spectra, and the rotational �ne structure which is associated with vibrational
spectra, depend on the rotational constants, and in turn these depend on the
e�ective mass. Likewise, vibrational frequencies also depend on the e�ective
mass, and so they too will be a�ected.
Di�erent isotopes may have di�erent nuclear spins and this can a�ect the pattern of intensities of lines arising from di�erent rotational states.
Solutions to exercises
E��C.�(a)
�e wavenumber of the fundamental vibrational transition is simply equal to
the vibrational frequency expressed as a wavenumber. �is is given by [��C.�b–
���], ν̃ = (1�2πc)(k f �m eff )1�2 , where m eff is the e�ective mass, given by m eff =
m 1 m 2 �(m 1 + m 2 ). It follows that k f = m eff (2πc ν̃)2 . With the data given the
following table is drawn up.
�
ν̃�cm−1
m eff �m u
k f �N m
−1
H ��F
�
H ��Cl
�
H ��Br
�
H ���I
4141.3
2988.9
2649.7
2309.5
0.9570
0.9796
0.9954
0.9999
967.0
515.6
411.7
314.2
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E��C.�(a)
�e terms (energies expressed as wavenumbers) of the harmonic oscillator are
given by [��C.�b–���], G̃(υ) = (υ + 12 )ν̃; these are wavenumbers and so can be
converted to energy by multiplying by hc to give E(υ) = (υ + 12 )hc ν̃. �e
ground state has υ = 0, and the �rst excited state has υ = 1. �e relative
population of these levels is therefore given by the Boltzmann distribution,
n 1 �n 0 = e−(E 1 −E 0 )�k T . �e energy di�erence E 1 − E 0 = hc ν̃, and hence n 1 �n 0 =
e−hc ν̃�k T . It is convenient to compute the quantity hc ν̃�k �rst to give
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (559.7 cm−1 )
1.3806 × 10−23 J K−1
= 805.3... K
hc ν̃�k =
It follows that n 1 �n 0 = e−(805.3 ... K)�T
(i) At ��� K, n 1 �n 0 = e−(805.3 ... K)�(298 K) = �.����
(ii) At ��� K, n �n = e−(805.3 ... K)�(500 K) = �.���
1
0
As expected, the relative population of the upper level increases with temperature.
E��C.�(a)
Taking y e = 0 is equivalent to using the terms for the Morse oscillator, which
are given in [��C.�–���], G̃(υ) = (υ + 12 )ν̃ − (υ + 12 )2 ν̃x e . �e transition υ ← 0
has wavenumber
∆G̃(υ) = G̃(υ) − G̃(0)
= [(υ + 12 )ν̃ − (υ + 12 )2 ν̃x e ] − [(0 + 21 )ν̃ − (0 + 12 )2 ν̃x e ]
= υ ν̃ − υ(υ + 1)ν̃x e
Data on three transitions are provided, but only two are needed to obtain values
for ν̃ and x e . �e ∆G̃(υ) values for the �rst two transitions are
1←0
2←0
ν̃ − 2ν̃x e = 1556.22 cm−1
2ν̃ − 6ν̃x e = 3088.28 cm−1
Multiplying the �rst expression by � and subtracting the second gives
3(ν̃ − 2ν̃x e ) − (2ν̃ − 6ν̃x e ) = ν̃
hence ν̃ = 3 × (1556.22 cm−1 ) − (3088.28 cm−1 ) = 1580.4 cm−1
E��C.�(a)
�is value for ν̃ is used in the �rst equation, which is then solved for x e to give
x e = 12 − (1556.22 cm−1 )�[2 × (1580.4 cm−1 )] = 7.65 × 10−3 .
Following the discussion in Section ��C.�(b) on page ���, D̃ 0 is given by the area
under a plot of ∆G̃ υ+1�2 against (υ + 12 ), where ∆G̃ υ+1�2 = G̃(υ + 1) − G̃(υ).
�e data are shown in the table and the plot in Fig. ��.�.
401
11 MOLECULAR SPECTROSCOPY
υ
0
1
2
3
4
G̃ υ �cm−1
1 481.86
4 367.50
7 149.04
9 826.48
12 399.80
∆G̃ υ+1�2 �cm−1
2 885.64
2 781.54
2 677.44
2 573.32
υ + 12
0.5
1.5
2.5
3.5
3 000
∆G̃ υ+1�2 �cm−1
402
2 000
1 000
0
0
5
Figure 11.6
10
15
20
υ + 12
25
30
�e data are a good �t to the line
∆G̃ υ+1�2 �cm−1 = −104.11 × (υ + 12 ) + 2 937.7
�is line intercepts the horizontal axis when
0 = −104.11 × (υ + 12 )max + 2 937.7
hence
(υ + 12 )max = 28.22
�e area under the line is simply the area of a triangle, 12 × base × height, which
in this case is 12 × (28.22) × (2 937.7) = 4.14 × 104 . �e dissociation energy
is therefore D̃ 0 = 4.14 × 104 cm−1 ; only modest precision is quoted because a
long extrapolation is made on the basis of few data points.
�e fact that the data fall on a good straight line indicates that the Morse levels
apply, in which case, according to [��C.�b–���], ∆G̃ υ+1�2 = ν̃ − 2(υ + 1)x e ν̃.
�is expression is rewritten
∆G̃ υ+1�2 = ν̃ − 2(υ + 12 )x e ν̃ − x e ν̃
which implies that a plot of ∆G̃ υ+1�2 against (υ + 12 ) will have slope −2x e ν̃ and
intercept (ν̃ − x e ν̃). Hence, using the slope of the plot already made
x e ν̃�cm−1 = − 12 (−104.11) hence
x e ν̃ = 52.06 cm−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
and then using the intercept
(ν̃ − x e ν̃)�cm−1 = 2 937.7
hence ν̃ = (2 937.7 cm−1 ) + (52.06 cm−1 ) = 2989.8 cm−1
�e depth of the well, D̃ e is then found using [��C.�–���], x e = ν̃�4 D̃ e rearranged to D̃ e = ν̃�4x e = ν̃ 2 �4ν̃x e . �e dissociation energy is D̃ 0 = D̃ e − G̃(0)
(Fig. ��C.� on page ���), hence
ν̃ 2
− G̃(0)
4ν̃x e
(2989.8 cm−1 )2
=
− (1481.86 cm−1 ) = 4.14 × 104 cm−1
4 × (52.06 cm−1 )
D̃ 0 = D̃ e − G̃(0) =
To within the precision quoted, both methods give the same result.
E��C.�(a)
E��C.�(a)
To convert to eV, the conversion 1 eV = 8065.5 cm−1 from inside the front cover
is used to give D 0 = 5.14 eV .
�e wavenumber of the transition arising from the rotational state J in the R
branch (∆J = +1) of the fundamental transition (υ = 1 ← υ = 0) is given
by [��C.��c–���], ν̃ R (J) = ν̃ + 2B̃(J + 1). In this case ν̃ = 2308.09 cm−1 and
B̃ = 6.511 cm−1 hence
ν̃ R (2) = (2308.09 cm−1 ) + 2 × (6.511 cm−1 ) × (2 + 1) = ����.� cm−1
�e vibrational frequency of a harmonic oscillator is given by [�E.�–���], ω =
(k f �m)1�2 ; ω is an angular frequency, so to convert to frequency in Hz, ν, use
ω = 2πν. �erefore 2πν = (k f �m)1�2 . Rearranging this gives the force constant
as k f = m(2πν)2
k f = (0.100 kg) × (2π × 2.0 Hz)2 = 16 N m−1
E��C.�(a)
where 1 N = 1 kg m s−2 and 1 Hz = 1 s−1 are used.
�e vibrational frequency, expressed as a wavenumber, of a harmonic oscillator
is given by [��C.�b–���], ν̃ = (1�2πc)(k f �m eff )1�2 , where m eff is the e�ective
mass, given by m eff = m 1 m 2 �(m 1 + m 2 ). Assuming that the force constants of
the two isotopologues are the same, ν̃ simply scales as (m eff )−1�2 . �e fractional
change is therefore
35
ν̃ Na 35Cl − ν̃ Na 37Cl
ν̃ 37
m
= 1 − Na Cl = 1 − � eff Na Cl �
ν̃ Na 35Cl
ν̃ Na 35Cl
m eff Na 37Cl
=1−�
1�2
22.9898 × 34.9688 22.9898 + 36.9651 1�2
×
� = 0.0107...
22.9898 + 34.9688 22.9898 × 36.9651
�e fractional change, expressed as a percentage, is therefore �.���% .
403
11 MOLECULAR SPECTROSCOPY
E��C.�(a)
�e wavenumber of the fundamental vibrational transition is simply equal to
the vibrational frequency expressed as a wavenumber. �is is given by [��C.�b–
���], ν̃ = (1�2πc)(k f �m eff )1�2 , where m eff is the e�ective mass, given by m eff =
m 1 m 2 �(m 1 + m 2 ). It follows that k f = m eff (2πc ν̃)2 ; for a homonuclear diatomic m eff = 12 m 1 . With the data given
k f = ( 12 × [34.9688 × (1.6605 × 10−27 kg)])
× [2π × (2.9979 × 1010 cm s−1 ) × (564.9 cm−1 )]2
= 328.7 N m−1
Note the conversion of the mass to kg.
Solutions to problems
P��C.�
(a) Figure ��.� shows plot of the total electronic energy (with respect to the
free atoms) as a function of the bond length for each of the hydrogen
halides. Calculations are performed with Spartan �� using the MP� method
with the �-���++G** basis set.
10
HI
HBr
8
6
HCl
4
V / eV
404
2
HF
0
–2
–4
–6
–8
Figure 11.7
50
100
150
200
250
300
350
R / pm
�e plot clearly shows that in going down the halogen group from HF to
HI the equilibrium bond length increases and the depth of the potential
well decreases. �e equilibrium properties of each molecule are summarized in the following table. �e force constants are computed in the
harmonic approximation using [��C.�b–���], ν̃ = (1�2πc)(k f �m eff )1�2 ,
with m eff = m 1 m 2 �(m 1 + m 2 ). It follows that k f = m eff (2πc ν̃)2 . �e
calculated bond lengths are in good agreement with the experimental
values, but the vibrational frequencies do not agree very well at all.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
property
R e �pm
R e, expt �pm
ν̃�cm−1
ν̃ expt �cm−1
k f �(N m−1 )
H 19F
H 35Cl
H 81Br
H 127I
91.7
127.3
141.3
161.2
91.680
127.45
141.44
160.92
4198.162 3086.560 2729.302 2412.609
4138.29 2990.95 2648.98 2309.01
993.7
549.8
436.7
342.9
(b) �e force constants decrease steadily down the series, as expected.
Figure ��.� shows a plot of V (x)�V0 as a function of x�a; the minimum at x = 0
is clearly rather ‘�at’.
0.0
−0.5
V (x)�V0
P��C.�
−1.0
Figure 11.8
−3
−2
−1
0
x�a
1
2
3
For a given potential the force constant is de�ned in [��C.�b–���] in terms of
the second derivative as k f = (d2 V �dx 2 )0 .
2
2
d
2a 2 2 2
V0 (e−a �x − 1) = V0 3 e−a �x
dx
x
d2
d
2a 2 −a 2 �x 2 2V0 a 2
3 2a 2 −a 2 �x 2
V
(x)
=
V
e
=
�−
+ 3 �e
0
dx 2
dx
x3
x3
x
x
�e second derivative, and hence the force constant, goes to zero at x = 0 on
account of the argument of the exponential term going to −∞; this dominates
the other terms. �us, for small displacements there is no restoring force and
harmonic motion will not occur.
�e potential is con�ning so it is expected that there will be quantized energy
levels. By loose analogy with the harmonic case the ground state wavefunction
is expected to have a maximum at x = 0 and then decay away to zero as x →
±∞. �e �rst excited state is likely to have a node at x = 0, increase to a
maximum at some positive value of x and then decay away to zero. �e function
will be odd with respect to x = 0, and so will show a symmetrically placed
minimum at a negative value of x.
405
406
11 MOLECULAR SPECTROSCOPY
P��C.�
(a) �e dissociation energy is D̃ 0 = D̃ e − G̃(0) (Fig. ��C.� on page ���), where
G̃(0) is the energy of the lowest vibrational term. For the Morse energy
levels given by [��C.�–���], G̃(υ) = (υ + 12 )ν̃ − (υ + 12 )2 x e ν̃, it follows
that G̃(0) = 12 ν̃ − 14 x e ν̃. �e conversion between cm−1 and eV is achieved
using 1 eV = 8065.5 cm−1 from inside the front cover. For �H ��Cl
hc D̃ 0 = hc D̃ e − G̃(0) = hc D̃ e − ( 12 ν̃ − 14 x e ν̃)
= (5.33 eV) − ( 12 × 2989.7 − 14 × 52.05) × [(1 eV)�(8065.5 cm−1 )]
= �.�� eV
(b) �e task is to calculate the values of ν̃ and x e ν̃ for the isotopologue �H ��Cl.
�e potential energy curve, and hence the value of the depth of the well
D̃ e , is the same for the two isotopologues.
In the harmonic limit the vibrational frequency is given by [��C.�b–���],
ν̃ = (1�2πc)(k f �m eff )1�2 , with m eff = m 1 m 2 �(m 1 + m 2 ). Assuming that
−1�2
the force constants of the two isotopologues are the same, ν̃ ∝ m eff .
From [��C.�–���] it is seen that x e = ν̃�4 D̃ e which rearranges to D̃ e =
−1�2
−1�2
ν̃�4x e . Because ν̃ ∝ m eff it follows that x e ∝ m eff also in order for D̃ e
to be una�ected by isotopic substitution. �us x e ν̃ ∝ m−1
eff .
m 1
ν̃ 2H 35Cl
= � eff , HX �
ν̃ 1H 35Cl
m eff , 2H 35Cl
1�2
ν̃ 2H 35Cl = (2989.7 cm−1 ) �
Similarly
= 2144.25 cm−1
hence ν̃ 2H 35Cl = ν̃ 1H 35Cl × �
m eff , 1H 35Cl
�
m eff , 2H 35Cl
1�2
1.0078 × 34.9688 2.0140 + 34.9688 1�2
×
�
1.0078 + 34.9688 2.02140 × 34.9688
m eff , 1H 35Cl
�
m eff , 2H 35Cl
1.0078 × 34.9688 2.0140 + 34.9688
= (52.05 cm−1 ) �
×
�
1.0078 + 34.9688 2.02140 × 34.9688
= 26.77 cm−1
x e ν̃ 2H 35Cl = x e ν̃ 1H 35Cl × �
Hence for �H ��Cl
hc D̃ 0 = hc D̃ e − G̃(0) = hc D̃ e − ( 12 ν̃ − 14 x e ν̃)
= (5.33 eV) − ( 12 × 2144.25 − 14 × 26.77) × [(1 eV)�(8065.5 cm−1 )]
= �.�� eV
P��C.�
�e term 14 x e ν̃ evaluates to 8.3 × 10−4 eV, so at the precision to which
hc D̃ e is quoted this term has no e�ect.
(a) �e dissociation energy D̃ 0 and the well depth D̃ e are related by D̃ e =
D̃ 0 + G̃(0) (Fig. ��C.� on page ���), where G̃(0) is the vibrational term of
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
the ground vibrational state. In the harmonic approximation G̃(0) = 12 ν̃,
so it follows that
ν̃ = 2(D̃ e − D̃ 0 ) = 2(D e �hc − D 0 �hc)
With the data given
ν̃ = 2[(1.51 × 10−23 J) − (2 × 10−26 J)]
�[(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )] = �.� cm−1
In the harmonic limit the vibrational frequency is given by [��C.�b–���],
ν̃ = (1�2πc)(k f �m eff )1�2 , with m eff = m 1 m 2 �(m 1 + m 2 ). For a homonuclear diatomic m eff = 12 m. It follows that k f = 12 m(2πc ν̃)2 .
k f = 12 × (4.0026) × (1.6605 × 10−27 kg)
× [2π × (2.9979 × 1010 cm s−1 ) × (1.5 cm−1 )]2
= 2.7 × 10−4 N m−1
�e moment of inertia is I = m eff R 2 = 12 mR 2
I = 12 × (4.0026) × (1.6605 × 10−27 kg) × (297 × 10−12 m)2
hence
B̃ =
=
= 2.93... × 10−46 kg m2 = 2.93 × 10−46 kg m2
ħ
4πcI
(1.0546 × 10−34 J s)
= �.��� cm−1
4π × (2.9979 × 1010 cm s−1 ) × (2.93... × 10−46 kg m2 )
(b) If the Morse energy levels are assumed G(0) = 12 ν̃ − 14 x e ν̃, and from
[��C.�–���] x e = ν̃�4 D̃ e . It follows that
D̃ e = D̃ 0 + 12 ν̃ − 14 x e ν̃ = D̃ 0 + 12 ν̃ − ν̃ 2 �16 D̃ e
�e result is a quadratic in ν̃ which is solved in the usual way
ν̃ 2 �16 D̃ e − 12 ν̃+(D̃ e − D̃ 0 ) = 0
With the data given
hence
1
ν̃ = 2
± [ 14 − (D̃ e − D̃ 0 )�4 D̃ e ]1�2
1�8 D̃ e
(D̃ e − D̃ 0 )�4D̃ e = [(1.51 × 10−23 J) − (2 × 10−26 J)]�[4 × (1.51 × 10−23 J)]
= 0.2497
D̃ e = (1.51 × 10−23 J)�[(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )]
= 0.760 cm−1
407
11 MOLECULAR SPECTROSCOPY
hence
ν̃ = 8 × (0.760 cm−1 ) × [ 12 ± ( 14 − 0.2497)1�2 ] = 2.93 cm−1 or 3.15 cm−1
With these values the anharmonicity constant is computed using x e =
ν̃�4 D̃ e
xe =
P��C.�
2.93 cm−1
= 0.96
4 × (0.760 cm−1 )
or
xe =
3.15 cm−1
= 1.04
4 × (0.760 cm−1 )
�e anharmonicity constant is expected to be < 1, so the plausible values
are ν̃ = 2.9 cm−1 and x e = 0.96 . �ese values are very approximate given
the data used to derive them.
�e data are shown in the table and the plot in Fig. ��.�.
υ
0
1
2
3
4
∆G̃ υ+1�2 �cm−1
2 143.1
2 116.1
2 088.9
2 061.3
2 033.5
1
2
υ+1
1
2
3
4
5
2 150
∆G̃ υ+1�2 �cm−1
408
2 100
2 050
Figure 11.9
�e data are a good �t to the line
From the slope
3
υ+1
4
5
∆G̃ υ+1�2 �cm−1 = −27.40 × (υ + 1) + 2 170.7
x e ν̃�cm−1 = − 12 × slope = − 12 (−27.40) hence
and from the intercept
ν̃�cm−1 = intercept = 2 170.70
hence
x e ν̃ = 13.7 cm−1
ν̃ = 2 170.7 cm−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P��C.��
�e data provided allow the calculation of two independent moments of inertia.
If it is assumed that the bond lengths are una�ected by isotopic substitution,
then it is possible to set up two equations and solve them simultaneously for
the CC and CH bond lengths.
Expressions for the wavenumbers of the lines in the P and R branches are given
by [��C.��a–���] and [��C.��c–���]; from these it follows that the spacing between the lines is 2B̃. �e rotational constant is given by [��B.�–���], B̃ =
ħ�4πcI; it follows that I = ħ�4πc B̃.
I H = ħ�4πc B̃ H
=
1.0546 × 10−34 J s
= 2.38... × 10−46 kg m2
4π × (2.9979 × 1010 cm s−1 ) × [(2.352�2) cm−1 ]
I D = ħ�4πc B̃ D
=
1.0546 × 10−34 J s
= 3.30... × 10−46 kg m2
4π × (2.9979 × 1010 cm s−1 ) × [(1.696�2) cm−1 ]
�e moment of inertia is de�ned as I = ∑ i m i r 2i , where r i is the perpendicular
distance from the atom with mass m i to the axis. In HCCH the axis passes
through the mid-point of the CC bond and is perpendicular to the long axis of
the molecule. �us
I H = 2m C (r CC �2)2 + 2m H (r CH + r CC �2)2
I D = 2m C (r CC �2)2 + 2m D (r CH + r CC �2)2
�ese are the two equations which need to be solved simultaneously. Finding
the solution is much simpli�ed by letting p = (r CC �2)2 and q = (r CH + r CC �2)2
to give
I H = 2m C p + 2m H q
I D = 2m C p + 2m D q
It follows that
q=
IH − ID
2(m H − m D )
(2.38... × 10−46 kg m2 ) − (3.30... × 10−46 kg m2 )
= 2.75... × 10−20 m2
2(1.0078 − 2.0140) × (1.6605 × 10−27 kg)
mD IH − mH ID
p=
2m C (m D − m H )
=
=
(2.0140) × (2.38... × 10−46 kg m2 ) − (1.0078) × (3.30... × 10−46 kg m2 )
2 × (12.0000) × (2.0140 − 1.0078) × (1.6605 × 10−27 kg)
= 3.65... × 10−20 m2
If follows that r CC = 2× p1�2 = ���.� pm , and r CH = q 1�2 −r CC �2 = q 1�2 − p1�2 =
���.� pm .
P��C.��
�e variable x is the displacement from the equilibrium separation R e . �e
fact that the potential is symmetric about R e means that �R� = R e and �x� = 0.
409
410
11 MOLECULAR SPECTROSCOPY
On the other hand �x 2 � is de�nitely non-zero, as was seen for the case of the
harmonic oscillator in Topic �E.
It follows straightforwardly that 1��R�2 = 1�R e2 .
�R 2 � is found in the following way
�R 2 � = �(R e + x)2 � = �(R e2 + 2xR e + x 2 )�
� ��� � � �� � � �
= �R e2 � + �2xR e � +�x 2 �
A
B
Term A is simply the average of a constant term, which is equal to the term itself,
in this case R e2 . Term B is rewritten 2R e �x� by taking constant terms outside
the averaging; this term is zero because �x� = 0. �erefore �R 2 � = R e2 + �x 2 �.
Using this 1��R 2 � is found in the following way
1
1
1
1
= 2
= 2×
2
2
�R � R e + �x � R e 1 + �x 2 ��R e2
≈
1
�x 2 �
�1
−
�
R e2
R e2
where to go to the last line the expansion (1+ y)−1 ≈ 1− y is used. �e resulting
expression includes the lowest power of �x 2 ��R e2 , as required
�1�R 2 � is found in the following way
�
1
1
1
1
�=�
�= 2 �
�
R2
(R e + x)2
R e (1 + x�R e )2
1
1
≈ 2 �1 − 2x�R e + 3x 2 �R e2 � = 2 ��1� − (2�R e )�x� + (3�R e2 )�x 2 ��
Re
Re
=
1
3�x 2 �
�1
+
�
R e2
R e2
On the penultimate line the expansion (1 + y)−2 ≈ 1 − 2y + 3y 2 is used. To go
to the �nal line the fact that �x� = 0 is used; the �nal expression has the lowest
non-zero power of �x 2 ��R e2 , as required.
It is evident that none of the averages are the same and that
P��C.��
�
1
1
1
�>
>
R2
�R�2 �R 2 �
�e rotational constant B̃ 0 is computed from
B̃ 0 = B̃ e − 12 a = (0.27971 cm−1 ) − 12 (0.187 × 10−2 cm−1 ) = �.����� cm−1
and similarly for B̃ 1
B̃ 1 = B̃ e − 32 a = (0.27971 cm−1 ) − 32 (0.187 × 10−2 cm−1 ) = �.����� cm−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e wavenumber of the lines in the P and R branches are given by [��C.��–���]
ν̃ P (J) = ν̃ 0 − (B̃ 1 + B̃ 0 )J + (B̃ 1 − B̃ 0 )J 2
ν̃ R (J) = ν̃ 0 + (B̃ 1 + B̃ 0 )(J + 1) + (B̃ 1 − B̃ 0 )(J + 1)2
In these expressions ν̃ 0 is the wavenumber of the pure vibrational transition. If
the Morse levels are assumed, and if it is assumed that it is the υ = 1 ← υ = 0
transition which is being observed, the wavenumber is given by [��C.�b–���],
ν̃ 0 = ν̃ − 2x e ν̃.
For the line in the P branch from J = 3
ν̃ P (J)�cm−1 = 610.258 − 2 × 3.141 − (0.27691 + 0.27877) × 3
+ (0.27691 − 0.27877) × 32 = ���.���
and for the corresponding line in the R branch
ν̃ R (J)�cm−1 = 610.258 − 2 × 3.141 + (0.27691 + 0.27877) × 4
+ (0.27691 − 0.27877) × 42 = ���.���
�e depth of the well, D̃ e is found using [��C.�–���], x e = ν̃�4 D̃ e rearranged to
D̃ e = ν̃�4x e = ν̃ 2 �4ν̃x e . �e dissociation energy is D̃ 0 = D̃ e − G̃(0) (Fig. ��C.�
on page ���), and for the Morse oscillator G̃(0) = 12 ν̃ − 14 ν̃x e .
ν̃ 2
− 1 ν̃ + 14 ν̃x e
4ν̃x e 2
(610.258 cm−1 )2 1
=
− (610.258 cm−1 ) + 14 (3.141 cm−1 )
4 × (3.141 cm−1 ) 2
D̃ 0 =
= 2.93 × 104 cm−1
P��C.��
To convert to eV, the conversion 1 eV = 8065.5 cm−1 from inside the front cover
is used to give D 0 = 3.64 eV .
�e features centred about ����.�� cm−1 are the P and R branches. From [��C.��a–
���] and [��C.��c–���] the �rst line in the R branch occurs at ν̃ + 2B̃, and the
�rst line in the P branch is ν̃ − 2B̃. �e separation of these two, �.��� cm−1 , is
therefore 4B̃.
(a) �e centre of the band is at the vibrational wavenumber, ν̃ = 2143.26 cm−1
(b) In the harmonic approximation the vibrational terms are G̃(υ) = (υ+ 12 )ν̃,
and so the lowest term is G̃(0) = 12 ν̃. �e molar zero-point energy is
therefore N A × hc × 12 ν̃
E zpe = (6.0221 × 1023 mol−1 ) × (6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
× 12 × (2143.26 cm−1 ) = ��.�� kJ mol−1
411
412
11 MOLECULAR SPECTROSCOPY
(c) �e harmonic frequency is given by [��C.�b–���], ν̃ = (1�2πc)(k f �m eff )1�2 ,
with m eff = m 1 m 2 �(m 1 + m 2 ). It follows that k f = m eff (2πc ν̃)2 . With the
data given
kf =
12.0000 × 15.9949
× (1.6605 × 10−27 kg)
12.000 + 15.9949
× [2π × (2.9979 × 1010 cm s−1 ) × (2143.26 cm−1 )]2
= 1856 N m−1
(d) As noted at the start of the answer, 4B̃ = 7.655 cm−1 , hence B̃ = 1.914 cm−1 .
(e) �e rotational constant is given by [��B.�–���], B̃ = ħ�4πcI, and the moment of inertia is given by m eff R 2 . It follows that R = (ħ�4πcm eff B̃)1�2 .
12.0000 × 15.9949
× (1.6605 × 10−27 kg) = 1.13... × 10−26 kg
12.0000 + 15.9949
R = (ħ�4πcm eff B̃)1�2
m eff =
=�
1.0546 × 10−34 J s
�
4π(2.9979 × 1010 cm s−1 )×(1.13... × 10−26 kg)×(1.914 cm−1 )
= 113.3 pm
Although the data are given to quite high precision the assumption that the
harmonic oscillator/rigid rotor models apply means that the derived values of
the bond length and so on are likely to have systematic errors which are higher
than the apparent precision of the data.
P��C.��
�e method of combination di�erences, described in Section ��C.�(b) on page
���, involves taking the di�erence between two transitions which share a common lower rotational level or a common upper rotational level. In the case of
O and S branches, which correspond to ∆J = −2 and ∆J = +2, respectively,
the two transitions which share a common lower level are ν̃ O (J) and ν̃ S (J):
these are the transitions from J to J − 2, and from J to J + 2. As is evident from
Fig. ��.��, the di�erence in wavenumber between these two transitions is the
interval indicated by the dashed arrow which is simply G̃(J + 2) − G̃(J − 2) for
the upper vibrational state (assumed to be υ = 1)
G̃(J + 2) − G̃(J − 2) = B̃ 1 (J + 2)(J + 3) − B̃ 1 (J − 2)(J − 1) = B̃ 1 (8J + 4)
hence
ν̃ S (J) − ν̃ O (J) = 8B̃ 1 (J + 12 )
�e two transitions sharing a common upper level are ν̃ O (J + 2) and ν̃ S (J − 2):
these are the transitions from J + 2 to J, and from J − 2 to J. As is evident from
Fig. ��.��, the di�erence in wavenumber between these two transitions is the
interval indicated by the dotted arrow which is simply G̃(J + 2) − G̃(J − 2) for
the lower vibrational state (assumed to be υ = 0). �is is the same interval as
above, with the exception that the rotational constant is B̃ 0 .
G̃(J + 2) − G̃(J − 2) = B̃ 0 (J + 2)(J + 3) − B̃ 0 (J − 2)(J − 1) = B̃ 0 (8J + 4)
hence
ν̃ S (J − 2) − ν̃ O (J + 2) = 8B̃ 0 (J + 12 )
1�2
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
J+2
J+2
J
J−2
Figure 11.10
υ=1
ν̃ O (J +2)
ν̃ S (J −2)
ν̃ S (J)
ν̃ O (J)
J
J−2
υ=0
11D Vibrational spectroscopy of polyatomic molecules
Answers to discussion questions
D��D.�
�e gross selection rule for vibrational Raman scattering is that for a particular
normal mode to be active the vibration must result in a change in the polarizability as the molecule vibrates about the equilibrium position. �e origin
of this rule is that such an change in the polarizability will result in the dipole
induced in the molecule by the electric �eld of the incident radiation being
altered as the molecule vibrates. It is this modulation of the dipole that results
in Raman scattering.
D��D.�
�e gross selection rule for infrared spectroscopy is that for a particular normal
mode to be active the vibration must result in a change in the dipole moment
as the molecule vibrates about the equilibrium position. �e origin of this rule
is that such an oscillating dipole is needed to stir the electromagnetic �eld into
oscillation (and vice versa for absorption).
Solutions to exercises
E��D.�(a)
According to [��D.�–���], a linear molecule has 3N − 5 normal modes, where
N is the number of atoms in the molecule. �ere are �� atoms in this linear
molecule, and so there are 3(44) − 5 = ��� normal modes.
E��D.�(a)
G̃ 1 (0) + G̃ 2 (0) + G̃ 3 (0) = 12 (ν̃ 1 + ν̃ 2 + ν̃ 3 )
E��D.�(a)
According to [��D.�–���], a non-linear molecule has 3N −6 vibrational normal
modes, where N is the number of atoms in the molecule; therefore H� O has
� normal modes. �e terms (energies expressed as wavenumbers) for normal
mode q are given by [��D.�–���], G̃ q (υ) = (υ q + 12 )ν̃ q , where υ q is the quantum
number for that mode and ν̃ q is the wavenumber of the vibration of that mode.
�ese terms are additive, so the ground state term corresponds to each mode
having υ q = 0
A mode is infrared active if the vibration results in a dipole which changes
as the molecule oscillates back and forth about the equilibrium geometry. A
413
414
11 MOLECULAR SPECTROSCOPY
mode is Raman active if the vibration results in the polarizability changing as
the molecule oscillates back and forth about the equilibrium geometry.
(i) �e three normal modes of an angular AB� molecule are analogous to
those of H� O illustrated in Fig. ��D.� on page ���; all three modes are
both infrared and Raman active.
(ii) �e four normal modes of a linear AB� molecule are analogous to those
of CO� illustrated in Fig. ��D.� on page ���. Of these, three are infrared
active (the asymmetric stretch and the doubly degenerate bend), and one
(the symmetric stretch) is Raman active. �e molecule has a centre of
symmetry, so the rule of mutual exclusion applies and no mode is both
Raman and infrared active.
E��D.�(a)
�e benzene molecule has a centre of symmetry, so the rule of mutual exclusion
applies. �e molecule has no permanent dipole moment and if the ring expands
uniformly this situation does not change: such a vibration does not lead to a
changing dipole and so the mode is infrared inactive .
�is kind of ‘breathing’ vibration does lead to a change in the polarizability, so
the mode is Raman active .
E��D.�(a)
�e exclusion rule applies only to molecules with a centre of symmetry. H� O
does not possess such symmetry, and so the exclusion rule does not apply .
E��D.�(a) With the exception of homonuclear diatomics, all molecules have at least one
infrared active normal mode. Of the molecules listed, HCl, CO� , and H� O
have infrared active modes.
E��D.�(a)
According to [��D.�–���], a non-linear molecule has 3N −6 vibrational normal
modes, where N is the number of atoms in the molecule; a linear molecule has
3N − 5 normal modes. All of the molecules listed are non-linear.
(i) H� O has N = 3 and hence � normal modes.
(ii) H� O� has N = 4 and hence � normal modes.
(iii) C� H� has N = 6 and hence �� normal modes.
Solutions to problems
P��D.�
Figure ��.�� shows a plot of V (h)�V0 as a function of hb 1�4 .
For a given potential the force constant is de�ned in [��C.�b–���] in terms of
the second derivative of the potential with respect to the displacement from
equilibrium; in this case h is the varaible which describes the displacement.
�us k f = (d2 V �dh 2 )0 .
4
4
d
V0 (1 − e−bh ) = V0 4bh 3 e−bh
dh
4
4
d2
d
V (h) =
V0 4bh 3 e−bh = 4V0 b �3h 2 − 4bh 6 � e−bh
2
dh
dh
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
1.0
0.8
V (h)�V0
0.6
0.4
0.2
Figure 11.11
0.0
−2
−1
0
1
2
hb 1�4
�e second derivative, and hence the force constant, goes to zero at h = 0. �us,
for small displacements there is no restoring force.
P��D.�
�e potential is con�ning so it is expected that there will be quantized vibrational energy levels. By loose analogy with the harmonic case the ground state
wavefunction is expected to have a maximum at h = 0 and then decay away to
zero as h → ±∞.
(a) Calculations on SO� are performed with Spartan �� using the MP� method
with the �-���++G** basis set. �e calculated equilibrium structure is
shown in Fig. ��.�� where it is seen that the two S–O bonds have equal
length, as expected. �e calculated bond length and angle agrees quite
well with the experimental values of ���.�� pm and ���.��○ .
(b) �e calculated values of the fundamental vibrational wavenumbers, and
illustrations of the displacements involved in the normal modes, are also
shown in Fig. ��.��. �ey correlate well with experimental values but
are about ��–�� cm−1 lower. SCF calculations o�en yield systematically
lower or higher values than experiment while approximately paralleling
the experimental to within an additive constant.
C2v, 119.26°
146.9 pm
A1, 489.086 cm−1
Figure 11.12
146.9 pm
A1, 1072.813 cm−1
B2, 1285.263 cm−1
415
416
11 MOLECULAR SPECTROSCOPY
11E
Symmetry analysis of vibrational spectroscopy
Answer to discussion question
D��E.�
Because benzene has a centre of symmetry the rule of mutual exclusion applies.
�erefore a particular normal mode will be observed in either the infrared or
in the Raman spectrum (or, possibly, in neither). �e most complete characterization of the normal modes therefore requires the observation of both kinds
of spectra.
Solutions to exercises
E��E.�(a)
E��E.�(a)
A mode is infrared active if it has the same symmetry species as one of the
functions x, y, and z; in this point group these span B1 + B2 + A1 . A mode is
Raman active if it has the same symmetry as a quadratic form; in this group
such forms span A1 + A2 + B1 + B2 . �erefore all of the normal modes are both
Raman and infrared active.
(i) H� O belongs to the point group C 2v . Rather than considering all � displacement vectors together it is convenient to consider them in sub-sets
of displacement vectors which are mapped onto one another by the operations of the group. �e x, y, and z vectors on the oxygen are not
mapped onto the displacements of the H atoms and so can be considered
separately. In fact, because these displacement vectors are attached to the
principal axis, they transform as the cartesian functions x, y, and z as
listed in the character table: that is as B1 + B2 + A1 .
Assuming the same axis system as in Fig. ��E.� on page ���, the two x
displacements on the H atoms map onto one another, as do the two y
displacements, as do the two z displacements: however, the x, y, and z
displacements are not mixed with one another. For the two z displacements the operation E leaves both una�ected so the character is �; the C 2
operation swaps the two displacements so the character is �; the σv (xz)
operation swaps the two displacements so the character is �; the σv′ (yz)
operation leaves the two displacements una�ected so the character is �.
�e representation is therefore (2, 0, 0, 2), which is easily reduced by inspection to A1 + B2 . For the two y displacements the argument is essentially the same, resulting in the representation (2, 0, 0, 2), which reduces
to A1 + B2 .
For the two x displacements the operation E leaves both una�ected so
the character is �; the C 2 operation swaps the two displacements so the
character is �; the σv (xz) operation swaps the two displacements so the
character is �; the σv′ (yz) operation leaves the two displacements in the
same position by changes their direction, so the character is −2. �e representation is therefore (2, 0, 0, −2), which is easily reduced by inspection
to A2 + B1 .
�e � displacements therefore transform as 3A1 + A2 + 2B1 + 3B2 .
�e displacements include translations and rotations. For the point group
C 2v , x, y, and z transform as B1 , B2 , and A1 , respectively. �e rotations
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
R x , R y , and R z transform as B2 , B1 , and A2 , respectively. Taking these
symmetry species away leaves just the normal modes as 2A1 + B2 .
A mode is infrared active if it has the same symmetry species as one of
the functions x, y, and z; in this point group these span B1 + B2 + A1 .
�erefore all of the normal modes are infrared active.
E��E.�(a)
(ii) H� CO is a straightforward extension of the case of H� O as both molecules
belong to the point group C 2v . �e H� C portion lies in the same position
as H� O, with the carbonyl O atom lying on the z axis (the principal axis).
�e analysis therefore includes three more displacement vectors for the
O, and as they are connected to the principal axis they transform as the
cartesian functions x, y, and z, that is as B1 + B2 + A1 . �e tally of
normal modes is therefore those for H� O plus these three in addition:
3A1 + B1 + 2B2 . All these modes are infrared active.
�e displacements include translations and rotations. For the point group C 2v ,
x, y, and z transform as B1 , B2 , and A1 , respectively. �e rotations R x , R y , and
R z transform as B2 , B1 , and A2 , respectively. Taking these symmetry species
away leaves just the normal modes as 4A1 + A2 + 2B1 + 2B2 . �ese correspond
to � normal modes, which is the number expected for CH� Cl� .
Solutions to problems
P��E.�
(a) CH� Cl has a C 3 axis (the principal axis) along the C–Cl bond, and three
σv planes, one passing along each C–H bond and containing the principal
axis. �e point group is therefore C 3v .
(b) �e molecule is non-linear and has N = 5, there are thus 3 × 5 − 6 = �
normal modes.
(c) �e task is to �nd the symmetry species spanned by the set of (x, y, z)
displacement vectors on each atom. �e (x, y, z) displacement vectors on
the Cl can be considered separately and, as these vectors are connected to
the principal axis, their symmetry species is simply read from the character table as E + A1 . �e (x, y, z) displacement vectors on the C behave
in the same way and so transform as E + A1 .
Consider the set of � (x, y, z) displacement vectors on the H atoms. Under the operation E these are all una�ected so the character is �; under
the C 3 operation they are all moved to new positions so the character is �.
Next consider one of the σv planes: the (x, y, z) vectors on the H atoms
which do not lie in this plane are all moved and so contribute � to the
character. Now consider the H atom which lies in the plane, and arrange
a local axis system such that z points along the C–H bond, y lies in the
plane and x lies perpendicular to the plane. �e e�ect of σv on (x, y, z)
is to transform it to (−x, +y, +z); two of the basis functions remain the
same and one changes sign, so the character is −1 + 1 + 1 = +1. �e
representation formed from the � (x, y, z) displacement vectors on the
H atoms is thus (9, 0, 1). �is is reduced using the reduction formula,
[��C.�a–���], to give 2A1 + A2 + 3E.
417
418
11 MOLECULAR SPECTROSCOPY
In total, the displacements therefore transform as 4A1 + A2 + 5E. Taking
away the translations, A1 +E, and the rotations, A2 +E, leaves the symmetry species of the vibrations as 3A1 + 3E . As expected, there are � normal
modes (recall that the E modes are doubly degenerate).
(d) A mode is infrared active if it has the same symmetry species as one of
the functions x, y, and z; in this point group these span A1 + E. All the
normal modes are infrared active.
(e) A mode is Raman active if it has the same symmetry as a quadratic form;
in this point group these span A1 + E. All the normal modes are Raman
active.
11F
Electronic spectra
Answers to discussion questions
D��F.�
A simple model for the energy of the HOMO–LUMO transition in a polyene
is discussed in Example �D.� on page ���. In this the energy levels of the π
electrons in a polyene are modelled by those of a particle in a one-dimensional
box of length L. If the polyene consists of n conjugated double bonds, the length
L may be written as L = nd, where d is the length of a single conjugated bond.
A molecule with n conjugated double bonds will have 2n π electrons which
will occupy the energy levels pairwise. �erefore the HOMO is the level with
quantum number n and the LUMO has quantum number n + 1. �e energy of
the HOMO–LUMO transition is therefore
E n+1 − E n = [(n + 1)2 − n 2 ]
h2
(2n + 1)h 2
=
8mL 2
8mL 2
For large n, (E n+1 − E n ) goes as n. However, L = nd, therefore overall (E n+1 −
E n ) goes as 1�n, and hence the wavelength of the transition goes as n. �us,
increasing the number of conjugated double bonds will increase the wavelength
of the absorption; that is, shi� it to the red.
�e intensity of the transition will depend on the square of the transition dipole
moment, given by
�
L
0
ψ n+1 µ̂ x ψ n dx = (2�L) �
L
sin[(n + 1)πx�L] µ̂ x sin[nπx�L] dx
0
where µ̂ x is the operator for the dipole moment along x, and the normalized
wavefunctions ψ n = (2�L)1�2 sin(nπx�L) are used. �e dipole moment operator is proportional to x, so setting aside the constants of proportion and using
(n + 1) ≈ n for large n, the required integral is
I = (2�L) �
As noted above, L = nd therefore
L
0
I = (2�nd) �
x sin2 (nπx�L) dx
nd
0
x sin2 (πx�d) dx
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e integral is of the form of Integral T.�� with k = π�d and a = nd; it evaluates
to n 2 d 2 �4. Taking into account the normalization factor gives the �nal result
nd�2. �e conclusion is therefore that the transition dipole moment increases
with n, the number of conjugated double bonds.
In summary, the expectation is that increasing the number of conjugated double bonds will increase the wavelength of the absorption and also the intensity
of the absorption.
D��F.�
�is is explained in Section ��F.�(a) on page ���.
D��F.�
�e wavenumbers of the lines in the P, Q and R branches are given in [��F.�–
���]. Consider �rst the P branch ν̃ P (J) = ν̃ − (B̃′ + B̃)J + (B̃′ − B̃)J 2 , and recall
that (B̃′ + B̃) � �(B̃′ − B̃)�. As J increases the lines move to lower wavenumber
on account of the term −(B̃′ + B̃)J. However, as J becomes larger still the term in
J 2 becomes proportionately more and more important. If (B̃′ − B̃) > 0 this term
contributes to an increase in the wavenumber of the lines, and for su�ciently
large J it will overcome the −(B̃′ + B̃)J term and cause the lines to start to move
to higher wavenumber as J increases further. �ere will therefore be a lowest
wavenumber at which any line appears: this is the band head. If (B̃′ − B̃) < 0
the term in J 2 simply causes the lines to move to lower wavenumber and no
band head is formed.
Next consider the R branch ν̃ R (J) = ν̃ + (B̃′ + B̃)(J + 1) + (B̃′ − B̃)(J + 1)2 .
As J increases the lines move to higher wavenumber on account of the term
+(B̃′ + B̃)(J +1). However, as J becomes larger still the term in (J +1)2 becomes
proportionately more and more important. If (B̃′ − B̃) < 0 this term contributes
to an decrease in the wavenumber of the lines, and for su�ciently large J it
will overcome the +(B̃′ + B̃)(J + 1) term and cause the lines to start to move
to lower wavenumber as J increases further. �ere will therefore be a highest
wavenumber at which any line appears: this is the band head. If (B̃′ − B̃) > 0
the term in (J + 1)2 simply causes the lines to move to higher wavenumber and
no band head is formed.
�e lines in the Q branch, ν̃ Q (J) = ν̃ + (B̃′ − B̃)J(J + 1), all appear at higher or
lower wavenumber than ν̃ depending on the sign of (B̃′ − B̃); no band heads
are formed.
If a parameter α is de�ned as α = B̃′ �B̃, so that B̃′ = α B̃, the wavenumbers of
the lines in the P and R branches can be written
[ν̃ P (J)−ν̃]�B̃ = −(1+α)J−(1−α)J 2 [ν̃ R (J)−ν̃]�B̃ = (1+α)(J+1)−(1−α)(J+1)2
�ese functions are plotted in Fig. ��.�� for representative values of α. If α < 1,
meaning that (B̃′ − B̃) < 0, the band head occurs in the R branch, but if α > 1,
meaning that (B̃′ − B̃) > 0, the band head occurs in the P branch.
Solutions to exercises
E��F.�(a)
�e Gaussian functions are written e−α x �2 and e−α(x−a�2) �2 , where the parameter α determines the width. To evaluate the normalizing factor for the function
2
2
419
11 MOLECULAR SPECTROSCOPY
20
P branch, α = 1.2
P branch, α = 0.8
R branch, α = 1.2
R branch, α = 0.8
10
[ν̃(J)P or R − ν̃]�B̃
420
0
−10
−20
0
2
4
Figure 11.13
6
8
J or (J + 1)
e−α x �2 requires the integral
10
12
2
�
+∞
−∞
e−α x dx
2
which is of the form of Integral G.� with k = α and evaluates to (π�α)1�2 . �e
normalizing factor is therefore N 0 = (α�π)1�4 . �e same factor applies to the
other Gaussian function as this is simply the same Gaussian shi�ed to a�2: the
area under the square of this function is the same.
�e transition moment is given by the integral
I = (α�π)1�2 �
+∞
−∞
+∞
= (α�π)1�2 �
−∞
+∞
= (α�π)1�2 �
= (α�π)
= (α�π)
1�2
1�2
−∞
+∞
�
−∞
+∞
�
−∞
xe−α x �2 e−α(x−a�2) �2 dx
2
2
xe−α[x �2+(x−a�2) �2] dx
2
2
xe−α[x �2+x �2−x a�2+a �8] dx
2
2
2
xe−α[x −x a�2+a �8] dx
2
2
xe−α(x−a�4) e−α a �16 dx
2
2
�e �nal equality above is veri�ed by expanding out the square and recombining the terms. Taking out the constant factor and then writing x as (x − a�4) +
a�4 gives
I = (α�π)1�2 e−α a �16 �
2
+∞
−∞
�(x − a�4)e−α(x−a�4) + (a�4)e−α(x−a�4) � dx
2
2
�e �rst term in the integral is an odd function, and so evaluates to zero. �e
second term is simply a shi�ed Gaussian and, as before, the integral is form of
Integral G.� with k = α and evaluates to (π�α)1�2
I = (α�π)1�2 e−α a �16 (a�4)(π�α)1�2 = (a�4)e−α a �16
2
2
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e ‘width’ w of a Gaussian can be de�ned as the distance between the values
of the coordinates (±w�2) at which the height falls to half its maximum value.
Because the maximum of the function is �, the value of α for a given width is
found by solving
2
1
= e−α(w�2) �2
2
α = (8 ln 2)�w 2
hence
�e exercise speci�es that the Gaussian should have width a, so α = (8 ln 2)�a 2 .
With this the transition moment becomes
(a�4)e−α a �16 = (a�4)e−(8 ln 2)a �16a = (a�4)e−(ln 2)�2
2
E��F.�(a)
2
= (a�4)e−(ln 2
1�2
)
2
= (a�4)(1�21�2 ) = a�(4 × 21�2 )
A simple model for the wavelength of the HOMO–LUMO transition in a polyene is discussed in Example �D.� on page ���. In this model the energy levels
of the π electrons in a polyene are modelled by those of a particle in a onedimensional box of length L. If the polyene consists of n conjugated double
bonds, the length L may be written as L = nd, where d is the length of a single
conjugated bond. A molecule with n conjugated double bonds will have 2n π
electrons which will occupy the energy levels pairwise. �erefore the HOMO
is the level with quantum number n and the LUMO has quantum number n+1.
�e energy of the HOMO–LUMO transition is therefore
E n+1 − E n = [(n + 1)2 − n 2 ]
h2
(2n + 1)h 2
=
8mL 2
8mL 2
For large n, E n+1 − E n goes as n. However, L = nd, therefore overall E n+1 −
E n goes as 1�n, and hence the wavelength of the transition goes as n. �us,
increasing the number of conjugated double bonds will increase the wavelength
of the absorption; that is, shi� it to the red. �e transition at ��� nm is therefore
likely to be from �, and that at ��� nm is likely to be from �.
E��F.�(a)
�e electronic con�guration of H� is σ 2g . �e two electrons are in the same
orbital and so must be spin paired, hence S = 0, and (2S + 1) = 1. Each σ
electron has λ = 0, thus Λ = 0 + 0 = 0, which is represented by Σ. Two electrons
with g symmetry have overall symmetry g × g = g. �e σ orbital is symmetric
with respect to re�ection in a plane containing the internuclear axis, therefore
the two electrons in this orbital are also overall symmetric with respect to this
mirror plane; this is indicated by a right-superscript +. �e term symbol is
therefore 1 Σ+g .
E��F.�(a)
�e electronic con�guration of Li� + is 1σ 2g 1σ 2u 2σ 1g . �e �lled orbitals, 1σ 2g and
1σ 2u , make no contribution to Λ and S, so can be ignored. �erefore, only the
single electron in the 2σ g orbital needs to be considered: it has λ = 0 and
s = 12 , hence Λ = 0 and S = 12 (giving a le� superscript of 2S + 1 = 2). �e
symmetry with respect to inversion is g and with respect to re�ection is +. �e
term symbol is therefore 2 Σ+g .
421
422
11 MOLECULAR SPECTROSCOPY
E��F.�(a)
�e electronic con�guration given is 1σ 2g 1σ 2u 1π 3u 1π 1g . �e �lled orbitals, 1σ 2g
and 1σ 2u , make no contribution to Λ and S, and so can be ignored. With three
electrons in a pair of degenerate π u orbitals, two of the spins must be paired
leaving one unpaired. �ere is another electron in a π g orbital. �ese two
electrons can be paired, giving S = 0 (a singlet, 2S + 1 = 1), or parallel, giving
S = 1 (a triplet, 2S + 1 = 3).
�e three electrons in the π u have overall symmetry with respect to inversion
(parity) u × u × u = g × u = u. �e remaining electron has g symmetry, so overall the state has symmetry u × g = u.
In summary, the multiplicity is � or � , and the parity is u .
E��F.�(a)
E��F.�(a)
(i) Allowed
(ii) Allowed
(iii) No allowed, ∆Σ = 2
(iv) Not allowed, + � −
(v) Allowed
To evaluate the normalizing factor for the function e−ax �2 requires the integral
2
�
+∞
−∞
e−ax dx
2
which is of the form of Integral G.� with k = a and evaluates to (π�a)1�2 . �e
normalizing factor is therefore N 0 = (a�π)1�4 . �e same factor applies to the
2
function e−a(x−x 0 ) �2 as this is simply a Gaussian shi�ed to x 0 : the area under
the square of this function is the same.
�e Franck–Condon factor is given by [��F.�–���] and involves the square of
integral of the product of the two wavefunctions
I = N 02 �
+∞
−∞
= (a�π)
1�2
+∞
e−ax �2 e−a(x−x 0 ) �2 dx = (a�π)1�2 �
2
�
+∞
−∞
+∞
= (a�π)1�2 �
−∞
2
−a[x 2 �2+x 2 �2−x x 0 +x 02 �2]
e
dx
−∞
e−a[x −x x 0 +x 0 �2] dx = (a�π)1�2 �
2
2
e−a[x �2+(x−x 0 ) �2] dx
+∞
−∞
2
2
e−a(x−x 0 �2) e−ax 0 �4 dx
2
2
the �nal equality above is veri�ed by expanding out the square and recombining
the terms. Taking out the constant factors gives
I = (a�π)1�2 e−ax 0 �4 �
2
+∞
−∞
e−a(x−x 0 �2) dx
2
as before, the integral is form of Integral G.� with k = a and evaluates to (π�a)1�2
I = (a�π)1�2 e−ax 0 �4 (π�a)1�2 = e−ax 0 �4
2
2
�e Franck–Condon factor is therefore I 2 = e−ax 0 �2 . As expected this factor is
a maximum of � when x 0 = 0, that is when the two functions are aligned, and
falls o� towards zero as x 0 increases.
2
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E��F.�(a)
�e Franck–Condon factor is given by [��F.�–���] and involves the square of
integral of the product of the two wavefunctions. �e region over which both
wavefunctions are non-zero is from L�4 to L: this is the domain of integration
I = (2�L) �
L
L�4
sin(πx�L) sin(π[x − L�4]�L) dx
applying the identity sin A sin B = 12 [cos(A − B) − cos(A + B)] gives
πx π(x − L�4)
πx π(x − L�4)
−
� − cos �
+
� dx
L
L
L
L
L�4
L
π
2πx π
= (1�L) � cos � � − cos �
− � dx
4
L
4
L�4
I = (1�L) �
L
cos �
Next the identity cos(A − B) = cos A cos B + sin A sin B is used
= (1�L) �
L
L�4
π
2πx
π
2πx
π
cos � � − �cos �
� cos � � + sin �
� sin � �� dx
4
L
4
L
4
Recognising that cos π�4 = sin π�4 = (2)−1�2 allows this factor to be taken
= (1�L)2−1�2 �
L
L�4
1 − cos �
2πx
2πx
� − sin �
� dx
L
L
Each term is now integrated and evaluated between the limits
= (1�L)2−1�2 �x −
L
2πx
L
2πx L
sin �
�+
cos �
��
2π
L
2π
L L�4
= (1�L)2−1�2 �L − L�4 +
L
2πL
2π(L�4)
� − sin �
� + sin �
�
2π
L
L
2πL
2πL�4
� − cos �
� ��
L
L
L
= (1�L)2−1�2 �3L�4 + � − sin(2π) + sin(π�2) + cos(2π) − cos(π�2)��
2π
3 + 4�π
= (1�L)2−1�2 (3L�4 + L�π) = 2−1�2 (3�4 + 1�π) =
4 × 21�2
+ cos �
E��F.�(a)
�e Franck–Condon factor is I 2 = (1�32)(3 + 4�π)2 ; numerically this is �.���.
�e wavenumbers of the lines in the P branch are given in [��F.�–���], ν̃ P (J) =
ν̃ − (B̃′ + B̃)J + (B̃′ − B̃)J 2 . �e band head is located by �nding the value
of J which gives the smallest wavenumber, which can be inferred by solving
dν̃ P (J)�dJ = 0.
d
�ν̃ − (B̃′ + B̃)J + (B̃′ − B̃)J 2 � = −(B̃′ + B̃) + 2J(B̃′ − B̃)
dJ
423
424
11 MOLECULAR SPECTROSCOPY
Setting the derivative to zero and solving for J gives
J head =
B̃′ + B̃
2(B̃′ − B̃)
A band head only occurs in the P branch if B̃′ > B̃.
E��F.��(a) Because B̃′ < B̃ a band head will occur in the R branch .
�e wavenumbers of the lines in the R branch are given in [��F.�–���], ν̃ R (J) =
ν̃ + (B̃′ + B̃)(J + 1) + (B̃′ − B̃)(J + 1)2 . �e band head is located by �nding
the value of J which gives the largest wavenumber, which can be inferred by
solving dν̃ R (J)�dJ = 0.
d
�ν̃ + (B̃′ + B̃)(J + 1) + (B̃′ − B̃)(J + 1)2 � = (B̃′ + B̃) + 2(J + 1)(B̃′ − B̃)
dJ
Setting the derivative to zero and solving for J gives
J head =
−(B̃′ + B̃)
B̃ − 3B̃′
−1=
′
2(B̃ − B̃)
2(B̃′ − B̃)
With the data given
J head =
E��F.��(a)
0.3540 − 3 × 0.3101
= 6.56
2(0.3101 − 0.3540)
Assuming that it is satisfactory simply to round this to the nearest integer the
band head occurs at J = 7 .
�e fact that a band head is seen in the R branch implies that B̃′ < B̃. It is shown
in Exercise E��F.�(b) that the band head in the R branch occurs at
�is rearranges to
J head =
B̃ − 3B̃′
2(B̃′ − B̃)
(��.�)
2J + 1
(��.�)
2J + 3
A band head at J = 1 might arise from a value of J determined from eqn ��.�
anywhere in the range 0.5 to 1.5, followed by subsequent rounding. Using these
non-integer values of J in eqn ��.� gives B̃′ in the range �� cm−1 to �� cm−1 .
B̃′ = B̃ ×
�e bond length in the upper state is longer than that in the lower state (a
longer bond means a larger moment of inertia and hence a smaller rotational
constant).
E��F.��(a) Assuming that the transition corresponds to that between the two sets of d
orbitals which are split as a result on the interaction with the ligands (Section ��F.�(a) on page ���), the energy of the transition is the value of ∆ o . Hence
˜ o = 1�(700 × 10−7 cm) = 1.43 × 104 cm−1 or �.�� eV . �is value is very
∆
approximate as it does not take into account the energy involved in rearranging
the electron spins.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E��F.��(a) A rectangular wavefunction with value h between x = 0 and x = a is normalized if the area under the square of the wavefunction is equal to �: in this case
1 = ah 2 , hence h = a −1�2 . For the wavefunction which is non-zero between
x = a�2 and x = b the height is h ′ = (b − a�2)−1�2 . �e region where the
wavefunctions are both non-zero is x = a�2 to x = a (because b > a); this is
the domain of integration. �e transition moment is
�
a
a�2
P��F.�
1�2
1
=�
�
a(b − a�2)
1�2
=�
Solutions to problems
P��F.�
1
�
a(b − a�2)
ψ i xψ f dx = �
1
�
a(b − a�2)
1�2
�
a
a�2
x dx
� 12 x 2 � a�2 = �
a
1
�
a(b − a�2)
3a 2
3
a3
= �
�
8
8 b − a�2
1�2
1�2
1
(a 2 − a 2 �4)
2
�e �rst transition is not allowed because it violates the spin selection rule,
∆S = 0; this transition has ∆S = 1. �e second transition is not allowed because
it violates the selection rule for Λ, ∆Λ = 0, ±1: this transition has ∆Λ = 2.
Figure ��.�� is helpful in understanding the various quantities involved in this
problem. �e pure electronic energy of the ground electronic state, that is the
energy at the equilibrium separation, is (as a wavenumber) T̃e (X). Likewise,
the pure electronic energy of the excited state is T̃e (B). �e vibrational terms
of the ground state (in the harmonic approximation) are G̃ X (υ X ) = (υ X + 12 )ν̃ X ,
with these terms measured from T̃e (X). Likewise, those of the excited state are
G̃ B (υ B ) = (υ B + 12 )ν̃ B measured from T̃e (B).
�e wavenumber of the �–� transition, ν̃ 00 , is therefore
ν̃ 00 = T̃e (B) − T̃e (X) + G̃ B (0) − G̃ X (0)
= T̃e (B) − T̃e (X) + 12 ν̃ B − 12 ν̃ X
�e di�erence T̃e (B) − T̃e (X) is the value quoted as �.��� eV; this is converted
to a wavenumber using the factor from inside the front cover
8065.5 cm−1 1
+ 2 × (700 cm−1 ) − 12 × (1580 cm−1 )
1 eV
= 4.936 × 104 cm−1
ν̃ 00 = (6.175 eV) ×
P��F.�
(a) �e photoelectron spectrum involves a transition from the ground state
of the molecule to an electronic state of the molecular ion. �e energy
needed for the transition is measured indirectly by measuring the energy
of the ejected electron, but in all other respects the spectrum is interpreted
in the same way as electronic absorption spectra.
425
426
11 MOLECULAR SPECTROSCOPY
T̃e (B)
Figure 11.14
T̃e (X)
ν̃ 00
υ′ = 0
υ=0
In HBr only the ground vibrational state of the ground electronic state
will be signi�cantly populated, so only transitions from this level need
be considered. In principle there can be transitions from this vibrational
level to a range of di�erent vibrational levels of the upper electronic state
(a υ = 0 progression) and the intensities of these transitions will be governed by the Franck–Condon factors. As described in Section ��F.�(c)
on page ���, if the two electronic states have similar equilibrium bond
lengths the υ = 0 → υ′ = 0 transition will be the strongest, and then the
intensity will drop of quickly for higher values of υ′ . On the other hand, if
the upper state is displaced to the le� or right, several transitions will have
signi�cant Franck–Condon factors and several lines in the progression
will be observed.
In the photoelectron spectrum of HBr the band centred at about �� eV
shows extensive structure which is interpreted as being due to several
lines of a vibrational progression. �is is consistent with this band being
due to the removal of a bonding electron, resulting in the upper state (of
the ion) having a signi�cantly longer bond length than the ground state.
P��F.�
(b) �e band at around ��.� eV shows two peaks: these are not due to vibrational �ne structure, but to spin-orbit coupling in the molecular ion,
speci�cally such coupling associated with the Br atom. In the ion there
are unpaired electrons, so spin-orbit coupling in manifested in the spectrum. Removal of a nonbonding electron from a lone pair on the Br is
not expected to give a change in equilibrium bond length, so only the
υ = 0 → υ′ = 0 transition has signi�cant intensity: there is no vibrational
progression.
�e intensity of a transition depends on the transition dipole moment, in this
case given by the integral
�
L
0
ψ n µ̂ x ψ 1 dx = −e(1�L) �
L
0
sin(nπx�L) x sin(πx�L) dx
where the normalized wavefunctions ψ n = (2�L)1�2 sin(nπx�L) are used, and
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
n = 2 or �. �e integral is most easily evaluated by �rst using the given identity
nπx πx
nπx πx
−
� − cos �
+
�� dx
L
L
L
L
0
L
(n − 1)πx
(n + 1)πx
= −e(1�L) � x �cos �
� − cos �
�� dx
L
L
0
I = −e(1�L) �
L
x �cos �
�e integral is of the form of Integral T.�� with a = L
�
�
L2
L2
�
= −e(1�L)�
[cos(n − 1)π − 1] +
sin(n − 1)π
2
2
�
(n − 1) π
(n − 1)π
�
�
�
�
L2
L2
�
−
[cos(n
+
1)π
−
1]
−
sin(n
+
1)π
�
2
2
�
(n + 1) π
(n + 1)π
�
�
For the case n = 2
I 1,2 = −e(1�L) �
L2
L2
16Le
[−1 − 1] − 2 [−1 − 1]� =
2
π
9π
9π 2
I 1,3 = −e(1�L) �
L2
L2
[1
−
1]
+
−
[1 − 1]� = 0
4π 2
16π 2
For the case n = 3
�us, as was to be shown, the transition dipole is non-zero for the transition
1 → 2, but zero for 1 → 3.
�is result is obtained much more simply by rewriting the integral using x =
(x − a�2) + a�2
I = −e(1�L) �
= −e(1�L) �
L
0
L
0
− e(1�L) �
sin(nπx�L) [(x − a�2) + a�2] sin(πx�L) dx
sin(nπx�L) (x − a�2) sin(πx�L) dx
L
0
sin(nπx�L) (a�2) sin(πx�L) dx
�e second integral is zero because the two eigenfunctions sin(nπx�L) and
sin(πx�L) are orthogonal for n ≠ 1. For the �rst integral the integrand is
a product of three functions which can all be classi�ed as odd or even with
respect to x = a�2. For n = 3, sin(nπx�L) is even, (x − a�2) is odd, and
sin(πx�L) is even: the integrand is therefore odd overall, and hence when
integrated over a symmetrical interval the result is necessarily zero.
For n = 2, sin(nπx�L) is odd, (x − a�2) is odd, and sin(πx�L) is even: the integrand is therefore even overall, and hence when integrated over a symmetrical
interval the result is not necessarily zero. What is not shown by this argument
is that the integral is non-zero: however, a quick sketch of the integrand shows
that it is negative everywhere, so the integral is non-zero.
427
11 MOLECULAR SPECTROSCOPY
P��F.�
�e overlap integral for two �s hydrogen (Z = 1) orbitals separated by a distance
R is
R 1 R 2 −R�a 0
S = �1 +
+ � � �e
a0 3 a0
where a 0 is the Bohr radius. �e transition moment, given as −eRS, is therefore
µ = −eR �1 +
hence
−µ�ea 0 =
R 1 R 2 −R�a 0
+ � � �e
a0 3 a0
R
R 1 R 2 −R�a 0
�1 +
+ � � �e
a0
a0 3 a0
Figure ��.�� shows a plot of −µ�ea 0 against R�a 0 . �e maximum occurs at
R�a 0 ≈ 2.1, and the transition moment tends to zero at large distances simply
because the overlap also goes to zero in this limit.
1.0
−µ�ea 0
428
0.5
0.0
0
2
4
6
8
10
R�a 0
Figure 11.15
P��F.��
�e adsorption at ��� nm is due to a π∗ ← π transition of the carbonyl group.
�e weaker adsorption at ��� nm is due to a π∗ ← n transition of the carbonyl
group.
11G Decay of excited states
Answers to discussion questions
D��G.�
Intersystem crossing (ISC) is the process by which the excited singlet state (S1 )
makes a radiationless transition to a triplet state, T1 . Following this process
spontaneous emission may occur from T1 down to the ground electronic state
– this is phosphorescence. Both ISC and the phosphorescent transition are spin
forbidden and so only occur relatively slowly, and to the extent that the spin
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
selection rule is broken. Spin-orbit coupling is one of the e�ects that leads to
this rule being broken, and it is observed that the rates of these spin forbidden
transitions is enhanced by the presence of heavy atoms which have signi�cant
spin orbit coupling.
In the present case, if the added iodide ion is able to interact with the chromophore for a signi�cant period of time (a time comparable with ISC or phosphorescence) then the spin-orbit coupling in the transient species may increase
the rate of ISC and/or of the phosphorescence, leading to an increase in the
intensity of the latter.
D��G.�
�is is described in Section ��G.� on page ���.
D��G.�
�e characteristics of �uorescence which are consistent with the accepted mechanism are: (�) it ceases as soon as the source of illumination is removed; (�)
the time scale of �uorescence, 10−9 s, is typical of a process in which the rate
determining step is a spontaneous radiative transition between states of the
same multiplicity; (�) it occurs at longer wavelength (lower frequency) than
the exciting radiation; (�) its vibrational structure is characteristic of that of a
transition from the ground vibrational level of the excited electronic state to the
vibrational levels of the ground electronic state; and (�) the observed shi�ing,
and in some instances quenching, of the �uorescence spectrum by interactions
with the solvent.
Solutions to exercises
E��G.�(a)
�is observed increase in the linewidth is a result of predissociation, as illustrated in Fig. ��G.� on page ���. Where the dissociative 5 Рu state crosses
the bound upper electronic state the possibility exists that molecules in the
upper electronic state will undergo radiationless transitions to the dissociative
state leading to subsequent dissociation. �is process reduces the lifetime of
the excited states and so increases the linewidth of the associated transitions
(lifetime broadening, see Section ��A.�(b) on page ���).
E��G.�(a)
(i) �e vibrational �ne structure of the �uorescent transition is determined
by the vibrational energy levels of the ground electronic state because the
transitions observed are from the ground vibrational level of the upper
electronic state to various vibrational levels of the ground electronic state.
(ii) No information is available about the vibrational levels of the upper electronic state because the spectrum only shows transitions from the ground
vibrational level of this state.
Solutions to problems
P��G.�
�e anthracene �uorescence spectrum re�ects the vibrational levels of the lower
electronic state. �e given wavelengths correspond to wavenumbers 22 730 cm−1 ,
24 390 cm−1 , 25 640 cm−1 , and 27 030 cm−1 ; these indicate spacings of the
vibrational levels of (24 390 cm−1 ) − (22 730 cm−1 ) = 1660 cm−1 , and, by
similar calculations, 1250 cm−1 , and 1390 cm−1 .
429
430
11 MOLECULAR SPECTROSCOPY
�e vibrational �ne structure in the absorption spectrum re�ects the vibrational levels of the upper electronic state. �e wavenumbers of the absorption
peaks are 27 800 cm−1 , 29 000 cm−1 , 30 300 cm−1 , and 32 800 cm−1 . �e vibrational spacings are therefore 1200 cm−1 , 1300 cm−1 , and 2500 cm−1 .
�e fact that the �uorescent transitions are all at longer wavelength (lower
energy) that the absorption transitions is consistent with the loss of vibrational
energy (by collision induced radiationless decay) a�er the initial excitation of
the molecule. �e vibrational �ne structure in the absorption and �uorescence
spectra do not mirror one another, suggesting that the bonding in the ground
and excited electronic states are dissimilar.
P��G.�
(a) �e resonant modes satisfy nλ�2 = L therefore λ = 2L�n and ν = nc�2L.
With the data given
ν=
n × (2.9979 × 108 m s−1 )
= n × 150 MHz
2 × (1.00 m)
(b) �e spacing of the modes is therefore ��� MHz .
P��G.�
�e peak power is energy/(duration of pulse)
Ppeak =
0.10 J
= �� MW
3.0 × 10−9 s
where 1 W = 1 J s−1 is used. Consider a total time of � s: because the repetition
frequency is �� Hz, there will be �� pulses in this time.
Pav =
total energy 10 × (0.10 J)
=
= �.� W
total time
1s
�e average power is very much less than the peak power.
Integrated activities
I��.�
�e electronic ground state of a closed-shell molecule transforms as the totally
symmetric irreducible representation. �is is because all of the orbitals are
doubly occupied by spin paired electrons. �e overall symmetry of a �lled
orbital is found from the direct product Γ(i) × Γ(i) , where Γ(i) is the irreducible
representation as which the orbital transforms. When multiplied out, such a
product always includes the totally symmetric irreducible representation and it
is this symmetry species which needs to be combined with the anti-symmetric
spin function for two spin-paired electrons.
(a) Ethene belongs to the point group D 2h : assume that the molecule lies
in the x y-plane, with the C–C bond along x. �e π bonding molecular
orbital transforms in the same way as the cartesian function z, that is as
B1u . �e π∗ anti-bonding molecular orbital transforms in the same way
as the cartesian function xz, that is as B2g . �e excited state (π)1 (π∗ )1
therefore has symmetry B1u × B2g = B3u (the direct product is found by
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
multiplying together the characters of the two irreducible representations,
as described in Section ��C.�(a) on page ���).
�e transition moment is given by ∫ ψ f ∗ µ̂ψ i dτ, and the symmetry species
of the integrand is found using the direct product; µ̂ transforms as x, y, or
z, which in this case is B3u , B2u , or B1u . �e integral is only non-zero if the
integrand transforms as the totally symmetric irreducible representation,
which is the case here when the component of the dipole is along x
� � �
B3u × B3u × Ag = Ag
ψf
ψi
µ̂ x
�us the π → π∗ transition is symmetry allowed , with the transition
dipole along x .
(b) A carbonyl group (as exempli�ed by that in methanal) is assumed to belong to the point group C 2v : assume that the H� CO fragment lies in the
xz-plane, with the C–O bond along z. �e π∗ anti-bonding molecular
orbital transforms in the same way as the cartesian function y, that is as
B2 .
A non-bonding electron on oxygen is usually considered to be in a �px
orbital which transforms as B1 . When an electron is promoted from n to
π∗ , the excited state has symmetry B2 × B1 = A2 . �e symmetry of the
integrand for the transition moment is
µ̂
� � �
A2 × Γx , y,z × A1 = A2 × Γx , y,z
ψf
ψi
Because Γx , y,z ≠ A2 this product is never A1 , so the integral is zero and
the transition is forbidden .
I��.�
�e energy of the HOMO, E HOMO , is reported in the table below, based on
calculations performed with Spartan �� using the DFT/B�LYP/�-��G* method.
�e experimentally determined energy of the I� –aromatic hydrocarbon charge
transfer bands is also given.
Figure ��.�� shows a plot of E HOMO against hν max ; the best-�t straight line is
also shown. �ere is a modest correlation between the two quantities.
hydrocarbon
hν max /eV
benzene
�.���
biphenyl
�.���
naphthalene
�.���
phenanthrene
�.���
pyrene
�.���
anthracene
�.���
E HOMO �eV
−6.70
−5.91
−5.78
−5.73
−5.33
−5.23
431
11 MOLECULAR SPECTROSCOPY
−5.5
E HOMO �eV
432
−6.0
−6.5
2.8
3.0
3.2
3.4 3.6 3.8
hν max �eV
Figure 11.16
4.0
4.2
R�2
d
R
Figure 11.17
I��.�
R�2
(a) �e geometry of the molecule is illustrated in Fig. ��.��.
�e moment of inertia about the axis parallel to the symmetry axis, that
is the axis passing through the middle of the molecule and perpendicular
to the plane of the molecule, is given by
I�� = � m i r 2i = 3m H d 2
i
where r i is the perpendicular distance from the√axis to mass m i . From
√
the diagram is follows that (R�2)�d = cos 30○ = 3�2, hence d = R� 3
√
I�� = 3m H (R� 3)2 = m H R 2
�e moment of inertia perpendicular to the symmetry axis is independent of where this axis is located provided it passes through the centre of
the molecule and lies in the plane of the molecule: one such convenient
choice is the dashed line shown in the �gure.
As expected, I�� = 2I� .
I� = 2m H (R�2)2 = 12 m H R 2
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(b) �e moment of inertia is related to the rotational constant through [��B.�–
���], B̃ = ħ�4πcI� ; note that the rotational constant B̃ is always associated
with the moment of inertia perpendicular to the symmetry axis. It follows
that I� = ħ�4πc B̃ and hence, using the result above, 12 m H R 2 = ħ�4πc B̃.
�is rearranges to give the following expression for R
R=�
=�
ħ
�
2πm H c B̃
1�2
1.0546 × 10−34 J s
�
−27
2π(1.67... × 10 kg) × (2.9979 × 1010 cm s−1 ) × (43.55 cm−1 )
1�2
= ��.�� pm
where the mass in kg is given by (1.0079 m u )×(1.6605×10−27 kg)�(1 m u )
= 1.67... × 10−27 kg. An alternative value for R is found from the other
rotational constant. Here C̃ = ħ�4πcI�� and hence m H R 2 = ħ�4πc C̃.
R=�
=�
ħ
�
4πm H c C̃
1�2
1.0546 × 10−34 J s
�
−27
4π(1.67... × 10 kg) × (2.9979 × 1010 cm s−1 ) × (20.71 cm−1 )
1�2
= ��.�� pm
(c) With the given value of R, the rotational constant is computed from B̃ =
ħ�4πcI� = ħ�2πm H cR 2
B̃ =
1.0546 × 10−34 J s
2π(1.67... × 10−27 kg) × (2.9979 × 1010 cm s−1 ) × (87.32 × 10−12 m)2
= ��.�� cm−1
�e other rotational constant is just half of this, C̃ = 21.94 cm−1
−1�2
(d) For a harmonic oscillator the vibrational frequency goes as m eff , where
m eff is the e�ective mass. �e value of the e�ective mass depends on the
vibration in question, but in this case as all the atoms are the same it is
reasonable to assume that the e�ective mass of H� + will be proportional
to m H , and that of D� + will be proportional to m D . It therefore follows
that
ν̃ D = ν̃ H �
m H 1�2
�
mD
= (2521.6 cm−1 ) × �
1.0079 1�2
� = ����.� cm−1
2.0140
�e rotational constants are inversely proportional to the moment of inertia, and for this molecule all the atoms are the same so the rotational
433
434
11 MOLECULAR SPECTROSCOPY
constant goes as m−1 .
B̃ D = B̃ H
mH
mD
= (43.55 cm−1 ) ×
I��.�
1.0079
= ��.�� cm−1
2.0140
A similar calculation gives C̃ D = 10.36 cm−1 .
−1�2
(a) For a harmonic oscillator the vibrational frequency goes as m eff , where
m eff is the e�ective mass; for a diatomic the e�ective mass is simply 12 m.
It therefore follows that
ν̃ 18O2 = ν̃ 16O2 �
m16 O 1�2
�
m18 O
= (844 cm−1 ) × �
15.9949 1�2
� = ��� cm−1
17.9992
(b) �e bond order of O� is �, and to form the anions electrons are added
to the anti-bonding π g molecular orbital, thereby decreasing the bond
order to �.� for O� – and � for O� � – . �e steady decrease in the bond order
is matched by the steady decrease in the vibrational frequency. �ere is
thus a correlation between bond strength and vibrational frequency, as
expected.
(c) �e observed vibrational frequency of O� bound to haemerythrin most
�–
closely matches that for O� � – , so of the alternatives o�ered Fe�+
� O�
seems the most likely.
(d) �e observation of two bands attributable to O–O stretching implies that
the O� is bound is such a way that the two oxygen atoms are no longer
equivalent. If this is the case, when the isotopologue ��O ��O is used,
two di�erent frequencies will result because the two ends of the oxygen
molecule are now distinguished. �is eliminates structures � and �.
I��.�
Expressed in terms of the absorbance A the Beer–Lambert law is given by [��A.�c–
���], A = ε[J]L. To convert the given absorbance into a molar absorption
coe�cient requires [J] in mol dm−3 , which is computed using the perfect gas
law, pV = nRT, rearranged to n�V = p�RT. Assuming that the quoted composition of CO� , ‘�.� per cent’, refers to a mole per cent, x CO2 = 0.021, so
p CO2 = 0.021×(1.00 bar) = 0.021 bar. In the calculation of n�V it is convenient
to use R = 8.3145 × 10−2 dm3 bar K−1 mol−1
n
p
0.021 bar
=
=
V RT (8.3145 × 10−2 dm3 bar K−1 mol−1 ) × (298 K)
= 8.47... × 10−4 mol dm−3
�e given expression for A(ν̃) is therefore converted to an expression for the
molar absorption coe�cient using
ε(ν̃) = A(ν̃)�[J]L = A(ν̃)�[(8.47... × 10−4 mol dm−3 ) × (10 cm)]
= (1.17... × 102 mol−1 dm3 cm−1 ) × A(ν̃)
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
2.0
200
A(ν̃)
1.5
1.0
100
0.5
0.0
ε(ν̃)�mol−1 dm3 cm−1
(a) Graphs of A(ν̃) and ε(ν̃) are shown in Fig. ��.��. �is band is likely to be
due to the fundamental transition of the antisymmetric stretching normal
mode of CO� , which has ν̃ 2 = 2349 cm−1 . �e broad features are the
unresolved P and R branches; no Q branch is expected for this mode. �e
principal contribution to the linewidth of an infrared transition is likely
to be pressure broadening.
2 300
Figure 11.18
2 350
ν̃�cm
2 400
−1
0
2 300
2 350
−1
2 400
ν̃�cm
(b) Expressions for the wavenumbers of the lines in the P and R branches are
given in [��C.��a–���] and [��C.��c–���]
ν̃ P (J) = ν̃ − 2B̃J
ν̃ R (J) = ν̃ + 2B̃(J + 1)
Here ν̃ = 2349 cm−1 . As described in Section ��B.� on page ���, because
��
O is a boson, only even rotational states are occupied (and, in fact, odd
rotational states in the �rst excited vibrational state of the anti-symmetric
stretch), so in the above expression J takes the values 0, 2, 4 . . .. �e
rotational constant is computed from [��B.�–���], B̃ = ħ�4πcI, where
I = 2m O R 2 . With the data given
I = 2 × (15.9949 m u )×(1.6605 × 10−27 kg)×(116.2 × 10−12 m)2 �(1 m u )
= 7.17... × 10−46 kg m2
B̃ =
1.0546 × 10−34 J s
4π×(2.9979 × 1010 cm s−1 ) × (7.17... × 10−46 kg m2 )
= 0.3903 cm−1
�e intensity of the transition from level J is proportional to the population of that level, taking into account the degeneracy
intensity ∝ (2J + 1)e−hc B̃�k T
Using these expressions the positions and intensities of the lines in the P
and R branches are computed and a synthetic spectrum constructed by
assuming a linewidth and lineshape (here arbitrarily taken as a Gaussian).
Two such spectra are shown in Fig. ��.��: in (a) the linewidth has been
chosen so that the lines are well resolved; in (b) a much wider line is used
so that the lines in the P and R branches merge into a broad contour. �e
spectrum in (b) is roughly similar to that in Fig. ��.��, but the asymmetry
between the two branches is not reproduced by the calculation.
435
436
11 MOLECULAR SPECTROSCOPY
(a)
2300
(b)
2320
2340
2360
~
v / cm−1
2380
2400 2300
2320
2340
2360
~
v / cm−1
2380
2400
Figure 11.19
(c) �e transmittance T is de�ned as I�I 0 and hence from the Beer–Lambert
law [��A.�–���], T = e−ε[J]L ; it follows that log T = −A. �e concentration
of CO� is computed as in (a) using [CO2 ] = x CO2 p atm �RT; both the
atmospheric pressure and temperature vary with the height, and therefore
so will the concentration. �e pressure varies with height according to
p(h) = p 0 e−h�H , where the scale height H is about ���� m. �e total
absorbance up to height h 0 is therefore given by integral
A=�
h0
0
= ε�
where h is in m.
ε[CO2 ] dh = ε �
h0
0
0
−h�H
h0 x
CO 2 p
RT
dh
x CO2 p 0 e
dh
R(288 − 0.0065h)
= εx CO2 p 0 �
h0
0
e−h�H
dh
R(288 − 0.0065h)
�is integral cannot be solved by hand, but some work with mathematical
so�ware and a typical value of ε shows that the absorbance exceeds � by
the time h 0 = 30 m. Over such a small height it is safe to assume that the
pressure and temperature are constant, in which case the calculation of
the absorbance is much simpler
x CO2 p
h0
RT
= [ε�(mol−1 dm3 cm−1 )]
A = ε[CO2 ]h 0 = ε
×
(3.3 × 10−4 ) × (1 bar)
× (h 0 �cm)
(8.3145 × 10−2 dm3 bar K−1 mol−1 ) × (288 K)
= (1.378 × 10−5 ) × [ε�(mol−1 dm3 cm−1 )] × (h 0 �cm)
�e transmittance is T = 10−A .
Figure ��.�� shows plots of the transmittance as a function of height for
some representative values of the molar absorption coe�cient. For the
maximum value of ε seen in this absorption band (refer to Fig. ��.��) the
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
transmittance drops to �.� a�er less than � m. For a value of ε more typical
of the average (say ε = 100 mol−1 dm3 cm−1 ), the transmission drops to
�.� a�er about �.� m. Even for values of ε typical of the extremities of
the band, the transmission has fallen to �.� within �� m or so. A surface
plot of the transmission as a function of both wavenumber and height is
shown in Fig. ��.��.
1.0
ε = 200 mol−1 dm3 cm−1
ε = 100
ε = 50
ε = 25
0.8
T
0.6
0.4
0.2
0.0
0
5
Figure 11.20
10
15
h 0 �m
20
25
30
~
v /cm−1
h0 /m
Figure 11.21
I��.��
In Problem P��A.� it is shown that the integrated absorption coe�cient for a
Gaussian line shape is A = 1.0645 ε max ∆ν̃ 1�2 , where ∆ν̃ 1�2 is the width at half
height. Interpolating, by eye, a smooth curve across the band centred at about
��� nm, gives ε max = 250 mol−1 dm3 cm−1 . �e molar absorbance drops to half
437
438
11 MOLECULAR SPECTROSCOPY
this value at about ��� nm, which is 3.70×104 cm−1 and at about ��� nm, which
is 3.23 × 104 cm−1 , giving a width of ���� cm−1 . �e integrated absorption
coe�cient is therefore
A = 1.0645 × (250 mol−1 dm3 cm−1 ) × (4700 cm−1 )
= 1.25 × 106 mol−1 dm3 cm−2
�e transition moment is given by ∫ ψ ∗f µ̂ψ i dτ; µ̂ transforms as x, y, or z, which
in this case is B1 , B2 , or A1 . �e integral is only non-zero if the integrand transforms as the totally symmetric irreducible representation, which is determined
by computing the direct product
µ̂ x
� � �
Γf × Γx , y,z × A1 = Γf × Γx , y,z
ψf
ψi
where that fact that the direct product with the totally symmetric irreducible
representation has no e�ect is used. �e only way this product can contain
the totally symmetric irreducible representation is if the irreducible representation of the �nal state, Γf , is equal to the irreducible representation of Γx , y,z .
�us, transitions from the A1 ground state to A1 , B1 , or B2 excited states are
allowed.
12
12A
Statistical
thermodynamics
The Boltzmann distribution
Answers to discussion questions
D��A.�
In terms of molecular energy levels the thermodynamic temperature is the one
quantity that determines the most probable populations of the states of the
system at thermal equilibrium, as discussed in Section ��A.�(b) on page ���.
�e equipartition theorem allows a connection to be made between the temperature as understood in statistical thermodynamics and the empirical concept
of temperature which arises in classical thermodynamics. Temperature is a
measure of the intensity of thermal energy, and is directly proportional to the
mean energy for each quadratic contribution to the energy (provided that the
temperature is su�ciently high).
D��A.�
�e population of a state is the number of molecules in a sample that are in
that state. �e con�guration of a system is a list of populations in order of the
energy of the corresponding states. For example, {N−3, 2, 1, 0, . . .} is a possible
con�guration of a system of N molecules in which all but three molecules are
in the ground state, two are in the next highest state, and one in the state above
that. �e weight of a con�guration is the number of ways a given con�guration
can be achieved, and is given by [��A.�–���]. When N is large (as it is for
any macroscopic sample), the most probable con�guration has a much greater
weight, that is it is more probable, than any other con�guration. Under such
circumstances it can be assumed that the con�guration adopted by the system
is this most probable con�guration.
Solutions to exercises
E��A.�(a)
�e Boltzmann population ratio is given by [��A.��a–���], N i �N j = e−β(ε i −ε j ) .
�is is rearranged to β = − ln(N i �N j )�∆ε , where ∆ε = (ε i − ε j ). Substituting
440
12 STATISTICAL THERMODYNAMICS
β = 1�(kT) and rearranging for T gives
∆ε
hc ν̃
=−
k ln (N 1 �N 0 )
k ln (N 1 �N 0 )
−34
(6.6261 × 10 J s) × (2.9979 × 1010 cm s−1 ) × (400 cm−1 )
=−
(1.3806 × 10−23 J K−1 ) × ln(1�3)
= 524 K
T =−
E��A.�(a) �e Boltzmann population ratio for degenerate energy levels is given by [��A.��b–
���], N i �N j = (g i �g j )e−β(ε i −ε j ) . �e rotational term of a linear rotor is given
by [��B.��–���], F̃(J) = B̃J(J + 1) and, as explained in Section ��B.�(c) on page
���, its degeneracy is given as g J = 2J + 1. �e rotational energy is related to
the rotational term as ε J = hc F̃(J). �erefore
N5 2 × 5 + 1
=
× e−hc B̃[5×(5+1)−0×(0+1)]�k T = 11 × e−30 B̃hc�k T
N0 2 × 0 + 1
using kT�hc = 207.224 cm−1 at 298.15 K (from inside the front cover)
−1
−1
N5
= 11 × e−30×(2.71 cm )�(207.224 cm ) = �.��
N0
E��A.�(a) �e Boltzmann population ratio is given by [��A.��a–���], N i �N j = e−β(ε i −ε j ) .
�is is rearranged to β = − ln(N i �N j )�∆ε , where ∆ε = (ε i − ε j ). Substituting
β = 1�(kT) and rearranging for T gives
∆ε
hc ν̃
=−
k ln (N 1 �N 0 )
k ln (N 1 �N 0 )
−34
(6.6261 × 10 J s) × (2.9979 × 1010 cm s−1 ) × (540 cm−1 )
=−
(1.3806 × 10−23 J K−1 ) × ln(10%�90%)
= 354 K
T =−
E��A.�(a) �e weight of a con�guration is given by [��A.�–���], W = N!�(N 0 !N 1 !N 2 !...),
thus
16!
W=
= 21 621 600
0! × 1! × 2! × 3! × 8! × 0! × 0! × 0! × 0! × 2!
E��A.�(a)
(i) 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40 320 .
(ii) Stirling’s approximation for x � 1 is given by [��A.�–���], ln (x!) ≈
x ln x − x. �is is rearranged to x! ≈ e(x ln x−x) , thus
8! ≈ e(8×ln 8−8) = 5.63 × 103 .
(iii) Using the more accurate version of Stirling’s approximation
8! ≈ (2π)(1�2) × 8(8+1�2) × e−8 = 3.99 × 104
E��A.�(a) �e Boltzmann population ratio is given by [��A.��a–���], N i �N j = e−β(ε i −ε j ) ,
where β = 1�(kT). At in�nite temperature β becomes zero, therefore the
relative populations of two levels N 1 �N 0 = e−0 = � .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Solutions to problems
P��A.�
(a) �ere is no con�guration in which the molecules are distributed evenly
over the states and which, at the same time, satis�es the constraint that
the energy is 5ε.
(b) �e energy of a con�guration is E�ε = N 1 + 2N 2 + 3N 3 . . ., and the weight
of a con�guration is given by [��A.�–���], W = N!�(N 0 !N 1 !N 2 !...). �e
con�gurations satisfying the total energy constraint E = 5ε are
N0
�
�
�
�
�
�
�
N1
�
�
�
�
�
�
�
N2
�
�
�
�
�
�
�
N3
�
�
�
�
�
�
�
N4
�
�
�
�
�
�
�
N5
�
�
�
�
�
�
�
W
�
��
��
��
��
��
�
Hence the most probable con�gurations are {�,�,�,�,�,�} and {�,�,�,�,�,�} .
P��A.�
�e energy of a con�guration is E�ε = N 1 + 2N 2 + 3N 3 . . ., and the weight of a
con�guration is given by [��A.�–���], W = N!�(N 0 !N 1 !N 2 !...). �ere are a very
large number of possible con�gurations, but one way of selecting an interesting
and diverse set of these is to consider con�gurations in which N 0 is ��, then ��,
then �� and so on.
N 0 N 1 N 2 N 3 N 4 N 5 N 6 N 7 N 8 N 9 N 10 W
�� � � � � � � � � �
� ��
�� � � � � � � � � �
� ���
�� � � � � � � � � � � � ���
�� � � � � � � � � � � ��� ���
�� � � � � � � � � � � ��� ���
�� � � � � � � � � � � � ��� ���
�� � � � � � � � � � � � ��� ���
�� � � � � � � � � � � � ��� ���
�� � � � � � � � � � � � ��� ���
�� �� � � � � � � � � � ��� ���
Of the ones listed in the table the con�guration with the greatest weight is
{12, 6, 2, 0, 0, 0, 0, 0, 0, 0, 0} .
�e Boltzmann population ratio is given by [��A.��a–���], N i �N 0 = e−βε i ,
where it is assumed that ε 0 = 0. It follows that ln N i �N 0 = −βε i = −iβε, and
hence βε = (ln N i �N 0 )�(−i). For i = 1, βε = (ln 6�12)�(−1) = 0.693; for
i = 2, βε = (ln 2�12)�(−2) = 0.896. Taking an average of these two values gives
βε = 0.795 and hence T = ε�(0.795k) .
441
442
12 STATISTICAL THERMODYNAMICS
With this value for the temperature the populations predicted by the Boltzmann distribution are
N i �N 0 = e−βε i = e−i βε = e−0.795×i
P��A.�
�erefore N 1 �N 0 = 0.452, N 2 �N 0 = 0.204, N 3 �N 0 = 0.092. For the most
probable distribution given above these ratios are N 1 �N 0 = 0.500, N 2 �N 0 =
0.167, N 3 �N 0 = 0, which are roughly comparable.
�e Boltzmann population ratio for degenerate energy levels is given by [��A.��b–
���], N i �N j = (g i �g j )e−β(ε i −ε j ) . Taking logarithms gives
ln(N i �N j ) = ln(g i �g j ) − β(ε i − ε j ) hence
β = − ln[(N i �N j )(g j �g i )]�∆ε
where ∆ε = (ε i − ε j ). Substituting β = 1�(kT) and rearranging for T gives
∆ε
hc ν̃
=−
k ln[(N 1 �N 0 )(g 0 �g 1 )]
k ln [(N 1 �N 0 )(g 0 �g 1 )]
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (450 cm−1 )
=−
(1.3806 × 10−23 J K−1 ) × ln [(30%�70%) × (2�4)]
= 420 K
T =−
�e populations of the electronic states do not correspond to the translational
temperature; therefore the electronic states are not in equilibrium with the
translational states.
P��A.�
�e Boltzmann population ratio is given by [��A.��a–���], N i �N j = e−β(ε i −ε j ) .
�e energy ε i is interpreted as that at height h, and ε j is interpreted as that at
height �, giving
N(h)�N(0) = e−β(m g h−m g×0) = e−βm g h
From the perfect gas law, pV = nRT, and because N ∝ n, it follows that p ∝ N,
and therefore p(h)�p 0 = N(h)�N(0). Because β = 1�(kT), m = M�N A , and
k = R�N A it follows that
βmgh =
mgh (M�N A )gh M gh h
=
=
=
kT
(R�N A )T
RT
H
where H = RT�M g is used. Hence
p(h)�p 0 = e−h�H
and so
p(h) = p 0 e−h�H
From the perfect gas law N = N�V ∝ p, therefore for O�
M gh
�
RT
(32.00 × 10−3 kg mol−1 ) × (9.807 m s−2 ) × (8.0 × 103 m)
= exp �−
�
(8.3145 J K−1 mol−1 ) × (298 K)
= 0.36
N(h)�N0 = exp �−
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
and for H� O
N(h)
M gh
= exp �−
�
N0
RT
(18.02 × 10−3 kg mol−1 ) × (9.807 m s−2 ) × (8.0 × 103 m)
= exp �−
�
(8.3145 J K−1 mol−1 ) × (298 K)
= 0.57
In these calculations the temperature is taken as ��� K and is assumed to be
constant with height, which is in fact not the case.
12B Partition functions
Answer to discussion questions
D��B.�
As discussed in Section ��B.�(b) on page ���, the symmetry number is the
number of indistinguishable orientations that the molecule can be rotated into.
�is factor is needed in order to avoid counting contributions to the rotational
partition function which are forbidden by symmetry considerations arising
from the e�ects of nuclear spin (Section ��B.� on page ���).
If the partition function is computed term-by-term then the symmetry number
is not needed because those terms which are forbidden are simply omitted.
However, in the high-temperature limit in which many terms are included, it
is convenient to allow all terms to contribute to the sum and then compensate
for those which should not have been included by division by the symmetry
number.
D��B.�
It is possible for there to be di�erent wavefunctions which have the same energy: such wavefunctions are said to be degenerate. If this is the case, for a
given ‘energy level’, that is a given value of the energy, there are several ‘states’
each of which is distinct but has the same energy.
�e partition function is computed as a sum over the states. However, because
degenerate states have the same energy, the sum may be computed as a sum over
energy levels, as long as the degeneracy g i of each level is taken into account.
q = � e−βε i = � g i e−βε i
states i
levels i
Solutions to exercises
E��B.�(a)
�e rotational partition function of a heteronuclear diatomic is given by [��B.��–
���], q R = ∑ J (2J + 1)e−β hc B̃ J(J+1) . �is is evaluated explicitly by summing
successive terms until they become too small to a�ect the result to a given level
of precision. �e partition function in the high-temperature limit is given by
443
12 STATISTICAL THERMODYNAMICS
[��B.��a–���], q R = kT�hc B̃. For the data given it follows that
k×T
(1.3806 × 10−23 J K−1 ) × T
=
−34
(6.6261 × 10 J s) × (2.9979 × 1010 cm s−1 ) × (1.931 cm−1 )
hc B̃
= (0.359... K−1 ) × T
qR =
�e values of q R computed in these two di�erent ways are compared in Fig. ��.�.
�e high temperature limit becomes accurate to within � % of the exact solution
at around 18 K .
10
qR
444
5
0
exact
high T limit
0
5
10
15
T�K
20
25
30
Figure 12.1
E��B.�(a)
�e partition function is given by [��B.�b–���], q R = ∑ J g J e−βε J , where the
degeneracy is given as g J = (2J + 1)2 , as explained in Section ��B.�(c) on page
���, and ε J is given by [��B.�b–���], ε J = hc B̃J(J+1). �is is evaluated explicitly
by summing successive terms until they become too small to a�ect the result
to a given level of precision.
�e partition function in the high-temperature limit is given by [��B.��b–���],
q R = (kT�hc)3�2 (π�ÃB̃ C̃)1�2 = π 1�2 (kT�hc B̃)3�2 , because for a spherical rotor
B̃ = Ã = C̃. Ignoring the role of the nuclear spin means that all J states are
accessible and have equal weight. For the data given it follows that
q R = π 1�2 �
= π 1�2 �
k × T 3�2
�
hc B̃
(1.3806 × 10−23 J K−1 )
� × T 3�2
−34
(6.6261 × 10 J s)×(2.9979 × 1010 cm s−1 )×(5.241 cm−1 )
= (0.0855... K−3�2 ) × T 3�2
3�2
�e values of q R computed in these two di�erent ways are compared in Fig. ��.�.
�e high temperature limit becomes accurate to within � % of the exact solution
at around 37 K .
E��B.�(a)
(i) CO is a heteronuclear diatomic and so has σ = 1 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
qR
30
20
10
0
exact
high T limit
0
10
20
30
T�K
40
50
Figure 12.2
(ii) O� is a homonuclear diatomic and so has σ = 2 .
(iii) H� S has a twofold rotational axis bisecting the H–S–H angle; rotation
about this axis interchanges two identical hydrogen atoms, therefore σ = 2 .
(iv) SiH� is tetrahedral and so has the same symmetry number as CH� : σ = 12 .
(v) CHCl� has a threefold rotational axis along the C–H bond; rotation about
this axis interchanges three identical chlorine atoms, therefore σ = 3 .
E��B.�(a) �e rotational partition function of an asymmetric rotor is given by [��B.��–
���], q R = (1�σ)(kT�hc)3�2 (π�ÃB̃C̃)1�2 , where σ is the symmetry number.
For a molecule with high symmetry the simplest was to determine the symmetry number is to count the total number of rotational symmetry operations, C n ,
listed in the character table of the point group to which the molecule belongs,
which in this case is D 2h . For this group the rotational operations are (E, C 2x ,
y
C 2 , C 2z ), and therefore σ = 4.
qR =
=
E��B.�(a)
1 kT 3�2
π 1�2
� � �
�
4 hc
ÃB̃C̃
1
(1.3806 × 10−23 J K−1 ) × (298.15 K)
�
�
4
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
3�2
π
�
�
(4.828 cm−1 ) × (1.0012 cm−1 ) × (0.8282 cm−1 )
1�2
= ���.�
�e vibrational partition function is given by [��B.��–���], q V = 1�(1−e−β hc ν̃ ),
where β = 1�kT. �e high-temperature approximation is given by [��B.��–
���], q V ≈ kT�hc ν̃.
k×T
(1.3806 × 10−23 J K−1 ) × T
=
hc ν̃
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (323.2 cm−1 )
= (2.15... × 10−3 K−1 ) × T
445
12 STATISTICAL THERMODYNAMICS
�e values of q V computed using these two di�erent expressions are compared
in Fig. ��.�. �e high temperature limit becomes accurate to within � % of the
exact solution at 4500 K .
10
qV
446
5
0
exact
high T limit
0
1 000
2 000
3 000
T�K
4 000
5 000
Figure 12.3
E��B.�(a) �e vibrational partition function for each mode is given by [��B.��–���], q V =
1�(1 − e−β hc ν̃ ), where β = 1�kT. �e overall vibrational partition function
is the product of the partition functions of the individual modes; the bend is
included twice as it is doubly degenerate.
hcβ =
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
(1.3806 × 10−23 J K−1 ) × (500 K)
= 2.87... × 10−3 cm
−1
q V1 = �1 − e−hc β ν̃ 1 �
Similarly
−3
cm)×(658 cm−1 )
�
q V2 = �1 − e−(2.87 ...×10
−3
cm)×(397 cm−1 )
�
q V3 = �1 − e−(2.87 ...×10
E��B.�(a)
−1
= �1 − e−(2.87 ...×10
−3
cm)×(1535 cm−1 )
−1
= 1.17...
= 1.46...
−1
�
= 1.01...
q V = q V1 × (q V2 )2 × q V3 = (1.17...) × (1.46...)2 × (1.01...) = 2.57
�e vibrational partition function for each mode is given by [��B.��–���], q V =
1�(1−e−β hc ν̃ ), where β = 1�kT. �e overall vibrational partition function is the
product of the partition functions of the individual modes, taking into account
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
the stated degeneracies.
hcβ =
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
(1.3806 × 10−23 J K−1 ) × (500 K)
= 2.87... × 10−3 cm
−1
q V1 = �1 − e−hc β ν̃ 1 �
Similarly
−1
= �1 − e−(2.87 ...×10
−3
cm)×(459 cm−1 )
�
q V2 = �1 − e−(2.87 ...×10
−3
cm)×(217 cm−1 )
�
q V4 = �1 − e−(2.87 ...×10
−3
cm)×(314 cm−1 )
�
q V3 = �1 − e−(2.87 ...×10
−3
−1
cm)×(776 cm )
q V = q V1 × (q V2 )2 × (q V3 )3 × (q V4 )3
−1
−1
�
−1
= 1.36...
= 2.15...
= 1.12...
= 1.68...
= (1.36...) × (2.15...)2 × (1.12...)3 × (1.68...)3 = 42.1 .
E��B.�(a) �e partition function is given by [��B.�b–���], q = ∑ i g i e−βε i , where g i is
degeneracy and the corresponding energy is given as ε i = hc ν̃ i . At T = 1900 K
βhc =
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
= 7.57... × 10−4 cm
(1.3806 × 10−23 J K−1 ) × (1900 K)
�erefore the electronic partition function is
q E = g 0 + g 1 × e−βε 1 + g 2 × e−βε 2
−4
−1
−4
−1
= 4 + 1×e−(7.57 ...×10 cm)×(2500 cm ) + 2×e−(7.57 ...×10 cm)×(3500 cm )
= 4 + 0.150... + 0.141... = �.���
(��.�)
�e population of level i with degeneracy g i is N i = (N g i �q)e−βε i , therefore
the relative populations of the levels are proportional to g i e−βε i , which are the
terms in eqn ��.�. �us the populations, relative the ground state are
N 0 �N 0 ∶ N 1 �N 0 ∶ N 2 �N 0 = 4�4 ∶ (0.150...�4) ∶ (0.141...�4)
E��B.�(a)
= 1 ∶ 0.0376 ∶ 0.0353
(i) �e thermal wavelength is de�ned in [��B.�–���], Λ = h�(2πmkT)1�2 .
Because the mass of a molecule m is m = M�N A and k = R�N A it follows
that
Λ=
h
hN A
=
[2π(M�N A )(R�N A )T]1�2 (2πMRT)1�2
447
448
12 STATISTICAL THERMODYNAMICS
Λ(300 K) =
(6.6261 × 10−34 J s) × (6.0221 × 1023 mol−1 )
[2π×(0.150 kg mol−1 )×(8.3145 J K−1 mol−1 ) × (300 K)]1�2
= 8.22... × 10−12 m = 8.23 × 10−12 m
Similarly, Λ(3000 K) = 2.60 × 10−12 m
(ii) �e translational partition function in three dimensions is given by [��B.��b–
���], q T = V �Λ 3 .
q T (300 K) = (1.00 × 10−6 m3 )�(8.22... × 10−12 m)3 = 1.78 × 1027
q T (3000 K) = 5.67 × 1028
E��B.��(a) �e translational partition function in three dimensions is given by [��B.��b–
���], q T = V �Λ 3 , where Λ is the thermal wavelength de�ned in [��B.�–���],
Λ = h�(2πmkT)1�2 .
qHT 2 V �Λ H2 3
Λ He
h�(2πm He kT)1�2
m H2 3�2
=
=
�
�
=
�
�
=
�
�
T
qHe
Λ H2
m He
h�(2πm H2 kT)1�2
V �Λ He 3
3
3
Because the mass of a molecule m is m = M�N A it follows that
qHT 2
M H2 3�2
2 × 1.0079 g mol−1
=
�
�
�
=
�
T
qHe
M He
4.00 g mol−1
3�2
= �.���
E��B.��(a) �e rotational partition function of a symmetric linear rotor is given by [��B.��a–
���], q R = kT�(2hc B̃), where the rotational constant is de�ned in [��B.�–���],
B̃ = ħ�(4πcI). �e moment of inertia of a diatomic is I = µR 2 , where R is the
bond length and µ = m A m B �(m A +m B ). For a homonuclear diatomic m A = m B
so it follows that µ = m B �2. Using m = M�N A , this becomes µ = M B �2N A .
I = µR 2 =
M B R 2 (0.01600 kg mol−1 ) × (120.75 × 10−12 m)2
=
2N A
2 × (6.0221 × 1023 mol−1 )
= 1.93... × 10−46 kg m2
kT
kT
4πcI
kT
4πcI
kT
qR =
=�
��
�=�
��
�= 2 I
2hc
ħ
4πcħ
ħ
ħ
2hc B̃
−1
−23
(1.3806 × 10 J K ) × (300 K)
=
× (1.93... × 10−46 kg m2 ) = ��.�
(1.0546 × 10−34 J s)2
E��B.��(a) �e rotational partition function of a non-linear rotor is given by [��B.��–���],
q R = (1�σ)(kT�hc)3�2 (π�ÃB̃C̃)1�2 , where σ is the symmetry number. NOF is
not centro-symmetric so σ = 1.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(i) At 25 ○ C which is 298.15 K
qR = �
=�
kT 3�2
π 1�2
� �
�
hc
ÃB̃C̃
(1.3806 × 10−23 J K−1 ) × (298.15 K)
�
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
�
3�2
π
�
−1
(3.1752 cm ) × (0.3951 cm−1 ) × (0.3505 cm−1 )
1�2
= 7.97 × 103
(ii) At 100 ○ C which is 373.15 K
qR = �
(1.3806 × 10−23 J K−1 ) × (373.15 K)
�
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
�
3�2
π
�
−1
(3.1752 cm ) × (0.3951 cm−1 ) × (0.3505 cm−1 )
1�2
= 1.12 × 104
Solutions to problems
P��B.�
�e partition function is given by [��B.�b–���], q = ∑ i g i e−βε i , where g i is degeneracy and the corresponding energy is ε i . �erefore the partition function
is
q = 1 × e−β0 + 1 × e−βε + 1 × e−β(2ε) = 1 + e−ε�k T + e−2ε�k T
�is is plotted in Fig ��.�. As expected, the partition function rises from a
value of � at low temperatures, where only the ground state is occupied, and
approaches a value of � at high temperatures, when all three states are nearly
equally populated.
P��B.�
As discussed in Section ��C.�(a) on page ���, the Morse oscillator has a �nite
number of bound levels between the ground level and the dissociation limit.
�e number of these is found by noting that E υ reaches a maximum value at
the dissociation limit, and therefore this limit is found by solving dE υ �dυ = 0
d
�(υ + 12 )hc ν̃ − (υ + 12 )2 hcx e ν̃� = hc ν̃ − 2(υ + 12 )hcx e ν̃
dυ
solving 0 = hc ν̃ − 2(υ max + 12 )hcx e ν̃ gives υ max = 1�2x e − 12
�e energy of the lowest state is E 0 = 12 hc ν̃ − 14 hcx e ν̃, therefore the energies
used to evaluate the partition function are
E υ′ = E υ − E 0 = �(υ + 21 )hc ν̃ − (υ + 12 )2 hcx e ν̃� − [ 12 hc ν̃ − 14 hcx e ν̃]
= υhc ν̃ − (υ 2 + υ)hcx e ν̃
449
12 STATISTICAL THERMODYNAMICS
3
q
450
2
1
0
5
10
15
kT�ε
20
25
30
Figure 12.4
�e partition function is evaluated from the sum
υ max
q VM = � e−(υhc ν̃−(υ +υ)hc x e ν̃)�k T
2
υ=0
De�ning the characteristic vibrational temperature as θ V = hc ν̃�k gives
υ max
q VM = � e−(υ−(υ +υ)x e )θ �T
2
V
υ=0
For the harmonic oscillator the partition is given by the exact expression [��B.��–
V
���], q VHO = (1 − e−θ �T )−1 .
Figure ��.� compares the partition functions for various values of x e with that
for the harmonic case. For the smallest value of x e the partition function is
initially larger than that for the harmonic oscillator. �is can be attributed
to more energy levels contributing at these temperatures as they are closer in
energy for the Morse oscillator than for the harmonic case. However, at higher
temperatures the partition function for the Morse oscillator starts to level o�
because there are a �nite number of levels, whereas for the harmonic case the
partition function continues without limit as there are an in�nite number of
levels.
�is behaviour is even more pronounced for x e = 0.05 and x e = 0.10. In these
two cases υ max is �� and �, respectively, and these values set the limiting hightemperature value of the partition function.
P��B.�
�e partition function is given by [��B.�b–���], q = ∑ i g i e−βε i , where g i is
degeneracy and the corresponding energy is given as ε i = hc ν̃ i , and β = 1�kT.
�erefore the electronic partition function is
q E = g 0 + g 1 × e−hc ν̃ 1 �k T + g 2 × e−hc ν̃ 2 �k T + g 3 × e−hc ν̃ 3 �k T
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
harmonic
x e = 0.02
x e = 0.05
x e = 0.10
qV
15
10
5
0
0
2
4
6
8
T�θ
10
12
14
16
V
Figure 12.5
(a) (i) At T = 298 K
hcβ =
(6.6261 × 10−34 J s)×(2.9979 × 1010 cm s−1 )
= 4.82...×10−3 cm
(1.3806 × 10−23 J K−1 )×(298 K)
q E = 5 + 1 × e−(4.82 ...×10 cm)×(4707 cm )
−3
−1
+ 3 × e−(4.82 ...×10 cm)×(4751 cm )
−3
+ 5 × e−(4.82 ...×10
−3
−1
cm)×(10559 cm−1 )
= 5 + 1.34... × 10−10 + 3.27... × 10−10 + 3.61... × 10−22
= �.��
(ii) At T = 5000 K
hcβ =
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
(1.3806 × 10−23 J K−1 ) × (5000 K)
= 2.87... × 10−4 cm
q E = 5 + 1 × e−(2.87 ...×10 cm)×(4707 cm )
−4
−1
+ 3 × e−(2.87 ...×10 cm)×(4751 cm )
−4
+ 5 × e−(2.87 ...×10
−4
−1
cm)×(10559 cm−1 )
= 5 + 0.258... + 0.764... + 0.239... = 6.262... = �.���
(b) �e population of level (term) i is N i �N = (g i × e−hc ν̃ i �k T )�q E
(i) T = 298 K
N 0 �N = 5�5.00... = �.��
N 2 �N = (3.26... × 10−10 )�5.00... = 6.54 × 10−11
451
452
12 STATISTICAL THERMODYNAMICS
P��B.�
(ii) T = 5000 K
N0
5
=
= �.���
N
6.262...
N 2 0.764...
=
= �.���
N
6.262...
�e partition function is given by [��B.�b–���], q = ∑ i g i e−βε i , where g i is
degeneracy and the corresponding energy is given as ε i = hc ν̃ i , and β = 1�kT.
Here g i = 2J +1, where J is the right subscript in the term symbol. At T = 298 K
hcβ =
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
= 4.82... × 10−3 cm
(1.3806 × 10−23 J K−1 ) × (298 K)
�erefore the electronic partition function is
q E = 1 + 3 × e−(4.82 ...×10 cm)×(557.1 cm )
−3
−1
−3
−1
+ 5 × e−(4.82 ...×10 cm)×(1410.0 cm ) + 5 × e−(4.82 ...×10 cm)×(7125.3 cm )
−3
+ 1 × e−(4.82 ...×10
−3
−1
cm)×(16367.3 cm−1 )
= 1 + 0.203... + 5.52... × 10−3 + 5.72... × 10−15 + 4.78... × 10−35
= �.���
Similarly, at T = 1000 K
hcβ =
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
= 1.43... × 10−3 cm
(1.3806 × 10−23 J K−1 ) × (1000 K)
�erefore the electronic partition function is
q E = 1 + 3 × e−(1.43 ...×10 cm)×(557.1 cm )
−3
−1
−3
−1
+ 5 × e−(1.43 ...×10 cm)×(1410.0 cm ) + 5 × e−(1.43 ...×10 cm)×(7125.3 cm )
−3
+ 1 × e−(1.43 ...×10
−3
−1
cm)×(16367.3 cm−1 )
= 1 + 1.34... + 0.657... + 1.76... × 10−4 + 5.92... × 10−11
P��B.�
= �.���
�e partition function is given by [��B.�b–���], q R = ∑ i g i e−βε i . Where g i is
degeneracy and ε i = hc F̃ i . �e rotational terms of a symmetric rotor is given
by [��B.��a–���], F̃(J, K) = B̃J(J + 1) + (à − B̃)K 2 , with J = 0, 1, 2, ... and
K = 0, ±1, ..., ±J. �us, the partition function is
� +J
�
∞
2�
�
q R = �(2J + 1)e−hc β B̃ J(J+1) � � e−hc β( Ã−B̃)K �
�K=−J
�
J=0
�
�
Using mathematical so�ware the terms are evaluated and summed until convergence is achieved to within the required precision.
In the high-temperature limit the partition function is given by [��B.��b–���],
q R = (kT�hc)3�2 (π�ÃB̃C̃)1�2 ; for a symmetric rotor B̃ = C̃, therefore q R =
(kT�hc)3�2 (π�ÃB̃ 2 )1�2 . With the given data
q R = (kT�hc)3�2 (π�ÃB̃ 2 )1�2 = (1.02... × 10 K) × T 3�2
�e two forms of the partition function are plotted in Fig. ��.�; the high temperature limit is accurate to within � % of the exact solution at 4.5 K .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
qR
30
20
10
0
exact
high T limit
0
2
4
6
8
10
T�K
Figure 12.6
12C Molecular energies
Answers to discussion questions
D��C.�
�is is described in Brief illustration ��C.� on page ���.
Solutions to exercises
E��C.�(a)
�e mean vibrational energy is given by [��C.�–���], �ε V � = hc ν̃�(e β hc ν̃ − 1);
this result is exact. �e equipartition value is �ε V � = kT, because there are two
quadratic terms for a harmonic oscillator. �ese two expressions for the energy
are plotted as a function of T in Fig. ��.�. �e value from the equipartition
theorem comes within � % of the exact value at 4.80 × 103 K .
8
�ε V � × 1020 �J
6
4
2
0
exact
equipartition
0
1 000
2 000
3 000
T�K
4 000
5 000
Figure 12.7
E��C.�(a) �e mean vibrational energy per vibrational mode is given by [��C.�–���],
453
12 STATISTICAL THERMODYNAMICS
�ε Vi � = hc ν̃ i �(e β hc ν̃ i − 1); this result is exact. �e overall vibrational energy
is the sum of the contributions from each normal mode, taking into account
the degeneracy of each
�ε V � = �ε V1 � + 2 × �ε V2 � + �ε V3 �
�e equipartition value is �ε V � = 4kT, because there are two quadratic terms
for a harmonic oscillator, and four modes in total. �ese two expressions for the
energy are plotted as a function of T in Fig. ��.�. �e value from the equipartition theorem comes within �% of the exact value at 1.10 × 104 K .
60
�ε V � × 1020 �J
454
40
20
0
exact
equipartition
0
2 000 4 000 6 000 8 000 10 000 12 000
T�K
Figure 12.8
E��C.�(a) �e mean vibrational energy per vibrational mode is given by [��C.�–���],
�ε Vi � = hc ν̃ i �(e β hc ν̃ i − 1); this result is exact. �e overall vibrational energy
is the sum of the contributions from each normal mode, taking into account
the degeneracy of each
�ε V � = �ε V1 � + 2 × �ε V2 � + 3 × �ε V3 � + 3 × �ε V4 �
�e equipartition value is �ε V � = 9kT, because there are two quadratic terms
for a harmonic oscillator, and nine modes in total. �ese two expressions for
the energy are plotted as a function of T in Fig. ��.�. �e value from the equipartition theorem comes within �% of the exact value at 6.85 × 103 K .
E��C.�(a) �e mean molecular energy is given by [��C.�a–���], �ε� = −(1�q)(∂q�∂β)V ,
where β = 1�kT, and q is the partition function given by [��B.�b–���], q =
∑ i g i e−βε i , where g i is degeneracy and the corresponding energy is given as
ε i = hc ν̃ i . At T = 1900 K
βhc =
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
= 7.57... × 10−4 cm
(1.3806 × 10−23 J K−1 ) × (1900 K)
�erefore the electronic partition function is
q E = g 0 + g 1 e−β hc ν̃ 1 + g 2 e−β hc ν̃ 2 = 4.29...
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�ε V � × 1020 �J
100
50
0
exact
equipartition
0
2 000
4 000
T�K
6 000
8 000
Figure 12.9
�erefore the mean energy is
1 ∂q E
hc
�ε E � = − E �
� = E �g 1 ν̃ 1 e−β hc ν̃ 1 + g 2 ν̃ 2 e−β hc ν̃ 2 �
q
∂β V q
=
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
4.29...
−4
−1
−1
× �1 × (2500 cm ) × e−(7.57 ...×10 cm)×(2500 cm )
+2 × (3500 cm−1 ) × e−(7.57 ...×10
= 4.03 × 10−21 J
−4
cm)×(3500 cm−1 )
�
E��C.�(a) �e mean energy of a molecule is given by [��C.�–���], �ε� = (1�q) ∑ i ε i e−βε i ,
where ε i = hc ν̃ i , β = 1�kT, and q is the partition function given by [��A.��–
���], q = ∑ i e−βε i . �erefore for the two-level system
�ε� =
=
0 + εe−βε
ε
hc ν̃
= βε
= hc ν̃�k T
−βε
1+e
e +1 e
+1
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (500 cm−1 )
e
(6.6261×10−34 J s)×(2.9979×10 10 cm s−1 )×(500 cm−1 )
(1.3806×10−23 J K−1 )×(298 K)
= 8.15 × 10−22 J
+1
E��C.�(a) �e mean molecular energy is given by [��C.�a–���], �ε� = −(1�q)(∂q�∂β)V ,
where β = 1�kT and q is the partition function. �e rotational partition function of a heteronuclear diatomic is given in terms of the rotational constant B̃
by [��B.��–���], q R = ∑ J (2J + 1)e−β hc B̃ J(J+1) .
1 ∂q R
1
�ε R � = − R �
� = R � hc B̃J(J + 1)(2J + 1)e−β hc B̃ J(J+1)
q
∂β V q J
455
12 STATISTICAL THERMODYNAMICS
�e terms of the sum above, and also of the sum needed to compute q R , are
evaluated and summed until the result has converged to the required precision.
�e equipartition value is �ε R � = kT. �ese two expressions for the energy
are plotted as a function of T in Fig. ��.��. �e value from the equipartition
theorem comes within � % of the exact value at 19.6 K .
2
�ε R � × 1022 �J
456
1
0
exact
equipartition
0
5
10
T�K
15
20
Figure 12.10
E��C.�(a) �e mean molecular energy is given by [��C.�a–���], �ε� = −(1�q)(∂q�∂β)V ,
where β = 1�kT, and q is the partition function given by [��B.�b–���], q R =
∑ J g J e−βε J . �e energy levels of a spherical rotor are given in [��B.�–���], ε J =
hc B̃J(J + 1) and, as is explained in Section ��B.�(c) on page ���, each has a
degeneracy g J = (2J + 1)2 . It follows that
q R = �(2J + 1)2 e−β hc B̃ J(J+1)
J
1 ∂q R
1
2 −β hc B̃ J(J+1)
�ε R � = − R �
� =
� hc B̃J(J + 1)(2J + 1) e
q
∂β V q R J
�e terms in the sum needed to compute q R and �ε R � are evaluated and summed
until the result has converged to the required precision. �e equipartition value
is �ε R � = 32 kT, because for this non-linear molecule there are three rotational
degrees of freedom. �ese two expressions for the energy are plotted as a function of T in Fig. ��.��. �e value from the equipartition theorem comes within
� % of the exact value at 26.4 K .
Solutions to problems
P��C.�
�e mean molecular energy is given by [��C.�a–���], �ε� = −(1�q)(∂q�∂β)V ,
where β = 1�kT, and q is the partition function given by [��B.�b–���], q =
∑ i g i e−βε i . For a symmetric rotor the rotational terms are given in [��B.��a–
���], F̃(J, K) = B̃J(J + 1) + (à − B̃)K 2 , with J = 0, 1, 2, ... and K = 0, ±1, ..., ±J;
the corresponding energies are hc F̃(J, K) and the degeneracy is (2J + 1). �e
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�ε R � × 1022 �J
6
4
2
0
exact
equipartition
0
5
10
15
T�K
20
25
30
Figure 12.11
rotational partition function is therefore given by
� +J
�
∞
2�
�
q R = �(2J + 1)e−β hc B̃ J(J+1) � � e−β hc( Ã−B̃)K �
�K=−J
�
J=0
�
�
�e mean energy is therefore
1 ∂q R
�ε R � = − R �
�
q
∂β V
=
=
� +J
�
1 �∞
−β hc B̃ J(J+1) �
−β hc( Ã− B̃)K 2 �
�
�
hc
B̃J(J
+
1)(2J
+
1)e
e
�
�
�K=−J
�
q R � J=0
�
�
� +J
�
∞
2 ��
�
+ �(2J + 1)e−β hc B̃ J(J+1) � � hc(Ã − B̃)K 2 e−β hc( Ã−B̃)K �
�K=−J
��
J=0
�
�
1 ∞
−β hc B̃ J(J+1)
�(2J + 1)e
q R J=0
+J
�
�
2�
�
× �hc � �B̃J(J + 1) + (Ã − B̃)K 2 � e−β hc( Ã−B̃)K �
�
� K=−J
�
�
�e terms in the sum in this expression, and the terms in q R , are evaluated and
summed until the value converges to the required precision. �e equipartition
value of the energy is �ε R � = 32 kT because there are three rotational degrees
of freedom. �e two expressions for the energy are plotted as a function of
temperature in Fig. ��.��. �e equipartition value is within �% of the exact
solution at 4.59 K .
P��C.�
H� O has three translational and three rotational degrees of freedom. It therefore follows from the equipartition principle that the molar internal energy is
U m = (3 + 3) × 12 RT = 3RT. �e constant-volume molar heat capacity is
therefore C V ,m = (∂U m �∂T)V = 3R.
457
12 STATISTICAL THERMODYNAMICS
1.0
�ε R � × 1022 �J
458
0.5
0.0
exact
equipartition
0
1
2
3
T�K
4
5
6
Figure 12.12
P��C.�
�e energy needed to raise the temperature of n mol of H� O by ∆T is equal
to the change in the internal energy which is ∆U = nC V ,m ∆T. In this case
∆U = (1.0 mol) × (3 × 8.3145 J K−1 mol−1 ) × (100 K) = �.� kJ .
�e energy levels for a spin in a magnetic �eld are given by [��A.�d–���], E m I =
−g I µ N B0 m I , where m I = +1, 0, −1. �ese energies are conveniently written as
E m I = −m I δ, with δ = g I µ N B0 . If the energy of the lowest state is de�ned as the
energy zero, then the three levels have energies E 0 = 0, E 1 = δ, and E 2 = 2δ.
�e partition function is
q = 1 + e−βδ + e−2βδ
�e mean molecular energy is given by [��C.�a–���]
1 ∂q
� �
q ∂β V
1
�−δe−βδ − 2δe−2βδ �
= ε gs −
1 + e−βδ + e−2βδ
δe−βδ + 2δe−2βδ
δe−βδ + 2δe−2βδ
= ε gs +
=
−δ
+
1 + e−βδ + e−2βδ
1 + e−βδ + e−2βδ
�ε� = ε gs −
In the last step the fact that the energy of ground state (the one with m I = +1)
is −δ is used. With the data given
δ = g I µ N B0 = 2.0 × (5.0508 × 10−27 J T−1 ) × (2.5 T) = 2.52... × 10−26 J
and assuming that T = 298 K
βδ = δ�kT = (2.52... × 10−26 J)�[(1.3806 × 10−23 J K−1 ) × (298 K)]
= 6.13... × 10−6
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�ε� = (−2.52... × 10−26 J)
(2.52... × 10−26 J)e−6.13 ...×10 + 2 × (2.52... × 10−26 J)e−2×(6.13 ...×10 )
1 + e−6.13 ...×10−6 + e−2×6.13 ...×10−6
−31
= −1.03 × 10 J
−6
+
−6
�e separation of the energy levels is very much smaller than kT, therefore the
three levels have almost equal populations giving a mean energy of very close
to zero.
�e partition function given by [��B.�b–���], q = ∑ i g i e−βε i , where g i is degeneracy, ε i = hc ν̃ and β = 1�kT. �erefore
q E = 2e0 + 2e−β hc ν̃ = 2 + 2e−hc ν̃�k T
�is function in plotted in Fig. ��.��.
4.0
3.5
qE
P��C.�
3.0
2.5
2.0
0
200
400
600
800
1 000
T�K
Figure 12.13
(a) �e ratio of the populations is given by [��A.��b–���]
At ��� K this ratio is
N i �N j = (g i �g j )e−β(ε i −ε j )
N 1 �N 0 = (g 1 �g 0 ) × e−β hc ν̃ = (2�2) × e−β hc ν̃
−
=e
(6.6261×10−34 J s)×(2.9979×10 10 cm s−1 )×(121.1 cm−1 )
(1.3806×10−23 J K−1 )×(300 K)
�us the populations expressed as a fraction of N are
= 0.559....
N 0 �N = N 0 �(N 0 + N 1 ) = 1�(1 + 0.559...) = �.���
N 1 �N = N 1 �(N 0 + N 1 ) = 0.559...�(1 + 0.559...) = �.���
459
460
12 STATISTICAL THERMODYNAMICS
(b) �e mean molecular energy is given by [��C.�a–���], �ε� = −(1�q)(∂q�∂β)V .
�us
1 dq E
e−hc ν̃�k T
hc ν̃
�ε E � = − E
= hc ν̃ ×
=
q dβ
1 + e−hc ν̃�k T e hc ν̃�k T + 1
=
P��C.�
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (121.1 cm−1 )
e
(6.6261×10−34 J s)×(2.9979×10 10 cm s−1 )×(121.1 cm−1 )
(1.3806×10−23 J K−1 )×(300 K)
= 8.63 × 10−22 J
+1
Mean values of any observable are given by a sum of the observed value over
all the possible states weighted by the probability of each state. �us, the mean
of the square of energy is given by �ε 2 � = (1�q) ∑ j ε j 2 e−βε j , where q is the
partition function given by [��A.��–���], q = ∑ i e−βε i .
It is useful to consider the �rst and second derivatives of the term e−βε j with
respect to β. �e �rst derivative is
d −βε j
e
= −ε j e−βε j
dβ
and hence the second is
d2
dβ
−βε j
=
2e
d
d −βε j
�−ε j e−βε j � = (−ε j ) ×
e
dβ
dβ
= (−ε j ) × (−ε j ) × e−βε j = ε j 2 × e−βε j
�is latter expression is used to rewrite the de�nition of �ε 2 � as
�ε 2 � =
=
�erefore
1
2 −βε
�ε e j
q j j
�
�
2
�
1
d2 −βε
1 d2 �
�� e−βε j � = 1 d q
� 2e j =
�
2 �
q j dβ
q dβ � j
� q dβ 2
�
�
�ε �
2 1�2
1 d2 q
= �
�
q dβ 2
1�2
�ese results are also used to �nd the root mean square of the deviation from
the mean as
��ε � − �ε� �
2
2 1�2
� 1 d2 q
1 dq �
=
�
2 −�
q dβ �
� q dβ
2 1�2
dq �
1 � d2 q
=
q 2 −� �
q � dβ
dβ �
2 1�2
For a harmonic oscillator the partition function is given by [��B.��–���], q =
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(1 − e−β hc ν̃ )−1 . �erefore
dq
−hc ν̃ e−β hc ν̃
=
= −hc ν̃ e−β hc ν̃ q 2
dβ (1 − e−β hc ν̃ )2
d2 q
−hc ν̃ e−β hc ν̃
−2hc ν̃ e−β hc ν̃ e−β hc ν̃
=
−hc
ν̃
�
+
�
dβ 2
(1 − e−β hc ν̃ )2
(1 − e−β hc ν̃ )3
= (hc ν̃)2 e−β hc ν̃ �
1
(1 − e−β hc ν̃ )2
+
= (hc ν̃)2 e−β hc ν̃ �q 2 + 2e−β hc ν̃ q 3 �
= q 2 (hc ν̃)2 e−β hc ν̃ �1 + 2e−β hc ν̃ q�
2e−β hc ν̃
�
(1 − e−β hc ν̃ )3
Using these results and making the substitution x = hc ν̃
�ε 2 � − �ε� =
2
1 d2 q
1 dq
�
2 −�
q dβ
q dβ
2
= q x 2 e−βx �1 + 2e−βx q� − �−x e−βx q�
= q x 2 e−βx + q 2 x 2 e−2βx =
hence
��ε 2 � − �ε� �
2 1�2
2
x 2 e−βx
x 2 e−2βx
+
−βx
1−e
(1 − e−βx )2
=
x 2 e−βx (1 − e−βx ) + x 2 e−2βx
x 2 e−βx
=
(1 − e−βx )2
(1 − e−βx )2
=
hc ν̃ e−β hc ν̃�2
1 − e−β hc ν̃
In terms of the vibrational temperature this is
kθ R e−θ �2T
1 − e−θ R �T
R
At high temperatures, T � θ R , the exponential in the denominator is approximated as 1 − θ R �T, and the exponential in the numerator goes to �, giving
(�ε 2 � − �ε�2 )1�2 = kT. At low temperatures, T � θ R , (�ε 2 � − �ε�2 )1�2 tends to
�. �is is as expected because if all the particles are in the ground state there is
no uncertainty in their average energy.
12D The canonical ensemble
Answer to discussion questions
D��D.�
An ensemble is a set of a large number of imaginary replications of the actual
system. �ese replications are identical in some, but not all, respects. For
example, in the canonical ensemble, all replications have the same number of
461
462
12 STATISTICAL THERMODYNAMICS
particles, the same volume, and the same temperature, but they need not have
the same energy.
Ensembles are useful in statistical thermodynamics because it is mathematically more tractable to perform an ensemble average to determine the (time
averaged) thermodynamic properties than it is to perform an average over time
to determine these properties. Recall that macroscopic thermodynamic properties are averages over the time dependent properties of the particles that compose the macroscopic system. In fact, it is taken as a fundamental principle
of statistical thermodynamics that the (su�ciently long) time average of every
physical observable is equal to its ensemble average. �is principle is connected
to a famous assumption of Boltzmann’s called the ergodic hypothesis.
D��D.�
In the context of ensembles, the thermodynamic limit is achieved as the number of replications Ñ approaches in�nity. In this limit, the dominating con�guration is overwhelmingly the most probable con�guration, and its properties
are essentially the same as those of the system.
Solutions to exercises
E��D.�(a)
It is essential to include the factor 1�N! when considering indistinguishable
particles which are free to move. �us, such a factor is always needed for gases.
In the solid state, particles are distinguished by their positions in the lattice
and therefore the particles are regarded as distinguishable on the basis that
their locations are distinguishable. For the cases mentioned, the factor 1�N!
is needed for all but solid CO.
Solutions to problems
P��D.�
p = kT �
∂ ln(q N �N!)
∂(N ln q − ln N!)
∂ ln Q
� = kT �
� = kT �
�
∂V T
∂V
∂V
T
T
�e ln N! term is volume independent and thus p = N kT �∂ ln q�∂V �T . �e
molecular partition function, q, for the perfect gas is just the translational partition function given by [��B.��b–���], q T = V �Λ 3 , where Λ is the thermal
wavelength which is independent of volume. �erefore
∂ ln(V �Λ 3 )
∂(ln V − ln Λ 3 )
� = N kT �
�
∂V
∂V
T
T
∂ ln V
N kT nRT
= N kT �
� =
=
∂V T
V
V
p = N kT �
where for the last step N = nN A and R = kN A are used.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
12E
The internal energy and entropy
Answer to discussion questions
D��E.�
�e statistical entropy is de�ned by Boltzmann’s formula, S = k ln W, in terms
of the number of con�gurations or microstates consistent with a given total
energy. �e thermodynamic entropy is de�ned by dS = dq rev �T that is, in
terms of reversible heat transfer.
�e concept of the number of microstates makes quantitative the qualitative
concepts of ‘disorder’ and ‘dispersal of matter and energy’ that are o�en used
to introduce the concept of entropy: a more ‘disorderly’ distribution of energy
and matter corresponds to a greater number of microstates consistent with the
same total energy. �e more molecules that can participate in the distribution
of the energy, the more microstates there are for a given total energy and hence
the greater the entropy.
�e molecular interpretation of entropy embodied in the Boltzmann formula
also suggests the thermodynamic de�nition. At high temperatures, where the
molecules of a system can occupy a large number of energy levels, a small
additional transfer of energy as heat will cause only a small change in the number of accessible energy levels, whereas at low temperatures the transfer of the
same quantity of heat will increase the number of accessible energy levels and
microstates signi�cantly. 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, as required by the thermodynamic de�nition.
D��E.�
�e entropy of a monatomic perfect gas is given by the Sackur–Tetrode equation [��E.�a–���]
S m = R ln �
Vm e5�2
�
NA Λ3
Λ = h�(2πmkT)1�2
Vm = RT�p
Because the molar volume appears in the numerator, the molar entropy increases with the molar volume. In terms of the Boltzmann distribution, this
relationship is expected: large containers have more closely spaced energy levels than do small ones, so more states are thermally accessible. Temperature
appears in the numerator of the expression (through the denominator of Λ),
so the molar entropy increases with the temperature. Again, this is consistent
with the Boltzmann distribution, because more states are accessible at higher
temperatures than at lower ones.
�e fact that diatomic and polyatomic gases have rotational and vibrational
modes of motion as well does not change the above arguments. �e partition functions of those modes are independent of volume, so the volume dependence of the entropy is as described above. At most temperatures, rotational modes of motion are active and contribute to the entropy, as expressed
in [��E.��a–���]; the contribution increases with temperature. Finally, most
vibrational modes contribute little if at all to the entropy, but as with rotation
the contribution increases with temperature.
463
464
12 STATISTICAL THERMODYNAMICS
D��E.�
�e temperature is always high enough for the mean translational energy to
be 32 kT, the equipartition value. �erefore, the molar constant-volume heat
capacity for translation is C VT ,m = 32 R.
When the temperature is high enough for the rotations of the molecules to
be highly excited (when T � θ R ) the equipartition value kT for the mean
rotational energy (for a linear rotor) can be used to obtain C VR ,m = R. For nonlinear molecules, the mean rotational energy is 32 kT, so the molar rotational
heat capacity rises to 32 R when T � θ R . At intermediate temperatures the total
heat capacity takes a value between that due to translation, 32 R, and 52 R (for a
linear molecule) when both translation and rotation contribute fully.
Molecular vibrations contribute to the heat capacity, but only when the temperature is high enough for them to be signi�cantly excited. For each vibrational
mode, the equipartition mean energy is kT, so the maximum contribution to
the molar heat capacity is R. However, it is unusual for the vibrations to be so
highly excited that equipartition is valid, and in general the contribution to the
heat capacity has to be calculated using [��E.�–���].
Solutions to exercises
E��E.�(a)
For atoms with �lled shells the only contribution to the entropy is translational.
�e standard molar entropy of a monatomic perfect gas is given by the Sackur–
Tetrode equation [��E.�b–���]
−
○
Sm
= R ln �
kTe5�2
�
p−○ Λ 3
Λ = h�(2πmkT)1�2
(i) Taking the mass of a He atom as 4.00 m u
Λ=
6.6261 × 10−34 J s
�2π(4.00×1.6605 × 10−27 kg)×(1.3806 × 10−23 J K−1 )×(298 K)�
= 5.05... × 10−11 m
−
○
Sm
= (8.3145 J K−1 mol−1 ) × ln �
= ��� J K−1 mol−1
1�2
(1.3806 × 10−23 J K−1 ) × (298 K) × e5�2
�
(105 N m−2 ) × (5.05 . . . × 10−11 m)3
(ii) Taking the mass of a Xe atom as 131.29 m u
Λ=
6.6261 × 10−34 J s
�2π(131.29×1.6605 × 10−27 kg)×(1.3806 × 10−23 J K−1 )×(298 K)�
= 8.82... × 10−12 m
−
○
Sm
= (8.3145 J K−1 mol−1 ) × ln �
= ���.� J K−1 mol−1
1�2
(1.3806 × 10−23 J K−1 ) × (298 K) × e5�2
�
(105 N m−2 ) × (8.82... × 10−12 m)3
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E��E.�(a)
For atoms with �lled shells the only contribution to the entropy is translational.
�e standard molar entropy of a monatomic perfect gas is given by the Sackur–
Tetrode equation [��E.�b–���]
−
○
Sm
= R ln �
kTe5�2
�
p−○ Λ 3
Λ = h�(2πmkT)1�2
−
○
It follows that S m
= A ln(T 5�2 m 3�2 ), where A is a constant. �erefore
−
○
−
○
Sm
(He, T1 ) − S m
(Xe, T2 ) = A ln(T1 m He ) − A ln(T2 m Xe )
5�2
−
○
−
○
If S m
(He, T1 ) = S m
(Xe, 298)
3�2
5�2
0 = A ln[T1 m He ] − A ln[(298 K)5�2 m Xe ]
5�2
3�2
hence T1 m He = (298 K)5�2 m Xe
5�2
E��E.�(a)
3�2
3�2
T1 = �
3�2
3�2
(298 K)5�2 × (131.29)3�2
�
(4.00)3�2
2�5
= 2.42 × 103 K
�e rotational partition function for a non-linear molecule is given by [��B.��–
���]
1 kT 3�2
π 1�2
qR = � � �
�
σ hc
ÃB̃C̃
For H� O the symmetry factor σ = 2. At ��� K
kT�hc =
(1.3806 × 10−23 J K−1 ) × (298 K)
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
= 207.1... cm−1
q R = 12 (207.1... cm−1 )3�2 ×
�
π
�
(27.878 cm−1 ) × (14.509 cm−1 ) × (9.827 cm−1 )
1�2
= 43.1... = ��.�
�e entropy is given in terms of the partition function by [��E.�a–���]
S m = [U m (T) − U m (0)]�T + R ln q
�is is the appropriate form for the rotational contribution; for the translational
contribution the ln term is ln qe�N. At ��� K kT�hc = 207 cm−1 which is significantly greater than any of the rotational constants, therefore the equipartition
theorem can be used to �nd U m (T): there are three rotational modes, therefore
U m (T) − U m (0) = 32 RT.
R
Sm
= ( 32 RT)�T + R ln q R = R( 32 + ln q R )
= (8.3145 J K−1 mol−1 ) × [ 32 + ln(43.1...)]
= ��.�� J K−1 mol−1
465
466
12 STATISTICAL THERMODYNAMICS
E��E.�(a)
Only the ground electronic state contributes to the electronic partition function, which is therefore simply the degeneracy of the ground state q E = g 0 .
For a given term the degeneracy is given by the value of J, which is the right
subscript: g 0 = (2J + 1) = (2 × 92 + 1) = 10. �e entropy is given in terms of the
partition function by [��E.�a–���]
S m = [U m (T) − U m (0)]�T + R ln q
�is is the appropriate form for the electronic contribution; for the translational
contribution the ln term is ln qe�N. In this case U m (T) − U m (0) = 0 as only
the ground state is considered
E��E.�(a)
S m = R ln q = R ln 10 = ��.�� J K−1 mol−1
�e contribution of a collection of harmonic oscillators to the standard molar
entropy is given by [��E.��b–���] (note that there is an error in the expression
in the text: the argument of the exponential term in the ln should be negative)
V
θ V �T
V
Sm
= R � θ V �T
− ln(1 − e−θ �T )�
e
−1
θ V = hc ν̃�k
�e following table shows the vibrational temperatures and the contribution to
the molar entropy for each of the normal modes
−1
ν̃�cm
���
���
����
����
����
����
����
����
����
E��E.�(a)
θ �K
���
���
����
����
����
����
����
����
����
V
θ �T
�.��
�.��
�.���
�.���
�.���
�.���
�.���
��.��
��.��
V
��� K
V
Sm
�R
�.���
�.���
�.�����
�.�����
�.�����
9.515 × 10−3
1.855 × 10−3
1.026 × 10−5
5.957 × 10−7
θ �T
�.��
�.��
�.���
�.���
�.���
�.���
�.���
�.���
��.��
V
��� K
V
Sm
�R
�.���
�.���
�.����
�.����
�.����
�.�����
�.�����
1.988 × 10−3
3.895 × 10−4
�e molar entropy is obtained by summing the contributions from each normal
V
V
mode. �us at ��� K S m
= 4.18 J K−1 mol−1 and at ��� K S m
= 14.3 J K−1 mol−1 .
�e equipartition value for C V ,m is expressed in [��E.�–���]: each translational or rotational mode contributes 12 R, and each active vibrational mode
contributes R.
(i) I� : three translational modes, two rotational modes (linear) and, because
the vibrational frequency of the molecule is rather low, one vibrational
mode: C V ,m �R = 3 × 12 + 2 × 12 + 1 = 72 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(ii) CH� : three translational modes, three rotational modes (non-linear), and
no active vibrational modes: C V ,m �R = 3 × 12 + 3 × 12 = 3 .
(iii) C� H� : three translational modes, three rotational modes (non-linear),
and no active vibrational modes: C V ,m �R = 3 × 12 + 3 × 12 = 3 . �ere
are four low-frequency normal modes which, if active, will contribute a
further 4R.
E��E.�(a)
�e equipartition value for C V ,m is expressed in [��E.�–���]: each translational or rotational mode contributes 12 R, and each active vibrational mode
contributes R. �e number of vibrational modes is (3N −6), which is � for NH�
and � for CH� . For NH� there are three translational modes, three rotational
modes (non-linear) giving C V ,m = 3R; if the � vibrations are included, C V ,m =
9R. For CH� there are three translational modes, three rotational modes (nonlinear) giving C V ,m = 3R; if the � vibrations are included, C V ,m = 12R.
γ = C p,m �C V ,m = (C V ,m + R)�C V ,m = 1 + R�C V ,m
γ NH3 = 1 + R�3R = 1.33
γ NH3 = 1 + R�9R = 1.11
γ CH4 = 1 + R�3R = 1.33
γ CH3 = 1 + R�12R = 1.08
no vibrational contribution
with vibrational contribution
no vibrational contribution
with vibrational contribution
�e experimental value for γ is �.�� for both gases: evidently the vibrational
modes are not active.
E��E.�(a)
�e partition function of this two-level system is
q = g 0 + g 1 e−β hc ν̃
where g 0 and g 1 are the degeneracies of the ground and �rst excited state,
respectively. �e mean energy is given by [��C.�a–���], �ε� = −(1�q)(∂q�∂β)V
g 1 hc ν̃ e−β hc ν̃
g 1 hc ν̃
=
g 0 + g 1 e−β hc ν̃ g 0 e β hc ν̃ + g 1
N A g 1 hc ν̃
hence U m = N A �ε� =
g 0 e β hc ν̃ + g 1
�ε� =
By de�nition C V ,m = (∂U m �∂T)V , therefore
C V ,m = �
=
=
∂U m
∂U m
dβ
∂U m
−1
� =�
�
=�
� ×
∂T V
∂β V dT
∂β V kT 2
1
g 0 hc ν̃e β hc ν̃
× N A g 1 hc ν̃
2
kT
(g 0 e β hc ν̃ + g 1 )2
N A (hc ν̃)2
g 0 g 1 e β hc ν̃
kT 2
(g 0 e β hc ν̃ + g 1 )2
For an electronic term the degeneracy is 2J + 1, hence g 0 = 2 × 32 + 1 = 4 and
467
468
12 STATISTICAL THERMODYNAMICS
g 1 = 2 × 12 + 1 = 2. With the data given
hc ν̃ = (6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (881 cm−1 )
= 1.75... × 10−20 J
at ��� K βhc ν̃ = hc ν̃�kT
= (1.75... × 10−20 J)�[(1.3806 × 10−23 J K−1 ) × (500 K)] = 2.53...
at ��� K βhc ν̃ = hc ν̃�kT
= (1.75... × 10−20 J)�[(1.3806 × 10−23 J K−1 ) × (900 K)] = 1.40...
C V ,m (500) =
=
N A (hc ν̃)2
8 e β hc ν̃
kT 2
(4 e β hc ν̃ + 2)2
(6.0221 × 1023 mol−1 )×(1.75... × 10−20 J)2
8 e2.53 ...
−1
−23
2
(4 e2.53 ... + 2)2
(1.3806 × 10 J K ) × (500 K)
= �.�� J K−1 mol−1
C V ,m (900) =
E��E.�(a)
(6.0221 × 1023 mol−1 )×(1.75... × 10−20 J)2
8 e1.40 ...
−1
−23
2
(4 e1.40 ... + 2)2
(1.3806 × 10 J K ) × (900 K)
= �.�� J K−1 mol−1
�e contribution of a collection of harmonic oscillators to the molar heat capacity is given by [��E.�–���]
θ V � e−θ �2T �
C V ,m = R � �
V
T
� 1 − e−θ �T �
2
V
2
�is function is plotted in Fig ��.��.
θ V = hc ν̃�k
�e following table shows the vibrational temperatures and the contribution to
the heat capacity for each of the normal modes
ν̃�cm−1
���
���
����
����
����
θ V �K
���
����
����
����
����
��� K
T�θ V
C V ,m �R
�.���
�.���
�.���
�.���
�.���� 6.593 × 10−3
�.����� 3.226 × 10−5
�.����� 2.233 × 10−5
T�θ V
�.���
�.���
�.����
�.����
�.����
��� K
C V ,m �R
�.���
�.���
�.����
6.980 × 10−3
5.725 × 10−3
�e heat capacity is obtained by summing the contributions from each normal
mode, taking into account the double degeneracy of the modes at ��� cm−1 and
��� cm−1 by counting each twice. �us at ��� K C V ,m = 14.95 J K−1 mol−1 and
at ��� K C V ,m = 25.62 J K−1 mol−1 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
1.0
0.8
C V ,m �R
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
T�θ
0.8
1.0
V
Figure 12.14
Solutions to problems
P��E.�
�e electronic levels of NO form a two-level system, an expression for the heat
capacity of which is derived in the solution to Exercise E��E.�(a).
C V ,m =
N A (hc ν̃)2
g 0 g 1 e β hc ν̃
g 0 g 1 e β hc ν̃
= N A k(βhc ν̃)2
2
β
hc
ν̃
2
kT
(g 0 e
+ g1 )
(g 0 e β hc ν̃ + g 1 )2
where g 0 and g 1 are the degeneracies of the ground and �rst excited state,
respectively. For NO g 0 = 2 and g 1 = 2. With the data given
βhc ν̃ = hc ν̃�k × T −1
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (121.1 cm−1 )
× T −1
1.3806 × 10−23 J K−1
= (174.2... K) × T −1
=
A plot of C V ,m as a function of T is shown in Fig. ��.��.
P��E.�
�e energy levels for a particle on a ring are given by [�F.�–���], E m = m 2 ħ 2 �2I
where I is the moment of inertia and m = 0, ±1, ±2, . . .. In the high-temperature
limit the partition function is well-approximated by an intergral
q R = � e−βE m = �
+∞
−∞
m
e−βm ħ �2I dm
2 2
�e integral is of the form of Integral G.� with k ′ = βħ 2 �2I; the value needed is
twice that given for G.� as that integral is from 0 to +∞.
2πI
q = � 2�
βħ
R
1�2
�ese energy levels contribute one quadratic contribution to the energy, that is
there is one rotational mode. In the high-temperature limit the equipartition
theorem applies and hence the internal energy is U m = 12 RT and C VR ,m = 12 R .
469
12 STATISTICAL THERMODYNAMICS
0.4
C V ,m �R
470
0.2
0.0
0
100
200
300
400
500
T�K
Figure 12.15
�e entropy is given in terms of the partition function by [��E.�a–���]
S m = [U m (T) − U m (0)]�T + R ln q
�is is the appropriate form for the rotational contribution; for the translational
contribution the ln term is ln qe�N. As has already been explained, U m (T) −
U m (0) = 12 RT, therefore
R
Sm
= ( 12 RT)�T + R ln q R = R( 12 + ln q R )
With the data given
qR = �
=�
2πI
�
βħ 2
1�2
=�
2πkTI 1�2
�
ħ2
2π(1.3806 × 10−23 J K−1 ) × (298 K) × (5.341 × 10−47 kg m2 )
�
(1.0546 × 10−34 J s)2
1�2
= 11.1...
R
Sm
= (8.3145 J K−1 mol−1 ) × [ 12 + ln(11.1...)] = ��.� J K−1 mol−1
P��E.�
�is calculation is for a particle on a ring. When used as a model for a rotating
CH� group a symmetry factor of � is needed, so that q R is one third of the value
R
calculated here, giving S m
= 15.1 J K−1 mol−1 .
�e characteristic vibrational temperature is de�ned as θ V = hc ν̃�k. It follows
that ν̃ = kθ V �hc, so the vibrational frequency for a characteristic temperature
of ���� K is
ν̃ =
(1.3806 × 10−23 J K−1 ) × (1000 K)
= 695 cm−1
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
�e following modes have vibrational frequencies of 695 cm−1 or less (the degeneracies are given in parentheses)
525(3) 578(3) 354(3) 345(4) 403(5) 525(5) 667(5)
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�us in total there are �� modes with characteristic vibrational temperatures
of less than ���� K.
P��E.�
�e contributions to C V ,m are three translational, three rotational and �� vibrational modes, giving a heat capacity of C V ,m = 32 R + 32 R + 28R = 31R
�e partition function of this three-level system is
q = 1 + e−βε + e−2βε
�e mean energy is given by [��C.�a–���], E mean = −(1�q)(∂q�∂β)V
ε e−βε + 2ε e−2βε
ε e βε + 2ε
=
1 + e−βε + e−2βε e2βε + e βε + 1
e βε + 2
hence U m = N A E mean = N A ε 2βε
e + e βε + 1
E mean =
�e entropy is given in terms of the partition function by [��E.�a–���]
S m = [U m (T) − U m (0)]�T + R ln q
NA ε
e βε + 2
+ R ln(1 + e−βε + e−2βε )
2βε
T e + e βε + 1
βε(e βε + 2)
= R � 2βε
+ ln(1 + e−βε + e−2βε )�
e + e βε + 1
=
P��E.�
where to go to the last line N A �T = Rβ is used. At high temperatures, βε → 0
and S m tends to R ln 3. At low temperatures, βε → ∞ and S m tends to �, as
expected.
�e data in Problem P��B.� �t very well to the terms F̃(J) = B̃J(J + 1) with B̃ =
10.593 cm−1 . �e rotational contribution to the entropy is given by [��E.�a–
���]
R
Sm
= [U m (T) − U m (0)]�T + R ln q R
where the internal energy is given by [��E.�a–���]
U m (T) − U m (0) = −
and the partition function is
N A ∂q R
�
�
q R ∂β V
q R = �(2J + 1)e−hc β B̃ J(J+1)
J
In order to compute the entropy down to low temperatures it is necessary to
evaluate the sums term by term rather than approximating them by an integral.
�e derivative of q R is
�
∂q R
� = −hc B̃ �(2J + 1)[J(J + 1)]e−hc β B̃ J(J+1)
∂β V
J
471
12 STATISTICAL THERMODYNAMICS
It is convenient to rewrite these expressions in terms of the characteristic vibrational temperature θ R = hc B̃�k: using this hc B̃ = kθ R and hcβ B̃ = θ R �T.
With these, the expressions for q R and its derivative become
q R = �(2J +1)e−θ J(J+1)�T
�
R
J
R
∂q R
� = −kθ R �(2J +1)[J(J +1)]e−θ J(J+1)�T
∂β V
J
�e internal energy is therefore
U m (T) − U m (0) =
Rθ R
−θ R J(J+1)�T
�(2J + 1)[J(J + 1)]e
qR J
�e sums are best evaluated using mathematical so�ware and the results are
expressed in terms of the dimensionless parameter T�θ R . �e result of such a
calculation is shown in Fig. ��.��
2
R
Sm
�R
472
1
0
0
1
2
3
T�θ
4
5
R
Figure 12.16
P��E.��
Contributions to the entropy from translation, rotation and vibration are expected. �e molecule has a doubly-degenerate ground electronic state, so this
will also contribute to the entropy. However, the excited electronic states are at
energies very much greater than kT at ��� K (kT at ��� K is �.��� eV), so their
contribution is negligible.
�e translational contribution to the standard molar entropy is given by the
Sackur–Tetrode equation [��E.�b–���]
T
Sm
= R ln �
kTe5�2
�
p−○ Λ 3
Λ = h�(2πmkT)1�2
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Taking the mass of F−2 as 38.00 m u
Λ=
6.6261 × 10−34 J s
�2π(38.00×1.6605 × 10−27 kg)×(1.3806 × 10−23 J K−1 )×(298 K)�
= 1.64... × 10−11 m
T
Sm
= (8.3145 J K−1 mol−1 ) × ln �
= 1.54... × 102 J K−1 mol−1
1�2
(1.3806 × 10−23 J K−1 ) × (298 K) × e5�2
�
(105 N m−2 ) × (1.64... × 10−11 m)3
�e rotational constant is given by [��B.�–���], B̃ = ħ�4πcI, with I = µR 2 and
µ = 12 m for a homonuclear diatomic.
B̃ =
=
ħ
ħ
=
4πcI 2πcmR 2
1.0546 × 10−34 J s
2π(2.9979 × 1010 cm s−1 )×[19.00×1.6605 × 10−27 kg)]×(190.0 × 10−12 m)2
= 0.491... cm−1
�e rotational contribution to the entropy is given by [��E.��a–���]; this hightemperature form is applicable at ��� K because this temperature is much higher
than the characteristic rotational temperature, θ R = hc B̃�k = 0.707 K.
R
Sm
= R �1 + ln
kT
�
σ hc B̃
= (8.3145 J K−1 mol−1 ) × �1+
ln
(1.3806 × 10−23 J K−1 ) × (298 K)
�
2(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (0.491... cm−1 )
= 52.7... J K−1 mol−1
�e characteristic vibrational temperature is
θ V = hc ν̃�k
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (450.0 cm−1 )
1.3806 × 10−23 J K−1
= 647.4... K
=
�e vibrational contribution to the standard molar entropy is given by [��E.��b–
���] (note that there is an error in the expression in the text: the argument of
the exponential term in the ln should be negative)
V
θ V �T
V
Sm
= R � θ V �T
− ln(1 − e−θ �T )�
e
−1
= (8.3145 J K−1 mol−1 )×
(647.4... K)�(298 K)
� (647.4 ... K)�(298 K)
− ln[1 − e(−647.4 ... K)�(298 K) ]�
e
−1
= 3.32... J K−1 mol−1
473
474
12 STATISTICAL THERMODYNAMICS
�e electronic partition function is q E = g 0 = 2, therefore the electronic contribution to the molar entropy is
E
Sm
= R ln q E = (8.3145 J K−1 mol−1 ) × ln 2 = 5.76... J K−1 mol−1
�e molar entropy is therefore
P��E.��
−
○
T
R
V
E
Sm
= Sm
+S m
+S m
+S m
= 1.54...×102 +52.7...+3.32...+5.76... = ���.� J K−1 mol−1
It is convenient to calculate the entropy using the approach set out in Problem
P��E.� in which the two quantities q (the partition function) and q̇ are de�ned
q = � e−βε j
q̇ = � βε j e−βε j
j
j
It is shown in the solution to that problem that the entropy can be written in
terms of these as
q̇
S m = R � + ln q�
(��.�)
q
As discussed in Section ��C.�(a) on page ���, the Morse oscillator has a �nite
number of bound levels between the ground level and the dissociation limit.
�e number of these is found by noting that E υ reaches a maximum value at
the dissociation limit, and therefore this limit is found by solving dE υ �dυ = 0
d
�(υ + 12 )hc ν̃ − (υ + 12 )2 hcx e ν̃� = hc ν̃ − 2(υ + 12 )hcx e ν̃
dυ
solving 0 = hc ν̃ − 2(υ max + 12 )hcx e ν̃ gives υ max = 1�2x e − 12
�e energy of the lowest state is E 0 = 12 hc ν̃ − 14 hcx e ν̃, therefore the energies
used to evaluate the partition function are
E υ′ = E υ − E 0
= �(υ + 12 )hc ν̃ − (υ + 12 )2 hcx e ν̃� − [ 12 hc ν̃ − 14 hcx e ν̃]
= υhc ν̃ − (υ 2 + υ)hcx e ν̃
�e partition function is evaluated from the sum
υ max
q = � e−(υhc ν̃−(υ +υ)hc x e ν̃)�k T
2
υ=0
De�ning the characteristic vibrational temperature as θ V = hc ν̃�k gives
υ max
q = � e−(υ−(υ +υ)x e )θ �T
2
V
υ=0
�e sum needed to compute the quantity q̇ is written, by analogy, as
υ max
q̇ = � (υ − (υ 2 + υ)x e )
υ=0
θ V −(υ−(υ 2 +υ)x e )θ V �T
e
T
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�ese results are used with eqn ��.� to compute the entropy. For comparison,
for a harmonic oscillator the entropy is given by [��E.��b–���] (note that there
is an error in the expression in the text: the argument of the exponential term
in the ln should be negative)
S m �R =
θ V �T
eθ V �T − 1
− ln(1 − e−θ �T )
V
Figure ��.�� compares the entropy for various values of x e with that for the
harmonic case. For the smallest value of x e the entropy is initially larger than
that for the harmonic oscillator. �is can be attributed to fact that the energy
levels are more closely spaced for the Morse oscillator than for the harmonic
oscillator. However, at higher temperatures the entropy for the Morse oscillator
starts to level o� because there are a �nite number of levels, whereas for the
harmonic case the entropy continues to increase without limit as there are an
in�nite number of levels. �is behaviour is even more pronounced for x e = 0.05
and x e = 0.10, with the plateau at high temperatures being evident.
�ese plots are, however, somewhat unrealistic. For a typical molecule θ V ≈
1000 K, so at ��� K T�θ V ≈ 0.1, and x e is around �.���. With these parameters
the contribution to the entropy determined using the Morse levels is 5.0 ×
10−4 × R; the result obtained using the harmonic levels is the same. �is is
because there is very little contribution from excited vibrational states, so the
small di�erence between these low-lying states for the Morse and harmonic
oscillators has no signi�cant e�ect on the partition function.
harmonic
x e = 0.02
x e = 0.05
x e = 0.10
3.0
S m �R
2.0
1.0
0.0
0
2
4
6
T�θ
8
10
V
Figure 12.17
P��E.��
�e partition function for a particle con�ned to a box of length X in one dimension is given by [��B.�–���]
q X = X�Λ
Λ = h�(2πmkT)1�2
�erefore the partition function for a particle con�ned to a two-dimensional
box of dimensions X and Y is
q XY = q X qY = XY�Λ 2 = A�Λ 2
475
476
12 STATISTICAL THERMODYNAMICS
where A is the area. Because there are two translational modes, the internal
energy of such a system is given by the equipartition theorem as U = nRT,
where n is the amount in moles. �e entropy of n moles is given by [��E.�b–
���]
S = U�T + N k ln qe�N
Ae
Ae
= nR + N k ln 2 = nR + nN A k ln 2
Λ N
Λ nN A
Am e
= nR + nR ln 2
Λ NA
where in the last step the molar area, A m = A�n, is introduced. �e molar
entropy is therefore
2D
Sm
= R + R ln
= R ln
Am e
Am e
= R ln e + R ln 2
Λ2 NA
Λ NA
A m e2
Λ2 NA
�e translational molar entropy in three dimensions is given by [��E.�a–���]
3D
Sm
= R ln
Vm e5�2
Λ3 NA
�erefore the molar entropy of condensation is
2D
3D
∆S cond. = S m
− Sm
A m e2
Vm e5�2
−
R
ln
Λ2 NA
Λ3 NA
Am Λ
= R ln
Vm e1�2
= R ln
P��E.��
If there are N binucleotides of four di�erent kinds then W = 4 N and
S = k ln W = N k ln 4 = (5 × 108 ) × (1.3806 × 10−23 J K−1 ) × ln 4
= 9.6 × 10−15 J K−1
12F
Derived functions
Answers to discussion questions
D��F.�
�is is discussed in Section ��F.�(c) on page ���.
D��F.�
�is is discussed in Section ��F.�(c) on page ���.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Solutions to exercises
E��F.�(a)
�e equilibrium constant for this dissociation reaction is computed using [��F.��–
���]
K=
g I2 kTΛ 3I2
e−hc D̃ 0 �k T
g I2 p−○ qIR2 qIV2 Λ 6I
where g I = 4 and g I2 = 1. �e various factors are computed separately. Λ is
given by [��B.�–���]
Λ = h�(2πmkT)1�2
Λ I2 =
6.6261 × 10−34 J s
[2π(253.8×1.6605 × 10−27 kg)×(1.3806 × 10−23 J K−1 )×(1000 K)]1�2
= 3.46... × 10−12 m
Λ I = 4.90... × 10−12 m
�e rotational partition function for a homonuclear diatomic in the high-temperature
limit is given by [��B.��a–���]
qR =
=
kT
2hc B̃
(1.3806 × 10−23 J K−1 ) × (1000 K)
= 9.31... × 103
2(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (0.0373 cm−1 )
�e vibrational partition function is given by [��B.��–���], q V = (1−e−hc ν̃�k T )−1
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (214.36 cm−1 )
(1.3806 × 10−23 J K−1 ) × (1000 K)
= 0.308...
V
q = (1 − e−0.308 ... )−1 = 3.76...
hc ν̃�kT =
�e dissociation energy is computed from the well depth (the conversion factor
from eV to cm−1 from inside the front cover is used), using the energy of the
ground state of the harmonic oscillator ε̃ 0 = 12 ν̃
D̃ 0 = D̃ e − ε̃ 0
8065.5 cm−1 1
− 2 × (214.36 cm−1 )
1 eV
= 1.23... × 104 cm−1
= (1.5422 eV) ×
hc D̃ 0 (6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (1.23... × 104 cm−1 )
=
kT
(1.3806 × 10−23 J K−1 ) × (1000 K)
−hc D̃ 0 �k T
e
= 17.7...
= e−17.7 ... = 1.96... × 10−8
477
478
12 STATISTICAL THERMODYNAMICS
With these results the equilibrium constant is computed as
K=
E��F.�(a)
42 × (1.3806 × 10−23 J K−1 ) × (1000 K) × (3.46... × 10−12 m)3
1 × (105 Pa) × (9.31... × 103 ) × (3.76...) × (4.90... × 10−12 m)6
× (1.96... × 10−8 )
= 3.72 × 10−3
�e Gibbs energy is computed from the partition function using [��F.�–���],
G(T) = G(0) − nRT ln q�N. As usual, the partition function is factored into
separate contributions from translation, rotation and so on. �e factor of 1�N
is usually taken with the translational contribution, so that, for example, the
rotational contribution to the Gibbs energy is −nRT ln q R , or −RT ln q R for the
molar quantity.
For a centro-symmetric linear molecule the rotational partition function in the
high-temperature limit is given by [��B.��a–���]
qR =
kT
2hc B̃
(1.3806 × 10−23 J K−1 ) × (298 K)
2(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (0.3902 cm−1 )
= 265.3...
=
R
Gm
= −RT ln q R
= −(8.3145 J K−1 mol−1 ) × (298 K) × ln(265.3...) = −13.83 kJ mol−1
�e vibrational partition function for each mode is given by [��B.��–���], q V =
1�(1−e−β hc ν̃ ), where β = 1�kT. �e overall vibrational partition function is the
product of the partition functions of the individual modes, taking into account
any degeneracy. In this case the contribution from the mode at ���.� cm−1 is
included twice.
hcβ =
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
(1.3806 × 10−23 J K−1 ) × (298 K)
= 4.82... × 10−3 cm
−1
q V1 = �1 − e−hc β ν̃ 1 �
= �1 − e−(4.82 ...×10
q V2 = �1 − e−(4.82 ...×10
q V3 = �1 − e−(4.82 ...×10
−3
−3
−3
cm)×(1388.2 cm−1 )
cm)×(2349.2 cm−1 )
cm)×(667.4 cm−1 )
−1
�
−1
�
−1
�
= 1.00...
= 1.00...
= 1.04...
q V = q V1 × q V2 × (q V3 )2 = (1.00...) × (1.00...) × (1.04...)2 = 1.08...
Hence
V
Gm
= −RT ln q V
= −(8.3145 J K−1 mol−1 ) × (298 K) × ln(1.08...) = −0.204 kJ mol−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E��F.�(a)
�e Gibbs energy is computed from the partition function using [��F.�–���],
G(T) = G(0) − nRT ln q�N. As usual, the partition function is factored into
separate contributions from translation, rotation and so on. �e factor of 1�N
is usually taken with the translational contribution, therefore the electronic
contribution to the Gibbs energy is −nRT ln q E , or −RT ln q E for the molar
quantity.
�e electronic partition function of this two-level system is
q E = g 0 + g 1 e−β hc ν̃
where g 0 and g 1 are the degeneracies of the ground and �rst excited state,
respectively. For an electronic term the degeneracy is 2J + 1, hence g 0 = 2 × 32 +
1 = 4 and g 1 = 2 × 12 + 1 = 2. With the data given
βhc ν̃ = hc ν̃�k × T −1
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 ) × (881 cm−1 )
× T −1
1.3806 × 10−23 J K−1
= (1.26... × 103 K) × T −1
=
at ��� K
q E = 4 + 2e−(1.26 ...×10 K)�(500 K) = 4.15...
3
E
Gm
= −RT ln q E
= −(8.3145 J K−1 mol−1 ) × (500 K) × ln(4.15...) = −5.92 kJ mol−1
at ��� K
q E = 4 + 2e−(1.26 ...×10 K)�(900 K) = 4.48...
3
E
Gm
= −(8.3145 J K−1 mol−1 ) × (900 K) × ln(4.48...) = −11.2 kJ mol−1
Solutions to problems
P��F.�
�e equilibrium constant for this reaction is given by [��F.��b–���]
K=
−
○
qCHD
q −○
3 ,m DCl,m −∆ r E 0 �RT
e
−
○
qCD
q −○
4 ,m HCl,m
It is convenient to consider the contribution of each mode to the fraction in the
above expression separately.
�e standard molar translational partition function is qm−○ = Vm−○ �Λ 3 , with Λ =
h�(2πmkT)1�2 , therefore qm−○ goes as m 3�2 . In the fraction all of the other constants cancel to leave
�
−
○
qCHD
q −○
m CHD3 m DCl
3 ,m DCl,m
�
=�
�
−
○
−
○
qCD4 ,m qHCl,m trans
m CD4 m HCl
=�
3�2
19.06 × 37.46 3�2
� = 0.964...
20.07 × 36.46
479
480
12 STATISTICAL THERMODYNAMICS
Assuming the high-temperature limit, the rotational partition function for a
heteronuclear diatomic is given by [��B.��b–���], q R = kT�hc B̃, and for a
nonlinear molecule by [��B.��–���], q R = (1�σ)(kT�hc)3�2 (π�ÃB̃C̃)1�2 ; for
CD� à = C̃ = B̃ and for CHD� C̃ = B̃. �e symmetry number is �� for CD� and
� for CHD� . In the fraction the terms in kT�hc cancel to leave
B̃ 3CD4
�
qCHD3 qDCl
12 �
�
� =
2
qCD4 qHCl rot 3 � Ã CHD3 B̃ CHD3 �
=
1�2
12
(2.63)3
�
�
3 (2.63) × (3.28)2
B̃ HCl
B̃ DCl
1�2
10.59
= 6.23...
5.445
�e vibrational partition function is given by [��B.��–���] q V = (1−e−hc ν̃�k T )−1 ,
which is conveniently expressed as q V = (1 − e−(1.4388 cm K)ν̃�T )−1 . �is term
is temperature dependent and so needs to be re-evaluated at each temperature.
�e vibrational partition function for CHD� and CD� is the product of the
partition function for each normal mode, raised to the power of its degeneracy.
For example
V
V
V
2
V
3
V
3
qCD
= q(2109
cm−1 ) × (q(1092 cm−1 ) ) × (q(2259 cm−1 ) ) × (q(996 cm−1 ) )
4
�e term ∆ r E 0 is computed as
∆ r E 0 = E 0 (CHD3 ) + E 0 (DCl) − E 0 (CD4 ) − E 0 (HCl)
To a good approximation it can be assumed that the pure electronic energy
of a species is una�ected by isotopic substitution, however the vibrational zero
point energy will be a�ected. For a harmonic oscillator the energy of the ground
state is 12 hc ν̃, therefore to compute the total vibrational zero point energy of
CHD� and CD� the contribution from each normal mode has to be taken into
account; a mode with degeneracy g contributes g × 12 hc ν̃.
E 0 (CHD3 )vib = 12 N A hc(2993 + 2142 + 3 × 1003 + 2 × 1291 + 2 × 1036)
= N A hc(6399 cm−1 )
E 0 (CD4 )vib = 12 N A hc(2109 + 2 × 1092 + 3 × 2259 + 3 × 996)
= N A hc(7029 cm−1 )
E 0 (HCl)vib = N A hc(1495.5 cm−1 )
E 0 (DCl)vib = N A hc(1072.5 cm−1 )
∆ r E 0 = N A hc(6399 + 1072.5 − 7029 − 1495.5) = N A hc(−1053 cm−1 )
�us the term −∆ r E 0 �RT evaluates as
−∆ r E 0 −N A hc(−1053 cm−1 )
=
RT
RT
= −(6.0221 × 1023 mol−1 ) × (6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
×
(−1053 cm−1 )
= (1515 K)�T
(8.3145 J K−1 mol−1 ) × T
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
With these expressions the equilibrium constant is evaluated using mathematical so�ware and the results are plotted as a function of temperature in Fig ��.��.
At ��� K K = 945 and at ���� K K = 36.9; the value of the equilibrium constant
is dominated by the symmetry factors and the e−∆ r E 0 �RT term.
800
K
600
400
200
300
400
500
600 700
T�K
800
900 1 000
Figure 12.18
P��F.�
In the absence of a magnetic �eld the ground state of the I atom, with term
symbol 2 P3�2 , has a degeneracy given by (2J + 1) = (2 × 32 + 1) = 4. When a
magnetic �eld is applied this level splits into four states characterised by M J =
+ 32 , + 12 , − 12 , − 32 . �e energy of these states is given, by analogy with [��A.��c–
���], by E M J = g µ B BM J , where the g-value is given as 43 , µ B is the Bohr magneton, and B is the applied magnetic �eld. �e energies are therefore E±3�2 =
±2µ B B, E±1�2 = ± 23 µ B B, giving the partition function as
q E = e2µ B B�k T + e(2�3)µ B B�k T + e−2µ B B�k T + e−(2�3)µ B B�k T
Because it is expected that µ B B � kT the exponentials are expanded to second
order; letting x = µ B B�kT gives
q E = [1 + 2x + 2x 2 ] + [1 + (2�3)x + (2�9)x 2 ]
+ [1 − 2x + 2x 2 ] + [1 − (2�3)x + (2�9)x 2 ]
= 4 + (40�9)x 2 = 4 �1 + (10�9)(µ B B�kT)2 �
As is seen, the linear terms cancel which is why it is necessary to expand to
second order. In the absence of a magnetic �eld q E = 4, and because the
⇀ � I, the electronic partition function appears squared
equilibrium is I� ���
in the numerator of the expression for K. �erefore
2
K(B) � 4 �1 + (10�9)(µ B B�kT) � �
=
K(0) �
4
�
2
= �1 + (10�9)(µ B B�kT)2 � ≈ 1 + (20�9)(µ B B�kT)2
2
481
482
12 STATISTICAL THERMODYNAMICS
where to go to the �nal expression only the squared term is retained.
For a change of �%
(20�9)(µ B B�kT)2 = 0.01
hence B 2 = 0.01 × (9�20) × (kT�µ B )2
9 1�2 (1.3806 × 10−23 J K−1 ) × (1000 K)
� ×
2000
9.2740 × 10−24 J T−1
= ��� T
B=�
�is is a very strong magnetic �eld which at present can only be generated by
special techniques and only then for a very short times.
P��F.�
�e standard molar Gibbs energy is computed from the partition function us−
○
−
○
ing [��F.�b–���], G m
(T) = G m
(0) − RT ln qm−○ �N A . As usual, the partition
function is factored into separate contributions from translation, rotation and
so on. �e factor of 1�N A is usually taken with the translational contribution.
�e standard molar translational partition function is given by qm−○ = Vm−○ �Λ 3 =
RT�p−○ Λ 3 . Taking the mass of Cl� O� as 2(35.45+16.00) = 102.9 m u , Λ is given
by [��B.�–���]
Λ = h�(2πmkT)1�2
=
6.6261 × 10−34 J s
[2π(102.9×1.6605 × 10−27 kg)×(1.3806 × 10−23 J K−1 )×(200 K)]1�2
=
(8.3145 J K−1 mol−1 ) × (200 K)
(105 N m−2 ) × (1.21... × 10−11 m)3 × (6.0221 × 1023 mol−1 )
= 1.21... × 10−11 m
RT
qm−○ �N A = −○ 3
p Λ NA
= 1.53... × 107
For a nonlinear molecule the rotational partition function is given by [��B.��–
���], q R = (1�σ)(kT�hc)3�2 (π�ÃB̃C̃)1�2 ; here σ = 2. If the rotational constants are expressed in frequency units this expression becomes
q R = (1�σ)(kT�h)3�2 (π�ABC)1�2
= 12 �
(1.3806 × 10−23 J K−1 ) × (200 K)
�
6.6261 × 10−34 J s
�
3�2
π
�
(1.31094 × 1010 Hz) × (2.4098 × 109 Hz) × (2.1397 × 109 Hz)
= 2.89... × 104
1�2
�e vibrational partition function for each mode is given by [��B.��–���], q V =
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
1�(1 − e−β hc ν̃ ), where β = 1�kT.
hcβ =
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
(1.3806 × 10−23 J K−1 ) × (200 K)
= 7.19... × 10−3 cm
−1
q V1 = �1 − e−hc β ν̃ 1 �
= �1 − e−(7.19 ...×10
−3
cm)×(753 cm−1 )
−1
�
= 1.00...
�e partition functions for the other normal modes evaluate to 1.02..., 1.12...,
1.66..., 1.00..., 1.05... in order of the given modes. �e overall vibrational partition function is the product of these individual contributions: q V = 2.03....
�e overall partition function is the product of these contributions from the
di�erent modes, therefore
−
○
−
○
Gm
(200)−G m
(0) = − RT ln qm−○ �N A
= −(8.3145 J K−1 mol−1 ) × (200 K)
×ln �(1.53... × 107 )×(2.89... × 104 )×(2.03...)�
Answers to integrated activities
I��.�
= −45.8 kJ mol−1
Note that there is an error in the question: the expression for ξ(β) should
include an additional factor of g(J). To make the notation more compact the
energy levels will be written ε J and the degeneracies g J ; derivatives with respect
to β will be assumed to be at constant V .
First, an expression for C V is developed.
U = −N
1 dq
1 d
−βε
= −N
� gJ e J
q dβ
q dβ J
1
= N � g J ε J e−βε J
q J
Noting that d�dT = −kβ 2 (d�dβ)
dU
dU
= −kβ 2
dT
dβ
�
�
d �1
�
� � g J ε J e−βε J �
= −N kβ 2
�
dβ �
q
� J
�
� −1 dq
�
1
�
�
= −N kβ 2 � 2
g J ε J e−βε J − � g J ε 2J e−βε J �
�
� q dβ J
�
q
J
�
�
�
�
�
�
1
2� 1 �
−βε J ′ � �
−βε J �
2 −βε J �
= −N kβ � 2 � g J ′ ε J ′ e
− � gJ εJ e
� gJ εJ e
�
�� J
� q J
� q � J′
�
�
�
CV =
483
484
12 STATISTICAL THERMODYNAMICS
�e numerator and denominator of the �nal term in the bracket are both multiplied by q, and then a factor of 1�q 2 is taken outside the bracket to give
�
�
�
�
−N kβ 2 �
−βε J ′ � �
−βε J �
2 −βε J �
�
CV =
− q � gJ εJ e
� g J′ ε J′ e
� gJ εJ e
�
�
2
q
�� J
�
�� J ′
�
J
�
�
�
2��
−N kβ �
−βε � �
−βε �
=
� g J ′ ε J ′ e J′ � g J ε J e J
q2 �
�
�
�
�
� J′
J
�
�
�
��
��
− � g J ′ e−βε J′ � g J ε 2J e−βε J �
� J′
�� J
��
�
�
�e product of the sums are next rewritten as double sums
CV =
�
−N kβ 2 �
�
�
�� g J g J ′ ε J ε J ′ e−β(ε J +ε J′ ) − � g J g J ′ ε 2J e−β(ε J +ε J′ ) �
2
� J, J ′
�
q
′
J, J
�
�
Taking a hint from the �nal result, consider the double sum
�(ε J − ε J ′ ) g J g J ′ e
2
J, J ′
−β(ε J +ε J ′ )
= � ε 2J g J g J ′ e−β(ε J +ε J′ ) + � ε 2J ′ g J g J ′ e−β(ε J +ε J′ ) − 2 � ε J ε J ′ g J g J ′ e−β(ε J +ε J′ )
J, J ′
J, J ′
J, J ′
�e �rst two sums only di�er by swapping the indices J and J ′ , so they are in
fact identical. Hence the last line may be written
= 2 � ε 2J g J g J ′ e−β(ε J +ε J′ ) − 2 � ε J ε J ′ g J g J ′ e−β(ε J +ε J′ )
J, J ′
J, J ′
Apart from an overall sign and a factor of 12 , these two terms are the same as
those in the bracket in the expression for C V above, hence
CV =
=
�
−N kβ 2 �
�
−β(ε J +ε J ′ )
2 −β(ε J +ε J ′ ) �
′ ε J ε J′ e
′ε e
�
�
g
g
−
g
g
�
�
J
J
J
J
J
�
q2 �
J, J ′
�
� J, J ′
N kβ 2
2
−β(ε J +ε J ′ )
�(ε J − ε J ′ ) g J g J ′ e
2q 2 J, J ′
which is the required expression.
For a diatomic βε J = βhc B̃J(J + 1) = hc B̃J(J + 1)�kT = θ R J(J + 1)�T, where
θ R = hc B̃�k; the degeneracy is g J = (2J + 1). For the molar quantity N A kβ 2 =
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
N A k�k 2 T 2 = R�k 2 T 2 . �e molar hear capacity is therefore given by
C V ,m �R =
1
1
k 2 T 2 2q 2
(hc B̃)2
× �[J(J + 1) − J ′ (J ′ + 1)]2 (2J + 1)(2J ′ + 1) e−θ [J(J+1)+J (J +1)]�T
=�
J, J ′
R
′
′
2
θR
1
�
T
2q 2
× �[J(J + 1) − J ′ (J ′ + 1)]2 (2J + 1)(2J ′ + 1) e−θ [J(J+1)+J (J +1)]�T
R
J, J ′
′
′
�is expression is used to generate the curves in Fig. ��.�� for particular pairs of
values of J and J ′ , that is just one term from the double sum. However, the term
for J = 0, J ′ = 1 is identical to that for J = 1, J ′ = 0, so the curves plotted in the
�gure are twice the value for the particular combination of J and J ′ indicated.
�is double sum is not a particularly e�cient method for computing the heat
capacity, but it can be evaluated using mathematical so�ware to give the curve
also shown in Fig. ��.��. For a plot up to T�θ R = 5 if is su�cient to consider
contributions from levels with J ≤ 10; this makes the calculation more tractable.
J = 0, J ′ = 1
J = 0, J ′ = 2
J = 1, J ′ = 2
J = 1, J ′ = 3
J = 0, J ′ = 3
total
C V ,m �R
1.0
0.5
0.0
0
1
2
3
4
5
T�θ R
Figure 12.19
I��.�
(a) In the high-temperature limit, the rotational partition function of an asymmetric rotor is given by [��B.��–���], q R = (1�σ)(kT�hc)3�2 (π�ÃB̃C̃)1�2 ,
where σ is the symmetry number. �e point group for ethene is D 2h
485
486
12 STATISTICAL THERMODYNAMICS
which contains the rotational operations (E, C 2x , C 2 , C 2z ); therefore σ = 4.
y
qR =
=
1 kT 3�2
π 1�2
� � �
�
4 hc
ÃB̃C̃
1
(1.3806 × 10−23 J K−1 ) × (298.15 K)
�
�
4
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
�
3�2
π
�
(4.828 cm−1 ) × (1.0012 cm−1 ) × (0.8282 cm−1 )
1�2
= ���.�
(b) Pyridine belongs to the point group C 2v which contains the rotational
operations (E, C 2 ); therefore σ = 2.
qR =
=
1 kT 3�2
π 1�2
� � �
�
2 hc
ÃB̃C̃
1
(1.3806 × 10−23 J K−1 ) × (298.15 K)
�
�
2
(6.6261 × 10−34 J s) × (2.9979 × 1010 cm s−1 )
�
3�2
π
�
−1
(0.2014 cm ) × (0.1936 cm−1 ) × (0.0987 cm−1 )
= 4.26 × 104
1�2
Molecules in motion
13
13A
Transport properties of a perfect gas
Answers to discussion questions
D��A.�
�e di�usion constant is given by [��A.�–���], D = 13 λυ mean . �e mean free
path λ decreases as the pressure is increased ([��A.�a–���]), so D decreases
with increasing pressure and, as a result, the gas molecules di�use more slowly.
�e mean speed υ mean increases with the temperature ([��A.�b–���]), so D also
increases with temperature. As a result, molecules in a hot gas di�use more
quickly than those when the gas is cool (for a given concentration gradient).
Because the mean free path increases when the collision cross-section σ of
the molecules decreases, the di�usion coe�cient is greater for small molecules
than for large molecules.
�e viscosity is given by [��A.��c–���], η = 13 υ mean λmN . �e mean free path
is inversely proportional to the pressure and N is proportional to the pressure, therefore the product λN , and hence the viscosity, is independent of
pressure. �e physical reason for this pressure-independence is that as the
pressure increases more molecules are available to transport the momentum,
but they carry it less far on account of the decrease in mean free path. �e
mean speed goes as T 1�2 (at constant volume) and so the viscosity increases
with temperature. �is is because at high temperatures the molecules travel
more quickly, so the �ux of momentum is greater.
Solutions to exercises
E��A.�(a)
�e rate of e�usion, r is given by [��A.��–���], r = pA 0 N A �(2πMRT)1�2 ; this
rate is the number of molecules escaping through the hole in a particular period
of time, divided by that time. �e mass loss ∆m in period ∆t is therefore ∆m =
∆t pA 0 N A �(2πMRT)1�2 × m, where m is the mass of a molecule. �is mass is
written m = M�N A and so it follows ∆m = ∆t pA 0 M 1�2 �(2πRT)1�2 . �is is
rearranged to give an expression for p
p=
∆m(2πRT)1�2
∆m 2πRT 1�2
=
�
�
∆tA 0
M
∆tA 0 M 1�2
2.85 × 10−4 kg
2π × (8.3145 J K−1 mol−1 ) × (673.15 K)
=
�
�
(400 s) × π × (2.5 × 10−4 m)2
0.100 kg mol−1
= 2.15 × 103 Pa
1�2
488
13 MOLECULES IN MOTION
E��A.�(a) �e rate of e�usion, r is given by [��A.��–���], r = pA 0 N A �(2πMRT)1�2 ; this
rate is the number of molecules escaping through the hole in a particular period
of time, divided by that time. In this experiment the pressure changes so the rate
of e�usion changes throughout the experiment; nevertheless, the rate is always
proportional to M −1�2 . �e two experiments involve comparing the time for
the same drop in pressure, therefore the only factor that a�ects this time is the
molar mass of the e�using gas. Because the rate is proportional to M −1�2 the
time for a given fall in pressure will be proportional to the inverse of this, that
is M 1�2 . It follows that
rate for gas A time for gas B
M B 1�2
=
=�
�
rate for gas B time for gas A
MA
�erefore
42 s
M N2 1�2
=�
�
52 s
MA
hence
M A = (28.02 g mol−1 ) �
52 2
� = ��.� g mol−1
42
E��A.�(a) �e rate of e�usion is given by [��A.��–���], dN�dt = pA 0 N A �(2πMRT)1�2 ;
this is the rate of change of the number of molecules. If it is assumed that the
gas is perfect, the equation of state pV = N kT allows the number to be written
as N = pV �kT, and therefore dN�dt = (V �kT)dp�dt. �e rate of change of
the pressure is therefore
dp
kT
pA 0 N A
RTA 0
A 0 RT 1�2
=−
=
−
×
p
=
−
�
� ×p
dt
V (2πMRT)1�2
V 2πM
V (2πMRT)1�2
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
α
�e minus sign is needed because the pressure falls with time. �is di�erential equation is separable and can be integrated between p = p i and p = p f ,
corresponding to t = 0 and t = t.
�
pf
pi
(1�p) dp = �
t
0
−α dt
hence
ln(p f �p i ) = −αt
�e time for the pressure to drop by the speci�ed amount is therefore
t = − ln(p f �p i )�α = ln(p i �p f )
= ln �
V 2πM 1�2
�
�
A 0 RT
8.0 × 104 Pa
(3.0 m3 )
2π × (3.200 × 10−2 kg mol−1 )
�
�
�
4
−4
2
7.0 × 10 Pa [π(10 m) ] (8.3145 J K−1 mol−1 ) × (298 K)
= 1.15 × 105 s = �.� days
1�2
E��A.�(a) For a perfect gas, the collision �ux Z w is [��A.�a–���], Z w = p�(2πmkT)1�2 .
�e number of argon molecule collisions within area A in time interval t is
therefore N = Z w At . �e mass m is written in terms of the molar mass M:
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
m = M�N A .
N = Z w At =
=
1�2
pN A
p
At =
At
1�2
(2πmkT)
(2πMkT)1�2
(90 Pa) × (6.0221 × 1023 mol−1 )1�2 × [(2.5 × 3.0) × 10−6 m2 ] × (15 s)
[2π × (0.03995 kg mol−1 ) × (1.3806 × 10−23 J K−1 ) × (500 K)]1�2
= 1.9 × 1020
collisions
E��A.�(a) �e di�usion constant is given by [��A.�–���], D = 13 λυmean , where λ is the
mean free path length λ = kT�σ p [��A.�a–���], and υmean is the mean speed
υmean = (8RT�πM)1�2 [��A.�b–���].
D=
=
1 kT 8RT 1�2
�
�
3 σ p πM
(1.3806 × 10−23 J K−1 )×(293.15 K)
3 × (3.6 × 10−19 m2 ) × (p�Pa)
8×(8.3145 J K−1 mol−1 )×(293.15 K)
�
π×(0.03995 kg mol−1 )
1
= (1.477... m2 s−1 ) ×
p�Pa
�
1�2
�e �ux of argon atoms J z is related to the di�usion coe�cient D and the
concentration gradient dN �dz by [��A.�–���], J z = −DdN �dz. From the
perfect gas equation, pV = N kT, the number density is expressed in terms
of the pressure as N = N�V = p�kT. With this, the concentration gradient is
written in terms of the pressure gradient: dN �dz = (1�kT)dp�dz, and hence
the �ow is J z = −(D�kT)dp�dz
−D dp
−1
(1.47... m2 s−1 ) × (1.0 × 105 Pa m−1 )
=
×
kT dz p�Pa (1.3806 × 10−23 J K−1 ) × (293.15 K)
1
= −(3.64... × 1025 m−2 s−1 ) ×
p�Pa
Jz =
p�Pa
�.��
1.00 × 105
1.00 × 107
D�(m2 s−1 )
�.��
1.48 × 10−5
1.48 × 10−7
J z �(m−2 s−1 )
−3.65 × 1025
−3.65 × 1020
−3.65 × 1018
(J z �N A )�(mol m−2 s−1 )
−60.6
−6.06 × 10−4
−6.06 × 10−6
E��A.�(a) �e thermal conductivity is given by [��A.��c–���], κ = νpD�T, where the
di�usion coe�cient D is given by [��A.�–���], D = λυ mean �3. �e mean free
path λ is given by [��A.�a–���], λ = kT�σ p, and the mean speed υ mean is
given by [��A.�b–���], υmean = (8RT�πM)1�2 . �e quantity ν is the number
of quadratic contributions to the energy, and this is related to the heat capacity
489
490
13 MOLECULES IN MOTION
by C V ,m = νkN A , hence ν = C V ,m �kN A . �e thermal conductivity is therefore
expressed as
κ=
hence κ =
νpD νpλυ mean C V ,m p kT 8RT 1�2 C V ,m 8RT 1�2
=
=
�
� =
�
�
T
3T
kN A 3T σ p πM
3σ N A πM
12.5 J K−1 mol−1
3 × (3.6 × 10−19 m2 ) × (6.0221 × 1023 mol−1 )
�
8 × (8.3145 J K−1 mol−1 ) × (298 K)
�
π × (3.995 × 10−2 kg mol−1 )
1�2
= 7.6 × 10−3 J K−1 m−1 s−1
E��A.�(a) �e thermal conductivity is given by [��A.��c–���], κ = νpD�T, where the
di�usion coe�cient D is given by [��A.�–���], D = λυ mean �3. �e mean free
path λ is given by [��A.�a–���], λ = kT�σ p, and the mean speed υ mean is
given by [��A.�b–���], υmean = (8RT�πM)1�2 . �e quantity ν is the number
of quadratic contributions to the energy, and this is related to the heat capacity
by C V ,m = νkN A , hence ν = C V ,m �kN A . �e thermal conductivity is therefore
expressed as
κ=
νpD νpλυ mean C V ,m p kT 8RT 1�2 C V ,m 8RT 1�2
=
=
�
� =
�
�
T
3T
kN A 3T σ p πM
3σ N A πM
Rearranging gives an expression for σ in terms of the thermal conductivity
σ=
C V ,m 8RT 1�2
�
�
3κN A πM
�e value of C p,m is given in the Resource section; C V ,m is found using C p,m −
C V ,m = R for a perfect gas.
σ=
(20.786 J K−1 mol−1 ) − (8.3145 J K−1 mol−1 )
3 × (4.65 × 10−2 J K−1 m−1 s−1 ) × (6.0221 × 1023 mol−1 )
�
8 × (8.3145 J K−1 mol−1 ) × (273 K)
�
π × (2.018 × 10−2 kg mol−1 )
1�2
�e value reported in Table �B.� on page �� is 0.24 nm2 .
= 0.0795 nm2
E��A.�(a) �e �ux of energy is given by [��A.�–���], J z = −κ dT�dz. �e value of the
thermal conductivity κ for Ar at ��� K is determined in Exercise E��A.�(a) as
7.6 × 10−3 J K−1 m−1 s−1 . As is seen in that Exercise, κ ∝ T 1�2 provided that
the heat capacity is constant over the temperature range of interest. It therefore
1�2
follows that κ 280 K = (280 K�298 K) κ 298 K = 7.40... × 10−3 J K−1 m−1 s−1 .
With these data the �ux is computed as
Jz = −κ dT�dz = −(7.40... × 10−3 J K−1 m−1 s−1 ) × (10.5 K m−1 )
= −0.078 J m−2 s−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E��A.�(a) �e �ux of energy is given by [��A.�–���], J z = −κ dT�dz, where κ is the
thermal conductivity and the negative sign indicates �ow of heat is towards
the lower temperature. �e rate of energy transfer is r = J z A, where A is the
cross-sectional area. �e temperature gradient is approximated as dT�dz =
∆T�∆z; because � W = � J s−1 it follows that �� mW K−1 m−1 is equivalent to
2.4 × 10−2 J K−1 m−1 s−1
r = J z A = −κA
∆T
∆z
= −(2.4 × 10−2 J K−1 m−1 s−1 ) × (1.0 m2 ) ×
= 103 J s−1 = 103 W
[(−15) − (28)] K
0.010 m
Hence a heater of power 103 W is required to make good the loss of heat.
E��A.��(a) �e viscosity η is given by [��A.��c–���], η = pMD�RT. In turn the di�usion constant is given by [��A.�–���], D = 13 λυmean , where λ is the mean free
path length λ = kT�σ p [��A.�a–���], and υmean is the mean speed υmean =
(8RT�πM)1�2 [��A.�b–���]. �e �rst step is to �nd an expression for η as a
function of temperature
pMD pM kT 8RT 1�2
M
8RT 1�2
1
8RM 1�2 1�2
=
�
� =
�
� =
�
� T
RT
RT 3σ p πM
3σ N A πM
3σ N A
π
1
=
−19
2
3 × (4.0 × 10 m ) × (6.0221 × 1023 mol−1 )
η=
�
8 × (8.3145 J K−1 mol−1 ) × (0.029 kg mol−1 )
�
π
1�2
T 1�2
= (1.08... × 10−6 kg K−1�2 m−1 s−1 ) × (T� K)1�2
where 1 J = 1 kg m2 s−2 has been used to arrive at the units on the �nal line. Using this expression the following table is drawn up (recall that 10−7 kg m−1 s−1 =
1 µP)
T�K
���
���
����
η�(kg m−1 s−1 )
1.79 × 10−5
1.87 × 10−5
3.43 × 10−5
η�(µP)
���
���
���
E��A.��(a) In the solution to Exercise E��A.��(a) it is shown that
η=
1
8RMT 1�2
�
�
3σ N A
π
hence
σ=
1
8RMT 1�2
�
�
3ηN A
π
491
492
13 MOLECULES IN MOTION
Recalling that 10−7 kg m−1 s−1 = 1 µP, the cross section is computed as
1
σ=
3 × (2.98 × 10−5 kg m−1 s−1 ) × (6.0221 × 1023 mol−1 )
�
8 × (8.3145 J K−1 mol−1 ) × (0.02018 kg mol−1 ) × (273 K)
�
π
1�2
= �.��� nm2
E��A.��(a) �e rate of e�usion, r is given by [��A.��–���], r = pA 0 N A �(2πMRT)1�2 ; this
rate is the number of molecules escaping through the hole in a particular period
of time, divided by that time. �e mass loss ∆m in period ∆t is therefore ∆m =
∆t pA 0 N A �(2πMRT)1�2 × m, where m is the mass of a molecule. �is mass is
written m = M�N A and so it follows ∆m = ∆t pA 0 M 1�2 �(2πRT)1�2 . Evaluating
this with the values given
∆m =
=
∆t pA 0 M 1�2
(2πRT)1�2
(7200 s) × (0.835 Pa) × π × ( 12 × 2.50 × 10−3 m)2 × (0.260 kg mol−1 )1�2
[2 × π × (8.3145 J K−1 mol−1 ) × (400 K)]1�2
= 1.04 × 10−4 kg = ��� mg
Solutions to problems
P��A.�
In the solution to Exercise E��A.��(a) it is shown that
η=
1
8RMT 1�2
�
�
3σ N A
π
hence
At ��� K and �.�� bar
σ=
σ=
1
8RMT 1�2
�
�
3ηN A
π
1
3 × (9.08 × 10−6 kg m−1 s−1 ) × (6.0221 × 1023 mol−1 )
�
8 × (8.3145 J K−1 mol−1 ) × (0.0170 kg mol−1 ) × (270 K)
�
π
1�2
= 6.00... × 10−19 m2
P��A.�
�e collision cross-section is σ = π(2r)2 , where r is the molecular radius of
NH� and d = 2r is the e�ective molecular diameter. With the value of σ
determined above d is found as 437 pm . A similar calculation at ��� K and
��.� bar gives σ = 4.21... × 10−19 m2 and d = 366 pm .
In the solution to Exercise E��A.�(a) it is shown that the di�usion constant is
given by
1 kT 8RT 1�2
D=
�
�
3 σ p πM
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
If the gas is assumed to be perfect then the equation of state pV = N kT can be
used to �nd the number density N as N = N�V = p�kT. �e collision cross
section is estimated as σ = π(2a 0 )2 where a 0 is the Bohr radius. A density of
� atom cm−3 corresponds to N = 1 × 106 m−3 .
1 kT 8RT 1�2
1
8RT 1�2
�
� =
�
�
3 σ p πM
3[π(2a 0 )2 ]N πM
1
=
3 × [π(2 × 5.2918 × 10−11 m)2 ] × (1 × 106 m−3 )
D=
8 × (8.3145 J K−1 mol−1 ) × (10 × 103 K)
�
�
π(1.0079 × 10−3 kg mol−1 )
1�2
= 1.37 × 1017 m2 s−1
�e thermal conductivity is given in terms of the di�usion constant by [��A.��c–
���], κ = νpD�T, which is rewritten using N = p�kT as κ = νN kD. For an
atom there are just three degrees of translational freedom, ν = 32 .
κ = νN kD = 32 ×(1 × 106 m−3 )×(1.3806 × 10−23 J K−1 )×(1.37... × 1017 m2 s−1 )
= �.�� J K−1 m−1 s−1
For a gas at ambient temperature and pressure a typical value for the di�usion
coe�cient is D = 1.5 × 10−5 m2 s−1 , and a typical value for the thermal conductivity is κ = 0.025 J K−1 m−1 s−1 . �e di�usion constant is much higher in
interstellar space when compared to ambient conditions because in interstellar
space the much higher temperature results in a higher mean speed, and the
much lower pressure results in a longer mean free path. Molecules move more
quickly and experience fewer collisions, resulting in more rapid di�usion.
Because κ ∝ N D and D ∝ 1�N , the value of the thermal conductivity is
una�ected by the change in number density in going from ambient pressure to
interstellar conditions. �e higher thermal conductivity in the latter is therefore
attributable to the higher mean speed.
�e kinetic theory of gases assumes that the rate of atomic collisions is very
high such that thermal equilibrium is established quickly. However, at such
a dilute concentration, the timescales on which particles exchange energy by
collision make this assumption questionable. In fact, atoms are more likely to
interact with photons from stellar radiation than with other atoms.
P��A.�
�e rate of e�usion, r is given by [��A.��–���], r = pA 0 N A �(2πMRT)1�2 . �e
area of the slit is A 0 = (10 mm) × (1.0 × 10−2 mm) = 0.1 mm2 = 1.0 × 10−7 m2 .
r=
pA 0 N A
(2πMRT)1�2
(p� Pa) × (1.0 × 10−7 m2 ) × (6.0221 × 1023 mol−1 )
[2π × (M� kg mol−1 ) × (8.3145 J K−1 mol−1 ) × (380 K)]1�2
(p� Pa)
= (4.27... × 1014 s−1 ) ×
(M� kg mol−1 )1�2
=
493
494
13 MOLECULES IN MOTION
For cadmium, r = (4.27... × 1014 s−1 ) × 0.13�(0.11241)1�2 = 1.7 × 1014 s−1 .
Hence there are 1.7 × 1014 atoms per second in the beam.
For mercury, r = (4.27... × 1014 s−1 ) × 12�(0.20059)1�2 = 1.1 × 1016 s−1 . Hence
there are 1.1 × 1016 atoms per second in the beam.
13B Motion in liquids
Answers to discussion questions
D��B.�
�e Grotthuss mechanism for conduction by protons in water is described
in Section ��B.�(a) on page ��� and illustrated in Fig. ��B.� on page ���. It
seems plausible that such a mechanism could also occur in the relatively open
hydrogen bonded structure of ice.
Solutions to exercises
E��B.�(a)
�e ion molar conductivity λ is given in terms of the mobility u by [��B.��–���],
λ = zuF, where z is the charge number of the ion (unsigned) and F is Faraday’s
constant; it follows that u = λ�zF. Note that 1 S = 1 C V−1 s−1 .
u Li+ =
3.87 mS m2 mol−1
= 4.01×10−5 mS m2 C−1 = 4.01 × 10−8 m2 V−1 s−1
(1)(96485 C mol−1 )
u K+ =
7.35 mS m2 mol−1
= 7.62×10−5 mS m2 C−1 = 7.62 × 10−8 m2 V−1 s−1
(1)(96485 C mol−1 )
u Na+ =
E��B.�(a)
E��B.�(a)
5.01 mS m2 mol−1
= 5.19×10−5 mS m2 C−1 = 5.19 × 10−8 m2 V−1 s−1
(1)(96485 C mol−1 )
�e ion molar conductivity λ is given in terms of the mobility u by [��B.��–���],
λ = zuF, where z is the charge number of the ion (unsigned) and F is Faraday’s
constant. Note that 1 S = 1 C V−1 s−1 .
λ = zuF = (1)×(7.91×10−8 m2 V−1 s−1 )×(96485 C mol−1 ) = �.�� mS m2 C−1
�e dri� speed s of an ion is given by [��B.�b–���], s = uE, where E is the
electric �eld strength. �is �eld strength is given by E = ∆��l where ∆� is the
potential di�erence between two electrodes separated by distance l.
∆�
25.0 V
= (7.92 × 10−8 m2 V−1 s−1 ) ×
l
7.00 × 10−3 m
−4
−1
−1
= 2.83 × 10 m s = ��� µm s
s = uE = u
E��B.�(a) �e Einstein relation, [��B.��–���], u = zDF�RT, gives the relationship between the mobility u, the charge number of the ion z, and the di�usion coe�-
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
cient D.
D=
E��B.�(a)
uRT (7.40 × 10−8 m2 V−1 s−1 ) × (8.3145 J K−1 mol−1 ) × (298 K)
=
zF
(1) × (96485 C mol−1 )
= 1.90 × 10−9 m2 s−1
�e temperature dependence of the viscosity η is given by [��B.�–���], η =
η 0 eE a �RT , where η 0 is viscosity in the limit of high temperature and E a is the
associated activation energy. Taking the natural logarithm gives ln η = ln η 0 +
E a �RT. Hence
ln η 1 − ln η 2 = (ln η 0 + E a �RT1 ) − (ln η 0 + E a �RT2 ) =
Rearranging gives an expression for the activation energy
Ea = R
ln (η 1 �η 2 )
(T1−1 − T2−1 )
= (8.3145 J K−1 mol−1 )
= ��.� kJ mol−1
Ea 1
1
� − �
R T1 T2
ln [(1.002 cP)�(0.7975 cP)]
[(293.15 K)−1 − (303.15 K)−1 ]
E��B.�(a) According to the law of independent migration of ions, the limiting molar
conductivity Λ○m of an electrolyte is given by the sum of the limiting molar
conductivities λ i of the ions present, [��B.�–���], Λ○m = ν+ λ+ + ν− λ− ; in this
expression ν+ and ν− are the numbers of cations and anions provided by each
formula unit of electrolyte. For each of the given electrolytes it follows that
Λ○AgI = λ Ag+ + λ I−
Λ○NaNO3 = λ Na+ + λ NO3 −
�ese expressions are manipulated to give Λ○AgI
Λ○AgNO3 = λ Ag+ + λ NO3 −
Λ○AgI = λ Ag+ + λ I−
= (Λ○AgNO3 − λ NO3 − ) + (Λ○NaI − λ Na+ ) = Λ○AgNO3 + Λ○NaI − Λ○NaNO3
= (13.34 + 12.69 − 12.16) mS m2 mol−1 = ��.�� mS m2 mol−1
Solutions to problems
P��B.�
�e temperature dependence of the viscosity η is given by [��B.�–���], η =
η 0 eE a �RT , where E a is the activation energy . Taking the natural logarithm gives
ln η = ln η 0 +E a �RT. A plot of ln η against (1�T) therefore has slope E a �R; such
a plot is shown in Fig. ��.�.
495
13 MOLECULES IN MOTION
θ�○ C
10
20
30
40
50
60
70
T�K
283
293
303
313
323
333
343
η�cP
0.758
0.652
0.564
0.503
0.442
0.392
0.358
1�(T�K)
0.003 53
0.003 41
0.003 30
0.003 19
0.003 10
0.003 00
0.002 92
ln (η�cP)
−0.277
−0.428
−0.573
−0.687
−0.816
−0.936
−1.027
0.0
−0.5
ln (η�cP)
496
−1.0
−1.5
0.0028
0.0030
0.0032
1�(T�K)
0.0034
0.0036
Figure 13.1
�e data are a good �t to a straight line with equation
ln (η� cP) = (1.2207 × 103 ) × 1�(T�K) − 4.5939
�e activation energy is computed from the slope
E a = R × (slope)
= (8.3145 J K−1 mol−1 )(1.2207 × 103 K) = ��.�� kJ mol−1
P��B.�
�e molar conductivity Λm is de�ned by [��B.�–���], Λm = κ�c, where κ is
conductivity and c is concentration. �e Kohlrausch law, [��B.�–���], gives the
variation of the molar conductivity with concentration as Λm = Λ○m − Kc 1�2 .
Hence a plot of Λm against c 1�2 has slope equal to −K and y-intercept equal to
the limiting molar conductivity Λ○m . In computing Λm the concentration needs
to be converted from mol dm−3 to mol m−3 . �e graph is shown in Fig. ��.�.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
κ�S m−1
13.1
13.9
14.7
15.6
16.4
c�mol dm−3
1.334
1.432
1.529
1.672
1.725
Λm �mS m2 mol−1
9.82
9.71
9.61
9.33
9.51
c 1�2 �(mol dm−3 )1�2
1.155
1.197
1.237
1.293
1.313
Λm �mS m2 mol−1
9.8
9.6
9.4
1.10
1.15
1.20
c
Figure 13.2
1�2
1.25
−3 1�2
�(mol dm )
1.30
�e two points corresponding to the highest concentrations seem to be anomolous
and are ignored in �nding the best-�t line, the equation of which is
Λm �mS m2 mol−1 = (−2.5262) × c 1�2 �(mol dm−3 )1�2 + 12.737
P��B.�
�erefore, K = 2.53 mS m2 (mol dm−1 )−3�2 and Λ○m = 12.7 mS m2 mol−1
(a) �e molar conductivity Λm is given by [��B.�–���], Λm = κ�c, where
κ is conductivity and c is concentration. �e Kohlrausch law, [��B.�–
���], gives the dependence of the molar conductivity on concentration
for strong electrolytes, Λm = Λ○m − Kc 1�2 . According to this law a plot of
Λm against c 1�2 will be a straight line with slope −K and y-intercept Λ○m .
Given that κ = C�R where C = 0.2063 cm−1 , the molar conductivity is
computed from Λm = C�cR. �e plot is shown in Fig. ��.�.
R�Ω
3 314
1 669
342
174
89
37
c�mol dm−3
0.000 50
0.001 0
0.005 0
0.010
0.020
0.050
Λm �mS m2 mol−1
12.45
12.36
12.06
11.85
11.58
11.11
c 1�2 �(mol dm−3 )1�2
0.022
0.032
0.071
0.100
0.141
0.224
497
13 MOLECULES IN MOTION
12.5
Λm �mS m2 mol−1
498
12.0
11.5
11.0
0.00
0.05
0.10
c
Figure 13.3
1�2
0.15
−3 1�2
�(mol dm )
0.20
0.25
�e data fall on a good straight line, as predicted by the Kohlrausch law,
and the equation for the best-�t line as
Λm �mS m2 mol−1 = (−6.6551) × c 1�2 �(mol dm−3 )1�2 + 12.5558
�us K = 6.655 mS m2 (mol dm−1 )−3�2 and Λ○m = 12.56 mS m2 mol−1 .
(b) �e law of independent migration of ions, [��B.�–���], allows the limiting molar conductivity to be calculated from the values for the individual ions, and this is then converted to the molar conductivity using the
Kohlrausch law
Λ○m = ν+ λ(Na+ ) + ν− λ(I− )
= (1) × (5.01 mS m2 mol−1 ) + (1) × (7.68 mS m2 mol−1 )
= 12.69 mS m2 mol−1
Λm = Λ○m − Kc 1�2
= (12.69 mS m2 mol−1 )
− [6.655 mS m2 (mol dm−1 )−3�2 ] × (0.010 mol dm−3 )1�2
= ��.�� mS m2 mol−1
�e conductivity is found using [��B.�–���], and the resistance using the
given cell constant
κ = cΛm = (0.010 mol dm−3 ) × (12.02 mS m2 mol−1 )
= (0.010 × 103 mol m−3 ) × (12.02 mS m2 mol−1 ) = ��� mS m−1
R=
C 0.2063 × 102 m−1
=
= 172 Ω
κ 120 × 10−3 S m−1
where 1S−1 = 1 Ω is used.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P��B.�
A spherical particle of radius a and charge ze travelling at a constant speed
through a solvent of viscosity η has mobility u given by [��B.�–���], u = ze� f ,
where f is the frictional coe�cient with Stokes’ law value f = 6πηa. Hence
a=
ze
(1) × (1.6022 × 10−19 C)
=
6πηu 6π × (0.93 × 10−3 kg m−1 s−1 ) × (1.1 × 10−8 m2 V−1 s−1 )
= 8.30... × 10−10 m = �.�� nm
�is is substantially larger than the �.� nm van der Waals radius of a Buckminsterfullerene (C�� ) molecule because the anion attracts a considerable hydration shell through the London dispersion attraction to the nonpolar solvent
molecules and through the ion-induced dipole interaction. �e Stokes radius
re�ects the larger e�ective radius of the combined anion and its solvation shell.
P��B.�
(a) �e initial concentration of AB is c AB . A�er a fraction α has dissociated
the concentration of AB is c = (1 − α)c AB and the concentration of A is
c A = αc AB , which is equal to the concentration of B, c B . �erefore, the
equilibrium constant K is given by
K=
(c A �c −○ ) (c B �c −○ ) c A c B
(αc AB )2
α 2 c AB
=
=
=
c�c −○
cc −○
(1 − α)c AB c −○ (1 − α)c −○
(b) As the solution becomes more dilute, the degree of dissociation increases
and, in the limit of in�nite dilution, α = 1. �is is a consequence of the
form of K derived in part (a): because K is constant, a decrease in c AB
requires an increase in α towards �.
�e concentration of ions in the solution scales directly with α, therefore
the conductivity, and hence the molar conductivity, will be proportional
to α: Λ m ∝ α. At in�nite dilution the molar conductivity takes the value
Λ m,l and α = 1, therefore α = Λm �Λm,l .
(c) Substitution of this expression for α into the equilibrium expression gives
K=
2
α 2 c AB
Λm
c AB
=
2
−
○
(1 − α)c
Λm,l [1 − (Λm �Λm,l )]c −○
2
α2
Λm
= 2
(1 − α) Λm,1 (1 − ΛΛm )
(1 − α)Λm
=
2
α 2 Λm,1
m
(1 − ΛΛm,1
)
Λm
m,l
=
1
1
−
Λm Λm,l
13C Diffusion
Answers to discussion questions
D��C.�
hence
See the text following [��C.�–���].
2
α 2 Λm,1
Λm
=
(1 − α)Λm (1 − ΛΛm )
m,l
hence
1
1 (1 − α)Λm
=
+
2
Λm Λm,l
α 2 Λm,l
499
500
13 MOLECULES IN MOTION
Solutions to exercises
E��C.�(a)
�e Einstein–Smoluchowski equation [��C.��–���], D = d 2 �2τ, relates the diffusion coe�cient D to the jump distance d and time τ required for a jump.
Approximating the jump length as the molecular diameter, then d ≈ 2a where
a is the e�ective molecular radius. �is is estimated using the Stokes–Einstein
equation [��C.�b–���], D = kT�6πηa, to give 2a = 2kT�6πηD.
Combining these expressions and using the value for viscosity of benzene from
the Resource section gives
τ=
=
2
d2
1
2kT
1
kT
=
�
� =
� �
2D 2D 6πηD
18D 3 πη
2
1
(1.3806 × 10−23 J K−1 ) × (298 K)
�
�
−9
2
−1
3
18 × (2.13 × 10 m s )
π × (0.601 × 10−3 kg m−1 s−1 )
2
= 2.73 × 10−11 s = 27.3 ps
E��C.�(a) �e root mean square displacement in one dimension is given by [��C.��a–���],
�x 2 �1�2 = (2Dt)1�2 , where D is the di�usion coe�cient and t is the time period.
For an iodine molecule in benzene, D = 2.13 × 10−9 m2 s−1
�x 2 �1�2 = (2Dt)1�2 = [2 × (2.13 × 10−9 m2 s−1 ) × (1.0 s)]1�2 = 6.5 × 10−5 m
= 65 µm
For a sucrose molecule in water, D = 0.5216 × 10−9 m2 s−1
�x 2 �1�2 = [2 × (0.5216 × 10−9 m2 s−1 ) × (1.0 s)]1�2 = 3.2 × 10−5 m
= 32 µm
E��C.�(a) �e root mean square displacement in three dimensions is given by [��C.��b–
���], �r 2 �1�2 = (6Dt)1�2 , where D is the di�usion coe�cient and t is the time
period.
�r 2 �
(5.0 × 10−3 m)2
t=
=
= 6.2 × 103 s
6D 6 × (6.73 × 10−10 m2 s−1 )
E��C.�(a) �e di�usion in one dimension from a layer of solute is described by [��C.��–
���]
2
n0
c(x, t) =
e−x �4D t
A(πDt)1�2
where c(x, t) is the concentration at time t and distance x from the layer, and
n 0 is the amount in moles in the layer of area A placed at x = 0. If the mass of
sucrose is m, then n 0 = m�M, where M is the molar mass (���.�� g mol−1 ).
c(x, t) =
2
m
e−x �4D t
1�2
MA(πDt)
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(0.020 kg) × e−(10×10 m) �4×(5.216×10 m s )t
c(10 cm, t)=
(342.30 g mol−1 )×(5.0 × 10−4 m2 )×[π(5.216 × 10−9 m2 s−1 )t]1�2
5
= [(9.12... × 102 mol dm−3 ) × (t� s)−1�2 ]e−4.79 ...×10 �(t� s)
−2
2
−9
2 −1
c(10 cm, 10 s) = (9.12... × 102 mol dm−3 ) × (10)−1�2 × e−4.79 ...×10 �(10)
5
= 0.00 mol dm−3
c(10 cm, 24 h) = (9.12... × 102 mol dm−3 )[24(3600)]−1�2 × e−4.79 ...×10 �[24(3600)]
5
= 0.0121 mol dm−3
Di�usion is a very slow process: a�er �� s the concentration at a height of �� cm
is zero to within the precision of the calculation. Even a�er �� hours only a very
small amount of the sucrose has moved up into the liquid.
E��C.�(a) �e thermodynamic force F is given by [��C.�b–���]
F =−
RT ∂c
� �
c ∂x T , p
Substituting c(x) = c 0 − αc 0 x into the above expression gives
F =−
RT
αRT
(−αc 0 ) =
c 0 − αc 0 x
1 − αx
�e constant α is found by noting that c = c 0 �2 at x = 10 cm = 0.10 m. Hence
c 0 �2 = c 0 − αc 0 × (0.10 m) and therefore α = 5.0 m−1 . At T = 298 K and
x = 10 cm the force is
F=
(5 m−1 ) × (8.3145 J K−1 mol−1 ) × (298 K)
= �� kN mol−1
1 − (5 m−1 )(10 × 10−2 m)
A similar calculation at x = 15 cm gives F = �� kN mol−1 . �e force is greater
at the larger distance, even though the gradient is the same.
E��C.�(a) �e thermodynamic force F is given by [��C.�b–���]
F =−
RT ∂c
� �
c ∂x T , p
Substituting c(x) = c 0 e−α x into the above expression gives
2
F =−
2
RT
(−2αc 0 xe−α x ) = 2αxRT
c 0 e−α x 2
�e constant α is found by noting that c = c 0 �2 at x = 5 cm = 0.05 m.
2
Hence c 0 �2 = c 0 e−α(0.05 m) and therefore α = ln 2�(0.05 m)2 = 277 m−2 �e
thermodynamic force at T = 293 K and x = 5.0 cm is
F = 2(277 m−2 )×(0.050 m)×(8.3145 J K−1 mol−1 )×(293 K) = ��.� kN mol−1
501
502
13 MOLECULES IN MOTION
E��C.�(a) �e root mean square displacement in three dimensions is given by [��C.��b–
���], �r 2 �1�2 = (6Dt)1�2 , where D is the di�usion coe�cient and t is the time
period. Hence,
t=
�r 2 �
(5.0 × 10−3 m)2
=
= 1.3 × 103 s
6D 6 × (3.17 × 10−9 m2 s−1 )
E��C.�(a) �e Stokes–Einstein equation [��C.�b–���], D = kT�6πηa, relates the di�usion coe�cient D to the viscosity η and the radius a of the di�using particle,
which is modelled as a sphere. Recall that 1 cP = 10−3 kg m−1 s−1 .
a=
kT
(1.3806 × 10−23 J K−1 ) × (298 K)
=
= �.�� nm
6πηD 6π × (1.00 × 10−3 kg m−1 s−1 ) × (5.2 × 10−10 m2 s−1 )
Solutions to problems
P��C.�
�ermodynamic force, F, is given by [��C.�b–���].
F =−
RT ∂c
� �
c ∂x T , p
where c is the concentration.For a linear gradation of intensity, that is concentration, down the tube
dc�dx = ∆c�∆x = [(0.050 − 0.100) × 103 mol m−3 ]�(0.10 m) = 500 mol m−4
RT dc
(8.3145 J K−1 mol−1 ) × (298 K)
=−
× (−500 mol m−4 )
c dx
c
1.23... × 106 N mol−1
=
(c� mol m−3 )
F =−
At the le� face, c = 0.100 mol dm−3 :
F = (1.23... × 103 kN mol−1 )�(0.100 × 103 ) = 12.4 kN mol−1
−1
= 2.1 × 10−20 N (molecule)
In the middle, c = 0.075 mol dm−3 :
F = (1.23... × 103 kN mol−1 )�(0.075 × 103 ) = 16.5 kN mol−1
−1
= 2.7 × 10−20 N (molecule)
Close to the le� face, c = 0.050 mol dm−3 :
F = (1.23... × 103 kN mol−1 )�(0.050 × 103 ) = 24.8 kN mol−1
−1
= 4.1 × 10−20 N (molecule)
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P��C.�
�e thermodynamic force F is given by [��C.�b–���]
F =−
RT ∂c
� �
c ∂x T , p
Substituting c(x) = c 0 (1 − e−ax ) into the above expression gives
2
F =−
RT
2axRTe−ax
2axRT
−ax 2
(2ac
xe
)
=
−
=
0
2
c 0 (1 − e−ax )
(1 − e−ax 2 )
(1 − e ax 2 )
2
2
�e �nal step involves multiplying top and bottom of the fraction by e ax . For
thermodynamic force for a = 0.10 cm−2 = 1000 m−2 and T = 298 K is
(5.0 MN mol−1 ) × (x�m)
(1 − e1000×(x�m)2 )
(8.2 × 10−18 N molecule−1 ) × (x�m)
=
(1 − e1000×(x�m)2 )
F=
F�kN mol−1
2 000
0
−2 000
Figure 13.4
−2.0
−1.0
0.0
x�cm
1.0
2.0
A plot of the thermodynamic force per mole against x is shown in Fig. ��.�. It
demonstrates that the force is directed such that mass is pushed by the thermodynamic force toward the centre of the tube to where the concentration is
lowest. A negative force pushes mass toward the le� (x > 0) and a positive
force pushes mass toward the right (x < 0).
At x = 0 the gradient of the concentration is zero, so the thermodynamic force
is also zero. However, as x approaches zero the modulus of the thermodynamic
force increases without limit on account of the concentration becoming smaller
and smaller.
P��C.�
�e generalised di�usion equation is [��C.�–���], where c is concentration, t
is time, D is the di�usion coe�cient and x is displacement.
∂c
∂2 c
=D 2
∂t
∂x
503
504
13 MOLECULES IN MOTION
An expression for c(x, t) is a solution of the di�usion equation if substitution
of the expression for c(x, t) into each side of the di�usion equation gives the
same result. �e proposed solution is
c(x, t) =
LHS
2
n0
e−x �4D t
1�2
A(πDt)
∂c n 0 e−x �4D t
x2
1
=
�
− �
2
1�2
∂t A(πDt)
4Dt
2t
2
RHS
2
−x 2 �4D t
∂2 c
∂ �
−x �
n 0 e−x �4D t
x 2
1
� n0 e
�
�
�
D 2 =D
�
�
=
D
��
� −
�
1�2
1�2
�
�
∂x
∂x � A(πDt)
2Dt �
2Dt
2Dt
A(πDt)
n 0 e−x �4D t
x2
1
=
�
− �
A(πDt)1�2 4Dt 2 2t
2
As required the LHS = the RHS, hence the proposed form of c(x, t) is indeed
a solution to the di�usion equation.
As t → 0 the exponential term e−x �4D t falls o� more and more rapidly, implying that in the limit t = 0 all the material is at x = 0. �e exponential function
dominates the term t 1�2 in the denominator.
2
P��C.�
As discussed in Section ��C.�(c) on page ��� the probability of �nding a molecule
in an interval dx at distance x from the origin at time t is P(x, t)dx, where
P(x, t) is given by
2
1
P(x, t) =
e−x �4D t
1�2
(πDt)
�e mean value of x 4 is found by integrating P(x, t)x 4 dx over the full range
of x, which in this case is � to ∞
�x 4 � = �
=
0
∞
x 4 P(x) dx =
∞
2
1
x 4 e−x �4D t dx
�
1�2
0
(πDt)
1
× 3 (4Dt)2 × (4πDt)1�2 = 12D 2 t 2
(πDt)1�2 8
where to go to the �nal line Integral G.� is used with k = 1�4Dt. Hence,
�x 4 �1�4 = (12D 2 t 2 )1�4 .
A similar calculation is used to �nd �x 2 �
�x 2 � = �
=
0
∞
x 2 P(x) dx =
∞
2
1
x 2 e−x �4D t dx
�
1�2
0
(πDt)
1
× 1 π 1�2 (4Dt)3�2 = 2Dt
(πDt)1�2 4
where to go to the �nal line Integral G.� is used with k = 1�4Dt. Hence,
�x 2 �1�2 = (2Dt)1�2 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e ratio of the �x 4 �1�4 to �x 2 � is
�x 4 �1�4 (12D 2 t 2 )1�4
12 1�4
=
=
�
� = 31�4
4
�x 2 �1�2
(2Dt)1�2
�e result is independent of the time.
P��C.�
�e probability of being n steps from the origin is P(nd) = N!�(N−N R )!N R !2 N
where N R is the number of steps taken to the right and N is the total number
of steps. Note that n = N R − N L and N = N R + N L , where N L is the number of
steps taken to the le�.
NR = N − NL = NL + n
NL = N − NR = NR − n
therefore P(nd) =
hence
hence
NL =
NR =
N−n
2
N+n
2
N!
N!
= N+n
N−n
N
�
�!
�
�!
� N−n
�! 2 N
[N − ( N−n
)]!
2
2
2
2
2
�e probability of being six paces away from the origin (x = 6d) is
Pexact (6d) =
N!
� N+6
�!
� N−6
�! 2 N
2
2
�is is the ‘exact’ value of the probability according to the random walk model.
In the limit of large N the probability density of being at distance x and time t
is given by [��C.��–���]
P(x, t) = �
2τ 1�2 −x 2 τ�2td 2
� e
πt
For the present case the value of x is taken as nd, and t�τ is taken as N because
the time to take N steps is N τ. With these substitutions
Plim (nd, N) = �
2 1�2 −n 2 �2N
� e
πN
�e following table compares the exact values of the probability with those
predicted for large N. �e discrepancy between the two values falls to less than
�.�% when N is greater than about ��.
505
506
13 MOLECULES IN MOTION
N
6
10
14
18
22
26
30
34
38
42
46
50
54
58
60
P��C.��
Pexact
0.015 6
0.043 9
0.061 1
0.070 8
0.076 2
0.079 2
0.080 6
0.081 0
0.080 9
0.080 4
0.079 7
0.078 8
0.077 9
0.076 8
0.076 3
Plim
0.016 2
0.041 7
0.059 0
0.069 2
0.075 1
0.078 3
0.079 9
0.080 6
0.080 6
0.080 2
0.079 5
0.078 7
0.077 8
0.076 8
0.076 3
100(Pexact − Plim )�Pexact
−3.79
5.09
3.51
2.30
1.55
1.07
0.75
0.54
0.38
0.27
0.19
0.13
0.08
0.05
0.03
�e Stokes–Einstein relation [��C.�b–���], shows that D ∝ T�η where D is
the di�usion coe�cient and η is the viscosity. �e temperature dependence of
viscosity is given by [��B.�–���], η = η 0 eE a �RT , it therefore follows that D ∝
Te−E a �RT . �e activation energy E a can therefore be determined from the ratio
of the di�usion constants at two temperatures
Solving for E a gives
Ea =
=
D T1 T1 e−E a �RT1 T1 ERa (1�T2 −1�T1 )
=
= e
D T2 T2 e−E a �RT2 T2
R
D T T2
ln 1
1�T2 − 1�T1 D T2 T1
(8.3145 J K−1 mol−1 )
(298 K) × (2.05 × 10−9 m2 s−1 )
× ln �
�
1�(298 K) − 1�(273 K)
(273 K) × (2.89 × 10−9 m2 s−1 )
= 6.9 kJ mol−1
�e activation energy associated with di�usion of therefore 6.9 kJ mol−1 .
Answers to integrated activities
I��.�
(a) �e di�usion equation is [��C.�–���].
�e proposed solution is
∂c
∂2 c
=D 2
∂t
∂x
c(x, t) = c 0 + (c s − c 0 )[1 − erf(ξ)]
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
where ξ(x, t) = x�(4Dt)1�2 and erf(ξ) = 2π−1�2 ∫0 e−y dy. Note that, by
2
de�nition, ∂[erf(ξ)]�∂ξ = 2π−1�2 e−ξ .
�e �rst step is to examine the initial and boundary conditions. At t = 0,
2
∞
ξ(x, 0) = ∞. Hence erf(ξ) = 2π−1�2 ∫0 e−y dy = (2π−1�2 )×( 12 π 1�2 ) = 1
and c(x, 0) = c 0 + (c s − c 0 )[1 − 1] = c 0 for 0 < x < ∞. At x = 0,
2
0
ξ(0, t) = 0. Hence erf(ξ) = 2π−1�2 ∫0 e−y dy = (2π−1�2 ) × 0 = 0 and
c(0, t) = c 0 +(c s −c 0 )[1−0] = c s for 0 ≤ t ≤ ∞. �erefore, this expression
for c(x, t) satis�es the initial and boundary conditions.
To determine whether or not the proposed solution solves the di�usion
equation it is substituted into the each side. For the LHS
ξ
2
∂c ∂(c 0 + (c s − c 0 )[1 − erf(ξ)])
∂[erf(ξ)] ∂ξ
=
= −(c s − c 0 )
∂t
∂t
∂ξ
∂t
1�2
2 ∂[x�(4Dt)
]
= −(c s − c 0 )(2π−1�2 e−ξ )
∂t
2
x
x(c s − c 0 ) x 2 �4D t
= −(c s − c 0 )(2π−1�2 e−x �4D t ) �
�=
e
1�2
3�2
(4D) t
(4πD)1�2 t 3�2
For the RHS
D
∂2 c
∂
∂[erf(ξ)] ∂ξ
=D
�−(c s − c 0 )
�
2
∂x
∂x
∂ξ
∂x
2D(c s − c 0 ) ∂[e−x �4D t ]
2D(c s − c 0 ) −x −x 2 �4D t
=−
=−
�
e
�
∂x
(4πDt)1�2
(4πDt)1�2 2Dt
2
x(c s − c 0 )
=
ex �4D t
(4πD)1�2 (t)3�2
2
�erefore the LHS is equal to the RHS and the proposed form of c(x, t)
does indeed satisfy the di�usion equation, as well as the initial and boundary conditions.
(b) Di�usion through aveoli sites (about � cell thick) of oxygen and carbon
dioxide between lungs and blood capillaries (also about � cell thick) occurs through about 75 µm (the diameter of a red blood cell). �us, the
range 0 ≤ x ≤ 100 µm is reasonable for concentration pro�les for the
di�usion of oxygen into water. Given the maximum distance, the longest
time is estimated using [��C.��–���].
tmax ≈
2
πxmax
π(0.1 × 10−3 m)2
=
= 3.74 s
4D
4(2.10 × 10−9 m2 s−1 )
�e plots shown in Fig. ��.� are with c 0 = 0, c s = 2.9 × 10−4 mol dm−3 ,
and D = 2.10 × 10−9 m2 s−1 .
507
13 MOLECULES IN MOTION
3.0
t = 0.05 s
t = 0.25 s
t = 0.75 s
t=2s
104 c�(mol dm−3 )
508
2.0
1.0
0.0
0.00
Figure 13.5
0.02
0.04
0.06
x�mm
0.08
0.10
Chemical kinetics
14
14A
The rates of chemical reactions
Answers to discussion questions
D��A.�
D��A.�
A reaction order for a particular species can only be ascribed when the rate
is simply proportional to a power of the concentration of that species. For
example, if the rate law is of the form υ = k r [A] a [B]b . . . an order is ascribable
to both A and B, but if the rate law is of the form υ = (k 1 [A])�(k 2 + k 3 [B]), an
order is ascribable to A, but not to B.
�is is discussed in Section ��A.� on page ���.
Solutions to exercises
E��A.�(a)
For a homogeneous reaction in a constant volume system the rate of reaction
is given by [��A.�b–���], υ = (1�ν J )d[J]�dt, which is rearranged to d[J]�dt =
ν J υ. In these expressions ν J is the stoichiometric number of species J, which
is negative for reactants and positive for products. For this reaction ν A = −1,
ν B = −2, ν C = +3 and ν D = +1.
For A
For B
For C
For D
d[A]�dt = ν A υ = (−1) × (2.7 mol dm−3 s−1 ) = −2.7 mol dm−3 s−1
d[B]�dt = ν B υ = (−2) × (2.7 mol dm−3 s−1 ) = −5.4 mol dm−3 s−1
d[C]�dt = ν C υ = (+3) × (2.7 mol dm−3 s−1 ) = +8.1 mol dm−3 s−1
d[D]�dt = ν D υ = (+1) × (2.7 mol dm−3 s−1 ) = +2.7 mol dm−3 s−1
�e rate of consumption of A is 2.7 mol dm−3 s−1 , the rate of consumption of
B is 5.4 mol dm−3 s−1 , the rate of formation of C is 8.1 mol dm−3 s−1 , and the
rate of formation of D is 2.7 mol dm−3 s−1 .
E��A.�(a) For a homogeneous reaction in a constant volume system the rate of reaction
is given by [��A.�b–���], υ = (1�ν J )d[J]�dt, where ν J is the stoichiometric
number of species J which is negative for reactants and positive for products.
For species C, which has ν C = +2, this gives
υ=
1 d[C]
1
=
× (2.7 mol dm−3 s−1 ) = 1.35... mol dm−3 s−1
ν C dt
+2
= �.� mol dm−3 s−1
510
14 CHEMICAL KINETICS
Rearranging [��A.�b–���] then gives
For A
For B
For D
d[A]�dt = ν A υ = (−2) × (1.35... mol dm−3 s−1 ) = −2.70... mol dm−3 s−1
d[B]�dt = ν B υ = (−1) × (1.35... mol dm−3 s−1 ) = −1.35... mol dm−3 s−1
d[D]�dt = ν D υ = (+3) × (1.35... mol dm−3 s−1 ) = +4.05... mol dm−3 s−1
�e rate of consumption of A is 2.7 mol dm−3 s−1 , the rate of consumption of
B is 1.4 mol dm−3 s−1 , and the rate of formation of D is 4.1 mol dm−3 s−1 .
E��A.�(a) As explained in Section ��A.�(b) on page ��� the units of k r are always such as
to convert the product of concentrations, each raised to the appropriate power,
into a rate expressed as a change in concentration divided by time. In this
case the rate is given in mol dm−3 s−1 , so if the concentrations are expressed
in mol dm−3 the units of k r will be dm3 mol−1 s−1 because
[A]
[B]
��� � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � �� ��� � � � � � � � � � � � �� � � � � � � � � � � � � ��� � � � � � � � � � � � �� � � � � � � � � � � � �
(dm3 mol−1 s−1 ) × (mol dm−3 ) × (mol dm−3 ) = mol dm−3 s−1
kr
�e rate of reaction is given by [��A.�b–���], υ = (1�ν J )(d[J]�dt), where ν J
is the stoichiometric number of species J. Rearranging gives d[J]�dt = ν J υ. In
this case ν C = +3, ν A = −1, and υ = k r [A][B] so
d[C]
= ν C υ = 3k r [A][B]
dt
d[A]
= ν A υ = −k r [A][B]
dt
�e rate of formation of C is therefore d[C]�dt = 3k r [A][B] and the rate of
consumption of A is −d[A]�dt = k r [A][B] .
E��A.�(a) �e rate of reaction is given by [��A.�b–���], υ = (1�ν J )(d[J]�dt). In this case
ν C = +2 so
υ=
1 d[C]
1
=
k r [A][B][C] = 12 k r [A][B][C]
ν J dt
+2
As explained in Section ��A.�(b) on page ��� the units of k r are always such as
to convert the product of concentrations, each raised to the appropriate power,
into a rate expressed as a change in concentration divided by time. In this
case the rate is given in mol dm−3 s−1 , so if the concentrations are expressed
in mol dm−3 the units of k r will be dm6 mol−2 s−1 because
[A]
[B]
[C]
��� � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � �� ��� � � � � � � � � � � � �� � � � � � � � � � � � � ��� � � � � � � � � � � � �� � � � � � � � � � � � � ��� � � � � � � � � � � � �� � � � � � � � � � � � �
(dm6 mol−2 s−1 ) × (mol dm−3 ) × (mol dm−3 ) × (mol dm−3 ) = mol dm−3 s−1
kr
E��A.�(a) As explained in Section ��A.�(b) on page ��� the units of k r are always such as
to convert the product of concentrations, each raised to the appropriate power,
into a rate expressed as a change in concentration divided by time.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(i) A second-order reaction expressed with concentrations in moles per cubic decimetre is one with a rate law such as υ = k r [A][B]. If the rate is
given in mol dm−3 s−1 then the units of k r will be dm3 mol−1 s−1 because
[A]
[B]
��� � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � �� ��� � � � � � � � � � � � �� � � � � � � � � � � � � ��� � � � � � � � � � � � �� � � � � � � � � � � � �
(dm3 mol−1 s−1 ) × (mol dm−3 ) × (mol dm−3 ) = mol dm−3 s−1
kr
A third-order reaction expressed with concentrations in moles per cubic
decimetre is one with a rate law such as v = k r [A][B][C]. �e units of k r
will then be dm6 mol−2 s−1 because
[A]
[B]
[C]
��� � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � �� ��� � � � � � � � � � � � �� � � � � � � � � � � � � ��� � � � � � � � � � � � �� � � � � � � � � � � � � ��� � � � � � � � � � � � �� � � � � � � � � � � � �
(dm6 mol−2 s−1 )×(mol dm−3 )×(mol dm−3 )×(mol dm−3 ) = mol dm−3 s−1
kr
(ii) If the rate laws are expressed with pressures in kilopascals then a secondorder reaction is one with a rate law such as υ = k r p A p B and a third-order
reaction is one with a rate law such as υ = k r p A p B p C . If the rate is given in
kPa s−1 then the units of k r will be kPa−1 s−1 and kPa−2 s−1 respectively.
��� � � � � � � � � � � �� � � � � � � � � � � �� � �
(kPa−1 s−1 ) × (kPa) × (kPa) = kPa s−1
kr
For second-order
E��A.�(a)
pB
��� � � � � � � � � � � �� � � � � � � � � � � �� � � �
(kPa−2 s−1 ) × (kPa) × (kPa) × (kPa) = kPa s−1
kr
For third-order
pA
pA
pB
pC
(i) In the rate law υ = k r1 [A][B]�(k r2 + k r3 [B]1�2 ) the concentration of A
appears raised to the power +1, so the reaction is �rst order in A, and
hence can be assigned an order with respect to A, under all conditions .
(ii) �e concentration of B does not appear as a single term raised to a power,
so the reaction has an inde�nite order with respect to B. However, if k r2 �
k r3 [B]1�2 , which might occur at very low concentrations of B, then the
term k r3 [B]1�2 in the denominator is negligible compared to the k r2 term
and so the rate law becomes
υ = k r1 [A][B]�k r2 = k r,eff [A][B] where k r,eff = k r1 �k r2
In this e�ective rate law the concentration of B appears raised to the power
+1, so under these conditions the reactions is �rst order in B. Similarly, if
k r2 � k r3 [B]1�2 , which might occur at very high concentrations of B, the
term k r2 in the denominator is negligible compared to the k r3 [B]1�2 term
and so the rate law becomes
υ = k r1 [A][B]�k r3 [B]1�2 = k r,eff [A][B]1�2
where k r,eff = k r1 �k r3
In this e�ective rate law the order with respect to B is + 12 .
To summarize, an order can only be assigned with respect to B if either
k r2 � k r3 [B]1�2 , in which case the order is +1, or k r2 � k r3 [B]1�2 , in
which case the order is + 12 .
511
512
14 CHEMICAL KINETICS
(iii) An overall order can be assigned only if all of the individual orders can
be assigned. Consequently the reaction can only be assigned an overall
order if k r2 � k r3 [B]1�2 or k r2 � k r3 [B]1�2 . �e overall order in these
two cases is +2 and + 32 .
E��A.�(a) �e gaseous species is denoted A and the order with respect to A as a. �e rate
law expressed in terms of partial pressure is then υ = k r p Aa . Taking (common)
logarithms gives
log υ = log k r + log p Aa = log k r + a log p A
where the properties of logarithms log(x y) = log x + log y and log x a = a log x
are used.
�is expression implies that a graph of log υ against log p A will be a straight line
of slope a, from which the order can be determined. However, because there
are only two data points a graph is not necessary so an alternative approach is
used.
If the initial partial pressure of the compound is p A,0 then the partial pressure
when a fraction f has reacted, so that a fraction 1 − f remains, is (1 − f )p A,0 .
Data are given for two points, f 1 = 0.100 and f 2 = 0.200. Denoting the rates
at these points by υ 1 and υ 2 and using the expression log υ = log k r + a log p A
from above gives the equations
log υ 1 = log k r + a log [(1 − f 1 )p A,0 ]
log υ 2 = log k r + a log [(1 − f 2 )p A,0 ]
Subtracting the second equation from the �rst gives
Hence
log υ 1 − log υ 2 = a log [(1 − f 1 )p A,0 ] − a log [(1 − f 2 )p A,0 ]
log �
υ1
(1 − f 1 )p A,0
1 − f1
� = a log �
� = a log �
�
υ2
(1 − f 2 )p A,0
1 − f2
where the property of logarithms log x −log y = log(x�y) is used and the factor
of p A,0 is cancelled.
Rearranging for a gives
a=
log �(9.71 Pa s−1 )�(7.67 Pa s−1 )�
log(υ 1 �υ 2 )
=
= �.��
log [(1 − f 1 )�(1 − f 2 )]
log [(1 − 0.100)�(1 − 0.200)]
E��A.�(a) Assuming perfect gas behaviour the total pressure is proportional to the total
amount in moles of gas present, provided that the temperature is constant and
the volume of the container is �xed. �e reaction � ICl(g) + H� (g) → I� (g) +
� HCl(g) involves the same number of gas molecules on both sides of the reaction arrow and therefore the total amount in moles of gas present does not
change as the reaction proceeds. Consequently there is no change in the total
pressure during the reaction. �is means that the composition of the reaction
mixture cannot be determined by measuring the total pressure in this case.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E��A.�(a) �e stoichiometry of the reaction shows that one mole of Br� is formed for
every two moles of NO formed. �erefore the rate of formation of Br� is half
the rate of formation of NO.
d[Br2 ] 1 d[NO] 1
=2×
= 2 × (0.24 mmol dm−3 s−1 ) = �.�� mmol dm−3 s−1
dt
dt
Solutions to problems
�e rate law is assumed to take the form υ 0 = k r [C6 H12 O6 ] a where υ 0 is
the initial rate and a is the order with respect to glucose. Taking (common)
logarithms gives
log υ 0 = log k r + log[C6 H12 O6 ] a = log k r + a log[C6 H12 O6 ]
where the properties of logarithms log(x y) = log x + log y and log x a = a log x
are used.
�is expression implies that a graph of log υ 0 against log[C6 H12 O6 ] will be a
straight line of slope a and intercept log k r . �e data are plotted in Fig. ��.�.
[C6 H12 O6 ] υ 0
�mol dm−3 �mol dm−3 s−1
1.00 × 10−3
5.0
1.54 × 10−3
7.6
−3
3.12 × 10
15.5
−3
4.02 × 10
20.0
log([C6 H12 O6 ]
/mol dm−3 )
−3.000
−2.812
−2.506
−2.396
log(υ 0
�mol dm−3 s−1 )
0.699
0.881
1.190
1.301
1.4
log(υ 0 �mol dm−3 s−1 )
P��A.�
1.2
1.0
0.8
0.6
Figure 14.1
−3.0
−2.8
−2.6
−3
log([C6 H12 O6 ]�mol dm )
−2.4
�e data fall on a good straight line, the equation for which is
log(υ 0 �mol dm−3 s−1 ) = 1.00 × log([CH6 H12 O6 ]�mol dm−3 ) + 3.692
513
14 CHEMICAL KINETICS
(a) Identifying the order a with the slope gives a = 1.00; that is, the reaction
is �rst order in glucose.
(b) �e intercept at log([C6 H12 O6 ]�mol dm−3 ) = 0 is log(υ 0 �mol dm−3 s−1 ) =
3.692, which corresponds to υ 0 = 4.92×103 mol dm−3 s−1 when [C6 H12 O6 ]
= 1 mol dm−3 . Because a = 1, the rate law is υ 0 = k r [C6 H12 O6 ]1 , which
is rearranged to give
P��A.�
kr =
υ0
4.92 × 103 mol dm−3 s−1
=
= 4.92 × 103 s−1
[C6 H12 O6 ]
1 mol dm−3
(a) Experiments � and � both have the same initial H� concentration, but
experiment � has an ICl concentration twice that of experiment �. Because
the rate of experiment � is also twice that of experiment �, it follows that
the rate is proportional to [ICl] and hence that the reaction is �rst order
in ICl.
Experiments � and � both have the same initial ICl concentration, but
experiment � has an H� concentration three times that of experiment
�. Because the rate of experiment � is approximately three times that of
experiment �, it follows that the rate is proportional to [H2 ] and hence
that the reaction is �rst order in H� .
�erefore the rate law is υ = k r [ICl][H2 ] .
(b) �e rate law υ = k r [ICl][H2 ] implies that a graph of υ 0 against [ICl][H2 ]
should be a straight line of slope k r and intercept zero. �e data are plotted
in Fig. ��.�.
Expt.
1
2
3
[ICl]0
�mol dm−3
1.5 × 10−3
3.0 × 10−3
3.0 × 10−3
[H2 ]0
/mol dm−3
1.5 × 10−3
1.5 × 10−3
4.5 × 10−3
[ICl]0 [H2 ]0
�mol2 dm−6
2.25 × 10−6
4.50 × 10−6
1.35 × 10−5
4
8
υ0
�mol dm−3 s−1
3.7 × 10−7
7.4 × 10−7
2.2 × 10−6
υ 0 �(10−7 mol dm−3 s−1 )
514
20
10
0
Figure 14.2
0
2
6
[ICl]0 [H2 ]0 �(10
−6
10
12
−6
mol dm )
2
14
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e data lie on a good straight line, the equation of which is
υ 0 �mol dm−3 s−1 = 0.163 × {[ICl]0 [H2 ]0 �(mol2 dm−6 )} + 6.19 × 10−9
Identifying the slope with k r gives k r = 0.16 dm3 mol−1 s−1 .
(c) �e initial reaction rate for experiment � is predicted from the rate law
υ 0 = k r [ICl][H2 ]
= (0.163... dm3 mol−1 s−1 ) × (4.7 × 10−3 mol dm−3 )
× (2.7 × 10−3 mol dm−3 ) = 2.1 × 10−6 mol dm−3 s−1
14B Integrated rate laws
Answers to discussion questions
D��B.�
D��B.�
In general the rate of a reaction depends on the concentration of the various
species involved. If all but one of the species is placed in large excess, such
that their concentrations do not vary with time, the rate law may reduce to the
simple form υ = k r′ [A] a , where A is the species not in excess. If this is the case,
the power a is the order with respect to A: if a = 1 the reaction is said to be
pseudo�rst order with respect to A, if a = 2 it is pseudosecond order. It is also
possible that under certain conditions a more complex rate law will simplify to
pseudo�rst or pseudosecond order. For example if the rate law is of the form
υ = (k 1 [A]2 )�(k 2 +k 3 [A]), when k 3 [A] � k 2 the rate law becomes pseudo�rst
order in A, but when k 3 [A] � k 2 , the rate law is pseudosecond order in A.
�e determination of a rate law is simpli�ed by the isolation method in which
the concentrations of all the reactants except one are in large excess. If B is
in large excess, for example, then to a good approximation its concentration
is constant throughout the reaction. Although the true rate law might be υ =
k r [A][B], we can approximate [B] by [B]0 and write
υ = k r′ [A] where
k r′ = k r [B]0
which has the form of a �rst-order rate law. Because the true rate law has been
forced into �rst-order form by assuming that the concentration of B is constant,
it is called a pseudo�rst-order rate law. �e dependence of the rate on the
concentration of each of the reactants may be found by isolating them in turn
(by having all the other substances present in large excess), and so constructing
the overall rate law.
In the method of initial rates, which is o�en used in conjunction with the
isolation method, the rate is measured at the beginning of the reaction for
several di�erent initial concentrations of reactants. Suppose that the rate law
for a reaction with A isolated is υ = k r [A] a ; then its initial rate, υ 0 , is given by
the initial values of the concentration of A and is written υ 0 = k r [A]0a . Taking
logarithms gives
log υ 0 = log k r + a log[A]0
515
516
14 CHEMICAL KINETICS
For a series of initial concentrations, a plot of the logarithms of the initial rates
against the logarithms of the initial concentrations of A should be a straight
lime with slope a.
�e method of initial rates might not reveal the full rate law, for the products may participate in the reaction and a�ect the rate. For example, products
participate in the synthesis of HBr, where the full rate law depends on the
concentration of HBr. To avoid this di�culty, the rate law should be �tted to the
data throughout the reaction. �e �tting may be done, in simple cases at least,
by using a proposed rate law to predict the concentration of any component at
any time, and comparing it with the data.
Because rate laws are di�erential equations, they must be integrated in order to
�nd the concentrations as a function of time. Even the most complex rate laws
may be integrated numerically. However, in a number of simple cases analytical
solutions are easily obtained and prove to be very useful. �ese are summarized
in Table ��B.� on page ���. Experimental data can be tested against an assumed
rate law by manipulating the integrated rate law into a form which will give a
straight line plot. If the data do indeed fall on a good straight line, then the data
are consistent with the assumed rate law.
Solutions to exercises
E��B.�(a)
Using [��A.�b–���], υ = (1�ν J )(d[J]dt), the rate of the reaction 2A → P is
υ = − 12 (d[A]�dt). Combining this with the rate law υ = k r [A]2 gives
−
1 d[A]
= k r [A]2
2 dt
hence
d[A]
= −2k r [A]2
dt
�is is essentially the same as [��B.�a–���], d[A]�dt = −k r [A]2 except with
k r replaced by 2k r . �e integrated rate law is therefore essentially the same as
that for [��B.�a–���], that is, [��B.�b–���] 1�[A] − 1�[A]0 = k r t, except with
k r replaced by 2k r . Hence for the reaction in question
1
1
−
= 2k r t
[A] [A]0
Rearranging for t gives
1
1
1
�
−
�
2k r [A] [A]0
1
1
1
=
�
3
−1 −1 × �
−3 −
−4
2 × (4.30 × 10 dm mol s )
0.010 mol dm
0.210 mol dm−3
t=
= 1.1 × 105 s or �.� days
E��B.�(a) �e integrated rate law for a second-order reaction of the form A + B → P is
given by [��B.�b–���],
ln
[B]�[B]0
= ([B]0 − [A]0 ) k r t
[A]�[A]0
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(i) In � hour the concentration of B falls from 0.060 mol dm−3 to 0.030 mol dm−3 ,
so the change in the concentration of B in this time period is −0.030 mol dm−3 .
It follows from the reaction stoichiometry that the concentration of A
must fall by the same amount, so the concentration of A a�er � hour is
[A]0
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
[A] = (0.080 mol dm−3 ) −(0.030 mol dm−3 ) = 0.050 mol dm−3
�e rate constant is then found by rearranging the integrated rate equation for k r and using the values of [A] and [B] at � hour, which corresponds to 1 h × (602 s h−1 ) = 3600 s.
1
[B]�[B]0
ln
([B]0 − [A]0 ) t [A]�[A]0
1
=
−3
��0.060 mol dm � − �0.080 mol dm−3 �� × (3600 s)
kr =
× ln
−3
−3
� �0.030 mol dm � � �0.060 mol dm � �
� �0.050 mol dm−3 � � �0.080 mol dm−3 � �
= 3.09... × 10−3 dm3 mol−1 s−1 = 3.1 × 10−3 dm3 mol−1 s−1
(ii) �e half-life of a particular reactant is the time taken for the concentration
of that reactant to fall to half its initial value. �e half-life of B is � hour
because it is given in the question that a�er � hour the concentration of
B had fallen from 0.060 mol dm−3 to 0.030 mol dm−3 , half the original
value.
�e initial concentration of A is 0.080 mol dm−3 so the half-life is the
time at which the concentration of A has dropped by �.��� mol dm−3 to
�.��� mol dm−3 . It follows from the stoichiometry of the reaction that the
concentration of B must also fall by �.��� mol dm−3 during this period,
so the concentration of B will be
[B] = 0.060 mol dm−3 − 0.040 mol dm−3 = 0.020 mol dm−3
Rearranging the integrated rate equation then gives
1
1
[B]�[B]0
ln �
�
k r ([B]0 − [A]0 )
[A]�[A]0
1
1
=
×
3.09... × 10−3 dm3 mol−1 s−1 (0.060 mol dm−3 ) − (0.080 mol dm−3 )
t=
× ln
E��B.�(a)
−3
−3
� �0.020 mol dm � � �0.060 mol dm � �
� �0.040 mol dm−3 � � �0.080 mol dm−3 � �
= 6.5 × 103 s or �.� hours
(i) �e integrated rate law for a zeroth-order reaction is given by [��B.�–���],
[A] = [A]0 − k r t where in this case A is NH� . If concentrations are
517
518
14 CHEMICAL KINETICS
expressed in terms of partial pressures, this becomes p NH3 = p NH3 ,0 − k r t.
Rearranging for k r and using p NH3 = 10 kPa when t = 770s with p NH3 ,0 =
21 kPa gives
p NH3 ,0 − p NH3 (21 × 103 Pa) − (10 × 103 Pa)
=
= 14.2... Pa s−1
t
770 s
= 14 Pa s−1
kr =
(ii) When all the ammonia has been consumed, p NH3 = 0. Rearranging the
rate law for t gives
t=
p NH3 ,0 − p NH3 (21 × 103 Pa) − 0
=
= 1.5 × 103 s
kr
14.2... Pa s−1
E��B.�(a) �e fact that the two half-lives are not the same establishes that the reaction is
not �rst-order because, as explained in Section ��B.� on page ���, a �rst-order
reaction has a constant half-life. For orders n ≠ 1 the half-life is given by [��B.�–
���], t 1�2 = (2n−1 − 1)�[(n − 1)k r [A]0n−1 ]. Denoting the two measurements by
t 1�2,i and t 1�2,ii and expressing concentration in terms of partial pressure gives
the two equations
t 1�2,i =
2n−1 − 1
n−1
(n − 1)k r p A,i
t 1�2,ii =
2n−1 − 1
n−1
(n − 1)k r p A,ii
�e second equation is divided by the �rst to give
t 1�2,ii
p A,i
=�
�
t 1�2,i
p A,ii
n−1
hence
log �
t 1�2,ii
p A,i
� = (n − 1) log �
�
t 1�2,i
p A,ii
where log x a = a log x is used. Rearranging for n gives
n=
log �t 1�2,ii �t 1�2,i �
log (p A,i �p A,ii )
+1=
log [(880 s)�(410 s)]
+ 1 = 2.00
log [(363 Torr)�(169 Torr)]
�erefore the reaction is second-order .
E��B.�(a)
For the reaction 2N2 O5 (g) → 4NO2 (g) + O2 (g) the rate, as given by [��A.�b–
���], υ = (1�ν J )(d[J]�dt), is
υ=
1 dp N2 O5
−2 dt
where concentrations are expressed in terms of partial pressures. It is given that
the reaction is �rst-order in N� O� , so υ = k r p N2 O5 . Combining this with the
above expression for υ gives
1 dp N2 O5
= k r p N2 O5
−2 dt
hence
dp N2 O5
= −2k r p N2 O5
dt
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�is has the same form, except with 2k r instead of k r , as [��B.�a–���], (d[A]�dt) =
−k r [A], for which it is shown in Section ��B.� on page ��� that the half-life and
the integrated rate law are
t 1�2 =
ln 2
kr
[A] = [A]0 e−k r t
�e expressions for the reaction in question are analogous, but with k r replaced
by 2k r .
ln 2
t 1�2 =
p N2 O5 = (p N2 O5 ,0 )e−2k r t
2k r
�e half-life is
t 1�2 =
ln 2
ln 2
=
= 1.03 × 104 s
2k r 2 × 3.38 × 10−5 s−1
�e partial pressures at the speci�ed times are calculated from the above integrated form of the rate law. Hence
t = 50 s
t = 20 min
p N2 O5 = (500 Torr) × e−2×(3.38×10
p N2 O5 = (500 Torr) × e−2×(3.38×10
−5
−5
s−1 )×(50 s)
−1
= ��� Torr
s )×([20×60] s)
= ��� Torr
E��B.�(a) �e reaction is of the form A + B → products. Assuming that it has rate law
υ = k r [A][B], the integrated rate law is given by [��B.�b–���]
ln
[B]�[B]0
= ([B]0 − [A]0 )k r t
[A]�[A]0
Suppose that a�er time t the concentration of A has fallen by an amount x so
that [A] = [A]0 − x. Because of the stoichiometry of the reaction the concentration of B must fall by the same amount, so [B] = [B]0 − x. �erefore
ln
Hence
Rearranging gives
Hence
([B]0 − x)�[B]0
= ([B]0 − [A]0 )k r t
([A]0 − x)�[A]0
([B]0 − x)[A]0
= e([B]0 −[A]0 )k r t
([A]0 − x)[B]0
[B]0 [A]0 − x[A]0 = [B]0 [A]0 e([B]0 −[A]0 )k r t − x[B]0 e([B]0 −[A]0 )k r t
x=
[B]0 [A]0 �e([B]0 −[A]0 )k r t − 1�
[B]0 e([B]0 −[A]0 )k r t − [A]0
=
[B]0 [A]0 (e λ − 1)
[B]0 e λ − [A]0
where λ = ([B]0 − [A]0 )k r t. Taking A and B as OH− and CH� COOC� H�
respectively, the concentrations at the speci�ed times are
519
520
14 CHEMICAL KINETICS
For t = 20 s
λ = ([B]0 − [A]0 )k r t
= �(0.110 mol dm−3 ) − (0.060 mol dm−3 )� × (0.11 dm3 mol−1 s−1 ) × (20 s)
= 0.11
x=
[B]0 [A]0 (e λ − 1) (0.110 mol dm−3 ) × (0.060 mol dm−3 ) × (e0.11 − 1)
=
[B]0 e λ − [A]0
(0.110 mol dm−3 ) × e0.11 − (0.060 mol dm−3 )
= 0.0122... mol dm−3
Hence the concentration of ester is
[B] = [B]0 − x
= (0.110 mol dm−3 ) − (0.0122... mol dm−3 ) = �.���� mol dm−3
For t = 15 min
λ = ([B]0 − [A]0 )k r t
= �(0.110 mol dm−3 ) − (0.060 mol dm−3 )� × (0.11 dm3 mol−1 s−1 )
x=
× ([15 × 60] s) = 4.95...
[B]0 [A]0 (e λ − 1) (0.110 mol dm−3 ) × (0.060 mol dm−3 ) × (e4.95 ... − 1)
=
[B]0 e λ − [A]0
(0.110 mol dm−3 ) × e4.95 ... ) − (0.060 mol dm−3 )
= 0.0598... mol dm−3
Hence the concentration of ester is
[B] = [B]0 − x
= (0.110 mol dm−3 ) − (0.0598... mol dm−3 ) = �.���� mol dm−3
Solutions to problems
P��B.�
�e concentration of B is given in the question as
[B] = n[A]0 (1 − e−k r t ) hence
[B]�[A]0 = n(1 − e−k r t )
�e concentration of A for a �rst-order reaction is given by [��B.�b–���],
[A] = [A]0 e−k r t
hence
�ese expressions are plotted in Fig. ��.�
P��B.�
[A]�[A]0 = e−k r t
�e �rst task is to convert the masses of urea into concentrations of ammonium
cyanate A. Because the only fate of the ammonium cyanate is to be converted
into urea, the mass of ammonium cyanate m A remaining at any given time
is equal to the original mass of ammonium cyanate minus the mass of urea,
m A = m A,0 − m urea . In this case m urea = 22.9 g. Dividing by the molar mass
of the ammonium cyanate, M A = 60.0616 g mol−1 , gives the amount of A in
moles, and division by the volume of the solution then gives the concentration
in mol dm−3 .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
B, n = 2
1.0
B, n = 1
[A]�[A]0 or [B]�[A]0
2.0
0.0
0
B, n = 12
A
1
2
3
4
5
kr t
Figure 14.3
t�min
0
20
50
65
150
t�s m urea �g
0 0.0
1 200 7.0
3 000 12.1
3 900 13.8
9 000 17.7
m A �g [A]�mol dm−3
22.9
0.381
15.9
0.265
10.8
0.180
9.1
0.152
5.2
0.087
�e order is determined by testing the �t of the data to integrated rate law
expressions. A zeroth-order reaction of the form A → P has an integrated rate
law given by [��B.�–���], [A] = [A]0 − k r t, so if the reaction is zeroth-order
then a plot of [A] against t will be a straight line of slope −k r . On the other
hand, a �rst-order reaction has an integrated rate law given by [��B.�b–���],
ln([A]�[A]0 ) = −k r t, so if the reaction is �rst-order then a plot of ln [A]�[A]0
against t will be a straight line of slope −k r . Finally, if the order is n ≥ 2 the
integrated rate law is given in Table ��B.� on page ��� as
kr t =
1
1
1
1
1
�
−
� hence
= (n−1)k r t+
n−1
n−1
n−1
n − 1 ([A]0 − [P])
[A]0
[A]
[A]0n−1
where to obtain the second expression the relation [P] = [A]0 − [A] is substituted and the equation rearranged. �is expression implies that if the reaction
has order n ≥ 2 a plot of 1�[A]n−1 against t will be a straight line of slope
(n − 1)k r .
�e data are plotted assuming zeroth-, �rst-, second-, and third-order in Fig. ��.�
and using the data in the table below. �e second-order plot shows a good
straight line, while the other three plots show the data lying on distinct curves.
It is therefore concluded that the reaction is second-order .
521
14 CHEMICAL KINETICS
[A]
�mol dm−3
0 0.381
1 200 0.265
3 000 0.180
3 900 0.152
9 000 0.087
t�s
ln
[A]
[A]0
0.000
−0.365
−0.752
−0.923
−1.482
1�[A]2
�dm6 mol−2
6.879
14.269
30.928
43.562
133.410
1�[A]
�dm3 mol−1
2.623
3.777
5.561
6.600
11.550
In an alternative approach visual examination of the concentration data indicates that the half-life is not constant, and comparison of the �–�� min data and
the ��-��� min data suggests that the second half-life is approximately double
the �rst. �at is, the half-life starting from half the initial concentration is about
twice the initial half-life, suggesting that the half-life is inversely proportional
to the initial concentration. According to [��B.�–���] the half-life of a reaction
with order n > 1 is given by t 1�2 ∝ 1�[A]0n−1 , so this result suggests that the
reaction may have n = 2 as this gives t 1�2 ∝ 1�[A]0 . Because it is suspected
on this basis that the reaction may be second-order, only the second-order plot
from Fig. ��.� is made, and the fact that it gives a good straight line con�rms
that the reaction is indeed second-order.
0.4
0.0
−0.5
0.3
0.2
−1.0
0.1
0
4 000
t�s
8 000
0
150
second-order
10
4 000
t�s
8 000
third-order
100
50
5
0
−1.5
[A]−2 �dm6 mol−2
0.0
�rst-order
ln([A]�[A]0 )
[A]�mol dm−3
zeroth-order
[A]−1 �dm3 mol−1
522
0
0
4 000
t�s
8 000
0
Figure 14.4
�e equation of the line in the second-order plot is
[A]−1 �dm3 mol−1 = 9.95 × 10−4 × (t�s) + 2.62
4 000
t�s
8 000
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Identifying the slope with (n − 1)k r as discussed above, and noting that n = 2
for a second-order reaction, gives k r = 9.95 × 10−4 dm3 mol−1 s−1 .
�e concentration of ammonium cyanate le� a�er 300 min, which is (300min)×
(60s�1 min) = 18000 s, is calculated using the integrated rate law for a secondorder reaction [��B.�b–���], [A] = [A]0 �(1 + k r t[A]0 )
[A] =
(0.381... mol dm−3 )
1 + (9.95... × 10−4 dm3 mol−1 s−1
× (18000 s) × (0.381... mol dm−3 )
)
= 0.0487... mol dm−3
Multiplication by the volume gives the amount in moles, and multiplication of
this by the molar mass gives the mass of A in g.
P��B.�
m A = MV [A] = (60.0616 g mol−1 )×(1.00 dm)×(0.0487... mol dm−3 ) = �.� g
�e order is determined by testing the �t of the data to integrated rate law
expressions. A zeroth-order reaction of the form A → P has an integrated rate
law given by [��B.�–���], [A] = [A]0 −k r t, so if the reaction is zeroth-order then
a plot of [A] against t will be a straight line of slope −k r . In this case, A is the
organic nitrile. On the other hand, a �rst-order reaction has an integrated rate
law given by [��B.�b–���], ln([A]�[A]0 ) = −k r t, so if the reaction is �rst-order
then a plot of ln [A]�[A]0 against t will be a straight line of slope −k r . Finally,
if the order is n ≥ 2 the integrated rate law is given in Table ��B.� on page ���
as
kr t =
1
1
1
1
1
�
−
� hence
= (n−1)k r t+
n−1
n−1
n−1
n − 1 ([A]0 − [P])
[A]0
[A]
[A]0n−1
where to obtain the second expression the relation [P] = [A]0 − [A] is substituted and the equation rearranged. �is expression implies that if the reaction
has order n ≥ 2 a plot of 1�[A]n−1 against t will be a straight line of slope
(n − 1)k r .
�e data are plotted assuming zeroth-, �rst-, second-, and third-order in Fig. ��.�.
�e second-order plot shows the best �t to a straight line, so it is concluded
that the reaction is likely to be second-order . However, the �rst-order and
third-order plots also give a reasonable �t to a straight line, so experimental
data over a wider range of concentrations would be needed to establish the
order with greater con�dence.
t�103 s
0
2
4
6
8
10
12
[A]
�mol dm−3
1.500 0
1.260 0
1.070 0
0.920 0
0.810 0
0.720 0
0.650 0
ln
[A]
[A]0
0.000
−0.174
−0.338
−0.489
−0.616
−0.734
−0.836
1�[A]
�dm3 mol−1
0.67
0.79
0.93
1.09
1.23
1.39
1.54
1�[A]2
�dm6 mol−2
4.444 × 10−1
6.299 × 10−1
8.734 × 10−1
1.181 × 100
1.524 × 100
1.929 × 100
2.367 × 100
523
14 CHEMICAL KINETICS
1.5
�rst-order
−0.5
0
5
10
3
t�10 s
second-order
1.5
1.0
−1.0
0
5
10
3
t�10 s
third-order
[A]−2 �dm6 mol−2
1.0
0.5
0.0
ln([A]�[A]0 )
[A]�mol dm−3
zeroth-order
[A]−1 �dm3 mol−1
524
2
1
0.5
0
5
10
0
3
t�10 s
0
5
10
3
t�10 s
Figure 14.5
�e equation of the line in the second-order plot is
[A]−1 �dm3 mol−1 = 7.33 × 10−5 × (t�s) + 0.652
Identifying the slope with (n − 1)k r as discussed above and noting that n = 2
for a second-order reaction gives k r = 7.33 × 10−5 dm3 mol−1 s−1 .
P��B.�
�e order is determined by testing the �t of the data to integrated rate law
expressions. A �rst-order reaction has an integrated rate law given by [��B.�b–
���], ln([A]�[A]0 ) = −k r t, or ln[A] − ln[A]0 = −k r t, so if the reaction is
�rst-order then a plot of ln [A] against t will be a straight line of slope −k r .
On the other hand, a second-order reaction has an integrated rate law given
by [��B.�b–���], 1�[A] − 1�[A]0 = k r t, which implies that if the reaction is
second-order then a plot of 1�[A] against t will be a straight line of slope k r .
�e data are plotted in Fig. ��.�. �e �rst-order plot shows a good straight line
while in the second-order plot the data lie on a curve. It is therefore concluded
that the reaction is �rst-order .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
t�min
30
60
120
150
240
360
480
ln(c�ng cm−3 )
6
�rst-order
1�(c�ng cm−3 )
1.43 × 10−3
1.61 × 10−3
2.42 × 10−3
3.42 × 10−3
6.58 × 10−3
1.67 × 10−2
4.17 × 10−2
second-order
0.04
5
0.02
4
3
0
ln(c�ng cm−3 )
6.550
6.433
6.023
5.677
5.024
4.094
3.178
1�(c�ng cm−3 )
7
c�ng cm−3
699
622
413
292
152
60
24
200
t�min
400
0.00
0
200
t�min
400
Figure 14.6
�e equation of the line in the �rst-order plot is
ln(c�ng cm−3 ) = −7.65 × 10−3 × (t�min) + 6.86
Identifying the slope with −k r as discussed above gives the �rst-order rate constant as k r = 7.65 × 10−3 min−1 .
�e half-life of a �rst-order reaction is given by [��B.�–���], t 1�2 = ln 2�k r .
P��B.�
t 1�2 =
ln 2
ln 2
=
= 91 min
kr
7.65 × 10−3 min−1
�e units of the rate constants show that both reactions are �rst-order, so their
rate equations are assumed to be
υ1 =
d[CH4 ]
= k 1 [CH3 COOH] and
dt
υ2 =
d[CH2 CO]
= k 2 [CH3 COOH]
dt
where [��A.�b–���], υ = (1�ν J )(d[J]�dt), is used to express the rates in terms
of rate of formation of CH� and CH� CO. �e ratio of the rate of ketene formation to the total rate of product formation is therefore
k 2 [CH3 COOH]
k2
4.65 s−1
=
=
k 1 [CH3 COOH] + k 2 [CH3 COOH] k 1 + k 2 (3.74 s−1 ) + (4.65 s−1 )
= 0.554...
525
526
14 CHEMICAL KINETICS
Because this ratio is independent of the CH� COOH concentration it will be
constant throughout the duration of the reaction and will be equal to the ratio of
ketene formed to total product formed. �e maximum possible yield of ketene
is therefore 0.554... or 55.4 % .
Similarly the ratio of the rate of formation of ketene and methane is
k 2 [CH3 COOH] k 2 4.65 s
=
=
= 1.24...
k 1 [CH3 COOH] k 1 3.74 s
P��B.��
�is ratio is independent of the CH� COOH concentration so it will be constant
throughout the duration of the reaction and equal to the ratio of the total ketene
and methane formed up to any given time. Hence [CH2 CO]�[CH4 ] = 1.24...
and is constant over time.
�e �rst task is to calculate the concentrations of the reactant A at each time.
�e stoichiometry of the reaction 2A → B means that the initial concentration
of A is twice the �nal concentration of B, [A]0 = 2[B]∞ . In addition, the
amount of A that has reacted at any given time is equal to twice the amount
of B that has been formed. It follows that
��� � � � � � � � � � � � � � � � � � � � � �
[A]0 − [A] = 2[B] hence
A that has reacted
[A] = [A]0 − 2[B]
Substituting [A]0 = 2[B]∞ from above gives [A] = 2([B]0 − [B]); this expression is used to calculate the concentration of [A] at each of the times.
�e order is determined by testing the �t of the data to integrated rate law
expressions. If the rate law is υ = k r [A]n , where n is the order to be determined, expressing υ in terms of the rate of change of concentration of [A] using
[��A.�b–���], υ = (1�ν J )(d[J]�dt) gives
υ=
1 d[A]
= k r [A]n
−2 dt
hence
d[A]
= −2k r [A]n
dt
Integrated rate laws are given in Table ��B.� on page ���, but care is needed
because these are for reactions of the form A → P but here the reaction is 2A →
B.
For n = 0, Table ��B.� on page ��� shows that a reaction A → P with rate law
υ = d[P]�dt = k r has integrated rate law A = A0 − k r t. To adapt this expression
for the reaction in question, the rate law for the reaction in the table is �rst
written as d[A]�dt = −k r using d[P]�dt = −d[A]�dt for a reaction of the form
A → P. �is rate law matches that found above, d[A]�dt = −2k r [A]n , for n = 0
except that k r is replaced by 2k r . �e integrated rate law will therefore be the
same except with k r replaced by 2k r , that is, [A] = [A]0 − 2k r t. �is expression
implies that if the reaction is zeroth-order a plot of [A] against t will give a
straight line of slope −2k r .
Similarly Table ��B.� on page ��� gives the integrated rate law for a �rst-order
reaction A → P with rate law υ = d[P]�dt = k r [A] as ln([A]0 �[A]) = k r t,
equivalent to [��B.�b–���], ln([A]�[A]0 ) = −k r t. By the same reasoning as
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
above the integrated rate law for the reaction will therefore be ln([A]�[A]0 ) =
−2k r t, implying that a plot of ln([A]�[A]0 ) against t will give a straight line of
slope −2k r .
Finally, if the order is n ≥ 2 the integrated rate law for a reaction A → P with
rate law υ = d[P]�dt = k r [A]n is given in Table ��B.� on page ��� as
kr t =
1
1
1
1
1
�
−
� hence
= (n−1)k r t+
n − 1 ([A]0 − [P])n−1 [A]0n−1
[A]n−1
[A]0n−1
where to obtain the second expression the relation [P] = [A]0 − [A] is substituted and the equation rearranged. Adapting this expression for the reaction
in question gives 1�[A]n−1 = 2(n − 1)k r t + 1�[A]0n−1 . �is expression implies
that if the reaction has order n ≥ 2 a plot of 1�[A]n−1 against t will be a straight
line of slope 2(n − 1)k r .
�e data are plotted assuming zeroth-, �rst-, second-, and third-order in Fig. ��.�.
�e �rst-order plot shows a good �t to a straight line, while the other plots are
curved, so it is concluded that the reaction is �rst-order .
t�min
0
10
20
30
40
∞
[B]
�mol dm−3
0.000
0.089
0.153
0.200
0.230
0.312
[A]
�mol dm−3
0.624
0.446
0.318
0.224
0.164
0.000
ln
[A]
[A]0
1�[A]
�dm3 mol−1
1.603
2.242
3.145
4.464
6.098
0.000
−0.336
−0.674
−1.025
−1.336
1�[A]2
�dm6 mol−2
2.57
5.03
9.89
19.93
37.18
�e equation of the line in the �rst-order plot is
ln([A]�[A]0 ) = −0.03361 × (t�min) − 1.896 × 10−3
Identifying the slope with −2k r as discussed above gives the �rst-order rate
constant as
P��B.��
k r = − 12 × (−0.03361 min−1 ) = 0.0168 min−1
�e order is determined by �tting the data to integrated rate laws. Table ��B.�
on page ��� gives integrated rate laws for the reaction A → P with rate law
υ = d[P]�dt = k r [A]n as
n=0
[A] = [A]0 − k r t
n≥2
kr t =
n=1
[A]0
k r t = ln
[A]
hence
hence
k r = ([A]0 − [A]) �t
kr =
1 [A]0
ln
t
[A]
1
1
1
�
−
�
n−1
n − 1 ([A]0 − [P])
[A]0n−1
hence
kr =
1
1
1
�
−
�
(n − 1)t [A]n−1 [A]0n−1
527
14 CHEMICAL KINETICS
0.0
[A]�mol dm−3
−0.5
0.4
−1.0
0.2
0.0
0
20
t�min
40
−1.5
0
40
30
4
20
t�min
40
third-order
[A]−2 �dm6 mol−2
second-order
6
20
2
0
�rst-order
ln([A]�[A]0 )
zeroth-order
0.6
[A]−1 �dm3 mol−1
528
10
0
20
t�min
40
0
0
20
t�min
40
Figure 14.7
where in the n ≥ 2 case, [A] = [A]0 − [P] is used.
�ese expressions imply that if the reaction is zeroth-order the quantity ([A]0 −
[A])�t should be a constant, equal to k r , while if it is �rst-order [ln([A]0 �[A])]�t
should be constant, and if the order is n ≥ 2 the quantity
([A]−(n−1) − [A]0n−1 )�[(n − 1)t]
should be constant. In this case A is cyclopropane and the concentrations are
expressed in terms of partial pressures. Results assuming n = 0, 1, 2 are shown
in the following table.
p A,0
�Torr
200
200
400
400
600
600
t�s
100
200
100
200
100
200
pA
�Torr
186
173
373
347
559
520
n=0
(p A,0 − p A )�t
�Torr s−1
0.140
0.135
0.270
0.265
0.410
0.400
n=1
[ln(p A,0 �p A )]�t
�s−1
7.26 × 10−4
7.25 × 10−4
6.99 × 10−4
7.11 × 10−4
7.08 × 10−4
7.16 × 10−4
n=2
(1�p A − 1�p A,0 )�t
�Torr−1 s−1
3.76 × 10−6
3.90 × 10−6
1.81 × 10−6
1.91 × 10−6
1.22 × 10−6
1.28 × 10−6
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P��B.��
�e values assuming n = 1 are approximately constant, while those assuming n = 0 or n = 2 are not. It is therefore concluded that the reaction is
�rst-order . �e average value of [ln(p A,0 �p A )]�t, and hence of k r , from the
table is 7.1 × 10−4 s−1 .
A reaction of the form A → P that is nth order in A has rate law υ = k r [A]n .
Combining this with [��A.�b–���], υ = (1�ν J )(d[J]�dt) gives
1 d[A]
= k r [A]n
−1 dt
hence
− [A]−n d[A] = k r dt
Initially, at t = 0, the concentration of A is [A]0 , and at a later time t it is [A].
�ese are used as the limits of the integration to give
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
Integral A.�
�
[A]
[A]0
hence
−[A]−n d[A] = �
t
0
[A]
1
[A]−(n−1) �
= k r t�
n−1
[A]0
0
t
k r dt
hence
1
1
1
�
−
� = kr t
n − 1 [A]n−1 [A]0n−1
(for n ≠ 1)
�is integrated rate law is equivalent to that given in Table ��B.� on page ���.
(a) When t = t 1�2 , [A] = 12 [A]0 . �erefore, on dividing through by k r , the
above integrated rate law gives
t 1�2 =
=
�
1
1
1 �
1
2n−1
1
−
=
�
−
�
n−1
n−1
n−1
1
(n − 1)k r � � [A]0 �
[A]0 � (n − 1)k r [A]0
[A]0n−1
2
2
−1
(n − 1)k r [A]0n−1
n−1
(b) �e time taken for the concentration of a substance to fall to one-third its
initial value is denoted t 1�3 . �us, at t = t 1�3 , [A] = 13 [A]0
t 1�3 =
P��B.��
=
�
1
1
1 �
1
3n−1
1
−
=
�
−
�
(n − 1)k r � � 1 [A]0 �n−1 [A]0n−1 � (n − 1)k r [A]0n−1 [A]0n−1
3
3
−1
(n − 1)k r [A]0n−1
n−1
�e stoichiometry of the reaction 2A + B → P implies that when the concentration of P has increased from � to x, the concentration of A has fallen to
[A]0 − 2x and the concentration of B has fallen to [B]0 − x. �is is because
each P that forms entails the disappearance of two A and one B. �e rate law
υ = d[P]�dt = k r [A]2 [B] then becomes
d[P]
= k r ([A]0 − 2x)2 ([B]0 − x) hence
dt
dx
= k r ([A]0 − 2x)2 ([B] − x)
dt
529
530
14 CHEMICAL KINETICS
where to go to the second expression [P] = x is used, which implies that d[P]dt =
dx�dt. �e expression is rearranged and the initial condition x = 0 when t = 0
is applied. �is gives the integrations required as
�
x
0
t
1
dx = � k r dt
2
([A]0 − 2x) ([B]0 − x)
0
�e right-hand side evaluates to k r t. �e le�-hand side is evaluated below.
(a) If [B]0 = 12 [A]0 the le�-hand side becomes
�
x
1
dx =
1
x
�
1
1
0 ([A]0 − 2x)2 ( [A]0 − x)
0 ([A]0 − 2x)2 × ([A]0 − 2x)
2
2
x
x
1
1
1
−3
−2
= � 2([A]0 − 2x) dx = ([A]0 − 2x) � =
−
2
2
2
2([A]
−
2x)
2[A]
0
0
0
0
dx
Combining this with the right-hand side from above gives the integrated
rate law as
1
1
−
= kr t
2([A]0 − 2x)2 2[A]20
(b) If [B]0 = [A]0 the le�-hand side is integrated using the method of partial
fractions described in �e chemist’s toolkit �� in Topic ��B. �e integrand
is �rst written as
1
A
B
C
=
+
+
2
2
([A]0 − 2x) ([A]0 − x) ([A]0 − 2x)
[A]0 − 2x [A]0 − x
where A, B, and C are constants to be found. �is expression is multiplied
through by ([A]0 − 2x)2 ([A]0 − x) to give
1 = A([A]0 − x) + B([A]0 − 2x)([A]0 − x) + C([A]0 − 2x)2
�is expression must be true for all x, so the values of A, B and C are most
conveniently found by substituting particular values of x.
When x = [A]0
When x = 12 [A]0
When x = 0
hence
1 = C(−[A]0 )2
hence
1 = A( 12 [A]0 ) hence
1 = A[A]0 + B[A]20 + C[A]20
=
C = 1�[A]20
A = 2�[A]0
2
1
[A]0 + B[A]20 +
[A]20
[A]0
[A]20
1 = 3 + B[A]20
hence
B = −2�[A]20
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e required integral is therefore
1
dx
0 ([A]0 − 2x)2 ([A]0 − x)
x
2
2
1
=�
−
+
dx
[A]20 ([A]0 − 2x) [A]20 ([A]0 − x)
0 [A]20 ([A]0 − 2x)
�
x
=
1
1
1
+
ln([A]0 − 2x) −
ln([A]0 − x)�
[A]0 ([A]0 − 2x) [A]20
[A]20
0
=�
x
1
1
1
+
ln([A]0 − 2x) −
ln([A]0 − x)�
2
[A]0 ([A]0 − 2x) [A]0
[A]20
1
1
1
+
ln[A]0 −
ln[A]0 �
2
2
[A]0 [A]0
[A]20
1
1
[A]0 − 2x
1
=
+
ln
−
2
[A]0 ([A]0 − 2x) [A]0
[A]0 − x
[A]20
−�
Combining this with the right-hand side integral found above gives the
integrated rate law as
1
1
[A]0 − 2x
1
+
ln
−
= kr t
[A]0 ([A]0 − 2x) [A]20
[A]0 − x
[A]20
14C Reactions approaching equilibrium
Answers to discussion questions
D��C.�
If the equilibrium position shi�s with pressure, a pressure jump can be used
to alter the rate of the reaction. For such an e�ect on equilibrium, the volume
change of the reaction must be non-zero.
Solutions to exercises
E��C.�(a)
�e relaxation time in a jump experiment is given by [��C.�a–���], τ = 1�(k r +
k r′ ). �is equation is rearranged for k r′ . It is convenient to convert τ to ms.
k r′ =
1
1
− kr =
− (12.4 ms−1 ) = 23.8 ms−1
τ
27.6 × 10−3 ms
E��C.�(a) �e equilibrium constant in terms of rate constants is given by [��C.�–���], K =
k r �k r′ . However because the forward and backward reactions are of di�erent
order it is necessary to include a factor of c −○ so that the ratio of k r , with units
dm3 mol−1 s−1 , to k r′ , with units s−1 , is turned into a dimensionless quantity.
�e equation required is
K=
k r c −○ (5.0 × 106 dm3 mol−1 s−1 ) × (1 mol dm−3 )
=
= 2.5 × 102
k r′
2.0 × 104 s
531
532
14 CHEMICAL KINETICS
Solutions to problems
P��C.�
�e expression for [A] in [��C.�–���] is di�erentiated
[A] =
k r′ + k r e−(k r +k r )t
[A]0
k r + k r′
′
hence
′
d[A]
= −k r [A]0 e−(k r +k r )t
dt
According to [��C.�–���], d[A]�dt = −(k r + k r′ )[A]+ k r′ [A]0 . To verify that the
two expressions for d[A]�dt are the same, the expression for [A] from [��C.�–
���] is substituted into [��C.�–���]
[A]
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
′
d[A]
k ′ + k r e−(k r +k r )t
= −(k r + k r′ ) � r
[A]0 � +k r′ [A]0
dt
k r + k r′
= −k r′ [A]0 − k r e−(k r +k r )t [A]0 + k r′ [A]0 = −k r [A]0 e−(k r +k r )t
′
′
�erefore the two expressions for d[A]�dt are the same and so the equation is
satis�ed.
P��C.�
(a) �e forward and backward reactions are
d[A]
= −k r [A]
dt
A→B
B→A
�e overall rate of change of [A] is therefore
d[A]
= +k r′ [B]
dt
d[A]
= −k r [A] + k r′ [B]
dt
�e stoichiometry of the reaction A � B means that the amount of B
present at any time is equal to the initial amount plus the amount of A
that has reacted. Hence [B] = [B]0 + ([A]0 − [A]). �is is substituted
into the above expression to give
d[A]
= −k r [A] + k r′ ([B]0 + [A]0 − [A])
dt
Rearranging and integrating with the initial condition that [A] = [A]0
when t = 0 gives
�
[A]
[A]0
d[A]
k r′ ([A]0 + [B]0 ) − (k r + k r′ )[A]
Performing the integration gives
Hence
t
0
dt
[A]
ln [k r′ ([A]0 + [B]0 ) − (k r + k r′ )[A]]
�
=t
−(k r + k r′ )
[A]0
1
k r′ ([A]0 + [B]0 ) − (k r + k r′ )[A]
ln
=t
−(k r + k r′ ) k r′ ([A]0 + [B]0 ) − (k r + k r′ )[A]0
Rearranging for [A] yields
[A] =
=�
k r′ ([A]0 + [B]0 ) + (k r [A]0 − k r′ [B]0 )e−(k r +k r )t
k r + k r′
′
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
(b) As t → ∞, the exponential term in the expression for [A] decreases to
zero and the concentrations reach their equilibrium values. �e equilibrium concentration of A is therefore
[A]eq =
k r′ ([A]0 + [B]0 )
k r + k r′
Noting from above that the concentration of B is given by [B] = [B]0 +
[A]0 − [A], the equilibrium concentration of B is therefore
[B]eq = [B]0 + [A]0 − [A]eq = [B]0 + [A]0 −
=
k r′ ([A]0 + [B]0 )
k r + k r′
(k r + k r′ )([B]0 + [A]0 ) − k r′ ([A]0 + [B]0 ) k r′ ([A]0 + [B]0 )
=
k r + k r′
k r + k r′
�is result is alternatively and more simply obtained by noting that at
equilibrium the rates of the forward and backward reactions are equal,
implying that
k r [A]eq = k r′ [B]eq
hence
P��C.�
[A]eq =
hence
k r [A]eq = k r′ ([B]0 + [A]0 − [A]eq )
k r′ ([B]0 + [A]0 )
k r + k r′
as before
(a) Application of [��A.�b–���], υ = (1�ν J )(d[J]�dt), to the forward and
backward reactions gives
Forward 2A → A2
Backward A2 → 2A
1 d[A]
−2 dt
1 d[A]
υ = k a′ [A2 ] =
2 dt
υ = k a [A]2 =
hence
hence
�e overall rate of change of A is therefore
d[A]
= −2k a [A]2
dt
d[A]
= 2k a′ [A2 ]
dt
d[A]
= −2k a [A]2 + 2k a′ [A2 ]
dt
If the deviation of [A] from its new equilibrium value is denoted 2x, so
that [A] = [A]eq + 2x, the stoichiometry of the reaction implies that
[A2 ] = [A2 ]eq − x. �ese are substituted into the above expression to
give
d[A]
2
= −2k a �[A]eq + 2x� + 2k a′ �[A2 ]eq − x�
dt
= −2k a �[A]2eq + 4x[A]eq + 4x 2 � + 2k a′ �[A2 ]eq − x�
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
�
= −2k a [A]2eq + 2k a′ [A2 ]eq −8k a x[A]eq + 2k a′ x + 8k a x 2
0
= − �8k a [A]eq + 2k a′ � x
Neglect
533
534
14 CHEMICAL KINETICS
In the third line the �rst two terms cancel because at equilibrium the rates
of the forward reaction k a [A]2eq and the backward reaction k a′ [A2 ]eq are
equal. �e last term is neglected because x is assumed to be small. Next,
because [A] = [A]eq + 2x it follows that d[A]�dt = 2 dx�dt. �is is
substituted into the above expression to give
2
dx
= − �8k a [A]eq + 2k a′ � x
dt
hence
dx
= − �4k a [A]eq + k a′ � x
dt
Rearranging, and integrating with the condition that x = x 0 when t = 0
gives
�
x dx
x0
hence
x
=�
t
0
x
= − �4k a [A]eq + k a′ � t
x0
1
where
= 4k a [A]eq + k a′
τ
− �4k a [A]eq + k a′ � dt hence ln
x = x 0 e−(4k a [A]eq +k a )t = x 0 e−t�τ
′
Squaring both sides of the expression for 1�τ gives
1
2
= �4k a [A]eq + k a′ � = 16k a2 [A]2eq + 8k a k a′ [A]eq + k a′2
τ2
k a′ [A 2 ]eq
��� � � � � � � � � �� � � � � � � � � �
= 16k a �k a [A]2eq � +8k a k a′ [A]eq + k a′2 = 16k a k a′ [A2 ]eq + 8k a k a′ [A]eq + k a′2
[A]tot
��� � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � � ��
= 8k a k a′ �2[A2 ]eq + [A]eq � +k a′2 = 8k a k a′ [A]tot + k a′2
In the second line, k a [A]2eq = k a′ [A2 ]eq is used; these quantities are equal
because as explained above the rates of the forward and backward reactions are equal at equilibrium. In the third line, the relationship [A]tot =
[A] + 2[A2 ] is used; this expression is valid at all stages of the reaction
including at equilibrium.
(b) �e result 1�τ 2 = 8k a k a′ [A]tot + k a′2 implies that a plot of 1�τ 2 against
[A]tot should give a straight line of intercept k a′2 and slope 8k a k a′ ; from
these quantities k a′ and k a are determined.
(c) �e data are plotted in Fig. ��.�.
[P]�mol dm−3
0.500
0.352
0.251
0.151
0.101
τ�ns
2.3
2.7
3.3
4.0
5.3
τ −2 �ns−2
0.189
0.137
0.092
0.063
0.036
�e data fall on a reasonable straight line, the equation for which is
τ −2 �ns−2 = 0.380 × ([P]�mol dm−3 ) + 2.87 × 10−4
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
0.20
τ −2 �ns−2
0.15
0.10
0.05
0.00
0.0
0.1
Figure 14.8
0.2
0.3
[P]�mol dm
0.4
−3
0.5
0.6
Identifying the intercept with k a′2 gives k a′2 = 2.87 × 10−4 ns−2 , hence
√
k a′ = 2.87 × 10−4 ns−2 = 0.0169... ns−1 = 1.7 × 107 s−1
Identifying the slope with 8k a k a′ gives
8k a k a′ = 0.380 dm3 mol−1 ns−2
hence k a =
0.380 dm3 mol−1 ns−2 0.380 dm3 mol−1 ns−2
=
8k a′
8 × (0.0169... ns−1 )
= 2.80... dm3 mol−1 ns−1 = 2.8 × 109 dm3 mol−1 s−1
�e equilibrium constant is given by [��C.�–���], K = (k a �k a′ )×(k b �k b′ )×
..., but it is necessary to include a factor of c −○ because the forward reaction
is second-order while the backward reaction is �rst-order.
K=
k a c −○ (2.80 ... dm3 mol−1 ns−1 ) × (1 mol dm−3 )
=
= 1.7 × 10−2
k a′
(0.0169... ns−1 )
It is noted that the points in Fig. ��.� do not lie on a perfect straight line,
and the intercept is closer to zero than some of the points are to the line.
In fact, mathematical so�ware gives the standard error in the intercept as
4 × 10−3 ns−2 , which is an order of magnitude larger than the intercept
itself. �is indicates that there is considerable uncertainty in the intercept
and therefore in the values of the rate constants and equilibrium constant
deduced from it.
14D The Arrhenius equation
Answers to discussion questions
D��D.�
�e temperature dependence of some reactions is not Arrhenius-like, in the
sense that a straight line is not obtained when ln k r is plotted against 1�T.
535
536
14 CHEMICAL KINETICS
However, it is still possible to de�ne an activation energy using [��D.�–���],
E a = RT 2 (d ln k r �dT). �is de�nition reduces to the earlier one (as the slope
of a straight line) for a temperature-independent activation energy. However,
this latter de�nition is more general, because it allows E a to be obtained from
the slope (at the temperature of interest) of a plot of ln k r against 1�T even if
the Arrhenius plot is not a straight line. Non Arrhenius behaviour is sometimes
a sign that quantum mechanical tunnelling is playing a signi�cant role in the
reaction. A reaction with a very small or zero activation energy, so that k r = A,
such as for some radical recombination reactions in the gas phase, has a rate
that is largely temperature independent.
Solutions to exercises
E��D.�(a)
�e relationship between the values of a rate constant at two di�erent temperatures is given by [��D.�–���], ln(k r,2 �k r,1 ) = (E a �R)(1�T1 − 1�T2 ). Hence,
taking T1 = 37 ○ C and T2 = 15 ○ C,
k r,2
Ea 1
1
= exp � � − ��
k r,1
R T1 T2
87 × 103 J mol−1
1
1
�
−
��
[37+273.15] K [15+273.15] K
8.3145 J K−1 mol−1
= �.���
= exp �
�e rate constant therefore drops to about 7.6 % of its original value when the
temperature is lowered for 37 ○ C to 15 ○ C.
E��D.�(a) As explained in Section ��D.�(a) on page ��� the fraction f of collisions that are
su�ciently energetic to be successful is given by the exponential factor e−E a �RT .
Rearranging f = e−E a �RT for T and setting f = 0.10 gives
T =−
Ea
50 × 103 J mol−1
=−
= 2.6 × 103 K
R ln f
(8.3145 J K−1 mol−1 ) × ln 0.10
E��D.�(a) �e Arrhenius equation is given by [��D.�–���], k r = Ae−E a �RT . In this case
k r = (8.1 × 10−10 dm3 mol−1 s−1 ) × exp �−
= 3.2 × 10−12 dm3 mol−1 s−1
23 × 103 J mol−1
�
(8.3145 J K−1 mol−1 ) × (500 K)
E��D.�(a) �e relationship between the values of a rate constant at two di�erent temperatures is given by [��D.�–���], ln(k r,2 �k r,1 ) = (E a �R)(1�T1 − 1�T2 ). Rearranging for E a gives
Ea =
−1
−1
−2
−3
R ln(k r,2 �k r,1 ) (8.3145 J K mol ) × ln �(2.67 × 10 )�(3.80 × 10 )�
=
1�T1 − 1�T2
1�([35 + 273.15] K) − 1�([50 + 273.15] K)
= 1.07... × 105 J mol−1 = ��� kJ mol−1
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e frequency factor is found by rearranging the Arrhenius equation [��D.�–
���], k r = Ae−E a �RT , for A. �e data for both temperatures gives the same result.
At T1
A = k r eE a �RT1
= (3.80×10−3 dm3 mol−1 s−1 )×exp
= 6.62 × 1015 dm3 mol−1 s−1
At T2
A = k r eE a �RT2
= (2.67×10−2 dm3 mol−1 s−1 )×exp
= 6.62 × 1015 dm3 mol−1 s−1
1.07...×105 J mol−1
(8.3145 J K−1 mol−1 )×([35+273.15] K)
1.07...×105 J mol−1
(8.3145 J K−1 mol−1 )×([50+273.15] K)
E��D.�(a) �e relationship between the values of a rate constant at two di�erent temperatures is given by [��D.�–���], ln(k r,2 �k r,1 ) = (E a �R)(1�T1 − 1�T2 ). Rearranging for E a , and using k r,2 �k r,1 = 3 because the rate constant triples between the
two temperatures, gives
Ea =
R ln(k r,2 �k r,1 )
(8.3145 J K−1 mol−1 ) × ln 3
=
= �� kJ mol−1
1�T1 −1�T2
1�([24+273.15] K)−1�([49+273.15] K)
Solutions to problems
P��D.�
�e de�nition of E a in [��D.�–���], E a = RT 2 (d ln k r �dT), is rearranged and
integrated.
Ea
d ln k r
=
RT 2
dT
P��D.�
hence
� d ln k r = �
Ea
dT
RT 2
�e le�-hand side integral is simply ln k r . If E a does not vary with temperature
then the integral on the right is evaluated by taking E a �R outside the integral
to give
Ea
1
Ea
ln k r =
dT = −
+c
�
2
R
T
RT
�is is [��D.�–���], ln k r = ln A − E a �RT, once the constant of integration c is
identi�ed with ln A.
�e Arrhenius equation [��D.�–���], ln k r = ln A−E a �RT, implies that a plot of
ln k r against 1�T should give a straight line of slope −E a �R and intercept ln A.
�e data are plotted in Fig. ��.�.
T�K
1 000
1 200
1 400
1 600
k r �dm3 mol−1 s−1
8.35 × 10−10
3.08 × 10−8
4.06 × 10−7
2.80 × 10−6
1�(T�K)
0.001 000
0.000 833
0.000 714
0.000 625
ln(k r �dm3 mol−1 s−1 )
−20.90
−17.30
−14.72
−12.79
537
14 CHEMICAL KINETICS
ln(k r �dm3 mol−1 s−1 )
538
−15
−20
0.0007
0.0009
1�(T�K)
0.0011
Figure 14.9
�e data fall on a good straight line, the equation for which is
ln(k r �dm3 mol−1 s−1 ) = (−2.165 × 104 ) × 1�(T�K) + 0.7457
Identifying the slope with −E a �R gives the activation energy as
E a = −slope × R = −(−2.165 × 104 K) × (8.3145 J K−1 mol−1 ) = ��� kJ mol−1
Identifying the intercept with ln A gives the frequency factor as
A = e0.7457 dm3 mol−1 s−1 = �.�� dm3 mol−1 s−1
�e units of A are the same as the units of k r .
P��D.�
�e Arrhenius equation [��D.�–���], ln k r = ln A−E a �RT, implies that a plot of
ln k r against 1�T should give a straight line of slope −E a �R and intercept ln A.
�e data are plotted in Fig. ��.��.
T�K
295
223
218
213
206
200
195
k r �dm3 mol−1 s−1
3.55 × 106
4.94 × 105
4.52 × 105
3.79 × 105
2.95 × 105
2.41 × 105
2.17 × 105
1�(T�K)
0.003 39
0.004 48
0.004 59
0.004 69
0.004 85
0.005 00
0.005 13
ln(k r �dm3 mol−1 s−1 )
15.08
13.11
13.02
12.85
12.59
12.39
12.29
�e data fall on a reasonable straight line, the equation for which is
ln(k r �dm3 mol−1 s−1 ) = (−1.642 × 103 ) × 1�(T�K) + 20.59
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
ln(k r �dm3 mol−1 s−1 )
15
14
13
12
0.0035
0.0040
0.0045
1�(T�K)
0.0050
Figure 14.10
Identifying the slope with −E a �R gives the activation energy as
E a = −slope × R = −(−1.642 × 103 K) × (8.3145 J K−1 mol−1 ) = ��.� kJ mol−1
Identifying the intercept with ln A gives the frequency factor as
A = e20.59 dm3 mol−1 s−1 = 8.75 × 108 dm3 mol−1 s−1
�e units of A are the same as the units of k r .
14E
Reaction mechanisms
Answers to discussion questions
D��E.�
In the pre-equilibrium assumption an intermediate is assumed to be in equilibrium with the reactants. For this assumption to apply it is necessary for the rate
at which the intermediate returns to reactants to be fast compared to the rate at
which the intermediate goes to products. �e rate-determining step is between
the intermediate and the products. �e rate at which the intermediate is formed
from the reactants is not relevant to the establishment of pre-equilibrium provided the other criteria are satis�ed.
In the steady-state assumption an intermediate is formed from the reactants
and the moment it is formed it goes on to products. �e step between reactants
and intermediate is therefore the rate-determining step, and necessarily the
concentration of the intermediate is low.
�e two approximations di�er in that in the steady-state approximation the intermediate is necessarily at low concentration, whereas in the pre-equilibrium
approximation this condition does not hold – indeed, the intermediate may accumulate. However, if the intermediate reacts immediately by either returning
to reactants or going on to products, then the pre-equilibrium assumption also
results in a low concentration of the intermediate.
539
540
14 CHEMICAL KINETICS
In the pre-equilibrium assumption the apparent rate constant is a product of a
rate constant and an equilibrium constant and so the activation energy may be
negative (Section ��E.� on page ���). For the steady-state approximation the
activation energy is positive.
D��E.�
Suppose that a reactant R can give alternative products P and Q by di�erent
reactions. If the rate constant for the formation of P is greater than that for
forming Q, then to start with more P will be formed. However, as time proceeds
it may be that the reverse reactions from P and Q back to R start to become
signi�cant, and eventually the reactions reach equilibrium. It may be that at
equilibrium the amount of Q exceeds that of P, even though initially the amount
of P exceeded that of Q.
If the relative proportions of the products are determined by the rate at which
they are formed, the reaction is said to be under kinetic control. If the amounts
are determined by the relevant equilibrium constants, the reaction is said to
be under thermodynamic control. �e latter will only occur if the reverse
reactions are signi�cant.
D��E.�
�e overall reaction order is the sum of the powers of the concentrations of
all of the substances appearing in the experimental rate law for the reaction;
hence, it is the sum of the individual orders (exponents) associated with a each
reactant. Reaction order is an experimentally determined quantity.
Molecularity is the number of reactant molecules participating in an elementary reaction. Molecularity has meaning only for an elementary reaction, but
reaction order applies to any reaction. In general, reaction order bears no
necessary relation to the stoichiometry of the reaction, with the exception of
elementary reactions, where the order of the reaction corresponds to the number of molecules participating in the reaction; that is, to its molecularity. �us
for an elementary reaction, overall order and molecularity are the same and are
determined by the stoichiometry.
Solutions to exercises
E��E.�(a)
�e steady-state approximation is applied to the intermediate species O.
d[O]
= k a [O3 ] − k a′ [O2 ][O] − k b [O][O3 ] = 0
dt
Rearranging for [O] gives
(k a′ [O2 ] + k b [O3 ]) [O] = k a [O3 ] hence
�e rate of decomposition of O� is
[O] =
k a [O3 ]
k a′ [O2 ] + k b [O3 ]
d[O3 ]
= −k a [O3 ]+k a′ [O2 ][O]−k b [O][O3 ] = −k a [O3 ]+[O]{k a′ [O2 ]−k b [O3 ]}
dt
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
because O� is consumed in steps � and �, but produced in step �. Inserting the
steady-state expression for [O] gives
d[O3 ]
k a [O3 ]{k a′ [O2 ] − k b [O3 ]}
= −k a [O3 ] +
dt
k a′ [O2 ] + k b [O3 ]
−k a k a′ [O2 ][O3 ] − k a k b [O3 ]2 + k a k a′ [O3 ][O2 ] − k a k b [O3 ]2
=
k a′ [O2 ] + k b [O3 ]
−2k a k b [O3 ]2
= ′
k a [O2 ] + k b [O3 ]
If step � is rate limiting, such that k a′ [O2 ][O] � k b [O][O3 ], and hence k a′ [O2 ] �
k b [O3 ], the denominator simpli�es to k a′ [O2 ] and hence
d[O3 ] −2k a k b [O3 ]2
=
dt
k a′ [O2 ]
E��E.�(a)
As required, the rate of decomposition of O� is second order in O� and order
−1 in O� .
�e overall activation energy for a reaction consisting of a pre-equilibrium
followed by a rate-limiting elementary step is given by [��E.��–���], E a = E a,a +
E a,b − E a,a′ , where E a,a and E a,a′ are the forward and reverse activation energies for the pre-equilibrium and E a,b is the activation energy for the following
elementary step. In this case
E a = (25 kJ mol−1 ) + (10 kJ mol−1 ) − (38 kJ mol−1 ) = −3 kJ mol−1
As explained in Section ��E.� on page ���, negative activation energies such as
this are possible for composite reactions.
E��E.�(a)
(i) A pre-equilibrium A2 � 2A between A� and A is described by the equilibrium constant K given by
K=
([A]�c −○ )2
[A]2
=
([A2 ]�c −○ ) [A2 ]c −○
hence
[A] = (Kc −○ [A2 ])1�2
�e equilibrium constant K is written in terms of rate constants using
[��C.�–���], K = (k a �k a′ ) × (k b �k b′ ) × .... However, in order to make
K dimensionless it is necessary in this case to include a factor of 1�c −○
because k a is a �rst-order rate constant with units s−1 while k a′ is a secondorder rate constant with units dm3 mol−1 s−1 . �us K = k a �k a′ c −○ , which,
on substituting into the above expression for [A] yields
ka
[A] = � ′ −○ c −○ [A2 ]�
ka c
1�2
ka
= � ′ [A2 ]�
ka
1�2
�is expression is alternatively obtained by noting that at equilibrium the
rates of the forward and reverse reactions are the same (provided that the
step to P can be ignored)
k a [A2 ] = k a′ [A]2
hence
ka
[A] = � ′ [A2 ]�
ka
1�2
541
542
14 CHEMICAL KINETICS
�e rate of formation of P is given by d[P]�dt = k b [A][B]; substituting
the above expression for [A] into this gives
1�2
1�2
d[P]
ka
ka
= k b [A][B] = k b � ′ � [A2 ][B] = k b � ′ � [A2 ]1�2 [B]
dt
ka
ka
(ii) �e net rate of change in the concentration of A is
d[A]
= 2k a [A2 ] − 2k a′ [A]2 − k b [A][B]
dt
In the steady-state approximation this is assumed to be zero
2k a [A2 ] − 2k a′ [A]2 − k b [A][B] = 0
Hence 2k a′ [A]2 + k b [B][A] − 2k a [A2 ] = 0. �is is a quadratic equation in
[A], for which the solution is
hence
[A] =
−k b [B] + �k b2 [B]2 + 16k a′ k a [A2 ]�
1�2
4k a′
where the positive square root is chosen in order to avoid obtaining a
negative value for [A]. �e rate of formation of P is given by d[P]�dt =
k b [A][B]; substituting the above expression for [A] into this gives
−k b [B] + �k b [B] − 16k a k a [A2 ]�
d[P]
= k b [A][B] = k b [B] ×
dt
4k a′
�
1�2 �
′
�
k b [B] �
�−k b [B] + k b [B] �1 + 16k a k a [A2 ] � �
=
�
�
2
4k a′ �
k b [B]2
�
�
�
1�2 �
2
2 �
′
�
k [B] �
16k a k a [A2 ]
= b ′ �
−1 + �1 +
� �
�
�
2
2
4k a �
k b [B]
�
�
�
2
′
2
1�2
Under certain circumstances this rate law simpli�es. If 16k a′ k a [A2 ]�k b2 [B]2 �
1 then
�
1�2 �
′
�
d[P] k b2 [B]2 �
�−1 + � 16k a k a [A2 ] � �
≈
�
�
2
′
2
dt
4k a �
k b [B]
�
�
�
≈
k b2 [B]2
16k a′ k a [A2 ]
�
�
′
4k a
k b2 [B]2
1�2
1�2
ka
= k b � ′ � [A2 ]1�2 [B]
ka
which is the same as the rate law derived in part (i) assuming a preequilibrium. �e condition 16k a′ k a [A2 ]�k b2 [B]2 � 1 corresponds to the
A2 ⇄ A + A steps being much faster than the step involving B and k b ;
this is precisely the situation corresponding to a pre-equilibrium because
the removal of A in the reaction with B is then too slow to a�ect the
maintenance of the equilibrium.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
On the other hand, if 16k a′ k a [A2 ]�k b2 [B]2 � 1 then the square root is
approximated by the expansion (1 + x)1�2 ≈ 1 + 12 x to give
k b2 [B]2 8k a′ k a [A2 ]
d[P] k b2 [B]2
16k a′ k a [A2 ]
1
≈
�−1
+
�1
+
×
��
=
×
2
dt
4k a′
k b2 [B]2
4k a′
k b2 [B]2
= 2k a [A2 ]
�is rate law corresponds to the step A2 → A + A being rate-determining:
once A has formed from A2 in this step it immediately goes on to form
product. �e factor of � arises because each molecule of A� that dissociates forms two A molecules which react with two B molecules to
form two molecules of product. Hence the rate of product formation is
twice the rate of A� dissociation. �is situation does not correspond to a
pre-equilibrium because the immediate removal of A by its reaction with
B does not allow A� and A to come to equilibrium.
Solutions to problems
�e concentration of I in the reaction mechanism A → I → P is given by
[��E.�b–���],
ka
�e−k a t − e−k b t � [A]0
[I] =
kb − ka
ka
kb
�is expression is plotted in Fig. ��.�� for [A]0 = 1 mol dm−3 , k b = 1 s−1 , and
various values of k a . �e line for k a = 10 s−1 corresponds to part (a) of the
question.
k a = 10 s−1
k a = 2 s−1
k a = 0.5 s−1
k a = 0.2 s−1
k a = 0.1 s−1
k a = 0.02 s−1
0.8
0.6
[I]�mol dm−3
P��E.�
0.4
0.2
0.0
0
1
2
3
4
5
t�s
Figure 14.11
If k b � k a , the concentration of I remains low and, apart from the initial induction period, approximately constant during the reaction. �us the steady-state
543
544
14 CHEMICAL KINETICS
P��E.�
approximation that d[I]�dt = 0 becomes increasingly valid as the ratio k b �k a
increases.
It is shown in Example ��E.� on page ��� that for the case of two consecutive
unimolecular reactions the concentration of the intermediate is greatest at a
time given by t max = (ln[k a �k b ])�(k a − k b ). �e half-life of a �rst-order reaction is related to the rate constant according to [��B.�–���], t 1�2 = ln 2�k r . �is
is rearranged to k r = ln 2�t 1�2 and used to substitute for the rate constants in
the expression for t max .
ln 2�t 1�2,a
1
ka
1
ln
=
ln
k a − k b k b (ln 2�t 1�2,a ) − (ln 2�t 1�2,b ) ln 2�t 1�2,b
t 1�2,b
1
=
ln
ln 2 �1�t 1�2,a − 1�t 1�2,b � t 1�2,a
t max =
P��E.�
Hence t 1�2 =
1
33.0 d
× ln
= ��.� d
ln 2 × [1�(22.5 d) − 1�(33.0 d)]
22.5 d
For the scheme A ⇄ B ⇄′ C ⇄ D the rates of change of the intermediates B and
C are
ka
kb
kc
k a′
kb
k c′
d[B]
= k a [A]−k a′ [B]−k b [B]+k b′ [C]
dt
d[C]
= k b [B]−k b′ [C]−k c [C]+k c′ [D]
dt
In the steady-state approximation, both of these expressions are equal to zero.
Furthermore, because D is removed as soon as it is formed, [D] = 0 and so the
expression for d[C]�dt becomes
k b [B] − k b′ [C] − k c [C] = 0
�e expression for d[B]�dt becomes
hence
hence
hence
hence
[B] =
(k b′ + k c )[C]
kb
k a [A] − (k a′ + k b )[B] + k b′ [C] = 0
(k b′ + k c )[C]
+ k b′ [C] = 0
kb
(k ′ + k b )(k b′ + k c ) − k b k b′
� a
� [C] = k a [A]
kb
k a k b [A]
k a k b [A]
[C] = ′
= ′ ′
′
′
(k a + k b )(k b + k c ) − k b k b k a k b + k a′ k c + k b k c
k a [A] − (k a′ + k b )
where on the second line the expression for [B] derived above is substituted in.
Finally, the rate of formation of D is
d[D]
k a k b k c [A]
= k c [C] − k c′ [D] = k c [C] = ′ ′
dt
k a k b + k a′ k c + k b k c
where [D] = 0 is used and the expression for [C] is substituted in.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P��E.�
�e equilibrium constants for the two pre-equilibria are
[(HCl)2 ]c −○
K 1 [HCl]2
hence [(HCl)2 ] =
2
[HCl]
c −○
[complex]c −○
K 2 [HCl][CH3 CH−CH2 ]
K2 =
hence [complex] =
[HCl][CH3 CH−CH2 ]
c −○
K1 =
�e factors of c −○ are needed to make K 1 and K 2 dimensionless. �e rate of
product formation is
d[CH3 CHClCH3 ]
= k r [(HCl)2 ][complex]
dt
K 1 [HCl]2 K 2 [HCl][CH3 CH−CH2 ]
= kr ×
×
c −○
c −○
kr K1 K2
=
[HCl]3 [CH3 CH−CH2 ]
c −○ 2
υ=
�us the reaction is predicted to be �rst-order in propene and third-order in
HCl, as required.
P��E.�
Applying the steady-state approximation to the intermediates OF and F gives
d[OF]
= k a [F2 O]2 + k b [F][F2 O] − 2k c [OF]2 = 0
dt
d[F]
= k a [F2 O]2 − k b [F][F2 O] + 2k c [OF]2 − 2k d [F]2 [F2 O] = 0
dt
On adding together these two equations the k b and k c terms cancel to give
2k a [F2 O]2 − 2k d [F]2 [F2 O] = 0
hence
[F] = �
hence
1�2
ka
[F2 O]�
kd
k a [F2 O] = k d [F]2
Steps a and b lead to the net consumption of one F� O, while steps d and c lead
to no net change. �e rate of consumption of F� O is therefore
−
1�2
d[F2 O]
ka
= k a [F2 O]2 + k b [F2 O][F] = k a [F2 O]2 + k b [F2 O] � [F2 O]�
dt
kd
k a 1�2
= k a [F2 O]2 + k b � � [F2 O]3�2
kd
�
kr
��� � � � � � � � � � �� � � � � � � � � � � ��
k r′
which is the required expression with k r = k a and k r′ = k b
14F
�
k a �k d .
Examples of reaction mechanisms
Answers to discussion questions
D��F.�
In the analysis of stepwise polymerization, the rate constant for the secondorder condensation is assumed to be independent of the chain length and to
545
546
14 CHEMICAL KINETICS
remain constant throughout the reaction. It follows, then, that the degree of
polymerization is given by [��F.��b–���], �N� = 1 + k r t[A]0 . �erefore, the
average molar mass can be controlled by adjusting the initial concentration of
monomer and the length of time that the polymerization is allowed to proceed.
As discussed in Section ��F.�(b) on page ���, chain polymerization involves
initiation, propagation, and termination steps. �e derivation of the overall
rate equation utilizes the steady-state approximation and leads to the following
expression for the average number of monomer units in the polymer chain
([��F.��–���]) �N� = 2k r [M][In]−1�2 where k r = k p (4 f k i k t )−1�2 , and where
k p , k i , and k t are the rate constants for the propagation, initiation, and termination steps respectively, and f is the fraction of radicals that successfully
initiate a chain. It is seen that the average molar mass of the polymer is directly
proportional to the monomer concentration, and inversely proportional to the
square root of the initiator concentration, and to the rate constant for initiation.
�erefore, the slower the initiation of the chain, the higher the average molar
mass of the polymer.
D��F.�
As temperature increases, the rate of an enzyme-catalyzed reaction is expected
to increase. However, at a su�ciently high temperature the enzyme denatures
and a decrease in the reaction rate is observed. Temperature related denaturation is caused by the action of vigorous vibrational motion, which destroys secondary and tertiary protein structure. Electrostatic, internal hydrogen bonding, and van der Waals interactions that hold the protein in its active, folded
shape are broken with the protein unfolding into a random coil. �e active site
and enzymatic activity is lost.
�e rate of a particular enzyme-catalyzed reaction may also appear to decrease
at high temperature in the special case in which an alternative substrate reaction, which has a relatively slow rate at low temperature, has the faster rate
increase with increasing temperature. A temperature may be reached at which
the alternative reaction predominates.
Solutions to exercises
E��F.�(a)
�e Michaelis–Menten equation for the rate of an enzyme-catalysed reaction
is given by [��F.��a–���], υ = υ max �(1 + K M �[S]0 ). Rearranging for υ max gives
υ max = υ �1 +
E��F.�(a)
KM
0.046 mol dm−3
� = (1.04 mmol dm−3 s−1 ) × �1 +
�
[S]0
0.105 mol dm−3
= �.�� mmol dm−3 s−1
Example ��F.� on page ��� gives the values K M = 10.0 mmol dm−3 and υ max =
0.250 mmol dm−3 s−1 for an enzyme concentration of [E]0 = 2.3 nmol dm−3 .
�e catalytic e�ciency is de�ned in the exercise as k b �K M , and υ max is related to
k b according to [��F.��b–���], υ max = k b [E]0 , hence k b = υ max �[E]0 . �erefore,
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
the catalytic e�ciency is
kb
υ max
0.250 × 10−3 mol dm−3 s−1
=
=
K M K M [E]0 (10.0 × 10−3 mol dm−3 ) × (2.3 × 10−9 mol dm−3 )
E��F.�(a)
= 1.1 × 107 dm3 mol−1 s−1
�e e�ective rate constant for the Lindemann–Hinshelwood mechanism is given
by [��F.�–���], 1�k r = k a′ �k a k b + 1�k a [A]. �e di�erence between the e�ective
rate constant at two pressures is therefore
1
1
1
1
1
−
=
�
−
�
k r,2 k r,1 k a [A]2 [A]1
hence
ka =
1�[A]2 − 1�[A]1
1�k r,2 − 1�k r,1
�e rate constant for the activation step, k a , is therefore
ka =
E��F.�(a)
1�(12 Pa) − 1�(1.30 × 103 Pa)
= 1.9 × 10−6 Pa−1 s−1
1�(2.10 × 10−5 s−1 ) − 1�(2.50 × 10−4 s−1 )
or 1.9 MPa−1 s−1 .
�e fraction of condensed groups at time t of a stepwise polymerisation is given
by [��F.��–���], p = k r t[A]0 �(1 + k r t[A]0 ). Hence, a�er �.�� h, or 5.00 h ×
(602 s)�(1 h) = 1.80 × 104 s,
p=
k r t[A]0
1 + k r t[A]0
(1.39 dm3 mol−1 s−1 ) × (1.80 × 104 s) × (10.0 × 10−3 mol dm−3 )
1 + (1.39 dm3 mol−1 s−1 ) × (1.80 × 104 s) × (10.0 × 10−3 mol dm−3 )
= 0.996... = �.���
=
�e degree of polymerisation in a stepwise polymerisation is given by [��F.��a–
���], �N� = 1�(1 − p).
�N� =
E��F.�(a)
1
1
=
= ���
1 − p 1 − 0.996...
�e kinetic chain length in a chain polymerisation reaction is given by [��F.��c–
���], λ = k r [M][In]−1�2 . �e ratio of chain length under the two di�erent sets
of conditions is therefore
−1�2
−1�2
λ 2 k r [M]2 [In]2
[M]2
[In]2
=
=�
��
�
λ 1 k r [M]1 [In]−1�2
[M]1
[In]1
1
=
1
× 3.6−1�2 = �.��
4.2
Solutions to problems
P��F.�
�e e�ective rate constant k r in the Lindemann–Hinshelwood mechanism is
given by [��F.�–���], 1�k r = k a′ �k a k b + 1�k a [A]. �is expression implies that
a plot of 1�k r against 1�[A] should be a straight line. �e data are plotted in
Fig. ��.��, using pressure as a measure of concentration.
547
14 CHEMICAL KINETICS
p�Torr
84.1
11.0
2.89
0.569
0.120
0.067
104 k r �s−1
2.98
2.23
1.54
0.857
0.392
0.303
1�(p�Torr)
0.011 9
0.090 9
0.346
1.76
8.33
14.9
1�(104 k r �s−1 )
0.336
0.448
0.649
1.167
2.551
3.300
4
3
1�(104 k r �s−1 )
548
2
1
0
0
5
10
1�(p�Torr)
15
Figure 14.12
�e data lie on a curve rather than on a straight line, so it is concluded that the
Lindemann–Hinshelwood mechanism does not �t these data.
P��F.�
Each molecule of hydroxyacid has one OH group and one COOH group (A),
so [OH] = [A]. Hence the given rate expression, d[A]�dt = −k r [A]2 [OH],
becomes
d[A]
1
= −k r [A]3 hence −
d[A] = k r dt
dt
[A]3
Integration of this expression, with the limits that the concentration is [A]0 at
time t = 0 and [A] at some later time t, gives
[A]
t
1
−
d[A]
=
k r dt
�
�
[A]3
[A]0
0
hence
hence
1
1
−
= 2k r t
2
[A]
[A]20
[A]
1
t
�
= k r t�0
2[A]2 [A]0
Rearranging gives
1
1
= 2k r t +
[A]2
[A]20
hence
[A]2 =
1
[A]20
=
2k r t + 1�[A]20 2k r t[A]20 + 1
To go to the �nal expression the top and bottom of the fraction are multiplied by
[A]20 . Taking the square root gives [A] = [A]0 �(2k r t[A]20 + 1)1�2 . As explained
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
in Section ��F.�(a) on page ���, the degree of polymerisation �N� is the ratio of
the initial concentration of A, [A]0 , to the concentration of end groups, [A], at
the time of interest. Hence
P��F.�
�N� =
[A]0
[A]0
=
= (2k r t[A]20 + 1)1�2
[A] [A]0 �(2k r t[A]20 + 1)1�2
�e Michaelis–Menten equation [��F.��a–���] is
υ=
υ max
1 + K M �[S]0
�is equation is plotted for �xed υ max with a range of K M values in Fig. ��.��,
and for �xed K M with a range of υ max values in Fig. ��.��.
1.0
υ�mmol dm−3 s−1
0.8
K M = 0.01 mol dm−3
K M = 0.1 mol dm−3
0.6
K M = 0.5 mol dm−3
0.4
0.2
0.0
0.0
0.2
Figure 14.13
P��F.�
0.4
υ max = 1 mmol dm−3 s−1
0.6
[S]0 �mol dm
−3
0.8
1.0
�e Lineweaver-Burk equation, [��F.��b–���], expresses the reciprocal of the
velocity as 1�υ = 1�υ max + (K M �υ max )(1�[S]0 ). �is expression implies that a
plot of 1�υ against 1�[S]0 will be a straight line of slope K M �υ max and intercept
1�υ max . Such a plot is shown in Fig. ��.��.
[ATP]�
µmol dm−3
0.6
0.8
1.4
2.0
3.0
υ�
µmol dm−3 s−1
0.81
0.97
1.30
1.47
1.69
1
[ATP]�(µmol dm−3 )
1.67
1.25
0.71
0.50
0.33
1
υ�(µmol dm−3 s−1 )
1.23
1.03
0.77
0.68
0.59
549
14 CHEMICAL KINETICS
1.0
K M = 0.1 mol dm−3
0.8
υ�mmol dm−3 s−1
υ max = 1 mmol dm−3 s−1
0.6
0.4
υ max = 0.5 mmol dm−3 s−1
0.2
0.0
0.0
υ max = 0.1 mmol dm−3 s−1
0.2
0.4
0.6
[S]0 �mol dm
Figure 14.14
0.8
−3
1.0
1�(υ�µmol dm−3 s−1 )
550
1.0
0.5
0.0
Figure 14.15
0.5
1.0
−3
1�([ATP]�µmol dm )
1.5
�e data lie on a good straight line, the equation of which is
1�(υ�µmol dm−3 s−1 ) = 0.48 × 1�([ATP]�µmol dm−3 ) + 0.43
�e intercept is identi�ed with 1�υ max so that
υ max =
1
= 2.32... µmol dm−3 s−1 = �.� µmol dm−3 s−1
0.43 µmol dm−3 s−1
�e slope is identi�ed with K M �υ max so that
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
�
K M = ( 0.48 s) × (2.32... µmol dm−3 s−1 ) = �.� µmol dm−3
slope
υ max
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
14G Photochemistry
Answer to discussion question
D��G.�
�e time scales of atomic processes are rapid indeed. Note that the times given
here are in some way typical values for times that may vary over two or three
orders of magnitude. For example, vibrational wavenumbers can range from
about ���� cm−1 (for H� ) to ��� cm−1 (for I� ) and even lower, with a corresponding range of associated times. Radiative decay rates of electronic states
can vary even more widely: times associated with phosphorescence can be in
the millisecond and even second range. A large number of time scales for physical, chemical, and biological processes on the atomic and molecular scale are
reported in Figure � of A. H. Zewail, ‘Femtochemistry: Atomic-Scale Dynamics
of the Chemical Bond’, Journal of Physical Chemistry A, ���, ���� (����).
Radiative decay of excited electronic states can range from about 10−9 s to
10−4 s, and even longer for phosphorescence involving ‘forbidden’ decay paths.
Molecular rotational motion takes place on a scale of 10−12 s to 10−9 s. Molecular vibrations are faster still, about 10−14 s to 10−12 s. Proton transfer reactions
occur on a timescale of about 10−10 s to 10−9 s, although protons can hop from
molecule to molecule in water even more rapidly (1.5 × 10−12 s).
Harvesting of light during plant photosynthesis involves very fast time scales
of several energy-transfer and electron-transfer steps in photosynthesis. Initial
energy transfer (to a nearby pigment) has a time scale of around 10−13 s to
5 × 10−12 s, with longer-range transfer (to the reaction centre) taking about
10−10 s. Immediate electron transfer is also very fast (about � ps), with ultimate
transfer (leading to oxidation of water and reduction of plastoquinone) taking
from 10−10 s to 10−3 s. �e mean time between collisions in liquids is similar
to vibrational periods, around 10−13 s.
Solutions to exercises
E��G.�(a)
�e e�ciency of resonance energy transfer η T is de�ned by [��G.�–���], η T =
1 − � F �� F,0 , and the distance-dependence of the e�ciency is given by [��G.�–
���], η T = R 06 �(R 06 + R 6 ), where R is the donor–acceptor distance and R 0 is a
constant characteristic of the particular donor–acceptor pair.
In this case a decrease of the �uorescence quantum yield by ��% implies that
� F = 0.9� F,0 . Hence the e�ciency is η T = 1 − � F �� F,0 = 1 − 0.9 = 0.1.
Rearranging [��G.�–���] for R, and taking R 0 = 4.9 nm from Table ��G.� on
page ���, gives
R = R0 �
1 − ηT
�
ηT
1�6
= (4.9 nm) × �
1 − 0.1 1�6
� = �.� nm
0.1
E��G.�(a) �e primary quantum yield is de�ned by [��G.�a–���], � = N events �N abs . In
this equation N abs is the number of photons absorbed and N events is, in this
case, the number of molecules of A that decompose, N decomposed . Rearranging
551
552
14 CHEMICAL KINETICS
gives
N abs =
N decomposed n decomposed N A (n formed �2)N A
=
=
�
�
�
In the �nal expression, n formed is the amount in moles of B that is formed. �e
stoichiometry of the reaction A → 2B + C implies that the amount of A that
decomposes is half the amount of B that is formed, n decomposed = n formed �2.
�e quantum yield is 210 mmol einstein−1 , or 0.210 mol mol−1 = 0.210, hence
N abs =
(n formed �2)N A (2.28 × 10−3 mol)�2 × (6.0221 × 1023 mol−1 )
=
�
0.210
= 3.27 × 1021
E��G.�(a) �e �uorescence quantum yield is given by [��G.�–���], � F,0 = k F τ 0 . �e
observed lifetime τ 0 is given by [��G.�b–���], τ 0 = 1�(k F + k ISC + k IC ), which
is written as τ 0 = 1�k r where k r = k F + k ISC + k IC is the e�ective �rst-order
rate constant for the decay of the excited state of the �uorescing species. For a
�rst-order process k r is related to the half-life according to [��B.�–���], t 1�2 =
ln 2�k r , and combining this expression with τ 0 = 1�k r gives t 1�2 = ln 2�(1�τ 0 ) =
(ln 2)τ 0 . Hence τ 0 = t 1�2 � ln 2.
Rearranging [��G.�–���] then gives
kF =
� F,0
� F,0
� F,0 ln 2
0.35 × ln 2
=
=
=
= 4.3 × 107 s−1
τ0
t 1�2 � ln 2
t 1�2
5.6 × 10−9 s
E��G.�(a) �e Stern–Volmer equation [��G.�–���] is � F,0 �� F = 1 + τ 0 k Q [Q], where � F
and � F,0 are the �uorescence quantum yields with and without the quencher.
�e rate of �uorescence υ, and hence the �uorescence intensity, is directly proportional to the �uorescence quantum yield according to [��G.�b–���], � =
υ�I abs . �erefore to reduce the �uorescence intensity to 50% of the unquenched
value requires � F = 12 � F,0 and hence � F,0 �� F = 2. Rearranging the Stern–
Volmer equation then gives
[Q] =
� F,0 �� F − 1
2−1
=
−9
τ0 kQ
(6.0 × 10 s)×(3.0 × 108 dm3 mol−1 s−1 )
= �.�� mol dm−3
Solutions to problems
P��G.�
�e quantum yield is given by [��G.�a–���], � = N events �N abs where N abs is
the number of photons absorbed and N events is, in this case, the number of
molecules of the absorbing substance that decomposed. �e latter is equal
to n decomposed N A , where n decomposed is the amount in moles of substance that
decomposed.
�e number of photons absorbed is found by noting that the energy transferred
by each photon is given by [�A.�–���], ∆E = hν = hc�λ. �erefore the total
energy absorbed is E abs = N abs hc�λ. �is energy is also given by E abs = f Pt,
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
where P is the incident power, t is the time of exposure, and f is the fraction
of incident radiation that is absorbed. In this case f = 1 − 0.257 = 0.743.
Combining these expressions gives
f Pt =
N abs hc
λ
N abs =
hence
f Ptλ
hc
�is is substituted into � = N events �N abs , together with N events = n decomposed N A ,
to give
�=
=
N events n decomposed N A n decomposed N A hc
=
=
N abs
f Ptλ�hc
f Ptλ
(0.324 mol)×(6.0221×1023 mol−1 )
(0.743) × (87.5 W) × (28.0 min) × (60 s)�(1 min)
(6.6261×10−34 J s)×(2.9979×108 m s−1 )
×
= �.��
(320 × 10−9 m)
Note that 1 W = 1 J s−1 . �e fact that the quantum yield is greater than �
indicates that each absorbed photon can lead to the decomposition of more
than one molecule of absorbing material.
P��G.�
(a) �e concentration of the excited dansyl chloride decays with time according to [��G.�a–���], [S∗ ] = [S∗ ]0 e−t�τ 0 , or [S∗ ]�[S∗ ]0 = e−t�τ 0 . �e rate
of �uorescence is given by υ = k F [S∗ ] so the rate of �uorescence, and
hence the �uorescence intensity I F , is proportional to [S∗ ]. �erefore
I F �I 0 = [S∗ ]�[S∗ ]0 , and hence
IF
[S∗ ]
= ∗ = e−t�τ 0
I 0 [S ]0
hence
ln �
IF
t
�=−
I0
τ0
�is expression implies that a plot of ln(I F �I 0 ) against t should be a straight
line of slope −1�τ 0 and intercept zero. �e data are plotted in Fig. ��.��.
t�ns I F �I 0
5.0 0.45
10.0 0.21
15.0 0.11
20.0 0.05
ln(I F �I 0 )
−0.799
−1.561
−2.207
−2.996
�e data lie on a good straight line that passes close to the origin. �e
equation of the line is
ln(I F �I 0 ) = −0.145 × (t�ns) − 0.081
Identifying the slope with −1�τ 0 gives
−
1
= −0.145 ns−1
τ0
hence
τ0 =
1
= 6.89... ns = �.� ns
0.145 ns−1
553
14 CHEMICAL KINETICS
0
−1
ln(I F �I 0 )
554
−2
−3
0
5
10
15
20
25
t�ns
Figure 14.16
(b) �e �uorescence quantum yield is given by [��G.�–���], � F,0 = k F τ 0 . �is
equation is rearranged for k F
kF =
P��G.�
� F,0
0.70
=
= 1.0 × 108 s−1
τ0
6.89... × 10−9 s
�e Stern–Volmer equation [��G.�–���] is � F,0 �� F = 1+τ 0 k Q [Q]. As explained
in Section ��G.� on page ���, the ratio τ 0 �τ, where τ is the lifetime in the
presence of the quencher, is equal to � F,0 �� F , so the Stern–Volmer equation
becomes
τ0
= 1 + τ 0 k Q [Q] hence
τ
kQ =
τ 0 �τ − 1
τ 0 [Q]
In order to use the equation to calculate k Q it is necessary to �nd τ 0 and τ. �is
is done as follows.
�e concentration of an excited species such as Hg∗ varies with time according
to [��G.�a–���], [Hg∗ ] = [Hg∗ ]0 e−t�τ . Rearranging and taking logarithms
gives
[Hg∗ ]
= e−t�τ
[Hg∗ ]0
hence
ln �
[Hg∗ ]
t
�=−
[Hg∗ ]0
τ
�e rate of �uorescence is υ = k F [Hg∗ ], so the �uorescence intensity I is proportional to the Hg∗ concentration. Hence I�I 0 = [Hg∗ ]�[Hg∗ ]0 , and from the
above equation a plot of ln(I�I 0 ) against t should therefore be a straight line of
slope −1�τ and intercept zero. �e �uorescence intensity data are given relative
to the value at t = 0 and therefore represent I�I 0 .
�e data are plotted in Fig. ��.��.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
p N2 = 0
t�µs
0.0
5.0
10.0
15.0
20.0
I�I 0
1.000
0.606
0.360
0.220
0.135
ln(I�I 0 )
0.000
−0.501
−1.022
−1.514
−2.002
p N2 = 0
−1
−2
0
10
t�µs
t�µs
0.0
3.0
6.0
9.0
12.0
I�I 0
1.000
0.585
0.342
0.200
0.117
−1
20
−2
ln(I�I 0 )
0.000
−0.536
−1.073
−1.609
−2.146
p N2 = 9.74 × 10−4 atm
0
ln(I�I 0 )
ln(I�I 0 )
0
p N2 = 9.74 × 10−4 atm
0
10
t�µs
20
Figure 14.17
�e data fall on good straight lines, the equations of which are
p N2 = 0
p N2 = 9.74 × 10−4 atm
ln(I�I 0 ) = −0.100 × (t�µs) + −4.18 × 10−3
ln(I�I 0 ) = −0.179 × (t�µs) + 7.02 × 10−5
�e slopes are identi�ed with −1�τ 0 and −1�τ respectively
τ0 = −
1
= 10.0... µs
−0.100 µs−1
τ=−
1
= 5.59 ... µs
−0.179 µs−1
�e rearranged form of the Stern–Volmer equation found earlier, k Q = (τ 0 �τ −
1)�τ 0 [Q] is then used to calculate k Q . �e concentration of the N� quencher
is calculated from the partial pressure using the perfect gas equation [�A.�–�],
pV = nRT.
n N2 p N2
=
V
RT
9.74 × 10−4 atm
1.01325 × 105 Pa
1 m3
=
×
×
1 atm
(8.3145 J K−1 mol−1 ) × (300 K)
103 dm3
[N2 ] =
= 3.95... × 10−5 mol dm−3
where 1 Pa = 1 kg m−1 s−2 and 1 J = 1 kg m2 s−2 are used. Hence
kQ =
τ 0 �τ − 1
(10.0... × 10−6 s)�(5.59... × 10−6 s) − 1
=
τ 0 [N2 ]
(10.0 × 10−6 s) × (3.95... × 10−5 mol dm−3 )
= 2.00 × 109 dm3 mol−1 s−1
555
556
14 CHEMICAL KINETICS
P��G.�
�e e�ciency of resonance energy transfer is given by [��G.�–���],
ηT = 1 −
�F
τ
=1−
� F,0
τ0
where the second expression comes from the fact that, according to [��G.�–
���], � F,0 = k F τ 0 , the lifetime is proportional to quantum yield. �e e�ciency
of resonance energy transfer in terms of donor–acceptor distance is given by
[��G.�–���], η T = R 06 �(R 06 + R 6 ). Equating the two expressions for η T gives
1−
R6
τ
1
R6
1
= 6 0 6 hence
= 1 + 6 hence R = R 0 �
− 1�
τ0 R0 + R
1 − τ�τ 0
R0
1 − τ�τ 0
�e distance required to give τ = 10 ps is therefore
R = (5.6 nm) × �
1
− 1�
1 − (10 × 10−12 s)�(1 × 10−9 s)
1�6
1�6
= �.� nm
Solutions to integrated activities
I��.�
�e rate of the forward and backward steps are
A→B
d[B]
= IA
dt
B→A
d[B]
= −k r [B]2
dt
�e overall rate of change of [B] is therefore d[B]�dt = I A − k r [B]2 . In the
steady state, d[B]�dt = 0, hence
k r [B]2 = I a
hence
[B] = �
I a 1�2
�
kr
�is concentration can di�er signi�cantly from an equilibrium distribution
because changing the illumination may change the rate of the forward reaction
without a�ecting the reverse reaction. �is is in contrast to corresponding
equilibrium expression, in which the ratio [B]�[A] depends on a ratio of rate
constants for the forward and reverse reactions as explained in Section ��C.�
on page ���.
I��.�
(a) �e expressions [A] = [A]0 − x and [P] = [P]0 + x are substituted into
the rate law to give
υ=−
d[A]
= k r [A][P] = k r ([A]0 − x)([P]0 + x)
dt
�e expression [A] = [A]0 − x implies that d[A]�dt = −dx�dt so the
expression becomes
dx
= k r ([A]0 − x)([P]0 + x) hence
dt
dx
= k r dt
([A]0 − x)([P0 ] + x)
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Integration of this expression, using x = 0 at time t = 0 gives
�
x
0
t
dx
= � k r dt
([A]0 − x)([P0 ] + x)
0
�e le�-hand side is evaluated using Integral A.�
�
x
0
Integral A.� with A = [A]0 and B = −[P]0
��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �
x
dx
dx
= −�
([A]0 − x)([P0 ] + x)
0 ([A]0 − x)(−[P0 ] − x)
=−
=
1
(−[P]0 − x)[A]0
ln �
�
(−[P]0 ) − [A]0
([A]0 − x)(−[P]0 )
1
[A]0 ([P]0 + x)
ln �
�
[A]0 + [P]0
[P]0 ([A]0 − x)
�e right-hand side is k r t, hence the integrated rate law is
1
[A]0 ([P]0 + x)
ln �
� = kr t
[A]0 + [P]0
[P]0 ([A]0 − x)
�e expression [P] = [P]0 + x is rearranged to x = [P] − [P]0 . �is is used
to replace x in the integrated rate law
1
[A]0 [P]
ln �
� = kr t
[A]0 + [P]0
[P]0 ([A]0 − [P] + [P]0 )
��� � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � �
[A]0 [P]
ln �
� = ([A]0 + [P]0 )k r t
[P]0 ([A]0 − [P] + [P]0 )
a
hence
hence
hence
[A]0 [P] = [P]0 ([A]0 − [P] + [P]0 )e at
[P]([A]0 + [P]0 e at ) = [P]0 ([A]0 + [P]0 )e at
��� � � � � � � � ��� � � � � � � � � �
[P] ([A]0 + [P]0 )e at (1 + [P]0 �[A]0 )e at
(1 + b)e at
=
=
=
at
at
[P]0
[A]0 + [P]0 e
1 + ([P]0 �[A]0 ) e
1 + be at
��� � � � � � � � � � � � ��� � � � � � � � � � � � � �
b
hence
b
(b) �e quantity [P]�[P]0 is plotted against at in Fig. ��.�� for various values
of b.
�e plots are sigmoid in shape because the reaction is initially slow because only a small amount of P is present. As more product is formed, the
rate of the reaction υ = k r [A][P] increases and the curve becomes steeper,
until the reaction slows down towards the end due to the reactant A being
used up.
�e curves level o� at di�erent values because [P]�[P]0 is being plotted.
In each case the �nal concentration of P is given by the initial concentration of A, because all the A is eventually converted to P, plus the concentration of P that was present at the start, that is, [P]∞ = [A]0 + [P]0 . �e
557
14 CHEMICAL KINETICS
100
b = 0.01
[P]�[P]0
558
b = 0.02
50
0
0
2
4
Figure 14.18
at
6
b = 0.1
8
10
�nal value of [P]�[P]0 is therefore
[P]∞ [A]0 + [P]0 [A]0
1
=
=
+1= +1
[P]0
[P]0
[P]0
b
Changing b therefore changes the �nal value of [P]�[P]0 .
�e integrated rate equation for a �rst-order process A → P is given in
Table ��B.� on page ��� as [P]�[A]0 = 1 − e−k r t . In order to facilitate
comparison to the autocatalytic reaction it is instructive to re-plot the autocatalytic curves as [P]�[A]0 rather than as [P]�[P]0 . Furthermore it is
convenient to consider ([P]−[P]0 )�[A]0 rather than [P]�[A]0 , because in
this way the plot re�ects the amount of P that is produced in the reaction
rather than including any P that was present at the start. �e expression
for [P]�[P]0 derived above is adapted to give ([P] − [P]0 )�[A]0
Hence
[P] (1 + b)e at
=
[P]0
1 + be at
hence
[P] =
(1 + b)e at [P]0
1 + be at
b
��� � � � � � � � ��� � � � � � � � � �
[P] − [P]0
[P] ��� � � � � � � � ��� � � � � � � � � � (1 + b)e at [P]0 �[A]0
=
− [P]0 �[A]0 =
−b
[A]0
[A]0
1 + be at
b(1 + b)e at
=
−b
1 + be at
b
�is expression is plotted against t in Fig. ��.�� for various values of b, taking a = 1 s−1 in each case. �e quantity [P]�[A]0 = 1−e−k r t is also plotted,
taking k r = 1 s−1 . As already noted, the autocatalytic curves are sigmoid,
in contrast to the �rst-order curve which is not. �e autocatalytic curves
with larger b, that is a greater initial amount of P relative to the initial
amount of A, reach their maximum value faster than those with smaller
b. �is is because, if less P is present to begin with, the autocatalytic step
is initially slower and the amount of P present builds up more slowly.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
1.0
([P] − [P]0 )�[A]0
�rst-order
b = 0.1
b = 0.03
b = 0.01
b = 0.002
0.5
0.0
0
2
4
6
8
10
t�s
Figure 14.19
(c) �e rate law found in part (a), [P]�[P]0 = (1 + b)e at �(1 + be at ), is rearranged to [P] = (1 + b)e at [P]0 �(1 + be at ) and di�erentiated to give an
expression for the rate.
υ=
d[P] d (1 + b)e at [P]0
=
�
�
dt
dt
1 + be at
(1 + be at ) × a(1 + b)e at [P]0 − (1 + b)e at [P]0 × abe at
(1 + be at )2
a(1 + b)e at [P]0
=
(1 + be at )2
=
�e maximum rate is found by di�erentiating υ with respect to t and
setting the derivative equal to zero
dυ (1 + be at )2 × (a 2 (1 + b)e at [P]0 ) − a(1 + b)e at [P]0 × 2abe at (1 + be at )
=
dt
(1 + be at )4
At the maximum, when dυ�dt = 0, the numerator of this expression is
zero
a 2 (1 + b)(1 + be at )2 e at [P]0 − 2a 2 b(1 + b)(1 + be at )e2at [P]0 = 0
Cancelling of terms followed by rearrangement gives
1 + be at − 2be at = 0
hence
be at = 1
hence
t = −(1�a) ln b
(d) As in part (a), [A] and [P] are written as [A]0 −x and [P]0 +x respectively.
�e rate law is
υ=
d[P]
= k r [A]2 [P] = k r ([A]0 − x)2 ([P]0 + x)
dt
�e expression [P] = [P]0 + x implies that d[P]�dt = dx�dt so the rate
law becomes
dx
= k r ([A]0 − x)2 ([P]0 + x) hence
dt
dx
= k r dt
([A]0 − x)2 ([P]0 + x)
559
560
14 CHEMICAL KINETICS
Integration of this expression, using x = 0 at time t = 0, gives
�
x
0
x
dx
= � k r dt
2
([A]0 − x) ([P]0 + x)
0
�e right-hand side is k r t. �e le�-hand side is evaluated using the method
of partial fractions described in �e chemist’s toolkit �� in Topic ��B. �e
fraction is expressed as a sum
1
A
B
C
=
+
+
2
2
([A]0 − x) ([P] + x) ([A]0 − x)
[A] − x [P]0 + x
where A, B, and C are constants to be found. �is expression is multiplied
through by ([A]0 − x)2 ([P] + x)
1 = A([P]0 + x) + B([A]0 − x)([P]0 + x) + C([A]0 − x)2
�e brackets are expanded and the terms are collected
1 = (C−B)x 2 +(A+B[A]0 −B[P]0 −2C[A]0 )x+(A[P]0 +B[A]0 [P]0 +C[A]20 )
Equating coe�cients gives the three equations
C−B=0
A + B[A]0 − B[P]0 − 2C[A]0 = 0
A[P]0 + B[A]0 [P]0 + C[A]20 = 1
�e �rst equation implies that C = B. Substituting this into the second
two equations gives
A − B([A]0 + [P]0 ) = 0
A[P]0 + B([A]0 [P]0 + [A]20 ) = 1
�e �rst equation of these two equations is multiplied by [P]0 and subtracted from the second to give
B([A]0 [P]0 + [A]20 ) + B([A]0 [P]0 + [P]20 ) = 1
Rearranging gives
B([A]20 + 2[A]0 [P]0 + [P]20 ) = 1
hence
B=C=
1
([A]0 + [P]0 )2
�is is substituted back into the equation A − B([A]0 + [P]0 ) = 0 from
above to give
A−
1
× ([A]0 + [P]0 ) = 0
([A]0 + [P]0 )2
hence
A=
1
[A]0 + [P]0
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�e required integral is therefore
kt = �
=�
=�
=�
=
1
dx
([A]0 − x)2 ([P]0 + x)
x
0
x
0
+
�
1
1
+
([A]0 + [P]0 )([A]0 − x)2 ([A]0 + [P]0 )2 ([A]0 − x)
1
� dx
([A]0 + [P]0 )2 ([P]0 + x)
1
ln([A]0 − x)
ln([P]0 + x)
−
+
�
2
([A]0 + [P]0 )([A]0 − x) ([A]0 + [P]0 )
([A]0 + [P]0 )2 0
x
1
1
[P]0 + x
+
ln
�
([A]0 + [P]0 )([A]0 − x) ([A]0 + [P]0 )2 [A]0 − x
−�
1
1
[P]0
+
ln
�
2
([A]0 + [P]0 )[A]0 ([A]0 + [P]0 )
[A]0
1
1
1
1
[A]0 ([P]0 + x)
�
−
�+
ln
[A]0 + [P]0 [A]0 − x [A]0
([A]0 + [P]0 )2 [P]0 ([A]0 − x)
Substituting x = [P] − [P]0 gives the integrated rate law as
kt =
1
1
1
�
−
�
[A]0 + [P]0 [A]0 + [P]0 − [P] [A]0
1
[A]0 [P]
+
ln
([A] + [P]0 )2 [P]0 ([A]0 + [P]0 − [P])
It is not possible to rearrange this equation to give a simple expression for
[P].
To �nd the time at which the rate reaches a maximum, the expression
for the rate, υ = k r [A]2 [P] = k r ([A]0 − x)2 ([P]0 + x), is di�erentiated
and the derivative is set equal to zero. �e chain rule for di�erentiation
implies that dυ�dt = (dυ�dx) × (dx�dt), hence
dυ
d
dx
�k r ([A]0 − x)2 ([P]0 + x)� ×
=
dt dx
dt
= �−2k r ([A]0 − x)([P]0 + x) + k r ([A]0 − x)2 �
dx
dt
Setting this equal to zero implies that
−2k r ([A]0 − x)([P]0 + x) + k r ([A]0 − x)2 = 0
or
dx
=0
dt
Because x = [P] − [P]0 , dx�dt = d[P]�dt = υ and so the solution dx�dt =
0 corresponds to υ = 0. �is represents a minimum rate rather than a
maximum and so is rejected. Examining the other solution gives
−2k r ([A]0 − x)([P]0 + x) + k r ([A]0 − x)2 = 0
hence
([A]0 − x) [([A]0 − x) − 2([P]0 + x)] = 0
561
562
14 CHEMICAL KINETICS
Hence
x = [A]0
or [A]0 − x = 2([P]0 + x)
Because x = [A]0 − [A], the solution x = [A]0 corresponds to [A] = 0.
From the rate law υ = k r [A]2 [P] this corresponds to v = 0 and therefore to
a minimum rate rather than a maximum. �e maximum rate is therefore
given by the second expression, which is rearranged to yield x = 13 ([A]0 −
2[P]0 ).
�is is then substituted into the integrated rate law from above
kr t =
=
=
1
1
1
1
[A]0 ([P]0 + x)
�
−
�+
ln
[A]0 + [P]0 [A]0 − x [A]0
([A]0 + [P]0 )2 [P]0 ([A]0 − x)
1
1
1
�2
−
�
2
[A]0 + [P]0 3 [A]0 + 3 [P]0 [A]0
+
[A]0 ( 13 [P]0 + 13 [A]0 )
1
ln
([A]0 + [P]0 )2 [P]0 ( 23 [A]0 + 23 [P]0 )
1
3 [A]0 + [P]0
[A]0
� −
+ ln
�
([A]0 + [P]0 )2 2
[A]0
2[P]0
�e time at which the rate is at a maximum is therefore
t=
1
1 [P]0
[A]0
� +
+ ln
�
2
k r ([A]0 + [P]0 ) 2 [A]0
2[P]0
(e) �e rate law is integrated as in part (d). Writing [A] = [A] − x and [P] =
[P]0 + x the rate law is
υ=
d[P]
= k r [A][P]2 = k r ([A]0 − x)([P]0 + x)2
dt
�e expression [P] = [P]0 + x implies that d[P]�dt = dx�dt. �erefore
x
t
dx
dx
= k r ([A]0 −x)([P]0 +x)2 hence �
=
k dt
�
dt
0 ([A]0 − x)([P]0 + x)2
0
�e right-hand side is k r t. �e le�-hand side is evaluated using the method
of partial fractions, as in part (d). �e fraction is expressed as a sum
1
A
B
C
=
+
+
([A]0 − x)([P]0 + x)2 [A]0 − x [P]0 + x ([P]0 + x)2
where A, B, and C are constants to be found. �e expression is multiplied
through by ([A]0 − x)([P]0 + x)2 .
1 = A([P]0 + x)2 + B([A]0 − x)([P]0 + x) + C([P]0 + x)
= (A − B)x 2 + (2A[P]0 + B[A]0 − B[P]0 − C)x
+ (A[P]20 + B[A]0 [P]0 + C[A]0 )
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Equating coe�cients gives
A−B = 0
2A[P]0 + B[A]0 − B[P]0 − C) = 0
A[P]20 + B[A]0 [P]0 + C[A]0 = 1
�e �rst equation implies that A = B. Substituting this into the second
two equations gives
A([A]0 + [P]0 ) − C = 0
A([P]20 + [A]0 [P]0 ) + C[A]0 = 1
�e �rst of these two equations is multiplied by [A]0 and added to the
second equation to give
A([A]20 + 2[A]0 [P]0 + [P]20 ) = 1
hence
A=B=
1
([A]0 + [P]0 )2
�is is substituted back into the equation A([A]0 + [P]0 ) − C = 0 from
above to give
1
× ([A]0 + [P]0 ) − C = 0
([A]0 + [P]0 )2
�e required integral is therefore
kr t = �
=�
=�
x
0
+
�
hence
C=
1
([A]0 + [P]0 )
1
1
+
2
([A]0 + [P]0 ) ([A]0 − x) ([A]0 + [P]0 )2 )([P]0 + x)
1
� dx
([A]0 + [P]0 )([P]0 + x)2
− ln([A]0 − x)
ln([P]0 + x)
1
+
−
�
2
2
([A]0 + [P]0 )
([A]0 + [P]0 )
([A]0 + [P]0 )([P]0 + x) 0
x
1
[P]0 + x
1
ln
−
�
([A]0 + [P]0 )2 [A]0 − x ([A]0 + [P]0 )([P]0 + x)
−�
1
[P]0
1
ln
−
�
2
([A]0 + [P]0 )
[A]0 ([A]0 + [P]0 )[P]0
=
1
[A]0 ([P]0 + x)
1
1
1
ln
+
�
−
�
2
([A]0 + [P]0 )
[P]0 ([A]0 − x) [A]0 + [P]0 [P]0 [P]0 + x
kr t =
1
[A]0 [P]
1
1
1
ln
+
�
−
�
2
([A]0 +[P]0 )
[P]0 ([A]0 +[P]0 −[P]) [A]0 +[P]0 [P]0 [P]
Substituting x = [P] − [P]0 gives the integrated rate law as
As in part (d) it is not possible to rearrange this equation to give a simple
expression for [P].
563
564
14 CHEMICAL KINETICS
�e time at which the rate reaches a maximum is found by the same
method as used in part (d).
dυ dv dx
d
dx
�k r ([A]0 − x)([P]0 + x)2 � ×
=
×
=
dt dx dt dt
dt
dx
2
= �2k r ([A]0 − x)([P]0 + x) − k r ([P]0 + x) � ×
dt
dx
= k r ([P]0 + x)(2[A]0 − [P]0 − 3x)
dt
At the maximum, this expression is equal to zero. �e solution dx�dt = 0
is discarded for the same reason as in part (d), and the solution x = −[P]0
is discarded because x must be positive. �e remaining solution is
2[A]0 − [P]0 − 3x = 0
hence
x = 13 (2[A]0 − [P]0 )
�is is substituted into the integrated rate law from above
kr t =
=
=
1
[A]0 ([P]0 + x)
1
1
1
ln
+
�
−
�
([A]0 + [P]0 )2 [P]0 ([A]0 − x) [A]0 + [P]0 [P]0 [P]0 + x
[A]0 ( 23 [A]0 + 23 [P]0 )
1
ln
([A]0 + [P]0 )2 [P]0 ( 13 [A]0 + 13 [P]0 )
+
1
1
1
�
− 2
�
[A]0 + [P]0 [P]0 3 [A]0 + 23 [P]0
1
2[A]0 [A]0 + [P]0 3
�ln
+
− �
([A]0 + [P]0 )2
[P]0
[P]0
2
Hence the maximum rate is reached at
I��.�
t=
1
2[A]0 [A]0 1
�ln
+
− �
k r ([A]0 + [P]0 )2
[P]0
[P]0 2
(a) Because the second step is rate-determining, the �rst step and its reverse
are treated as a pre-equilibrium because the rate of reaction of A− with AH
to form product is assumed to be too slow to a�ect the maintenance of the
pre-equilibrium. As explained in Section ��E.� on page ��� it follows that
K=
k a [BH+ ][A− ]
=
k a′
[AH][B]
hence
[A− ] =
k a [AH][B]
k a′ [BH+ ]
�e rate formation of product is υ = d[P]�dt = k b [A− ][AH]. �e expression for [A− ] is substituted into this to give
υ = k b [A− ][AH] = k b
k a [AH][B]
k a k b [AH]2 [B]
[AH]
=
k a′ [BH+ ]
k a′ [BH+ ]
�e same result is alternatively derived using the steady-state approximation. Applying the steady-state approximation to A− gives
d[A− ]
= k a [AH][B] − k a′ [BH+ ][A− ] − k b [A− ][AH] = 0
dt
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
�is expression is rearranged
[A− ] =
k a [AH][B]
′
k a [BH+ ] + k b [AH]
Substituting this into the rate of formation of product gives
υ = k b [A− ][AH] = k b
k a [AH][B]
k a k b [AH]2 [B]
[AH]
=
k a′ [BH+ ] + k b [AH]
k a′ [BH+ ] + k b [AH]
Finally it is noted that because the second step is rate-determining the
rate of conversion of A− to products, k b [A− ][AH], is much slower than
the rate of its reversion to reactants, k a′ [A− ][BH+ ].
k a′ [A− ][BH+ ] � k b [A− ][AH] hence
k a′ [BH+ ] � k b [AH]
�e term k b [AH] is therefore neglected in the denominator of the rate
law which then becomes υ = k a k b [AH]2 [B]�k a′ [BH+ ] as before.
(b) Because the second step is rate-determining, the formation of HAH+ from
HA and H+ forms a pre-equilibrium in which the rates of the forward and
backward steps are considered to be equal because the gradual removal of
HAH+ to form products is assumed to be too slow to signi�cantly a�ect
the maintenance of the equilibrium. Equating the rates of the forward
and backward steps gives
k a [HA][H+ ] = k a′ [HAH+ ] hence
[HAH+ ] =
k a [HA][H+ ]
k a′
�e rate of product formation is equal to the rate of the second step
I��.�
v = k b [HAH+ ][B] = k b
k a [HA][H+ ]
ka kb
[B] =
[HA][H+ ][B]
′
ka
k a′
A polymer consisting of N monomer units has a molar mass of N M 1 , where M 1
is the molar mass of a single monomer unit. �e mean molar mass is therefore
�N M 1 � = M 1 �N�, and likewise the mean square molar mass and mean cube
molar mass are
�M 2 � = �(N M 1 )2 � = M 12 �N 2 � and
�M 3 � = �(N M 1 )3 � = M 13 �N 3 �
�e task is therefore to �nd �N 2 � and �N 3 �.
It is supposed that each monomer has one end group A with which it can join
to another monomer. In a polymer, only the terminal monomer unit in the
chain has a free end group.
�e probability PN that a polymer consists of N monomers is equal to the
probability that it has N − 1 reacted end groups and one unreacted end group.
�e fraction of end groups that have reacted is p and the fraction of free end
groups remaining is 1 − p, so the probability that a polymer contains N − 1
reacted groups and one unreacted group is p N−1 × (1 − p).
565
566
14 CHEMICAL KINETICS
It is convenient to begin by evaluating the average value of N.
∞
∞
∞
�N� = � N PN = � N p N−1 (1 − p) = (1 − p) � N p N−1
N=1
N=1
N=1
To evaluate the sum, it is noted that N p N−1 corresponds to the derivative of p N .
Hence
∞
� Np
N−1
N=1
d N
d ∞ N
d
�p + p2 + p3 + ...�
p =
�� p � =
dp
dp
dp
N=1
N=1
∞
=�
�e expression in square brackets is a geometric series with �rst term p and
common ratio p; the sum to in�nity of this series is therefore p�(1 − p). Hence
∞
� Np
N−1
N=1
d
p
(1 − p) + p
1
�
�=
=
2
dp 1 − p
(1 − p)
(1 − p)2
=
�e average value of N is therefore
∞
�N� = (1 − p) � N p N−1 = (1 − p) ×
N=1
1
1
=
2
(1 − p)
1− p
�is is the same result as [��F.��a–���] which is derived in Section ��F.�(a) on
page ��� by a di�erent method. However the approach used here is more easily
generalised to �nd an expression for �N 2 � and �N 3 �.
∞
∞
∞
�N 2 � = � N 2 PN = � N 2 p N−1 (1 − p) = (1 − p) � N 2 p N−1
N=1
N=1
N=1
2 N−1
�e sum ∑∞
is evaluated by noting that N p N−1 is the derivative of
N=1 N p
N
p
∞
�N p
N=1
2 N−1
∞
∞
= � N × N p N−1 = � N ×
N=1
=
∞
N=1
d
�p � N p N−1 �
dp N=1
d N
d ∞
N
p =
� Np
dp
dp N=1
N−1
�e sum ∑∞
was already evaluated above; its value is 1�(1 − p)2 .
N=1 N p
Hence
∞
d
1
1+ p
2 N−1
=
�p ×
�=
�N p
2
dp
(1
−
p)
(1
− p)3
N=1
�e mean value of N 2 is therefore
∞
�N 2 � = (1 − p) � N 2 p N−1 = (1 − p) ×
N=1
1+ p
1+ p
=
(1 − p)3 (1 − p)2
�e mean value of N 3 is evaluated in a similar way, using the result already
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
2 N−1
deduced that ∑∞
= (1 + p)�(1 − p)3 .
N=1 N p
∞
∞
∞
�N 3 � = � N 3 PN = � N 3 p N−1 (1 − p) = (1 − p) � N 2 × N p N−1
N=1
∞
N=1
= (1 − p) � N 2
N=1
= (1 − p)
∞
N=1
d N
d
2 N
p = (1 − p)
�N p
dp
dp N=1
(1+p)�(1−p)3
��� � � � � � � � � � � � ��� � � � � � � � � � � � �
∞
d
d
1+ p
�p � N 2 p N−1 � = (1 − p) �p ×
�
dp N=1
dp
(1 − p)3
= (1 − p) ×
p2 + 4p + 1 p2 + 4p + 1
=
(1 − p)4
(1 − p)4
Using the results �M 2 � = M 12 �N 2 � and �M 3 � = M 13 �N 3 � deduced at the start of
the question, together with the expressions for �N 2 � and �N 3 �, the mean square
and cube molar masses are
�M 2 � =
M 12 (1 + p)
(1 − p)2
(a) �e required ratio is
�M 3 � =
M 13 (p2 + 4p + 1)
(1 − p)3
�M 3 � M 13 (p2 + 4p + 1)�(1 − p)3
M 1 (p2 + 4p + 1)
=
=
�M 2 �
M 12 (1 + p)�(1 − p)2
(1 + p)(1 − p)
(b) �e average number of monomers per polymer, that is the chain length,
is given by [��F.��a–���], �N� = 1�(1 − p). �is expression is rearranged
to p = 1 − 1��N�. �is is then substituted into the expression derived in
(b) to give the ratio in terms of chain length.
�M 3 � M 1 (p2 + 4p + 1)
(1 − 1��N�)2 + 4(1 − 1��N�) + 1
=
=
M
1
�M 2 �
(1 + p)(1 − p)
(1 + [1 − 1��N�])(1 − [1 − 1��N�])
=
M 1 (6�N�2 − 6�N� + 1)
2�N� − 1
567
15
15A
Reaction dynamics
Collision theory
Answers to discussion questions
D��A.�
Reactions between complex molecules might be expected to have strong steric
requirements (small steric factors) as a result of the reaction requiring a particular orientation and approach of the reacting parts of the molecule: the
more complex the molecules, the smaller the fraction of collisions which are
potentially reactive.
In the RRK theory of unimolecular reactions molecular complexity plays a different role in that it governs the distribution of energy in the excited molecule.
In this theory the rate constant for the unimolecular decay of an energized
molecule A* is given by [��A.��–���],
k b (E) = �1 −
E ∗ s−1
� kb
E
Here E ∗ is the minimum energy that must be accumulated in a bond for it to
break, E is the total energy, and s is the number of modes of motion (modelled
as harmonic oscillators) that the molecule possesses. �e term in parentheses is
less than �, therefore the expression implies that the more complex the molecule
(the greater s), the smaller the rate constant becomes.
D��A.�
To the extent that real gases deviate from perfect gas behaviour, they do so
because of intermolecular interactions. Interactions tend to be more important
at high pressures, when the size of the molecules themselves is not negligible
compared to the average intermolecular distance (mean free path). Attractive
interactions might enhance a reaction rate compared to the predictions of collision theory, particularly if the parts of the molecules that are attracted to each
other are the reactive sites. Similarly, repulsive interactions might reduce the
frequency of collisions compared to what would be predicted for perfect gases.
570
15 REACTION DYNAMICS
Solutions to exercises
E��A.�(a)
�e collision theory expression for the rate constant is given in [��A.�–���].
kr = σ NA �
8kT
�
πµ
1�2
e−E a �RT
= (0.36 × 10−18 m2 ) × (6.0221 × 1023 mol−1 )
�
8 × (1.3806 × 10−23 J K−1 ) × (650 K)
�
π(3.32 × 10−27 kg)
× e−(171×10 J mol )�[(8.3145 J K
−1
3
= 1.0 × 10−5 mol−1 m3 s−1
−1
1�2
mol−1 )×(650 K)]
�e units are best resolved by realising that (8kT�πµ)1�2 is a speed, with units
m s−1 . Note that �.�� nm2 is 0.36 × 10−18 m2 .
E��A.�(a) As described in Section ��A.�(b) on page ���, the reactive cross section may be
estimated from the (non-reactive) collision cross sections of A and B: σest =
1�2
1�2
1
(σA + σB )2 . �e steric factor is given by the ratio of the experimental
4
reactive cross section, σexp , to the estimated cross section
P=
σexp
9.2 × 10−22 m2
=
σest [(0.95 × 10−18 m2 )1�2 + (0.65 × 10−18 m2 )1�2 ]2 �4
= 1.2 × 10−3
E��A.�(a) In the RRK theory the rate constant for the unimolecular decay of an energized
molecule A* is given by [��A.��–���],
k b (E) = �1 −
E ∗ s−1
� k b = (1 − x)s−1 k b
E
where x = E ∗ �E. For a non-linear molecule with � atoms there are 3N − 6 =
3 × 5 − 6 = 9 normal modes, so s = 9. �is expression is rearranged for x to give
x = 1 − [k b (E)�k b ]1�(s−1)
= 1 − [3.0 × 10−5 ]1�(9−1) = �.��
E��A.�(a) In the RRK theory the rate constant for the unimolecular decay of an energized
molecule A* is given by [��A.��–���],
k b (E)
E ∗ s−1
= �1 −
�
kb
E
where E ∗ is the minimum energy needed to break the bond, and E is the energy
available from the collision. With the data given
k b (E)
200 kJ mol−1
= �1 −
�
kb
250 kJ mol−1
10−1
= 5.12 × 10−7
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
E��A.�(a) �e collision frequency is given by [�B.��b–��], z = σ υ rel p�kT, where σ is the
collision cross-section, given in terms of the collision diameter d as σ = πd 2 ,
and υ rel is the mean relative speed of the colliding molecules. �is speed is
given by [�B.��b–��], υ rel = (8kT�πµ)1�2 , with µ = m A m B �(m A + m B ). For
collisions between like molecules µ = m�2 and υ rel = (16kT�πm)1�2 .
σ υ rel p πd 2 p 16kT 1�2
π 1�2
=
�
� = 4d 2 p �
�
kT
kT
πm
mkT
= 4 × (380 × 10−12 m)2 × (120 × 103 Pa)
z=
π
�
�
(17.03 × 1.6605 × 10−27 kg) × (1.3806 × 10−23 J K−1 ) × (303 K)
1�2
= 1.12 × 1010 s−1
To con�rm the units of z it is useful to recall that 1 J = 1 kg m2 s−2 and 1 Pa =
1 kg m−1 s−2 .
�e collision density between identical molecules is given by [��A.�b–���]
Z AA = σ �
4kT 1�2 2
� N A [A]2
πm
where [A] is the molar concentration of the gas. In turn, this is expressed in
terms of the pressure using the perfect gas equation to give [A] = n A �V =
p A �RT.
1�2
1�2
p2A
4kT 1�2 N A2 p2A
π
2 πkT
2
�
=
2d
�
�
=
2d
�
�
p2A
πm
R2 T 2
m
k2 T 2
mk 3 T 3
= 2 × (360 × 10−12 m)2 × (120 × 103 Pa)2
Z AA = πd 2 �
π
�
�
(17.03 × 1.6605 × 10−27 kg) × (1.3806 × 10−23 J K−1 )3 × (303 K)3
1�2
= 1.62 × 1035 m−3 s−1
�e above expression shows that z ∝ pT −1�2 , but at constant volume p ∝ T,
therefore the overall temperature dependence is z ∝ T 1�2 . �e percentage
increase in z on increasing T by �� K is therefore
3131�2 − 3031�2
= 0.0163... = �.�%
3031�2
Similarly the �nal expression for the collision density shows Z AA ∝ p2 T −3�2
which, with p ∝ T, gives Z AA ∝ T 2 T −3�2 ∝ T 1�2 . �is is the same dependence
as z, so the same percentage increase will result.
E��A.�(a) �e collision theory expression for the rate constant is given in [��A.�–���].
In this expression, the factor e−E a �RT is identi�ed as the fraction of collisions f
571
572
15 REACTION DYNAMICS
having at least kinetic energy E a along the �ight path. For example with E a =
20 kJ mol−1 and T = 350 K
Ea
20 × 103 J mol−1
=
= 6.87...
RT (8.3145 J K−1 mol−1 ) × (350 K)
f = e−6.87 ... = 1.04 × 10−3
A similar calculation gives f = 0.069 at T = 900 K. With E a = 100 kJ mol−1
the result is f = 1.19 × 10−15 at T = 350 K, and f = 1.57 × 10−6 at T = 900 K.
E��A.�(a) �e method for calculating the fractions is shown in the solution to Exercise
E��A.�(a). For E a = 20 kJ mol−1 and T = 350 K it is found that f = 1.03...×10−3
and increasing the temperature to ��� K gives f = 1.25...×10−3 . �e percentage
increase is
100 ×
(1.25... × 10−3 ) − (1.03... × 10−3 )
= ��%
1.03... × 10−3
A similar calculation gives an increase by �.�% at ��� K. With E a = 100 kJ mol−1
the result is ���% at T = 350 K, and ��% at T = 900 K.
Solutions to problems
P��A.�
�e collision theory expression for the rate constant is given in [��A.�–���]
8kT
kr = σ NA �
�
πµ
∗
1�2
e−E a �RT
Here σ ∗ is interpreted as the reactive cross-section, related to the collision
cross-section σ by σ ∗ = Pσ, where P is the steric factor. Comparison of the
above expression for k r with the Arrhenius equation, k r = Ae−E a �RT , gives
1�2
the frequency factor as A = σ � N A (8kT�πµ) ; this is rearranged to give
∗
an expression for σ . It is convenient to express the given frequency factor
2.4 × 1010 dm3 mol−1 s−1 as 2.4 × 107 m3 mol−1 s−1 . �e mass of a CH� radical
is 15.03 m u , therefore the reduced mass of the collision is µ = 12 × 15.03 m u =
1.24... × 10−26 kg.
σ∗ =
=
A
πµ 1�2
�
�
N A 8kT
2.4 × 107 m3 mol−1 s−1
π(1.24... × 10−26 kg)
�
�
8 × (1.3806 × 10−23 J K−1 ) × (298 K)
6.0221 × 1023 mol−1
1�2
= 4.34... × 10−20 m2 = 0.043 nm2
�e units are best resolved by realising that (8kT�πµ)1�2 is a speed, with units
m s−1 .
To estimate the collision cross-section assume that d is twice the C–H bond
length and compute σ = πd 2 = π(2 × 154 × 10−12 m)2 = 2.98... × 10−19 m2 . �e
steric factor is P = σ ∗ �σ = (4.34... × 10−20 m2 )�(2.98... × 10−19 ) = �.�� .
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P��A.�
�e collision theory expression for the rate constant is given in [��A.�–���]
kr = σ NA �
8kT
�
πµ
1�2
e−E a �RT
�e maximum value for the rate constant is when E a = 0. �e collision cross
section is taken as σ = πd 2 = π(308 × 10−12 m)2 = 2.98... × 10−19 m2 . �e
mass of a CH� radical is 15.03 m u , therefore the reduced mass of the collision
is µ = 12 × 15.03 m u = 1.24... × 10−26 kg
kr = σ NA �
�
8kT
�
πµ
1�2
= (2.98... × 10−19 m2 ) × (6.0221 × 1023 mol−1 )
8 × (1.3806 × 10−23 J K−1 ) × (298 K)
�
π(1.24... × 10−26 kg)
1�2
= 1.64 × 108 mol−1 m3 s−1
�e units are best resolved by realising that (8kT�πµ)1�2 is a speed, with units
m s−1 .
For a second-order reaction the integrated rate law is [��B.�b–���], 1�[CH3 ] −
1�[CH3 ]0 = k r t. Suppose that initially an amount in moles n 0 of C� H� is introduced into the vessel, and that a fraction α dissociates. �e amount of C� H�
remaining is n 0 (1 − α) and the amount of CH� produced is 2n 0 α. �e total
amount of gas is n 0 (1+α), therefore the mole fraction of CH� is 2α�(1+α) and
hence the partial pressure of CH� is 2α p tot �(1 + α). �e molar concentration
corresponding to this pressure is found using the perfect gas law as
[CH3 ] =
n CH3 p CH3
2α p tot
=
=
V
RT
RT(1 + α)
With the data given this evaluates as
[CH3 ] =
2 × 0.1 × (100 × 103 Pa)
= 7.33... mol m−3
(8.3145 J K−1 mol−1 )×(298 K)×(1 + 0.1)
1
If recombination proceeds to ��%, the amount of CH� remaining is 10
of the
initial. �e time for this to take place is found by solving
Hence
9
9
=
[CH3 ]0 k r (7.33... mol m−3 ) × (1.64 × 108 mol−1 m3 s−1 )
= �.� ns
t=
P��A.�
10
1
−
= kr t
[CH3 ]0 [CH3 ]0
�e collision theory expression for the rate constant, including the steric factor
P, is given in [��A.��–���]
k r = Pσ N A �
8kT
�
πµ
1�2
e−E a �RT
573
574
15 REACTION DYNAMICS
As described in Section ��A.�(b) on page ���, the collision cross-section between A and B may be estimated from the collision cross sections of A and
1�2
1�2
B: σ = 14 (σA + σB )2 . From the Resource section the cross section for O� is
�.�� nm2 . No values are given for the ethyl and cyclohexyl radicals, so these will
be approximated by the values for ethene (�.�� nm2 ) and benzene (�.�� nm2 ),
respectively. �e reactive cross sections are therefore
σethyl = 14 [(0.40)1�2 + (0.64)1�2 ]2 = 0.512... nm2
A similar calculation gives σhexyl = 0.616... nm2
�e mass of O� is 32.00 m u , that of the C� H� radical is 29.06 m u , and that of
the C� H�� radical is 83.15 m u . �e reduced mass of the O� –C� H� collision is
µ=
m O2 m C2 H5
32.00 × 29.06
=
× (1.6605 × 10−27 kg) = 2.52... × 10−26 kg
m O2 + m C2 H5 32.00 + 29.06
For the O� –C� H�� collision the reduced mass is 3.83... × 10−26 kg.
Taking the activation energy as E a = 0, the steric factor is given by
P=
kr
πµ 1�2
�
�
σ N A 8kT
For this calculation it is convenient to express the rate constants in units of
m3 mol−1 s−1 . For the reaction with C� H�
P=
4.7 × 106 m3 mol−1 s−1
(0.512... × 10−18 m2 ) × (6.0221 × 1023 mol−1 )
�
π(2.52... × 10−26 kg)
�
8 × (1.3806 × 10−23 J K−1 ) × (298 K)
1�2
= �.���
A similar calculation for the reaction with C� H�� gives P = 0.043 .
15B Diffusion-controlled reactions
Answers to discussion questions
D��B.�
In the cage e�ect, a pair of molecules may be held in close proximity for an
extended period of time by the presence of other neighbouring molecules, typically solvent molecules. Such a pair is called an encounter pair, and their
time near each other is called an ‘encounter’ as opposed to a simple collision.
An encounter may include a series of collisions. Furthermore, as a result of
collisions with neighbouring molecules, an encounter pair may pick up enough
energy to react, even though the pair may not have had enough energy when
�rst formed.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
Solutions to exercises
E��B.�(a)
For a di�usion-controlled reaction the rate constant is approximated by [��B.�–
���], k d = 8RT�3η, where η is the viscosity. Recall that 1 P = 10−1 kg m−1 s−1 ,
so that 1 cP = 10−3 kg m−1 s−1 . �erefore the rate constant is
kd =
8 × (8.3145 J K−1 mol−1 ) × (320 K)
3 × (0.89 × 10−3 kg m−1 s−1 )
= 7.97... × 106 m3 mol−1 s−1 = 8.0 × 106 m3 mol−1 s−1
�e half-life of a second-order reaction is given by [��B.�–���], t 1�2 = 1�k r [A]0 .
�e initial concentration is �.� mmol dm−3 which is �.� mol m−3 . With the data
given
E��B.�(a)
t 1�2 =
1
= �� ns
(7.97... × 106 m3 mol−1 s−1 ) × (1.5 mol m−3 )
�e second-order rate constant for a di�usion-controlled reaction is given by
[��B.�–���], k d = 4πR∗ DN A , where R∗ is the critical distance and D is the
di�usion constant. As explained in the text D is the sum of the di�usion constants of the two species. �e value of D is estimated using the Stokes–Einstein
equation, D = kT�6πηR, and with the data given separate values of D are
computed for the two species. �e critical distance is taken as R ∗ = R A + R B .
k d = 4π(R A + R B )(D A + D B )N A
kT
1
1
= 4πN A (R A + R B )
�
+
�
6πη R A R B
= 4π × (6.0221 × 1023 mol−1 ) × (655 + 1820)
×
(1.3806 × 10−23 J K−1 ) × (313 K) 1
1
�
+
�
−1 −1
−3
655
1820
6π × (2.93 × 10 kg m s )
= 3.04... × 106 m3 mol−1 s−1
�e initial concentrations are [A] = 0.170 mol dm−3 = 0.170 × 103 mol m−3
and [B] = 0.350 mol dm−3 = 0.350 × 103 mol m−3 . �e initial rate is therefore
d[P]
= k d [A][B]
dt
= (3.04... × 106 m3 mol−1 s−1 )
× (0.170 × 103 mol m−3 ) × (0.350 × 103 mol m−3 )
= 1.81 × 1011 mol m−3 s−1
Using [��B.�–���], k d = 8RT�3η, the rate constant is
kd =
8 × (8.3145 J K−1 mol−1 ) × (313 K)
= 2.37 × 106 m3 mol−1 s−1
3 × (2.93 × 10−3 kg m−1 s−1 )
�is value would result in a signi�cantly slower initial rate, casting doubt therefore on the validity of the approximations used.
575
576
15 REACTION DYNAMICS
E��B.�(a)
�e second-order rate constant for a di�usion-controlled reaction is given by
[��B.�–���], k d = 4πR∗ DN A , where R∗ is the critical distance and D is the
di�usion constant. As explained in the text, D is the sum of the di�usion
constants of the two species, therefore in this case D is twice the value given.
With the data given
k d = 4π × (0.5 × 10−9 m) × (2 × 6 × 10−9 m2 s−1 ) × (6.0221 × 1023 mol−1 )
= 4.5 × 107 m3 mol−1 s−1
E��B.�(a) For a di�usion-controlled reaction the rate constant is approximated by [��B.�–
���], k d = 8RT�3η, where η is the viscosity.
(i) For water
kd =
8 × (8.3145 J K−1 mol−1 ) × (298 K)
= 6.61 × 106 m3 mol−1 s−1
3 × (1.00 × 10−3 kg m−1 s−1 )
In sorting out the units it is useful to recall 1 J = 1 kg m2 s−2 .
(ii) For pentane
kd =
8 × (8.3145 J K−1 mol−1 ) × (298 K)
= 3.0 × 107 m3 mol−1 s−1
3 × (2.2 × 10−4 kg m−1 s−1 )
Solutions to problems
P��B.�
∗
To simplify the notation the dependence of [J] and [J] on x and t will not
∗
be written explicitly. �e proposed solution, [��B.�–���], [J] = [J]e−k r t , is
substituted into the right-hand side of [��B.�–���]
D
∗
∂ 2 [J]
∂2
∗
− k r [J] = D 2 [J]e−k r t − k r [J]e−k r t
2
∂x
∂x
∂ 2 [J] −k r t
=D
e
− k r [J]e−k r t
∂x 2
�e solution is now substituted into the le�-hand side of [��B.�–���]
∗
∂[J]
∂
∂[J] −k r t
= [J]e−k r t =
e
− k r e−k r t [J]
∂t
∂t
∂t
�e le�-and right-hand sides are now set equal
D
∂ 2 [J] −k r t
∂[J] −k r t
e
− k r [J]e−k r t =
e
− k r e−k r t [J]
2
∂x
∂t
�e term k r e−k r t [J] cancels to give
D
∂ 2 [J] −k r t ∂[J] −k r t
e
=
e
∂x 2
∂t
hence
D
∂ 2 [J] ∂[J]
=
∂x 2
∂t
As speci�ed in the problem, [J] is a solution of [��B.�–���] when k r = 0, and
indeed this is precisely the di�erential equation which has just been generated.
SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY
P��B.�
∗
It is �rst convenient to compute the derivative of [J] with respect to t and its
second derivative with respect to x.
∗
t
∂[J]
∂
=
�k r � [J]e−k r t dt + [J]e−k r t �
∂t
∂t
0
∂[J] −k r t
∂[J] −k r t
= k r [J]e−k r t +
e
− k r e−k r t [J] =
e
∂t
∂t
(��.�)
∗
t
∂ 2 [J]
∂2
= 2 �k r � [J]e−k r t dt + [J]e−k r t �
2
∂x
∂x
0
t ∂ 2 [J]
∂ 2 [J] −k r t
−k r t
e
dt
+
e
= kr �
∂x 2
∂x 2
0
(��.�)
Recall that [J] is a solution to [��B.�–���] with k r = 0
∂[J]
∂ 2 [J]
=D
∂t
∂x 2
�is is used to substitute for ∂ 2 [J]�∂x 2 in eqn ��.�
D
∗
t ∂[J]
∂ 2 [J]
∂[J] −k r t
= kr �
e−k r t dt +
e
2
∂x
∂t
∂t
0
where the factor of D has been taken over to the le�. Next use of made of the
∗
result in eqn ��.�, ∂[J] �∂t = (e−k r t )∂[J]�∂t to rewrite the last expression as
D
∗
∗
∗
t ∂[J]
∂ 2 [J]
∂[J]
= kr �
dt +
2
∂x
∂t
∂t
0
∗
∗
= k r �[J] (t) − [J] (0)� +
∗
= k r [J] (t) +
∗
∂[J]
∂t
∗
∂[J]
∂t
∗
where the initial condition that [J], and hence [J] , must be zero at t = 0 is
used. Rearranging the �nal equation gives the required di�erential equation,
∗
thus demonstrating that the proposed form of [J] is indeed a solution.
∗
∗
∂[J]
∂ 2 [J]
∗
=D
− k r [J]
∂t
∂x 2
15C Transition-state theory
Answers to discussion questions
D��C.�
D��C.�
�is is described in Section ��C.�(b) on page ���. If the solvent were altered
to one with a lower dielectric constant the interaction between the ions would
be greater and this would be manifested in an increased value of A, and hence
steeper slopes for the plots of the rate constant against ionic strength.
�e discarded mode would be the anti-symmetric stretch in which the B–C
distance lengthens and the A–B distance decreases.
577
578
15 REACTION DYNAMICS
Solutions to exercises
E��C.�(a)
�e variation of the rate constant with ionic strength is given by [��C.��–���],
lg k r = lg k r○ + 2Az A z B I 1�2 ; at ��� K and for aqueous solutions A = 0.509. In
the absence of further information assume z A = +1 and z A = −1. Rearranging
for lg k r○ gives
lg k r○ = lg k r − 2Az A z B I 1�2
= lg(12.2 dm6 mol−2 min−1 ) − 2