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Student Solutions Manual
to accompany
Physical Chemistry
Fifth Edition
Ira N. Levine
Chemistry Department
Brooklyn College
City University of New York
Brooklyn, New York
.
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McGraw-Hill Higher Education $2
A Division of The McGraw-Hill Companies
Student Solutions Manual to accompany
PHYSICAL CHEMISTRY, FIFTH EDITION
IRA N. LEVINE
Published by McGraw-Hill Higher Education, an imprint of The McGraw-Hill Companies, Inc.,
1221 Avenue of the Americas, New York, NY 10020. Copyright © The McGraw-Hill Companies,
Inc., 2002, 1995, 1988, 1983, 1978. All rights reserved.
No part of this publication may be reproduced or distributed in any form or by any means, or stored
in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies,
Inc., including, but not limited to, network or other electronic storage or transmission, or broadcast
for distance learning.
This book is printed on acid-free paper.
1234567890
BKM BKM 0321
ISBN 0-07-239360-2
www.mhhe.com
To the Student
The purpose of this solutions manual is to help you learn physical chemistry.
This purpose will be defeated if you use this manual to avoid working homework
problems. You cannot learn how to play the guitar solely by reading books titled
“How to Play the Guitar” or by watching other people play the guitar. Rather, most of
your time is best spent actually practicing the guitar. Likewise, you won’t learn how to
solve physical chemistry problems solely by reading the solutions in this manual.
Rather, most of your time is best spent actually working problems.
Do not look up the solution to a problem until you have made a substantial
effort to work the problem on your own. When you work a problem, you learn a lot
more than when you only read the solution. You can learn a lot by working on a
problem even if you don’t succeed in solving it. True learning requires active
participation on your part. After you have looked up the solution to a problem you
could not solve, close the solutions manual and work through the problem on your
own.
Use the solutions manual as an incentive to work problems, not as a way to
avoid working problems.
Ira N. Levine
INLevine @brooklyn.cuny.edu
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Before using this manual, please read the preceding note To the Student.
Chapter |
1.1
(a)
F.
(b)T.
(c)T.
(@)F.
(e)F. (A mixture of ice and liquid water has
one substance.)
(c) open; nonisolated.
1.2
(a)
Closed, nonisolated;
1.3
(a)
Three.
1.4
So that pressure or composition differences between systems A and B won't
cause changes in the properties of A and B. Such changes can then result only
from a temperature difference between A and B.
15
(a)
(b) Three; solid AgBr; solid AgCl, and the solution.
g
19.3
(b) open, nonisolated;
Ikg
(100 cm)’
cm? 1000g_
= 19300
(1m)’
troy oz 480 grains
lpound
kg/m’
e/
453.59¢g
m = pV= (19.3 g/cm?)(10° cm’) = 1.93 x 10’ g
(1.93 x 10’)($9.65/g) = $1.86 x 10°
1.6
(a)
ee
(b) Pe
(cha
(dale
1.7
(a)
32.0.
1.8
100 g of solution contains 12.0 g of HCI and 88.0 g of water.
(b) 32.0 amu.
(c) 32.0.
(12.0 g)(1 mol/36.46 g) = 0.329 mol;
4.885 mol.
LS
(a)
(d) 32.0 g/mol.
ny 9= (88.0 g)(1 mol/18.015 g) =
xXyc) = 0.329/(0.329 + 4.885) = 0.0631;
12.0 g/mol
ee
ara
6.022 x10*°
atoms/mol
=
nyo =
1.99x10
293
~ g/atom
Xy,9 =1- Xa = 0.9369.
1.10
18.0 g/mol
(b)
<=
6.022 x 10°
molecules/mol
(a)
T.
(c) F. (d) T.
(b) T.
—______ = 2.99x 10" g/molecule
(e) F.
( T.
(a) (5.5 m*)(100 cm)*/(1 m)? = 5.5 x 10° cm*. (b) (1.0 x 10° Pa)(1 bar)/
(10° Pa) = 1.0 x 10‘ bar. (c) (1.0 x 10° Pa)(1 atm)/(101325 Pa) = 0.99 x 10°
atm.
(d)
6
3
=
pm econ omeA stake
cm a!O9e
elim
The system pressure is less than the barometric pressure by 304.3 torr —
202.1 torr = 102.2 torr. Hence Psystem = 754.6 torr — 102.2 torr = 652.4 torr.
113
(a)
P=Pye Shug =Pu,08Mu,0»
3
SO Pughig =PH,0/n,0.
joan
Ce sie
OOS),
Me) ACS
y
0.997 g/cm
3 Ot
where the vapor pressure of water was neglected.
(b)
Use of P = pgh and Eq. (2.8) gives P as
(13.53 g/cem*)(978 cm/s*)(30.0 x 2.54 cm) ——Latm____ =
1013250 dyn/cm?
0.995 atm
1.14
For m constant, n is constant, so (1.18) becomes PV/T = nR = const,
which is (1.17).
1.15
(a)
n= (24.0 g)(1 mol/44.0 g) = 0.545 mol.
P=nRT/V=
(0.545 mol)(82.06 cm?-atm/mol-K)(273. 1 K)/(5000 cm’) = 2.44 atm.
(b)
V=nRT/P = (1 mol)(82.06 cm?-atm/mol-K)(298 K)/(1 atm) =
24500 cm’. One ft® = (12 in.) = (12 x 2.54 cm) = 28300 cm’.
Percent error = [(28300 — 24500)/24500] x 100% = 16%
1.16
Use of P;V\/T, = P2V2/T2 gives P2 = (V\/V2)(T2/T))P; = (V;/2V;)(3T;/T))Pi =
1-5P; = 1.5(0.600 bar) = 0.900 bar:
1.17
P=nRT/V =mRT/MV, so M = mRT/PV and
ie (0.0200 g)(82.06 cm*-atm/mol- K)(298.1 K)
(24.7/760) atm (500 cm’)
= 30.1 g/mol
The only hydrocarbon with molecular weight 30 is C2He.
1.18
Atthis T and P, N; is a gas that behaves nearly ideally. From PV = nRT =
(m/M)RT, we get m/V = PM/RT, so
_ PM _ [(725/760) atm](28.01g/mol) _
~ RT
(82.06cm?-atm/mol-K)(293 K) —
0.00111 g/cm*
since 0.967 bar = 0.967(750 torr) = 725 torr.
ISS
BR
82.025
1.0000
81.948
3.0000
81.880
5.0000
cm°-atm/mol-K
atm
Plotting these data and extrapolating to P = 0, we find limp_,0(PV/nT) =
82.06 cm*-atm/mol-K.
(PV/nT)/(cm’-atm/mol-K)
82.10 + -
1.20
The P/p values are 715.3, 706.2, and 697.1 cm’ atm/g. A plot of P/p vs. Pisa
straight line with intercept 721.. cm” atm/g. We have PV = mRT/M, so
M = RT/(P/p), and
M _ (82.06 cm*-atm/mol- K)(273.15 K) =31.0, g/mol
721., cm* atm/g
The only amine with molecular weight 31 is CH3NH)p.
(P/p)/(cm’-atm/g)
725
720
TMS
710
705
P/atm
1.21.
Use of mor = PV/RT gives nyo
(4.85x10° Pa)[1600(10-* m)*} _
1.867 mol
(8.314 m° - Pa/mol - K)(500 K)
The reaction is 2NH;3 = N2 + 3H>. Let x moles of Nz be formed. The numbers
of moles of NH3, N2, and Hp present at equilibrium are 1.60 — 2x, x, and 3x,
respectively. Thus mo/mol = 1.60 — 2x + x + 3x = 1.867, and x = 0.133. Then
n(N2) = 0.133 mol, n(H2) = 0.409 mol, n(NH3) = 1.33 mol
1.22
Boyle’s law and Charles’ law apply under different conditions (constant T, m
vs. constant P, m); such equations cannot be combined.
1.23.
Consider the processes
(Pi, Vi, Ti, m)) ——~
(Pi, Vand
1, ms) +
(Po-V2-1o512)
For step (a), P and T are constant, so V\/m, = V,/mp. For step (b), m is constant,
so P\V,/T| = P2V2/T>. Substitution for V, in this last equation gives P2V/T> =
P,V\m2/T\m, or P2V2/m2T2 = P\V\/myT}.
1.24
P)=xP.
Noo, = (30.0 g)(1 mol/44.0 g) = 0.682 mol.
Xco, = 0.682/(0.682 + 0.625) = 0.522.
4
No, = 9.625 mol.
Poo, = 9.522(3450 kPa) = 1800 kPa.
1.25
(a)
At constant temperature, P2 = P;V\/V2 for each gas. Therefore
_ (20.0 kPa)(3.00 L)
Hy
4.00L
Py, =15.0 kPa
_ (10.0 kPa)(1.00 L)
CH,
4.00L
Poy 2:0 KPa
Prot = 15.0 kPa + 2.5 kPa = 17.5 kPa
(b)
P; = n,RT/V and Prot = MoT/V. Hence Pi/ Prot = ni/Nior = XiWe get xy, = 15.0 kPa/17.5 kPa = 0.857 and Xcu, = 2-5 kPa/17.5 kPa =
0.143.
1.26
P(O2) = 751 torr — 21 torr = 730 torr. The equation P\V,/T = P2V2/T2 gives
V> =
Y=
1.27
V,P\T2/P2T
and
(36.5 cm?*)(730 torr)(273 K)
(760 torr)(296 K)
=32.3cm>
ae
When a steady state is reached, the pressures in the two bulbs are equal. From
P, = Po, we get m)RTj/V\= n2RT2/V2. Since V; = V2, we have n,T) = 272. Thus
n,(200 K) = (1.00 mol — n;)(300 K); solving, we get n; = 0.60 mole in the
200-K bulb and n> = 0.40 mole in the 300-K bulb.
1.28
1.29
We have PV = nRT = NRT/Na, So N/V = PNa/RT and
DS tyeaP Gk AOE
N
-]
(6.02x107? 23 mol!)P
V
(82.06 cm?-atm/mol-K)(298 K)
(a)
For P = | atm, we get N/V= 2.5 x ae cm”;
(b)
for P = (1/760)10~ atm, we get N/V= 3.2 x 10.°em.:
(c)
for P = (1/760)10"'' atm, N/V = 3.2 x 10° cm™.
atm
Substitution in PV = niokT gives Mot = 0.01456 mol. Also
Mot = M, + M2 = nM,
+ noM>
0.1480 g = nye(4.003 g/mol) + (0.01456 mol — nye)(20.18 g/mol).
Nye = 0.00902 mol, ne = 9.00554 mol
Xte = 0.00902/0.01456 = 0.619, mute = 0.0361 g
1.30
the
The downward force of the atmosphere on the earth’s surface equals
,
weight W of the atmosphere, so P = W/A = mg/A and m = AP/g = 4nr’Ple
5
where ris the earth’s radius and P = 1 atm = 101325 N/m”. Thus
oa 47(6.37
(6.37 10°
ues 2 Nina) Z eens te
X10° m) m)?(1.
9.807 m/s
1.31
(a)
Multiplication of both sides of the equation by 10° bar gives
P = 6.4 x 10° bar.
(b)
1.32
460K.
(c) 1.2x 10° bar. (d) 312K.
Take one liter of gas. This volume has m = 1.185
g= my, +mMo,.
We have not = PV/RT =
(1.000 atm)(1000 cm?*)/(82.06 cm>-atm/mol-K)(298.1 K) = 0.04087 mol.
Not = Ny, +No, =My,/Myn, +Mo,/Mo, =
my, /(28.01 g/mol) + (1.185 g—- my, /(32.00 g/mol) = 0.0487 mol.
Solving, we get my, = 0.862 g, hence mo, = 0.323 g. Then ny, =
0.0308 mol and Bis 0.0101 mol; x9, = 0.0101/0.0409 = 0.247.
1°33
(a)>
Use of P=
P gives Py, = 0.78(1.00 atm) = 0.78 atm,
Po, =(0.21 atm,
(b) V=3000ft*.
Pa,=0.0093 atm,
Foo, = 0.0003 atm.
1 ft=12in. = 12 x 2.54 cm = 30.48 cm.
V = (3000 ft*®)(30.48 cm)*/ft® = 8.5 x 10’ cm®. mot = PV/RT =
[(740/760) atm](8.5 x 10’ cm?)/[(82.06 cm°-atm/mol-K)(293 K)] =
3.44x 10° mol.
ny, = Xn, Mor = 0.78(3.44 x 10° mol) = 2.63 x 10° mol.
my, = (2680 mol)(28.0 g/mol) = 75 kg.
Similarly, mo, = 23 kg,
Mar = 1.5 ke, Mco, = 4.5 x 1; g. We have P = mto/V=
(99.; kg)/(8.5 x 10’ cm*) = 0.00117 g/cm’.
1.34
f(x) is zero at the two points where fis a local minimum and where fis a
local maximum.
maximum
135
f(x) is negative for the portion of the curve between the
and the minimum.
dy/dx= 2x +1 Atx
=) the slope 18 2¢1) aie
1.36
(a) 6x2e* — 6x3e**; (b) — 30xe™2" ; (c) I/x (not 1/2x); (d) 1/(1 — x);
(e) I(x + 1) —x/(x + 1)? = W(x +1); (f) 2e 7/1 — &™); (g) 6 sin 3x cos 3x.
1.37
(a)
(b)
y=2/(1—x) and dy/dx = 2(1 — x)’.
d(xe*)/dx = 2xe* + 3x°e™;
Vd? = 2e* + 6xe™* + 6xe™* + 9x76 = 2e* + 12xe* + 9x7e™*.
B(x
(c)
Reminder.
1.38
(a)
dy = (10x —3 — 2/x’) dx.
Work the problems before looking up their solutions.
x
x
0.01
0:955m
0.1
0.794
0.00001
0.0001
0.001
¢0.993.1495.0:999 Ise 0.99988
This indicates (but does not prove) that the limit is 1.
(by
(xy
TOO
271?
OG
2720
1 Oy eel Oe
nO
oleae tote 2 1o27)
Hie:
2.1 L826
This suggests that the limit is e = 2.7182818....
1.39
(a)
Results on acalculator with 8-digit display and 11 internal digits are:
Ay/Ax = 277, 223.4, 218.88, 218.44, 218.398, 218.393, 218.4 for Ax =
LOR elOgu Og al Ose 10°, 10°, 10°’, respectively. The best estimate is
pac
Sie\ey
(b)
dy/dx = 2xe* and at x = 2, dy/dx = 218.3926.
A BASIC program for part (a) is
5 CX =0.1
10 FORN=1T07
20 X=2
30 CY = EXP((X + CX)*2) - EXP(X*2)_
40 R=CY/CX
1.40
(a) axcos(axy);
(b) —2byz sin (by’z);
(d) 0; (e) ae? /y(e™ + he
50 PRINT “DELTAX=";CX;
“ RATIO=";R
60 CX =CX/10
70 NEXT N
80 END
(c) -(x“ly’)e,
1.41
(a) nR/P;
(b) -2P/nRT?.
1.42
Equation (1.30) gives dz = 2axy’ dx + 3ax°y? dy.
1.43
Partial differentiation of z = x°/y° gives
U5x OE eee
ee
Oep ~3x40
dee a
az _9(_3x?)__ 15x!
aan
hal
kl
Spaaniyie
ne
Sales
dz toes 7
y* ’ dydx
15a
dz
y*
dx dy
ody y?
P is a function of n, T, and V, so dP = (0P/dn)ry dn + (OP/0T)y,n aT +
1.44
(OP/0V)rn AV. Partial differentiation of P = nRT/V gives (dP/dn)r,v =
RT/V = P/n (where PV = nRT was used), (OP/0T)y,n = nR/V = P/T, and
(OP/0V)rn = _nRT/V’ = —P/V. Substitution into the above equation for dP
gives the desired result. (Note that from P = nRT/V, we have
InP=Inn+InR+1nT—In V, from which dln P=dInn+dInT—d\n V follows at once.)
(b)
Approximating small changes by infinitesimal changes, we have
dn=An=0,
dt=AT=1.00K,
dV=AV=50cm’.
The original pressure is P = nRT/V = 0.8206 atm. Then AP = dP =
(0.8206 atm)[0 + (1.00 K)/(300 K) — (50 cm*)/(30000 cm*)] =
0.00137 atm.
(c)
The accurate final pressure is (1.0000 mol)(82.06 cm>-atm/mol-K) x
(301.00 K)/(30050 cm’) = 0.82197 atm. The accurate AP is
0.82197 atm — 0.8206 atm = 0.00137 atm.
1.45
1.000 bar = 750 torr = (750 torr)(1 atm/760 torr) = 0.987 atm.
Vm = V/n = (nRT/P)/n = RT/P =
(82.06 cm*-atm/mol-K)(293.1 K)/(0.987 atm) = 2.44 x 10* cm*/mol.
1.46
(a)
Division by n gives (P + a/V,~. (Vm — b) = RT.
(b)
The units of b are the same as those of Vm, namely, cm?/mol.
P and a/V,. have the same units, so the units of a are bar - cm®/mol’.
1.47
O = (1/Vn)(OVn/OT)p = (1/Vm)(c2 + 2¢3T— csP), where Vm is given by (1.40).
K = — (1/Vn)(OVn/OP)r = —A/Vin)(—Ca — C57) =
(c4 t+ CsT)/(C) + C2T + Gi — c4P —csPT).
1.48
(a)
p = m/V = (m/n)I(V/n) = M/Vm, 80 Vm = M/p =
(18.0153 g/mol)/(0.98804 g/cm*) = 18.233 cm*/mol.
(b)
K = —(1/Vin)(OVn/OP)r and dVin/
Vm = —K AP at constant T. Integration
gives In (Vin2/Vin1)= —K(P2 os Prat constant T.
K = (4.4 x 107! Pa ')(101325 Pa/1 atm) = 4.45 x 10° atm”! and
In[Vm2/(18.233 cm?/mol)] = -(4.46 x 10° atm™')(100 atm — 1 atm) =
0.0044, so Vm2/(18.233 cm*/mol) = e°° = 0.9956 and Vinz =
18.15 cm?/mol.
1.49
(a)
At constant P, the equation PVm = RT gives Vm = aT, where a = R/P is a
positive constant. The isobars on a Vp, vs. T diagram are straight lines
that start at the origin and have positive slopes. (As P increases, the slope
decreases.)
(b)
For Vm constant, PVm = RT gives P = bT, where b = R/V» is constant. The
isochores on a P vs. T diagram are straight lines that start at the origin
and have positive slopes.
Partial differentiation of
1.50
V=nRT(1 + aP)/P gives (OV/0T)p»=
nR(1 + aP)/P. The equation of state gives nR(1 + aP) = PV/T, so
(OV/OT)pn = V/T. Then @ = (1/V)(OV/0T)p.» = IT.
Partial differentiation of V = nRT(1/P + a) gives (OV/0P)rn =
_nRTIP? = -[PV/(1 + aP)\/P° =-V/P(1 + aP), where the equation of state
was used. Then k = —(1/V)(OV/0P)7 = 1/P(. + aP).
(b)
Solving the equation of state for P, we get P = nRT/(V— ank7T), partial
differentiation gives (0P/07T)y = nR/(V — anRT) + an’R°TI(V — anRTY =
PIT + aP’/T, where P = nRT/(V— anRT) was used. From (a), we have O/k
= P(1 + aP\/T = PIT + aP”/T, which agrees with Eq. (1.45).
1.51
For small AT, we have
Folate
Wop VAL),
Since @ is an intensive property, we can take any quantity of water. For ! g,
the equation V = m/p gives V = 1.002965 cm? at 25°C, | atm and
V = 1.003227 cm’ at 26°C, 1 atm.
1
1.003227 cm? -1.002965cm* _ 9 gqgn6 K-'
Hence a =
26°C -25°C
1.003 cm*
Similarly, k = -(1/V)(OV/0P)7 = —(/V)(AV/AP)r. At 25°C and 2 atm,
we calculate V = 1.002916 cm’ for 1 g of water.
3
3
1
: 1.002916cm* —1.002965cm™ _ 4 9419-5 atm7!
Thus
k=-—
1.003 cm”
2 atm —latm
_a_ ae 26x10*K
ae
Eq. (1.45) gives (22)
ane
—4
K-49
1.52
(a)
<1 0.
=i
5.3 atm/K
atm 4
dD pak Val Vei= 0d Toei Vaca, Coe
= (LIV=)(OVicl
In(Vin2/Vm1) = 0(T2 — T;) at constant P, where the T dependence of & was
neglected over the short range of T.
In(Vm2/18.2334 cm* mol') co
(4.576 x 107/K)(2.00 K). In(Vm2/em? mol") = In 18.2334 + 0.0009152
and V2 = 18.2501 cm?/mol.
(b)
K=—-(1/Vn)(OVp/OP)r; dVn/Vm =—K dP; In(Vm2/ Vii) = —K(P2 — P)) at
constant 7, where the P dependence of k was neglected.
In(Vmn2/ 18.2334 cm? mol!) = -(44.17 x 10° bar™')(199 bar) and Vino =
18.074 cm*/mol.
153
(a)
Drawing the tangent line to the 500-bar Vp-vs.-T curve at 100°C, one
finds its slope to be (20.8 cm*/mol — 17 cm*/mol)/(300°C — 0°C) = 0.013
cm?/mol-K = (Vm/OT)p at this T and P. The figure gives Vm = 18.2
cm’/mol at 500 bar and 100°C, so a= (1/V,3)(OV,,/0T) p=
(0.013 cm*/mol-K)/(18.2 cm*/mol) = 0.0007 K™'.
(b)
Drawing the tangent line to the 300°C V,,-vs.-P curve at 2000 bar, one
finds its slope to be -0.0011 cm*/mol-bar. The figure gives Vn =
20.5; cm*/mol at this T and P, so K = —-(1/Vm)(OVm/OP)r =
(0.0011 cm?/mol-bar)/(20.5 cm*/mol) = 5 x 10 bar”.
1.54
Equation (1.45) gives o/« = (dP/0T)y = (AP/AT)y, so
a
aS
ie
AP ~— AT =—_——_—__
(6K) =22atm; P=23atm
K
4.7x10 ”° atm
(SR)
(a)
As P increases, the molecules are forced closer together; the decrease 1n
empty space between the molecules makes it harder to compress the
substance, and k is smaller.
(b)
Most substances expand as T increases. The increased space between
molecules makes it easier to compress the substance, and x increases.
1.56
K = -(1/V)(OV/0P)7 = —A/V)(AV/AP)r and AP = —AV/Vx at constant T.
(a)
For a 1% volume decrease, AV = —0.01V and we have AP = 0.01 V/VK =
0.01/K = 0.01/(5 x 10° atm') = 2000 atm.
Lod
(b)
AP = 0.01/« = 0.01/(1 x 107 atm”) = 100 atm.
(a)
Ht Qt l +44 +++
ye U4 SOF
B+) =25.
(b)
ve (bis + bis + bie) = big + Dis + dio + b2g + bas + bre.
(c)
1.58
n
Se Ca; = Ca, + Cay + °° + Cay = C(A, +
n
me
(a; + bj) = (ay + bi) + (az + bz) +
tte
bet bn,
2+ °° + gy
ore
Gi ED:
+ (Gn + bn) =
— eee
ie
OED
The left side of (1.51) is 7", 0", aibj = Di, (Gib + aib2 + + ibm) =
Dasa aj (by + bp + + bm) = Sia a; (b; + bz + + Dm) =
Pe a, ye bj, which is the right side of (1.51).
n
n
159
(a)
(b)
OV
s5V)dV=(Ve+ 5V°/3)|;° = (4 — 40/3) — (9 + 45) = -190/3.
{,Vv! dV=InV\5
mth eet oe
Nee
09 5.
2
(c)
ibe eave tizve | =04G4)=4
(d)
Let z= x°. Then dz= 3x° dx and ee cos x° dx = (1/3) pe cos z dz=
(1/3) sin z |"”* = (1/3)[sin(m'/8) - 0] = -0.2233.
1.60
(a)
~a' cos ax + C.
(b)
=@ 4COS Axim
(c)
Differentiation of the (b) answer gives -a
(d)
AMEE.
l= COs ay ad)
:
ae
:
1.61
(a) Function;
1.62
In (b) and (c).
1.63
(Aye ese
5 Cu (b) dx
1.64
(a)
[berax
(b) number;
ey
=i)
“+a
LG)
cosatm+a
=f]
:
T Sin am.
(c) number.
x,’ Ax. For Ax = 0.1 and x; at the left end of each
subinterval, 0” x;7 Ax =0.1[2” + (2.1)° + (2.2) ++ + (2.9)'] =
6.085. For Ax = 0.01, we get 6.30835. For Ax = 0.001, we get 6.33083.
The exact value is (x°/3) 3 =D I/o
(b)
si
6 303s
[be * de=0.01fe +6-OO™ 46° OO 4... 46°") = 0.74998. A
BASIC program for part (a) 1s
1
45 X=X+DX
15 FORJ=1T03
50 NEXT |
PAO) DK ee 2
55 PRINT DELTAX=
253DX=1/N
60 N=10*N
30 S=0
65 NEXT J
S5eFOR
t= iFOuN
70 END
40
1.65
OFN=.0
"Dx? SUM="s
S=S+ X*X*DX
(b)
log (4.2 x 10'”°) = log 4.2 + log 10'”°° = 0.62 + 1750 = 1750.62.
In (6.0 x 10°°”°) = 2.3026 log (6.0 x 107°") =
2.3026 log 6.0 + 2.3026 log 10°°” = 1.79 — 460.52 = 458.73.
(c)
log y =—138.265¢
(a)
ty = 10). 20 = 107 ey
= 04a
10
1.66
(d)
In z= 260.433 = 2.3026 log z,; log z= 113.10,
(a)
5, since 2° = 32.
(b)
0.
(c)
26° = 8; logi08 = logio(26)* = x logio26;
SAPS
EO
1026 =
x = log108/log
0.90309/1.41497 = 0.6382
1.67
(a) intensive;
(Cc) intensive;
(b) extensive;
(d) intensive;
(e) intensive;
(f) intensive;
1.68
pt of
One finds that a plot of PVm vs. P is approximately linear with an interce
to O2 in
PV = 58.90 L atm/mol at P = 0. The ideal-gas law PV, = RT applies
the limit of zero pressure, sO
.cm? - atm/mol
58900
_ PV...
ee
T = lim —"= ee ee
R
p30
ae
82.06 cm? - atm/mol-K
PV,,/(L-atm/mol)
59.04
59.02
59.00
58.98
|11
()tf
58.96
58.94
5s.d2
58.90
58.88
i
=
SO)
(J
inn)
5
Si
GGG
&
oe
—
0
200
1
4_1
400
i
aS
600
800
FA
el Cer a
1000
Tees
1200
P/torr
1.69
(a) T.
(j) F.
(b) F.
(c) F.
(d) T.
(6) Perth)
(2)
ee)
ee(he
Chapter 2
Paes
(a) T. (b) F.
2.2
(a) J; (b) J; (c) m; (d) N; (e) m/s,
2.3
(a)
1J=INm=lkgms’m=1
kgm’ s~
(b)
1 Pa=1N/m>=1 kgm s?m?
=lkgm's~
(ec)
1 L= 10° cm = 10° (107 my = 10° m’
(Ameen
2.4
(ene
(f) kg.
ekeamt ce
ts
2
=mg Ax = (0.155 kg) (9.81 m/s )(10.0 m) = 15.2J
mg dx
(a) w=J? Fdxy=J;
(hime
(c)
SAK | Ky Kpa hy 152 J.
4mv* = K and v= (2K/m)'” = [2(15.2 JO.155 kg)]'” = 14.0 m/s,
since |
J= 1 kg m’/s*.
jee)
P = F/A = me/A = (0.102 kg)(9.81 m/s*)/(1.00 m*) = 1.00 N/m” = 1.00 Pa.
2.6
(ayer
Dad
(a)
(Dye
(chal
(yee
ey
oeDaal ge(2 ar,
area = length x height = (V2 — V\)P) = (5000 — 2000)cm°(0.230 atm) =
690 cm? atm. Wyey = —area = —(690 cm* atm)(8.314 J/82.06 cm” atm) =
69.9 J.
(b)
2.8
w.., =-]7 P dV =-P(V, -V,)=ete.
w,, =]?
PdV =-P(V, -V,) = -(275/760) atmx(875 — 385) cm" =
~177cm* atm (8.314 J/82.06 cm? -atm)=-—18.0J.
29
(a)
The area under the curve is the sum of the areas of a rectangle and a right
triangle. The rectangle’s area is (V2 — V;)P2 = (2000 — 500)cm°(1.00 atm)
= 1500 cm’ atm.
The triangle’s area is Y2(base)(altitude) = “(V2 — V,)(P, — P2) =
(2000 — 500)cm*(3.00 — 1.00)atm = 1500 cm* atm.
Thus Wey = —3000 cm’ atm (8.314 J/82.06 cm*-atm) = -304 J.
(b)
Replacement of y and x with P and V in the straight-line equation gives
(P — Pi /(V— Vi) = (P2 — Pi)(V2— Vi).
w= -|; PdV= al Ae ee
a) Vers Vi) |aaa
=
= P.(V; =V, IP, — PRVWVo nV MC2Vs -ViV2)-— (AV, -V
=P(V, —V,
4 (8 = 13 )\V5 -V ) = as inia),
2.10
(a)
(b)
4
P
1
o>
2
E
1
A
2
»>-
V
pagOI
V
Neglecting the dependence of specific heat on 7, we equate the heat gained by
the water to that lost by the metal. The heat gained by the H20 is
(24.0 g)(1.00 cal/g-°C)(10.0°C) = 240 cal.
Thus 240 cal = (45.0 g)cmetai(70.0 — 20.0)°C and Cmetai = 0.107 cal/g-°C.
es.) iile mG) eka (Celie)
aly
2.12
(AlAl
213
Only (c).
2.14
cal 4.184J lday th
tOnw
(a) 220010: foal adil talaayes 07 Vee
day
lIcal
24hr 3600s
(b) (6 x 10°)(107 J/s)(3600 s/hr)(24 hr/day)(365 days/yr) = 2 x fo
215
Since the process is cyclic, AU = 0. Hence g = -w = -145 J.
2.16
(a) The bulk kinetic energy acquired by the 167-ft fall is converted into
internal energy, thereby warming the water by AT. The bulk kinetic energy
equals the potential-energy decrease mg Ah. The AU for a temperature increase
of AT can be set equal to the heat g = mcp AT that would be needed to increase
the temperature by AT, since the expansion work 1s negligible. Therefore
mg Ah = mcp AT and
ap a SAN _ (9.80 m/s" (167 x12x2.54x10° m)_ Ical
=
4.184]
(1.00 cal/g-°C)(10* g/kg)
Cp
2s
(b) mg Ah = (2.55 x 10° cm*)(1.00 g/em*)(1 kg/10° g)(9.80 m/s”)(50.9 m) =
Peg 10)
Dali
We have 0 = AU, + AU2 = q, + Ww) + q2 + W2 = qi + q2 (since the wall is ngid);
therefore g2 = —q\.
2.18
This notation might seem to imply that g and w are state functions, which is
not so. There is no such thing as the change in heat for a system. There 1s only
an amount of heat transfer for a process.
2319
Cool the water to some temperature below 25°C and then do enough stirring
work to raise its T to 30°C.
2.20
V = Yk = ¥2(125 N/m)(0.100 m)° = 0.625 J = 0.149 cal. AY =
0.149 cal
(71;C)
+ m2C2)AT and AT =
(1.00 cal/g - °C)(112.g) + (20 g)(0.30 cal/g - °C)
AT = 0.00126 °C and the final temperature is 18.001°C
Jaga
(a)
OO
OE
ett Che
Cbaeh
hue
Vinee
dE xy + dKpis. + mg dh + 0, so dE sys = —mg dh — dKpis.
(b)
dE xy = dq + dWimey = 0 + AWirey, SO AWirrey = dE sys. = —mg dh — dKpist. But
mg dh = (mg/A)A dh, where A is the piston’s area. Since mg/A = Pex, and
A dh = dV, we have mg dh = Pex, dV and dWimey = —Pext dV — dK pist-
2.22
2
Eq. (2.33) gives Winey = — [°Pay AV — AK pise = —Pa [dV -0 =
—P,,,(V. — V,) = —(0.500 bar)(4.00 dm*) = —2.00 dm? bar.
1 dm? = 1000 cm’ and 1 bar = 750 torr = (750/760) atm = 0.987 atm,
SO Wirev = -1974 cm? atm x (8.314 J/82.06 cm?-atm) = —200 J.
2.23
(a) T; (b) F; (c) F.
2.24
All except force, mass, and pressure.
2.25
(a)
From the equation AH = qp.
(b)
It can mislead one into thinking that heat is a state function.
2.26
No. For example, in a cyclic process, AH is zero but g need not be zero, since q
is not a state function.
2.27
AH = qp = 0 for the entire system. Since H is extensive,
H = H, + H2 and
AH
= AH, + AH2
= qi + G2. Since
AH = 0,7 q; + q2 = 0.
2.28
(a) T; (b) T.
2.29
(a)
Cpm= Cp/n and Cp = nCpm = (586 g/16.04 g mol!)(94.4 J/mol-K) =
3.45 kI/K.
(b)
(10.0 carat)(0.2 g/carat) = 2.00 g and Cp =nCpm=
(2.00 g/12.01 g mol7')(6.115 J/mol-K) = 1.018 J/K.
cp = Cp/m = (1.018 J/K)/(2.00 g) = 0.509 J/g-K.
2.30
vp = Vim=(mIVy! =p"! = (0.958 g/cm*)|= 1.044 cm’/g.
2.31
(a) U;, (b) H.
2.32
uyr = (OT/OP)y,
AT = J; wyr aP, and AT ~ pyr AP for H constant. Thus AT =
(0.2 °C/bar)(—49 bar) = —10 °C. The final temperature is approximately 15°C.
17
2.33
Eq. (1.35) gives (QUm/OVm)r = (QU m/OP)H(OP/OVm)r. Partial differentiation ofP
= RT/Vm gives (OP/OVm)r = -RT/V,, = —P°/RT. Hence, (QUm/OVm)r =
—(QU p/OP)r P°URT.
2.34
(a)
(OUm/0Vm)r= (6.08 J/mol-atm)(1 atm)*/(82.06 cm°-atm/mol-K)(301 K) =
(b)
2.46 x 107 J/cm’.
Doubling P multiplies (0U,,,/0V,,)7 by 4 to give 9.84 x 10~ J/cm’.
(a)
Use of (1.34), (1.32), (2.64), and (2.53) gives
-l= (= (= (=) = EE
OP
(b)
),\0H
);\ OT
)p
(OH/0P),
ag (2.65) follows.
Partial differentiation of H = U + PV and use of (2.63), (1.35), (1.44),
and (2.65) give
(=) -(3) ‘i ) 4V -(4| (=) “ret
OPE
ony?
oP ye
OVE) COPD),
—Cp Wy, = Cy pw, Vk— PVk+ Vand the desired result follows.
Jashs)
(a)
py, = (OT/OV), =-(QU/OV),/Cy.
(OU/OV), is intensive, since it is the
ratio of the changes in two extensive quantities. Cy is extensive.
Therefore, doubling the size of the system at constant 7, P, and
composition will double Cy, will not affect (QU/dV), , and hence will cut
u, inhalf. Therefore, 1, is neither intensive nor extensive, since it 1s
not independent of the size of the system and 1s not equal to the sum of
the p,’s of the parts of the system.
2.36
(ABE
2.37
(a)
aD) ar, (Cab
)elee(e) a
Since Tis constant, AU = 0 and AH = 0. (U and H of a perfect gas depend
2
2
on Tonly.) w= ff PdV = —nRT { V' dV =-nRT In (V2/V,) =
—(5.00 mol) (8.314 J/mol-K)(300 K) In (1500/500) = -13.7 kJ.
AU =q+w=0,sog
=-w = 13.7 kJ.
(b)
Since U and H are state functions, AU and AH are still zero. The work w
1S Zero.
T)=T, =300K. P>=nRT>/V> =
(1.00 mol)(82.06 cm?-atm/mol-K)(300 K)/(49200 cm’) =
2.38
0.500 atm. (An alternative solution uses P2V = P,V).)
(b)
Y= CemlCvm= (Cym+ R)/Cyvm = 2.5R/1.5R = 1.667.
P, = nRT,/V, = 1.00 atm. For a reversible adiabatic process with Cy
constant, Eq. (2.77) gives P,V," = P2V2" and P2 = (V,/V>)‘P, =
(24.6 L/49.2 L)'! "(1.00 atm) = 0.315 atm.
T> = P2V>/nR = 189 K.
(c)
P/ atm
0.5
isotherm
adiabat
(0)
V/L
0
2.39
(a)
24.6
49.2
q = O since the process is adiabatic.
Cy n= Com
R= 25Ris
I= Pi Vink =100 Ke
TT VilVoeei= (LOOK )(4,00)e t= 279 Ka
dU = Cy dT and AU = Cy AT, since Cy is constant. Thus
AU = (0.0400 mol)2.5(8.314 J/mol-K)(119 K) = 98.9 J.
AU=q+w=w=98.9J;_ AH = Cp AT = 138%].
(b)
From P)V," = P2V2", we get P\(nRT,/P\)' = P2(nRT2/P2)", which becomes
PIT) =P} 7," Hence (T1/T2)’ = (PoP)
and P2/P\ = (TWiT).
y = Cpm/Cym = 3.5R/2.5R = 1.40. P2/(1 atm) = (298 K/100 K)' 40
(2.98)>> = 0.0219, so Pz = 0.0219 atm = 2.22 kPa.
2.40
(a)
n=0.500 mol.
0.800 bar = 0.789 atm.
T> =P2V2/nR = 769.2 K.
7) = P)V\/nR = 384.6 K,
dU =CydT and AU = Cy AT
(0.500 mol)(1.5R)(384.6 K) = 2.40 kJ.
=nCy AT =
AH = Cp AT =
(0.500 mol)(2.5R)(384.6 K) =4.00 kJ. w= - |Pdv =
~P(V> — V;) = -(0.789 atm)(2000 cm>)(8.314 J/82.06 cm?-atm) =
-1.60kJ. gq =qp=AH =4.00kJ.
=
(b),
w=0
at constant VV.
Tp=PivVi/nR
= 210, Keel
= 324k
AU= Cy ABR= 0.675 kI- “AH =! Cp AT SailsKI
AU =9w = g;
SO. gd = 0.07 orn.
2.41 “ (a) Pe (Dy gleea(cjer.
2.42
(a) Process;
(b) system property;
(e) system property;
2.43
(d) PF; (e) EF:
C,,= dq ,/dT,,.
(f) system property;
2.44
(a)
(d) process;
(g) system property.
(a) dqpr > 0 and dT,, = 0, so Cp, = ©.
(c) dU =0 = dp, + dWpr;
Hence C,, =.
(c) process,
ddpr = —dWpr;
(b) —c°, since dqp, < 0.
Wp, < 0 and dq,,>0.
dT,,=0.
(d) 0, since dqp, = 0.
Heat is required to melt the benzene, sog>0.
AH =q,>0. The
constant-P work is w = —P AV; since the benzene expands on melting,
w <0. The volume change is slight, so |w| << |g| and AU = q + w =q;
hence AU > 0.
(b)
The same as (a) except that w > O since the system contracts on melting.
(c)
gq =O for this adiabatic process. w is negative for an expansion. We have
AU=q+w=w,so
AU <0. AH=AU + A(PV) = AU + nR AT. We have
dU =Gy di. where dU <= Orand Cy= 0: hence al <— Vand AT <0:
Therefore AH < 0.
(d)
With T constant, we have AH = 0 = AU for the perfect gas. w is negative
for the expansion. AU = 0 = q + w, so gq = -w and qs positive.
(e)
g=0,w=0,AU=q+w=0,AH=AU
+ A(PV) =AU +nR AT=0.
AT is zero because HW, = 0 for the perfect gas.
(f)
For Joule-Thomson throttling, AH = 0. Since by7 = (OT/OP) ;, =O tora
perfect gas, T is constant. Hence AU = 0, since dU = Cy dT = 0. The
process is adiabatic, so q = 0. Hence w = AU — q = 0; this also follows
from the equation w = PV; — PV; in the text, since T) = 7}.
(g)
q>0
for heating. From dqp= Cp dT, it follows that AT > 0, since
Cp >0. From dU = Cy dT, it follows that AU > 0, since Cy > 0.
From PV = nRT, it follows that AV > 0. Hence
AH = AU + A(PV) = AU + nR AT > 0.
20
w =—P AV <0.
2.45
2.46
(h)
q<Qan
ATd
<0. Hence AU<0.
AH = AU +nR AT <0.
(a)
q = 0 (since adiabatic), w = 0 (since constant V), AU = qtw=0.
(b)
w = 0 (const. V). The combustion is an exothermic process that releases
heat to the surrounding bath; hence g <0. AU=q+w<0O.
(c)
q = 0 (since adiabatic), w = 0 (expansion into vacuum), AU = g + w=0.
w=Osince
dw=-PdV=0.
The process is adiabatic, so q = 0. The volume change is being neglected, so
w=Q.
Hence AU=q+w=0.
AH=AU+A(PV)
=0+ VAP, since Vis
constant. Thus
ATE (18 cm9
2.47
(a)
1.987 cal/mol-K
Oratin ee
82.06 cm?-atm/mok K
pe
8 Ocal 16)
2
dgp
= Cp dT: gp = ifCpdT = J,n(a+bT)dT =
n{a(T7 — T,) + “2b(T5,— T; )]. Hence
q = (2.00 mol)[6.15 cal/mol-K)(100 K) +
(0.00310 cal/mol-K’)(400° — 300°)K’]. g = 1447 cal
Z
w=-] PdV =-PAV=-nRAT=
—(2.00 mol)(1.987 cal/mol-K)(100 K) = —397 cal
AU = q + w= 1447 cal — 397 cal = 1050 cal. AH = q, = 1447 cal
(b)
dw =—PdV=Oandw=0.
AU and AH are the same as in (a), since the
final and initial temperatures are the same as in (a) and U and H are
functions of T only for a perfect gas. Hence AH = 1447 cal and AU =
1050 calg._ AU =q+w=q= 1050 cal.
2.48
(a)
q = (79.7 cal/g)(18.015 g) = 1436 cal
w=-[ PdVv SPAVStPmpy ipl) =
cal__
-(1 atm)(18.0 g)(1.000 cm/g — 1.0905 cm/g) —1287
82.06cm~ atm
w =0.039 cal;
AU=q+w=1436cal;
21
AH =q,=
1436 cal.
(b)
2
Op = iECpdl
=Cp AT= (1.00 cal/g-°C)(18.01 g)(100 °C) = 1801 cal
w=-PAV=
(1 atm)(18.0 g)(1.044 cm'/g — 1.000 cm’/g) —L287cal__
82.06cm”™ atm
w = -0.019 cal
AU=q+w=
(c)
1801 cal.
AH =q, = 1801 cal
q = (18.015 g)(539.4 cal/g) = 9717 cal
V;
_ (1mol)(82.06 cm? - atm/mol -K)(373.15 K)
¥
l atm
= 30620 cm’
V; = (18.01 g)/(0.958 g/cm’) = 19 cm*, AV= 30600 cm?
w = -P AV =-(1 atm)(30600 cm?) Sienna ea eal
2.06 cm? atm
AU =q + w= 8976 cal;
2.49
AH =q, =9717 cal
For Cym independent of T, we have dU = Cy dT and AU = Cy AT= 1.5nR AT;
also dH = C, dT and AH = (Cy + nR)AT= 2.5nR AT.
(a)
AT = 200 K and substitution of numerical values gives AU = 6240 J and
AH = 10400 J.
(b)
Use of RV
=nRiccivésd> = 29275 Krand’/
y= 243.7 Ko sori — 48.5 K.
We tind AU = 1520 J and Ab=2450);
(c)
Since AT = 0. we have AU =0 and AH =0
2.50
No, since g and w are not state functions.
2.51
1 dm? = (10°! m)? = 10° m? = 10°°(107 cm)’ = 1000 cm*
(a)
8.314 J/mol-K
w = —-P AV =-(1 atm)(20000 cm’)
=-—2.02, kJ
82.06 cm*-atm/mol-K
T, = P\V\/nR = 121.9 K, T2 = P2V2/nR = 243.4 K
Gp = Cp AT= (2.00 mol)3.5(8.314 J/mol-K)(121.g K)
q=7.09kJ;
(b)
AU=qtw=5.054kJ;
w = 0 since Vis constant.
gy=Cy AT;
AH =q,=7.09 kJ
T goes from 2437 K to 121.9 K ?
so AT = —121.3 K and q = (2.00 mol)2.5(8.314 J/mol-K)(-121., K) =
—5.065kJ; AU=q +w=—5.06; kJ; AH=C, AT=-7.09 kJ
22
(c)
Since Tis constant, AU=0=AH.
w= -[' RdVes= ah nRT dV/V=
—nRT In (V2/V;) = 1.404 kJ. AU=0=q + wand g =-w =-1.40, kJ.
For the cycle, AU=0 =AH, g=7.90 kJ — 5.06; kJ — 1.40, kJ = 0.62 kJ,
w = —-2.02¢ kJ
+0 + 1.403 kJ = -0.62 kJ. On the P-V diagram, step (a) is a
horizontal line, step (b) a vertical line, and step (c) a hyperbolic line.
2.52
(a) Kinetic;
2.53
At the low temperature of 10 K, the l-atm gas density is rather high, the
(b) kinetic;
(c) kinetic and potential;
(d) kinetic and potential.
average intermolecular distance is rather small, and intermolecular interactions
are of significant magnitude. These interactions make Cp,» deviate from the
monatomic ideal-gas Cp.m.
2.54
(a)
Vea, = NRT/P = 24500 cm*. Each hypothetical cube has a volume Veute =
(24500 cm°*)/(6.02 Ss 102} = 4.1 x 10°°° cm? and an edge length
(4.1 x 107° cm’)? = 3.4 x 107 em =34 A.
(b)
The distance between the uniformly distributed molecules equals the
distance between the cube centers, which is 3.4 x 10°’ cm.
(c)
At 40 atm and 25°C, Vyas = 610 cm’, Veube = 1.0 x 10-7! cm? and the
distance between centers is 10 A.
2.55
At 300 K and | atm, the contribution of intermolecular interactions to Cp and
Cy is small and can be neglected. At room T, Cyvin is negligible for diatomic
molecules that are not heavy. We thus consider only Cy», and Cyrot.
2.56
(a)
Cym=
CVtrm = 3R/2;
Gem =CvmtR=SR/2
(b)
Cyv.m = CVitem ae Cy rot,m = 3R/2 +
R=5R/2;
Cp.m = 7R/2.
The contribution of intermolecular interactions to Cp of the liquid can be
estimated by taking Cp.tiqg — Cp.gas, Since intermolecular interactions are quite
small in the gas. Figure 2.15 shows that Cp.jiq — Cp.gas 18 positive. Therefore
CPintermot = (OUintermo/OT)p iS positive and Uintermo! Must increase as T increases
at constant P. (Recall from Sec. 2.11 that Uintermo! 18 negative. An increase in
Uintermot Means a less negative Uintermo: and corresponds to a decrease in
intermolecular attractions as T increases.)
23
2500)
(a)
Using q = mc AT and w = power x time, we have
g = (27 1b)(454 g/lb)(1 cal/g-°C)(100°C) = 12.3 x 10° cal
w = (746 J/s)(3600 s/hr)(2.5 hr) = 67.1 x 10° J
12.3 x 10° cal = 67.1 x 10° J and I cal = 5.5 J
(b)
Using V = mgh to find the work needed to raise a one-pound weight by
one foot, we have
1 ft-lb = (454 2)(980 cm/s”)(12 x 2.54 cm)(1 J/10’ ergs)
1 ft-lb = 1.356 J and 772 ft-lb = 1047 J
q = (454 g)\(1 cal/g-°C)(1°F2°C/°F) = 252 cal
252 cal = 1047 J and 1 cal = 4.15 J
2.58
(a)
(b)
T = 273.15° + 1.8° = 274.95 K, which has 4 significant figures.
Calculations of 1/T to 4 significant figures gives 1/T= 0.003637 Ken:
Ignoring significant figures, we have log 4.83 = 0.68395, log 4.84 =
0.68485, log (4.83 x 107°) = log 4.83 + log 10°° = 20.68395,
log (4.84 x 10°°) = 20.68485. The numbers 4.83 and 4.84 differ by | in
their third significant digit and their logs differ by | in the third
significant digit after the decimal point. The logs of 4.83 x 10° and
4.84 x 10° differ by 1 in the third significant digit after the decimal
point. The portion of the log that precedes the decimal point should not
be considered (since this portion comes from the power of 10) and the
log should have as many significant digits after the decimal point as there
are significant digits in the number. Thus the logs should be expressed as
0.684, 0.685, 20.684, and 20.685.
2.59
(c)
(210.6 K)! — (211.5 K)! = 0.004748 K — 0.004728 K™! =
0.000020 K™!.
(a)
Solving (1.39) for P, we get P = nRT/(V — nb) - an’/V*. For a reversible
2
2
2)
?
isothermal process, w = -| Payee -{ [nRT/(V — nb) — an*/V"| dV=
—nRT \n (V — nb) i - an’lV|; = nRT In [(V; — nb)/(V2 — nb)] +
an*(1/V, — 1/V2). Fora =0 = b, we get w =nRT In (V,/V2), which is
(b)
OTE
w = (0.500 mol)(8.314 J/mol-K)(300 K) In N +
(1.35 x 10° em® atm/mol’)(0.500 mol)?(8.314 J/82.06 cm?-atm) x
24
[1/(400 cm*)- 1/(800 cm*)], where N =
[400 cm’ - (% mol)(38.6 cm’/mol)]/[800 cm? — (4% mol)(38.6 cm?/mol)].
w = -895.7J + 42.73 =-853 J. Wideat = nRT In (V;/V2) = 864. J.
2.60
(a) -8.0K.
2.61
PV =constant holds only when T is constant, and T is not constant in a
reversible adiabatic perfect-gas expansion.
2.62
dq = Cp aT holds only in a constant-P process, and P is not constant in a
reversible isothermal perfect-gas expansion.
2.63
Genevieve erroneously applied an equation for a reversible adiabatic process to
an irreversible adiabatic process.
2.64
dU = CydT.
(b) 5.00 J/K
T= PV/nR and dT= (P/nR) dV. Since dV> 0, dTis positive.
Hence dU is positive and U increases.
2.65
(a) Intensive; kg/m’.
(d) Extensive; J/K.
(b) Extensive; J. (c) Intensive; J/mol. ;
(e) Intensive; J/kg-K.
(g) Intensive; Pa = N/m’.
2.66
(f) Intensive; J/mol-K.
(h) Intensive; kg/mol.
(i) Intensive; K.
Insertion of the dimensions of each physical quantity in the equation gives
energy
energy
hemp tar (press.)
temperature
temperature
(temp.)”
Equating the powers of temperature on each side of the equation, we get
—1 =1-—mandm
=2. Also, n= 1, since the product pressure x volume has the
dimensions (force/length’)length = force x length = energy.
2.67
Y = Cpn/Cy.m = 1.13. Also, the gas is nearly ideal under these conditions, so
Crpteteva th
lience
y= (Cy aa Gyy,
and Gym—inU.13 = 15 cal/mol-Kie
=
i+ Ki Cyipai lose
RiCy., = 0.13
Cp, = Cyn tk = 17 cal/mol-k.
25
2.68
(a) m°/K; extensive.
(b) K7!: intensive.
(c) m?/mol-Pa; intensive
(d) J/m*: intensive.
2.69
(a)
False; AH is the change in a state function.
(b)
False; for a perfect gas, Cy is a function of T only, but need not be
independent of T.
(c)
False; the system must be closed.
(d)
False: for an isothermal process, 7 must remain constant throughout the
process.
(e)
False; U must be replaced by AU.
(f)
False; For a perfect gas, U depends on T only, but this is not true for
other kinds of systems. An isothermal pressure change changes the
average intermolecular distance and hence changes the contribution of
intermolecular interactions to the internal energy.
(g)
False; see the first example in Sec. 2.8.
(h)
Tue:
(i)
False; for example, T drops in a reversible adiabatic perfect-gas
expansion.
(j)
(k)
False; the entire path of the process must be specified.
False; for example, a real gas expanding adiabatically into vacuum can
undergo a change in T.
(I)
True.
(m)
False; work can be done.
(n)
False; when heat flows into a system of ice and liquid water in
equilibrium, T remains constant as long as some ice remains.
(0)
False.
(p) True.
26
Chapter 3
3.1
(a) T; (b) T; (c) T; (d) F.
3.2
(a) ey = 1 —TeITy = 1 — (273 K)/(1073 K) = 0.746
(b) rey = —Wmax/yy » —Wmax = 0.746(1000 J) = 746 J,
~dc min = 1000 J- 746 J = 254)
S23
€rev = 1 —Tc/Ty = 0.90 = 1 — (283 K)/Ty and Ty = 2830 K
3.4
w=-2.50kJ.
e=-w/qy = 0.45 = 2.50 kJ/qu and qy = 5.56 kJ.
AU=0=q+t+w=qce+t+ Gy +w=qct5.56
kJ -2.50 kJ and q- =-3.06 kJ.
SPS
(a)
Since a Carnot cycle is a cyclic process, the first law gives AU=q +w=
0,so0= qc + qy +w
[Eq. (1)]. Since a Carnot cycle is reversible, we
have AS = fdey (T =0 = qc/Tc + qu/Tu, which can be rearranged to
Qc!qy =-Tc/Ty
(Eq. (2)]. Use of Eqs. (1) and (2) gives Kjey =
c/w =— qcl(Gc + qu). Division of numerator and denominator by qgives Krey = —I/(1 + qx/qc)
= -1/C1 — Ty/Tc) = To (Ty - Tc). Also, Erey =
—qu!lw = qul(qc + Gu) = V(qgcelqu + 1) = W(-T C/T + 1) = To/(Ty - Tc).
(b)
From (a), &ey = 1/(1 — Tc/Ty); the denominator is less than | and greater
tnamOsso ey oo
(c)
&ey = Ty (Tu — Tc) = (293 K)/(20 K) = 15 =|@,, Ww, so |q;, |=
1S5w = 15 J. (This indicates why heat pumps are attractive devices for
heating homes in winter.)
(d)
Krey = Tc/(Ty - Tc), which goes to 0 as Tc goes to 0 K.
27
3.6
Le
VAS
the area enclosed
Combining the shaded areas using the marked signs, we get
by the cycle.
APY
at the same
For the anti-Clausius device removing heat from the cold reservoir
device that
rate the heat engine deposits heat in this reservoir, we have acyclic
ir is
has a positive net work output; the net energy flow into the cold reservo
zero, so the energy input for the work output must come entirely from heat
device
extracted from the hot reservoir. The combined system is thus a cyclic
that converts heat completely into work with no other effects and therefore
violates the Kelvin—Planck statement.
An anti-Kelvin—Planck device converts the heat input from the hot
reservoir completely into work. We use this work output to run a heat pump
and we run the heat pump at a rate that uses up all the work output of the antiKelvin—Planck device. The combined system is then a cyclic system that
transfers heat from the cold to the hot reservoir with no net work input, thereby
violating the Clausius statement. This completes the proof.
3.8
(ajeee(b) I
(c)e deme)
(eyo)
(abe
(2) ee eter) EE
(k) T.
aye)
(a)
AS\ap = q/T = (1560 cal)/(87.3 K) = 17.9 cal/K.
28
ee) ele
(b)
(5.00 g)(1 mol/39.95 g) = 0.125 mol.
q = —(0.125 mol)(1560 cal/mol) = —195 cal.
AS = q/T= (-195 cal)/(87.3 K) = —2.24 cal/K.
3.10
AS = J? dqre/T = J? (CpIT) aT = J? n(alT + b) aT =
n{a In (T2/T;) — b(T2 — T;)]. Thus
AS
= 2.00 mol
moll 6.15 molK
C2 jn 300K
499 4 9.00310
—S2! 100OKk |= |=4.1l6cal/K
4.
ape
Jl
For melting the ice to liquid water at O°C, AS, = Gre/T =
(18.015 g)(79.7 cal/g)/(273.15 K) = 5.26 cal/K. For heating the water from 0 to
100°C at 1 atm, AS, = J? Cp dTIT= Cp In (T2/T))
= (18.01 g)(1.00 cal/g-K) In (373.15/273.15) = 5.62 cal/K.
For vaporization of the liquid to vapor at 100°C and | atm, AS, = qye/T = 26.04
cal/K. For the isothermal expansion of the vapor (assumed ideal), Eq. (3.29)
gives ASq = nR In (V2/V,) = (1.00 mol)(1.987 cal/mol-K) In 2 = 1.38 cal/K.
We have for the overall process:
AS = 5.26 cal/K + 5.62 cal/K + 26.04 cal/K + 1.38 cal/K = 38.30 cal/K.
aul4
Zero, since the process is cyclic.
3.13
For an ideal gas with Cym = 1.5R at all T’s, Eq. (3.29) gives
AS = Cy In (T2/T;) + nR In (V2/V,) = nR[1.5 In (T2/T)) + In (V2/V,)].
(a)
(2.50 mol)(8.314 J/mol-K)(1.5 In 1.5 + In 0.75 ) = 6.66 J/K
(b)
(2.50 mol)(8.314 J/mol-K)(1.5 In 1.2 + In 1.5) = 14.1 J/K
(c)
(2.50 mol)(8.314 J/mol-K)(0 + In 1.474) = 8.06 J/K
3.14
Ofsincedg., =—0:
O15
Use the reversible path in Fig. 3.7. For the first step, AS; = Cp In (T2/T;) =
(10 g)(1.01 cal/g-K) In (273.1/263.1) = 0.38 cal/K.
For the second step AS? = Gre/T = —(79.7 cal/g)(10 g)/(273.1 K) = —2.92 cal/K.
For the third step, AS3 = Cp In (T2/T,) = (10 g)(0.50 cal/g-K) In (263.1/273.1) =
29
—0.19 cal/K. For the complete process,
AS = 0.38 cal/K — 2.92 cal/K — 0.19 cal/K = —2.73 cal/K.
3.16
For the step 1 > 2 in Fig. 3.4b: g > 0, w < 0, AU = 0 (since U depends only on
T for a perfect gas), AS = qx/Ty > 0. For the step 2 > 33g 0a <0;
AU <0, AS = 0 (since reversible and adiabatic). For the step 3 > 4: q < 0,
p 1:g=0,w>0, AU>0,
U = 9-/Tc <0. For theste452
= 09AS
Wie OFA
AS
= 0.
Ss Uy!
(a)
m\c\|AT\|
= mC2|AT|
(200 g)(0.0313 cal/g-K)(120°C — x) = (25.0 g)(1.00 cal/g-K)(x — 10°C)
iS SVU
(b)
AS =Cp In (T)/T)). For Au, AS is
(0.0313 cal/g-K)(200 g) In (305.1/393.1) = —1.59 cal/K
(c)
For the water, AS is
(1.00 cal/g-K)(25.0 g) In (305.1/283.1) = 1.87 cal/K
(d)
3.18
-1.59 cal/K + 1.87 cal/K = 0.28 cal/K
AS =ngR In [(ng + np)/na] + npR In [(1¢a + nb)/no]
Na oO mole
= 02
sno!
AS = (8.314 J/mol-K)[{2.50 In (2.81/2.50) + 0.3125 In (2.81/0.312s)] =
8.14 J/K
3.19
AS for the unmixing is the negative of AS in Eq. (3.33),
so AS = ngR In Xq + npR In xp = (Ng + Ny)R(Xa In Xq + Xp In XH).We have
tal
aa
0.00100 g
eimol
=. 09943553)
a
-5
ae
AS = (1.837 x 10° mol)(1.987 cal/mol-K)(0.758 In 0.758 + 0.242 In 0.242) =
2.02 x 10° cal/K = -8.45 x 10° J/K
Reminder:
Do not look up the solution to a problem until you have made a serious
effort to solve it.
30
3.20
To carry out the change of state reversibly, we put the part at 7> in contact with
a heat reservoir whose temperature is infinitesimally less than T> and wait until
heat dq flows into the reservoir. Then we remove the reservoir from this part of
the system and put the part at 7, in contact with a heat reservoir whose
temperature is infinitesimally greater than 7, and wait until heat dq flows into
the system part at 7. Since these two heat flows are reversible, we can use dS
= dqrey/T to write the system’s entropy change as dS = dq/T, — dq/T>. (The
entropy changes of the reservoirs are irrelevant to dS of the system.)
3.21
(a) F; (b) F; (c) T; (d) T; () T; () F; (g) T; (h) F.
3.22
Using dS = ddye\/T, we have for the signs of AS:
(a)
edgier = 0 SOAS
(C)igedGrey
0:
(b)
0 SO Nor)
ddrey > 0 so AS > 0.
(dredge
0, sOvNS = 0.
(e)
This is an irreversible process in an isolated system, so AS > 0.
(f)
This is an irreversible adiabatic process, so AS > 0. This also follows
from Eq?(3.29) with dl= 0 and VV;
(oy
edd) > Orso AS >.0,
(h)
dgin= Vesa AS <0.
(i)
This is an irreversible process in an isolated system, so AS > 0.
(j)
This is an irreversible process in an isolated system, so AS > 0.
ASuniv 18 O for a reversible process and positive for an irreversible process.
Hence the AS,niyv values are:
(a) 0;
(f) positive;
(i) positive;
(g) 0;
(h) 0;
qu/Ty, 90, qc/Tc, 0.
(b)
(b) 0;
(c) 0;
(d) 0;
(e) positive;
(j) positive.
3.23.
(a)
0,0, 0, 0, since each step
is reversible.
3.24
Assume that dS = dq;e/T is the differential of a state function such that ASyniy >
0 for any process. Let an anti-Clausius device exist. Such a device extracts heat
q > 0 from a cold reservoir and deposits an equal amount of heat in the hot
reservoir, with no other effects. For one cycle of such a device, AShot res. = g/Tu,
AScotdtes.
= —G/Tc, and AS¢evice = 0 (Since the process is cyclic). We have ASyniy
= q/Ty — q/Tc = q(T — TH)/TcTx < 0, which violates the assumption that ASyniv
> 0. Hence an anti-Clausius device cannot exist. This completes the proof.
31
3.25
—qcelay = TI's and tc/ty =4e/qy - Thus /200.00°M = q?/q, But |g| / |gue| = 7/(273.16 K). Hence the Melvin temperature t is given by
1/200.00°M = T°/(273.16 Ky’.
(a)
(b)
3.26
For the steam point, T = 373.13 K and we find t = Boao
For the ice point, T = 273.15 K and t = 199.99°M.
= —qralq3
€rev = 1 + qc/qu and g = 1 — rey = — gclqu. We have g(T, 73)
and g(T), T2) = —qi/qze = 4i/42a, 80 8(T2, T3)8(T1, T) = —-qi/q3.
3.27
(a)
Substitution ofx= 3 gives 0.002736 as the probability that x = 3. Hence |
_ 0.002736 = 0.9973 is the probability that an observation is within 3
standard deviations from the mean.
(b)
Substitution of x = 10° gives the desired probability as p =
8x 107 x 67 95*!0"
Tf e = 10”, then taking logs gives z log e = y; hence
y =-0.5 x 10! x 0.434 = -2.2 x 10'' and we have
p=8x
3.28
(a)
10° x 10722*10" ~ 107 22%10"
The probability p of observing a deviation = 10° standard deviations in
are probability of not observing such a
one observation is re
deviationdnputials is (=al0sele \ ron (lea10 Wu daes0
asthe
probability of not observing such a deviation is 0.5 and the probability of
)=
observing such a deviation is 0.5. We have In 0.5 = n In (1 — ty
-n x 1072*!°" | where Eq. (8.36) was used. We get n = 0.7 x (2
There are 3 x 10’ seconds in a year, so it takes
(Ociex 1Oceeee
\/(B x 10’) years of measurements at the rate of one
per second to reach the 50% probability.
(b)
We want | - (1 —p)" 20.99. Since p = 2, we want
0.01 > 1/2" or 2" > 100. The minimum value of nis 7.
3.29
(a) q; (b) T.
3.30
In one minute of operation, w| = (1000 x 10° J/s)(60 s) = 6 x 10'° J. Then e =
0.40 = |wV/qu and qu = (6 x 10'° J)/0.40 = 15 x 10'° J in one minute. Hence qc
32
= 15x 10'°J-6 x 10!° J=9 x 10'°J per minute. Use of g = mcp AT gives
m = (9 x 10'” J)/(4.184 J/g-K)(10 K) = 2 x 10” g. The density of water is 1
g/cm’, so 2 x 10° cm* = 2 million liters are used per minute.
3.31
(a)
q cannot be calculated, since g depends on the path and the path is not
specified.
(b)
The path-dependent quantity w cannot be calculated.
(c)
dU = Or die=viGuadl
MAU —ancyedi=
na
bly dt =
na(T, — 7) + Yanb( Ty - i) = (4.00 mol)(25.0 J/mol-K)(500 — 300)K +
¥(4.00 mol)(0.0300 J/mol-K*)(500° — 300°)K? = 29.6 kJ.
(d)
AH = AU + A(PV) = AU +nR AT =
29600 J + (4.00 mol)(8.314 J/mol-K)(200 K) = 36.3 kJ.
(e)
Brome 2.9) 7As
iF (Cy/T) dT + nR In (V2/V,). But Cy/T=
nCyw
i = niall + 5), somsen
iF (a/T + b) dT + nR In (V2/V;) =
na |n (T2/T) + nb(T2 — T\) + nR In [(nRT>/P2)/(nRT,/P)] =
(4.00 mol)(25.0 J/mol-K) In (500/300) +
(4.00 mol)(0.0300 J/mol-K’)(500 — 300)K +
(4.00 mol)(8.314 J/mol-K) In [(500 K)(2.00 atm)/(300 K)(3.00 atm)] =
78.6 J/K.
BEEI
(a) Rev;
6yaKh)
(a)
Since U and S are extensive, the 10 g has the higher U and the higher S.
(b)
The vapor; the vapor.
(c)
The 40°C benzene; the 40°C benzene.
(d)
If the system at 300 and 310 K is adiabatically enclosed, it will
spontaneously go to the state at 305 K. Since S increases in a
spontaneous adiabatic process, the 305-K system has the higher S. Since
q = 0 and w is negligible for this process, the two systems have the same
(b) irrev;
(c) irrev;
(d) irrev;
i.
(e)
Neither; the l-atm gas.
33
(e) irrev;
(f) irrev;
(g) rev.
3.34
PdV+VdP=d(PV),
VdV=d(V),
ddrev/T = dS, dqp = dH, and
dw,e/P = —dV, so b, c, d, f,h, andj = 0.
3555
(a)
3.36
The second law of thermodynamics is “hardly ever” violated.
3:37,
No. (A pilot plant using this method has been built.)
3.38
Suppose we could prepare a reservoir at Tc = 0 and could reduce the engine’s
temperature to absolute zero. Then erey would become equal to 1. The Carnot
GP; k, Ot
(b)
Cp.m;
R
cycle would convert the heat gy completely into work. But this would violate
the Kelvin—Planck statement of the second law. Hence we can’t achieve
absolute zero. (See also Sec. 5.11 in the text.)
3.39
(a)
AH = qr = 0.
(b)
Suppose the final state consisted of ice at 0°C, with no liquid present. A
hypothetical path to attain this state is to warm the supercooled liquid
from —10°C to 0°C and then to freeze all the liquid at 0°C. AH for
warming the liquid is (1.00 cal/g-K)(10.0 g)(10.0 K) = 100 cal and AH
for freezing all the liquid is -(79.7 cal/g)(10.0 g) = —797 cal. The overall
AH is -697 cal, which is not 0. Hence the equilibrium state is not ice at
0°C. If the equilibrium state were all ice below 0°C, AH would be even
more negative than —697 cal, and hence this is not the equilibrium state.
Therefore the equilibrium state must consist of ice and liquid water in
equilibrium at 0°C. To satisfy the condition that AH = 0, the mass mice of
ice produced must satisfy -(79.7 cal/g)mice = —100 cal and mice = 1.25 g.
The mass of liquid remaining is 10.0 g — 1.25 g = 8.75 g.
(c)
A reversible path for the process is
liq. at 10°C —*-> lig. at O°C —*-9 mice + mig at O°C
AS = AS, + AS, = (10.0 g)(1.00 cal/g-K) In (273/263) —
(79.7 cal/g)(1.25 g)/(273 K) = 0.373 cal/K — 0.365 cal/K =
0.008 cal/K, where (3.30) and (3.25) were used.
34
3.40
(a) J/K;
(b) J/mol-K;
3.41
(a) False.
(b) False.
3.42
(a)
(c) J; (d) Pa=N/m’*;
(c) True.
(d) False.
(e) no units;
(f) kg/mol.
(e) True.
False. (For example, a nonideal gas expanding into vacuum can undergo
a change in JT—the Joule experiment.)
(b)
rue:
(c)
False; S increases in an irreversible process in an isolated system.
(d)
False. For example, a Carnot cycle is reversible, has AV = 0 and w ¥ 0.
(e)
True, since S is a state function.
(f)
False, e.g., AS # 0 for (the irreversible) adiabatic expansion of an ideal
gas into vacuum.
False (unless T and P are constant).
False. (A counterexample is the melting of ice at 0°C.)
False. (Reversibility and constant pressure are required.)
False. (A counterexample is the melting of ice at 0°C.)
False. (Only the system has to return to its initial state.)
35
Chapter 4
te (e) be (ae
4.1
(a) ee(b) cE mca)
4.2
Processes a and b are reversible, isothermal, and isobaric and have AS = qplT.
(a)
AG = AH —T AS = qp — T(qgp!T) = 9. AA = AU -T AS = qp + w - T(qp/T)
=w=-J? PdV=-PAV=
—(1 atm)[(36.0 g)/(1.000 g/cm’) — (36.0 g)/(0.917 g/cm’)] =
(3.26 cm? atm)(8.314 J/82.06 cm*-atm) = 0.330 J.
(b)
(c)
As in (a), AG = 0 and AA = w =—-P AV= —P(Vgas — Vig) = —PVeas =
_nRT = —-(0.50 mol)(8.314 J/mol-K)(353.2 K) = -1.47 kJ.
Since the gas is perfect, the final T is 300K. AH = 0 and AU = 0,
since U and H depend only on T for a perfect gas. From (3.29),
AS = nR In (V2/V;) = (0.100 mol)(8.314 J/mol-K) In (6.00/2.00) =
0.913 J/K. AG=AH-T AS =0 - (300 K)(0.913 J/K) = —274 J.
AS =-274 J. Processes a and b are reversible and so have
AA = AU-T
ASuniv = 0. For process c, there is no change in the surroundings, so
OS SOUT
NSS
D) BG oun (Clie @ pial:
4.3
(a) nye
4.4
Setting dU=0 in dU =T dS — P dV, we get
0=T dS, —PdV,,
so
(0S/O0V),, = (OS/0V), =PIT.
4.5
N=V,x=S,y=P,
For dH = TdS + VdP = M dx + N dy, we have M =T,
and (dM/dy), = (ON/0x), gives (OT/OP)s = (0V/0S)p. The equations in (4.45) are
derived similarly from Eqs. (4.35) and (4.36).
4.6
(a)
Use of (QU/0V)7 = AT/x — P gives
(=) _ (3.0410 K~')(303.15 K)
ee
ov
if
4.52x10~° atm!
36
—latm = 2040 atm
(b)
Use of tyr = ViaT— 1)/Cp gives pyr as
18.1cm?/mol
Oot nGlaKls
=
il
tLe tx
(1.987 cal)/(82.06 cm*-atm) = -0.0221 K/atm
4.7
From (4.53), Cp.m— Cym = TVmoe/K.
Vin = M/p = (119.4 g/mol)/(1.49 g/cm’) = 80.1 cm*/mol.
TV m0C/« = (298 K)(80.1 cm?/mol)(0.00133 K7!)?/(9.8 x 107 atm7!) =
(431 cm?-atm/mol-K)(8.314 J/82.06 cm?-atm) = 44 J/mol-K.
Then Cym = (116 — 44) J/mol-K = 72 J/mol-K.
4.8
Division of (4.30) by n gives (OH,,/0T)p = Cp.m. For (b), we use (4.48) divided
by n; for (c), (4.47); for (d), (4.49)/n; for (e), (4.50)/n; for (f), (4.53)/n;
for (g), we use (0A/0V)7 = —P, which follows from (4.35). Substitution of
numerical values gives:
(a)
(0H,/0T)p = Cp.m = 40 cal/mol-K;
(b)
(OHm/OP)r = Vm — TV m& = 50 cm*/mol — (298 K)(50 cm?/mol)(10~> K7')
= 35 cm’/mol = (35 cm?/mol)(1.99 cal/82 cm?-atm) = 0.85 cal/atm-mol:
(QU/0V); = AT/k — P = (10° K')(298 K)/(10~ atm!) — 1 atm =
3000 atm = (3000 atm)(1.99 cal)/(82 cm?-atm) = 70 cal/cm?;
(c)
(d)
(OSp/0T)p = Cp m/T = (40 cal/mol-K)/(298 K) = 0.13 cal/mol-K?:
(e)
(0Sp/OP)7 = —AVm = —(10-* K7')(50 cm*/mol)(1.99 cal)/(82 cm?-atm) =
—0.001> cal/mol-K-atm;
(ff)
Coe Creel ncik=
(40 cal/mol-K) — (298 K)(50 cm*/mol)(10° K~*)/(10~ atm”) =
40 cal/mol-K — (150 cm?-atm/mol-K)(1.99 cal/82 cm?-atm) =
40 cal/mol-K — 3.6 cal/mol-K = 36 cal/mol-K;
(g)
(dA/0V)7 = —P =—(1 atm)(1.99 cal/82 cm*-atm) = -0.024 cal/em’,
where (1.19) and (1.21) were used.
4.9
(a)
The Gibbs equation dU = T dS — P dV becomes dU; =T dS; — P dVr at
constant T. Division by dP7 gives dUr/dPr =
T dSr/dP; — P dV;/dPr or (OQU/0P)7 = T(0S/0P)7 — P(OV/OP)z . But the
Maxwell equation (4.45) gives (0S/0P)r = — (0V/0T)p, so (QU/OP)z =
—~T(OV/0T)p— P(OV/OP)7 = -TVa. + PV«.
37
4.10
(b)
From (1.35) we have (dU/dP)r = (QU/0V)r (OV/OP)r =
_(QU/AV)7«V. Substitution of (4.47) gives the desired result.
(a)
Atconstant P, dUp = T dSp — P dVp. Division by dTp gives (QU/OT)p =
T(OSIOT)p — P(OV/OT)p = Cp — PVa, where (4.31) and (4.39) were used.
(b) Cp = (OHIOT)p = [AU + PV)/OT]p = (QU/AT)p + POOVIAT)p,
So (QU/0T)p = Cp- P(OV/OT)p = Cp- PVa
4.11
At constant T, we have dH7 = T dS; + V dPr. Division by dV7 gives
(OH/0V)7 = T(0S/0V)7 + V(OP/OV)7 = T(OP/OT)y + V(OP/O0V)7 =
aT/K — 1/K, where (4.45), (1.42), (1.44), and (1.32) were used.
4.12
Liquids. Solids.
4.13
Differentiation followed by use of (4.51) gives
GT | al
ar
4.14
1 (36) CS
A
Ha TS
SR
=T dS —-PdV becomes dU, =T dS, . Division by dT,
and use of (4.29) gives Cy =(0U/0T)y =T(0S/0T),. At constant P, (4.34)
At constant V, dU
is dH p=T dS>p and division by dTp gives Cp = (0H /0T)p =T(0S/0OT)p.
4.15
From (2.63), ly = -(QU/0V)7/Cy. Use of (4.47) gives the desired result.
Reminder:
4.16
Don’t look up the solution to a problem until you
have made a serious effort to solve it.
Vin = RTIP + BRT.
= (1/Vm)(OVm/OT)p = (A/V )(R + bPR)/P =
RU + BPYPVm. K = —(/Vm)(OVm/OP)r = RT/P?Vm(QU/OV)7 = ATK — P = (1 + bP)P — P = bP”. ‘Cpm—Cv.m =
TVm0/K =
(R + bPRY/IR = R(1+bP). pyr = WaT — 1)/Cp=
(V/Cp)[RT(1 + bP)/PVm — 1] = (V/Cp)\(PVm/PVm — 1) = 0.
38
4.17
Seon haan cmapnion iene |.
-1 2%)-() --of™)
= {%)
Ci eee cl at Ole) >
DHE he
2
4.18
(a)
(b)
paw
VemiVgtet Vole) tbo 2K
(QV/OT)p = aVo + 2bVo(T— 273 K) and (0° V/OT*) = 2bVo.
(8Cp MOP) =-T(0°Vm/OT)p = —2bVmoT, 80 (OCp.m/OP)r =
~2(0.78 x 1078 K)(298 K)(200.6 g/mol)/(13.595 g/cm?) =
(6.85 x 10° cm? K™ mol"!)(1.987 cal)/(82.06 cm?-atm) =
~1.66 x 10° cal/mol-K-atm.
ACpm = (0Cp.m/OP)7 AP = (-1.66 x 10° cal/mol-K-atm)(10* atm) =
—0.02 cal/mol-K.
Cpm = 6.66 cal/mol-K — 0.02 cal/mol-K =
6.64 cal/mol-K.
Cp —- Cy = TVo2/«. As in Prob. 1.47, & = (1/Vin)(c2 + 2c3T— csP) and
K = (c4 + c5TV Vm. SO Cp — Cy = nT(co + 2¢3T— c5P)M(ca + €5T), since
4.19
ViV5. =:
(QU/OV)7 = ATI — P = T(c2 + 203T— csP)M(c4 + c5T) — P.
(0S/0P)7 = —QV = —n(c2 + 2c3T— csP).
Wye = (Vin CpmOT — 1) = (T/Cpm)(C2 + 2€3T — CsP) — VielCrm.
(OS/0T)p = CplT.
(OG/0P)r = V.
(OVIOT)s = 1/(OTIOV)s = -1/(0P/0S)y = “OS/OP)y =
~(AS/OT)KOTIOP)y = —-(Cy/IT)(OP/0T)y = —CvW/AT, where
4.20
(1.45) was used. Hence Os = —Cyk/TVa.
(b)
For a perfect gas, « = 1/T and k = 1/P [Eqs. (1.46) and (1.47)], so Os
_Cyl[PV = —CyinRT = —Cym/RT = V'(OVIOT)s, so dV/V =
—(Cym/RT) aT at constant S, and In (V2/V}) = (Cv,m/R) In (T)/T2) =
In (T,/T, )""® ;hence VoVi = (T,/T,)%"" or T2/T) = W/VEE
39
(c)
4.21.
(AV/OP)s = (OV/OT)s(AT/OP)s. From (a), (8V/0T)s = —Cyk/aT. From
(4.44), (OTIOP)s = (QV/9S)p = [(AS/V)p} |= ((AS/9T)p(OTIOV)p} |=
(Cp/T) \(OV/OT)p = (T/Cp)aV = ATV/Cp. So (OV/OP)s =
(—CyK/aT)(ATV/Cp) = —CykV/Cp; Ks =-V'(OV/OP)s = Cyk/Cp.
(dH/dP)r = V— T(OV/OT)p. For an ideal gas, V = nRT/P and (OV/0T)p = nR/P,
so (OH/0P)7 = nRT/P — T(nR/P) = 0.
4.22
At25°C and 1 atm, Vm = RT/P = 24500 cm*/mol and Uintermoim = —@/Vm =
~(10° to 10”)(cm®-atm/mol”)/(24500 cm*/mol) =
(40 to —400)(cm?-atm/mol)(1.99 cal/82 cm?-atm) = -—1| to-10cal/mol,
where (1.19) and (1.21) were used. At 25°C and 40 atm,
V., = RT/P = 610 cm?/mol and Uintermolm ~ —40 to 400 cal/mol.
4.23
(a)
AU vapm ~ —Uintermoim- From Fig. 4.4, Uintermol.m = —35) kJ/mol at Vn = 107
cm’/mol.
(b)
AUvap.m = AHvapm— ACPVim)vap-
VWmgas — Vmatiq = RT/P — 107 cm*/mol =
25200 cm?/mol. AU\ap.m = (6400 cal/mol)(4.184 J/cal) —
(1 atm)(25200 cm?/mol)(8.314 J)/(82.06 cm>-atm) = 24.2 kJ/mol.
4.24
(a)
a=(1.34~x 10° cm° atm/mol’)(8.314 J/82.06 cm*-atm) =
1.36x10° Jem /mol’.” U= Usa t+ Onemolmie Ua
(lo Rl
alV, + const = (12.5 J/mol-K)T— (1.36 x YOAd cm?/mol)/Vm +const,
where U;;m 1s the molar molecular translational energy.
(b)
For both the liquid and the gas, Utr.m = (3/2)RT = 1090 J/mol.
Vin,liq = M/p = (39.95 g/mol)/(1.38 g/cm’) = 28.9 cm*/mol.
Ey AMO
ee
Aa
~(1.36 x 10° J cm*/mol’)/(28.9 cm?/mol) = 4710 J/mol.
Uintennolmgat = CVa gaspee OuOL,
(c)
AU m.vap = Uintermol,m,gas a Uintermol.m,liq
4.7 kJ/mol.
4259
(a) ¢lee(D) el;
40
=
—19 J/mol
+ 4710
J/mol
=
4.26
Since T is constant, we have AA = AU — T AS and AG = AH —T AS.
From Prob. 2.49c, AH = 0 and AU = 0. From Prob. 3.13c, AS = 8.06 J/K.
So AA = 0 — (400 K)(8.06 J/K) = —3220 J and AG = -3220 J.
4.27
Each process is isothermal, so AA = AU — T AS and AG = AH — T AS. (For g,
w, AU and AH, see the answers to Prob. 2.44.)
(a)
AU=q+wand AS =q/T, so AA = w. Since w < 0, we have AA <0.
Since
AH = gq,AG=q-q=0.
(b)
The same as (a), except that w > 0 and AA > 0.
(d)
AU=0O=AH. AA =-T AS =-T(q/T)
=-q =w <0.
Also, AG =-T AS <0.
(e)
AU=0=
AH. For this irreversible adiabatic process, AS is positive and
AA = -T AS is negative. Also, AG = -T AS is negative.
(f)
The same’as (e).
4.28
Each is zero, since the process is cyclic.
4.29
(a)
AG=AH-—T AS
= AH -T(AH!T) = 0, as it must be for a reversible
(equilibrium) process at constant 7 and P.
(b)
From Prob. 3.15, AS = —2.73 cal/K. For the reversible path in Fig. 3.7,
AH for each step is:
AH, = (1.01 cal/g-K)(10.0 g)(10 K) = 101 cal,
AH) = -(79.7 cal/g)(10.0 g) = —797 cal,
AH; = —(0.50 cal/g-K)(10.0 g)(10 K) =—50 cal.
The overall process therefore has AH = —746 cal. For the isothermal
process at —10°C, we have AG = AH — T AS =
—746 cal —(263.15 K)(—2.73 cal/K) = —28 cal.
4.30
Since T is constant and U and H of ideal gases depend on T only, AU = 0 and
AH = 0. From (3.33), AS = —0.200 mol R In 0.400 — 0.300 mol R In 0.600 =
2.80 J/K. AA=AU-T AS =-T AS = -(300 K)(2.80 J/K) = -840 J.
AG = AH -T AS = -T AS = -840 J.
4l
4.31
Let the path be
a
b
(27°C, 1 atm) > (100°C, 1 atm) — (100°C, 50 atm) where step (a) is isobaric
and step (b) is isothermal. For a liquid, V varies slowly with 7 and P.
(a)
Use of (4.63) with Cp, a, and V assumed constant gives
AH, = Cp(T2 — T;) = (18.0 cal/K)(73 K) = 1310 cal = 5.50 kJ
AHp= (Ves iVa)(P?—
Pi) =
[18.1 cm? — (373 K)(18.1 cm*)(3.04 x 107 K"')](49 atm) x
(1.987 cal)/(82.06 cm*-atm) = 19 cal = 0.080 kJ
AH = AH, + AHp = 1.33 kcal = 5.58 kJ.
(b)
AU = AH — A(PV). Since V changes only slightly,
A(PV) = V AP = (18.1 cm*)(49 atm) = 887 cm” atm =
(887 cm? atm)(1.987 cal/82.06 cm?-atm) = 21 cal = 90 J.
AU= 1330 cal — 21 cal = 1.31 kcal = 5.49 kJ.
(c)
Use of (4.60) [or (4.61) and (4.62)] with Cp, a, and V assumed constant
gives AS, = Cp In (T2/T;) = (18.0 cal/K) In (373/300) = 3.92 cal/K =
16.4 J/K;
AS, = -OV(P2 — P}) =
3.04 x 10 K“)(18.1 em?)(49 atm) —L287cal___
82.06cm”
atm
—0.0065 cal/K = -0.027 J/K
AS = AS, + AS» = 3.91 cal/K = 16.4 J/K
4.32
dG= —S dT + V dP = V dP at constant T.
AG = V AP at constant T and V.
AG =
—2028
0.997 g/cm
4.33
io 99
1.987 cal eal
aU eee
82.06cm” atm
Vin = RT/P + BRT + cRTP.
Use the path
a
= 302 J
= (1/Vm)(OVn/OT)p = (1/Vm)(R/P + DR + CRP).
b
(Pi, 11) > (Pi, T2) > (P2, T2)
where step (a) is isobaric and step (b) is isothermal.
Vin — TVmO& = RT/P + BRT + cRTP — (RT/P + bRT + cRPT) = 0.
42
Equation (4.63) gives AHma = J7? Cp.m AT = Cp.m (12-1);
AHm.» = 0. Then AHm = Ama + AHm.b = Cp.m(T2 — T2), where Cpm is assumed
constant. From (4.60) with Cp, constant: ASma = Cp.m In (T2/T));
ASm.b= —> (RIP + bR + CRP) dP = R In (P\/P2) + bR(P; — P2) +
YacR( P” — P37); ASm= ASma + ASma = ete.
4.34
From Prob. 4.20c, (@V/0P)s = —Vks and AVs = —Vks APs =
~V«(Cy/Cp) APs. From (4.54) and preceding data:
AVs = -(18.1 cm>)(4.52 x 10°°/atm)(17.72/17.99)(9.00 atm) = -7.25 x 10° cm?
We have Vena = 18.1 cm? — 0.007 cm? = 18.1 cm’.
From Prob. 4.20a, (OV/0T)s = asV and AV; = asV ATs, so
ATs = AVs/OsV = —AV;(aT/Cyk).
ATs = 0.00725 cm
HEC SAIE IS NOUNS nem:
82.06cm* atm
(17.72 cal/K)(4.52x107>/atm)
0.0202 K, and Trina = 30.02°C.
We have dU = dq+dw = dw=—-P, dV,
=-P;(0V/0P). dP, = PsVKs dP,
since (OV /0P); =—VKs. Approximating Vand Ks as constants, we have
AU = VK, J? Ps dP; =V«5 4(P; - PB). Using AV; =—Vks APs (given
earlier in this problem), we have AU = —(AVs / APs (Py —P*)=
[(0.00725 cm3)/(9.00 atm)]0.5(100 — 1) atm? = 0.040 cm* atm = 0.00097 cal =
0.0040 J.
4.35
We have dU = (QU/dT)y dT + (QU/0V)r dV = Cy dT + (QU/OV)r AV.
Integrating and using a path similar to that in Fig. 4.5 but with V as the vertical
V
axis, we get AU = . Cy dT+ We (QU/0V)r dV. Using
(QU/AV)r = al V2 = an’/V’ for a van der Waals gas, we have AU =
iy Cy aT - an?/V> + an’/V\. If Cy is approximately constant over the
temperature interval, then AU = CUT? - T1) + an’(1/V, — 1/V2) for a van der
Waals gas.
4.36
(a) ale Dye
(CG)
(OT,
(), Deja
43
(faye
4.37
4.38
Set dS = 0, dP = 0, and dnjz; = 0 in Eq. (4.76). Then set dT =0, dV =0,
and dnjzi = 0 in (4.77).
For a closed system, dU = dq + dw. For a mechanically reversible process in a
closed system with P-V work only, dw =—P dV and dU = dq — P dV. Equating
this expression for dU to that in (4.74), we get dq = T ds + 3; w; dn; under the
conditions stated in (4.73).
4.39
(a) F; (b) F; .(c),T;
4.40
Use of (4.88) gives the following results.
4.41
(d) F-
(a)
Uy,o(solid) = LH,0ciquid)
(b)
Usucrose(solid) = Usucrose(in solution)
(c)
Hether(in water phase) = Wether(in ether phase) and Uwater(in ether phase) = Uwater(in water phase)
(d)
Ly, o(oria) = HH,01in soin.)
The more-stable phase at the given T and P has the lower i.
(a) HO(g);
w’s;
(b) neither; the two phases are in equilibrium and have equal
(ce) H,O(J);
higher p;;
(d) CoH 120¢(s); substance 7 flows out of the phase with the
(e) neither;
(f) CeHi20¢(aq);
(g) H2O(g), since p = Gm for a
pure substance.
4.42
Hy,0) =HH,0¢)
=Gmn,0) = Gmn,0u) > Since
= Gm for a pure substance.
Multiplication by n gives Gyo) = Gyo) Or AG = 0.
4.43
Youu,
4.44
Hx, +3¥y, =2Hnn,» where (4.98) was used.
4.45
a= Ano, /Vo, = (7.10 mol — 6.20 mol)/(—2) = —0.45 mol.
——
1.
Vo>
=-—5,
viCO,
=.
Vio
44
=.
4.46
(a) No;
4.47
(a)
(b) no;
(c) no;
(d) yes;
(e) yes;
(f) no.
Heat is needed to vaporize the liquid, so q is positive. Hence AH =
qp > 0. For this reversible isothermal process, AS = g/T > 0. Since the
process is reversible, ASyniy = 0. Also, AG = AH - T AS = g -—q =0, as it
must be for a reversible process at constant 7 and P.
(b)
q is positive. AU =q+w-=q>0. For this irreversible isothermal
process, Eq. (4.8) gives dS > dq/T and AS > q/T. Since q is positive, AS is
positive. Since the process is irreversible, ASuniy iS positive. Finally, AA =
AU —TAS. Since AU = q and T AS > q, we have AA < 0. [This also
follows from (4.22) with Wpy = 0.]
4.48
Only processes (a) and (d) are reversible [assuming the surroundings in (d) are
only infinitesimally warmer than the system] and so (a) and (d) have ASyniy = 0.
As to AU, AH, AS, AA, and AG, the following 1s true:
(a)
All are zero, since the process is cyclic.
(b)
The process is adiabatic, so g = 0. Since V is constant, w = 0. The system
is closed. Hence AU = q + w= 0. There is no reason for any of the others
to be zero.
4.49
(c)
q = 0, but w # 0. Hence AU # 0. Section 2.7 gives AH = 0. There is no
reason for any of the others to be zero.
(d)
AG = 0 for this reversible constant-7-and-P process. None of the others is
zero.
From (4.47), (QU/0V); = aT/« — P. Both « and T are always positive. For
liquid water between 0°C and 4°C at | atm, @ is negative and hence (dU/dV)7
is negative.
4.50
4.51
(a)
nu, stoichiometric coefficient;
(b) mu, chemical potential;
(c)
xi, extent of reaction;
(e)
kappa, isothermal compressibility,
(a)
Closed system at rest in the absence of external fields.
(d) alpha, thermal expansivity;
45
(f) rho, density.
(b)
Closed system (at rest in the absence of external fields), reversible
process, P-V work only.
(c)
System (at rest in the absence of fields) in mechanical and thermal
equilibrium, P-V work only.
4.52
(a) J;
4.53
(a)
(b) Jmol! K7:
(d) J/mol.
(c) J/K;
The chemical potential of substance i is the same in every phase in which
i is present, and this condition holds for each substance.
(Dye
(c)
4.54
= 0)
dG=0isa
T and P.
valid equilibrium condition only for systems held at constant
G=H-TS. Atconstant T and P, dG = dH -T dS = dqp — T dS. Solving for
dS, we get dS = dqp/T— dG/T.
4.55
(a) and (b): From dG = -S dT+ VaP, we get Sm = —(0Gn/0T)p and Vm =
(OG p/OP)r .
(c) Hm=Gm+ TSm = Gm—- T(0Gp/9T)p.
(d)
Um=Hm—PVm = Gm—- T(OGm/OT)p — P(OGn/OP)r(e)
Cpm=(0H,/OT)p and partial differentiation of (c) gives Cp.m =
(0G.,/0T)p — (0G.f0T)p—1(0-G,
/oT )p == 110
G, JO)
(f)
eer n= Cp me TV,,0°/K and the results of (b), (e), (g), and (h) give
Cym = —T(0’GmlOT)p + T(OG/OP)[(OGu/OP)
1} “(0° Gm/OT OP)? x
(OGml/OP)
(0° GOP?) = —T(0’ GOT?) p + T(0°G OT OP(O’
Gal OP”).
(g)
O&=(1/Vn)(OV,/0T)p and the result of (b) gives a =
(h)
(0° Gm/OTOP)/(OGm/OP)r.
Kk =—(OVp/OP)7/Vm = —(0°G/OP*)
1(OG m/OP)r.
4.56
Thesame since S is a state function.
4.57
(a)T;
(b) S.
46
4.58
(a)
Geea(boene and pos (cje@me(d)1S.¥(e) Sav. (1G:
4.59
(a)
False; the equation holds only for ideal gases.
(b)
True.
(c)
False; the system must be held at constant T and P.
(d)
False; the system must be held at constant T and P.
(e)
True; wpy = —w = q— AU and if q is positive then wpy > —AU.
(f)
ue:
(g)
False; there is no law of conservation of free energy.
(h)
False; ASyniv is positive for an irreversible process, but AS of the system
can be positive, negative, or zero.
(i)
True.
(n)
False; the system must be isolated or adiabatically enclosed.
(0)
False.
(j) False.
(k) True.
(1) False.
47
(m) False.
Chapter 5
5.1
(a) F; (b) F; (ce) F.
S72
(a) F;) (b) Toa(c)a Ts yd),
ae:
AH; =2H
5.4
(a)
(b)
yr 4,00) +24 mrsor(e) ~ 2H m.r.H,5() ~ 344mr.0>(2)°
The stoichiometric coefficients are doubled, so Eq. (5.3) gives
2(-319 kJ/mol) = -638 kJ/mol.
4(-319 kJ/mol) = -1276 kJ/mol. (c) —1(-319 kJ/mol) = 319 kJ/mol.
sh)
(a) F; (b) T; (c) T.
5.6
(a)
C(graphite) + 2Cl2(g) > CCLG):
(b)
4N2(g) + 3 H)(g) + 2C(graphite) + O2(g) > NH>CH>COOH(s).
(c)
+H2(g)
> H(g). (d) Na(g)> Nafg).
Shifl
The 25°C reference form is the form most stable at 25°C and | bar. The
elements that are liquid at 25°C and | bar are Hg and Br. Those that are
gaseous at 25°C and 1 bar are He, Ne, Ar, Kr, Xe, Rn, Ho, Fo, Clo, N2, Oo.
5.8
(a) C4Hio(g) + 44 O2(g)
4CO2(g) + SH20(¢ )
(b) C2H5OH(¢ ) + 302(g) — 2CO2(g) + 3H20(£
)
ye)
(aye la (b)) Toe (c) el (Gd) el aa(e)ake
5.10
(a) [2(—285.830) + 2(—296.830) — 2(—20.63) — 3(0)] kJ/mol = —1124.06 kJ/mol
(b) [2(—241.818) + 2(—296.830) — 2(—20.63) — 3(0)] kJ/mol = —1036.04 kJ/mol
(c) [-187.78 + 4(0) — 2(294.1) — 2(90.25)] kJ/mol = —956.5 kJ/mol
48
5.11
(a)
CoH1206(c) + 602(g)
6CO2(g) + 6H20( 2).
AH? 59¢/(kJ/mol) =
6(—393.509) + 6(—285.830) — (-1274.4) — 6(0) = -2801.6.
AH? = AU® + (An,/mol)RT = AU® + (6 — 6)RT = AU® = -2801.6 kJ/mol.
(b)
(0.7805 g)(1 mol/180.158 g) = 0.004332 mol. The heat flowing out of the
bomb is (2801.6 kJ/mol)(0.004332 mol) = 12.137 kJ. The water mass is
(2500 cm*)(0.9973 g/cm*) = 2493 g. Cp of the steel bomb plus
surrounding water is (14050 g)(0.450 J/g-°C) + (2493 g)(4.180 J/g-°C) =
1.674 x 10* J/°C. So 12137 J = (1.674 x 10° J/°C)At and At = 0.725°C.
fea 24. 030,6 +0725 6 — 24./5).C:
5.12
Initially, no, = PW/RT = (30 atm)(380 cm°)/R(297.2 K) = 0.47 mol. At the end,
No, = 0.47 mol — 6(0.004332 mol) = 0.44 mol, ny,o4) = 9.026 mol, neo, =
0.026 mol. The heat capacity C.ys of the system is gotten by adding the heat
capacity Ccon of the bomb contents to the heat capacity of the steel bomb and
surrounding water. The gases are heated at constant V, so we use Cy.m of the
gases. Appendix data gives Coon = (0.44 mol)(29.36 — 8.31)(J/mol-K) +
(0.026 mol)(37.11 — 8.31)(J/mol-K) + (0.026 mol)(75.29 J/mol-K) = 12.0 J/K.
Coyst = 12.0 JK + 1.674 x 10° J/K = 1.675 x 10° J/K. We get At = 0.725°C and
fe. Miso €,
5.13.
(a)
For the benzoic acid run, A-U =
(—26.434 kJ/g)(0.5742 g) + (0.0121 g)(—6.28 kJ/g) = Lie Aayaal
and (5.8) gives —15.254 kJ = —Cx+p(1.270 K) and Cx+p = 12.01 kJ/K.
(b)
For the naphthalene run, (5.8) gives A,U = —(12.01 kJ/K)(2.035 K) =
24.44 kJ. The contributions of the combustion wire and the naphthalene
to AU, are (0.0142 g)(-6.28 kJ/g) + (0.6018 g)A-Unapn, Where A-Unaph iS
per gram of naphthalene. Then
24.44 kJ = 0.089 kJ + (0.6018 g)A-Unapn and A-Unaph = 40.46 kJ/g.
A-U° = A-Umnaph = (40.46 kJ/g)(128.17 g/mol) = —5186 kJ/mol. The
combustion reaction C;9H(s) + 1202(g) — 10CO2(g) + 4H2O( ¢ ) has
An,/mol = 10 — 12 =—2 and A,H° = A,U° —2RT=
—5186 kJ/mol — 2(8.314 x 10° kJ/mol-K)(298 K) = -5191 kJ/mol.
5.14
U.) = Vit = (8.412 V)(0.01262 A)(812 s) = 86.2 J.
A,U0g = —U- = -86.2 J. If we neglect the difference between U of the
49
standard states and U of the states in the calorimeter, then AU jg is A,U298 per
mole, where “per mole” means for A€ = | mol. The reaction as written
involves a coefficient of 3 for B, so AE = | mol corresponds to An(B) = 3 mol.
We have n(B) = (1.450 g)/(168.1 g/mol) = 0.008626 mol, and
nSO2T 2299.98 kd/mol
INGoe, gu
0.008626 mol
Then An,/mol = 6 — 2 = 4 and AH 39, = AU 9g + Ang RT/mol =
~29.98 kJ/mol + 4(0.0083145 kJ/mol-K)(298.15 K) = —20.06 kJ/mol.
5.15
(a)
With Vig neglected,
AH 593 — AU 39, = (An¢/mol)RT=
(—1.5)(8.3145 J/mol-K)(298.15 K) = -3718.5 J/mol.
(b)
ABs:
—
AU
50.
=
PAV
=
Vo
EEN
Viens
=
(1 atm)(18 cm3/mol)(8.3 J/82 cm*-atm) + (An,/mol)RT =
1.8 J/mol -3718.5 J/mol = -3716.7 J/mol.
§.16
For (CH3)2CO(Z ) + 402(g) > 3CO2(g) + 3H20(£ ), Eq. (5.6) gives:
AH 50g =; V; A ¢ A p98,1 <
3 A ; A x98,c0,(g) are) A ¢ A 398.4,0(0) 7 A ¢ A 398(CH,),CO(6) hss A 5 A 198.0,(g)-
Use of Appendix data for CO2 and H20 gives
—1790 kJ/mol = 3(—393.509 kJ/mol) + 3(—285.830 kJ/mol) — 4(0) —
A ¢ A x98 (cH,), core) and A ¢ A398 (cH,),COCé) = —248 kJ/mol. The formation
reaction 1s 3C(graphite) + 3H, (g) + 05 (g) — (CH;),CO(¢) and has
An ,/mol =—3.5. Neglecting the volumes of condensed phases, we have
AUx93 =A ¢ H 393 —A ¢ (PV)30g =A ¢ H 393 — RT An, /mol =
—248 kJ/mol — (0.008314 kJ/mol-K)(298.1 K)(—3.5) = —239 kJ/mol.
5.17
NH2CH(CH3)COOH(s) + 2 Ox) — 3CO2(g)
+ SH20( 2) + + N2(g).
For this reaction, Equation (5.6) gives
AH 593 = 3 A ¢ H 498 co, (gy + 3-5 A pH y98,4,0(6) + 3A A 98.N
(2) —
3.75 A HH, ) — A ¢ A298 atanineis): Use of Appendix data for CO, and
H,O gives — 1623 kJ/mol = 3(—393.51 kJ/mol) + 3.5(—285.83 kJ/mol)+
50
O) Us
A ¢ F398, alanine(s) and A; F498 atanine(s) = —558 kJ/mol. The
formation reaction is 5N,(g) 7 th (g) + 3C(graphite) + O,(g) >
NH,CH(CH; )COOH(s) and has An,/mol = —5.
Then Ap593 = Apt 59, — RT An,/mol =
—558000 J/mol — (8.314 J/mol-K)(298.1 K)(—-5) = -546 kJ/mol.
5.18
Let the reactions be numbered (1), (2), (3), (4). Taking —(2) + (4) (3), we
get the desired formation reaction Fe(s) + ’202(g) > FeO(s).
Hence A; 55, (FeO) = [-37 — 94 + ¥2(135)]kcal/mol = -63'% kcal/mol.
Taking —(1) + 3(4) - $(3), we get 2Fe + 202 — Fe 703, so
AyHog(Fe2Os) = [117 + 3(-94) - 3(-135)] kcal/mol = 196 kcal/mol.
spy
Let the reactions be numbered (1), (2), and (3). If we take
14(1) + (3) + 34(2), we get the desired reaction. Hence AH°/(kJ/mol) =
0.25(-1170) + 0.50(-72) + 0.75(-114) = 414.
5.20
For reaction (1), AH 593 = Xi Vi ApH 39g, = —1560 kJ/mol =
2(-393% kJ/mol) + 3(—286 kJ/mol) — AT 598.c,H,(g) — 7(0), where the data
of reactions (2) and (3) were used to give AyH59, of CO2(g) and H20(¢), and
where A;¢H° of the stable-form element O2(g) is zero. Solving, we find
Ar A
5.21
(a)
598,.c,H¢(e)
=
ee
kJ/mol.
Vm=RT/P + b. Eq. (5.16) gives Hmia(T, P) — Hmre(T, P) =
J? [T(0Vm/OT)p — Vil AP” = Sp [RT/P’ — (RTIP” + b)] dP” =
_b J? dP’ =-bP
8.314)
(b) -bP =-(45 cm*/mol)(1 atm) 82.06cm> —— atm =-4.6 J/mol
5.22
(a) From Eq. (5.17), Hin208 = 9-
5|
(b)
(OH/0T)p = Cp and with the T dependence of Cp neglected, AHm =
kct (28.824 J/mol-K)(10 K) = 288.2 J/mol.
Cp.m AT; thus A, 308 =a
(c)
From Sec. 5.4, An H,0(6) = ArH y,0(6 + HT ces + +S, 04(g) =
—285.83 kJ/mol at 25°C.
(d)
Use of AHm= CpmAT gives for H2O(¢): Hin303 = 1 in.20g + Cp.m(10 K)
= —285830 J/mol + (75.291 J/mol-K)(10 K) = —285.08 kJ/mol.
5.23
(a) T; (b) T; (c) F; (d) F,;
5.24
Eq. (5.19) with AC} assumed constant gives AH ;,- AH; =AG p(ig=1T)).
(See the Prob. 5.10 solution for AH 59 .)
(a)
AC+/(/mol-K) = 2(75.291) + 2(39.87) — 2(34.23) — 3(29.355) = 73.80
AH 34) = —1124.06 kJ/mol + (0.07380 kJ/mol-K)(72 K) =
—1118.75 kJ/mol
(b)
AC%/J/mol-K) = 2(33.577) + 2(39.87) — 2(34.23) — 3(29.355) = -9.63
AH 34) = —1036.04 kJ/mol + (-0.00963 kJ/mol-K)(72 K)
AH 34) = —1036.73 kJ/mol
(c)
5.25
AC, =58.55 J/mol-K,
AH 35) =—-952.3 kJ/mol
The formation reaction is + Ha(g) + +Cla(g) — HCl(g). From Example 5.6,
AH;, — AH7, = Aa(T2 -T;) + LAB(T; - 7?) + 4Ac(T} -T;) +
tAd(T; —T,').We have Aa/(J/mol-K) = 30.67 — +(27.14) - + (26.93) ES OS
Ab/(J/mol-K’) = -0.007201 — + (0.009274) ~ 5 (0.03384) = -0.028758;
Ac/(I/mol-K*) = 1.3925 x 10°; Ad/(J/mol-K*) = -1.5455; x 10°. From the
Appendix, A¢H59. = —92307 J/mol. Then A¢Hjo99/(J/mol) =
92307 + 3.635(1000 — 298.15) + 1(-0.028758)(1000° — 298.15°) +
3 (1.3925 x 10°)(1000° — 298.15*) -1(15.455 x 10°)(1000* - 298.15*) and
Are jo99 = —102.17 kJ/mol.
52
rabal|
25.665 + 0.013045(T/K) — 3.8115 x 10°°(T/K)° — 5.5756 x 107! '(T/K)’.
5.29
23.272 + 0.013246(T/K) — 3.5603 x 10°°(7/K)’ + 2.0154 x 10°/(T°/K’) gives a
sum of squares of residuals of 0.0140 compared with 0.0920 for the cubic
polynomial.
5.30
S31
Equation (5.33) is alee = Cp m (Mow), 80 4 = (0.62 J/mol-K)/(10.0 K) =
0.00062 J/mol-K*. At 10 K, Eq. (5.35) gives 5°,(Tiow ) — Sino = Sm(Tiow)=
fl (C% ,/T) AT = Cm (Tiow)/3 = (0.62 J/mol-K)/3 = 0.21 S/mol-K. At
6.0 K, C%, ,=aT® = (0.00062 J/mol-K*)(6.0 K)° = 0.134 J/mol-K and
5°, = C> 9/3 = (0.134 J/mol-K)/3 = 0.045 J/mol-K.
5.32
(a)
We use (5.29) and J fix) dx = Jt) flx) dx + Jy fx) dx. Equations
(5.31), (5.33), and (5.35) give S*.ox = a(10 K)'/3 = Cp miox/3 =
[c(10 K)? + d(10 K)*J/3. Then S%,99x = 333¢ K? + 3333d K* +
(208 (en? dT?) aN # N3q (e/T + f Fat hT*) dT +
(1450 J/mol)/(200 K) + J300K (i/T + j + kT + IT*) aT=
(2666 K*)c + (40833 K*)d + 2.3026¢ + (180 K)f+ (19800 K*)g +
(2664000 K yk+ 7.25 J/mol-K + 0.4055i + (100 K)j + (25000 K’)k +
(6333523 KYL.
(b)
From (0Hm/0T)p = Cp.m, we have Hy
— Amo = (e@ee aT forthe
solid at T’. Use of the Debye T? law (5.31) gives
Hee,
He), =) id pam
As noted in part (a),
IPS aly di 0:K)a;
@= Cp mio k/C10 K)°, so
Also,
Hey x—Hmo= LUO KC pmo x= (2-5 Kyle Ky? + d(10 K)’].
H
200 Kea Hn.10 K =alei. Cp, rie) dT +
leas Cire (s) dT,
AH in.tus =
1450 J/mol, and H®,399— Hin200 = $300k Cp,m(2) aT. Substitution of the
Kale
expressions for Cp, and integration gives H\300 — 4 mo = (40000
53
+ (645000 K°)d + (180 K)e + (19800 K’)f+ (2664000 K*)g +
(3.9996 x 10° K*)h + 1450 J/mol + (100 K)i + (25000 K’)j +
(6333333 K*)k + (1.625 x 10’ K*)I.
aha)
(a)
The first trapezoid has its parallel sides at a and a + w and its area is
Ya(fo +
f\)w. The second trapezoid’s area is ¥2(f, + f2)w; the third has area
YA(f, + fh)w; . . .; the last has area 2(f,_; + f,)w. Addition of these areas
gives (fo + 2f, + 2f2
(b)
+: -- + 2fn-1 +fn)w, which is the desired result.
With n = 10, the trapezoidal rule gives Fess dx =
O.1[Ma(1y? + (1.1)! + (1.2) +--+ + (1.9)! + (2)"] = 0.693771.
With n = 20 the trapezoidal rule gives
0.05['4(1)! + (1.05)! + (1.1) +--+ + (1.95)! + 4/2] = 0.693303.
The n = 10 Simpson’s rule estimate is
Sie
+ 4/1.1 + 2/1.2 + 4/1.3 + 2/1.4 + 4/1.5 + 2/1.6 + 4/1.7 +
2/1.8 + 4/1.9 + 1/2] = 0.693150. The exact value is J? x7! dx = In (2/1) =
0.693147.
5.34
(a)
isK (Cem) aT ~5[¥2(0.83)/15 + 1.66/20 + 2.74/25 +-- - + 8.02/55 +
¥2(8.62)/60] cal/mol-K = 5.70; cal/mol-K. Next, Ji? (Cp.m/T) dT =
10[12(8.62)/60 + 9.57/70 + - - - + 15.42/180 + ¥2(16.02)/190] cal/mol-K =
13.767 cal/mol-K. Finally, J13,°* § (Cp.m/T) aT =
7.64 [¥2(16.02)/190 + 2(16.50)/197.64] cal/mol-K = 0.64; cal/mol-K.
Addition of the three contributions gives ees * (Cpm/T) dT = 20.115
cal/mol-K.
(b)
fame GomiDo)Die +(5)[0.83/15 +4(1.66)/20 + 2(2.74)/25 +
4(3.79)/30 + 2(4.85)/35 + 4(5.78)/40 + 2(6.61)/45 + 4(7.36)/50 +
8.02/55] cal/mol-K = 4.993 cal/mol-K. Next, J£°K (Cp.9/T) aT =
S[/(8.02)/55 + ¥2(8.62)/60] cal/mol-K = 0.724 cal/mol-K. Next,
60 K (Cpm/T) aT ~! (10)[8.62/60 + 15.42/180 + 4(9.57/70 +
10.93/90 + 11.97/110 + 12.83/130 + 13.82/150 + 14.85/170) +
2(10.32/80 + 11.49/100 + 12.40/120 + 13.31/140 + 14.33/160)] cal/molK = 12.91, cal/mol-K. Next, [13° K(Cpm/T) aT =
10[2(15.42)/180 + 2(16.02)/190] cal/mol-K = 0.85 cal/mol-K.
54
Finally, {127646 (Cp./T) dT =~ - - = 0.64; cal/mol-K. Addition of the
contributions gives 20.119 cal/mol-K.
(c)
The trapezoidal rule gives Jeeork (Cp.m/T) dT = 0.245 cal/mol-K;
[282% (Cpm/T) aT = 5.469 cal/mol-K; J3¢6 x (Cen!T)aT =
0.245 cal/mol-K. Addition gives |7? 64, (Cp.m/T) dT =
5.963 cal/mol-K.For the gas, ]3¢9*, (Cp.m/T) dT = 0.603 cal/mol-K and
§328:15K (Cp ./T) dT = 0.613 cal/mol-K. Addition gives
[222% (Ce pf din 1.215cal/mol-K.
5.35.
The graph of Cpy/T versus T is
(Cp p/T)/(cal/mol-K”)
OD
0.10
0.05
ae
0.00
0
es
50
cae
eet
ee
ae
=
150
100
ce
es
200
T/K
This graph amounts to a total of 0.15 x 200 = 30 “squares,” where each
“square” corresponds to | cal/mol-K? in the vertical direction and | K in the
g
horizontal direction. Weighing the total area and then cutting out and weighin
the area under the curve, one finds that the area under the curve weighs 0.669
AS
times the total-area weight, so there are 20.1 “squares” under the curve and
= 20.1 cal/mol-K.
5.36
(a)
S°,.9, would be increased by a for graphite, by b for H2(g), and by c for
O.2(g). The formation reactions C(graphite) + 2H2(g) > CHu(g),
55
H2(g) + + 02(g) — H,O(£), and C(graphite) + O2(g)
CO2(g) and the
fact that AS° of a reaction is independent of any convention show that
Si..29g Would be increased by a + 2b for CH4(g), by b + ic for HO (£),
and by a + c for CO2(g).
20),
(b)
No change, as verified by
(a)
The Appendix gives S),59, = 69.91 J/mol-K.
(b)
Equation (4.61) with the T dependence of Cp,, neglected gives
Seria
S aor 11 Gp
a+ c + 2(b + $c) =(acn2by-vc=
alll (Lo) 11), S05
2 44g) 00-7 BOL
0.
bee
(75.291 J/mol-K) In (348.1/298.1) = 81.58 J/mol-K.
(c)
Equation (4.62) with the P dependence of a and V,, neglected gives
Sm(298 K, 100 bar) = Sv 493 — &Vm AP, So Sm(298 K, 100 bar) =
69.91 J/mol-K — (0.000304 K7')(18.1 cm*/mol)(99 bar) x
(8.134 J/83.14 cm® bar) = 69.86 J/mol-K.
(d)
Use the path in Fig. 4.5. The calculations of parts (b) and (c) of this
problem give AS = AS, + AS; = 11.67 J/mol-K — 0.05 J/mol-K =
11.62 J/mol-K. So $,,(348 K, 100 bar) = (69.91 + 11.62) J/mol-K =
81.53 J/mol-K.
5.38
5.39
(a)
[2(69.91) + 2(248.22) — 2(205.79) — 3(205.138)] J/mol-K =
—390.73 J/mol-K.
(b)
[2(188.825) + as in (a)] J/mol-K = —152.90 J/mol-K.
(c)
[109.6 + 4(191.61) — 2(238.97) — 2(210.761)] J/mol-K = —23.4 J/mol K.
With the temperature dependence of AC}, neglected, Eq. (5.37) gives AS =
AS 59g + AC In (T2/298 K). The AC%,’s are calculated in the Prob. 5.24
solution and the AS59.’s are calculated in Prob. 5.38.
(a)
AS 359 = -390.73 J/mol-K + (73.80 J/mol-K) In (370/298) =
—374.76 J/mol-K.
(b)
AS 379 =—152.90 J/mol-K + (—9.63 J/mol-K) In 0/298)
—154.98 J/mol-K.
56
(c)
AS379 =-23.4 J/mol-K + (58.55 J/mol-K) In (370/298) =
—10.7 J/mol-K.
5.40
Differentiation of (5.36) gives d AS°/dT = Dv; dS}, ,/dT = Dj viC>p maha
AC, /T. Integration gives AS°(72) — AS°(T;) = IF (NG, /Dae
5.41
(ayeeErom Eq, GO aAS a= AS ct lsgey (ACe/I) dl=
~173.01 J/mol-K + Ji000K (Aa/T + Ab + T Ac +T Ad) dT =
~173.01 J/mol-K + Aa In (1000/298.1) + Ab(1000 — 298.1)K +
(1/2) Ac{(10007 — 298.17)K? + (1/3)Ad[(1000° — 298.1°)K* = -173.01
J/mol-K + (-39.87 J/mol-K)(1.21003) + (0.11744 J/mol-K)(701.85) +
0.5(—9.8296 x 10°°)(911136)J/mol-K +
(1/3)(2.8049 x 10°8)(9.7351 x 10°)J/mol-K = -174.51 J/mol-K.
(b) AStoo9 = ASSog + AC}998 In (1000/298.15) =
—173.01 J/mol-K + (-13.367 J/mol-K)(1.21016) = —189.19 J/mol-K.
The approximation that AC;
is independent of T is a poor one.
5.42
Vin = RTIP + RTf(T) and (OVn/0T)p = R/P + Rf +RTF’. Substitution in (5.30)
gives Smid(T, P) — Smre(T, P) = JO LRAT) + RIF ’(T)) dP = RPT) + Tf'(T)I.
5.43
A ¢Gy9g = Ay Hyg — T A ¢S3og - The formation reaction is
C(graphite) + /202(g) + No(g) + 2H2(g) > CO(NH)2)2(c). Appendix data gives
for this reaction, AS° = >; viS },; = 104.60 J/mol-K — 5.740 J/mol-K —
V(205.138 J/mol-K) — 191.61 J/mol-K — 2(130.684 J/mol-K) =
456,69 J/mol-K = AyS3og urea»Then A ,G39g = -333.51 kJ/mol —
(298.15 K)(-0.45669 kJ/mol-K) = —197.35 kJ/mol.
5.44
(a)/
(AG? =AH* —T AS;=
1124.06 kJ/mol — (298.15 K)(-0.39073 kJ/mol-K) = -1007.56 kJ/mol.
AG; = -1036.04 kJ/mol — (298.15 K)(0.15290 kJ/mol-K) =
57
AG? =-956.5 kJ/mol — (298.15 K)(-0.0234 kJ/mol-K)
~990.45 kJ/mol.
= —949.5 kJ/mol.
(b)
AG; (kJ/mol) = 2(-237.129) + 2(-300.194) — 2(-33.56) — 3(0) =
—1007.53.
AG; /(kJ/mol) = 2(-228.572) + 2(-300.194) — 2(—33.56) —3(0)
=-990.41.
AG? /(kJ/mol) = -120.35 + 4(0) — 2(328.1) — 2(86.55) =
—949.6.
5.45
(a)
AG37) = AH 379 — T AS379 =
~1118.75 kJ/mol — (370 K)(-0.037476 kJ/mol-K) = —980.09 kJ/mol.
(b) AG349 =-1036.73 kJ/mol — (370 K)(-0.15498 kJ/mol-K) =
—979.39 kJ/mol.
5.46
(c)
AGiy) = -948.3 kJ/mol.
(a)
Gir = Hm20g — TSin29g = 0 — (298.15 K)(205.138 J/mol-K) =
—61.16 kJ/mol.
(b)
Using the result of Prob. 5.22c, we get Gi193 = Hin29g — TSin208 =
—285830 J/mol — (298.15 K)(69.91 J/mol-K) = —306.67 kJ/mol.
5.47
Landolt-Bornstein data give 70.7 kJ/mol.
5.48
AHSo = 2(-46.11 kJ/mol) — 0 — 3(0) = — 92.22 kJ/mol.
AH
5000 =
AH
558 +);
Vi( Jel
ore
|
He
ont) =
=—9)
22
kJ/mol
ae 2(98.18
kJ/mol)
— 56.14 kJ/mol — 3(52.93 kJ/mol) = -110.79 kJ/mol.
5.49
The reaction is +N2(g) + 3 H2(g) — NH3(g). The Appendix gives A ¢H 49. =
—46.11 kJ/mol. Use of (5.43) gives A ¢Gy99 = 46110 J/mol +
(2000 K)[—242.08 — 5 (223.74) - 3 (-161.94)](J/mol-K) = 179.29 kJ/mol.
5.50
AH 59g + T Li Vil Gur — Hn20g/TIi = AH 39 + T Di canny ellOe
i>,
Wyle
crete
=
AH
563 aP AG;
=
58
AH
59, =
AG;.
AG?" —AGT™ = AH} —T AS? — (AH®™ —T AS2™) =
sypil
-T(AS*" — AS3"™), since the difference between 1-bar and 1-atm
enthalpies of a solid or liquid is negligible and H° of a gas is independent
of P. Since the effect of a slight change in P on S of a solid or liquid is
negligible, we have AS7* — AS#'™ = AS 7) — AS*™. =
Measee VASE
Pe Soery ) (0:1 094° T/moleR) cases Vi =
(0.1094 J/mol-K)An,/mol and
AGS — AG#™ =-T(0.1094 J/mol-K)An,/mol.
(b)
Ha(g) + 102(g) > H20( 4). AyG353— AyG 398 =
—(298 K)(0.1094 J/mol-K)(—1.5) = 48.9 J/mol.
a4
H
H
|
|
a
gaseous
»
H—C—C—O—H(g)
—
atoms
—-
hemes
H
(a)
tH —
H
H
H
|
|
—— Hq}
€——_ OC
|
|
H
H
AH,/(kJ/mol) = 5(415) + 344 + 350 + 463 = 3232
AH,/(kJ/mol) = -[6(415) + 2(350)] = -—3190
AH = AH, + AH» = 42 kJ/mol
See)
(b)
AH/(kcal/mol) = [6(—3.83) + 2(—12.0)] — [5(—3.83) + (-12.0) + (-27.01)]
= 11.17, so AH = 11.2 kcal/mol = 46.7 kJ/mol.
(c)
AH/(kJ/mol) = 2(41.8) — 99.6 -(-41.8 — 33.9 — 158.6) = 51.1 kJ/mol.
(a)
3C(graphite) + 4H2(g) + + 02(g) —>»
CH30CH2CH3(g).
3C(g) + 8H(g) + O(g) ——
Nels = AH, +AH,.
Appendix data give
AH,/(kJ/mol) = 3(716.682) + 8(217.965) + 249.170 = 4142.94. Bond
energies give
AH , (kJ/mol) = -[8(415) + 344 + 2(350)] = 4364. So A, H®/(kJ/mol)
= 4143 — 4364 =-221.
59
(b)
A,A°/(kJ/mol) = 4.184[2.73 + 2(-12.0) + 8(-3.83)] =—-217.2
(ec)
A,H°/(kJ/mol) =-41.8 — 99.6 — 33.9 —41.8
=-217.1
In the Benson—Buss bond-contnbution method, the carbonyl! group is treated
as a unit and no explicit contnbution is made for the C—O bond. The effect
of this bond is absorbed into the contributions for bonds to the carbonyl
carbon. The bond contribution to S>,
59g of the F—CO bond is 31.6 cal/mol-K,
so the total bond contnbution 1s 2(31.6 cal/mol-K) = 63.2 cal/mol-K for
COF»2(g). In addition, the quantity R In o must be subtracted to allow for
molecular symmetry. For COF2, the symmetry number o is 2, since there are
two indistinguishable onentations of the molecule (obtained by 180° rotation
about the CO bond). The symmetry correction is —R In 2 = —1.38 cal/mol-K
and the predicted S* 493 is 61.8 cal/mol-K.
AH 59g of vaporization refers to a change from liquid at 1 bar and 25°C to
vapor at | bar and 25°C. A path to accomplish this is the following 25°C
isothermal path:
liq(1 bar) a liq(23.8 torr) = gas(23.8 torr) a gas(1 bar)
AH® = Ay) + AHm2 + AHm3. As noted in Sec. 4.5, a modest change in
pressure from | bar to 24 torr will have only a very slight effect on H and S of
a liquid, so we can take AH, ;= 0. Also, since the vapor is assumed to behave
ideally, its H depends only on T, and AH,, 3= 0. Thus AH® = AHm2 = 10.5
kcal/mol. For comparison, Appendix data give AH® = (241.818 + 285.830)
kJ/mol = 10.519 kcal/mol. Next, AS° = AS}
+ ASm2 + ASm3. To a good
approximation, AS;,; = 0. Equations (3.25) and (3.29) and Boyle’s law give
ASm2 + ASm3 = AHm2/T + R In (P\/P2) = (10500 cal/mol)/(298.1 K) +
(1.987 cal/mol-K) In (23.8/750) = 28.34 cal mol” K7' = AS°. The Appendix
data give AS® = (188.825 — 69.91) J/mol-K = 28.42 cal/mol-K. Finally,
AG® = AH® — T AS° = 10.5 kcal/mol — (298.1 K)(0.0283¢ keal/mol-K) =
2.03 kcal/mol. The Appendix gives AG° = 2.045 kcal/mol.
5.56
We use the 25°C path (where M is methanol)
2
3
4
M(“, 1 bar) >M(2£, 125 torr) >M(g, 125 torr)
M(g, | bar)
M(ideal gas,
60
1 bar). For this path, AH = Hy yy(g)— Him my = Ammgy—Aa — (A mya)
= A;Hyg)—
Sp Hus), where H;, is the standard-state enthalpy of the
elements needed to form methanol. Since a moderate change in P has little
effect on thermodynamic properties of a liquid, we have AH, = O (and AS, = 0).
AH> = 37.9 kJ/mol.
AH; = 0 and AH, =0, where we neglected nonideality of
the gas. Then AH = AH, + AH? + AH; + AH4 = 37.9 kJ/mol and A; Hy,,)=
—238.7 kJ/mol + 37.9 kJ/mol = —200.8 kJ/mol.
We have AS;
=0, AS2 = AH2/T
= (37900 J/mol)/(298 K) = 127. J/mol-K, AS3 = R In (V2/V;) = R In (P\/P2) =
(8.314 J/mol-K) In (125/750) = —14.90 J/mol-K [where (3.29) was used], and
AS, = 0, where the gas is approximated as ideal. Then AS = (127. — 14.9)
J/mol-K = 112.2 J/mol-K. So Si, y(,)= 126.8 J/mol-K + 112.2 J/mol-K =
239 J/mol-K.
PAS
(a)
A, Hy. = (2n+2)boy +(#—Dbec
(b)
Breaking the formation reaction into two steps, we have
nC(graphite) + (n + 1)H2(g)
> nC(g) + (2n + 2)H(g) — CrH2n+2(8). So
A, H 29 = AH, + AH, = nA , H x9g{C(g)] +(2n+ 2)A , H y9g{H(g)] =
(n—-l)Dece — (2n + 2) Dey (c)
Equating the expressions in parts a and b, we have forn = 1:
Aboy = A ris lCig
+ 4A , H y9,[H(g)]—4D(CH) and
HyoglC(g)]+ A y H x9g(H(g)] —D(CH) . For n = 2,
bey = 4A,
6bey + bcc = 2A , H y9g(C(g)] * 6A , H y9g[C(g)] — Dec — 6Dex Substitution of the result for bcy gives after cancellation:
bec = 0.5A , H y1C(g)] - Dec
(d) J; (e) m’/mol; (f) K.
5.58
(a) Pa: (b) J; (c) J/mol-K;
~ Pesky
If AH? is independent of T, then d AH°/dT= 0 and (5.18) gives
AC*, =0. Then (5.37) gives AS; = AS;, .Q.E.D.
5.60
(a)
Equation (4.45) gives (AV/OT)p = -(0S/0P)r. The 3rd law gives lim7_49 AS
= () for an isothermal pressure change in an equilibrium system. Hence
(0S/aP)r 3 0 as T > 0, and a = (1/V)(OV/0T)p — 0 as T— 0.
61
5.61
(b)
a=(1/V)(OV/0T)p =1/T [Eq. (1.46)], which goes to % as T > 0.
(a)
Nonzero.
(b)
Nonzero. ArH 5, refers to formation of the substance from elements in
their stable forms at 298 K. The 298 K stable form of chlorine is Cl2(g).
not Cl(g), and AH 5o. for /2Cl2(g) — Cl(g) 1s not zero.
(c)
Zero, since Cl2(g) is the stable form of an element.
(d)
Nonzero. (Entropies are zero at 0 K.)
(e)
Zero.
(f)
Zero, since 350 K formation of N2(g) from its stable-form element(s) is a
process in which nothing happens.
(g)
Zero.
(h)
Zero, since heat-capacities go to zero as T goes to zero, as shown by the
Debye equation (5.31).
(i)
5.62
Nonzero.
The 25°C reaction (step c) is CHy(g)
+ 202(g) > CO2(g) + 2H2O( ¢ ) and has
AH = [-393.509 + 2(—285.83) — (—74.81) — 2(0)] kJ = -890.4 kJ for burning 1
mole of CHy. We have AH, = AH, + AH, = 0 = -890 kJ + AH; and AH; = 890
kJ. From the Appendix, the heat needed to vaporize 2 mol of H2O is
2(—241.8 + 285.8) kJ = 88 kJ. In step b, the products are being heated from
25°C to the flame temperature 7. The heat required to do this is 88000 J +
(1 mol)(54.3 J/mol-K)(T— 298 K) + (2 mol)(41.2 J/mol-K)(T— 298 K) +
2(3.76 mol)(32.7 J/mol-K)(T— 298 K) = 890000 J. So T — 298 K = 2096 K;
T = 2400 K.
5.63
5.64
(a)
n-C4Hj0(g), since larger molecules have larger entropies.
(b)
H2O(g), since gases have higher entropies than the corresponding liquids.
(c)
H2(g), which has larger molecules than H(g).
(d)
CyoHg(g).
(a)
AH" = apis positive, since heat is needed to vaporize the liquid.
For liquid — gas, AS° > 0.
62
(b)
AH° is positive since energy is needed to break the bond. AS° is positive,
since the number of moles of gas is increasing.
(c)
gis negative when the vapor condenses, so AH® is negative.
For gas —> solid, AS° < 0.
(d)
(COOH))(s) + 202(g)
2CO2(g) + H2O( 2 ). We can expect the reaction
to be exothermic, as is generally true for combustion reactions. Hence,
AH° <0. Also, AS° > 0, since the number of moles of gases is increasing.
(e)
AS° <0, since the number of moles of gases is decreasing. All species are
gases, and the Table 20.1 bond energies yield AH® =
(2711 — 2834) kJ/mol = -123 kJ/mol, so AH® < 0.
5.65
10° MW = 10” J/s. Let y = the absolute value of the heat of combustion of the
fuel burned per second. Then 0.39 = (10° J/s)/y and y = 2.6 x 10° J/s.
10* Btu/Ib = 1.06 x 10’ J/lb. Then (2.6 x 10° J/s)/(1.06 x 107 J/Ib) = 245 Ib/s of
coal burned, which is 15000 pounds per minute, 21 million pounds per day,
and 7.7 x 10° pounds per year.
5.66
(a)
True, since AH = gp =0.
(b)
False. There is a thermodynamic standard state at each temperature.
(c)
False.
G = H—TS. Although the conventional H is taken as zero at
25°C, S is zero at O K and is not zero at 25°C.
5.67
To find S°,59g from (5.29), we need to measure (a) Cp, of the solid from a
,,, (c) the melting point, and
very low T up to the melting point, (b) Afusf7
(d) Cp ,, for the liquid from the melting point to 298 K. If there are any solid—
solid phase transitions, we need AH ,, and the temperature of each such
transition. To find ArH 59, we need AH 59, of combustion of the hydrocarbon.
A,G° is found from AyH°— T A,S°.
63
Chapter 6
6.1
For a pure substance, G = nGm = np and AG =n Aun. For an isothermal process
in an ideal gas, Au = °(T) + RT In (P2/P°) — [w°(T) + RT In (P\/P*)] =
RT |n (P2/P,) and AG = nRT In (P2/P\) =
(3.00 mol)(8.314 J/mol-K)(400 K) In [(1.00 bar)/(2.00 bar)] = -6.92 kJ.
6.2
(a) T; (b) T; (c) T.
6.3
(a)
Py, = (0.440)(1767 torr) = 777 torr; Peo, = (0.310)(1767 torr)
Pi=xiP.
= 548 torr; Po = 442 torr.
Cy LVM Ree tO
K> Shh (bie ik
= (7771750)°/(548/750)°(442/750) = 3.41.
P pl
AG° =-RT In K*, = (8.314 J/mol-K)(1000 K) In 3.41 =
—10.2 kJ/mol.
(b)
Kp =(Pio,/Pso, Po, )P° = KeP° and Kp = Kp/P° = 3.41/(1 bar) =
3.41 bar’.
(GME
6.4
KSe Keo (RT CARP? oes =
3.41[(82.06 cm?-atm/mol-K)(1000 K)(1 mol/1000 cm?)/(0.987 atm)?
= 284.
A+B < 2C+3D. If 10.0 mmol of C is formed, then 5.0 mmol of A and
5.0 mmol of B must have reacted and (3/2)(10.0 mmol) = 15.0 mmol of D is
formed. At equilibrium, na = 10.0 mmol, ng = 13.0 mmol, nc = 10.0 mmol, np
= 15.0 mmol; mo. = 48.0 mmol.
Xp = 073125.
Pe=226 tom,
x, = na/nio = 0.208,
xg =0.271,
Px=xaP =(0.208)(1085 tom) = 220 tor,
Pp=339 torr,
xc =0.208,
Pe = 294 torn:
Kp= (Pc IP°)(Pp /P°)'/(P/P°)(Pp/P°) =
(226/750)*(339/750)°/(226/750)(294/750) = 0.0710.
AG®° =-RT In Kp=
—(8.314 J/mol-K)(600 K) In 0.0710 = 13.2 kJ/mol.
6.5
Not = PV/RT = [(231.2/760) atm](1055 cm*)/(82.06 cm?-atm/mol-K)(323.7 K)
= 0.01208 mol. Let x mol of Br react to reach equilibrium. At equilibrium,
64
nxo/mol = 0.01031 - 2x, ng, = 0.00440 —x, nnopr=2x, Mot =
0.01471 — x= 0.01208; x= 0.00263. So mxo = 0.00505, ng,, = 0.00177,
nnosr = 0.00526; xno = NNo/Mo = 0.418, Xp, = 0.1465, xNopr = 0.435.
Pxo = xnoP = (0.418)(231.2 torr) = 96.6 torr, Pp, = 33.9 torr, Pros: = 100.6
torr,
P°=750torr.
K> = (PyopiP°)/(PNo/P°) (Pp, /P°) = 24.0.
AG? =-RT In K* = (8.314 J/mol-K)(323.7 K) In 24.0 = -8.55 kJ/mol.
6.6
(Py/P°)'I (Py, /P°) = (0.12)"(720)(750) = 2.7 x 10°, which is less than K%.
The system is not at equilibrium; Pyig) and hence nyig) Must increase to reach
equilibrium.
6.7
Consider the reaction aA
_ Cre
Gale
+ bB
paw
ce
UenIay pe kee
=cC + dD.
Use of P; = x;P gives
Picea
CP gee
=
K, CEP
ay a
2)(2 - 3)3 - 4)(4 - 5) = 2880.
6.8
abe jU+l) =(-
6.9
(a) 120 (by SP areas
6.10
AG? = -RT In Ky = —(8.314 J/mol-K)(298.1 K) In 0.144 = 4.80 kJ/mol. If
(fer
(do Fe (eye
ee) er, eh
AH? is assumed constant over the range 25°C to 35°C, then Eq. (6.39) applies
and In(0.321/0.144) = [AH°/(8.314 J/mol-K)][(298.1 Kye 08 koe:
AH° = 61.2 kJ/mol. AG° = AH® —T AS®° =
4800 J/mol = 61200 J/mol — (298.1 K)AS®° and AS®° = 189 J/mol-K.
6.11
We plot In K> versus 1/T. The data are
InK,
-1.406
0.688
1.601
Phfsie')
T'/K'
0.002065
0.001873
0.00179,
0.00174,
65
= -11399x + 22.085
Cease
2.5
N
In Kp
1.5
l
OS
0
-0.5
-]
-1.5
{EYSCS TERYVc CE)
CAG
PAAAR
I
AAA
Ce
ARATE
Ame
0.0017
PN
SV TS PT
0.0018
TS
LET
Gt
0.0019
LY)
SV
PEE
0.0020
1_J
0.0021
Tie
The plot is very nearly linear with slope
[2.500 — (—1.500)]/(0.001718 — 0.002069)K' =-11400 K; -AH°/R =
—11400 K; AH® = (11400 K)(8.314 J/mol-K) = 94.3 kJ/mol = 22.6 kcal/mol in
this range of T. Then AG.,,
=— RT In K, = -(8.314 J/mol-K)(534 K) In 1.99
= —3.055 kJ/mol = —730 cal/mol. From AG® = AH® — T AS°, AS;,, =
(AH® — AG°)/T = (94800 + 3055)(J/mol)/(534 K) = 183 J/mol-K =
43.7 cal/mol-K.
(b)
AH° = 22.6 kcal/mol = 94. kJ/mol, as in (a). (The near linearity of the
plot shows that AH° is essentially constant over this range of T.)
AG574 =—-RT In K =-10.7 kJ/mol = -2.55 kcal/mol.
A S574 = (AH 534 — AG55,)/T= 183 J/mol-K = 43.8 cal/mol-K.
6.12
Appendix data give AG 59, = 37.2 kJ/mol.
AG° =-RT In K%. 37200 J/mol =
—(8.314 J/mol-K)(298.1 K) In Kp 99g and K > 599, = 3.0x 10°.
AE Rie
celui Kee (Tein K p(T) =(AH*/R)1/T> —1/T)).
dln K%,/dT=
In K apo. =
In (3.0 x 10°’) - [(87900 J/mol)/(8.134 J/mol-K)][(400 K)! — (298.1 K)"'} =
~5.99 and K% yop ~ 0.0025.
6.13
AG® = -RT In Kp = 72400 J/mol = ~(8.314 J/mol-K)(600 K) In Kp and Kp
= 4.9, x 107’ at 600 K. Integration of the van’t Hoff equation (6.36) with AH°
assumed constant gives In(Kp/K, )= el TINUE) AUT ERE
In[26/(4.97 x 10-’)] = -[(217900 J/mol)/(8.314 J/mol-K)][1/T2 — 1/(600 K)]
and T> = 1010 K.
66
AG° =-RTInK § = -RT[Ina+ bIn(T/K)—c/(T/K)]. Equation (6.36)
gives AH° = RT7dInK/dT = RT?[b/T +(cK)/T*] = bRT+ (cRK).
Then AS° =(AH°-AG°)/T =bR+ R{Ina+bIn(T/K)} and
AC? = dAH°/dT =bR.
(b) At 300 K, AH? = (8.314 J/mol-K)[(—1.304)(300 K) + 7307 K] =
57.5 kJ/mol. At 600 K, AH? = 54.2 kJ/mol.
6.14
(a)
6.15
Substitution of AH°(T) = AH°(T;)
+ ACp(Ti)(T— T)) into (6.37) gives
In[K p(T2)/K p(1)] =R! [? AH (7, \ieo die
(A/T, -/T,)+
dT = R'AH°(T,)
R ACT)? WT -T/T*)
Re AG;@) (nG5/T)) Phi:
For No04(g) — 2NO2(g), AG3og = 4.73 kJ/mol, In K > 293 = 1.908, and K >19g
609 =
= 0.148. AH io, = 57.2 kJ/mol; AC>..93 =-2.88 J/mol-K. In K >
1.908 + (57200 J/mol)(8.314 J/mol-K) '(1/298.1 — 1/600)K7! +
(8.314 J/mol-K) '(-2.88 J/mol-K)[In(600/298) + 298/600 — 1] =
Dos 2andeK yy
6.16
aloe lee
(a)
Substitute (6.28) into (6.14).
(b)
In[K>(T2)/K »(T1)] = In K p (72) — In K
p(T) =
~AH°(T>)/RT2 + AS°(T2)/R + AH°(T| WRT, — AS°(T)/R =
AH°(T,)R (A/T, — 1/T2) if we take AH°(T2) = AH®(T)) and eC
i=
AS C1):
6.17
(a)
When 7) in the AH® expression in Example 5.6 is replaced by T and this
expression is inserted into (6.37), the right side of (6.37) becomes
R'{? (That
T w+ pAb +3 AcT +1lAdT*)dT , where w =
AH; —T, Aa- 17? Ab - 17; Ac -+T,* Ad . After integration, we have
) -T,°)+
+4Ac(Ty
LAD(T, -T,
R-'[AaIn(T, /T,) + WT, |-Tz+')
1 Ad(T; -T;)]= In[K°(T,)/Kp(7;)) (Eq. (1)]. Let T, = 298.15 K. We
have
67
SUNS aon SINGop SIAR
eer 2(-137.168) — 0] kJ/mol =
we
~514382 J/mol and In K > 59, = 207.50. Using data from Example 5.6,
find w = —558488 J/mol for T; = 298.15 K. We can thus find Kol)
using the preceding Eq. (1).
Substitution in Eq. (1) of part (a) gives In K (1000 K) = 207.50 +
(1/8.3145)(-48.25 — 1314.69 +41.21 — 14.93 + 2.28) = 47.01 and
(b)
K°(1000 K) =3 x 10”.
6.18
Equation (5.19) gives
=AH;, +
f
T JdT’
(2
+(2g-c)
—b)T’a+
AH; =AH; +J7. (2e(2e-a\(T -T,)+(f -5b\T” -T7)+5(2g—c)(T° —T,’). Substitution in (6.37)
and integration gives In KU) hie Gadlidhae
RAH? + (a-2e)T + (1b-f)T) + 5(c- 2o\Te Ce Ty4)rt
R™'(2e — a) In (TyT1) + f-Lb)(T2 - THR + §(28 — Ty -T; WR.
6.19
From (6.25), In K, = In K> — (An/mol) In T + const. Differentiation and use of
(6.36) and (5.10) give
dinK; dinKp
dT
6.20
dT
seAn/mol _ AH® _ AnRT/mol _ AU®
je
Rie
RI
Ri
5
From (6.27), In K, = In K» — (An/mol) In (P/bar). Differentiation with respect
to T and use of (6.36) gives (0 In K,/0T)p = d In K p/dT = AH?°/RT”. Partial
differentiation with respect to P gives (since Kp is independent of P):
(0 In K,/OP)7 = —(An/mol)/P.
6.21
(a) T; (b) F.
6.22
Since P; = n,RT/V, the partial pressures are proportional to the moles. Let
zV/RT moles of CO react. The equilibrium partial pressures are then Pco =
342 torr—2,
Fo, = 351.4 torr—z,
Pooc, =z. Thus
68
439.5 torr = 342.0 torr —z + 351.4 torr — z +z and z = 253.9 torr.
Kp= (253.9/750)/(88.1/750)(97.5/750) = 22.2.
6.23
AG°/(kJ/mol) = —394.359 + (—879) — 2(-619.2) =-35.
In K,
=—AG°/RT =
(35000 J/mol)/(8.314 J/mol-K)(298 K) In Kp = 14.1,
Kp =1.3 x 10°.
Let 2z moles of COF; react. The equilibrium amounts are n(COF>) = | — 2z,
n(CO2) =1+2z,
n(CF,4) =z. Because Kp is so large, we have 2z = 1 and z =
Vy, Hence, n(CO>) = 1.50 mol, n(CF4) = 0.50 mol, and
ye
1.3 x 10
6.24
tear reid WestieVed
ae)
_
n(COF2) =7 x 10~ mol
[n(GOE,
)/rice li
In K> =-AG°YRT= 1.258. K% = 0.284.
(a)
Let z moles of A react. The equilibrium amounts are na = 1 -z,
2c USC Oftias— \eives
2 e= 22, Np = oe, alld Mente
Mat
0.284=
TET T EY ss
l-z
ees
2(1+ z)
P;
227 1200
750
0.533 = sae 2a
(l1—z7)? \750
3339 Zz2 0333 1,
4 eos
Both times we took the positive square root, since z and | — z’ are
positive.
na = 0.622 mol =ng,
nc = 0.756 mol = np.
(b) The equilibrium amounts are na = 1-—z,
have
0.284 =
np= 2
Noe 2 Samp eWwe
[22/(3 + 22)]* (PIP*)*
[((Q— zB + 2z)][2- zVGB+ 2z)]
0.00693, — 24/1 — 2)(2 - 2)(3 + 22) =0
z lies between 0 and 1, and trial and error gives the desired root as z = 0.530.
Hence na = 0.470, ng = 1.470, nc = 1.060 = np.
6.25
Appendix AyG®° data give AG 39, = —2.928 kJ/mol. In Kp =-AG°/RT=
1.18lfand K, =3.26. Kp = [P(HD)/P°]*/[P(H2)/P° ][P(D2)/P?] =
69
[P(HD)]°/P(H2)P(D2) = [n(HD)RT/V}"/[n(H2)RT/V) [n(D2)RT/V] =
(n(HD)]?/n(H>)n(D>). Let x mol of H2 react to reach equilibrium. The
equilibrium amounts are then n(H2) = (0.300 — x)mol, n(D2) = (0.100 — x)mol,
n(HD) = 2x mol. Substitution in the Kp expression gives after simplification:
0.227x? + 0.400x — 0.0300 = 0. The quadratic formula gives x = 0.072 and
_1.83. Since we started with no HD, x must be positive. So x = 0.072, and
n(H>) = (0.300 — x)mol = 0.228 mol, n(D2) = 0.028 mol, n(HD) = 0.144 mol.
6.26
(a) K%, = [P(NH3)/P°}/[P(N2)/P°}LP(H2)P°P = P(NH3)(P°)"/P(N2)[P(H2))
Use of P; = n,RT/V gives K‘> = [n(NH3)"/n(No)n(Ho) \(P°V/RT)
Now K°,(RT/P°V) =
36[(82.06 cm?-atm/mol-K)(400 K)/(750/760)atm(2000 cm’)]° = 9960
mol. Let x mol of N2 react to reach equilibrium. The equilibrium
amounts are (0.100 — x) mol of N2, (0.300 — 3x) mol of H2, and 2x mol of
NH3. So 9960 = (2x)/(0.100 — x)(0.300 — 3x)? = 4x7/3°(0.1 — x)(0.1 — x)
= 4x’/27(0.100 — x)’. Taking the square root of both sides, we get +99.8 =
0.385x/(0.100 — x)’. Since x must be positive (we started with no NH3),
the negative sign is rejected and 0.01 — 0.2x + x’ = 0.00386x. The
quadratic formula gives x = 0.082 (the other root exceeds 0.1, which is
impossible). Then n(N2) = 0.018 mol, n(H2) = 0.054 mol, and n(NH3) =
0.164 mol.
(b)
At equilibrium, n(N2)/mol = 6.200 — x, n(H2)/mol = 0.300 — 3x, and
n(NH3)/mol = 0.100 + 2x. Then
9960 = (0.100 + 2x)7/(0.200 — x)(0.300 — 3x)° = W. To make W equal
9960, 3x must be close to 0.3. For an initial guess of x = 0.09, we find W
= 26400, which is too big. For x = 0.08,
W = 2600. Repeated trial and
error (or use of the Solver in a spreadsheet) gives x = 0.0865. Then n(N2)
= 0.113 mol, n(H2) = 0.0405 mol, n(NH3) = 0.273 mol.
6.27
The equilibrium amounts are ny, = 4.50 mol — E, ny, = 4.20 mol — 3&,
Nyy, = 1.00 mol + 2€. The requirement that each of these amounts be positive
gives €< 4.50 mol, & < 1.4 mol, § >-0.5 mol. So -0.5 mol < € < 1.4 mol.
Hence ny, lies between 4.50 mol — (—0.50 mol) and 4.50 mol — 1.4 mol; that
is, 3.1 mol < ny, <5.0 mol. Also, 0 < ny <1.7 mol and 0 < Nyy, < 3.8 mol.
70
6.28
(a)
Use of P, =x,P gives Kp =(Py/P°)’ (P,/P°) = P2/P,P° =
CrP) aleon PP: =i (xen =e CP) Po
(b)
With this definition of z, the equation in (a) becomes
z = = /(1—xpg), so
xg + 2g — 2 =0. The quadratic formula gives
xg =[-z+(z? +4z)'/*]/2
[Eq. (1)], where the negative root was
discarded since xg must be positive.
(ec)
z2=49.3(1 bar)/(1.50 bar) = 32.9. Substitution in Eq. (1) of part (b) gives
Xp = 2x0 = 0:97 Lisa Xo, = I — 0.971 = 0.029:
(d)
AGS, =4.73 kJ/mol = —RTInK@ and K° =0.148. Soz=
0.148(750 torr)/[(2 x 760 torr) = 0.0730. Equation (1) of part (b) gives xg
= Xyo, = 0.236, 80 Xy,o, = 0.764
6.29
K+, = [P(SO3)/P°]°/[P(SO2)/P°}"[P(O2)/P°] =
{[n(SO3)]°/[n(SO>)]’n(O2)}(P°V/RT), where P; = n)RT/V was used. We find
P°VIRT = 0.00222 mol, so [n(SO2)}/[n(SO2)]*n(O2) = 1540 mol’. Calculation
with the initial mole numbers shows that to reach equilibrium, more SO3 must
be formed. Let x moles of O react to reach equilibrium. The equilibrium mole
numbers are n(SO2)/mol = 0.00265 — 2x, n(O2)/mol = 0.00310 — x, and n(SO3)
= 0.00144 + 2x. Then 1540 = (0.00144 + 2x)°/(0.00265 — 2x)7(0.00310 — x) =
W. An initial guess of x= 0.001 gives W = 13340, which is too big. Repeated
trial and error (or use of the Solver in a spreadsheet) gives x = 0.000632. So
n(SO>) = 0.00139 mol, n(O2) = 0.00247 mol, and n(SO3) = 0.00270 mol.
6.30
xa = (1.000 mol/6.000 mol) = 0.1667, xg =0.5000, xc = 0.3333. Pa=
xaP = 0.1667(1.000 bar) = 0.1667 bar, Pg =0.5000 bar, Pc = 0.3333 bar.
K*> = 0.3333/(0.1667 x 0.5000) = 4.000. Let z moles of A react to reach the
new equilibrium position at P = 2.000 bar. For this equilibrium, a = | — z,
ng =3-Z, nc=2+zand
x, =(1 —z)/(6—2Z), xp = (3 —2)/(6—-2), xc=
(2+z)/(6-z).
K*, = 4.00 =
Pa=xaP = [(1—z)/(6-z)](2,bar),
Pp =etc.
(24 ZO
az
[d= z)/(6 — z) 213 —z)/(6— z)I2
92 —36z+12=0,
z= 0.367 and 3.63.
We have na = | —z, so z cannot exceed 1. Therefore z = 0.367.
na = 0.633 mol, ng = 2.633 mol, nc = 2.367 mol.
7I
6.31
AG 39g /(kJ/mol) = -267.8 + 0 — (305.0) = 37.2. AG? =-RT In Kp.
37200 Jmol = (8.314 J/mol-K)(298 K) Inkk fags MK pag 13 01x 105
AH 39g /(kJ/mol) = 287.0 + 0 — (-374.9) = 87.9. Eq. (6.39) gives In Kp.soo *
In (3.0 x 1077) + [(87900 J/mol)/(8.314 J/mol-K)][(298 Ky! - (S00 K)"'] =
0.699.
Kp599 = 0.50. The mole fractions are Xpc), = X¢), = ¥ and Xpc), =
L ~ Xpcy, — Xe), = 1 - 2x. We have Kpso9 = 0.50 = (xPIP°)I[(1 — 2x)P/P°] =
x/(1 — 2x), since P = P° = 1 bar. x = 0.50 — 1.00x, and the quadratic
formula gives x = 0.365 = pc, = Xc1,> Xpci, = 1 - 2x = 0.263.
6.32
(a)
Let No +3H2 ce 2NH; be called reaction I. Since reaction (a) has its
coefficients equal to one-half those of reaction I, we have K pa) =
(Kies he = (36)? = 6)
(De Keer = Uke 0028,
6.33
For n > iso, AG j999 = —670 cal/mol and we find K > 1000 = 1.40 = Kpya, where
A = n-pentane, B = isopentane, C = neopentane. For n — neo,
AGjo99 = 4900 cal/mol and Kp jo99 = 9.085 = Kea. Eqs. (6.45) and (6.46) give
Xa =Xn = W/(1 + 1.40 + 0.085) = 0.40,
xp = Xiso = 1.40/2.485 = 0.56,
XC = Xneo = 0.085/2.485 = 0.034.
6.34
AG 600 /(kJ/mol) = 1059.72 + 0 — 71.74 = 987.98.
In K >6099 = (987980 J/mol-K)/(8.314 J/mol-K)(6000 K) = -19.804 and
IKLei ORNs
6.35
AG° =-RT
In K>; AG wrong =-RT In K, P, wrong
fe]
*
AG° — AG‘ wrong =
2500 J/mol = —RT In (Kp/K ;P, wrong ). We get In (K'p/K
p wrong) = —1.00 and
Kp wrong/Kp = 2.7. The error is a factor of 2.7.
72
6.36
(a)
L(g) e 2I(g). Let n* be the number of I, moles before dissociation, and
let z moles of Ip dissociate to give 2z moles of I, leaving n* — z moles of
I,. The total number of moles at equilibrium is n* — z + 2z =n* + z. Then
x; = 2v/(n* + z) and ci (n* — z)/(n* + z). We have n* = P*V/RT and n*
+ z= PV/RT, so z= PV/RT — n* = (P — P*)V/RT. Use of these
expressions for z, n* + z and n* gives x; = [2(P — P*)V/RT]/(PV/RT) =
2(P — P*)/P and x1, = [P*V/RT — (P — P*)V/RT]/(PV/RT) = (2P* — P)/P.
(Dek
(P/P°)?/( xX), PIPoy= |2(R— P*)/P°}/{(2P* = P)/P?)| =
4(P — P*)’/(2P* — P)P?°.
(c)
Weuse the result of (b) to calculate the following Kp values:
K p973 = 4(0.0624 — 0.0576)7/[2(0.0576) — 0.0624]0.987 = 0.00177;
K
P1073
OO
2s
Kp.1173
=
K
0.0493;
pi
=
Ou
25
We plot In K vs. 1/T. The data are
InK,
-6.34
OE
OO
-+4.49
-3.01
—-1.757
Ee
TI
TS
One finds a straight line with slope
[—1.19 — (-6.73)]/(0.00075 — 0.00105)K” =-18800 K. Eq. (6.40) gives
AH? = (18800 K)(8.314 J/mol-K) = 156 kJ/mol = 37.4 kcal/mol.
y = -18850x + 13.054
>
7/.
te SF
cee
0.00075
eal a
|
pe PR NE
tt tin Ns es (Ses Hl
0.00095
0.00085
a
a
0.00105
ik
6.37
=11,(P/bar)"' /(P/atm)”* =
K°/K*™ = (11,(P/bar)” VII, (P/a]tm)”
II; (atm/bar)”' . For aA + bB Rese
ies dD,
|]; (atm/bar)”' = (atrn/bar)*4
(atm/bar)*”"™*! = (760 torr/750.062 torr)” =
73
(1.01325)""™.
S
6.38
At equilibnum, ny, /mol= 1 — x, ny, /mol = 3 — 3x, Nyy,/mol = 2x;
mol/mol = 4 = 2x. Py, = Xy, P= (My, /mor)P = [(1 — 2/4 - 2x))P,
Py, = (BU -x)/(4-2x)|P, Py, = [2x/(4 - 2x)]P. Kp =
a
(Pap Po) (Pam pey( Py (P°), = (P°y' P(Aeon) Ox)
(P°/P)*(16/27)(2 - xyx7/(1 - x)’ =K{.
(lx) =
x must be between 0 and 1. Taking the
positive square root of each side gives (P°/P)(4/27")(2 — x)x/(1 - x) =4 4 a 2
or 4(2x — x2/(1 — x)? = (27K>)"? (P/P®) = s. Hence 8x — 4x? = s — 2sx +.x’s and
(4 +s)x* — (25 + 8)x+s=O0or x’ — 2x + s/(s + 4) = 0. The quadratic formula
gives x= {2 +[4—4s/(s +4)]'7}/2 = 1 + [1 — ss + 4)]’”. Since x is less than
1, we take the minus sign: x = 1 — [1 -s/(s + 4)]'”.
6.39
(a) No;
(b) no;
(e) no;
(c) no; (d) no;
(f) no;
(g) yes;
(h) no;
(i) yes.
6.40
(b).
6.41
(a)
See Fig. 6.11. results for n(CH4)/mol in order of increasing pressure are
0.00061, 0.053, 0.690, 1.355, 1.515, 1.62, 1.71.
(b)
The NIST-JANAF tables give at 1200 K:
AGA = -77.92 kJ/mol,
AG? =-74.77 kJ/mol, K; = 2.46 x 10° and K> = 1.80 x 10°. One changes the
values of T, P, K; and K> in the Fig. 6.9 spreadsheet. With the guessed values
of the equilibrium amounts taken as equal to the starting amounts, the Solver
might not find a solution. The large values of K; and K2 indicate that one or
both of the reactants CH, and H20 are present in small amounts at equilibrium.
The 0.01 bar 900 K equilibrium amounts in Fig. 6.10 have small amounts of
the reactants, so it is a good idea to use the 0.01 bar 900 K equilibrium
amounts as the initial guess for the 1200 K problem.
6.42
Suppose one equilibrium constant K; is very small with the value 1 x 10°. If
the Solver changes the mole numbers and finds a calculated value of 8 x 10°
for K,, then the calculated and true values of K, differ by 7 x 10; which is
less than the Solver’s default precision of 1 x 10m and the Solver will declare
74
it has found a solution, even though the calculated K is 8 times the true K and
at least some mole numbers must be greatly in error.
6.43
Let € additional moles of CO be formed when reaction 1 reaches equilibrium
and let €; additional moles of CO2 be formed when reaction 2 reaches
equilibrium. The equilibrium amounts are then n(CHy) = 1 — & — &,
n(H2O) = 1 — €; — 2&, n(CO2)=1+&, n(CO)=2+ &,
n(H2) = 1 + 3&; + 4& (where the unit mol is omitted). The conditions that each
of these amounts be positive give the inequalities: (I) 1>€, +€,;
(lilacs mnLeon
le (LVOaGy
emmy) Scat ac. > =le
Addition of (I) and (III) gives 2 > €, (VI). Addition of (I) and (IV) gives
E, <1.5 (VI). From (I), (IV), (VD, and (VII), we have — 2G, <2, and
~1 <& < 1.5, which can be used to help set limits on the mole numbers.
6.44
(a)
CO occurs only in reaction 1 and CO; occurs only in reaction 2.
Therefore, we can use the changes in amounts of CO and CO; to find the
extents of reaction. CO goes from an initial amount of 2 mol to 3.3236
mol at equilibrium and has a stoichiometric coefficient of 1. Therefore &
= 1.3236. CO goes from 1 mol to 0.6758 mol, so & = -0.3242 mol.
(b)
Now CHy is the only species that occurs only in reaction |. CH, has
stoichiometric coefficient —1 and goes from | mol to 0.00061 mol, so E
= 0.99939 mol, which is quite different than in (a).
6.45
(a) (1) N> < 2N; (2) O & 20; (3) No +02 © 2NO. The reaction
N +Op — NO+Ois
N +0 © NO is—1(1)— 4(2)+ 4(3). The reaction
Seo).
1
6.46
]
|
(b)
Check your results against the mole fractions at 4000 K in Fig. 6.5.
(a)
dinK>,/dT= AH°/RT*. Appendix data give AH® = 87.9 kJ/mol > 0,
so K% decreases as T decreases and the equilibrium shifts to the left.
(b)
As V decreases at constant T, the pressure increases and the equilibrium
shifts to the side with fewer moles of gas, i.e., it shifts to the left.
75
(c)
This removal decreases
P,,,_ , So to restore equilibrium, the reaction must
5
shift to the left.
(d)
Constant T and V addition of He does not affect the partial pressures of
PCls, PCl3, or Clr, and there is no shift.
(e)
To keep P constant as He(g) is added, V must be increased. Since
P; = n;RT/V, this volume increase will decrease each partial pressure P;
by the same percentage. Since the reaction has more moles of products
than of reactants, the numerator of the reaction quotient Qp will decrease
more than the denominator. Therefore the equilibrium will shift to the
right to make Qp again equal to Kp.
6.47
From Prob. 6.19, d In K(/dT = AU°/RT *- if AU® is positive, then K~ increases
as T increases. Since c; = nj/V and V is held fixed, the mole numbers n; undergo
changes proportional to the changes in the concentrations, and the equilibrium
shifts to the right if AU® > 0.
6.48
(a)
In QO, = In (T]i (x)” ] = Zi In)“ = Ev; In x; = Z; Vi In (1 Mor) =
Ev; Inn; — (2; Vi) In (nj + n2 +--+: +). We have O In Q,/dn; =
vn; — AnI(nj + nz + - ~~) mol = (vj — xj An/mol)/nj.
(b)
Case I: suppose substance j is a reactant. Then v; is negative. To get a
shift to the left, we want 0Q,/dn; > 0, so that addition ofjwill make Q, >
K,. We thus want v; — x; An/mol > 0 and v; > x; An/mol. Since v; is
negative, An must be negative. Division of the inequality by the negative
quantity An reverses its direction to give vj/(An/mol) < x;. The fact that
An is negative means that the total moles of reactants exceeds the total
moles of products, in agreement with condition (1). Case II: suppose
substancej is a product. Then v; is positive. We want 0Q,/dn; < 0, so that
addition ofjwill make Q, < K, and shift the equilibrium to the nght. We
thus want v; — x,An/mol < 0 or v; < xjAn/mol. v; 1s positive, so An must be
positive and division by An gives x; > vj/(An/mol). The fact that An is
positive means that the product total moles exceed the reactant total
moles.
(c)
The left side of the reaction has the greater sum of the coefficients, so
constant-7-and-P addition of NH3 will never produce more NHs3. For N>,
condition (2) is Xn, > (—1)/(-2) = %, so for Xn, > 0.5, addition of N> at
76
constant T and P produces more N>. For Ho, condition (2) is
Xy,> (—3)/(—2) = 1.5, which can never be satisfied.
6.49
(a)
K,= (1/5)7/(3/5)(1/5)° = 8.333. If x is the extent of reaction to reach
the new equilibrium position, the new equilibrium amounts are n(N2) =
3.1 — x, n(H2) = 1 — 3x, n(NH3) =1 + 2x and
*
[(1+ 2x) (5.1-2x)]?
[3.1-—x)/(5.1-—2x)][0 — 3x) /(5.1- 2x)}°
The condition that the mole numbers be positive gives —0.5 < x < 0.333.
Use of the Solver gives x = -0.0005438, so n(N2) = 3.1005438, riot =
5.1010876, x(N2) = 0.607820.
(b)
K,= (0.4)? /(0.2)(0.4)° = 12.5. Ifx is the extent of reaction to reach the
new equilibrium position, the new equilibrium amounts are
n(N>) = 14-—x, n(H2) = 4 —-3x, n(NH3) = 4 + 2x and
0
[(4 + 2x) /(20-2x)]’
~~ [02 — x) /(20 — 2x)][(4 - 3x) (20 - 2x)?
The condition that the mole numbers be positive gives —2 < x < 1.333.
Use of the Solver gives x = 0.12608, so n(N2) = 11.8739, not = 19.7478,
x(N2) = 0.6013.
6.50
The reaction rate at lower temperatures is too slow to make the reaction
economically practical.
6.51
AG° =-RTIn Kp =—2.303RT log Kp =
~2.303(8.314 J/mol-K)(500 K)[7.55 — (4830 K)/(500 K)] = 20.2 kJ/mol.
din K3/dT =AH°RT. AH? = RT din Kp/dT =2.303RT d log Kp/dT=
2.303RT?(4830 K)/T? = 2.303(8.314 J/mol-K)(4830 K) = 92.5 kJ/mol.
AS° = (AH? — AG°T = (72.3 kJ/mol)/(500 K) = 144.6 J/mol-K.
Cp =(0H/OT)p. ACp = (0 AH°/0T)p = 0.
6.52
AG* refers to a change from pure, separated reactants, each in its standard
state, to pure, separated products, each in its standard state. AG® is not the
change in G that occurs in the actual reaction mixture. The reacting system
does not contain substances in their standard states.
77
6.53
K>, is a function of T only. Therefore only (d) will change Kp.
6.54
(a)
only and are
AH° = ;v;H;>, ,.Enthalpies of ideal gases depend on T
(b)
unaffected by pressure changes or by mixing with other ideal gases; so
the observed AH per mole of reaction does equal the reaction's AH”.
Entropies of ideal gases depend strongly on P and the gases in the
reaction mixture are not at | bar partial pressures. Therefore, the
entropies in the mixture do not equal the standard-state entropies and AS
per mole of reaction differs substantially from AS®.
(c)
6.55
Since AS, # AS®, it follows that AG, # AG®.
Let the superscripts 750 and 1000 denote standard states based on 750 torr and
on 1000 torr, respectively.
Kee
Ko ed =[(Pyo, POPU
(CA TE
code
oe 3d a
Sty
Cay
NEN:
(The gases are unaware of what choice of standard state has made, so the
equilibrium partial-pressure ratio Pyo,/Py,o, 18 independent of the standard
state choice and cancels.) The Appendix AyG° data give A;G’7 =
4.73 kJ/mol and use of AyG*'° =-RT In K3"° gives Kp” = 0.14. Then
K%'°° = (750/1000)0.14g = 0.11).
6.56
(a)
Since AH° <0, the relation d In K,/dT = AH°/RT” shows that Kee
decreases as T increases; thus the equilibrium amount of the cis isomer
increases as T increases.
(b)
We have AG° = AH® — T AS° = — T AS° in the high-T limit. Since
AS° > 0, we have AG° < 0 at high T and K» > 1 at high T. Thus the highT limit has more trans isomer than cis isomer.
(c)
Even though the percentage of cis isomer continually increases as T
increases, it is still possible to have more
% cis
trans than cis at high T. In the graph at the
right, the % cis continually increases with
78
T but always remains below its asymptotic T = © limit of 43%.
(d)
AG° = AH° —-TAS°, where AS° and AH® are constant. In this equation,
the coefficient —AS° of T is negative, so AG® decreases as T increases. As
noted in (a), K » decreases as T increases. We have AG°/T = — R In Kp,
and since K > is decreasing as T increases, AG°/T is increasing as T
increases, which is the opposite behavior as shown by AG*. (If this seems
puzzling, note that AG° is negative.)
(e)
Yes. For the reverse reaction, trans — cis, the results of (d) show that
AG? increases and K> increases as T increases. (The behavior of Kp is
determined not by AG° but by AG°/T.)
6.57
0 = (d/dm) Dj (i — mx; — bY” = Di AHx)Oi— mx — b) =
2) xpi t+ 2m Dux, +2b xs
[b=Ouxoi-m pox
|
0 = (d/db) Xi (y; — mx — bY? = -Li (i — mx; — b) = —Li yi + m Lixi+ nb, since
”,b=nb; thus [b= (Liyi—m Lx,)/n }. Equating the two bracketed
equations for b and solving the resulting equation for m, we find
m = (n Dixyi- Lixi Di yd/(n plea — (>; x;)’]. Substituting this expression for m
into b = (Yi yj —m L;x;))/n and using a common denominator for the terms on
the right, we obtain the equation given in the text.
6.58
(a)
nj — Ni,0 = An; = vie, SO Nieq = Nio + vit.
ny, =3 mol —3§
(b)
nn, =
1 mol—- Ge
My, = es
G=DinG,, (7, Pd = Li uw; (T, Pi) = Lini{w; (T) + RT In (P;/P°)), where
the sum goes over N2, H2, and NH3, the n;’S are given in (a), and the P;’s
are: Py = [(3 mol — 3&)/(4 mol — 2&)]P,
Pyu, = [2E/(4 mol — 2&)]P,
Py, = (ny, /Mo)P = [(1 mol — &)/(4 mol - 2§)]P. H= Lin, ; (T,Pi).
Since H of an ideal gas is independent of P, H ehe= Heesand
H = YinjH*, ;, where the n,’s are given in (a).
(c)
Results at € = 0, 0.2, 0.3, 0.4, 0.6, 0.8, 1.0 are: G/(kJ/mol) = 284.73,
288.45, -289.06, -289.20, -288.17, -286.07, -277.21;
Hi(kJ/mol) = 23.55, 3.60, -6.37, -16.34, -36.29, -56.23, -76.18;
TSI(kJ/mol) = 308.28, 292.05, 282.69, 272.86, 251.88, 228.83, 201.03.
79
10 HN =5.91
38 PN = ((1—X)/(4-2*X))*P
12 HH = 5.88
14 HA =-38.09
16 G1 =-97.46
18 G2 =-66.99
20 G3 =-144.37
22 T =500
24 R=8.314E-3
26 PO=1
28 P=4
30 FOR X=0TO1 STEP 0.1
32 NN=1-X
34 NH =3-3*X
36 NA =2*X
40 PH = ((3-3*X)/(4—2*X))*P
42 PA = (2*X/(4—2*X))*P
44 IF PA =0 THEN PA= 1E-7
46 IF PN = 0 THEN PN = 1E-7
48 IF PH =0 THEN PH = 1E-7
50 GN = G1 + R*T*LOG(PN/PO)
52 GH = G2 + R*T*LOG(PH/PO)
54 GA = G3 + R*T*LOG(PA/PO)
56 G=NN*GN + NH*GH + NA*GA
58 H = NN*HN + NH*HH + NA*HA
60 TS=H-G
62 PRINT “Xl=”";X;“G=”;G;
“H=:H;“TS=";TS
64 NEXT X
66 END
6.59
(a)
Any reaction with An = 0; e.g., H2(g) + Clo(g)
— 2HCI(g).
(b)
From (6.36), to have K, independent of T requires that AH° = 0. Two
mirror-image species have the same energy, so AH®
d-CHFCIBr(g)
is zero for
— /-CHFCIBr(g). [Actually, because of the
nonconservation of parity, two mirror-image molecules have an
extremely tiny, experimentally undetectable energy difference. ]
6.60
(a)
The system discussed near the end of Sec. 6.6 with ny, = 3.00 mol,
ny, = 1.00 mol, nyy, = 1.00 mol has xy, = 3.00/5.00 = 0.600. If the
reaction N2 + 3H2 «— 2NH; shifts slightly to the right with 0.100 mol of
Nz being consumed, the new amounts are ny, = 2.90 mol, m,, = 0.70
mol, nyy, = 1.20 mol, and xy, = 2.90/4.80 = 0.604. Thus x, has
increased even though ny, has decreased.
(b)
Xj =NINo
ax; = dnjlnio — (nin ” ) dno. We have dn; = v; d& and dnyot =
d ND; yi ay dn; = pa Vj dE = (dj vj) dé = (An/mol) dé. SOx,
(Vino) AE — (xi/ntor)(An/mol) d& = n« [v; — x(An/mol)] dE.
80
=
6.61
Rodolpho is nght. The equations cited by Mimi show that
(d/dT)(RT \n Kp) =AS°, so that the sign of AS° determines whether T In Ke
increases or decreases as T increases, but the sign of (d/dT)(T In Kp) can
differ from the sign of (d/dT)(In Kp)
6.62
(b).
6.63
(a) False.
(b) False; G is minimized only if the system is held at constant T and P.
(c) True; see Eq. (6.4).
(d) False; see Sec. 6.6.
(f) True; see Fig. 6.8.
(g) True.
(i) False.
(e) False; See Fig. 6.8.
(h) False; this is true only if An = 0.
() True.
(k) True; from (6.13) and (6.19), K, depends on P® but Kp does not.
(1) False.
Reminder:
(m) False;
(n) True.
Use the solutions manual as an incentive to work
problems, not as a way to avoid working problems.
8|
Chapter 7
7.1
(a) F; (b) F.
thy?
(a)
p= 1,r=0,a=0;f=c-p+2-r—a=
c= 2 (water and sucrose),
The degrees of freedom are T, P, and sucrose mole fraction.
(b)meoseepa=l7=0,a=0;
ne
f=4 The degrees of freedom are 7, P, Xsucrose;
Xribose-
(c)
c=3 (sucrose, ribose, water), p = 2 (solid sucrose, the solution);
f=3-24+2-0-0=3(7,P, Xribose). Note that Xsucrose in the solution
saturated with sucrose is fixed.
(d)
c= 3 (sucrose, ribose, water), p = 3 (solid sucrose, solid ribose, the
solution); f=3-—3+2-0-0=2 (T and P).
(e)
c=1 (water), p =2;f=1-—-2+2-0-0=1. The degree of freedom can
be taken as either T or P. (Once T is fixed, P is fixed at the vapor
pressure of water.)
(f)
c=2
[T and Xsuctose (or P
p= 2; f=2-2+2-0-0=2
(water, sucrose),
ANGE encrace)|:
Te)
(g)
c=2,p=3;f=2-3+2-0-0=1
(either T or P).
(h)
c= 2, p—3%f=
2 3-4 2 = (either Fore).
(a)
f=Cina—p +2 and p = Cina + 2 —f. The smallest possible fis zero, SO Pmax
= Cd +2:
(b)
7.4
If
p= 10, then cing must be at least 8.
(q)erLO 2 Hi + Ole HPO, © Ht LPOseE Osea
pi apd
BPO 7a
aha PO}
LoHO > aie
=
s
Bat
EPO, aaee
HPO; ;
teeta
ee EO, ae
. The equilibrium conditions are:
My por: = Hye + Pypo2- > Uypoz = Myr + Hpo:-- Phe electroneutrality
condition is Xi
=X a SX XPO;
+ 2X poz
+ 3x
=
PO,
. There are 7
species (H2O, H*, OH’, H3PO4, Hx2PO;, HPO 4 , PO} ) and Eg. (7.10)
gives
Cing = 7
—-4-—1=2.
Hence f=2—1+253.
82
(b)
X,+ = X,- and Ay = x, (assuming that no solid precipitates out of
the solution). We shall neglect the ionization of water. (Its inclusion
would not change
f.) The electroneutrality relation is Xe t+ Xea
x
Br
tp
]
=
and is not an additional condition since it follows from the
two preceding equations. The species are K, Br, Na‘, Cl’, and H20.
Equation (7.10) gives cing = 5 -O-— 2 =3. Hence
Ticks)
(a)
Cing =
3-0-0 =3 and
f=3-—1+2=4.
f=3-1+2=4.
A reasonable choice is T, P, and
two of the mole fractions.
(b)
2NH3 < N2 + 3H2, $0 Hy, +34y, =2Myn, and r= 1. Hence cing=
Ble
0 = 2and f=2 — l=
3
and
one mole fraction. (the other
two mole fractions are determined by the reaction equilibrium condition
and yy x, =1.
2Uyn, = Hn, + 3H, and xy, = 3xy, (since all the N2 and H2 come from
decomposition of NH3). Hence cing = 3-1-1=1
andf=1—1+2=2.
[In (b) and (c), the catalyst was not counted in finding c. If the catalyst is
considered to be a species whose amount can be varied, then c and Cina
are each increased by 1. If the catalyst is a solid, then p is increased by |
f isunchanged; if the catalyst is in the gas phase, then fis increased
and
by 1.] T and P.
(d)
We have r= 1. The mole fractions satisfy xy, +X, =1 and the condition
that all the
N come from N> does not give an additional relation between
mole fractions’ fimc— p+2—r-a=—2-14+2=)-0=2-
(e)
We have the reaction equilibrium CaCO3(s)
T and P.
ie CaO(s) + CO2(g) and the
phase equilibria CaCO3(s) = CaCO3(g) and CaO(s) = CaO(g). There is
1 reaction equilibrium condition. The phase equilibrium conditions have
already been taken into account in deriving the phase rule; see Eqs.
(7.3)-(7.6). There are no relations between mole fractions. There are
three species (CaCO3, CaO, COz) and Cina = 3 —1-—0=2. Hence
fe 2538-2
'S) lerT.
83
7.6
Weshave the teactionsI) HON 2H’
1CNeand (2H. Ops Hf Ole
Let n, ,,.and n, ,,, denote the moles of H coming from reactions | and 2,
respectively. Stoichiometry gives 1, 44+ = "Non- and n, \+ = Noy > $0
Ua
we As
(a)
c=2 (HO
(b)
c=4(H,O, NaCl, Na’, Cl), p =2; r= 1, since we have the equilibnium
H™
=n
= ip
Non-
1,H* “s do H*
Hg
and NaCl),p = 2,f=2-2+2-0-0=2
(T and P).
NaCl(s) a Na*(aq) + CI (aq); a = 1, since we have the electroneutrality
condition x Nav Gq) ae. Cla(aqy So f=4-2+2-1-1=2.
7.8
(a)
c=4,p=1,r=2 [HO < H*+OH and 2H,0 © (H)0)], anda= 1
(4
Cadi,
(b)
(a)
adie):
di
c=5,p=1,r=3 [the reaction in (a) and H.O + (HO);— (H20)s],
(= NO
1.9
=)
r-a=4 =) 4-2-7
= Xoy- ef =O -P+2—
1,
Here, the equilibrium conditions P® = p? = P’ =..- are eliminated.
Instead of specifying 1 pressure, we must specify p pressures (where p 1s
the number of phases). This increases
fby p — 1, sof=cina—D+2+p—1
= Cig tel.
(b)
Here, the equilibrium conditions of Eqs. (7.3) to (7.6) are eliminated.
There are c(p — 1) such conditions, so fis increased by c(p — 1) and f=
c-p+2-r-a+cp—-c=cp-pt+2-r—-—a=Cn—pt+t2+cp—c.
7.10
(aya
Gent
(a) Liquid;
7.12
(a)
Ub) To otc) Ea (d) bo (e)
(b) gas;
(c) liquid;
ys
(d) liquid;
ato)
(hh) ste)
(e) gas.
c=1,p=2;f=1-—2+2—-0-0=
1, as in Prob. 7.26.
(D)tec—ily
pi ls feeee
(c) c=)
84
pee
eee
OO)
7.13
(a)
Treat the vapor as an ideal gas. The liquid’s volume is negligible
compared to the container’s volume. If the equilibrium vapor pressure is
reached before all the liquid vaporizes, then the gas will be at 23.76 torr,
and
(23.76/760) atm (10000 cm°)
bE Nowe
= 0.01278 mol
BS RT (82.06 cm>- atm/mok K)(298.15 K)
Meas = (0.01278 mol)(18.015 g/mol) = 0.230 g
AV
There are 0.230 g of vapor and 0.130 g of liquid.
(b)
With V = 20000 cm’, we would get 0.460 g of vapor if the equilibrium
vapor pressure of 23.76 torr were reached before all the liquid vaporized.
However, there is only 0.360 g of water present initially, so all the liquid
vaporizes to give a system with only vapor present.
7.14
(a) Gas;
P/ atm
7:
i LOR
ae repepeticne
pemn
ed
7.15
7.16
TA/
E
s
(b) solid;
(d) liquid;
(e) solid;
(f) liquid;
(g) gas.
(c) gas;
°
Isothermally and reversibly increase the pressure to the pressure at which ice
melts at -10°C. Then reversibly freeze the water at constant T and P. Finally,
reversibly and isothermally reduce the pressure on the ice to | atm.
(a)
Ar, which is larger and so has greater intermolecular attractions.
(b)
HO, due to hydrogen bonding.
(a)
saan
(b)
(c) C3Hs, which 1s larger.
2) cal/mol-K = 87 J/mol-K = AvapHn/(319.4 K), and
Avapim =-9.7 kcal/mol = 28 kJ/mol.
AvapSm = 4.5R + R In 319.4 = 20.4 cal/mol-K = 85.4 J/mol-K =
Naot (319-4 K) and AvapHm = 6.52 kcal/mol = 27.3 kJ/mol.
85
7.18
7.19
The hydrogen bonding increases the degree of order in the liquid and hence
decreases S of the liquid. Therefore, AS for the transition liquid — gas 1S
increased.
(a)
The T-H-Erule gives Avapllmnbp.co (Sle) Kj
(8.314 J/mol-K)(4.5 + In 81.7) = 74.0 J/mol-K and AyapHm,.nbp.co = 6.05
kJ/mol. Similarly we get 55.7 kJ/mol for anthracene, 168 kJ/mol for
MgCl, and 295 kJ/mol for Cu.
(b)
Experimental values in kJ/mol (from Barin and Knacke, Reed et al., and
the AIP handbook) are: CO—6.04, anthracene—S6.5, MgCl.—156, and
Cu—304.
7.20
Assuming ideal vapor and using (3.29) we have AS, = const.=
ASm1 + ASm.2 = AvapSimabp +R In (Vi,/Vimi) = AvapSm.nbp +
Rin Vi, —R In (RTppp/1 atm) = AvapSmabp — R In (Tnbp/K) + R In Vi, —
R In (R K/1 atm) = const.,
ke) const, Hk In Ve _
so AyapSmnbp = R In (Tipp
R In (R K/1 atm) = R In (Tnbp/K) + const’., which is the T-H-E rule.
7.21
(a) T; (b) T; (c) F; (d) T; (©) F, @) F.
7.22
Assuming ideal vapor, neglecting Vinjiq iN Vm.gas — Vig, and neglecting the T
dependence of AyapHm, we use the integrated Clausius—Clapeyron equation
(7.21). If state 1 is the normal boiling point 34.5°C and state 2 is 25.0°C, then
In (P2/760 torr) = —[(6380 cal/mol)/(1.987 cal/mol-K)] x
[(298.1 K)' - (307.65 K) '] = -0.332s = In (P2/torr) — In 760 and
In (P2/torr) = 6.301; P2/torr = 545; P2.= 545 torr.
7.23
(a)
Integration of dP/dT = AH/(T AV) gives P2 — P, = (AH/AV) In (T2/T)) if
AH and AV are assumed constant. From Prob. 2.48, for one gram AH =
79.7 cal and AV = (1.000) 'cm? — (0.917)! cm? = -0.091 cm®. We have
Blan
79.7 cal
+
—0.091cm”
82.06cm* atm Wy 22-15 K
1.987 cal
86
n———__— = 154 atm
PASE
If AS and AV are assumed constant, then P2 — P; = (AH/T; AV)(T2 — T});
P, =latm+
(b)
82.06 cm? atm
79.7 cal
eee
ee
oe
(273.15 K)(-0.091cm3) 1.987 cal
OOK) ea aaaca
)
(
With 272.15 K replaced by 263.15 K, we get P2 = 1350 atm if AH and
AV are assumed constant or 1325 atm if AS and AV are assumed constant.
(c)
7.24
AH and AV change a lot for this large AP.
Use Eq. (7.24). For 1 g, AfusV = (1g)/(13.690 g/cm*) — (1 g)/(14.193 g/cm’) =
0.00259 cm?.
(a)
3
eR,
99 ii
(Ty -T,) =(234.25 Ky 202?
2.82 cal
82.06cm~ atm
K
T> = 234.25 K + 0.52 K = 234.77 K = -38.4°C
Tao
(b)
Replacing 99 atm by 499 atm in the above equation, we get AT’= 2.60 K
and I> = 236.85 K =—36.3-C.
(a)
Avapffmave = 40.7 kJ/mol and In(760 tort/P) =
—[(40700 J/mol)/8.314 J/mol-K][1/(351.5 K) — 1/(298.2 K)] = 2.49.
We get P, = 63 torr, which is pretty close to the true value 59 torr.
(b)
dP/dT = AHI(T AV) = AHI/TV gas. The true Voas 18 less than Vgas,ideal, SO the
true dP/dT is greater in magnitude than the dP/dT used in the calculation
in (a). Therefore the change in P calculated allowing for nonideality will
be greater in magnitude and the 25°C calculated vapor pressure will be
less than 63 torr, likely bringing it closer to the true value 59 torr.
7.26
Since IT = const. X P, we have In (JT/K) = In const + In P. The slope of an
In(IT/K) vs. T' curve therefore equals the slope of an In P vs. 1/T curve,
which equals —AsupHn/R, according to the Clausius—Clapeyron equation.
(b)
14)
-2.18x 104 K = —AgupHp/(8.314 J/mol-K) and Asupm = 181 kJ/mol.
Equation (7.21) gives
760 torr
M3 J6torr
l
Vals
1.987 cal/molK \373.15K
AH m = 10.2 kcal/mol = 42.7 kJ/mol
87
|
298.15K
7.28
AH m = (539.4 cal/g)(18.015 g/mol) = 9717 cal/mol.
(a)
Equation (7.21) gives
In
1
1
9717 cal/mol
PB
1.987 cal/mol-K Ger eee z]
760torr
In(P/760 torr) = 0.667, P/760 torr = 1.950, P = 1480 torr
(b)
These)
(a)
1
9717 cal/mol (1
, 446 torr ___
Toulon
T60NOTun a1 .987 calimolo Katara
WERT RON 10n Ke el =35 8)6 ne Saas
A plot of In(P/torr) vs. 1/T has slope —AH,,)/R.
—0:2934
-1.2986
-2.4214
In(P/torr)
Ke
24.204
25456
26.799
281d
10°/T
Okey An)
y = -7376.7x + 18.469
In (P/torr)
-2.5
So
0.0024
ee
ee
ee
0.0025
ee
SS
0.0026
ES
ee
0.0027
eee
0.0028
0.0029
Tk
The plot is linear with slope [0.76 — (—2.50)]/(0.00240 — 0.00284,)K"! =
—7380 K = —AH,,)/(1.987 cal/mol-K) and AH, = 14.66 kcal/mol =
61.3 kJ/mol.
(b)
—
Equation (7.21) gives
‘
B
1.845 torr
___
61300 J/mol
1
*
8.3145 J/mo-K \433.15K
In (P/1.845 torr) = 0.8240, P = 4.206 torr
(Alternatively, the graph can be extrapolated.)
88
l
413.15K
(c)
Use of Eq. (7.21) gives
760torr
14600 cal/mol
1
1
"7.845torr
1.987 cal/mok K
IM
AIBNS.K
T6280
350 -G
(The true normal boiling point is 356.6°C. The error arises because AH,
is not constant over the long temperature interval from 140 to 350°C.)
(d)
Setting up a spreadsheet like Fig. 7.8, one finds the regression analysis of
the In P versus 1/T data gives intercept b = 18.4706 and slope —7377.53
(corresponding to A,,,H, = 61.34 kJ/mol), and these values give a sum
of squares of residuals of P experimental versus P calculated of 7 x 10ue
When the Solver is run to minimize this sum, we get 18.4326 and
~7362.4 as the intercept and slope, with the sum reduced to 6 x [On eis
gives Avapffm = 61.2 kJ/mol. When this value is used in (b), we get P =
4.200 torr.
7.30
(a)
A plot of In(P/torr) vs. 1/T has slope -AH,)/R.
In(P/torr)
10°/T
In (P/torr)
2.283
6.5295
3.544,
6.1293
4.6521
5.7753
5.6330
5.4600
0.0060
0.0062 0.0064 0.0066
Ter?
Ke
6.5
6.0
a5)
5.0
4.5
4.0
335
3.0
2:0
2.0
0.0054
0.0056
0.0058
The plot is linear with slope (5.82 — 2.08)/(0.00540 — 0.00660)K™' =
~3120 K = —AHp/(8.3145 J/mol-K) and AHn = 25.9 kJ/mol =
6.20 kcal/mol.
(b)
Equation (7.21) gives In(P/279.5 torr) =
—[(25900 J/mol)/(8.314 J/mol-K)][(198 Ky = (838k el and
Pe OL) tor
89
Ts
Trouton’s rule is AH,,/T= 10.5 R. At the normal boiling point,
AV = Vin.gas = Vmitiq = Vingas = RI/P=RIp/Cpatm). Fora small change in P, we
have dP/dT = AP/AT and the reciprocal of Eq. (7.18) becomes AT/AP =
(AH m/T) 'AVm = (10.5R) RTp/(1 atm) = Tp/(10 +atm).
7.32
(a)
The 0°C path solid — liquid — gas shows that AsupH/m = Atusllm + Avapltm
= 51.07 kJ/mol. (Pressure changes have no significance since P has little
effect on H.)
(b)
The Clapeyron equation is dP/dT = AH,/(T AVm). For the solid—vapor
line, AHm = Asupffm and AV = Vin.gas —
Vso. = RT/P — Mgo1/Psoi. =
(82.06 cm?-atm/mol-K)(273 K)/[(4.585/760)atm] — (18.0 g/mol)/
(0.92 g/cm) = 3.716 x 10° cm?/mol — 20 cm*/mol = 3.716 x 10°
cm?/mol, assuming ideal vapor. (dP/dT)so1-gas
= AHm/(T AV) =
(51070 J/mol)/(273.16 K)(3.716 x 10° cm*/mol) =
(5.031 x 10° J/cm?-K)(82.06 cm?-atm/8.3145 J) =
4.966 x 10~ atm/K = 0.3774 torr/K. For the liquid—vapor line,
AH m= AvapH{m = 45060 J/mol; AVm = 3.716 x 10° cm*/mol;
(dP/dT)jig-vap = 9.3330 torr/K. For the solid—liquid line,
AH = 6.01 kJ/mol, AVm = M/Ptiq — M/Psotia =
(18.015 g/mol)(1.000 cm?/g — 1.0905 cm’*/g) = —1.630 cm*/mol, and
dP/dT = -1.012 x 10° torr/K.
TERE
(a)
At 25°C, In (P/torr) = 18.3036 — 3816.44/(298.15 — 46.13) = 3.1602 and
P = 23.58 torr. At 150°C, the Antoine equation gives In (P/torr) = 8.1810
ands? = 35/2 torr:
(b)
The Clausius—Clapeyron equation is d In P/dT= AH,,/RT”, with ideal
vapor assumed and the liquid’s volume neglected. Differentiation of the
Antoine equation gives at 100°C: d In P/dT= B/(T/K + CyK =
3816.44/(373.15 — 46.13)°K = 0.035687/K = AHm/RT? =
AH p/(8.3145 J/mol-K)(373.15 K)? and AHnvap = 41.315 kJ/mol.
7.34
(a)
We make the approximations that the solid and liquid volumes are
negligible compared to the vapor volume, that AH of vaporization and
sublimation are constant, and that the vapor is ideal. Then Eqs. (7.21) and
(7.22) apply and show that In P varies linearly with 1/T. We plot In P vs.
90
1/T for the solid and join the two points by a straight line; we do the same
for the liquid. At the triple point, the solid and liquid vapor pressures are
equal, so the intersection point of the two lines gives the triple point; this
is found.to bePi=1.5.¢torra?
(b)
= 200 K.
Equation (7.21) gives In (10.0 torr/1.00 torr) =
(AgubHm/ 1.987 cal mol” K7')(1/177.0 K = 1/195.8 K) and AjupbHm =
8430 cal/mol. Also In (100.0 torr/33.4 torr) =
AvapHm/1.987 cal mol! K7')(1/209.6 K — 1/225.3 K) and AvapHm =
6550 cal/mol.
TESS)
Ags/m/(cal/mol) = 8430 — 6550 = 1880.
To use the Clausius--Clapeyron equation to find the solid’s 1200°C vapor
pressure, we need the enthalpy AjupH of sublimation of the solid and we need
to know the solid’s vapor pressure at some particular temperature T~ We shall
take T’as the triple-point temperature Ty, since the solid and liquid vapor
pressures are equal at 7, The triple-point temperature differs only slightly
from the normal melting point, so we take T;, = 1452°C = 1725 K. We use Eq.
(7.21) to find the liquid’s vapor pressure at T,, = 1725 K; this equals the solid’s
vapor pressure at 1725 K. Equation (7.21) gives for the liquid—vapor
equilibrium: In (P2/P1) = —-AHm(1/T2 — 1/T)) and
7
1.00
torr
ap lm
1
1.987 cal/mok K
2078K
Yo
0.100 torr
a
1
1879K
which gives AyapHm = 89.79 kcal/mol. Use of (7.21) for the liquid between 7;,
= 1725 K and 1606°C gives
ee
A!
Lig
0-100 torr ___89790.cal/mol
l
1725K
1.987 cal/mol-K\1879K
which gives P,, = 0.0117 torr. The paths solid — gas and solid — liquid — gas
at the triple point give AjupH = AfusHT + Avapff. Thus AsupH = 4.25 kcal/mol +
89 79 kcal/mol = 94.04 kcal/mol. Now we use (7.21) for the solid:
pa
bene
0.0117 torr
li
pigunns4040‘eal/inol
1.987 cal/mo-K
(1473 K
wsinary
1725K
|= 4.694
P=1.1
x 10~ torr
7.36
When the two forms are in equilibrium, their chemical potentials are equal. We
thus want to make Gmai = Gmgr- For each form, dGm = —Sm dT + Vm dP =
V,, dP at constant T. Solids are nearly incompressible, so we neglect the
91
change in Vm, with P. For each form, AGm = Vm AP or Gm(P2) = Gm(P1) +
V., AP, where P; = 1 bar. Setting Gmai(P2) = Gm,gr(P2), we have
Gmai(P1) + Vidi AP = Gmge(P1) + Vingr AP and
AP = [GmeP1) — Gmai(P1)V/(Vm.di —
Ver) = Ay G
gi/(Vinai -— Vmgr), since Pi =
1 bar = P°. Using Vn = M/p, we get AP =
[(-6070 J/mol)/(-1.97 cm*/mol)](82.06 cm? atm)/(8.314 J) = 30400 atm and
Pz = 30400 atm.
Dead
When gray (g) tin is more stable than white (w), we have Uy < Hy and Gm <
Gmw. We have dGm = —Sm dT + Vm dP = —Sm GT at the constant P of 1 bar.
Appendix data show that Sm > Sm, SO Gmw decreases faster than Gm» as T
increases, and Gm, increases more rapidly than Gm.» as T decreases. At the
temperature Teg with Gm. = Gm, the two forms are in equilibrium. Below Teg,
we have Gm < Gmw and gray tin is more stables Let 7; = 25°C, 73 = Tq, _and
AT = Teg — 25°C. We have AGmn.g = Gm,g(Teq) — Gm,g(25°C) and AGmw =
Gmw(Teq) — Gmw(25°C), 80 Gm,g(Teq) = Gm,w(Teq) becomes Gm,g(25°C) + AGm.g
= Gmy(25°) + AGmw Or AGmg — AGmw = Gmw(25°C) — Gme(25°C) =
0 — 0.13 kJ/mol, where Appendix data were used. Since AGp = =|? SredT at
constant P, we have AGm,g — AGmw = — Jf (Smg — Smw) aT =
(Smw — SmgMT2 — T1) = (51.55 — 44.14)(J/mol-K)(T2 — 25°C), if we neglect the
T dependence of Smg — Smw. Then —130 J/mol = (7.41 J/mol-K)(T— 25°C) and
Teg — 25°C = -17.5 K = -17.5°C, 80 Teg = 7 5es
7.38
In Example 7.7, Viner —
Vm,ai WaS assumed independent of pressure. At 25°C
and | atm, Vingr= 5.34 cm?/mol and Vina = oa! cm?/mol. Since graphite is
more compressible than diamond, as P increases, Vm,gr Will decrease more
rapidly than Vin.gi and the difference Vin.pr—
Vmai Will decrease as P increases.
Since this volume difference occurs in the denominator of the expression for
Pz —P,, P2— P; will be greater than calculated in Example 7.7 and P2 will be
greater than 15100 bar.
1h
Cp = (OH/0T)p = T(0S/0T)p. Since Cp > 0, both H and S increase as T increases.
A first-order transition has AH > 0 and AS > 0, and so we see a sudden jump in
H and in S at the transition temperature 7;,,. A second-order transition has AH
= 0 and AC? # 0, so there is no sudden jump in H but there is a sudden change
92
in the slope 0H/dT = Cp at T,,;. A lambda transition has AH = 0 and if Cp > ©
at T;; the H vs. T curve has a vertical (infinite-slope) inflection point at Tis.
Since (0S/0T)p = Cp/T, the S-vs.-T curves resemble the H-vs.-T curves.
7.40
Lambda
Second order
First order
At T=0, w =Oand
o, =7/r=1.
If r=0, 6,=-w/w = -1. This is a highly ordered state; with Cu and Zn
atoms interchanged, each Cu is surrounded only by Zn atoms.
If r=w, then o, =0.
In the upper (7 = 0) diagram, w = 0 and o, = 1. In the lower diagram,
comparison with the upper diagram gives r= 16, w = 16, and 6, =
(16 — 16)/(16 + 16) = 0.
At
T=0,n,p =n, ando,=2-1=
i
(f)
As T > ©, nyp = Yanp and Os = 2(%) -1=0.
(g)
In the upper (T = 0) figure, n,p = Np and o, = 1. In the lower figure:
np = 4(3) +2 + 4(3) +2 + 403) + 2+7=49, and ny =
where
2424241424+143414342434+14+2+142+1=29,
the counting was done by looking at the neighbors of the atoms in rows,
2, 4, 6, and 8. Hence o, = 2(29)/49 — 1 = 0.18.
(h)
From Sec. 7.5, both curves have infinite slope at 7:
O¢
Os
Ty.
Th
93
7.41
At 0 K, the most stable arrangement in the crystal has the negative end of each
molecule next to the positive end of the adjacent molecule. As T increases, the
crystal becomes more disordered until there is a 50-50 chance for the negative
end of one molecule to be next to the positive end of the adjacent molecule.
7.42
(a)
At T=T, we have t=0. For T <T,, we have t > 0. As the positive
0.01285 goes to zero and Cp/(J/mol-K) goes
quantity t goes to zero, t © =1
to B. Likewise, s is positive for T >7, and s ~ — 0 as the positive
quantity s goes to 0. Hence Cp = 456 J/mol-K at 7) .
(b)
For T<T,, 0Cp/0T =(0Cp/dt)(dt/ dT) = —(Cp /dt)/T, =
~(A’/o)[-ar-*! + D(0.5—a)t°?* + Ed -a)t “]/T,. We have
a= -0.013, so -a-1<0,
-0.5-a<0,
and-—a>0. As the positive
quantity t goes to 0, 1/r°*! 41/0 =ce and 1/1°°** —5 co, The quantity
1/1%*! =1/1° goes to infinity faster than 1/t°°** = 1/t°” and the
7; from
coefficient A’ is positive, so 0Cp/0T — +°° as t > 0 and T
below. Similarly for s.
7.43
Since ice is the stable phase at —20°C, we have ice < Usc tig, Where SC =
supercooled. The phase equilibrium condition gives [ice = Wvapor above ice and
Use liq = Uvapor above sc liq therefore
Uvapor above ice S Lyapor above sc liq- Use of
uu; =p; + RT In (P/P°) (Eq. (6.4)] for the vapors gives
b; +RT
In (Pi.above ice) = LA; + RT I|n (Pi.above sc liq) sO
In (Piabove ice) <In (Piabove sc liq) and Pi above ice < Piabove sc liq:
7.44
dP/dT = AH/(T AV). Since V of the solid or liquid is negligible compared to V
of the gas, we have to high degree of accuracy, AVsotid-svapor = AViiq—vapor- AS
shown in Prob. 7.32, Asubm is greater than AyapHm by Afusm, which is not
negligible. Therefore
AH soiia
eas > AAtiggas and (dP/dT)soliagas >
(dP/dT)iigsgas- The solid—vapor line has greater slope.
7.45
The pressure due to 10 cm of water is given by (1.9) as
P= pgh =
[1.00 (10° kg)/(10~ m*)](9.80 m/s)(0.10 m) = 980 N/m? =
(980 J/m°)(82.06 x 10° m? atm)/(8.314 J) = 0.009, atm, where (2.7), (2.14),
(1.19) and (1.20) were used. The total pressure 10 cm below the surface is
94
1.009; atm and Eq. (7.24) and data in the example that follows it give
0.009, atm =
THOT
7.46
15
79.7 cal 82.06cm? atm T, — 273.15 K
273.15K
1.987 cal
—0.091cm*>
KEE 3SC10F-K:
Neglecting the change in water density with pressure and the change in g with
depth, we have as the pressure at 3000 m: P= pgh=
(1.0 g/em?)(1 kg/10° g)(10° cm*/1 m*)(9.8 m/s*)(3000 m)=
(2.9 x 107 Pa)(1 atm/1.01 x 10° Pa) =290 atm. P exceeds the 350°C vapor
pressure of water and the stable phase is liquid water.
7.47
Let sc denote supercooled liquid water. All equations in this problem are for
~10°C. The phase equilibrium condition and Eq. (6.4) ZIVE Mice =
solidi shi (he 950 torr/750 torr) and psc= vapor above sc=
=
Lvapor above ice=f,
torr/P;). But
uw; + RT In (P;/750 torr). Subtraction gives Bice — Use= RT |n (1.95
=
Prob. 4.29b gives Hice — Msc = Gmiice — Gmsc = (-2.76 cal/g)(18.0 g/mol)
_49.7 cal/mol. (This relation holds at 1 atm, but the effect of pressure on liquid
e
and solid thermodynamic properties is slight and can be ignored.) Therefor
49.7 cal/mol = RT In (1.95 tort/P;) =
and
(1.987 cal/mol-K)(263.1 K) In (1.95 torr/P;). We find 1.95 torr/P; = 0.9093
P, = 2.14 torr. (The experimental value is 2.15 torr.)
7.48
torr
0.01°C is the triple-point temperature, so ice has a vapor pressure of 4.585
at 0.01°C.
7.49
(a)
above liquid A;
The initial state (state 1) has partial pressures Pa and Pg
2)
the liquid is subject to a pressure P = P, + Pg. The final state (state
has partial pressures Pa + dPa and Pg + dPx above liquid A, which
s in
experiences a pressure P + dP, where dP = dP, + dPyx. Let the change
be
HEIOG and vapor A chemical potentials on going from state 1 to 2
2
~ and du(A°*). The oe equilibrium conditions for states | and
du(A ‘)
are [;(A ‘)= 1i(A®) and pofA’ ) = pp(A%)=pi(A‘) + ey Ve
dG A Ae) =
u)(A%) + du(A*). So gies we Ha We have du(A ‘) =
dT VVn(A ea wale iV‘(A ‘)dP, since T is constant. Also,
BSA
95
(AS) = 1°(AS) + RT In (Pa/bar) and du(A%) = (RT/Pa) dP at constant T.
(b)
7.50
Hence Vm(A ‘) dP = (RT/Pa) dP = Vm(A*) dPa. Q.E.D.
dPa/Px = [Vm(A ‘)/RT] dP. Integration with V,,(A “) assumed constant
gives In (Pao/Pa.1) = [Vm(A_)/RT\(P2- P1). So In(Pa.2/23.76 torr) =
(18.1 cm?/mol)(1 atm)/(82.06 cm?-atm/mol-K)(298 K) = 0.000740 and
Px,2/23.76 torr = 1.000740, Pa, = 23.78 torr.
Use the 25°C path:
a
b
liq(1 bar) >
liq(23.766 torr) >
c
— vap(1 bar).
vap(23.766 torr)
We have dGm = — Sm dT + Vm dP = Vm dP at constant T. For the liquid, Ge
Vm AP, and
82.06 cm? atm
== 1:8:J/mol
Step b is a constant-7-and-P equilibrium process, so AG» = 0. For step c,
AG, = |? Vm dP = |; (RUIP) dP = RT In (Po/Pi) =
(8.3145 J/mol-K)(298.15 K) In (750.062/23.766) = 8557.2 J/mol.
AG, + AG, + AG; = 8555.4 J/mol. From the Appendix, AG/(J/mol) =
=228572'+ 237129 = 8557.
7.51
(a)
Trouton’s rule is AH,/T; = 10.5R = c. Substitution into Eq. (7.22) gives
In (P/atm) = -(c/R)(T;/T) + c/R = —(10.5R/R)T;,/T + 10.5R/R =
10.5(1 —T,/T).
Teo2
P/atm = exp{10.5[1 — (353.25 K)/T]J}.
(b)
P/atm = exp [10.5(1 — 353.25/298.15)] = 0.1436 and P = 109 torr.
(c)
In (620/760) = 10.5(1 — 353.2s5/T) and T= 346.5 K = 73.4°C.
(a)
Use of (7.21) gives at 0°C:
In (4.926/4.258) =
—[AH p/(1.987/cal/mol-K)][(274.15 K)"! — (272.15 K)"] and AHm.273 =
10.80 kcal/mol = 45.20 kJ/mol. Similarly, AHjn373 = 9.89 kcal/mol =
41.37 kJ/mol. Then ASm.273
= AH m.273/T = (10800 cal/mol)/(273.1 K) =
39.5 cal/mol-K = 165 J/mol-K;
111 J/mol-K.
AS)373
= AHm373/T = 26.5 cal/mol-K =
AG m= AH, —-T AS, = 0, as it must be for constant-7-and-
P equilibrium processes. The calculated 100°C AHpy is slightly in error,
since the approximations used to derive Eq. (7.21) are less accurate the
higher the vaporization temperature and the closer the temperature is to
96
the critical point; for example, H2O(g) is denser and hence less ideal at |
atm and 373 K than at 5 torr and 0°C.
(b)
AH? and AS° are for the process liq — ideal gas at 1 bar. Let P be the 0°C
vapor pressure. A convenient path is the following 0°C path:
2
3
4
liq(P°) > liq(P) > vap(P).— vap(P°) — ideal vap (P*)
Since moderate pressure changes have little effect on liquid
thermodynamic properties, we can take AH; = 0, AS; = 0. Gas
nonideality is slight at 1 bar, so we take AH4 = 0, AS4 = 0. Since the
vapor is nearly ideal and H of an ideal gas depends on Tonly, we take
AH; = 0. Thus AH3,, = AHm 2= 45.20 kJ/mol. Use of (3.25), (3.29), and
Boyle’s law gives AS57, = ASm2 + ASm3 = AHm2/T+ R In (P/P*) =
(45200 J/mol)/(273 K) + (8.314 J/mol-K) In (4.58/750) = 123.2 J/mol-K.
7.53
(a)
dna=dn',—ad&,
dng=dn, —b dé, dng=edé&,
dng=fd§.Then
Y; p; dni = Ua(dn®, — a d&) + Up(dn; — b d&) + Wee d& + Upfdé =
— bie + ee + flr) d&= pa dn’, +Up dnz,
ua dn’, + tp dn, + (aa
since the equilibrium condition for the reaction is Deval =
—aua — bp + ele + flr = 0.
(b)
dG=-S dT +VdP +>; pj dn; =—S dT + VdP + Us dn‘, + Up dna;
at constant T, P, andn},dG = Ua dn, and Wa = (0G/dn ’,)r.P.n'y-
7.54
(a) and (c). See Fig. 7. 1a.
7.55
(a) T; (b) T; (c) F; (@) F; ©) T; © F
97
&) 1. th) B®
F-
Chapter 8
(b) Pa K!? m°/mol’ and m?/mol;
(c) m°/mol.
8.1
(a) Pa m°/mol’ and m*/mol;
8.2
As p > 0, Vm 2 &. In Eq. (8.2), -b can be neglected compared with Vm to
give
P=RT/Vm—alV> = (1/Vin)(RT— a/Vm). AS Vm > 2, RT — a/Vm goes to
RT and P > RT/Vm. (Note that it’s harder to use (1.39) to get the limiting
behavior since both P and a/V2, go to 0.) In (8.4), the terms B/Vm, C/Vi,, --.
each go to 0, giving PVm = RT. When (8.3) is solved for P and b is neglected
compared with Vm, we get P = RT/Vm—a/V2,T'\” = (1/Vm)(RT— al/VinT '*) >
REV
8.3
(a)
n=(28.8 g)/(30.07 g/mol) = 0.958 mol;
Vin = V/n = (999 cm*)/(0.958 mol) = 1043 cm*/mol.
(82.06 cm?- atm/moFK)(298. 1K) ,
pe
1043 cm?/mol
[
be 186 cm?/mol
a 1.06 x10* cm*/mol?
1043 cm?/mol
(1043 cm?/mol)?
=O
atm
The ideal-gas P is P = RT/Vm = 23.5 atm.
(b)
B' = B/RT= (-186 cm*/mol)/(82.06 cm>-atm/mol-K)(298.1 K) =
0.00760 atm”;
C' = (C — B’YR’T’ = (1.06 x 10* cm*/mol? — 1867 cm*/mol*)/
(82.06 cm*-atm/mol-K)?(298.1 K)* =-4.01 x 107 atm~.
Vin = [(82.06 cm?-atm/mol-K)(298.1 K)/(16.0 atm)] x
[1 — (0.00760 atm™')(16.0 atm) — (4.01 x 10° atm™)(16.0 atm)"] =
1327 cm*/mol. V=nVm = (0.958 mol)Vm = 1272 cm’. Also,
Vavidealsl oy
8.4
Eq. (8.4) gives
cm?/mol and Viel
40> cm’,
P = RT(1/Vim + BIV> + GIVE. +--+). Substitution into the nght
side of (8.5) gives PVm = RT[1 + B'RT(1/Vm + B/V2, + C/V>, +--+) +
OVA
OMCs + 2B/V> +---)+---]. We compare this expression for PVm
with (8.4); equating the 1/V,, coefficients, we get B = B'RT. Equating the Aaa
98
coefficients, we get C = B'RTB + C'R°T’ = BVR’°T’ + C'R°T’ =
Ral (Bas G:
8.5
V_.=RT/P+B = (82.06 cm’-atm/mol-K)(200 K)/(1 atm) — 47 cm*/mol =
16.36 L.
8.6
(a)
Bi =(B, + By/2 =-387 cm’/mol. Let gas | be CHg. Then x; =
0.0300/0.1000 = 0.300 and x27 = 0.700. The equation at the end of Sec.
8.2 gives B = (0.300)°(-42 cm*/mol) + 2(0.300)(0.700)(-387 cm*/mol) +
(0.700)°(-732 cm3/mol) = 525 em*/mol. V/not = 10000 cm*/mol.
P = [(82.06 cm?-atm/mol-K)(298.1 K)/(10000 cm*/mol)] x
[1 — (525 cm?/mol)/(1000 cm?/mol)] =r
(b)
s5 2 ath
With B,7 = —180 cm*/mol, we get B = 438 cm*/mol and P = 2.34 atm.
Also, Pideat = RT/(V/niot) = 2.45 atm.
8.7
n = (74.8 g)/(30.07 g/mol) = 2.48 mol.
Vm = V/n = (200 cm*)/(2.483 mol) =
80.4 cm?/mol.
(a)
p=
10.6K Perit
at me K)(310.6
R _(@2.06cm 3_atm/molRL
Von
(b)
80.4 cm” /mol
Equations (8.18) and (8.2) give
2
K)’ (305.4
K)” ee 50.105 cm® atm mol7
ON
a Ber (BU0G eminatu mee
64(48.2 atm)
eraennls.
ie (82.06 cm°-atm/mok K)(305.4 K) eee Cr
3)
8(48.2 atm)
_ (82.06 cm?-atm/mol K)(310.6K) _ 5.50 10° cm°-atm/mol?
(80.4 cm*/mol)?
80.4 cm?/mol — 65.0cm?/mol
P = 1655 atm — 851 atm = 804 atm
(c)
Equations (8.20), (8.21), and (8.3) give
_ 0.4275(82.06 cm?-atm/mol- K)’ (305.4 K)""
rR
48.2 atm
9.73107 cm®-atm-K'/mol?
3
+e 0.08664(82.06 cm’ - atm/mol-K)(305.4 K) wah.
48.2 atm
99
3/mol
(d)
P = RTI(Vm—b) — alVm(Vm + b)T '” = 720 atm — 548 atm = 172 atm
Interpolation gives at 372°C: B =-179 cm*/mol + (7.5/20)(22 cm’/mol)
= —171 cm*/mol and C = 10119 em°/mol’.
Bia |(82206 cm?-atm/mol-K)(310.6 K)/(80.4 cm?/mol)] x
[1 — (171 em?/mol)/(80.4 cm’/mol) + (10119 cm°/mol2)/(80.4 cm*/mol)’]
= 139 atm. Note: The observed pressure 1s 135 atm.
8.8
Mor = 0.2000 mol, V/ntor = 3500 cm*/mol.
(a)
(b)
l) =
P=RT/Vm = (82.06 cm*-atm/mol-K)(313.1 K)/(3500 cm°/mo
7.34 atm.
Let CH, be gas 1. For CoHy, a; = 27R°T? /64P. = 4.56 x 10° cm°
atm/mol2, b, = RT,/8P, = 58.3 cm*/mol. For CO:, we find az = 3.61 x
10° cm® atm/mol” and b2 = 42.9 cm*/mol. Also, x; = 0.0786/0.200 =
0.393,
(c)
8.9
x) = 0.607. From the last paragraph of Sec. 8.2,
a = [(0.393)°(4.56 x 10°) + 2(0.393)(0.607)(4.56 x 10° x 3.61 x 102je7
(0.607)°(3.61 x 10°)}(cm®-atm/mol*) = 3.97 x 10° cm°-atm/mol’;
b = 0.393(58.3 cm/mol) + 0.607(42.9 cm*/mol) = 49.0 cm*/mol.
P = (82.06 cm?-atm/mol-K)(313.1 K)/[(3500 — 49)cm*/mol] (3.97 x 10° cm®-atm/mol’)/(3500 cm*/mol)” = 7.12 atm.
Z=PVp/RT; P = ZRTIVm = 0.9689(82.06 cm*-atm/mol-K) x
(313.1 K)/(3500 cm*/mol) = 7.11 atm.
P =RT/Vm + RTB(TYV2, + RIC(TYV3,. (OP/Vm)r = 0 = -RTIV:, —
2RTBIV?, —3RTCIV’,. (@°P/OV2, r= 0 = 2RTIV>, + ORTBIV), + 12RTCIV>,.
RIAV. 8+ ORT BY met OR Le = On eC}
2RT.V2 .+ ORT,BV,,. + 12RT.C =0
(2)
Subtract twice equation (1) from equation (2) to get
B=—3C/V,,.. at T= T..
Substitution of this expression for B into equation (1) gives C(T;) =
V2 /3. Also, B(T.)
=—3C(TeVinc =— Vie. We have Z; = PeVnc /RT. =
Lee BCTV Vmet CIV
8.10
ea 1 a1
aan 3.
(a) From (8.18), b = RT,/8P. = (82.06 cm’-atm/mol-K)(150.9 K)/8(48.3 atm)
= 32.0cm*/mol; a= 27R°T//64P.=
100
27(82.06 cm?-atm/mol-K)*(150.9 K)7/64(48.3 atm) =
1.34 x 10° cm® atm/mol*
(b)
Comparison of (8.9) with (8.4) gives as the van der Waals estimate:
B=b-—ad/RT.
Byox = 32.0 cm*/mol — (1.34 x 10° cm®-atm/mol’)/
(82.06 cm’-atm/mol-K)(100 K) = -131 cm*/mol. Also, B20 k =
-49.6cm*/mol, B3o0x =—22.4 cm*/mol, Bsoo x =—0.7 cm°/mol,
B,o00 x =15.7 cm*/mol. Agreement with experiment is fair.
8.11
We have AyapUm = @/Vm = ap/M and AvapHm, =~ ap/M + RT. For N> at its normal
boiling point, Avapf{m ~ (1.35 x 10° cm® atm/mol*)(0.805 g/cm*)/(28 g/mol) +
RT = (3.9 x 10* cm? atm/mol)(1.987 cal/82.06 cm? atm) +
(1.987 cal/mol-K)(77.4 K) = 1.1 kcal/mol. For HCI, Avapm =
(3.65 x 10° cm® atm/mol”)(1.193 g/cm*)/(36.5 g/mol) + RT=
2.89 kcal/mol + 0.37 kcal/mol = 3.3 kcal/mol. For H20,
Avapffm =
(5.46 x 10° cm® atm/mol’)(0.96 g/em*)/(18 g/mol) + RT =
7.8 kcal/mol. Agreement with experiment is fairly good.
8.12
Vin
T in cell D2 is changed. The new graph shows we need to start at a smaller
value, so the value in A9 is reduced until the graph starts at a substantially
positive value for P. An initial Vm of 84 cm°/mol is found to be appropriate.
0 and
The maximum in P is close to 12 atm, so the vapor pressure is between
12 atm; averaging these values, we enter an initial guess of 6 atm in C3. The
values in columns A and B show that 6 atm corresponds to 2900 cm*/mol and
to a value between 84 and 89 atm, so we use 2900 and 86 cm*/mol as the initial
guesses for V) and V. The Solver constraints for C3, E3, and G3 need to be
appropriately modified. The Solver gives 3.01 atm, 6375 cm*/mol, and 86.0
cm?/mol as the vapor pressure, and molar volumes.
8.13
From Table 8.1 and Eqs. (8.20) and (8.21):
= 29.68 cm*/mol,
b = 0.08664(82.06 cm?-atm/mol-K)(304.2 K)/(72.88 atm)
a = 0.42748(82.06 cm?-atm/mol-K)(304.2 K)°5/(72.88 atm) = 6.375 x 10’
Fig. 8.6
K"? cm® atm/mol’. These values are entered into Cl and El of the
shows no
spreadsheet and the temperature is changed in D2. The new graph
in A9. One finds
minimum, indicating that we must start at a smaller Vm Value
P is at 48 atm, so the
50 or 55 cm’/mol to be suitable. The local maximum in
the initial guess.
vapor pressure is between 0 and 48 atm, and we use 24 atm as
101
The column A and B values show 24 atm to correspond to molar volumes of
about 60 and 760 cm*/mol, so we use these as the initial Vm guesses. The
Solver constraints for C3, E3, and G3 need to be appropriately modified. The
Solver gives 38.7 atm, 57.2 cm?/mol, and 390 cm*/mol.
8.14
Table 8.1 and Eq. (8.18) give
b = (82.06 cm?-atm/mol-K)(369.8 K)/8(41.9 atm) = 90.5 cm?/mol,
a = 27(82.06 cm?-atm/mol-K)*(369.8 K)7/64(41.9 atm) = 9.27 x 10°
cm° atm/mol” When the van der Waals expression for P is substituted into
(8.24), we get PV? -V,,) =] Ye(RT (Vn —b)—alV,)} dV,, , which gives
ve
Ve
(pe scoiaoad RT |n Yel? +a
Ve
m
m
m
eal!
-—
Ven Ve
(8.25)vdW
m
The Fig. 8.6 spreadsheet is modified by changing the a and b values, changing
(in an efficient way) the formulas in B9, B10,... to the van der Waals
expression for P, using the right side of Eq. (8.25)vdW (given above) in C4,
and using the right side of the second equation of (8.2) with V,, replaced by
V. or V,”, in E4, and G4, respectively. The Solver constraints for C3, E3, and
G3 need to be appropriately modified. The graph shows P is too high at 95
cm*/mol, and an appropriate starting Vm is 130 cm*/mol. The local maximum
in P is at 22 atm, and 11 atm is a reasonable initial guess for the vapor
pressure. The values in columns A and B show 11 atm corresponds to about
145 and 1885 cm*/mol, and these are reasonable initial guesses for Vm. The
Solver gives 16.65 atm , 141.5 cm*/mol, and 1093 cm*/mol.
8.15
(a)
m=0.480 + 1.574(0.153) —0.176(0.153)° = 0.717.
2
2 J, 717],
fe 0.42748(82.
8(82.06 cm?- 2 atm/molK)’(369.8K)"
41.9 atm
_{298-ISK
369.8K
v2))°
a = 1.08 x 10’ cm® atm/mol”
(b)
We change the a value in C1 and delete the T'” factors in the
denominators of the formulas in cells C4, E4, G4, B9, B10,... (do this in
an efficient way). The Solver gives 9.47 atm, 97.6 cm*/mol, and 2144
cm°/mol.
102
8.16
(a)
Using wo = 0.153, we find k = 0.604. From T, = 369.8 K, P, = 41.9 atm,
and k = 0.604, we get b = 56.35 cm?/mol and a = 1.133 x 10’ atm cm°/mol’.
(b)
When the Peng—Robinson expression for P is substituted into (8.24), we
get P(V? -Vi)= ie{RT(V,, —b) —al[Vn (Vp, +b) + D(V,, —b)}} dV, . Using
the given integral with s = 2b and c = —b’, we get
sei
m
ar Vn= -fu(a +(1-¥2)b Vi +d+2)b |a
Vie
m
bs
\Vo td +y2)b V.+0—V2)b
The Fig. 8.6 spreadsheet is modified by changing the a and b values, changing
(in an efficient way) the formulas in B9, B10,... to the Peng—Robinson
expression for P, using the nght side of Eq. (PR) (given above) in C4, and
using the right side of the Peng—Robinson equation with V,, replaced by Ve
or V., in E4, and G4, respectively. An appropriate starting value is 85
cm?/mol in A9. The maximum in P is at 18 atm, and we take 9 atm as the
initial guess for the vapor pressure. This P corresponds to 86 and 2260
cm?/mol, which are the initial guesses for the molar volumes. The Solver gives
9.38 atm, 2143 and 86.1 cm*/mol.
8.17
(a)
For
P=RTMV,,—b)—a/[Vn Vin +b)T 7),
a
R
fees
TieoP PT:
Vib. DV AVeuEDLE
aT Jy.
%
|
a
——
Va Tae aie
Neat
3a
eV (Velnb)
Ie2
Then, using [foo + b)y"' dv = b | In[v/(v + b)], we get
A
oe
ene
3a
Va( Ve
Pe
b) les Gee we
ea ee Be
ORD eV Ve
and addition of P AV gives the desired result.
(b)
Example 8.1 gave V,, =1823 cm*/mol and V,, = 100.3 cm°/mol. Then
PV” -VL) = (10.85 atm)(1723 cm’/mol) = 1.869 x 10* cm? atm/mol
=1894 J/mol. We have
Narr
Behe
x10° cm® atm K'”/mol*) 1 1823(100 + 63) _
we 3(1.807
(69.7 cm /mol(298.15.K)) a) 9100.3018 23 + 63)
1.13 x 10° cm? atm/mol = 1.146 x 10° J/mol. Then A,,,H =
103
(11460 + 1894) J/mol = 13.3 kJ/mol. (The experimental value is 14.8
kJ/mol.)
8.18
Use of (8.20) and (8.21) gives a = 3.77 Xx 10° cm® atm K!/mol? and b = 92.4
cm?/mol. The Fig. 8.6 spreadsheet with a, b, and T changed can be used. One
finds a minimum pressure of —14.5 atm at 210 cm’/mol.
8.19
= RT
(Vm — RT¢ /8P.)
(P+ 27R°T7/64P.V2
[P+ (27T2/64P, V2)(64P2V_ /9TZ Vm —Vnc!3) =(8PVinc/3T,)T
(P+3PIV2\(Vn —Vinc!3)=8P-VineTy/3
(P. +3/V,\(V, —1/3)=8T,/3
8.20
(a)
P=RTlMVm—b)- alTV>.. The critical-point conditions (OP/0Vm)r = 0
and (0*P/OV,.); =0 aS to
RT AV, —6)? =2a/T.Vz,, and REAV, 2sb) =3a/T Vane
Cc
Division of the first equation by the secondeve Vee bo =2V ne
and
Vin. = 3b. Substitution in the first equation gives RT./4b7 = DneTh
TT. and
= (2/3)(2a/3bR)" * Substitution in the Berthelot equation at the critical
point gives
“
~2 2/20)" _ 23/38 me NORGE RE
2b
3\3bR
9b?2\ 2a
DAelem Poise
Division of this P, equation by the preceding 7, expression gives
P./T, = R/8b, so.
a3
Babe 21GP
RR
ork
Pay,me
(Bye z. ase
Di Rigi 8Peal hetabove Peexpression gives
gPoR. 64 P.
2 oe1/2 Re pl/2 2) a 1/2 -3-0375
TEN
(c)
Toe
em
Substitution of R = 8P.Vm,/3T, and the above expression for a and b into
the Berthelot ee
peekOIE RET
ei AMR,
64 P. TV?
followed by division by P.Vm ¢«gives
3 Aa
SP.
104
OAS VSINTOR
ia
8.21
are
Ol
ShVewT a\l SPV.
fe
STEEP
Vg >
From (8.29), P, = 87,/3(V, -+) - 3/V,.
T, = TIT; = 310.6/305.4 = 1.017:
V, = VinlVine = V/NVm.e = (200 cm?)/(2.488 mol)(148 cm?/mol) = 0.543. Then
P, = 8(1.017)/3(0.543 — 0.333) — 3/(0.543)° = 2.75 = P/P, and P = 2.75P.. =
1324 atm.
8.22
B = (82.06 cm*-atm/mol-K)(150.9 K)(48.3 atm) '[0.597 — 0.462e° 70071509 KyT)
= [153.1 — 118.429: 8”) cm?/mol. At T = 100, 200, 300, 500, and 1000 K,
we get (in cm*/mol) —187.5, 47.7, -15.3, 6.3, 21.5. Agreement with experiment
is very good.
8.23
As noted in Sec. 8.8, at T and P°, H'! — Hm = JP [T0OVm/0T )p — Vl AP.
Differentiation of (8.5) gives (0VWdT)p = RP (1+ B'P+C'P+---)+
RTP | (PB” + P°C’’ +---), where the prime means differentiation with
respect to T. Use of (8.5) gives T(0Vm/0T)p — Vm = REERN PB CLP Co).
SoH! SHa= |) RUB pee Cy)
de
Ri, [Pee
es 8(ps) Gee
sid 5, = JF [OVa/0T)p— RIP] dP =
jf [RWB + CP +--+ -) + RT(BY + PC’ +---)] dP = R(B' + TB” )P° +
1R(C’ + TC")(P°)’+-- -. Finally, Gy — Gm = (Hy — Hm) — T(Sim— Sm)- Use
of the preceding results gives G = Gi==RIIBIP? + ME
8.24
(a)
ey,
Comparison of (8.9) with (8.4) gives the van der Waals estimates of the
virial coefficients as B = b — a/RT, C=b’,... . Use of (8.6) gives
B' = b/RT—a/R°T’,... . Differentiation gives B’ =—b/RT* + 2a/R°T°,
_.. . Substitution in the results of Prob. 8.23 gives H'4 — Hp =
RT?(-b/RT* + 2a/R°T°)P° +- - - = (2a/RT— b)P° + - - - and gives
Sees (aIRT Po + eee.
105
|.
(b)
From (8.18), a = 27R°T2/64P, = 5.50 x 10° cm® atm/mol” and b =
RT, /8P, = 65.0 cm*/mol. Then H'\ — Hm = (0.987 atm) x
[2(5.50 x 10° cm® atm/mol’)/(82.06 cm?-atm/mol-K)(298 K) 65.0 cm3/mol] = (379 cm?-atm/mol)(1.987 cal/82.06 cm’ atm) =
9.2 cal/mol. Also ies ayng—A (yall) 10° cm® atm/mol*)/
(82.06 cm>-atm/mol-K)(298 K)"](0.987 atm) =
(0.74 cm? atm/mol-K)(1.987 cal/82.06 cm? atm) = 0.018 cal/mol-K.
These are substantially smaller than the experimental values.
8.25
(a)
The Berthelot equation when solved for P and multiplied by Vp/RT is
alRT *Vm. Use of (8.8)
a
eb
GIR Vet
Pave Rk Viyl VD)
with x = b/Vm gives PVmn/RT= 1 + (b- alRT°)/Vm + BV, tees,
Comparison with (8.4) gives the Berthelot estimates of the vinal
C= b’, ej Use 0ff(8:6) gives
coefficients as B = b — alRT*,
. Differentiation gives B* =—bIRT? + 3a/R°T*.
B' =DIRT - alR°T?, ...
= (3a/R? —b)P° +---and
Use of Prob. 8.23 results gives He oH
SiS = (di Rigs)
(b)
Poe
Substitution of the a and b expressions of Prob. 8.20a gives
H® — Hp = (81RT?/64P,T* — RT,/8P,)P° + +++ and S'\ — Sm=
(27RT? /32P,T*)P° +---.
(c)
Substitution in the results of (b) gives S — Sm =
(27/32)(1.987 cal/mol-K)(305.4 K/298 K)*(0.987/48.2) = 0.037 cal/mol-K
and H'’ — Hm» = 15 cal/mol. Agreement with experiment is excellent.
(d)
gi — Sm = (27/32)(1.987 cal/mol-K)(430.8/298)°(1/77.8) =
0.065 cal/mol-K.
8.26
(a)
We use the following isothermal path (rg = real gas, ig = ideal gas):
l
2
3
4
rg(Vin) > rg(Vm = 2) > ig(Vm = 09) > 1g(Vm) > 19(Vm.ia)
From dAy = —Sm dT —P dVm, we get (dAn/OVm)r = —P. Then AA) =
aly bidVa = oer dV, oe AAn> = Osince AU —
and AS;
Oa
noted in Chap. 5); AAm3 =—J'" Pia dV{ =-J" (RTIV,) dV’; AAma =
106
—[¥" (RTIV,) dVq =-RT In (Vi8IVin)- Amia(T, P) - An(T, P) =
AAPIEE KAM
(b)
AA, 3 NAM, S63 (RORT/V. ) dVl= REIN
S/V ay
Solving (8.3) for P and substituting in the result of (a), we get A‘ —An=
[% [RTIV. —b)— al Vo (Vi. + b)T'? — RTIV,, | Vg, — RT In (ViE/Vm) =
[RT In(V.. — b) - (a/T"?)(—-1/b) In(1 + b/ Vy, ) — RT In Vy, J =
RT In(V'2/Vm) = [RT In(1 — b/ Vp,) + (a/bT"*) INL + BIV,, UP =
RT In(V"2/Vm) = RT In(1 = b/Vmn) + (a/bT'”) In + B/Vim) — RT In(V,/Vin)
(c) Si —Sm=-@/0T), (Aim — Am) =
_R In (1 — b/Vm) + (a/2bT *”) In(1 + b/Vm) +R In(ViC/Vin).
U4 — Um = (A — Am) + T(S'S — Sm) = Gal2bT"*) In(1 + B/Vmn).
8.27
Differentiation gives dB/dT = (RT;/P.)(-0.462)(-0.7002T./T*)ee
=
0.3235(RT2/P.T) e970" _ From (8.6), B’ = B/RT and dB'/dT = (dBIdT)/RT
_— B/RT?. Substitution of T, = 305.4 K, P, = 48.2 atm, and T= 298 K in the
equations for B and dB/dT gives B = -182 cm’/mol and dB/dT = 1.186
cm?/mol-K. Then B* = B/RT = -0.00744 atm”! and dB'/dT =
(1.186 cm>/mol-K)/(82.06 cm?-atm/mol-K)(298 K) +
(182 cm3/mol)/(82.06 cm?-atm/mol-K)(298 K)° = 7.35 x 1 Onmatmnan Kae
Fron Probas235 i
teen
REce ecBdd ie
(1.987 cal/mol-K)(298 K)°(750/760)atm(7.35 x 10> atm! K~) = 13 cal/mol
and S4 — S,= RP°(B' + T dB'/dT) =
(1.987 cal/mol-K)(0.987 atm)[-0.00744 + 298(7.35 x 10>)Jatm™ =
0.028 cal/mol-K.
8.28
8.29
© xen a Cole
OU)
fo=-0, f'@=0-x), fC
=1, fFOVas fF(ONE2) Oras ear
be ef0)
eo (eye.
foc) = 14 (4-0) + (x — OY/2! 43-2 OV BI 4---=Lextx tet:
) FON aS GIS, .
fey ey eeepeneh
As ipoee ix,
(ey HOE Tea ae ras (eae
ene /2iee eel ya a2 =O 2 een sae bon =
Tye ee
- --@-DV4+(x
+(x- YB
1/2 )(x—1
107
es
OES Ls FAO inl
iefan O)
hls
8.30
FO=IEa Ge ere
Salextx2sex3i+--:
8.31
(didx\(sin x) = cosx=1 — 3x7/3! + 5x'/5)—-
8.32
x is in radians; 35° = 0.610865 rad. Substitution in (8.35) gives sin 35° =
0.5736.
8.33
The function 1/(x? + 4) has singularities at x = 2i and x = —21 (where + 4=
0). The distance between the origin (point a) and either of these singularities is
2,s0b=2.
8.34
(a)
5p 1"/n! =2.7166667,
= Dax 12! ene /Al
Deg l"/n! = 2.7182818,
es
Lj2o1"/n! =
2.7182818. The exact value is e = 2.7182818.
(b)
For x= 10, we find 1477.6667, 12842.305, and 21991.482 for n =5, 10,
and 20. The exact value is e!° = 22026.466. (For n = 30, one finds
22026.464.)
A BASIC program is
70
10 INPUT “X=";X
20 FORM=5TO20
80 NEXT N
30 S=1
90 PRINT "M=";M;" SUM=";S
40 NF=1
100 NEXT M
50 FORN=1TOM
110 END
60
8.35
STEPS
S=S + XAN/NF
NF =N*NF
From Sec. 8.3, fe= 1.67, =30(353 K)905.K une = 2.7 Vn =
2.7(78 g/mol)/(0.81 g/cm*) = 260 cm*/mol. From Sec. 8.4, Z, is usually
between 0.25 and 0.30, so we take Z,. = P.Vm,-/RTc = 0.275. Thus
P. =0.275
RT,
(82.06 cm*-atm/mol-K)(565 K)0.275 _
Vine
260 cm*/mol
49 atm
(The experimental values are 562 K, 259 cm?/mol, 48 atm.)
108
8.36
Use the isothermal path: liq(I bar) = liq(23.766 torr) re vap(23.766 torr) _
ideal vap(23.766 torr) bs ideal vap(1 bar). For step (a), Prob. 7.50 gives
AGma = —1.8 J/mol. For steb (b), AGm,» = 0, since this is a reversible
constant-7-and-P process. For step (c), Prob. 8.24a gives AGn,, =
AHme— I ASmc = (alRT — b)P =
5.47x10° cm®-atm/mol?__4, <cm? |
23.766 atm
(82.06 cm>-atm/mol-K)(298K) mol | 760
3
ayes
mol
eee =0.6 J/mol
82.06cm°
atm
For step (d), Prob. 7.50 gives AGmq@ = 8557.2 J/mol. The net AG,, 1s 8556.9
J/mol compared with 8555.4 J/mol in Prob. 7.50 and 8557 J/mol from data in
the Appendix.
8.37
(a)
Use of Eqs. (4.52) and (8.5) gives pyr = (1/Cp,m)[T(OVmn/0T)p — Vin] =
Cpl {T (RIP) + B'P + CP? +--+) +(RTIP)(P dB'/dT + P* dC '/dT +
Pa GaP eae
fe I=(RTP)\UB
(RT7/Cp m)(dB'/dT + P dC'/dT + P’ dD'/dT + ---)
lim wor = (RT?/Cpm)dB'/dT
—>0
(b)
P=
Equations (4.47) and (8.4) give (QU/0V)r = T(OP/0T)y —
B
RT ee
Riel ae
B
R
T| —|
Velen
1+—+---
ale
|4+—|
+ :::
——
Veve dh
ee
Fe hoig ae
8.38
From (8.4), Vm — Vi? = (RT/P)(1 + B/Vn + C/V, +: ++) —RTIP =
RT(B/PVm + C/PV2, +--+). In the limit P > 0, we have PV, = RT and
B, since Vm —>eeas P > 0.
Vm - Vid + B+ C/Vm +--+
8.39
(a)
As found in Prob. 8.24a, the van der Waals equation gives B' =
BIRT — alR2T2 and BY = -b/RT? + 2a/R°T°. Substitution in the equation
of Prob. 8.37a with all terms but the first neglected gives pyr =
(2a/RT— b)/Cpm as the low-P estimate of wyr.
109
(b)
At the inversion temperature, the low-P
[jz is zero, So
(2a/RT— b)/Cp m= 0 and T;,p_50 = 2a/bR = 2(1.35 x 10° atm cm® mol™)/
(38.6 cm/mol)(82.06 cm?-atm/mol-K) = 852 K, where N> a and b values
in Sec. 8.4 were used. The low-P 298-K estimate of yz 18 Wy7 =
2(1.35x10°
6
atmcm’
6
-2
mol ~ ) ake
(82.06 cm>- atm/moFK)(298 K)
(10.3cm? eee
a
1
Ba
6.96 cal/mo-K
ee = (0.250 K/atm
82.06cm* atm
where Cp is from the Appendix and (1.19) and (1.21) were used.
Agreement with the experimental [17 1s good, but 7; is poorly predicted.
8.40
(a)
The larger size of a neon atom means that Ne has greater intermolecular
attractions than He, so Ne has a greater a, a greater T., and a greater
Avapf1m. Ne atoms are larger so Ne has a greater b value.
(b)
C3Hsg has greater intermolecular attractions and larger size, and so has the
greater a, T;, AyapHm, and b.
(c)
Due to hydrogen bonds, H20 has greater intermolecular attractions, and
so has the greater a, T,, and AyapHm. The H2S molecule is larger than the
H20 molecule, so H2S has the larger b.
8.41
From (8.18), b = RT,/8P,. = (82.06 cm>-atm/mol-K)(190.6 K)/8(45.4 atm) =
43.1cm*/mol; a =27R°T 2/64P, = 2.27 x 10° cm® atm mol. Vmo =b + RTIP
= 43.1 cm*/mol + (82.06 cm?-atm/mol-K)(273 K)/(100 atm) = 267 cm?/mol.
Vint = b+ RTMP + a/V>4)=
(82.06 cm>- /mol-K)(273 K)
100 atm + (2.27 x 10° cm® atm/mol?)/(267 cm?/mol)”
Vn,1 = 213 cm*/mol. Vn2= b+ RTP + a/V2,,) = 192 cm3/mol; Vm3 = 182
43.1cm?/ mol +
cm?/mol. Successive calculations give (in cm?/mol): 176,3L 725 1702169. 168)
167, 166.6, 166, and 166. From Fig. 8.1a, for CHy at 100 atm and 0°C, Z=
0.785 = PVm/RT and Vm = 0.785RT/P =
0.785(82.06 cm’-atm/mol-K)(273 K)/(100 atm) = 176 cm?/mol.
A BASIC program is
110
8.42
10 DIM V(100)
50° FOR=s1
TO 99
20 INPUT "A=";A
22 INPUT "B=";B
24 INPUT "P=";P
26 INPUT "T=":T
30 R = 82.06
40 V(1)=B+R*T/P
60 V(l+1) =B + R*TAP + A/V(I)A2)
70 PRINT"I="sI:" V=";V(1+1)
80 IF ABS(V(I+1) — V(I))/V(I) < 1E-5
THEN STOP
90 NEXT |
95 END
T, = T/T, = (286 K)/(190.6 K) = 1.50 and P, = P/P. = (91 atm)/(45.4 atm) =
2.00. At these T, and P, values, Fig. 8.10 gives Z=0.83 = PV,/RT and V,;, =
ZRTIP. = 0.83(82.06 cm>-atm/mol-K)(286 K)/(91 atm) = 214 cm*/mol.
8.43
(a)
Asin Prob. 7.33b, d In P/dT= (1/P)(dP/dT) = 0.035687/K and
dP/dT= (0.035687/K)(1 atm) = 0.035687 atm/K at 100°C.
(b)
AV» = (82.058 cm>-atm/mol-K)(373. 15 K)/(1 atm) — 452 cm*/mol —
19 cm*/mol-= 30149 cm?/mol. Then AHm + (T AVm)(dP/dT) =
(373.15 K)(30149 cm*/mol)(0.035687 atm/K) =
(4.0148 x 10° cm?-atm/mol)(8.3145 J/82.058 cm?-atm) = 40.689 kJ/mol.
8.44
Cells for B and C are specified. One can set each of these to zero for the initial
guess. The experimental P and V,, values are entered into columns A and B.
Formulas to calculate P from Eq. (8.4) are entered into column C. In column
D, the squares of the deviations between column-A and column-C pressures
are calculated. The sum of the squares of the deviations is calculated 1n a cell,
and the Solver is set up to minimize this sum by varying B and C. The result is
B = -83.12 cm*/mol and C = 3330 cm®/mol’, with an excellent fit to the data.
8.45
(a) F;
(b) F.
Chapter 9
9.1
(a) mol/m*;
(b) mol/kg;
(c) no units.
9.2
Gis
9.3
(a)
n;=c;V= (0.800 mol/L)(0.145 L) = 0.116 mol.
(b)
(145 g) x 10.0% = 14.5 g;
(c)
m,=nwa =n,/(w — w;) = n/(w — n,M;), where w is the mass of the
solution, w; is the mass of the HCl, n; is the number of moles of HCI, and
Cie
0.398 mol.
M,; is the HCI molar mass. Solving for n;, we get nj = mjw/(1 + mM;j) =
(4.85 mol/kg)(0.145 kg)/[1 + (4.85 mol/kg)(0.03646 kg/mol)] = 0.598
mol HCI. Alternatively, a solution with 1000 g of solvent has 4.85 mol
HCI, which is 176.83 g of HCl. The weight percent of HCI is
[176.83/(1000 + 176.83)]100% = 15.03%. 15.03% of 145 g is 21.79 g of
HCl, which is 0.598 mol HCl.
9.4
(a)
All quantities involved are intensive, so we can use any convenient
amount of solution. Let us take 1 dm? of solution. This amount of
solution has 8.911 mol of CH3;OH, which is 285.5 g of CH3OH. Since the
solution is 30% CH3OH, the solution’s mass is (100/30)(285.5 g) = 951.8
g. Its density is (951.8 g)/(1 dm*) = 0.9518 g/cm’.
9.5
(b)
m;=n/wa = (8.911 mol)/(951.8 g — 285.5 g) = 0.01337 mol/g =
13.37 mol/kg.
(c)
PcH,on =™Mcy,on/V = (285.5 g/d dm*) wey 8)sl ep
Let us take 1000 g of solution. This contains (0.800 %)(1000 g) =
8.00 g of NH; and 1000 g — 8 g = 992 g of water. This is 0.4697 mol of NH3
and 55.06 mol of water. Then m; = nj/wa = (0.4697 mol)/(992 g) =
0.000474 mol/g = 0.474 mol/kg. Also, x; = 0.4697/(0.4697 + 55.06) = 0.00846.
9.6 The solution’s mass is m = pV = (1.2885 g/cm?)(1000 cm’) = 1288.5 g. The
CsCl mass is (2.296 mol) x (168.358 g/mol) = 386.6 g. The solvent mass is
112
1288.5 g — 386.6 g = 901.9 g. The molality is (2.296 mol)/(0.9019 kg) =
2.546 mol/kg.
Dad,
All quantities involved are intensive, so we can take any amount of solution.
Take an amount of solution that contains one kg of water and hence contains
1.506 mol of KI. The solution’s mass is 1000 g + (1.506 mol)(166.00 g/mol) =
1250.0 g. The solution’s volume is V = m/p = (1250.0 g)/(1.1659 g/cm’) =
1072.1 cm*. The molarity is (1.506 mol)/(1.0721 L) = 1.405 mol/L.
9.8
cj = n;/V = nj/(w/p) = pn,/w, where w is the solution’s mass. Because the
solution is very dilute, we have w = w; + Wa = Wa = NAMa, where A is the
solvent. Then c; = pnj/naMa. Also, x; =n/(na + nj) = n/a, SO Cj =
ox/Ma. For the molality, we have m; = nwa =nj/naMa = x;/M,. From
c; = px/Ma and m; = x/Ma, we get c; = pmi.
De
As noted after Eq. (1.4), Ma = M, X 1 g/mol, so mp = np/naM a =
np/(naM,.a g/mol) = np/(naM, al0~> kg/mol) = (1000 np/naM,.a) mol/kg.
(cler: y(d)o0. (ey 1
9.10
(a)ee (Daron
9.11
V=nvV, + Ve=
Dab
aaa)
(alga)
Te
(0.500 mol)(18.63 cm?/mol) + (55.51 mol)(18.062 cm*/mol) = 1011.9 cm’.
9.12
Cp=nCp,
+ Bes. . An amount of 0.1000-mol/kg solution that contains
1000 g of solvent has 0.1000 mol (which is 5.844 g) of NaCl and has an NaCl
weight percentage of [5.844/(5.844 + 1000)]100% = 0.581%. Taking 0.581%
of 1000 g gives 5.81 g of NaCl in the 1000 g of solution, which is 0.0994 mol
of NaCl. The percent water is 100 — 0.581 = 99.419%. The HO mass is 994.19
g, which is 55.186 mol H2O. So Cp = (55.186 mol)(17.992 cal/mol-K) +
(0.0994 mol)(—17.00 cal/mol-K) = 991.22 cal/K.
3
V= nV, + n,Vq = 307.09 cm’ = [(72.061/18.0153) mol](16.488 cm*/mol) +
[(192.252/32.0422) mol] Voy,on and Vou,ou = 40.19 em’/mol.
113
9.14
Take solutions that each contain 1000 g of H2O (constant nyo ). Let w be the
solution mass. For the 12% solution: (1000 g)/w = 0.88000 and w = 1136.36 g;
this solution contains 136.36 g of CH3OH, which is 4.2557 mol of CH30H; the
solution has V = (1136.36 g)/(0.97942 g/cm’) = 1160.24 cm’. For the 13%
solution: (1000 g)/w = 0.87000 and w = 1149.43 g; this solution has 149.43 g
of CH3OH, which is 4.6636 mol of CH30H; the solution’s volume is V =
(1149.43 g)/(0.97799 g/cm’) = 1175.30 cm®. We have Voy.on =
(AV/AncH oH)7.Ping.g = (15.06 em’)/(0.4079 mol) = 36.92 em’/mol. To find
Vis,0 » We now take solutions with 100 g of CH30H (constant noy.o4 )and do
the calculations the same way as for the constant ny,9 solutions. For the 12%
solution, we find nyo = 40.706 mol and V = 850.84 cm’. For the 13%
solution, we find nyo = 37.148 mol and V = 786.54 cm’. Hence Vii,0 =
(AV/ Ano) Pincuson = (-64.30 cm?)/(-3.558 mol) = 18.07 cm*/mol.
915
(a)
Vof MgSO, at a given composition equals the slope of Fig. 9.3 at that
composition. The slope is zero at the minimum, which occurs at Nyoso, =
0.07 moles in 1000 g of water, which is a molality of 0.07 mol/kg.
(b)
Infinite dilution corresponds to Nugso,
0. Drawing the tangent line to
the curve at n MgSO, = 0, one finds its slope
P to be —3.s5 cm°/mol, which is
V of MgSOg.
(c)
Drawing the tangent line to the curve at nyggo, = 9.05 mol, one finds its
3
—_
17
—_
17
1001.69, cm? = (55.509 mol) Viz, + (0.05 mol)(-0.54 cm?/mol) and
Vino = 18.045 cm*/mol.
9.16
— waci=
co
V (Na‘)+V (Cl),
Vino,=¥
(Kk )+ V_(NO;),
and
Vyano,= V ~(Na*)+ V°(NO3). Then Vic, = V"(K*) + VV(Cl")=
Vino, + Vnaci — Vwano, = 38.0 + 16.6 — 27.8) cm’/mol = 26.8 cm*/mol.
9.17
Substitution of G =); nj; into
G = U + PV — TS gives the desired result.
114
—
9.18
A =(0A/0n, )F Pine where A is the Helmholtz energy of the solution, n; is the
number of moles of 7 in the solution, 7 and P are temperature and pressure, and
nz; Indicates that all mole numbers except 7 are held fixed.
9:19
H=U+
PV. Partial differentiation gives
(OH/8n; Jr. p.n,,,
9.20
(a)
=(OU/ON; Vr p.n,,, + PCOV/ON; Vr, p.n,,, »80 H, = U, + PV,.
G=Dinp= Din; + RT In (P/P°)). Taking P; = P° and P2 = P; in
(4.65), we have for n; moles of pure gas i: G*¥(T,P,,n;) — G*(T, P°, ni)
=njRT J" P'dP = njRT In (P/P°). So G¥(T, Pi, ni) = G(T, P®, ni) +
niRT In (PJP?) = n;G* ,(T, P°) + n:RT In (P/P°) = n,[p; + RT In (P/P*)),
since G* ;=; for a pure substance. Hence the above sum for G of the
mixture: becomes G =)
(b)
S= -(O/0T)p,
7.G?(1,.P.1,).
G= Li [-(OG*/0T)p,,J= Xi S¥(T, P,,n;), where (4.51)
was used.
(e)
Ha=G +78 =), Gt + PQ S* aia(G*
SS) = 2) (7. nj), where
P; is absent since H is independent of P for an ideal gas.
(d)
pp, Li H* = Li CQH*/OT),, = Li Ch, (T, ni).
Cp= (OH/OT)p,, = (O/T)
U=H-PV=)>;
Dey
(e)
a
= 2D; (A *—- PV)=
H* —PQi nRTIP) =D; (A *t- nRT)
1).
Assuming the mixture is ideal, we have Cp = ae Ch (7, n,)=
CB.o, + Cé.co, ="0,CP.m.o, + "co, CP,m,co, =
(0.100 mol)(29.355 J/mol-K) + (0.300 mol)(37.11 J/mol-K) = 14.07 J/K.
9.21
Taking d/0T of (9.32), we get (QA ni,G/OT)p,,= (OT)p,,Lini(G, - Gi) =
OT)pin,1= Li ni(-S; + SA) = Limi SH; —LiniS; =
Yi nil(OG,/OT)pn,OG,
S* — § =—AnixS, where (9.30) was used.
9.22
n(H20) = 1.119 mol and n(C2HsOH) = 0.977 mol.
x(C2HsOH) =
0.977/2.087 = 0.468. For this composition, Fig. 9.9 gives V (H,O) =
115
16.8 cm’/mol, V(C,H,OH) =57.0cm*/mol. V= V,n, + V,n, =
(16.8 cm?/mol)(1.11 mol) + (57.0 cm*/mol)(0.977 mol) = 74.3 cm’.
9.23
We draw the tangent line at Xethano! = 0.4. This line intersects the Xethanol =
0 axis at -1.1 cm’/mol = Vio — Viu,0 = Vu,o — 18.05 cm*/mol, and
Vito = 16.95 cm?/mol. The tangent line intersects the Xethano! = 1 axis at
3
—l.15 cm’/mol
Vee
9.24
Sa
= V,ethanol _ Vxm, ethanol =V.ethanol
3cm
—~ 58.4 cm*/mol, and
mol.
The pure molar volumes are Vi,4,9 = M/p = (18.015 g/mol)/(0.99705 g/cm’)
= 18.07 cm*/mol; V* cy,oH = 40.71 cm*/mol. We plot AmixV/n vs. x4,
(similar to Fig. 9.7).
(a)
Drawing the tangent line at x9
=0, we find it intersects the x49 = 1
line at —3.6 cm?/mol. (With a reasonable choice of scale, the intersection
occurs off the paper and can be calculated by extrapolation.) Hence, at
X4,0 =9, Vio — Vi.u,0 =—3.6 cm*/mol = Vj,9 — 18.1 cm’/mol and
Vit,0 = 14.5 cm*/mol. Of course, ie
Ve cujon = 40.7 cm?/mol at
Xy,0 =9.
Amix V/n
0.4
(b)
x(H,O)
0.6
The tangent line at Xy,0 = 0.4 intersects the x9
=0
line at
-0.5 cm’/mol and intersects X49 = 1 at-1.6 cm’/mol, so
116
VoH,OH 7 Vin.CH,OH = -0.5 cm’/mol and V,CH,0H = 40.2 cm?/mol; also,
Vii,0 = 16.5 cm*/mol.
(c)
9.25
Vicu,on ~ Vm,cu,on = —!-45 cm’/mol and V,CH,OH = 39.2; cm*/mol;
Vio — V¥.n,0 =-0.7 cm’/mol and Vy,9 = 17.4 cm*/mol.
(a)
pif =O + (3/2)cng\? + 2kng for naMa = 1 kg.
Va = (OVidng)7
(b)
For mpg = | mol/kg, we have ng = | mol in a solution with naMag = | kg.
(c)
V =n,V, +ngVz, andn,Va =V —npV, =a + bnp + Chea + kt, = npp
Qin = d= 5 Cee kn;,. Since na = (1 kg)/Ma, we get Vie
ic
V, = 16.6253 cm*/mol + 1.5(1.7738 cm’/mol*”)(1.0000 mol)!" +
2(0.1194 cm?/mol’)(1.0000 mol) = 19.5248 cm*/mol.
(Ma/1000 g)(a— 1 cng” — kng) for nsMa = 1 kg.
(d)
(e)
(f)
mp = np/naMa = np/(1 kg) = np/kg and ng = mp kg. Substitution in the
results of (a) and (c) gives the desired equations.
Vis,o = (18.0152 g/mol)[1002.96 cm* — 1(1.7738 cm?/mol*”) x
(1 mol/kg)®” kg*? — (0.1194 cm*/mol’)(1 mol/kg)” kg”]/(1000 g) =
18.050 cm?/mol.
In the infinite-dilution limit, mp goes to zero and Vy in (d) becomes
V;° = b = 16.6253 cm’*/mol.
9.26
(a)
z= AmxV/n = (V— V*)/(na + np), So V— V* = (na + np)z and
V=(natnp)zt+ V* =(natnp)z+ naVan
(b)
+ nevi
n = V, for
4),,.7,p
Taking (0/dn,4),,,.7,p Of the result of (a) gives (OV/O
the left side and z + (na + np) (Oz/On,),,, + Vn. for the nght side. (Note
that V* , is constant at fixed T and P and is independent of na.)
Equating these expressions, we get Ve =r
(c)
(ddonny,, + Vana:
Equation (1.35) gives (dz/dn,),, = (dz/dx, )(0xg/dn,),, - Since Xp =
np/(na + np), we have (0xp/0N 4) ng = —np/(na + np) = —xp/(na + Np) =
_xp/n. The result of (b) gives (02/dn4),, =(Va —Vik.4 — z)/n.
Substitution of these expressions for dx,/dn, and dz/dn, into
117
dz/dn ,= (dz/dxp)(Oxp/dna) gives dz/dxp =(V* , — V4 +2)/xp. The ng
subscript can be omitted from (dz/dx, ),,, because z = (V— V*)/n is an
intensive quantity and so is a function of xg only and is independent of
the size of the system and hence of npg.
(d)
The plotted quantity AnixV/n is symmetric in A and B. Since the intercept
at xp =.01S V, — V*,, the A-B symmetry means that the intercept at
va — OHS Ve cil Gaiaa Buti
= Ouse —i8
9.27
The tangent to the AnixH/n versus Xj4,59, Curve at Xy,59, = 0.4 intersects
X,50, = 9 at 13.9 kJ/mol = AA gig¢
4,0 and intersects X4,,59, = 1 at-16.4
kJ/mol = AA gre 1,50, - Drawing the tangent at x}, 59, = 0.333, we find
AF site 4,80, = —9-5 kJ/mol and AH git y,80, = 22-6 kJ/mol.
9.28
NAH
SS eH:
(0 A nix
H/ON8g )r pn,
=
(OH/dn,
ee
ae (OH
*/Ong
)r pn,
.
But by definition, H, = (H/dng);.p,,- Also, H* =nx H* , (T, P) +
NgHy g(T,P), so (GH*/0ng)r p.n,= Hh.g- Therefore (0A, ;,H/Ong)7
pn, =
El — HX , = Adgitrs, where (9.38) was used.
9729
Subtraction of A;Hyxacys) from the apparent A, H° values give the following
integral heats of solution per mole of NaCl vs. xnaci:
AFiintNacV/(kJ/mol)
1.874
2.347
3.016
3 hl
XNaC!
0.1
0.0625
0.03846
0.019608
As in the last example in Sec. 9.4, multiplication of AHintNaci by xNaci gives
Amixf/n:
(AmixH/n)/(J/mol)
187.4
146.7
116.0
72.77
0
XNaCl
0.1
0.0625
0.03846
0.019608
0
We plot AmixH/n vs. xnacj and draw the tangent at xnac) = 0.05. The tangent line
intersects Xnac) = O at 69 J/mol = Hino _ HyH,0 and intersects xnac; = 0.1 at
198 J/mol. Extrapolation gives the intersection at XNacr= 1 as
[69 + 10(198 — 69)]J/mol = 1.36 kJ/mol = Hyac} — H* naci-
118
200.0
0
0.02
0.04
0.06
0.08
0.1
x(NaCl)
9.30
The NBS tables give A ;H 39. = —92.307 kJ/mol for HCl(g) and A ;H 50g =
~121.55 kJ/mol for HCI in 1 mol of H2O. Thus AHiint uci = —29.24 kJ/mol for a
=
solution with xc) = 0.500. From (9.37), Amixf = nuci AHintuci and Amix/nior
xuci AHint uci = —14.62 kJ/mol at xuc1 = 0.5. Use of further data gives (values in
kJ/mol)
XHCI
0.500
0.400
0.333
0.286
0.250
0.200
AHin. uc)
29.24
40.36
-48.65
-53.17
—56.18
-60.61
Nee Ano
14.62)
ol 615
16.22)
15 18
-14.05
-12.12
The
We plot AmixH/mot VS. XHCI and draw the tangent line at xc) = 0.30.
intercepts at xyc) = 0 and xyc1 = 1 give AH git n,0 = —8.3 kJ/mol and
AA aise acl
= 32
kJ/mol.
Amixf/n
119
9.31
(a) F; (b) T; (c) T; (d) F;
SREY
No. Models show that the A-B intermolecular forces differ from the A-A and
B-B intermolecular forces.
9753
(a) mle Dlg
9.34
The vapor is ideal and the pressure dependence of p of the liquid can be
neglected.
9.35
peniest eo
C). He (d)eb;
a(e)
mOl-and Miet = 1-06) MOl
gat) a
ether= U4 12 ano
0 4 oe
Equation (9.44) gives AmixG = (1.987 cal/mol-K)(293.1 K) x
[(1.280 mol) In 0.5412 + (1.085 mol) In 0.4588] =—-950 cal.
Amixlt = 0)
AnixV= 0,
AmixG — Omit — 1DAmsS, SO Amino = (950 Caly@9s-) Ky = 3.24
cal/K.
9.36
(a)
Men = (100.0 g)/(78.11 g/mol) = 1.280 mol. nyo = 1.085 mol.
Xben = 1.280/(1.280 + 1.085) =0.5412. x4) =0.4588. Pren =
0.5412(74.7 torr)= 40.4 torr. Pio,=0.4588(22.3 torr)= 10.2 torr.
(b)
Prot= (40.4 + 10.2) torr = 50.6 torr. Use of P; = x; P gives
Xben = (40.4 torr)/(50.6 torr) = 0.798.
9.37
P=
Phex + Poct = Xhex, plaremit
le
Xs
Oe0e
nee) lace Xnex ere les Pe
(666 torr — 354 torr)/(1836 torr — 354 torr) := 0.2)
Phe = ne, hf = 021111836 torr = 287 torn ee
(387 torr)/(666 torr) = 0.581;
9.38
ee
ICE
oe
= Poet) =
O21
0 780:
Xocty = 0.419.
Pi= x; P= x;P¥. 0.555(95.0 torr) =0.305P*,, and P*,= 173 torr.
P= x; P¥ +x)P¥ = 95.0 torr= 0.305P*,, + 0. COBY fees
0.305(173 torr) +oe
695 Pe, and Pi», = 60.8 torr. We assumed an ideal
solution, ideal vapor, and the pressure independence of LF.
120
9.39
(QypeRbea ie a= py PA SO x2 =n P#/Ps Also, P'= Pa + Po=
DP etal
ex'sPEAS) 1x) P, so,x8 =
Pa LOX ibe ct Dede, Piece P*)/ (oc, PETP® cl
(b)
Yo),
If Bis benzene, then P3/P& = 74.7/22.3 = 3.35 and the equation in (a)
gives xp.y = 3.35 xy/(1 + 2.35 xp). We get
vD
08
0
A
TONS
ORE
OT
ra
Xp
0
0.2
0.4
0.6
0.8
l
The plot resembles the upper curve in Fig. 9.18b. For toluene (t), we find
x? = 0.299 x!/(1 — 0.701 x; ) and
9.40
yen
0
0.070ae
016698
0:310=
*0:5457%
al
x
0
0.2
0.4
0.6
0.8
1
For an ideal solution, V= V* = n,V*, + n,V*, =n\Mj\/p} + n2M2/pz =
m,/p* + m,/p% = (33.33 g)/(0.8790 g/cm’) + (33.33 g)/(0.8668 g/cm’) =
76.37 cm’. Then p = m/V = (66.66 g)/(76.37 cm’) = 0.8729 g/cm’.
9.41
(a) AmixCp = Cp— Cf = (QHIOT)p,, - (QH*/0T)>p,,, = (O/T)p,,,(H - H*) =
(0 A, H/0T)p,,= 9, since AmixH = 0 for an ideal solution.
(b)
9.42
Cp= CB = Ch m1 + MC¥mo = [(100/78.11) mol](136 J/mol-K) +
(100/92. 14)mol(156 J/mol-K) = 343 J/mol-K.
At xp = 0.50, the tangent is horizontal and intersects the xg = 0 and xp = | axes
at —0.415 kcal/mol = ta — WX = Up — YZ; the calculated result is
WW; — W* = RT In 0.5 = (1.987 cal/mol-K)(298.1 K) In 0.5 =-0.411 kcal/mol.
At xp = 0.25, the tangent line’s intercept at xg = 0 is 0.189 kcal/mol =
Ua — Ux and that at xp = | is -0.815 kcal/mol = Lp — pZ; the calculated values
are Ua — w* = RT In 0.75 =-0.170 kcal/mol and pp — us = RT In 0.25 =
—0.821 kcal/mol.
121
9.43
From (9.30), (9.31), (9.42), and (9.28): S; = —(0p,/0T)p,,, = —(OBF/OT)
pn,Rinx;= S4;-Rinx.
V, = —(0p,/0P);,, = (On#/0P); = Vinit
A, = wit TS, = pt + RT In x + TS5,, —RTInx)= Hj, (since we = Gy).
9.44
p= wo +RT In (P/P°) = wo + RT In (xP/P?) = wo + RT In (PIP?) + RT In x.
For x; = 1, this equation becomes p* = ph; + RT In (P/P°), so
Wj= W* + RTIn x;.
9.45
Substitution of Equation (9.42) into the phase-equilibrium condition py’ = uP
gives WX + RT In x,° = pt + RT In xP so xo = x? Similarly, x% = ae
9.46
(a)
The use of G=nju; + naa, for the solution gives G of the final state in
the dilution process as G2 = njli2 + NA2WA2. For the initial state of the
dilution process, G; = nil, + NAMA
+ (Ma2—Nai) UX.
AG of the
dilution process is G2 — G; and use of these expressions for G2 and G,
gives the desired result.
(b)
Equating the right sides of (9.69) and (9.70) and using (9.71), we get
(9.72).
9.47
(a)
The solvent A is ethanol and Raoult’s law gives Pa = ah P=
Pen = 0.9900(172.76 torr)'= 171.03 torr.
(177295
(b)
Ina
Poa =P
Po =
17 1103) torr = 6.92 tore:
gas mixture, P} = x;P, so x}, = 171.03/177.95 = 0.9611, and
Ken) = 0.92117 71.95 = 0.03880.
(c)
Pi=Kjx! , 80 Ke = Poni x¢y) = (6.92 torr)/0.0100 = 692 torr.
(d) Pa = x4 PX = Pen = 0.9800(172.76 torr) = 169.30 torr; Pon = Ken x!,. =
(692 torr)(0.0200) = 13.84 torr.
P = (169.30 + 13.84) torr = 183.14 torr.
Xeth = Pe/P = 169.30/183.14 = 0.9244 and xo, = Pon/P = 13.84/183.14
= 0.0756.
9.48
(a)
Pon = xg,P= 0.9794(438.59 torr) = 429.56 torr; Pen =
(1 — 0.9794)(438.59 torr) = 9.03 torr.
122
(b)
Raoult’s law for the solvent chloroform gives Pon = x1,,P*,, and
P*,, = (429.56 torr)/0.9900 = 433.90 torr.
9.49
(c)
Pet = Ken x), and Ken = (9.03 torr)/0.0100 = 903 torr.
(a)
At xcs, = 1 in Fig. 9.21b, P = 424 torr.
(b)
At x1,, =0.40 in Fig. 9.21a, the curves give P = 266 torr and Pen =
Seon, SOM
9.50
= Eee — Oo
We have K; = P,/ x for very dilute solutions of 7. For CS» as solute, we draw
the line tangent to the P(CS2) curve at the point ibe = 0. Since P; = K;x;, the
slope of this tangent line equals Kcg, ; the slope equals the intercept of the
tangent line with the ae = 1 line. We find Keg, = 1.25 x 10° torr in acetone as
solvent. Drawing the tangent line to the P(ac) curve at the point te = 1 and
finding its intercept with Xces = 0, we get Kx. = 2.0 x 10° torr in the solvent
CS2.
oSt
From (9.62), pw; — ee = RT \n (K;/P°) = (8.314 J/mol-K)(308 K) In (145/750)
=——4 21 kKI/mol.
9.52
(a)
ny =8.14x 10° mol, Ny,9 = 5-55 mol; ip = 1.47 x 10°.
Ky, = Py, /xy, = (1.00 atm)/(1.47 x 10) = 6.82 x 10° atm.
(b)
xy, = Py, /Ky, = (10.00 atm)/(68200 atm) = 1.47 x 10™.
ny, = 8.14 x 10° mol and my, = 1.64 mg.
Eee)
xy, = Py,/Ky, = (0.780 x 760 torr)/(5.75 x 10’ torr) = 1.03 x 10°,
Xo, = (0.210 x 760 torr)/(2.95 x 10’ torr) = 5.41 x 10%.
no, and ny, are
negligible compared to nyo = 5.55 mol, so ny, = (1.03 x 10°)(5.55 mol) =
5.72 x 10° mol and no, = (5.41 x 10°°)(5.55 mol) = 3.00 x 10” mol.
my, = 1.60 mg and mo, = 0.960 mg.
123
9.54
(a)
/OP) »,-But
V,=(0n,/0P);, =(0/0P)r_, (Hy + RT Inx,) = (Op;
(9.31) shows that (Op; /0P);
=.
.s0 Va=Ve.Since,, fonan
ideally dilute solution is a function of T and P only, its derivative
(Ou;/0P);,, is a function of T and P only, and (dy; /0P);,,, and V,°
(which equals (On? /0P rn, ) are independent of concentration for
concentrations in the ideally dilute range. Therefore V, (which equals
V,°) equals its limiting infinite-dilution value V2
(b)
Using the intercept method of Fig. 9.7, we draw the tangent line to the
curve at Xethano| = 1. The tangent line intersects the Xethanoi = 0 vertical axis
at 4.7 cm*/mol = Vio —Vinn,o. since Vino = 18.0 cm*/mol at 20°C,
we get Vis,0 = 13.3 cm’/mol in ethanol, in rough agreement with Fig.
on
9°55
Equations (9.30), (9.28), and (9.59) give: Si = —(Op;/0T) pn, =
—(0/0T)
py, (fb? + RT In x) = S; —R Inx;. The infinite-dilution limit of this
equation gives Si = CS: + R In x;)”. We have i =Gi+TS;
57S,
= uw; + RT In x;
aye x; = H;. The infinite-dilution limit of this equation gives
—oo
Hi; =e
9.56
Equations (9.17), (9.66) and (9.67) give AmixV=Diea n(V; — V*,) +
na(V, — V* 4) = Disa n(V,° — V* ,). Equations (9.68) and (9.66) give
AmixH=Disa n( H; — H*;)+na(H, — H*,)=Diean(H; -H*,).
ey
O= 2; VM; = 2, vj,(u; + RT In Kieg) = 2; Vib; + RT Z; In (Xieq )
PSNG Fe
RT In I]; (x;.4)"'
=AG°
+ RT InK,, and AG° =-RT InK,, where (1.68) and
1,eq
(1.67) were used.
124
9.58
(a)
Bs = P,/K;, so as K; increases, ie decreases. The solubility of Oz in water
decreases as T increases.
(b)
(H?° — AH; RT? = 0 In KJ/OT)p = (A In KJAT)p.
A In K; = In (3.52 X 10’) — In (2.95 x 10’) = 0.177.
Hee —~ H;” =-(8.314 J/mol-K)(298 K)’0.177/(10 K) = -13.1 kJ/mol.
(c)
The log of (9.62) gives RT In Kj/bar = py! — pj” = G; —G; and we
have G, — G, = (8.314 J/mol-K)(298 K) In (44100 bar/1 bar) =
26.5 kJ/mol.
(d)
9.59
AS =(AH —AG)/T, so 5, — 5,” = [(-13100 — 26500)/298] J/mol-K =
5
_133 J/mol-K.
The log of Eq. (9.62) reads In K; = In P® + (n>! — n°? RT. Partial
+
differentiation with respect to T gives (0 In Kj/0T)p = (et 24 ae WRT*
(ape! BT)p - dus’ /dTVRT = (Gi — Ge RIC # GSE Se DIRRE
ov
CEP
oy
=e,
DEP
~,!
SOG
IP
sOR
h
2S
MER? aoe eae
;
differentiation of In K; with respect to P gives (0 In K/0P)r = (0 uo! /OP)r=
Vi IRT= VeaiiRT (since 4," depends only on 7).
9.60
nbp,ben =
Trouton’s rule is AvapHmanbp/Tp = 21 cal/mol-K. We find Ayapfm.
for the vapor
7420 cal/mol; AvapHm.nbp.to! = 8060 cal/mol. Equation (7.21) gives
pressure of pure benzene at 120°C: In [P¥.,/(1 atm)] =
and
(7420 cal/mol)(1/353.2 K — 1/393.1 K)/(1.987 cal/mol-K) = 1.073
P*,, = 2.92 atm. Similarly, we find P%,, = 1.29 atm. Phen = Sed et
0.68(2.92 atm) = 1.99 atm.
Pro = 0.32(1.29 atm) = 0.41 atm.
P= 2.40 atm.
cy of
xv, = Pren/P = (1.99 atm)/(2.40 atm) = 0.83. We assumed the accura
gases, an ideal
Trouton’s rule, the T independence of AH of vaporization, ideal
solution, and the pressure independence of p*.
9.61
png
AmnG=G—G* =G—napt —ngpy. Then pa = (0G/0N4)7
+
(0/0n 4), p.ng (NA ux +npyy + AmixG) = wx + RT In xa
125
naRT(OIn xa/Ona) Pe
B RT(O In xp/Ona) ny = MA + RT In xa +
naRT{I/na — 1/(na + ng)] + nBRT[-1/(na + np)] = WX + RT In xa +
RT — xaRT — xpRT = wx + RT In Xa, since xa + xp = 1. (The partial derivatives
of the logs are found as in Prob. 9.64b.)
9.62
(a) Use of (AG/OP)r= V gives AG, = |& v* dP’ + J% V* dP’. We have
AG> = 0 for this constant-7-and-P equilibnum process. AG3 = AH3 —
T AS3 =—-T AS3 = —naRT In (P¥/Pa) — npRT In (P3#/Pg), where (3.29) and
Boyle’s law were used. AG4 = 0 (see the end of Sec. 6.1). AGs = 0
(constant-7-and-P equilibrium process). AG¢ = IRs+P Vasp dP’.
(b)
They are small because G of a liquid varies only slowly with P.
(c)
With AG, and AGg¢ neglected, AG = AnixG = AG3 = naRT In (Pa/ PX) +
npRT In (P,/P%).
(d)
9.63
Use (9.51).
Eq. (9.77) becomes AnixG/(ma + Ng) = RT[xa In a. P/ P=) +xp In Cee! Heep lh
At Xeth = 0.200, AmixG/(na + ng) = (8.314 J/mol-K)(318 K) x
{0.200 In [(0.1552)454.53/172.76] + 0.800 In [(1 — 0.1552)454.53/433.54]} =
—731 J/mol. At Xeth = 0.400, 0.600, 0.800, we get AmixG/mior
= —1034 J/mol,
—1554 J/mol, —997 J/mol, respectively. At xen = 0 and 1, AmixG/niot = 0.
9.64
(a) From (9.35), AmixS == (0 AppixGlOT)
py,= 24 R In Xa — ng R In Xp (na + ng)xaxp(OW/0T)p. From (9.33), AH mix = AmixG + T AmixS =
(na + Np)xaxplW— T(OW/0T)p]. From (9.34), AmixV = (0 AmixG/OP )7
=
(na + nNg)Xaxp(OW/OP)r.
(b)
AnixG = G- G*, so G = G* + AnixG = napx(T, P) + ngps (T, P) + AmixG.
Then pa = (0G/On4)7
ping = WE + (0 AnixG/ON
4)z pn, » We have In x, =
In na — In (na + np) and In xg = In ng — In (na + ng), SO 0 In xa/Ona =
I/na — I/(ng + ng) and O In xp/dna = —1/(na + np). Also, xa xp(na + np) =
nanp/(na + ng). Using these equations, we find La = ux + RT In xa +
naRT [1/nag — 1/(na + ng)] — ngRT/(na + ng) +
W(T, P)[np/(na + ng) — nanp/(na + npg)]= wk + RT In xy + W(T, P) x2.
126
Since AmixG is symmetric in A and B, by analogy to ta we have
Up = pe + RT In xg + WT, P)
(c)
x4.
Equating a, in the solution to pa in the vapor above the solution, we get
ust RE In xa Wx3 = Ws oa; + RT In (Pa/P°). Also, for pure liquid A
in equilibrium with its vapor at T: WY = WA eas + RT In (P¥/P°).
Subtraction of this equation from the preceding one gives RT In xa +
Wx? = RT In (Pa/ P¥), so In (Pal Pt) = In x4 + Wx3/RT; so Pal Pk =
exp(In xa)exp(W x5 IRD), and Pa=xa Pr exp(W x; /RT). By symmetry,
Pp = Xp P§ exp(Wx4/RT).
9.65
AyapG° involves isothermal conversion of liquid at 1 bar to vapor at 1 bar. We
use this isothermal path:
liq(P®) + lig(P,) > vap(P;) > vap(P?)
AG, = 0, since moderate pressure changes have little effect on liquid
thermodynamic properties. AG? = 0, since this is an equilibrium process at
constant T and P. Assuming ideal vapor, we have AG3 = AH; — TWAS; =—T AS3
= —nRT In (P,/P°), where (3.29) and Boyle’s law were used. Therefore AyapG°
= AGm.1 + AGm2 + AGm3 = —RT In (P;/P°) and P; = P° exp (—AyapG°/RT).
9.66
(a) UORHORVa, G(byeUnention Vim (c)s otic:
9.67
(a) All;
(b) ideal;
(f) ideal;
(g) ideal.
9.68
9.69
(c) ideal and ideally dilute;
(d) all;
(e) ideal;
(a)
Neither; neither.
(b)
CCl, — Raoult’s law; CH30H — Henry’s law.
(c)
CH;0H — Raoult’s law; CCl4 — Henry’s law.
(d)
Both — Raoult’s law.
(a)
True, since the equilibrium condition at constant T and P is minimization
of G. (b) False; see Fig. 9.5. (c) False. Ina liquid solution,
intermolecular interactions are large. (d) T. (e) F. (f) T.
127
Chapter 10
10.1
(a) T; (b) T; (©) T; @)T; (e) F.
10.2.
(a) No.
(b) Yes.
(c) Yes. y; = aj/x; and a; in (10.3) depends on p;.
(d) Yes.
10.3.
From (10.4) and (10.7), uw; =y,* (7, P)+
RT
a,;; > x;
Ina,,;. As x; 41, y,;; 1
and
1. As x; decreases (at constant 7 and P) from 1, Eq. (4.90) shows
that w; decreases. The equation p;
=—,;*+ RT Ina,, then shows that this
decrease in 1; means that a,,; decreases from its limiting value of | as x;
decreases. Hence a,, can never be greater than 1. The definition (10.3) shows
that a,; can never be negative. Hence 0 < a); < 1.
10.4
(a)
0G*/dn; = (0/dn,)(G — G*) = AG/An; — OG4/An; = wy; — w'? , where all
derivatives are at constant T, P, njzi.
(b)
p= pw; + RT In yix;= (pw; + RT In x) + RT Iny; = tee + RTIn y;, so
bi pie = RT In y; and use of the result of (a) gives the desired equation.
(c)
(0G*/dng),,. =(0/ Ong [ng + nc GE] =GE + (ng +nc)dGE/ On» (Eq.
(a)]. We have (0G,,/Ong),,, = (OG;,/ OXp )(OXg /Ang),.. Also,
x3 /Ong =(0/
Ong )[ng
(ng tno) ']= (ng +nc)—ng (ng tne)? =
(ng +c —N,)/@t,
+Nc) =Ne its + ne) =e (ng +Nc), So
(OG;,/ Ong), = (OG/OXp)xc M(ng +nc) [Eq. (b)]. Use of Eqs. (b) and
(10.106) in Eq. (a) gives RT Iny,, =Gi + x¢(OGE/Oxg)>p.
10S
ee(ael
10.6
(a)
(bal
yii= x) P/x!
PX.
Yicnt = (1 — 0.138)(304.2 torr)/(1 — 0.200)(295.1 torr) = 1.11.
128
Yieth = 0.138(304.2 torr)/0.200(102.8 torr) = 2.04.
Acht = YichXchi = 1.11(0.800) = 0.889; ajeth = 2.04(0.200) = 0.408.
(b)
y= p; + RT Ina; = p* + RT In jj, So Wj — W* = RT In ay,.
Meth — W*, = (8.3145 J/mol-K)(308.1 K) In 0.408 = —2300 J/mol.
Mon — WA, = RT In 0.889 = -301 J/mol.
(c)
Egs. (9.32), (10.7), and (10.4) give AmixG = Di nj (Uj — w*).=
> njRT In ay; = (8.3145 J/mol-K)(308.1 K)[(0.200 mol) In 0.408 +
(0.800 mol) In 0.889] = —700 J. Alternatively, AmixG =
(0.200 mol)(—2300 J/mol) + (0.800 mol)(—301 J/mol) = -701 J.
(d)
For an ideal solution, yj; = 1 and aj; = x;, So
AmixG = (8.3145 J/mol-K)(308.1 K) x
[(0.200 mol) In 0.200 + (0.800) In 0.800] = —1280 J.
10.7
(a)
Cie Pips =o FIPS
di w= 0,020(0,02 kPa)/(19.92 kPa) = 0.176.
Viw = Aw/Xw = 0.176/0.300 = 0.586.
At hp = (1 — 0.696)(5.03 kPa)/(2.35 kPa) = 0.651;
Viho = 0.651/(1 — 0.300) = 0.930.
(b)
Since water is the solvent, Yi1,w = Yiw = 0.586. Example 10.2 in Sec. 10.3
gives Yiihp = Une / Kyp ) Yinp> where Knp = (Php/ wi )”. We plot hs P/ Sin
versus an and extrapolate to saa = 0. We find Knp = 0.62 kPa. (Use of a
spreadsheet shows that a cubic or quartic polynomial gives an excellent
fit to the data; the fitted equation has an intercept of 0.61 for the cubic
polynomial and 0.62 for the quartic.) So Yihp = (2.35 kPa/0.62 kPa)0.930
— ie ae
(c)
ail,hp = 3.5(0.700)
Aiw = Alw = 0.696.
=
We have AmixG = G — G* = Lin(pi- we) = Dinkwi- B) i)=
>, n; RT \n ay;. An amount of solution with not = 1 mol has 0.300 mol of
water and 0.700 mol of H2O2, which is 5.405 g of water and 23.81 g of
H5O>. The water wt. % is [5.405/(S.405 + 23.81)]100% = 18.50%; so 125
g of solution has 0.1850(125 g) = 23.1 g of water and 101.9 g of H2O2;
thus ny = 1.284 mol and mp = 3.00 mol. Then AnixG =
(8.314 J/mol-K)(333.1 K)[(1.284 mol) In 0.176 + (3.00 mol) In 0.651] =
—9.75 kJ.
129
10.8
(a)
Let b=3.92. Equation (10.22) with B = Hg and A = Zn and state | being
pure Hg gives In Yup = —J? [xzo/(1 — X2zn)] d In (1 — bxzn) =
b azn [x/(1 — x)(1 — bx)] dx = [b/(1 — b)][-In(l — x) + b' In (1 - bx)] 52 =
[b/(b — 1)] In (1 — xzn) — [1/(6 -— 1)] In C1 - bxzn) =
(3.92/2.92) In (1 — xz,) — (1/2.92) In (1 — 3.92xz,).
(b)
10.9
Yuzn = 1 — 3.92(0.0400) = 0.843; aizn = Yu.znXzn = (0.843)(0.0400) =
0.0337. In Yue = (1/2.92) {3.92 In 0.960 — In [1 — (3.92)(0.0400)]}} =
0.00360; Yue = 1.0036. ayg = 1.0036(0.9600) = 0.9635.
From Eq. (10.12), G’ = RT Yj nj In y,;. So G'/n = RT Dj x; In yi =
RT(Xac IN Yiac + Xchi IN Yichi). Walues of y; listed in Sec. 10.3 give at 35.2°C:
10.10
er
0
0.200
0.336
0.506
Grin
Ome 42 8) 9641
—~133.1-.
+138.37—102:6 4 cal/mol
Xac
0.815
0.940
1
Guy
e698
24m
0
(a)
0.082
0.709
cal/mol
Inthe second example in Sec. 10.3, it was shown that |¥ , ;/Y,;=P7/K;
for i # A. In the infinite-dilution (x, —
1) limit, we have Yui —
1
[Eq. (10.10)] and y, ; > 7; ; in this limit, the boxed equation becomes
l/y;; =P#/K,
Equating the two expressions for P*/K, , we get
Vignes ae
(b)
From Fig. 10.3a, Yichi > 0.50 as xc
> 1. Therefore Yuen = 2.07,4) in
acetone.
10.11
(a)
Eq. (10.13) gives Fi,0 =4n,0Pi,o and Ayo = (23.34 torr)/(23.76 torr) =
9523)
(b)
ay,9 = 22.75/23.76 = 0.9575. An amount of solution with | kg of
solvent has 2.00 mol of sucrose and 55.51 mol of H2O, so XH.90 =
55.51/(55.51 + 2.00) = 0.9652. Then ay 5 = Yu,0%H,0 = 0.9575 =
Yu,0 (0.9652) and yy,9 = 0.9920.
an
130
10.12
All.
10.13
For an amount of solution containing | kg of water, ngycrose = 1.50 mol and
Ny,0 = Soe Pmolso x40 = So oiyo 7.01, —10.9737 and x4.1=10.0263: From
ae cl0.24)eVirie = YasueXA = 1.292/0.9737 =. 18327. Theniapac=Vitstexsuc =
19327,(010263) = 0.034984 lsondi suc =) Vmsuc iene, ) = 1.29201, 50y="1,94.
10.14
Equations (9.31) and (10.24) give V2, = (Ou, /OP); = (Op), ,/0P), =
Vij = V,~, where an equation on page 284 in the text was used. From (9.30)
and (10.24), S” ==’, /oT)p =—(On;, /OT)p —R In Mam? =
Si; -R In Mam® = (S;,+ R In x)” — R In Mam® = [S, + R In (x/Mam)]”,
where an equation on p. 284 was used. We have x/Ma = nj/moMa; in the
infinite-dilution limit, no, = na and x/M, =nj/naMa = nj/wa = m;, where wa 1S
the solvent mass and m; is the solute molality. So Sj, = [S; + R In (mj/m°)]”.
Finally, (9.28), (10.24), the above relation between S-seand Seand the
relaliounly BiH D.284) vives
me ll etna
us, + RT In Mam® + T(S;,; —R In Mam’) = wi,;+ TS wa ie
10.15
pe ee
we, + RT In (yici/c°) = Wi; + RT In yusx. In the limit x4 > 1, Eqs. (10.29)
and (10.10) show that the activity coefficients go to 1; so this limit gives
He ei
ae
ln (xic°/c;)_. We have x; = nj/mo = n/n and c; = nV =
n,/V* for xa near 1. Hence, (xic°/c))” = con Vx Ining = CVAAS Mee = Baa +
RT In V* ,c’ . Substitution of this result in (10.29) gives yj =
wi; + RT In VE co + RT In (ycici/c") = Wi, + RT In (VE yy, ;¢; ). Comparison
ac) V4 - We have
with (10.23) shows that V* ,y.,¢; = Yusxi and Yo = (%)/V
X; = Nj Nop = WAN WANot = WAM Not = PV {M;/No, (where w is mass) and Vi.4
=1V
fuNagsSO-2i Vo.
c= Oamna/Nior = PamixXa. Then Yi = (4/VE AC)14 =
(pamixalci)(Ymi/xa) = (Pamilci)Y¥m,i. AlSO, YeiCi = PAMYm,i and (Y-,ic¢;/c°)e° =
Pa(miym,/m°)M°, SO AciC° = Padmim® and a, = (Pam"/c")am,i-
131
10.16
We use Eg. (10.22) written for Convention I and with A changed to B and B
changed to C. Let state 1 be pure C, so ¥i,c.1 = 1 and xc = 1. Then (10:22)as
In vic2= —J? [xg /(—xg)] dIny,, [Eq. (a)] at constant T and P. We have
In Yip = (W/RT)xe, so dIny,, =2(W/RT)x¢ dx¢ at constant T and P. Also,
the integrand in Eq. (a) is xg/(1— xg) = (l- xc )/X¢ . Hence Eq. (a) becomes
InYicg
=—-2(WIRT) J? xc) die =-20WIRT)(Xe —F x0) ||? =
~2WIRT)(X¢.9 — 4x24 -1+4) = -2(W/RT)[I= xp. -$ (1 xg)" -4]=
(W/RT)xp 5, and RT Iny;c = Wx;,, where the unneeded subscript 2 was
dropped
10.17
Equation (10.106) gives RT In, = (0/dng)7
pp.[Wngnc (ng +Mc)1=
Wn l(ng +nc)—Wngncl(ng tc)? =Wine(ng +c) —Ngnc (ng +N)” =
Wnél(ng +c)? = Wxe.
10.18
In the limit x,y; = 0 and x,, = 1, Fig. 10.3 gives ¥jchi = 0.50. Hence
In Vichi = (W/RT) x2, becomes In 0.50 = (W/RT) and W/RT = —0.693. For Xen =
0.494, the simple-solution model gives Iny,,. = (-0.693)(0.494)* = -0.169 and
Yiac = 0.844. Also, Iny;.41 = (-0.693)(0.506)" = -0.177 and Yicn = 0.837. The
true values are 0.824 and 0.772.
10.19
(a)
From (10.12) and the definition of Gi in (10.32), we have Gi =
RT (Xe NY icht + Xhep INYi hep) and use of the data gives G£ i(J/mol) =
90.7, 197.99224,°1879 and 824tatien = Onl 0G90.5,.0.7, and 0%:
(b)
We designate three cells for A, Az, and A3. A suitable initial guess is zero
for each of these, corresponding to an ideal solution. The xcn) values are
put in column A, the Ge values in column B, the Redlich—Kister values
in column C, and the squares of the deviations of the column C values
from the column B values in column D. We sum these squared deviations
in a cell and use the Solver to minimize this sum by varying Aj, A>, and
A3. The Solver gives A; = 0.334656, Az = -0.021853, and A; = 0.038433.
The fit is quite good, with deviations of typically 0.5 J/mol.
132
(c)
At xp =Xch = 0, the equations of Prob. 10.23 give In ¥j.ch) = A}
—A2 + A3 =
0.39494
Xchl = 0.4,
and Vichi = 1.48; also, In Yihep = 0 and Yithep =
Ati
we get In Yi,chi = 0.11188 and yych = 1.118; In Yihep = 0.06163 and Yihep =
1.064.
10.20
For the two-parameter fit, the spreadsheet of Prob. 10.19b can be used by
setting A3 = 0 and omitting A3 from the By Changing Cells box of the Solver.
One finds A; = 0.340293 and Az = 0.021853. The fit is much poorer than for
the 3-parameter function, with the sum of squares of the deviations equal to 58
J*/mol? as compared with 1.44 J */mol” for the three-parameter function. For the
four-parameter function, we modify the spreadsheet by adding a cell for Ag,
modifying the formulas in column C, and including Aq in the By Changing
Cells box. The sum of squares of deviations is reduced to 1.42 J*/mol* , which
is not a significant improvement over the 3-parameter function.
10.21
From (10.12) and the definition of G& in (10.32), we have Gi =
RT (Xen NY, cht + Xac NM Y1,ac) and use of the data gives G£ /(J/mol) = -179,
402, -557, -579, 429, -292, —103 at xcn = 0.918, 0.8, 0.664, 0.494, 0.291,
0.185, 0.060. The spreadsheet of Prob. 10.19 can be used with the data revised.
We take B as chloroform. The two-parameter fit is poor with a sum of squares
of deviations equal to 904 J*/mol? and deviations of 3 to 22 J/mol. The threeJ’/mol’parameter fit is fair, with a sum of squares of deviations equal to 258
J’/mol’ and deviations of | to 10 J/mol. The four-parameter fit is very good,
with a sum of squares of deviations equal to 32 J’/mol’ and deviations of 1 to 3
J/mol. The four parameters are A; = —0.869763, Az = 0.22272, A3 = 0.10898,
Ag = 0.14581. The activity coefficients can be found from equations like those
in Prob. 10.23, but since the four-parameter version of these equations was not
given, this calculation is omitted.
10.22
Comparing GE = xgxcRTLA + Ap (%p — Xc) + Ag (Xp —Xc)” + Ay(XB — XC) +
.-] with GE = xox,RT[Al + Ay (Xe — Xp) +Ag(%e — XB) +AU (Xe — XQ) + --,
we see that A’ = A,, Aj
=—A), A; = A;, Ay =—Ay, ete.
133
10.23
xg and xc are not independent. Substitution of xc = 1 — xg in (10.32) gives
G> = (xg — xp) RT[A, + A,(2x_ —1) + A; (2xg -1)’], 80
(GE /0xg) = (1—2x_)RT[A, + Ay (2xg —1) + A,(2x_ -1)7]+
(xp — xg )RT[2A, +4A,(2x, —1)]. Using xp = 1 — xc, we have
Invyip = (RIG
x0 (0G, /ox, |=
oe arene, (2x)
Ae 218)
Xe (2x¢ —1[A, + A, (1- 2x0) + 4,(1- 2x¢)7]+
(1— xc) x@[2A, + 4A; (1- 2x¢)] =
(A, +3A, +5A3)x~ — (4A, +16A;)x0 +12A;xé . (The tedious algebra can be
done by certain calculators or by symbolic algebra programs such as Maple V
or Mathematica. If you don't have access to such resources, you are excused
from doing the algebra.) Interchanging B and C in the final expression for
In Y¥; 3 requires that the sign of Az be changed, as shown in Prob. 10.22, and
this gives Iny,¢.
10.24
(a)
At infinite dilution of B, we have xc = | and the first equation of Prob.
10.23 with A3 = 0 becomes Iny;, = A, — A). With xg = | and A; = 0, the
second equation of Prob. 10.23 becomes In y;~ = A; + A). Adding and
subtracting these equations, we get the desired results.
(b)
Let B be CCly. From the results of (a), A} = 0.5(1.129 + 1.140) = 1.134
and Az = 0.5(1.140 — 1.129) = 0.005. The equations of Prob. 10.23 then give
InYicca, = (A + SAy) Gc, =4A5Xsicl, = 0.1826 and Yjcc = 1.20;
Inyisic, = (4 BoAD cay +4A5Xoci, =, 0.407 and! ¥; ey
10.25
10.26
le 50.
(a) KCI>K*+CI; v,=1, v-=1, u=1, z=-1.
(b) MgC
> Mg**+2C
l, r; v,=1, v_=2, z,.=2 z=-1.
(c) "MgSO, —"Mg*" + S027? vial, Val ye? 2
(d)
Ca3(PO4)
4 3Ca*
2 4+2PO7°; v,=3, V_=2) z,=2)> 7 =-3)
(e)
Fora
V4 =n.
1:1 electrolyte, z, = |_| = 1. KClis a 1:1 electrolyte.
ye (Vo )t verve):
134
WP
Wie?
(a) y. =¥77"?.
(Cy.
2 (i=)
10:27)
Up = Vat
10283
(Vey =a)
(a)
(b)
) w= Q)” Qn”.
ys)
eV See
(vs) =1'- 1" and vz = I.
(vs)? = 1) - 2° and ve.= 1.587,
(cave = 7
(d)ui(vs),= 3. 2 andvs=2.951.
10.29
With
V,=V_,we
have
(ee
en
(v.)*
= (v,)"* (Va)
—
(v.)’* (v,)"* =(v,)?"* .From(v,)*"* =(v,)”*,we havevi =V,.
10.30 From Prob. 10.28(b), (v..)” =4 and (10.52) gives aygc), == 42 (m,/m°)’.
10.31
(a)
(b)
Equation (10.58) gives
rN
l
(0.018015 kg/mol)2(4.800 mol/kg)
23.76 torr = 0.990
— 20.02 torr
From Eq. (10.56), aa = P,/P% = (20.02 torr)/(23.76 torr) = 0.843.
One kg of water contains 55.51 mol of H2O. There is no significant ion
pairing in the KCI solution, so x4 = 55.51/(55.51 + 4.80 + 4.80) = 0.853.
Equation (10.5) gives Ya = da/xa = 0.843/0.853 = 0.988.
10.32.
(c)
ay =0.843. xq = 55.51/(55.51 + 4.80) = 0.920. Ya = 0.843/0.920 =
0.916.
(a)
In¥n=am/m? + b(m/m°y + c(mim°?)° + d(m/m°y* +
[a (alm° + bm’/m° + cm? /Im® + dm?/m™) dm’ = 2am/m? +
(3/2)b(mim°y + (4/3)c(m/m?)? + (5/4)d(mim?°)’.
(b)
Substitution of m/m° = 6.00 gives log Ym = 0.45415 and Ym = 2.85.
One kg of H2O has 55.51 mol H20, and x(H20) = 55.51/(55.51 + 6.00) =
0.902;. Equation (10.24) gives Yi = Ymi/%a = 2700) 0202s= 3,16;
135
10.33
(a)
Atconstant T and P,0 =n, dua +n; duj =
na(-ORTMav dm; — RTMavm; do) + ni[(VRT/m;) dm; + VRTd In yi] and
0 = (n/m; -— OnaAMa) dm; ~ n~»Mam; db + n; d In y;. We have m; = n/naMa.
Substitution of n4Ma = n,/m; into the preceding equation and division by
n; gives d In y; = do + [(@ — 1)/m;] dm; at constant T and P.
(b)
=
(ux —paVRTMavm, = -RT In Yx,axa/RTMavm, = —In (Yx.aXa)/Mavmi.
At high dilution, y;,4 > 1 and In (yx,4xq) — In xa. Equation (8.36) gives
for x, near 1: In x, =x, — 1. At high dilution, there is no ion pairing and
the electrolyte gives vn; moles of ions. Hence xa = 1 — vnj/nio =
1 —vnj/na. We have In x, = —vn,/na and @ > (vnj/na)/Mav(n/naMa) = 1.
(c)
Integration from the infinite-dilution state (where y; = 1 and
@= 1) toa
solution with molality m gives In yi(m) = o(m) — 1 + HG —1)/m,] dm,.
10.34
Use of w, =m}'? and dw, = (1/2)m,''* dm, in the integral in (10.59) gives this
integral as J [( —1)/m,]2m!/? dw, =25*” (0 -D/w,]dw,.
10.35
m(CI) = [0.0100 + 2(0.0050)]mol/(0.100 kg) = 0.200 mol/kg.
m(K*) = 0.100 mol/kg. m(Mg?*) = 0.070 mol/kg. m(SO2-)= 0.020 mol/kg.
jk
Sai
1[0.200 + 0.100 + 27(0.070) + (—2)°0.020] mol/kg =
0.330 mol/kg.
10.36
For the electrolyte M, X, with stoichiometric molality m; (and no ion
eee
y)
pairing), we have m, = v4m; and m_ = v_m,. Hence I, = 5 (z.V 4m; =f z°v_m;) =
zm,(v,z, 2
2
+v_z-)=
42, lel (va + vm; = ay |<.| vm;, where (10.65) and
(10.45) were used.
10.37
(aja
(b)
= +[2°(0.02) + 1°(0.04)] mol/kg = 0.06 mol/kg. Then log y+ =
~0.51(2)(1){(0.06)"7/[1 + (0.06)"?] -0.30(0.06)} = 0.182; ys = 0.657.
Im = 4+[27(0.02) + 17(0.04) + 27(0.01) + 27(0.01) + 37(0.005) + 1°(0.015)]
mol/kg = 0.13 mol/kg.
log y: =
—0.51(2)(1){(0.13)'7/[1 + (0.13)'7] - 0.30(0.13)} =-0.230;
136
ys = 0.588.
10.38
(c)
logy. =-0.51(2)*{ ... } = 2(-0.182) =-0.364; y, = 0.432;
log y. =-0.51(1)*{ ... } = (0.182) =-0.091 and y_ = 0.811.
(a)
[,/m°= 1[27(0.001) + 1°(0.002)] = 0.003.
(b)
10.39
log 0.888 = -0.51(2)(1)(0.003)'7/[1 + 0.328(a/A)(0.003)"”};
1.08298 = 1 +.0.01797(a/A) and a = 4.62 A.
Ip/m° = 1[2°(0.01) + 17(0.02)] = 0.03; log y+ = -0.51(2)(0.03)"7/
[1 + 0.328(4.62)(0.03)"7] =-0.140 and y. = 0.725.
From (10.61), we have at 25°C: Acn,on/An,o =
(OcH,on/Pn,0)
Er ano/Er,cjon)” = (0.787/0.997)'" (78.4/32.6)"" = 3.31
and Boy,ou/By,o = (0.787/0.997)'"(78.4/32.6)'” = 1.378, where data
following (10.66) were used. Using the Ayo and By,o values after (10.66),
we get Acyon = 3.89 (kg/mol)'” and Boy,on = 4.52 x 10° (kg/mol)"” m”.
Substitution in (10.66) gives for CH30H at 25°C:
In ys = -3.89z, |z-| Im/m?)'7/[1 + 0.452(a/A)(In/m®)'"]. Substitution of Im/m?®
= 0.0200 and a/A = 3 gives In ys = -0.462 and y: = 0.630.
10.40
b =0.76235. At molality 0.1 mol/kg, J = (1/2)[17(0.2) + (-2)°(0.1)] = 0.3 and
c = 0.9896. At molality 1 mol/kg, /=3 and c = 0.9944. At 0.1 mol/kg, we find
logio Y: = 0.3665 and y+ = 0.430. At 1 mol/kg, logio y+ = —0.6826 and y+ =
0.208.
10.41
(a)
Cells are designated for q and b. The values of J, and y+ are entered into
columns A and B and c is calculated in column C. The Meissner
equations are used to calculate log y+ and ys in columns D and E, and the
squares of the deviations from experimental values are calculated in
column F. The Solver is set up to make the deviation for the J = 0.3 value
equal to zero by varying q. We get g = 3.77. This q gives good values of
y+ at the lower molalities but very poor values at high molalities (42 and
1229 at 5 and 10 mol/kg).
(b)
The Solver is changed to minimize the sum of the squares of the
deviations for the first five values. We get g = 2.123. The predicted yz
137
values at 5 and 10 mol/kg are 3.36 and 19.7, better than in (a), but still
pretty poor.
10.42
(a)
Eight single electrolyte solutions: NaCl, NazSO4, KCl, K2SO4, MgCl,
MgSOgq, CaClz, CaSO4.
(b)
Sixteen two-electrolyte solutions that consist of all possible pairs with
one ion in common:
NaCl— Na2SO4, KCI— K2SO4, MgClo—- MgSOa,
CaClo— CaSO4; NaCI-KCl, NaCl-MgClz, NaCl-—CaClz, KCI-MgClz,
KCl-CaCl, MgCly-CaCl2; NazSO4—-K2SO4, NazSO4—MgSOq, Na2SO4CaSOx4, KxSO,-MgSOu4, K2SO4—CaSO4, MgSO4—CaSOxg.
(c)
Foreach solution, we get 9 for the pair of unlike ions of the three ions
and we get w for the three ions present in the solution. For example, from
NaCl-Na2SOug, we get 8¢150, and Wnaciso, > from NaCI-KCl, we get
Ona.k and Wna.k.cl-
10.43
Withee
Ven WEAVE Vie. Vict Vee 2 Vand Vai VietVe ==
as (7a)
nrl|—
becomes me = Ope les (lea a)
10.44
ys = OY+.
Substituting the m, and m_ expressions preceding (10.76) into (10.46) and
using (1.68), we find
i= fb; + RT In {(yz)"(v.m{m°)"* [v_-(1-a)v,J (m,/m’
)- } = we? +
RTIn {(y4.)° "+ (mlm? )"*""- (v,)"* (v_)- [1--@)v,/v_]- J= po +
VRT In{y.a"*” (v,)"*” (v_)’-" [1--a@)v,/v_]-"
mm’ }=o +
VRTIn{v, 7.0" [l—-d— a)v,/v_]’-"’m,/m’ }, which is (10.77) and (10.76).
10.45
From the m, and m_ equations preceding (10.76) and from m, =V4m, and
m= =Vv_m,;, we have m,/m> =a and m_/m™ = 1 — (1 —)v,/v_. Then
(m,/m?
)**" (m_Im?)*-y, =a" {1 (1 — a)v,/v_]’-"’y.., which equals Yt
eanclOw
10.46
7s
Use of the reaction-equilibrium condition yp = by + LL, Eq. (10.74), and Eq.
(10.38) gives G = nala + (V4n; — np), + (V_n; — Np) + nyp(y + PL) =
NaMa + (V4Py + V_LL)n; = Nala + nib.
138
10.47 (a) m(Pb**) = 0.100 mol/kg - 0.43(0.100 mol/kg) = 0.057 mol/kg; m(NO3)
= 0.200 mol/kg — 0.043 mol/kg = 0.157 mol/kg;
m(PbNO; ) = 0.043 mol/kg.
In = 5[4(0.057) + 1(0.157) + 1(0.043)] mol/kg = 0.214m?°.
(b)
log y+ =—0.257 and y+ = 0.553. Equation (10.77) gives
yi. = (0.57)?[1 — 0.43(1/2)}?70.553 = 0.390.
10.48
As T increases, the water dielectric constant decreases, which increases the
interaction forces between ions, thereby increasing ion pairing.
10.49
For sucrose(aq),
A ,G° =-1551 kJ/mol = A, H° — TA,S° =-2215 kJ/mol -
(298.1 K)A,S°
and A,S®° =—2227 J/mol-K.
A,S° = SS
°
elem ?
where the
entropy of the elements is S,,.,,elem = 125, [C(graphite)] + 11S, [H2(g)] +
5.5S5- [O2(g)] = 2635 J/mol-K, where Appendix data were used. Then ——
—2227 J/mol-K + 2635 J/mol-K = 408 J/mol-K.
AG3oq/(kI/mol) = -237.129 — 0 — (157.244) = -79.885. AH 59g /(kJ/mol)
= -285.830 — 0 — (-229.994) = -55.836.
AS3o9q ((J/mol-K) = 69.91 — 0 — (-10.75) = 80.66.
(b) AG3og (kJ/mol) = -237.129 — 394.359 — 2(0) — (-527.81) = 103.68.
AH 59g /(kJ/mol) = -285.830 — 393.509 — 2(0) — (677.14) = -2.20.
AS5og /(I/mol-K) = 69.90 + 213.74 — 2(0) — (-56.9) = 340.5
10.50 (a)
10.51
(a)
Use of (10.91), (10.92), and (10.93) gives
A ,G°/(kJ/mol) = 65.49 + 2(-111.25) = -157.01.
A , H° kJ/mol) = 64.77 + 2(-207.36) = -349.95.
(b)
10.52
S°J/mol-K) = -99.6 + 2(146.4) = 193.2.
AH°/(kJ/mol) = —240.12 — 167.159 — (411.153) = 3.87.
For this reaction, AG}o /(kJ/mol) = -1010.61 — (—261.905 — 744.53) = 4.18
(which corresponds to K° = 5.4.)
139
10.53
From (10.91) and (10.85), A ,G®° [HNO3(ai)] = A,G° [H*(ao)] +
A ,G°[NO; (ao)] = A ,G°[NO; (ao)] . The NBS tables do not satisfy this
equality and so contain an error.
10.54
Since the two ways of writing the reaction refer to the same process, AS® must
be the same for each. Hence S°(H*) + 5*(OH’)
= 5, (H350)-=
5 (HO, ease (OH) — 25310). and) § (HO. (aq)Jims sl yak
S? (H,O) =0 + 69.91 J/mol-K = 69.91 J/mol-K at 25°C.
10.55
)Bi,+V_we) =v, (-OM)/0T) p+
p(v,
S° =-(0u°/0T) p = -(0/0T
v_(-dp°/0T) p=v,5S,+Vv_S°. Subtraction of S;, from each side gives
A,S; =v, A,S2 +v_A,S°.Then A,H; = A,G; +T A,S; =
vi(A,G, +T Ne Se eave Ne Ceetale A ,S*)=v1 A,Hi +v_A,H_.
10.56
(a)
In Eg. (10.81), Ym.isat and Misa are for P = P° = 1 bar. Since the solution is
very dilute and nonionic, we can approximate Ym.isat aS 1 and (10.81)
gives A ,G59g[O2(aq)] = 0 — RT In 0.00115 = 16.8 kJ/mol.
(b) x; = P,/K; = (1 bar)/(30300 bar) = 3.30 x 10° = nj/mo = n;/ny,9 and
Nc,H, = "1,0 3.30 x 10) = (55.51 mol)(3.30 x 107) = 0.00183 mol in
1000 kg of water, so mc, = 0.00183 mol/kg. Then (10.81) gives
A ¢ G3og [C2H6(aq)] = -32.82 kJ/mol - RT In (0.00183) = -17.2 kJ/mol.
10.57
Use of Eq. (10.25) for 1; gives the following results. S; = —(ON,/0T) pn, =
—(pm.i/9T) p— (O/0T)
p, [RTIn(Y,,,m/m® = Sp; — Rin (Y,,;m,/m*)—
RT
In Ym/OT)p.»,-
Vi = (OW/OP)p»,= (OW{/OP)p+
RT On
Yn i/OP) py, = Ving + RTO, JOP).
Hy = bit TS; = Wry +
TS,,;— RT*(Olny,,
0D) p,, = H,,.mit — RE (OIny,,./00) p,.
140
10.58
From (10.91) with i= HCl, -131.23 kJ/mol = A ,G°[HCl(aq)] =
A ,G° [H"(aq)] + A ,G° [Cl (aq)] = 0 + A-G°[Cl(aq)] and A ,G° [Cl-(aq)] =
—131.23 kJ/mol. Similarly, (10.93) gives A ,H° [CI (aq)] = -167.16 kJ/mol.
Use of T A, S° [HCl(aq)] = A,H° [HCl(aq)] - A ,G° [HCl(aq)] gives
A ,S° [HCl(aq)] = -120.5; J/mol-K = S°[HCI(aq)] - 152, [H2(g)] 182 [Clo(g)] = S°[HCl(aq)] — 176.88 J/mol-K, so S° [HCl(aq)] =
56.36 J/mol-K = S°[H*(aq)] + S°(CI'(aq)] = S° [Cl(aq)].
10.59
From (10.84), A ,G° [i(A)] = A -G®° ((*) — VRT In (V2¥m.+.sat!Misa/M°), SO
A Gog [KCl(aq)] = 409140 J/mol — 2RT In [1(0.588)4.82] = 414.30 kJ/mol.
Equation (10.91) gives -414.30 kJ/mol = A ¢ Grog [K*(aq)] — 131.23 kJ/mol and
A , G39g[K*(aq)] = -283.07 kJ/mol. Equation (9.38) at 1 bar gives
AH SireKcyag) = 17-22 kI/mol = H™ [KCl(aq)] - H;,[KCI(s)] =
H° (KCl(aq)] — H°,[KCl(s)] = H° [K*(aq)] + H [Cl (aq)| =H A[KCI(s)].
Subtraction and addition of the standard enthalpies of K, 5Ch, ande
onthe
right side of the last equation gives 17.22 kJ/mol = A ; H 59g [K*(aq)] +
A ¢ A 9g (Cl (aq)] — A ¢ A x93 [KCI(s)] = A ; H p98 [K*(aq)] — 167.16 kJ/mol +
436.75 kJ/mol and AH ¥ 50g (K*(aq)] = -252.37 kJ/mol. To find S° [K*(aq)],
we use the fact that AG, for dissolving KCI in water equals
A , Gy9g [KCl(aq)] — A ¢ Gyog [KCI(s)] = (-283.07 — 131.23 + 409.14) kJ/mol =
—5.16 kJ/mol. Since
AH 59, for the solution process is 17.22 kJ/mol, we have
AS° = (AH° — AG°\T = 75.06 J/mol-K for dissolving KCI in water. Then 75.06
Imol-K = S°[KClK(aq)] — S°, [KCI(s)] = S°[K*(aq)] + S°[CI(aq)] -
S° [KCl(s)] = S°[K*(aq)] + 53.36 J/mol-K — 82.59 J/mol-K and S° [K*(aq)] =
104.3 J/mol-K.
10.60
As the charge increases, the ion binds more H2O molecules to itself, thereby
increasing the degree of order in the solution and decreasing S of the solution
and S,.
141
10.61
(a)
From (10.104), In 62 = J7? (WVm/RT— 1/P) dP =
Jy? (/P + B' + C'P + D'P? +---— 1/P)
dP =
| (BGP DLP aS) dP=B
Pee P22
Py id to
Dropping the unnecessary subscript 2 on @ and P, we get the desired
equation.
(b)
Comparison of (8.9) with (8.4) gives B = b—a/RT,
C= b*,...; use of
(8.6) gives B' = (bRT —a)/R°T’, C’ = (2abRT —a°)/R°T",.. . and
substitution into the result of (a) gives the desired result.
10.62 (a)
From Eq. (8.18), b = RT,/8P, = 42.9 cm*/mol and a = 27R°T2/64P. =
3.61 x 10° cm® atm mol. At 75°C, RT = 28570 cm?*-atm/mol and Prob.
10.61b gives In @ =~[(2.38 x 10°)/(8.16 x 10* atm)](1 atm) —
[(4.18 x 10')/(1.33 x 10'® atm’)](1 atm)? = -0.00292 and = 0.997
(compared to exper = 0.997). Replacing | atm by 25 atm in the preceding
equation, we get In
= -0.0749 and 6 = 0.928 (compared to Qexper =
0.92).
(b)
6; = OF (T, P) = 0* (75°C, 25 atm) = 0.928, where the result of (a) was
used.
10.63
fj = 0)P; = 0:x;P = 0.928(0.100)(25.0 atm) = 2.32 atm.
From (10.102) and (10.103), Gn = = n° + RT In (f/P°) = ° + RT In (OP/P°)
= 2° + RT In (P/P°) + RT In @. For the corresponding ideal gas, @ = 1 and Gee
= pl = wo + RT In (P/P?). So Gm= Git + RT In 6 and In 6 = (Gm — G'9 RT.
10.64 (a)
(b)
10.65
(a)
Equation (4.65) gives AG = J> VdP = J? (nRT/P) dP =
nRT In (P2/P;) = (1.000 mol)(8.314 J/mol-K)(273.15 K) In 1000 =
15.69 kJ = 3.75 kcal.
AG=nAp=n[p° + RT In (f/P°) — W° — RT In (f/P°)] = nRT In (fo/f,) =
NRT In ($2P2/o,P)) = (1.000 mol)(8.314 J/mol-K)(273.15 K) x
In [1.84(1000)/0.9996(1.000)] = 17.1 kJ = 4.08 kcal.
From (10.104), Ino equals the area under the V,,/RT— 1/P vs. P curve
from 0 to 120 atm. The data are:
142
10ratrn
CVa/RES
1/Py AGse =4570
Platm
24.78 © —5.07e
45.42
(6
20
40
60
=
Gemma
bae-4102
100
120
shud
10° atm (V,/RT— 1/P)
P/atm
80
-0.0040
(V,/RT— 1/P) atm
pe
E
-0.0045 =
:
-0.0050
-0.0055
LE
fC
i)
SIE TELEMANN
0
20
40
+60.
80
=
"HOO | 120
P/atm
Plotting the graph, cutting it out and weighing it, we find that the weight
of the area between the —0.0040 horizontal line and the curve 1s 0.534
times the weight of the rectangular area shown in the figure. The
rectangular area shown equals (0.0020 atm ')(120 atm) = 0.240, so the
area between the —0.0040 line and the curve is (0.534)(0.240) = 0.128.
Adding in the area (0.0040)(120) = 0.480 between the y = 0 axis and the
0.0040 line, we get an area of —0.608 as the value of In
=
jioam (y_/RT— 1/P) dP. So In $ = -0.608 and 6 = 0.544. Also, f= oP =
0.544 x 120 atm = 65.3 atm.
(b)
Problem 8.38 gives B = limp_49 (Vm — V0) = limp—o (Vm — RT/P). The
data in part (a) give limp_s0 (Vn/RT— 1/P) = -0.00464 atm’. Hence
B(—50°C) = (0.00464 atm !)(82.06 cm?-atm/mol-K)(223 K) =
—85 cm*/mol.
10.66 (a)
B= [(82.06 cm*-atm/mol-K)(126.2 K)/(33.5 atm)] x
Meo
12,
4ge
143
cm fmol;
(b) B'=B/RT=-5.7 x 10“ atm”. At | atm, Ing =B'P =
(-5.7 x 107“ atm™)(1 atm) =—5.7 x 107 and $ = 0.99943. At 25 atm,
In 6 =-0.014; and @ = 0.9859.
10.67
(a)
Bi, = Ben/RT = (-1040 cm*/mol)/RT = -0.04048 atm! =
5.326 x 10° torr. Bi, =—0.0569s atm! = -7.497 x 10™ torr”. In the
mixture: In ben = In o%,(T, P) = BoP =
(-5.326 x 10> torr !)(301.84 torr) = -0.01608 and ocni = 0.9841;
In Qcar = (-7.497 x 10° torr ')(301.84 torr) = —0.02263 and car = 0.9776.
In the pure vapors: In 4, = (—5.326 x 10~°)(360.51) = -0.01920 and
o*, = 0.9810; In 6%,Car =-0.01599; o%,,Car = 0.9841.
(b)
OFP* SOV ig = Or
fi= ¥, x; f* and dix? P= 5x;
P/ x;O*P* . Then Yicn1
= (0.9841)(0.6456)(301.84 torr)/0.5242(0.9810)(360.51 torr) = 1.034 and
Vicar = (0.9776)(0.3544)(301.84 torr)/0.4758(0.9841)(213.34 torr) =
1.047.
(c)
Here we take ; and * equal to | and ¥j,;= x? BE xiP*. We get Yichi =
1.031 and Yj.car = 1.054.
10.68
At constant T and P, we have du; = d(u* + RT In x;) = RT d |n x; =
(RT/x;) dx;. Therefore Ds n; di; =i]
di (nj/x;) dx; feel >; (niNtor/Ni) dx; =
RT not >; dx;. From >; x; = 1, it follows that ); dx;
=0. Hence );; n; du; = 0 at
constant T and P. This completes the proof.
10.69 There are 2Nq = 12 x 107° ions in 1000 cm? of solution. With uniform
distribution, each ion is in the center of a cube of volume
(1000 cm*)/(12 x 107") = 8.3 x 10°” cm? and edge length (8.3 x 10-7? cm*)!? =
9x 10° cm=9
10.70
(a)
A, and this is the nearest-neighbor distance.
Because A-B attractions are weaker than A-A and B-B attractions, U and
H of the solution will be higher than U and H of the corresponding ideal
solution. Hence Amx/? >0= A,
(b)
H".
Because A-A and B-B attractions are stronger than A-B attractions and
the molecules have similar sizes and shapes, A molecules in the solution
144
will tend to surround themselves preferentially with other A molecules
(and similarly for B molecules). Hence the degree of order in the solution
is greater than in an ideal solution (where there is complete randomness
in distribution of A and B molecules), and S of the solution is less than
Son
10.71
a
CARS <aNOS
(c)
eG = Anti es aimy Sonsince Amma =gAtnih
we have AmixG > AG".
band /
Am asealAvens
(a)
An increase in z, increases the attractions between the positive ion and
the negative atmosphere that surrounds it, thereby stabilizing the solution
and hence lowering Hl, and lowering Y,.
(b)
An increase in ionic diameter decreases the closest distance of approach
between positive and negative ions, decreasing the attractions between
them and hence increasing py, and increasing Y,.
(c)
An increase in ionic strength means an increase in the attractions between
each cation and the anions in its atmosphere, which lowers w, and hence
lowers Y+.
(d)
An increase in solvent dielectric constant decreases the interactions
between positive and negative ions, increasing [, and increasing Y;.
(e)
As T increases, the kinetic energy of random ionic motion increases,
which tends to distribute the ions more randomly and to thereby break up
the ordered ionic atmospheres. This destabilizes the solution and
increases pt, and y,.
10.72
(a)
Li (sin) = iv) and use of (10.51) gives wy; + VRT In (viyim/m?*) =
us” +RT In (P/P°), so In (Pi/P°) = (Me, — By? WRT+ In (veyimilm®)*
and P; = P° exp [(W,; — Hi WRT\(vsyimlm°)’ = K(vsyimj/m°)’. For
HCl, v, = 1 =v., v=2, and vs = 1, so Pi = Kiym/m°y’.
(b)
K;= P° exp [(wi,; — Bi /RT).
we? = Wicugy: Pm = B°UH™(aq)] +
w°(Cl(aq)]. we; — Wy? = 4,G° [H"(aq)] + 4,G°
[CI (aq)] A ,G° [HCl(g)] = [0 — 131.228 - (-95.299)] kJ/mol = -35.929 kJ/mol.
K; = (1 bar) exp [(-35929 J/mol)/RT] = 5.08 x 10°” bar.
145
P; = (5.08 x 107” bar)[0.80(0.100 mol/kg)/(1 mol/kg)]}* = 3.25 x 107’ bar
= 0.00024 torr.
10.73
10.74
(a)
The following become | in the limit xy — I: Yiw, Yu.w, Yue, Ym.E-
(b)
Only ye becomes |.
(a)
True. When uy; = pH; in (10.3), then a; = 1.
(b)
False. For example, for Convention I, with x; < 1, yj might be such that
(pesllbace
(c)
Miuemsee Lg.410.3).
(d)
messee Ege (105):
(e)
False.
(f)
False.
146
Chapter I |
(a) T:
(b) T.
(a) F; (b) F; (c) F; (d) F. The H’ from H:20O ionization can make m(H*)
exceed m if mis extremely small.
(b).
11.4
(a)
m; = nwa = (4.603 g)(1 mol/46.026 g)/(0.5000 kg) = 0.2000 mol/kg. We
neglect H” from H20. Let x = m(H") = m(HCOO). Then (11.15) gives Kz
= 1.80 x 107 mol/kg = y-x*/(0.2000 mol/kg — x). With the initial
approximation y+ = 1, we find x = 0.0059; mol/kg. Hence I, =
0.0059, mol/kg and the Davies equation gives y+ = 0.92. Use of this y+ in
the K, equation gives x = 0.00649 mol/kg. Then I, = 0.00649 mol/kg, y+ =
0.919, x = 0.0064, mol/kg.
(b)
m(KC]) = (0.1000 mol)/(0.500 kg) = 0.200 mol/kg. The KCI contribution
to I, is 0.200 mol/kg and the H” and HCOO contribution to J, is (from
part a) about 0.006 mol/kg, so J, = 0.206 mol/kg. The Davies equation
gives y+ = 0.745. Use of this y+ in the Kz equation gives x = 0.0079
mol/kg. This gives a revised I, of 0.2079 mol/kg, which gives y+ = 0.74s.
We then obtain x = 0.0079 mol/kg.
(c)
If xis the moles of HCOOH that ionize per kilogram, then m(H”’) = x and
m(HCOO) = x + 0.400 mol/kg, since the KHCOO
contributes 0.400
mol/kg of HCOO™. So 1.80 x 10~ mol/kg =
v2 x(0.400 mol/kg + x)/(0.200 mol/kg — x). With the initial
approximation /,, = 0.400 mol/kg (due to the salt), the Davies equation
gives Y: = 0.739. With x in the numerator and denominator neglected
compared with 0.400 and 0.200 mol/kg, the Ka equation gives x =
1.69 x 10~* mol/kg. This x is small enough to neglect its contribution to
I and to neglect it compared with 0.200 mol/kg.
(a)
A BASIC program 1s:
147
1OsG=i
75 XX=X
15 INPUT "KA";KA
80 GOTO 60
20 INPUT "M1";M1
85 PRINT "M=";M;" M(H+)=";X
25 INPUT "M2";M2
90 NEXT M
30 INPUT "DM";DM
95 STOP
35 PRINT "KA="KA
300 X=(—KA+SQR(KA*KA+4*KA*M*
40 FOR M=M1 TO M2 STEP DM
G*G))/(2*G*G)
45 GOSUB 300
310 RETURN
50 XX=X
400 |=X
60 GOSUB 400
410 S=SQRiI)
65 GOSUB 300
420 LG =-0.51*(S/(1+S)-0.3*1)
70 IF ABS(X-XX)/X<0.0002
430 G=10“LG
THEN 85
(b)
440 RETURN
Atextremely low molalities, the H’ from the ionization of water cannot
be neglected. At high molalities, the Davies equation is not accurate.
For H)S(aq) < H*(aq) + HS“(aq), AG3og = [12.08 + 0 — (-27.83)] kJ/mol =
39910 J/mol = -(8.3145 J/mol-K)(298.15 K) In K°, so K° = 1.02 x 107”.
For HS (aq) is H'(aq) + S*"(aq), AG35og = (0 + 85.8 — 12.08) kJ/mol =
73.72 kJ/mol = -RT In K° and K° = 1.2 x 10°.
For this very dilute solution of an extremely weak acid, we cannot neglect H”
from the H2O ionization. Example 11.3 in Sec. 11.3 gives m(H’) =
(Ky + mKq)'” = [1.00 x 107'* + (1.00 x 10°)(6.2 x 10°7'°)}!? mol/kg =
1.27 x 10 mol/kg.
11.8
2H,O < H30* + OH. The only significant contribution to I» is from the NaCl
so I, = 0.20 mol/kg. The Davies equation gives log yz = 0.127 and
Ys = 0.746 for H;O* and OH. Equation (11.12) gives 1.00 x 107'* =
(0.746) [m(H30*)/m°]? and m(H30*) = 1.34 x 1077 mol/kg.
148
’
11.9
(a)
Setting 7/K equal to 310.15 in the equation of Prob. 11.38 gives log K,,
= —13.618 and K°, = 2.41 x 10°'*. We can take J, = 0 and y: = 1.
Equation (11.12) gives 2.41 x 10°'* = [m(H30*)m°? and m(H3;0°) =
1.55 x 10°’ mol/kg.
(b)
AG5og /(kJ/mol) = 0 — 157.244 + 237.129 = 79.885.
—AG5og/RT = -32.225;
In Ky, (25°C) =
K,, (25°C) = 1.01 x 10°'*. AHSog/(kJ/mol) =
0 — 229.994 + 285.830 = 55.836. Assuming AH® to be independent of T
and using (6.39), we get In Kj, (37°C) = -32.225 +
[(55836 J/mol)/(8.314 J/mol-K)][1/(298.1 K) — 1/(310.1 K)] = -31.354
and K° (37°C) = 2.42 x 107".
11.10
In this extremely dilute HC! solution, the ionization of H,O cannot be
neglected. 2H2O = TOs
Oleh
m(H30*)m(OH_/m™, since y+ = 1 in
this very dilute solution. If y moles of HzO ionize per kilogram, then m(OH) =
y mol/kg and, since HCI is a strong acid, m(H30"*) = (y + 1.00 x 10°°) mol/kg.
Therefore K°, = 1.00 x 10°'* = (y + 1.00 x 10)y and y* + (1.00 x 10™)y 1.00 x 10°'* = 0. The quadratic formula gives the positive root as y =
9.51 x 10°. So m(H30°) = (y + 1.00 x 10°) mol/kg = 1.05 x 107’ mol/kg.
11.11
HX +H,O < H,0°+X™.
Ka = yzm(H,0* )m(X~)/m(HX).
m(H30°*) =
m(X_) = 0.0100 mol/kg, since the H;0° from the ionization of water 1s
negligible. In = 0.0100 mol/kg. The Davies equation gives log y+ = —0.0448
and y+ = 0.902. Hence Ka = (0.902)°(0.0100 mol/kg)*/[(0.200 - 0.010) mol/kg]
= 4.28 x 10° mol/kg.
11.12
a(H,0) = y(H2O)x(H20) ~ x(H20), since H2O is an uncharged species and its y
should be close to | in this fairly dilute solution. In 1 kg of H2O there are 55.5
mol of HO. There is no significant ion pairing, so n(Na‘) = n(CI) = 0.50 mol
in | kg of water. Therefore, x(H2O) =
(55.5 mol)/(55.5 mol + 0.50 mol + 0.50 mol) = 0.982 = a(H20).
11.13
C,H,0;
+H,O.HC,H,0, +OH
149
xe — 26HC,H30,)a(OH) _ Ky. _ 1.00 x 107"
Mepliispaiom
t ASG Ee Oma O)) ak
=
where K; = a(H,0" )a(C,H,0;) /a(HC,H,0,)a(H,O)
bah
and Eq. (11.11) were
used. If we neglect the OH from the ionization of water, then m(HC2H302) =
m(OH). The solution is reasonably dilute and the OH and C,H,O, ions have
the same charge, so the Debye—Hiickel (or Davies) equation gives y(OH) =
y(C,H,O,); these activity coefficients cancel in K,. We take a(H2O) = | and
y¥(HC,H,0,)= 1. Let
y= m(OH )/m?°. Then
5.7, x 107! = y°(0.10 — y) = y°/0.10 and y = 7.6 x 10°, so m(OH) =
7.6 x 10° mol/kg. Equation (11.12) gives 1.00 x 107'* =
yzm(H,0* )m(OH~ y/m°?. We have I, = 0.10 mol/kg and the Davies equation
gives log ys = —0.107 and yz = 0.78. We find m(H30*) = 2.2 x 10° mol/kg.
11.14 (a) H,S+H,O © H;0+HS;
HS-+H,O < H30+S7-. We neglect the
H;O0° from water. Because K, of HS” is much less than K, of H>S, the
H;0° from the second ionization step is much less than that from the first
step. Because of the smallness of K, of HS, the ionization of HS~ does
not change the HS’ molality to any significant extent. Hence we can take
m(H30") = m(HS_). Let =m HsOeyn « Neglecting activity coefficients,
we have 1.0 x 1077 = y*/(0.10 — y) = y°/0.10 and y = 1.0 x 10“: hence
m(H30*) = 1.0 x 10“ mol/kg = m(HS_). Let z = m(S*)/m°. For the
ionization of HS™: 1 x 107'’ = (1.0 x 10~)z/(0.00010 — z) =
(0.00010)z/0.00010 = z and m(S*) = 1 x 10-'” mol/kg.
(b)
From part (a), m(H30*) ~ 1 x 107+ mol/kg = m(HS_), so J, = 1 x 107%
mol/kg. The Davies equation gives log y+ = —0.0050 and y. = 0.988 for
H30° and HS’. Hence 1.0 x 107’ = (0.988)"y’/0.10 and y = 1.0, x 107:
hence m(H30*) = m(HS’) = 1.0; x 107 mol/kg. The Davies equation
gives log y(H30") = -0.0050 and y(H30*) = 0.988; also y(HS~) = 0.988:
y(S*) = 0.954. So
1 x 10°'” = (0.988)(1.01 x 10)(0.954)z/0.988(1.01 x 10“) and
m(S*-) = 1 x 107!” mol/kg.
11.15
AG®(kJ/mol) = -587.0 + 454.8 + 128.0=-4.2.
K® = 5.44.
150
In K°
=-AG°/RT= 1.694;
11.16
Cua +S0;,_ diCuSO, (aq). Neglecting ionic association, I, = 0.20 mol/kg
and the Davies equation gives log y+ = —0.5080 and y+ =~ 0.310. Let
m[CuSO,(aq)]/m° = z. Use of the initial estimate of y+ in K,, for the association
reaction gives 230 = z/(0.310)°(0.0500 — z)’ and z’ — 0.1452z + 0.0025 = 0. The
roots are found to be z = 0.0200 and 0.125. Since z cannot exceed 0.05, we
have m[CuSO,(aq)] = 0.0200 mol/kg and m(Cu**) = 0.0300 mol/kg. This gives
an improved estimate of J, = 0.12 mol/kg; the Davies equation gives an
improved estimate of y+ = 0.354. Then 230 = 2/(0.354)°(0.0500 ~ zy and
2’ —0.1347z + 0.0025 = 0; we find z = 0.0222. Hence m(Cu**) = 0.0278 mol/kg
and I, =~ 0.1112 mol/kg. The Davies equation gives y+ ~ 0.361 and this leads to
z= 0.0226. Another repetition gives y+ = 0.363 and z = 0.0227. Further
repetition is clearly unnecessary and Y: = 0.363 and m(Cu**) = 0.0273 mol/kg.
We have &@ = 0.0273/0.0500 = 0.546. Since v, = v_ = 1, Eq. (10.77) reduces to
(see Prob. 10.43) YL = ay+ = 0.546(0.363) = 0.198. (The experimental Yi 1S
Wee bile)
11.17
A BASIC program is
2 INPUT “MO"; MO
360 X1=(-—B+RT)/2
5 INPUT "K"; K
365 X2=(—B-RT)/2
10 INPUT "NUPLUS"; NP
370 IF X1>0 THEN 410 ELSE
X=X2:GOTO 440
15 INPUT "NUMINUS"; NM
IF X1<MO THEN X=X1 ELSE X=X2
204INPUT “ZPLUS<; Zr.
410
25 INPUT "ZMINUS"; ZM
440 MM=MO0*NP-X
30 NU=NP+NM
450
IF ABS(MM—MP)/MP<0.0002
THEN 490
333 X=0:MP=NP*MO
455 MP=MM
335 IM=.5*(ZP*2*(NP*MO-X)+
ZMA2*(NM*MO—X)+(ZP+ZM)*
460 GOTO 335
(ZP+ZM)*X)
490 PRINT "ION PAIR MOLALITY";X
340 GOSUB 600
500 STOP
345 B= -—GG/(K*GP*GM)-NU*MO
600 S=SQR(IM)
350 C=NP*NM*MO0*MO
610 CC=-(S/(S+1)—.3*IM)*.51
355 RT=SQR(B*B-4*C)
6207 LE=ZP-ZE2GG
151
630 LM=ZM*ZM*CC
640 LI=(ZP+ZM)*(ZP+ZM)*CC
650 GP=10ALP
11.18
660 GM=104LM
670 GG=10ALI
680 RETURN
The true values of K,, and K, are entered into designated cells. The initial
values [H*] = 1 x 107’, [OH] = 1 x 10°’, [OF] = 0, [HOT] = 1 x 107 are
entered into designated cells. The K,, and K, values calculated from [H’][OH]
and [H*][OIl}/[HO]], respectively, are entered into designated cells. The total
IO molarity calculated from [HOI] + [OT] is entered into a cell. The relative
(not absolute) errors for the two equilibrium constants are entered into
designated cells. 10'' times [H*] — [OH] - [OT] is entered into a cell. (Since
the ionic concentrations are all very small, the Excel Solver will consider the
electroneutrality condition to be satisfied within its designated precision even
when the charge is significantly unbalanced. Hence the 10'! factor.) The Solver
is set up to make the relative error in K, equal to zero by changing the four
concentrations subject to the constraints that the K,, relative error be zero, the
total [IO] be 0.0001, that 10'' times [H*] —- [OH] - [OF] be zero, that all
concentrations be nonnegative, that [H"] exceed [OH], and that [HOI] and
[OT] each not exceed 0.0001. We click Options in the Solver Parameters box
and check Use Automatic Scaling; this helps the Solver solve problems
involving quantities of greatly different magnitudes. The Excel 2000 Solver
gives [H*] = 1.11 x 10°’, [OH] = 9.02 x 10°, [QU]
520g) al Oye
11.19
(a)
The equilibria are HX + HO
vesH30° + X” and 2H;O
e H30+0OH.
With activity coefficients taken as 1, we have:
(1) Ka = m(X_)m(H30*)/m(HX)
and (2) K, = m(H30*)m(OH).
Electroneutrality gives (3) m(H30*) = m(OH ) + m(X_).
(b)
Conservation of X gives (4) m= m(HX) + m(X).
(c)
Using Eq. (2) to eliminate m(OH_) and Eq. (4) to eliminate m(X°), we
find that Eqs. (1) and (3) become: (1)’ K, = m(H30*)m(X)/[m — m(X )]
and (3)’ m(H30*) = K,/m(H30*) + m(X°). Let y = m(H30°). Solving
(3)’ for m(X-) and substituting into (1)’, we get
y + Kay’ “(Keen ye KK, = 0.
152
11.20
In the cubic equation of Prob. 11.19, we set Ka = 2.3 x 107"! K,, = 1.0 x 107",
and m = 1.0 x 10~ (where units are omitted). We designate a cell for y and
enter an initial guess into this cell. In another cell, we enter the formula for the
left side of the cubic equation and set up the Solver to make this cell equal to
zero subject to the constraint that y (the H® concentration) lie between 10’ and
10~*. We find that for any initial guess for y in this range, the Solver declares it
has found a solution without changing the initial guess. This is because the left
side of the cubic equation is extremely close to zero for y values in this range.
We therefore multiply the expression for the left side of the cubic equation by
10° and require that this quantity be zero. With an initial guess of 1 x 10-7, the
Solver gives y= 1.11 x TOme
11.21
(a)
ForHX+H,O
asH30° + X,
Kma is given by (11.14) with the m°’s
omitted and where the y’s are y,,’s. Letting + and — indicate the H3O° and
X ions, we have
Sipeap a (Vern
m,a
mente NGG eke hfepeittic )
Yeux CHx/Y mux
. Substitution
Hx
of ¥¢ic/'Ymi; = Pa for i= +, i=—, and i = HX gives K-a/Km.a = (Pa) /Pa =
Pa.
(b)
c;=n/V and m;=nj/wa, so cm; = wa/V. In a dilute solution, the
solution’s volume is approximately equal to the volume of pure solvent,
so cm; = W/V
11.22
= Pa.
(c)
Substituting cj/m; = Pa into Ye ici = PaYm,iMi ZIVES Ye = Ymii-
(a)
From Sec. 10.8, €,,4 = 78.4 for water at 25°C. For a 1:1 electrolyte, be
.602
C)?
x107"?
(1)(1)(1
4n(8.854x107!2C?/N-m?)78.4(4.5x10'° m)(1.381x 107° J/K)(298.1K)
(b)
b = 1.588
K, = (4/3)n(4.5 x 107!° m)*(6.022 x 10°? mol™')e!?** =
0.00112(10 dm)*/mol = 1.12 dm*/mol
z,=2and |z_| =1,so0 b = 2(1)(1.588)
=3.176 and we find K, =5.51
dm?/mol.
(c)
b=6.352 and K, = 132 dm’/mol.
(d)
b=9.528 and K, = 3160 dm?/mol. The K,’s have the correct magnitudes.
153
11.23 Vn = M/p = (58.44 g/mol)/(2.16 g/cm*) = 27.06 cm?/mol. From (11.23), In a; =
(P — Po)Vm/RT = (P — 1 bar)(27.06 cm*/mol)/(83.14 cm?-bar/mol-K)(298.1 K)
= 0.00109, bar /(P — 1 bar). At 1, 10, 100, and 1000 bar, we get In a; = 0,
0.0098, 0.108, and 1.09, and we get a; = 1, 1.01, 1.11, and 2.98, respectively,
11.24 K,,
=5.38 x 10° mol’/kg’ = yz m(Ag*)’. With the initial approximation
y+ = 1, we get m(Ag’) = 0.00733 mol/kg, which gives J, = 0.0073 mol/kg. The
Davies equation gives Y: = 0.914. This yi gives m(Ag’) =
0.00802 mol/kg, which gives I, = 0.00802 mol/kg and y: = 0.91).
This ys gives m(Ag’) = 0.00805 mol/kg as the solubility.
11.25
CaF (s) fa Ca**(aq) + 2F (aq).
m(F-) = 2m(Ca**) and Ky =
3.2 x 10"! mol*/kg? = y4m(Ca**)y* [m(F)]° = 4y3 [m(Ca?*)]?.
With y+ = 1, we get m(Ca?*) =20x
107% mol/kg. This gives I, =
5142.0 x 104 mol/kg) + 4.0 x 107 mol/kg] = 6.0 x 10~ mol/kg. The Davies
equation gives 7 = 0.94. With this y. we get m(Ca’*) = 2.1 x 107 mol/kg.
11.26 For BaF,(s) < Ba’*(aq) + 2F (aq), AG%og =
[-560.77 + 2(-278.79) — (-1156.8)] kJ/mol = 38.45 kJ/mol =-RT In K:,,
and Kp = 1.84 x 10°’ mol*/kg’.
11.27 (a)
(b)
KC\(s) — K*(aq)+Cl"(aq). AG3og =
(-283.27 — 131.228 + 409.14) kJ/mol = —5.36 kJ/mol =
-RT In K;,. In K?, = (5360 J/mol)/(8.3145 J/mol-K)(298 K) =
2.16. Ki, =8.7 and Ksp = 8.7 mol*/kg’.
Ks = yim(K* )m(Cl) =z (am, (cum,
)= 2002 m? =(y*)? m2.
(yi) = K,,/m) =8.7/(4.82)° and y = 0.61.
11.28 Ca** +SO} < CaSO,(aq). The 2.08 g is 0.0152g mol of CaSOy.
Let z mol/kg of of CaSO, ion pairs be formed. Then 190 =
154
zy; (0.0152, —z)* or z? — (0.0305, +1/190y2 )z + 0.000233, =0.
The quadratic formula gives
z=0.0152, + 1/380y2 + 4 (0.000322/y; + 0.0000277/y4)'”.. Initially we take
y+ = 1 and get z = 0.00856 (where the minus sign is used, since z can’t exceed
0.015). Thus, our initial estimates are m(Ca**) = 0.0067, = m(SO%_); these
give I, ~ 0.0269 mol/kg. The Davies equation then gives log y. = —0.0271 and
y+ = 0.536. Use of this y+ in the above equation for z gives z = 0.00536 and
m(Ca**) = 0.0099, mol/kg. Then J, = 0.0397 mol/kg, log y+ = -0.315,
y+ = 0.485. This y+ gives z = gives z = 0.00485 and m(Ca°*) = 0.0104; mol/kg.
Then J, = 0.0417 mol/kg, log y+ = —0.320, y+ = 0.478. This gives z = 0.00478,
In =~ 0.0420 mol/kg, y+ = 0.477. This gives z = 0.00477, which 1s the converged
value. Thus m(Ca°*) = m(SOz,_ ) = 0.0105, mol/kg and
K,, =Yzm(Ca** )m(SO{ )= (0.477)°(0.0105; mol/kg)* = 2.5; x 10~ mol’/kg”.
11.29
CaCO;(calcite) < CaO(s) + CO2(g). AGSog =
(604.03 — 394.359 + 1128.79) kJ/mol = 130.40 kJ/mol = —RT In K°.
In K° = —(130400 J/mol)/(8.3145 J/mol-K)(298.15 K) = —52.60.
K°=14x107%
= P(CO,)/P°, where we took the solid’s activities to be 1 and
assumed ideal vapor. Therefore P(CO2) = 1.4 x 10°°> bar.
11.30
Taking the activities of the solids as | and assuming ideal gases, we have
K° = P(CO2)/P(CO) = 1.15 = n(CO2)/n(CO). Since the initial n(CO2)/n(CO)
value exceeds 1.15, the equilibrium position lies to the left. Let z moles of CO2
react to reach equilibrium. Then 1.15 = (5 — z)/(3 + z) and z= 0.72 mol.
The equilibrium amounts are n(Fe304) = 2 2 MoOl.- nO i= 312 nol
n(FeO) = 1.84 mol, n(COz2) = 4.28 mol.
11.31
K° = P(CO>)/P° = (183 torr)/(750 torr) = 0.244.
(a)
(5.0 g)/(100.1 g/mol) = 0.050 mol CaCO3. We have Nco, =
(0.244 bar)(4000 cm?)/(83.14 cm?-bar/mol-K)(1073 K) = 0.0109 mol.
Hence the equilibrium composition is 0.039 mol of CaCO3, 0.0109 mol
of CaO, and 0.0109 mol of CO.
155
(b)
The 0.0050 mol of CaCO; is not enough to give 0.0109 mol of CO2;
therefore all the CaCO; dissociates to give 0 mol of CaCO3, 0.0050 mol
of CaO, and 0.0050 mol of CO3.
11.32
At the pressures involved, we can assume ideal-gas behavior. Initially, n(CaO)
= 0.00892 mol. K° = 0.244 = Peq(CO2)/(1 bar) and Peq(COz) = 0.244 bar =
183 torr.
(a)
The pressure of 125 torr is less than the equilibrium pressure 183 torr
when all species are present, so no reaction occurs and the final amounts
are n(CaCO3) = 0, n(CaO) = 0.00892 mol, and n(COz) =
(125/760) atm (4000 cm’)/RT = 0.00747 mol.
(b)
Peq(CO2) = 183 torr and neg(CO2) = (183/760) atm (4000 cm* RT =
0.0109 mol. The initial number of moles of COz is (235/760) atm (V/RT)
= 0.0140 mol, so (0.0140 — 0.0109) mol = 0.0031 mol of CO; react.
Hence neg(CaCO3) = 0.0031 mol; neg(CaO) = (0.00892 — 0.0031) mol =
0.0058 mol.
(c)
The initial number of CO moles is n(CO>2) = (825/760) atm (4000 cm°)/
RT = 0.0493 mol. If the pressure P.g(COz) = 183 torr were reached, then
(0.0493 — 0.0109) mol = 0.0384 mol of CO would have reacted. But
only 0.00892 mol of CaO is present initially, so only 0.00892 mol of CO?
react. The final amounts are thus n(CaO) = 0, n(CaCO3) = 0.00892 mol,
n(CO>) = (0.0493 — 0.00892) mol = 0.0404 mol.
L133
Eq. (11.4) holds: AG® =—RT In K° = 0 + 0 — 2(—95.299 kJ/mol) = 190.598
kd/MolaSO Keont= "alex 1Onee
11.34
The Lewis—Randall rule gives 9; = 6* (450°C, 300 bar). N2 + 3H2 pd 2NH3.
Let z mol of N> react to reach equilibrium. At equilibrium, ny, =1-z,
My, = 3-32, and ny, =2z.
[(3 — 32/4 —2z)|P,
Pi=xP and Py =[(1-z)/(4-22)]P,
P =
Pay, = [22/(4 — 2z)]P. The left side of (11.30) is
(4.6 x 10°°)[(0.91)7/(1.14)(1.09)*]"! = 8.2 x 10°. The right side is
(Pyu,/P)?
1(Py /P UP, IP’) = (PIP?) “(214
22)1G 232)6],—z) =
8.2 x 10°. Since P/P° = 300, we must solve 4z7(4 — 2z)7/3°(1 — z)* = 7.38.
Taking the square root gives 2z(4 — 2z)/(1 — z)’ = 14.12, where the negative
156
sign in front of 14.12 is rejected, since z and 4 — 2z must be positive. Use of the
quadratic formula to solve 18.1277 — 36.24z + 14.12 =0 gives z= 1.47 and
0.530. Since z must be less than | (to keep ny, > 0), we have z = 0.530 and
ny, = 0.470 mol, ny, = 1.41 mol, myy, = 1-06 mol.
11:35
N> + 3H — 2NH3. AGig = 2(6.49 kcal/mol) = 12.98 kcal/mol =—RT In K°.
In K° = -(12980 cal/mol)/(1.987 cal/mol-K)(700 K) = -9.332.
8.85 x 10°. The Lewis tule is 6; = *(T, P). For NH3, P, =
K°=
(500 atm)/(111.3 atm) = 4.49 and T, = (700 K)/(405.6 K) = 1.73. For No, P,; =
500/33.5 = 14.9 and T, = 700/126.2 = 5.55. For H2, P, = 500/(12.8 + 8) = 24.0
and T, = 700/(33.8 + 8) = 16.7. The Newton graphs of @ (the full reference is in
Sec. 10.11) give Oxy, = 0.86, On, = 1.26, and ,, = 1.16. The left side of Eq.
26)*]= 2.35 x 10~. Let 2w
'(1.16)
(11.30) equals (8.85 x 10°)/[(0.86)7(1.
moles of NH3 decompose. The equilibrium amounts are n(NH3)/mol = | — 2w,
n(N>)/mol = w, and n(H2)/mol = 3w.
mo/mol = 1 + 2w. Use OL B= 7,P cives
tor Ede (tie 30):
[((1—2w)/(1 + 2w)]? (507)?
[w/(1 + 2w)][Bw/ (1+ 2w)]° (507)*
1631w* = (1 —2w)(1 + 2w)* = [(1 — 2w)(1 + 2w)]? = (1 — 4’). Let z= w.
2.35x10* =
Then 1631z? = (1 —4z)° = 1 — 8z + 162” and 1615z°+ 8z— 1 =0. We find z=
0.0225. Then w =z! = 0.150 and n(NH3) = 0.700 mol, n(N2) = 0.150 mol, and
n(H>) = 0.450 mol.
11.36
In Example 11.1, we found Ip, = 1.96 x 10° mol/kg = m(H’). For this In,
logio YHX) = 0.1(1.96 x 10-*) and y(HX) = 1.00045, which when used in Eq.
(11.14) makes no significant change in m(H"). In Example 11.2, J, = 0.200
mol/kg, so logi9 Y(HC2H302) = 0.0200 and y(HC2H302) = 1.047. Equation
(11.14) becomes
1.75x10~° mol/kg =
(0.746)? m(H,0* )(0.200 mol/kg)
1.047(0.100 mol/kg)
and m(H’) = 1.65 x 10° mol/kg, 5% larger than the answer in Example 11.2.
1137,
(a)
Assuming ideal vapor and taking the solid’s activities as 1, we have
K° = P(CO)/P°. Equation (11.32) gives [d In K°/0(1/T)]p = —AH°/R. We
157
plot In K° vs. 1/T. The data are:
In: Kom
[OelG
BA8 5) 82237002
[Ogi
9.7
1-4
O932
0.046
8.57
Kk"!
y = -19988x+ 17.127
i
eat:
0.00
-0.50
l
Bie at00.
-1.50
-2.00
-2.50
-3.00
-3,5Q0
eee
8.89
OHHH
0.00085
tt
tt
0.00090 0.00095 0.00100
Cae
|
ee
Pes
ee
0.00110
The slope is [0.00 — (-3.50)]/(0.000857 — 0.001031)K™' =
~2.01 x 10* K = -AH°/R and AH® = 167 kJ/mol. At 1073 K,
AG® = -RT In K° = -(8.314 J/mol-K)(1073 K) In (183/750) =
12.6 kJ/mol. Then AG° = AH® — T AS° and AS;5,3 =
[(167000 — 12600) J/mol]/(1073 K) = 144 J/mol-K.
(b)
11.38
From the extrapolated graph, we read In K° = 1.5 at 10°/T= 7.855, so K°
= 4.5 and P(CO2) = 4.5P° = 4.5 bar.
Substitution of T/K = 298.15 into the equation gives log K°, = -13.998 and Ke
= 1.003 x 10°'*. We have AG° = -RT In K° =
—(8.3145 J/mol-K)(298.15 K) In (1.003 x 107!*) = 79.91 kJ/mol.
(Cf. Prob. 10.50.) We have (0 In K°/0T)p = AH°/RT- =
2.302585(0 log K°/0T)p. Differentiation of the K°, equation in the text (which
is for P = | bar) gives (0 log K?./0T)p = (24746.26 K)/T° —
405.8639/2.302585T+ 0.48796/K — 0.000237 1(27/K”) = 0.03376 K~ at 25°C.
Hence AH?® = 2.302585RT*(0.03376 K') = 57.45 kJ/mol at 25°C.
AS593 = (AH® — AG°)/T = [(57450 — 79910) J/mol]/(298 K) =-75.3 J/mol-K.
11.39 AV? =-5.4 cm’/mol + 1.4 cm’/mol - (18.015 g/mol)/(0.997 g/cm?) =
22.1 cm*/mol. (0 In K°/OP)r = -AV?/RT and J? dn K° = -[ 1(AV°/RT) aP at
158
constant T. Neglecting the pressure dependence of AV°, we have In (K3/K; ) =
par/(1-00 x 107'*)] =
~(AV°/RT)(Pz — Pi) and In [Kop
(22.1 cm?/mol)(83.14 cm?-bar/mol-K)
'(298 K) ‘(199 bar) = 0.178 =
Ingk mate 32.236-and Keay = U19,x 10m
11.40
(a)
As in Prob. 11.39, with the P dependence of AV° neglected, we have
—(AV°/RT)(P2 — P;) = 1n (K3/K, ) =1n 1.191 = 0.1748 and AV° =
—0.1748(83.14 cm?-bar/mol-K)(298.1 K)/(399 bar) = -10.9 cm*/mol.
(b)
'(298 K)'(P2 — P1)
In 2 = 0.693 = (10.9 cm?/mol)(83.14 cm?-bar/mol-K)
and P> — P; = 1.58 kbar, so P2 = 1.58 kbar.
11.41
(a)
Since 1, = wx, we have H3 = H*,. Eq. (10.27) gives Hy; = 7.
Hence AH° = v,H* , + Liza Vi; and substitution in (11.32) gives
the desired result.
(b)
(c)
K IK im=(Vxa%a)* Wien Voici) IV eae WADE GE ue
yea (Ve iCi¥mitt:) |=Miza Oa )’' =p, where b = Dien Vi
=
From (b), K,, =K,p4? and In K;, =In K? —bInp, + In (const),
where const involves c° and m°. Hence (0In K,,/OT) p=(0In K7/0T) p—
b(0 In palOT)p = (0 In K2/0T) p+ boa = AH*/RT”, where we used the
result of part (a) and used (0 In p/0T) p= (0/0T) p(Iinm—InV) =
~(1/V)(QV/0T) p=—0. Thus (0 In K2/0T)p = AH™/RT° — boy =
AH IRT — Oa Died Vie
(d)
From (11.32) with the concentration scale used, AH?/RT? = AH™/RT° —
Oa Dea Vi = VaH® ,/RT + Liew (v;H7/RT? —v,0., ). Since AH; =
V,H* 4+ Lien V)Hi,, it follows that H2,= H —RT*a, fori# A.
11.42
(c).
11.43 AG” -AG°= vai
[2.,
+ v(H*)a[H* at a(H*) = 1077] -
py (H")] = VCH) {BHT at a) = 10-'] - 1°(H")}. We
ae Vikty + V(H"
159
have p; = W? + RTIn aj, so p[H" at a(H") = 107] = 1°(H*) + RT In 107 =
°(H") — 16.118RT. AG” —AG° =-16.118v(H‘)RT.
11.44
AG® = 2(4.83 kJ/mol) = 9.66 kJ/mol.
Q = (Pyy,/P°)?/(Py,/P° )(Py,/P’)” and
use of P; = x,P gives Q = [(1/7)3}°/{(2/7)3][(4/7)3]° = 0.0425. (AG/0E)rp =
AG° + RT In Q = 9660 J/mol + (8.3145 J/mol-K)(500 K) In 0.0425 =
—3.47 kJ/mol. Since (0G/0€ )r.p < 0, the reaction proceeds to the right.
11.45
As found in Prob. 6.48, constant T and P addition of j will produce morej
when x; > v,/A |Vv |where v; and A |V7 | have the same sign.
v; and A |Vv | are
small integers and typical values of vj/A lv | are W2e 2/1 1/3, 2/5. etens ince
the solution is dilute andj is a solute, we have xj << 1, and x; will not exceed
vi/A |Vv |. Hence the answer is no.
11.46
(a)
The log of Eq. (9.62) gives In(K,/P°) = (MU;¢— Hj) /RT
=-AGYRT,
which is the desired equation.
(b)
For O2(ag)
> O2(g), AG5og (kJ/mol) = 0 — 16.4 = -16.4, and the
equation in (a) gives In(K;m/P°) = -AG°/RT = (16400 J/mol)/RT = 6.62:
K;.,/P° = 747 and K;m = 747 bar. For CH4(aq) > CH,(g),
AG jog /(kJ/mol) = -50.72 + 34.33 =-16.39;
O61
11.47
(a)
0K
In(K;m/P°)
=-AGYRT =
= 44ibar:
False. Intermolecular interactions between He and the gases in the
reaction may change the gases’ fugacity coefficients and so shift the
equilibrium.
(b)
True. See Fig. 4.9.
(Cymeralser See Pqr(lialo),
(d)
False. AG° # AG in the reaction mixture.
(e)
False. The molality-scale standard state is a state in solutio
n.
(f)
False. See (e).
(g)
False.
160
Chapter 12
12.1
(a) T; (b) T.
12.2
(a) T; (b) F.
12.3
The contribution of sucrose to the vapor pressure can be neglected.
P =P, = a, P*. Since the solution is reasonably dilute, we can take aa = Xa
and P=
x, Px. In 100 g of solution, there are (98.00 g)/(18.015 g/mol) = 5.440
mol of water and (2.00 g)/(342.3 g/mol) = 0.00584 mol of sucrose. Then x, =
0.99893 and P, = 0.99893(1074.6 torr) = 1073.4 torr.
12.4
Ves
(aye
(by,
(c) Fo (dele
(eel
eat
a(e)) &
Neu, = (0.226 g)(1 mol/72.15 g) = 0.003132 mol. mc,y,, = (0.003132 mol)/
(0.01645 kg) = 0.1904 mol/kg. kp= MyRT#7/AgsHma =
(84.16 g/mol)(8.3145 J/mol-K)(279.62 K)*/(31.3 J/g)(84.16 g/mol) =
2.08 x 10* K g/mol = 20.8 K kg/mol. AT;=—-kym = -(20.8 K kg/mol) x
(0.1904 mol/kg) = -3.96 K. Ty= 6.47°C — 3.96°C = 2.51°C. We assumed an
ideally dilute solution and that only pure cyclohexane freezes out.
12.6
m = —ATy/ ky= (0.112 K)/(1.860 K kg/mol) = 0.0602 mol/kg = n/wa =
n,/(0.0980 kg), so n; = 0.00590 mol of maltose. M; = w;/n; =
(2.00 g)/(0.00590 mol) = 339 g/mol and M,,; = 339.
12.7
(a)
AT;= -kpmp = —kpnp/Wa = —kywa/Mpwa and Mz = —kpw p/AT;wa.
(b)
For 100 g of solution, the 3% solution has wg = 3.00 g and wa =
97.00 g; Mg =-(1.86 K kg/mol)(3.00 g)/(-0.169 K)(97.0 g) =
0.3404 kg/mol = 340.4 g/mol. For the 6, 9, 12, and 15% solutions, we find
Mg = 337.3, 334.5, 331.6, and 328.2 g/mol. Plotting Mg vs. wt. % maltose,
we get a nearly linear graph that extrapolates to Mg = 343 g/mol at 0 wt.
os
161
B44
= -1,0033x
+ 343.43
ae Peer
eae,
penne ene
|
i
\
D
4
6
'
1
Mp
Om an]
(emt
AG
Wt. % maltose
12:8
1Bp = (UT*
PAR) i ey pL Ade
(1/T#)[1 — ATy/T# + (AT;/T#) +++ -].
12.9
(a) murea = [(1.00 g)/(60.06 g/mol)]/(0.200 kg) = 0.0833 mol/kg;
kp=—-ATjl Murea = (0.250 K)/(0.0833 mol/kg) = 3.00 K kg/mol.
my = -AT;/ky=(0.200 K)/(3.00 K kg/mol) = 0.0666; mol/kg =
ny/(0.125 kg) and ny = 0.00833 mol. My = wy/ny =
(1.50 g)/(0.00833 mol) = 180 g/mol. M,.y = 180.
(bee
Anda
0, Rip ky = (200 g/mol)(1.987 cal/mol-K)(285 K)/
(3000 g K/mol) = 10.8 kcal/mol = 45.0 kJ/mol.
12.10
Let U denote CO(NH2)2.
AT;(A + Z) =-kpm,,
AT;(A + U) = —kpmy; so 1.65
= AT;/(A + Z)/AT;(A + U) = m,/my = (nz/wa)(nu/wa) = nz/ny and nz = 1.65ny
= 1.65(0.679 g)/(60.05 g/mol) = 0.01866 mol.
29.0 g/mol;
Mz = (0.542 g)/(0.0186, mol) =
M,z= 29.0.
12.11 ky =M,RTEIA,,,Hp,4 = (18.0153 g/mol)(8.3145 J/mol-K) x
(373.15 K)°/(40660 J/mol) = 512.9; K g/mol = 0.51295 K kg/mol.
12.12
We have 0.00313. mol of CjoHg and mg = (0.00313, mol)/(26.6 g) =
[el SaxulOwe mol/g. Hence k, = (0.455 K)/(1.18 x 107 mol/g) = 3860 K g/mol.
Equation (12.19) gives AvapHm = (119.4 g/mol)(8.314 J/mol-K) x
(334.8 K)/(3860 K g/mol) = 28.8 kJ/mol = 6.89 kcal/mol.
162
12.13
(a)
We have Yaxa = aa and AT; = [R(T*)?/AtusE{m,a] In aa =
(kp/Ma)(-OM avi)
(b)
= —kpovm;.
In 96.0 g of water we have 4.00 g (or 0.02295 mol) of K,SO4. Hence
m; = (0.02295 mol)/(0.0960 kg) = 0.239 mol/kg. Then
(0.950 K)/(1.86 K kg/mol)3(0.239 mol/kg) = 0.712.
12.14
= —AT;/kyvm; =
Equation (12.17) gives Dien Mm; = (2.37 K)/(14.1 K kg/mol) = 0.168 mol/kg; in
100 g of bromoform, we have 0.0168 total moles of solutes. Let P and P2 be
phenol and its dimer. Then 2P z P>. The 2.58 g is 0.0274 mol of phenol. Let
2z mol of phenol react. At equilibrium, np/mol = 0.0274 — 2z, Np, =Z, and Mio =
0.0274 — z. Then 0.0168 = 0.0274 — z and z = 0.0106. So np = 0.0062 mol and
np, = 0.0106 mol. The molalities are mp = (0.0062 mol)/(0.100 kg) = 0.062
mol/kg and mp, = 0.106 mol/kg. Thus Ky = (0.106 mol/kg)/(0.062 mol/kg)” =
27.6 kg/mol.
12-15
Dien mj = -AT;/ky = (0.70 K)/(5.1 K kg/mol) = 0.137 mol/kg = (Map + Man)/
(0.300 kg); SO Mnap + Nan = 0.0412 mol. We have 6.0 g = Mnap(128.2 g/mol) +
Nan(178.2 g/mol) = Mnap(128.2 g/mol) + (0.0412 mol — mpap)(178.2 g/mol). We
get Mnap = 0.027 mol and nan = 0.014 mol.
12.16
(a)
In 100 g of solution, ng = (8.00 g)/(342.3 g/mol) = 0.02337 mol and
mg = (0.02337 mol)/(0.092 kg) = 0.254 mol/kg.
AT; =—(1.860 K kg/mol)(0.254 mol/kg) = -0.472s K.
(b)
In aq = (Afusflma/R)U/TF — W/Tp) =
[(6007 J/mol)/(8.3145 J/mol-K)](1/273.15 — 1/272.665)/K = —0.00470
and a, = 0.9953;.
(c)
1ZA7
False.
12.18
(a)
100 g of solution has na = (92.00 g)/(18.0153 g/mol)
= 5.1068 mol of H>0, so xa = 5.1068/(5.1068 + 0.02337) = 0.99544 and
VW, = Anixn = 0,99931/0.99544 = 0.99987.
= -(In aa)/Mam; = 0.00470/(0.018015 kg/mol)(0.254 mol/kg) = 1.027.
T=cRT= (0.282 mol/L)(1 L/10° cm’)(82.06 cm?-atm/mol-K)(293.1 K)
=O.) 5 alm.
163
(b)
12.19
Ve y,0 = (18.015 g/mol)/(0.998 g/cm*) = 18.05 cm*/mol.
In ayo =—TIV* ,/RT =-{(7.61 atm)(18.05 cm*/mol)}/
(82.06 cm?-atm/mol-K)(293.1 K)] = 0.00571; ay,9 =0.99431. Xy,0 =
55.508/(55.508 + 0.300) = 0.99462. Yy, = 0.99431/0.99462 = 0.99969.
cg = IVRT = (6.1/760)atm/(82.06 cm°-atm/mol-K)(273 K) =
Boe x 10°’ mol/cm?. One cm’ of solution contains 3.58 x 10°’ mol of the
protein and contains 0.0200 g of protein; so Mg = (0.0200 g)/(3.53 x 107’ mol)
= 56000 g/mol; M,.3 = 56000. This is approximate because the solution is not
actually ideally dilute.
12.20
We have 5.55 moles of water and 0.00292 moles of sucrose. Equation (12.26)
gives IT = [(82.06 cm>-atm/mol-K)(298.1 K)/(18.07 cm’*/mol)] x
(0.00292 mol)/(5.55 mol)] = 0.712 atm. We have II = pgh and h = II/pg =
x10’ erg
a 0.712 atm
oe ___ 8.134
=736cm
(1.00 g/em*)(980.7 cm/s”) 82.06cm* atm
12.21
II = pgh. In the infinite-dilution limit, p of the solution goes to p of water, and
we Shall use p of water in the following calculations. For the first solution, II =
(0.996 g/cm*)(980.7 cm/s”)(2.18 cm) = 2129 erg/cm? and IT/pz =
(2129 erg/cm*)/(0.00371 g/cm’) = 5.74 x 10° erg/g. The other solutions give
Il/pg values of 6.29 x 10°, 7.18 x 107 and 8.10 x 10° erg/g. A plot of []/ps vs.
Pp is nearly linear and extrapolates to 4.54 x 10° erg/g at Op = 0. This intercept
equals RT/Mg and Mg = (8.314 x 10’ erg/mol-K)(303.1 K)/(4.54 x 10° erg/g) =
5.55 x 10° g/mol; the number average molecular weight is 55500.
Il/Pg
8.50E+05
8. 00E+05
erg/g
7.50E+05
y = 31924x + 453459
I
7.00E+05
6.50E+05
6.00E+05
5.50E+05
5.00E+05
4.50E+05
|
!
|
Ne
|
!
OE
2
|
|
eee
164
SPU
a ee
!
aes
2
ee a
12.22
TI/pp ~ RT/My + A2RTPp + Az RTP3 = RT/Mp + ArRTPp + AZM ,RTP3/4 =
[((RT/Mp)” + A>(MgRT)'”pp/2]° and taking the Square root of both sides gives
the desired result.
12.23
(a)
Dies c; = IWRT = (7 atm)/(82.06 cm*-atm/mol-K)(310 K) =
0.000275 mol/cm? = 0.275 mol/dm* = Orat
Hoe > Therefore Cg
=~
0.138 mol/dm? as compared with 0.15 mol/dm..
(b)
We shall approximate the concentration by the molalities. Then
Dies C; = [2(0.460) + 3(0.034) + 2(0.019) + 2(0.009)] mol/dm? =
1.078 mol/dm? = 0.001078 mol/cm*. Then I =
(82.06 cm>-atm/mol-K)(293 K)(0.001078 mol/cm*) = 26 atm.
12.24
We have V; = n;RT/P, so V; is proportional to n; and to x;. Then M, =
>; xM; and M,,, = (0.78)(28.01) + 0.21(32.00) + 0.01(39.95) = 29.0.
12.25
12.26
I = -(RT/V* ,)In ag = -(RT/VX 4(Mavi) = (ORT/V* ,MavniinaMa =
oRTvnina Vi 4.
(a)
a1 =Mar. Assuming ideally dilute solutions, we have px (Pi, T) +
RT In xa, = pe(Pit U1, 7) + RT In x2 and pi (P; + 11, TF) - pe(Pi, 1) =
RT \n xa — RT In xa2 = —RTxp,) + RTxp.2, where (12.11) was used. From
(12.22), w*(Pi + TI, T) — wi (Pi, T) = Vi, U1, where the pressure
dependence of V* , is neglected. Hence Vi,,I] = RT(xp,2 — xB,1).
(b)
xp2=0.100/(0.100 + 55.51) = 0.00180 and xp, =
0.0200/(0.0200 + 55.51) = 0.000360. IT=
(82.06 cm?-atm/mol-K)(298.1 K)(0.00144)/(18.07 cm?/mol) = 1.95 atm.
sp A) At equilibrium, the equality sign holds in (4.12) and —P™ dV™ — P® av’ =
dA* + dA® = -P* dV" +5; we dns - PP dV + Yi wh dn?, so Di wi dni* +
gli dn® = 0. Let dn moles of substance jmove from phase « to phase B.
Then dn‘ = dn and dn‘ =—dn, we have FL
165
ne dn =O
and pi = ee
12.28 f=Ccina—p+2=2-p+2=4-pandp=4-f.
The minimum fis 0, so the
maximum p is 4 in a binary system.
12.29
(a)
Yes. The horizontal tie line at 7; intersects the boundanes of the twophase region at ven = 0.23 (point L) and at xz = 0.68 (point Q).
(b)
No. The overall mole fraction is different for different points on the tie
line.
12.30
As noted in Sec. 12.6, the upper curve is the P-vs.- ie curve and the lower
curve is the P-vs.-x, curve. We take B as acetone. Table 10.1 gives P and the
ce and x, values that correspond to each P value. Plotting these points, we get
a phase diagram with a minimum in P at Xacetone = 0.4.
12.31
The width between the vertical axes is 602
mm and xg = 0.3 lies at 18 mm.
We draw a vertical line at 18 mm, and then a horizontal line that goes from the
liquid line at 18 mm to the vapor line; the intersection with the vapor line is at
452 mm (corresponding to xg = 0.75) and gives the result of the first
distillation step; drawing a horizontal line from the liquid line at 454 mm, we
find it intersects the vapor line at 58/2 mm, corresponding to xg = 0.97.
12.32
12.33
(a)
The width of the figure is 60 mm and x = 0.72 corresponds to 43.5 mm.
Drawing a vertical line at 43 mm and then a horizontal line from the
intersection of this vertical line with the liquid line, we find the
horizontal line intersects the vapor line at 52’ mm, which corresponds to
Xp = 2/2/00 =0:895.
(b)
in is given by point D as (36 mm)/(60 mm) = 0.60. Point G gives
xp = (48% mm)/(60 mm) = 0.81.
The point giving the system’s state lies on the horizontal line at 7, and cannot
lie to the left of point Q, since all points to the left of Q have some liquid
present. Point Q corresponds to xp = (41 mm)/(60% mm) = 0.68,
so xg =
np/(ng + Nc) = 0.68; ng > 0.68np + 0.68nc; nc < 0.47n, = 0.47(2.0
0 mol)
0.94 mol.
nc <0.94 mol.
166
=
12.34
XB overall = 4.00/7.00 = 0.571. Point L gives xz = (14 mm)/(60% mm) = 0.23.
Point Q gives oe = (41 mm)/(60% mm) = 0.68. The lever rule gives
n'(0.57; — 0.23) = n°(0.68 — 0.57,) and n” = 3.1n
= 3.1(7.00 mol — n”) and
n’ =5.3 mol; n'=7.00 mol - 5.3 mol = 1.7 mol. n}, = 0.23(1.7 mol) = 0.39
mol; ng = (1.7 -0.39) mol = 1.3 mol. ng =0.68(5.3 mol) = 3.6 mol;
eral
12.35
g nol.
=
=
P=
Pren + Prot =
*
hen
aben +
Cie
Sole tol a
ok
=
Aten denen at (
=e
X ben )Pr1=
Pal
4
(P*.,— P%))Xben. The graph is linear, so we need only two points to plot it:
iy
12.36
tol
Bet?
Shortie
oand
P= Beben
=
Tae
torr at Xben
=
i
(a)
As found in the Prob. 12.31 solution, the first vapor has a ai) eich:
(b)
A horizontal line starting from x, = 0.30 intersects the liquid line at xe
= 0.03.
(c)
When n' =n’, the horizontal tie line is bisected by the vertical line at xg =
0.30. By trial and error, we find that the tie line with equal halves is the
one that joins x, = 0.11 with xp = 0.50.
12.37
Regina is correct. Going horizontally in the two-phase region varies the overall
mole fraction (whose value depends on the sizes of the two phases present),
but does not change the mole fractions in either of the phases present.
12.38
A horizontal line at 80°C in Fig. 12.17b intersects the curve at Wpic values of
0.07. for phase « and 0.69 for phase B. We have 10 g = 0.076” +
0.69(20 g — m%) and m® = 6.2 g; then m! = 20 g- 6.2 g = 13.8 g. (Alternatively,
the lever rule could be used.) The @ phase therefore has 0.076(6.2 g) = 0.5 g of
nicotine and 5.7 g of water. The B phase has 0.69(13.8 g) = 9.5 g of nicotine
and 4.3 g of water.
1239
Drawing a horizontal tie line at 80°C, we find it intersects the water-poor
portion of the curve at Wnic = 0.69. Drawing the vertical line at weight fraction
Wnic = 0.5 and using the lever rule (12.41), we find (20 aur mm) =
167
m"P Ge mm) and the mass of the water-poor phase is m”? = 45 g. The mass of
nicotine in the water-poor phase is 0.69(45 g) = 31 g and the mass of water in
this phase is 45
12.40
(a)
g-31 g= 14g.
Let @ be the water-poor phase and f the water-rich phase.
(0.400 — 0.375)m* = (0.89 - 0.40)m* = (0.89 — 0.40)(10.0 g — m”);
0.025m" = 4.90 ¢ — 0.49m" and m* = 9.51g, m? = 0.49 g. Then Myo =
0.375(9.5; g) = 3.6 8, Mpnenot = (9.5 — 3.6) g = 5.9 g; iar = 0.89(0.49 g)
= 0.44 g; m®_phenol = 0.05 g.
(b) 0.375 = mf o/m™, 0.89 = my o/m> = (4.00 g- myo /(10.0 g- m").
The first equation gives Mi ,0 = 0.375m”" and substitution in the second
equation gives 0.89 = (4.00 g — 0.375m")/(10.00 g — m"). Solving, we get
m” = 9.5 g, and the problem is completed as in (a).
12.41
Let o be the octanol-rich phase. We assume the small amount of DDT does not
affect the water (w) and octanol (oc) mole fractions in the phases. We use the
mole fractions given on p. 362 of the text. We have a total of 4.44 mol of water
and 0.0763 mol of octanol (CgsH,7OH, molecular weight 130.2), for a total
number of moles of 4.513 mol. The total moles of water equals the sum of the
water moles in each phase, so 4.44 mol = (1 — 0.793)n* + 0.993(4.51g —n") and
n® = 0.0577 mol. So n° = 4.46 mol. Then n. = 0.207(0.0577 mol) = 0.0119
mol, and m,, = 0.215 g. We are assuming no volume change on mixing, so we
shall take the volume of each phase as the sum of the volumes of the water
and octano] that went into that phase (neglecting the small contribution of
DDT). The contribution of water (density 1.00 g/cm’) to the volume of a is
thus 0.22 cm*. We have hee = (0.793)(0.0577 mol) = 0.0453 mol, and mS. =
5.96 g; from the density, the volume of this octanol is 7.1; cm*. The volume of
phase o is then V* = (0.22 + 7.18) cm* = 7.4 cm’. For phase B, n® =
(0.993)(4.46 mol) = 4.43 mol; m', = 79.8 g; V® = 79.8 cm’. Also, n®, =
(0.007)(4.46 mol) = 0.031 mol; m®. = 4.06 g; V8 = 4.9 cm?. Then V’ = 84.7
cm?*. The 0.100 g of DDT is 0.000282 mol. We have Koy ppr = 8.1 x 10° =
168
por _ Bopr/V" _ (0.000282
~nppr) (7-4 cm?) and n8), = 4.0 x 10°'° mol,
cbor
noone
which is 1.4
12.42
10°’ g.
nb yp /(84.7 cm?)
n5p = 0.000282 mol, which is 100 mg.
We have Kow.napth = 2.0 X 10°. In Eq. (12.45), & corresponds to the octanol-rich
phase. The quantity in parentheses in the equation preceding (12.45) is the
desired AG°. If we make the approximation that the activity coefficient ratio is
approximately equal to | (which may not be accurate), we have
AG° =—-RT In Kownapth = (8.314 J/mol-K)(298 K)3.30 = -8.2 kJ/mol.
12.43
The diagram should bear some resemblance to Fig. 12.19. The phases present
in each area are given in Fig. 12.19. On the horizontal line, three phases are
present: pure solid B, pure solid C, and liquid solution. On this line, fis zero,
as discussed on p. 364. In the two-phase areas, fis 1 (the temperature). In the
one-phase liquid-solution area, fis 2.
12.44
Let B = benzene and A = cyclohexane. Equation (12.46) becomes
7, Sli
REF (AfusHm.p) In xp] = (278.6 K)/(1 — 0.233 In xg) and
T, = (279.7 K)/(1 — 0.884 In xa), where T, and T, are the left- and nght-hand
curves in Fig. 12.19. We get
xt
if
Te
oer
01
278.6 271.9
ODwie<
264.8
03
257.2
cs
T,
1g,
Oea0.3
Bom 202
D1) 7233.6.
0.9
I
Ost 3
255.98 27197
0.4
249.0
0.5
2399
06
229.6
192.7
Plotting the curves, we find they intersect at 71/2 mole percent cyclohexane
and —58°C.
12.45
The halt at 84.3 wt. % Zn must correspond to a compound with melting point
595°C. (It could not be a eutectic halt since its temperature is too high for this.)
The compound’s empirical formula is found as follows: (84.3 g)/(65.38 g/mol)
169
= 1.289 mol Zn; (15.7 g)/(24.305 g/mol) = 0.646 mol Mg. The Zn:Mg mole
ratio is 2:1 and the compound’s empirical formula is MgZn2. The phase
diagram is of the type shown in Fig. 12.25. The eutectic temperatures are
345°C and 368°C. One eutectic composition is 97 wt. % Zn (corresponding to
368°C). It is difficult to tell which curve the reading at 50% belongs to, so all
that can be said is that the second eutectic composition 1s close to 50 wt. % Zn.
12.46
(a)
The phase diagram is the same type as Fig. 12.26c with line PM lying at
0.1°C. The Liquid (4) is a solution of salt in water, B is H20, the
compound is NaC] - 2H2O, and A is NaCl. Since the diagram goes up to
only 100°C, the right-hand melting-point curve (which is almost vertical
below 100°C) does not reach the NaCl axis. The compound lies at 62 wt.
% NaCl.
(b)
20°C lies above the peritectic temperature of 0.1°C. As the solution
evaporates, we move horizontally to the nght on the diagram, eventually
reaching the £ + A (solution + solid NaCl) region; eventually, pure NaCl
is obtained.
(c)
This system lies initially in the @ + A (solution + solid NaCl) region.
When the peritectic temperature of 0.1°C is reached, solid NaCl - 2H2O
begins to form. The 80 wt. % composition lies to the nght of MN in Fig.
12.26c, and the system stays at 0.1°C until all the liquid disappears,
leaving a mixture of the two solids NaC] and NaCl - 2H2O. Then this
mixture is cooled to -10°C. No solid H2O (ice) is present at —10°C.
12.47
A cooling curve that corresponds to l.s. > l.s. +a — a@ — a+ 8 shows three
breaks. A cooling curve that corresponds to l.s. > 1.s.+Q@—1s.+a+B 3
a. + B shows a break followed by a eutectic halt. A cooling curve at the eutectic
composition shows no break and one halt. Similarly for curves to the right of
the eutectic composition.
12.48
For
>
€+B—4B+A2B+
4
4B+A>B, a break followed by a (eutectic)
halt. For cooling a liquid with the eutectic composition, a halt. For @ 3 @ +
A2B> € +B+A,B—>B+A>
aB,
break and then
a halt. For 2 > @+A—£+A+A2B— £+A.3B—- ¢+B+A,B—>B
+ AdB, break, halt, halt. For
£>+A-—
¢ +A.B+A
ADB, break,
-—
halt.
A2B — A + A>B, break, halt.
170
For? >
2€+A5¢0+4+A4¢
12.49
Just as Fig. 12.25 resembles two Fig. 12.19 diagrams placed side by side, the
Bi-Te diagram resembles two Fig. 12.22 diagrams placed side by side:
a is a solid solution of BizTe;3 in Bi. f is a solid solution of Bi in BizTe3. y is a
solid solution of Te in BizTe3. 5 is a solid solution of Bi2Te3 in Te.
12.50
The intersection of the miscibility gap with the phase transition loop does not
include the composition corresponding to the minimum, so we get a
liquid (€)
Fe
Au
171
combination of Figs. 12.24 and 12.21. Phase a is a solution of Au in Fe. Phase
B is a solution of Fe in Au.
12.51
Let A = HNO; - 3H20 and B = HNO; - H20.
id (€)
x(HNO3)
12.52
12.53
(a)
Equation (12.46) gives In xg = [(147 J/g)(128.2 g/mol)/
(8.314 J/mol-K)][1/(353 K) — 1/(298 K)] = -1.185 and xg = 0.306.
(b)
The same as (a); xg = 0.306.
(c)
In xg = [(162 J/g)(178.2 g/mol)/(8.314 J/mol-K)][1/(489 K) — 1/(333 K)]
= —3.326 and xg = 0.036.
(1000 g)(1 mol/78.1 g) = 12.8 mol. The 522°C point on the solubility curve
CE is at pen = 0.47 = (12.8 mol)/(12.8 mol + Nnaph)- We get Mnapn = 14 mol.
172
12.54
160°C
H5O
12.55
LizSO,4
We change the formula in cell HI, choose a temperature and initial guesses for
cells D2 and D3, and run the Solver. Repeating at several temperatures, one
finds that at each temperature, the two phases have equal mole fractions,
indicating complete miscibility at all temperatures. This makes sense because
(Sec. 10.5), a negative W means the B-C interaction energy is lower (more
negative) than the average of the B-B and C-C interaction energies, and (since
the excess entropy of mixing is assumed to be zero), G* is negative.
12.56
We change the formula in cell H1, choose a temperature and initial guesses for
cells D2 and D3, and run the Solver. Repeating at several temperatures, one
finds that there is partial miscibility at 240 K and at temperatures below 240 K,
and there is complete miscibility at 260 K. Trying various temperatures
between 240 K and 260 K, we get the cntical solution temperature as 254 K.
12.57
(a)
We get a lower critical solution temperature of about 200 K. (Note that as
T decreases, W becomes more negative, favoring miscibility (Prob.
12255).
(b)
12.58
We get a lower critical solution temperature of about 265 K.
The relation is quite simple. In (b), we get a lower critical solution temperature
of about 174 K, and this satisfies the relation.
173
12.59
When (12.48) and the three other similar equations for the other three chemical
potentials are substituted into (12.47), the terms that involve uj and pe are
ntuk+ntuttnbus+nbut = (n& + nb )ws+(ng +ng)wx, and the total number
of moles of D and of E in the system are constant.
12.60
(a)
Higher acetone concentrations correspond to tie lines that are closer to
point K in Fig. 12.33. Consider moving from the tie line below FH to the
FH tie line. In doing so, the left end (water-rich phase) of the tie line
moves up a smaller amount than the right end (ether-rich phase) does, so
there is a greater enrichment of acetone in the ether-rich phase than in the
water-rich phase as the acetone concentration increases. Hence Kew,ac
ether-rich
increases, since C,,
(b)
is in the numerator of Kew.ac-
The right end of each tie line is higher than the left end, meaning a
greater concentration of acetone in the ether-rich phase, so Kew,ac > 1.
12.61
Pure water (wat) lies at the point xwat = 1, Xeth = 0; pure ether (eth) lies at
Xwat = 0, Xeth = 1; pure acetone (ac) lies at xwar = 0, Xen = 0. States with xa, = 0
correspond to the line xwat + Xeth = 1 Or Xen = 1 — Xwat. This line has intercept 1
and slope —1; the two-phase region is bounded on one side by part of this line.
The plait point K is richer in water than in ether. The phase diagram is
ether
1
Xeth
two phases
one phase
water
ne 0
Xwat
174
1
12.62 Draw DA, DB, and DC in Fig. 12.32a. Let Ar(ABC) be the area of triangle
ABC and let h be the altitude of this triangle. We have 1hBC = Ar(ABC) =
Ar(ABD) + Ar(BDC) + Ar(CDA) = +(AB- DG) + 1(BC - DE) +
1(CA- DF) = 4BC(D
+ DE G
+ DF). Hence h= DG + DE + DF.
12.63
(a)
We drop perpendiculars from point F to each of the three sides of the
triangle and take the ratio of each perpendicular distance to the triangle’s
height. We get for phase F: xa, = 0.06, Xwat = 0.92, Xeth = 0.02. Dropping
perpendiculars from point H, we get the phase-H composition as:
iis O36, Se SOME ce
UI.
(b)
FG =30%mmand GH = 22mm. The lever rule gives 3014 ng =
22'2 ny = 22'2 (40 mol — ng); then ng = 17 mol and ny = 23 mol. Phase F
has 0.06(17 mol) = 1.0 mol of acetone, 0.92(17 mol) = 15.6 mol of water,
and 0.02(17 mol) = 0.34 mol of ether. Phase H has 5.3 mol of acetone, 1.2
mol of water, and 16.¢ mol of ether.
12.64
(a)
The diagram resembles Fig. 12.33 with ether replaced by ethyl acetate.
(b)
The overall composition is Xac = 0.20, Xethy: = 0.40, Xwar = 0.40. This point
lies in the two-phase region and is just a bit below one of the tie lines
graphed in (a). Drawing in a tie line through this point, we find it
intersects the binodal curve at points corresponding to the following
compositions. Phase a: ~*wat
xv,, = 0.915, x, = 0.06, sein = 0.023.
Phase B: x®_,wat = 0.260, x?ac = 0.233, x®,,ethyl = 0.502. We have
0.915n™ + 0.260(0.50 mol — n“) = 0.20 mol; so n* = 0.107 mol and
n? = 0.393 mol. (Alternatively, the lever rule could be used.) Then n%,,
=
wat
0.915(0.107 mol) = 0.098 mol, no = 0.0066 mol, Nethy! = 0.0025 mol;
ni = 0.102 mol,
nb. = 0.093; mol,
nie = 0.197 mol. Multiplication
by the molar masses gives my, = 1.76 g, m,, =0.38 2, mony = 0.22 g;
m?, = 1.842, mi =5.43g, mi, =17.4¢.
12.65
The overall composition of the system is given by point G, so the overall mole
fraction xq of A equals the distance GS to the side opposite vertex A. Points F
175
and H give the compositions of the two phases F and H in equilibrium, and
xi = FR, x = HT. Equation (12.49) with a = F, B = H becomes
n' (GS — FR) = nt (Wi
GS) and |n*/n4 =HK/GM|,The figure gives sin 0
= HK/GH = GM/FG.So
HK/GM
becomes n‘/n"' = GH/FG
or FGn* = GHn".
= GH/FG
and the boxed equation
| Lot
ive)
12.66
Be)
4
Y
QO
(a)
Since the solution in beaker B has a lower vapor pressure than that in A,
the evaporation rate from the A solution exceeds that from the B solution
and the equilibrium state will have beaker A empty, its liquid having
vaporized from A and condensed in B. The vapor pressure in the box will
be that of the diluted salt solution in B.
(b)
The vapor pressure of the B solution is lower than that of the A solution.
Liquid will evaporate from A and condense in B until the sucrose
molalities are equalized. Let a mass z of HO vaporize from A. Equating
molalities, we have (0.01 mol)/(100 g — z) = (0.03 mol)/(100 g + z) and z
= 50 g. Beaker A ends up with 50 g of water and 0.01 mol of sucrose;
beaker B ends up with 150 g of water and 0.03 mol of sucrose.
12.67
on the horizontal axis, the areas are relabeled as follows: ].s. becomes vapor,
l.s. + & becomes vapor + & (where @ is a dilute solution of liquid A in the
solvent liquid B), I.s. + B becomes vapor + B (where f is a dilute solution of
liquid B in liquid A). The horizontal line is a mixture of phase @, vapor, and
phase B. The region & + B is a mixture of the two immiscible liquids o and 6.
12.68
Yes. For example, an endothermic chemical reaction might occur. Another
example is adding a salt to a mixture of water and ice at 0°C (see Sec. 12.10).
176
12.69
(a)
c=2,p=1,
f=c—-p+2-r—a=2-1+2-0but
-083
since ,
Pis
held fixed, there are two degrees of freedom (temperature and xq).
(b)
c=2,p=2; f=2—-2+2-0-0=2, but since
P is held fixed, there is
one degree of freedom. Once Tis fixed, the mole-fraction composition of
each phase is fixed.
(ec)
c=2,p=3;
f=2-3+2-0-0=1,
but since
P is held fixed, there are
no degrees of freedom. All intensive variables (7, P, mole fractions) are
fixed.
12.70
Assuming ideal vapor, we have P; = XivapP = Y 1 eekP*. But for an azeotrope,
xP = x;,80 x;P = y,,x,P* and y;= P/P*. Yjethanoi = (760 torr)/(581 torr) =
l
1.31 and Yjethyi acetate = (760 torr)/(631 torr) = 1.20.
12.71
(a)
For each substance, u* = uw and p* + RT In Veer
RT |n ye xP, 50");
the et
For phase a, x,, 1s very close to | and
Yow = 1. Hence (I)x* = y} xP and y?, = x%/x® = 0.999595/0.00300 =
333) Lorphasc Oo yin, = andy i x emi (Lx, SO yo
xP_/xt,, = 0.99700/0.000405 = 2460.
(b)
This solution is phase §. With ideal vapor assumed, P; = y, ;x; P*, so
Pren = 1(0.997)(95.2 torr) = 94.9 torr and P,, = 333(0.00300)(23.8 torr) =
PSS tOr
(c)
eel Low) LOTT:
This solution is phase &
P=
12.72
ar eth bent
Pben = 2460(0.000405)(95.2 torr) = 94.8 torr;
(0;999595)(23.8: torr = <23.8 tor
aes 8.6 tor,
Let b stand for benzene and « and f be the two liquid phases. pp) = uP and
lie Rianey,
x, = Uy tT
In Ye pXp 80 Yi,x- = i,x!. Above solution
a, P® = y%,x@P*. Above solution B, P® = y},xpP%. Use of y%,x% = y® x?
gives P™ = PP.
12.73
Assuming an ideally dilute solution, we have —0.64°C = —kymg and mg =
(0.64 K)/(1.86 K kg/mol) = 0.344 mol/kg.
177
(a)
AT» = kpmp = (0.513 K kg/mol)(0.344 mol/kg) = 0.176 K. The normal
boiling point of water is 99.974°C (Sec. 1.5 and Fig. 7.1), so Tp =
100.15°C.
(b)
P=P, = x,PX. We have mg = np/Wa = Np/naMa = xp/Ma since the
solution is dilute. Then xg ~ mpMq = (0.344 mol/kg)(0.01801 kg/mol) =
0.0062. So P = 0.9938(23.76 torr) = 23.61 torr.
(c)
Il ~ xpRT/V* ,= 0.0062(82.06 cm? atm/mol-K)(293 K)/(18.0 cm*/mol)
=soeoratin:
(b) Same.
(c) Same.
(d) Same.
(e)
Different.
(f) Different.
12.74
(a) Same.
12.75
(a)
Halse, see Hig. 12220)
(b)
(c)
(d)
Mimies Sco nie alae,
(e)
True as xg increases in the solution, Mg in the solution must increase
False. See line FHK in Fig. 12.19.
False. The eutectic solution in Fig. 12.19 freezes entirely at one
temperature.
[Eq. (4.90)]. So 3 in the vapor in equilibrium with the solution must
increase. Since Lp, = Uz (T)+ RT In (Pp/P°), if Wg increases at constant
T, then Pg must increase.
(f)
True.
Reminder:
(g) False.
(h) True.
(i) False.
(j) True.
(k) False.
Use this manual to check your work, and not to
avoid working the problems.
178
Chapter 13
13.1
(a) T; (b) T.
13.2.
N/m.
13.3.
(a)
V=(4/3)nr andr = (3V/4n)'? = [3(1.0 cm*)/4r]'? = 0.62 cm.
A= 4nr’ =42(0.62 cm)’ = 4.8 cm’.
(b)
The volume of one particle is (4/3)m(300 x 10° cm)’ = 1.13 x 107!® cm’.
The number of particles is (1.0 cm?)/(1.13 <10e° cm?) = 8.8,x 10!>. The
area of one particle is 4m(300 x 10° cm)? = 1.13 x 107'° cm”. The total
area of the particles is (8.84 x 10!°)(1.13 x 107!° cm?) = 1.0 x 10° cm’.
13.4
dWrey = y d4 and Wyey = Y M = (73 dyn/cm)(3.0 cm?) = 220 ergs.
13.5
y=Yo(1- T/T,)''”. At 0°C, 26.5 mN/m = Yo[1 — (273.1 K)/(523.2 K)]'"”” =
0.4057yo and Yo = 65.3 mN/m. At 50°C, y =
(65.3 mN/m){1 — (323.1 K)/(523.2 K)]'’” = 20.2 mN/m.
13.6
Yo = (37.8)°9(523.2)'°(0.432/0.252 — 0.951) dyn/cm = 69.3 dyn/cm.
y= Yo(1 — T/T)''”? = (69.3 dyn/cm)[1 — (273.1 K)/(523.2 K)]''” =
28.1 dyn/cm. The percent error is 100(28.1 — 26.5)/26.5 = 6.0%.
13.7
_y/l, = [(50 dyn/cm)/(10 cm)](82.06 cm? atm)/(8.314 x 10’ ergs) =
4.9 x 10° atm.
13.8
(a) F; (b) T.
13.9
pP%— P® = 2y/R = 2(73 dyn/cm)/(0.040 cm) = 3650 erg/cem? =
(3650 erg/cm*)(82.06 cm? atm)/(8.314 x 10’ ergs) = 0.0036 atm = 2.7 torr.
p* = P + 2.7 torr = 762.7 torr.
179
13.10
y= $(0.7914 ~ 0.0012)(g/cm*)(980.7 cm/s*)(3.33 cm)(0.0175 cm) =
22.6 dyn/cm, where (13.12) was used.
13.11
h = 2y cos O/(pg — Pa)gr = 2(490 dyn/cm) cos 140°/
(13.59 — 0.001)(g/cem*)(980.7 cm/s*)(0.0175 cm) = -3.22 cm.
13.12 h = 2y/(pg — padgr = 2(52.2 dyn/cm)/(0.3383 g/cm*)(980.7 cm/s*)(0.0175 cm) =
18.0 cm.
13.13
(a)
For 0=0, we have a hemispherical interface (as in Fig. 13.9b). The
volume of the liquid above the meniscus is the difference in volume
between a cylinder and a hemisphere and equals (tr’)r — 2(4/3)nr° =
tr/3 = (mr*)(r/3). Hence, this extra liquid has volume and mass equal to
those of a cylindrical column of liquid of height r/3, and we must replace
h by h + r/3 in the equation for y.
(b)
A+7/3 =3.33 cm + (0.0175 cm)/3 = 3.336 cm. Replacing 3.33 cm by
3.336 cm, we get y = 22.6 dyn/cm.
13.14
y= 5(Pp - Pa)ghir: and y = +(Pp - Padghar2.
hy — hz = [2y/(Pp — Pag] x
(1/r) — Wrz) and y = +(pp — padi — ha)rire/(r2 -— n) = 4(0.900 g/cm?) x
(980.7 cm/s*)(1.00 cm)(0.0600 cm)(0.0400 cm)/(0.0200 cm) = 53.0 dyn/cm.
13.15
C2=Cc=
Gy a y)/b and (0 In c2/OY)r = (1/c2)(Ac2/dY)r = (1/c2)(—1/b) = —|/bep.
(dy/d In c2)7 = —bc2 and Eq. (13.34) gives 21) = bce2/RT = (y* — y)/RT.
13.16
21)
=—(1/RT)(0y/0 In m/m°);. We plot y vs. In m/m°. The data are
y/(dyn/cm)
In (m/m°)
61.3
4.452
Se
-4.157
56.1
-3.913
BPE,
-3.597
47.2
-3.199
For m = 0.020 mol/kg, In (m/m°) = —3.912. Drawing the tangent line at this
point, we find its slope to be —-12 dyn/cm. Hence P(1) =
(12 dyn/cm)/(8.314 x 10’ ergs/mol-K)(294 K) = 4.9 x 107!° mol/em?.
180
y/(dyn/cm)
In (m/m°)
13.17
Egs. (10.4) and (10.51) give p; = ph; + RT In a; = pw; + VRT In (viy+mj/m°). We
therefore have In a; = v In (vzy:m,/m°). So d In az = v d In (v+y2m2/m*) =
v d{In v+ + In(y2m2/m°)] = v d In (y2m2/m°) and (dy/d In a2)r =
v'[dy/d In (y2m2/m°)]r. Substitution in Eq. (13.33) gives the desired result.
13.18
Equation (13.31) and nj“/n;* = x;*/x;" give
ia) =
45x107° re
5
100 cm
13,19
A=U-—TS
ae= mol ~—-=|
0.10
45x10 ~ mol
=-3.0.x 107° mol/em
0.90
bi dnid+ V+y
and dA = dU —-TdS-—SdT=TdS—-Pd
dV +y d+ dj py dn;. At constant T, V, and nj,
T dS —S dT=-S dT—P
dHand (AA/OA)
dA =y
7yn, = Y-
13.20
(a)
From Fig. 13.11, VW’ =/zo and V? = A(b — zo); Eq. (13.19) with i replaced
— 9)= 11 + zt (ch -cf) — cPV,
by l reads n? =n - c) Azo- cPA(b
since Ab = V. Setting ny = 0 and solving for Zo, we get Zo =
(PV —n, )/(cP - cf')A.
(b) Substitution of V*=zo and V’ = A(b — zo) into (13.19) gives n? =n; +
Az (e? =)
— cPV . Substituting the zo equation of part (a), dividing by
A, and using Ty) = n°/7, we get the equation in the text.
13.21
(a)
n=
Venice cPy8
ie sa: + noe +n? = GO
+n; .Since the volume of
the interphase region is very small, we can take V = Vouk + vo Then
(Vi— Vian) + ne =
n; = cl Veweree’
Viglen —c?) + cPV +n}. Since
cP << c*, we have nj= Viixcr + chV + 7° = Dero chy +N).
Replacing i by 1, we get the second equation.
(b)
Inthe Prob. 13.20b equation we have (crs EC)
[(CP =e) =
(Cen) (c.) = cf ley = (777 bux! Vouk (ny bunk! Vouk ) = MP butk/My bulk «
Use of this relation and the results of (a) gives
Via) =e ny mae $F — (Mount MP
pun! bu )) =
]
s
4
a
a
ae [n; * 1 (7; bux/™, buk I -
13.22
Vin = M/p = (284.5 g/mol)/(0.94 g/cm*) = 303 cm*/mol. The volume per
molecule is (303 cm*/mol)/(6.02 x 107°/mol) = 5.03 x 10°** cm?. The crosssectional area of a molecule is 20 A’. Let @ be the molecule’s length. Then
5.03 x 10°? cm? = (20 x 10°'° cm?)# and € =2.5 x 10°’ cm=25A.
13.23
(a)
(b)
1[4057 (10° cm)*]t = 4.8 cm? and t = 2.37 x 10°’ cm = 24 A.
Vn=M/p = (885.4 g/mol)/(0.90 g/cm*) = 984 cm?/mol =
(6.02 x 107° mol” )4(2.37 x 107’ cm) and 4= 6.9 x 10° cm’.
(c)
There is one molecule, which is (6.02 x 102
moles, per each
6.9 x 10° cm” area, so P21) = n/4 = (6.02 x 107/mol)!/
(6.9 x 10°? cm?) = 2.4 x 107° mol/cm’.
13.24
(a)
Weplot 1/v vs. 1/P. The data are
10°v'/(gicem*)
9.90
7.35
6.54
1072 /atme
428.6
0.085.090)
6.17
6.06
6.02
eaisOe OO mo 55
The straight-line fit is fairly good. The intercept is 0.0056; g/cm® =
L/Dinon andi yeaa
oy, cm’*/g. The slope is 0.0150 atm g/cm? =
1/Omonb, and b = [(177 cm*/g)(0.0150 atm g/cm*)]"! = 0.38 atm”.
182
v//(g/em’)
0.00
0.10
0.20
0.30
P"/atm'
(b)
To keep the units simple, we rewrite the Freundlich isotherm as
We k(PIP*)‘, where P* = | atm. We plot log [v/(cm?/g)] vs. log (P/P*).
The data are
log [v/(cm*/g)]
log (P/P*)
2.217 2.220
91525981593
2.004 2.134 2.185 2.210
0.544 pel 000 ©le2235g17410
y = 0.2693x + 1.8593
;
log [v/(cm’/g)]
220
295
2.20
215
2.10
2.05
2.00
1.95
1.90
(lime
1.85
OR
ea
Eee
40.00:
Ome
ee
eae
ee eee
Some
Lael
ea
As 1.6
log (P/P")
The straight-line fit is rather poor in that the points show a pronounced
curvature at high P. (Recall from the text that the Freundlich isotherm
doesn’t work at high P.) Ignoring the three points at high P, we draw a
straight line through the remaining points. The intercept is 1.86 =
log[k/(cm*/g)], SOK =a
2 cm’/g. The slope is 0.269 = a.
183
(c)
For the Langmuir isotherm, Vmon? = c = 67 cm> /(g atm) and v =
(67 cm*/g)(P/P*)/(1 + 0.38P/P*). With P/P* = 7.0, one gets v = 128
cm’/g. For the Freundlich isotherm, v = (72 cm?/g)(7)°7? alg cm*/g.
13.25
(a)
v=rIn(sP*)+rIn (P/P*), where P* = | atm. A plot of v vs. In (P/P*) is
linear with slope r and intercept r In (sP*).
(b)
The data are
v/(cm*/g)
In (P/P*)
101
“136
153
M62
"165.
166
1.253 2.303 2.815 3.246 3.512 3.669
As with the Freundlich isotherm in Prob. 13.24b, the points at high
pressures show a strong curvature, indicating that the Temkin isotherm
doesn’t apply at high P. (Note that it predicts v — % as P 4 .) We
ignore the three high-pressure points and draw a straight line through the
remaining points. The slope is 33.4 cm*/g = r. The intercept 1s 59 cm*/g =
(33.4 cm’/g) In (sP*) and s = 5.85 atm”.
vi(cm/g)
180
160
140
120
100
80
60
40
y = 33.30x + 59.29
ee
Le
el
i
Pra
SSS]
Ove
Se
SSS
See
li
21
O55
SSeS
5g)
eee
See
eee
Coes
ee
Soe
In (P/P*)
13.26
We have | + bP =cP/v, so a plot of P/v vs. P is linear.
13.27
(a)
The curve resembles Fig. 13.17b, indicating formation of more than a
monolayer. We have a type II isotherm and the BET equation is
appropriate.
184
v/(cm’/g)
E
\
1
Ua
ey
100
200
ao
0.0
0
t
i]
Af
ay es
300
400
1
A
1
‘
600
700
eH
500
P/torr
(b)
We plot P/v(P* — P) vs. P/P*, where (since 77 K is the normal boiling
point) P* = 760 torr. The data are
24969. 98 e318
10°[P/v(P* — P)\/(g/cm’)
0.07 winl 64am
10°P/P*
7 Agel
10°[P/v(P* — P\/(g/cm’)
10°P/P*
mele
AS 6a S i
27 Ome Se) aeOSES
25
el One
293
atl
OnLO2L6
Oa
The low- and medium-pressure points fit a straight line well, but the three
points at high pressure deviate greatly from this line (the BET isotherm
often works poorly at high pressures) and will be ignored. We draw a
straight line of slope m and intercept b through the first six points. We
find m=
exhaveuii=aol/O mon =! Omen
SO) Donat iter by! = 0.842 cm’/g: also. c =
1.17 g/cm® and
ANC i= ek)Den
b=0.01, g/cm’. W
+0.0166
y = 1.1661x
(1 = mUmon) |= 7 x 10'.
OMe
;
+
LS.46
eee
eon
ee eae
2
0.6
0.8
1.4
[e2,
1
0.8
0.6
0.4
On2
0
0.4
0.2
185
P/P*
(c)
n=(l atm)(0.842 cm*V/RT= 3.76 x 10° moles of N> adsorbed per gram
of sample. This is 2.26 x 10!” molecules, so the surface area of the
powder is (2.26 x 10!°)(16 x 107'° cm’) = 36000 cm? = 3.6 m’.
13.28
For 6 << 1 in Eq. (13.35), we have 1/6 >> | and 1/bP + 1 >> 1, so that
1/bP >> 0 and hence 1/bP >> | and bP << 1. Therefore bP can be neglected in
the denominator of (13.36) to give v = cP.
13.29
Let N be the number of adsorption sites. The rate of desorption of A is
proportional to the number 0,4N of adsorbed A molecules and equals kg a0 aN.
The A adsorption rate is proportional to the gas-A partial pressure Pa and to
the number (1 — 8, — 98)N of unoccupied sites. Hence kg, a8aN =
kaaPa(1 — 0,4 — Op)N and Oa = (Ka,a/ka.a)Pa(1 — 94 — OB) = baPa(l — Oa — Op),
where ba = kg. a/ka.a. Similarly, 09g = bgPp(1 — 8, — 9g). Division gives 0p/0,4 =
bpPp/baPa. Hence, 94 = baPa(] — Oa — bgPpO8a/baPa) and 8,4 =
baPa/(1 + baPa + bpPs). Similarly, 0g = bgPp/(1 + baPa + bgPp). The fraction
of occupied sites is 8, + Og and equals V/Um, SO V/Um = 8,4 + Og =
(baPa + bgPpyV/(1 + baPa + bpPs).
13.30
(a)
From (13.41),dIn P= —(AH ,/RT*) dT at constant 9; integration gives
In (P2/P,) = (< AH, >/R)(A/T, -1/T,), where < AH, > is an average
AH_, over the temperature range.
In (0.03/0.0007) =
[<AH, >/(8.314 J/mol-K)][(873 K) |- (773 K)'] and <AH, > =
—210 kJ/mol.
13.31
(b)
For @=0.10 and 500 to 600°C, <AH, > = R[(873 K)' — (773 K)'}! x
In (23/8) = —59 kJ/mol. For 600 to 700°C, <AH, > =
R((973 K)! — (873 K) ‘J! In (50/23) = —55 kJ/mol.
(a)
Multiplication of (13.40) by vc gives Pc/(P* — P) = (0/Vmon) X
[1+ (c= I)P/P*] and )/Umon = PP*c/(P* = P)(P* + Pe — PY.
(b)
For P << P*, we have P* — P = P* and V/Vmon = PP*c/P*(P* + Pc) =
Pcl(P* + cP) = (P*)'cP/[{1 + c(P*) 'P], which is the form of the
Langmuir isotherm (13.35).
186
13.32
This suggests that the CO is chemisorbed in two different forms, as shown in
Fig, 13815:
13.33
(a)
Taking activity coefficients as 1, we have K, = [L,]/[L]". Conservation of
matter gives c = [L] + n[L,], so K- = [Li]/(c — n[L,])" = x/(c — nx)", where
x=[L,];
(c—nx)" =x/K, and c = nx + (x/K,)".
(b)
Pa=iLilie=(C— nL )\/a=
na c.
(c)
With K° = 10°, we have K, = 10°°/(c°)””, where c° = | mol/dm’. The
equations of (a) and (b) give c = 50x + c°(xic?)? 7/104, f=1-50x/c, and
[L] = fc. For various assumed values ofx = [L,], we calculate c, etc., as
xIc°
0
10°c/c°
0
10°n[LrV/c° 0
0
10°[LV/c?
1
f
10)%
4.37
5x10"?
4.37
1.000000
xIc°
[0s ee OR
10°c/c°
2316 ae 51.9
50.0
10°n[L,/c° 15.8
91-54
1.70
POpevicome
0329550137
f
nary
6.31
0.00050
6.31
0.99992
10°
10m
1033
COTM
te = 12.6
0.050
0.500
5.00
7160
6.92) = a724a&
0.603
0.935
0.993
re
166
158
8.13
0.049
ior
508
500
S20
0.016
The plots of [L] and n[L,] have the form of Fig. 13.21b.
13.34
(d)
From the graph of f,we find that c = 0.00015 mol/dm? at f = 0.5.
(a)
log z = 0.4343 In z and y = [73.0 - 12.99 In (ac + 1)] dyn/cm. We have
d\lnc=c_ldc and (dy/d In c) = c(dy/dc) = (12.99 dyn/em)ac/(ac + 1);
Eq. (13.34) with R = 8.314 x 10’ ergs/mol-K gives Py, =
(5.37 x 10°'° mol/em’)ac/(ac + 1).
(b)
This resembles the Langmuir isotherm; c corresponds to P and I,1)
corresponds to v.
(c)
With a = 19.64 dm’*/mol, we get
187
10° EGV/(molicm?).
c/(mol/dm*)
Olgas
102
2
2a/(mol/cm’)
10;
0
2.664)-3'56 04 28)
0.059" 01002
2"
tae
e4e7ome
0.4
5. Ole
S07
Orly
al.
F
sme
E
As
E
SP Ae
2|
1
i=
O
i
0
SS
ee
(eit
0.2
1
0.4
ee
a
| eS
0.6
1
0.8
1
—|
l
c/(mol/dm’)
13-3588)
WdkliP.= (Vi 4/RT) dP at constant T. Integration gives In (Pa 2/Pa\) =
(Vi 4/RT)(P, — P,), where we neglected the P dependence of hesaL el
state 1 be the bulk state and state 2 be a drop of liquid with radius r.
Equation (13.9) gives the extra pressure experienced by the drop due to
its curvature as P2 — P; = 2y/r. Hence In (P/Ppux) = (V,,/RT)(2y/r) and
Bee Ppa CXDA2yVaiERE)
(b)
vi = M/p = (18.015 g/mol)/(0.998 g/cm") = 18.05 cm*/mol. The result of
part (a) gives P,/(17.535 torr) =
2(73 dyn/cm)(18.05 cm?/mol)
(1.00 10-° cm)(8.314x 10’ erg/mol-K)(293.1 7
and P. = 17.726 torr.
188
Chapter 14
14.1
(a) Yes; (b) no.
14.2
6 = 0 at infinity).
(a) T; (b) T; (c) T (assuming
14.3
F = |0,Q>|/4neor’ =
60x107!°C)(1.60107'°C)
2(1.
gee sh
go
Ras ee
ee eee ANG al Oni
4n(8.854x107!? C? N7! m77)(1.0107"° m)?
14.4
14.5
(a)
E= |Q\|/4neor? = (1.60 x 107'? C)(8.99 x 10° N m* C)/(2.0 x 107° m)’
(b)
= 3.6 x 10!° V/m.
0.90 x 10!° V/m.
2 — 0) = (Q/4mte9)(A/r2 — I/r}) =
1.60x10°°C
1
1
________________|=-3.60V
|
C? N' m7) 4.0x107°m_—_2.0x107"" 4n(8.854x10?
14.6
(ayers
14.7
(a)
(b)
(Dyck.
(3.00 mol)(2 x 96485 C/mol) = 5.79 x ere”
(0.600 mol)(-96485 C/mol) = 5.79 x 10° C.
14.8
(a) T.
14.9
Equation (14.23) gives pj" —H;” = -(96500 C/mol)(-0.1 V) = 10° J/mol.
14.10
(ayer
14.11
£= br — 01 = dcu- dew’ = (cu — Ocuso, (ag) ) + (Ocuso4(aq) — 9 2nS04 (a9) )+
aD) el S(O) RL:
4(aq) — 0Zn) + (bzn — Ocw) = 0.3
(0 zns0
189
V-0.1V+0+02V=04
V.
14.12
(a) T; (b) T; (c) T; (d) F.
14.13
(a) 2;
14.14
AG39g /(kJ/mol) = 0 + 4(0) + 2(-111.25) — 97.89 — 65.49 — 2(-237.129) =
14.15
14.16
(b) 1;
(c) 2;
(d) 6;
(e) 2.
88.38.
£ =-AG°/nF = —(88380 J/mol)/2(96485 C/mol) = -0.458 V.
(a)
The NaCl changes the ionic strength and hence changes the activity
coefficients and the activities. Therefore £ changes.
(b)
By definition,
=—AG°/nF. The NaCl doesn’t change AG° and so
doesn’t change £°..
The half-reactions are In > In** + 3e” and 2e7 + Hg2SO,(s)
> 2Hg + SOny ,
The cell reaction is 2In(s) + 3Hg2SO«(s) > 2In**(aq) + 6Hg(/) + 38077 (aq).
Taking the activities of the pure solids and the pure liquid as 1, we have for the
activity quotient: Q = (a,)*(a_)° = 2737°(y.m/m°)>, where + and — refer to
In**(aq) and SO?" (aq), respectively, and Eq. (14.46) was used. The Nernst
equation gives £= £ — (RT/6F) In (a) (ay =
& — (RTI6F) In ((108)'y.mjlm°}° = & — (SRTI6F) In (2.5508y.mj/m°), where i
refers to In2(SO4)3.
14.17
(a)
Equation (14.49) applies. We plot the left side (1.s.) of (14.49) vs.
(m/m°)'". We have L.s. =
£+ [2(8.314 J/mol-K)(333.15 K)/(96485 C/mol)] In (m/m°) =
£ + (0.05741 V) In (m/m°). The data are
Ls./V
(mim°)'?
0.1985
0.03162:
0.1993
0.04472
0.2008
«0.07071
0.2104
+~—«0.31622
Plotting the three points at high dilution and drawing a straight line
through them, we find the intercept to be & = 0.1966 V.
190
ls
IN
0.202
0.201
0.200
0.199
i}
H1
0.198
OTLOF,
FJ
Fa
OS)
Gt
Spa
7117
Sal
(LO
aL
y(t
0.196
Saranbemnetoeens
0
(b)
cory
(mim?)
ea
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08
From Eq. (14.49), (2RT/F) In y. = & —1.s. and In ys =
(17.417 V')(0.1966 V — 1.s.) = 3.424 — 17.41;(1.s/V). At 0.005 mol/kg,
In yz = 3.424 — 17.417(0.2008) = -0.073 and y. = 0.930. At 0.1 mol/kg,
In ys = —0.2405 and y. = 0.786.
14.18
(a)
In@=B'P = BP/RT = (14.0 cm*/mol)(750/760)atm/
(82.06 cm>-atm/mol-K)(298.1 K) = 0.000565 and @ = 1.00056s.
f = OP = 1.00056; bar.
(b)
The half-reaction is Hy
2H* + 2e and the error in fis
(RT/nF) In f(H2) — (RT/nF) In P(H2) = (RT/nF) In (f7P) = (RT/nF) In 9 =
[(8.3145 J/mol-K)(298.15 K)/2(96485 C/mol)]0.000565 = 0.0000073 V.
14.19
Equation (14.39) shows that £ depends only on the chemical potentials 4; of the
species involved in the cell’s chemical reaction, so must be independent of
what metals are used for the terminals.
14.20
(a) Doubled;
(b) squared;
(c) doubled, since In Q? =2-In QO;
(d) unchanged.
14.21
(a)
Equation (14.21) applied to Cu < Cu’*(aq) + 2e (Cu) gives fig, =
We
aa)
+ 2
e (Cu)
_ Substitution of (14.19) for ft of each species gives
Hou = Mey2*¢aqy + 2F (aq. CuSOg4) + 2[H.- cy) — Fo(Cu)] and
(Cu) — o(aq. CuSO4) = [Ha2* (ag) —Uc, + 2M
191
cayV2F.
(b)
By analogy to the result of (a), we have o(Zn) — o(aq. ZnSO«) =
[H732* (aq) lea
(c)
2H.
zn) \2F.
For e(Cu’) < e (Zn), Eqs. (14.21) and (14.19) give fi,-fe n= eee
~
(d)
Hen)
— Fo(Cu’) =
[cy
? Me (gq dF
e (Zn)
— Fo(Zn), and (Cu’) — (Zn) =
Substitution in (14.25) and use of 4; = 0(aq. CuSO) — o(aqg. ZnSOx)
gives £= see ae —Uoq, + 2M cy MAF + 4) [M2n2* (ag) it
Zn a 2.
Eres (Moy (aq) Bret
Zt
Zn) V2F + [H.- Zn)
=
H-cuy
I/F
x
Monet (ag) + Moy) /2F, since B-cu) 7 Hzicu)
(Sec. 14.3). Substitution of 4; = ww; + RT In a; for each species gives
f=ty+
(He 2 ia
~ Mn — Hen?* (ag)
(RT/2F)(1n ee ray In Ci
£, - AG°/2F— (RT/2F) |n [a
since
14.22
In ee
oy )/2F +
eh
In ace VS
the (aq) acy 12+ (ag)
az, ], which is (14.51),
=—AG°/2F.
For electrolyte i, p; = w; + RT In a; (Eq. (10.4)] and p; = v4p, + v_p =
V,M, +V.RT In a,+ v_u2 +Vv_RT In a_ (Eqs. (10.38) and (10.4)]. Equating
these two expressions for 1; and using 4; = v,u° + v_p° [Eq. (10.44)], we
get RT In a; = v,RT In a, + v_RT In a_ = RT In [(a,)”* (a_)’- ], soa; =
CY GE
14.23
(a) F;
14.24
(a)
(b) T.
AGyog/(kJ/mol) = 2(-4.7) + 3(0) — 3(65.49) — 2(0) = -205.9.
AG° = -2.06 x 10° J/mol.
& = -AG°/nF = (2.06 x 10° J/mol)/6(96485 C/mol) = 0.356 V.
(b)
The left half-reaction must be an oxidation and is Fe
right half-reaction is Cu** + 2e~ 3 Cu. Then
0.34 V — (-0.04 V) = 0.38 V.
192
> Fe>* + 3e ; the
& = Ep -F, =
14.25
(a)
At 252G8
221,978
Sea
aI
esl Fgene= ERT, tie 0.268
Vand fFonesuch = —1-710 V.
(b)
Use of © =F, —; gives at 43°C; —0.80 V = 2
icnpareit —Fcalome and
1.70 V =F ionpareit. — Fnonesuch: Mddition of these equations gives
0.90
e=
14.26
V= ERenparel = fen aie
A neich
BOWtNE cell nonpareil |nonesuch,
a cenparel = —0.90 V
FP =F, -F,=0.222
V-0=0.222 V. We have £=£ —(RT/nF) In Q so
In Q = (nF/RT)(& — 24). We shall write the cell reaction using the smallest
integers as coefficients, as in Eq. (14.34).
(a)
(b)
14.27
In Q = 2(96485 C/mol)(0.222 V + 1.00 V)/(8.314 J/mol-K)(298.1 K) =
95.1 andQ=2x 10".
With £= 1.00 V, we get In Q = -60.6 and Q = 5 x 107”.
The emf is given by Eq. (14.45) and the following paragraph as = £2 +
(RT/2F) |n [P(H2)/P°], where a(H")a(CI) = a(HCl) = 1. We have
In [P(H2)/P°] = (2F/RT)(é —&) = (77.85 VE
(a)
A= 0222.)),
In [P(H2)/P°] = (77.85 V~')(-0.300 V — 0.222 V) = 40.64 and
P(H») = 2 x 107'® bar.
14.28
(b)
In [P(H>)/P°] = 6.07 and P(H2) = 434 bar. (This answer is only
(a)
The left half-reaction is an oxidation, so the half-reactions are
approximate, since at this high pressure we should use the fugacity,
rather than the pressure, of Hz.)
Fe2* — Fe** +e” and lh + 2e — 21. Multiplying the left half-reaction by
2 and adding it to the right one, we get as the cell reaction:
2Fe** +1, > 2Fe** + 21.
(MEN) =-0.236 V.
2=2 —{Ri/nF) nO
(b)
eA ES ipape Woe yaa
(c)
= 0.236 V — [(8.314 J/mol-K)(298.1 K)/2(96485 C/mol)] x
In [(1.20)?(0.100)7/(2.00)° 1] = -0.164 V.
E= dr — b, < 0, SO Or < oy and the left-hand terminal is at the higher
potential.
193
(d)
The negative value of £ indicates that the spontaneous cell reaction is in a
direction opposite to that written in (a). Hence the half-reaction
Fe** +e — Fe”* occurs spontaneously in the left half-cell. Electrons
therefore flow into the left terminal from the load.
14.29
(a)
The left half-reaction is an oxidation, so
Cue Cucmeres
2e + Hg,SO,(c)
> 2Hg + SOF
Cu + Hg,SO,(c)
> 2Hg +Cu** +SO4
14.30
(b)
F=f,
-F, =0.615 V — 0.339 V = 0.276 V. Equation (14.46) gives
(c)
a(Cu?* )a(SO2) = (yzm/m°) = [0.043(1.00)]° = 0.00185. The activities
of the solids are 1 and £= 0.276 V — (RT/2F) In 0.00185 = 0.357 V.
With y. = 1, we would get £= = 0.276 V.
The cell reaction is Zn + 2AgCl(c) BI
2Ag + 2Cl and
£° = 0.222 V — (-0.762 V) = 0.984 V. Using Eq. (14.46), we have
f= F —(RT/2F) In [a(Zn**)a(Cl)’] = & - (RT/2F) In [4(0.0100y.)°] =
0.984 V — [(8.314 J/mol-K)(298.15 K)/2(96485 C/mol)] In[4(0.0100)°(0.708)°}
=i love Ve
14.31
Addition of (1) Cr°* + e° > Cr** and (2) Cr** + 2e" > Cr gives
(3) Cr’* + 3e7 > Cr. Hence AG; = AG; + AG;, which becomes —n3FF; =
-n\ FE, — n2FF3; 80 &, = (mE, + nF; yng = [1(-0.424 V) + 2(-0.90 V)}/3 =
—0.74 V.
14.32
£=£ —(RTInF) In [a(Zn**)/a(Cu**)]. The ZnSO, solution has Ip/m° =
+[2°(0.002) + 27(0.002)] = 0.00800 and the Davies equation gives
log y(Zn**) = -0.163; y(Zn?*) = 0.68g. Also, In/m® = 0.00400 for CuSO,(aq);
log y(Cu**) = -0.119; y(Cu?*) = 0.76. Use of ami = Yumi/m® gives
a(Zn**) = 0.683(0.00200) = 0.00138 and a(Cu2*) = 0.00076.
So £= 0.339 V — (-0.762 V) — (RT/2F) In (0.00138/0.00076p) = 1.093 V.
14.33
(a) PtlAg| AgCl(c)| KCK(aq)| HgsClo(c) |Hg |Pv.
194
(b) Pt|H»|H2SO.(aq) |Hg2SO.(c) |Hg |Pt’
(or we can use a Pb |PbSO, half-cell).
14.34 Pt’|H>|HCl(aq)| Clo |Pt,
Pt| H|HCl(aq) |AgCl(c) |Ag| Pr’,
Pt |Hz |HCl(aq) |Hg2Clo(c) |Hg |Pr’,
Pt| Ag| AgCl(c) |HCl(aq) |Hg2Cla(c) |Hg |Pr’.
14.35
(a)
The half-reactions are Ag > Ag*(0.01m°) +e and Ag’(0.05m°)
+e >
Ag. The cell reaction is Ag*(0.05m°) > Ag*(0.01m°). We have & =
F,, —&, = 0. Then £= 0 - (RT/F) In [y+,(0.01)/y+,.x(0.05)] =
(0.02569 V){—1.6094 + In (y+.1/Y+.2)] = 0.04135 V —
(0.02569 V) In (yx.1/Y+.r). We have Im, = 0.0100 mol/kg and In.r =
0.0500 mol/kg. The Davies equation for Ag” gives log y., = —0.0448 and
Ys. = 0.902; also, log Y+,.2 = 0.08555 and y,.x = 0.821. Hence
f= 0.04135 V — (0.02569 V) In (0.902/0.821) = 0.0389 V.
14.36
(b)
£=0r-—
6, >0, SO Or > Or.
(c)
The electrons flow from low to high potential and hence flow into the
right terminal.
£=-[(8.314 J/mol-K)(358.1 K)/2(96485 C/mol)] In (2521/666) = —0.0205 V,
where Eq. (14.60) was used.
14.37
(a) T; (b) F.
14.38
(a)
[Ag+Cl — AgCl(s) +e] x2
Cl
He Gl5(s)-2en—>2Heme
) +2Hg
> 2AgCl(s)
2Ag+Hg,Cl,(s
(b)
(RT 2F) In [a(AgCl)a’(Hg)/a*(Ag)a(Hg2Cls)] =e posites (date
activities of the solids can be taken as 1. Sof=/ =F, -F, =
G=27
0.2680 V — 0.2222 V = 0.0458 V.
= 0.0458, V.
(Cc)
=e
(d)
Since £=£/, we have dAOT = 0F-/0T. So AS° = nF(0Z°/0T)p =
2(96485 C/mol)(0.000338 V/K) = 65.2 J/mol-K. AG® = Ske =
195
~2(96458 C/mol)(0.0458 V) =—-8840 J/mol.
AH° = AG° + T AS® =
—8840 J/mol + (298.1 K)(65.2 J/mol-K) = 10.6 kJ/mol.
14.39 F =F, -F, =0-(-0.01 V)=0.01 V. AG? =-nFF° =-RT In K° and
In K° = nFP-IRT = 2(96485 C/mol)(0.01 V)/(8.314 J/mol-K)(298.1 K) = 0.8.
K° =2.
14.40
(a)
For the reaction 2Na* + H) > 2Na + 2H’, we have Fy, =fp —F) =
~2.714 V —0 =-2.714 V. Then AG‘o, = -—nFF = -2(96485 C/mol) x
(2.714 V) = 5.237 x 10° J/mol = 2(0) + 2(0) -2 A Gog (Na*) — 2(0) and
A ;G3og(Na*) = -2.619 x 10° J/mol = -261.9 kJ/mol.
(b)
For the reaction 2Cl + 2H? > Cl, + Hp, we have £59. = 0 — (1.360 V) =
—1.360 V. Then AGj, = —2(96485 C/mol)(-1.360 V) =
2.624 x 10° J/mol = 0+ 0- 2 A ¢Gy9g(Cl) — 2(0) and A -Gyog (Cl) =
1.312 x 10° J/mol = -131.2 kJ/mol.
(c)
For the reaction Cu** + H) > Cu + 2H*, £59, = 0.339 V -0 = 0.339 V
and AG5og = -2(96485 C/mol)(0.339 V) = -6.54 x 10* J/mol =
0 + 2(0) — A -Gyog (Cu’*) - 0 and A ¢ Grog (Cu**) = 6.54 x 10* J/mol =
65.4 kJ/mol.
14.41
2e° + PblI(c) > Pb + 20(aq)
Pb — Pb** (aq) + 2e°
PbI,(c) — Pb** (aq) + 21 (aq)
P= 0.365 V— (01126 V) = -0,239.V. eAGl == Ee —
-2(96485 C/mol)(-0.239 V) = 4.61 x 10° J/mol. In K},=-AG°/RT =
~(46100 J/mol)/(8.314 J/mol-K)(298.1 K) = -18.6 and K°, =8 x 10°.
14.42 (a) £ = 1.360 V- 1.078 V =0.282 V. AG® = -2(96485 C/mol)(0.282 V) =
~54400 J/mol. In K° =-AG°/RT =
(54400 J/mol)/(8.314 J/mol-K)(298.1 K) = 21.95 and K° = 3 x 10°.
196
(b)
£=0.282 V,n= 1, and AG? = -27200 J/mol.
In K° = 10.93 and K° =
6 x 10°.
(c)
The half-reactions are 2(Ag + Cl! > AgCl +e) and Cl; + 2e — 2CI.
F = 1.360 V — 0.222 V = 1.138 V.
-2.196 x 10° J/mol.
(d)
In K° = 88.59 and K° = 3 x 10”.
This is the reverse of (c), so AG® = 2.196 x 10° J/mol and K° =
Gx10
(e)
AG®° =-2(96485 C/mol)(1.138 V) =
jot= 350108 |
The half-reactions are 2(Fe** Pe hee
2 eV
P=20 440 N = 007 1aV=—1
e ) and Pew
2e => Fe,
AG a2 557 X 10° J/mol,
In K° =-94.2, and K°=1x 10".
14.43
14.44
PIV = 0.23646 — 5.1144 x 10“(t/°C) —2.0628 x 10°((t/°C)> —
1.0808 x 10%(#/°C)’.
Cell 1 is Pt| Fe?*(aq), Fe**(aq) ::Fe**(aq) |Fe| Pt’.
Cell 2 is Pt| Fe2*(aq), Fe**(aq) ::Fe**(aq) |Fe |Pr’.
Cell 3 is Fe| Fe**(aq) ::Fe**(aq) |Fe. All have the same cell reaction. For
example, Cell 3 has the half-reactions Fe’* + 2e” > Fe and Fe > Fe™* + 3e,
and multiplication of these half-reactions by 3 and 2, respectively, gives the
overall reaction as 3Fe** > Fe + 2Fe**. For Cell 1, =-0.44 V-0.77 V=
-1.21V, nf =2(-1.21 V)=-2.42V, AG° =-nF = 233 kJ/mol. For Cell
IF = 0,04 V 30.77 V =—0.8L V5 nee = 3(-0. 81) 2.43 V,, AG =
_nFP = 234 kJ/mol. For Cell 3, & =-0.44 V — (0.04 V) =-0.40 V, nf =
6(-0.40 V)=-2.40 V;, AG® = 232 kJ/mol. AG° must be the same in all three
ous
cases, since all the cells have the same reaction. (Because of the spontane
reaction 2Fe**(aq) + Fe(s) > 3Fe**(aq), an Fe** |Fe electrode is not
reproducible.)
14.45
(a)
The half-reactions are Fe > Fe?* + 2e” and 2(Fe** + e° — Fe”*). The cell
reaction is Fe + 2Fe** > 3Fe”*.
(b)
AG° =-nFF = -2(96485 C/mol)[0.771 V — (-0.440 V)] =
~2.33, x 10° J/mol. AS° = nF(0E°/0T)p = 2(96485 C/mol) x
(0.00114 V/K) = 220 J/mol-K. AH® = AG? PAS es
~2.33, x 10° J/mol + (298.1 K)(220 J/mol-K) = —1.68 x 10° J/mol.
197
14.46
Equations (14.67) and (14.68) give at 10°C:
= 0.23643 V —
(4.8621 x 10 V/K)(10 K) — (3.4205 x 10° V/K*)(10 K)* +
(5.869 x 10° V/K*)(10 K)° = 0.23123 V. Then AG? =
—2(96485 C/mol)(0.23123 V) = 4.4621 x 10° J/mol. Equation (14.69) gives at
10°C: AS° = 2(96485 C/mol)[-4.8621 x 10~ V/K +
2(-3.4205 x 10° V/K’)(10 K) +3(5.869 x 10°” V/K*)(10 K)’] =
~106.69 J/mol-K. Then AH® = AG° + T AS° =
4.4621 x 10* J/mol + (283.15 K)(-106.69 J/mol-K) = -7.4830 x 10* J/mol.
From (14.66) and (14.67), AC; = 2FT[2c + 6d(T— To)] = 2(96485 C/mol) x
(283.15 K)[2(—3.4205 x 10° V/K’) + 6(5.869 x 107° V/K?)(10 K)] =
~354.55 J/mol-K.
14.47
For. the cell Ag |Ag:l |Agl(c) |Ag, we have the half-reactions
Ag — Ag’ +e
and Agl(c)
+e — Ag +T; the cell reaction is
Agl(c) > Ag* +I. Equation (14.63) gives & = RT In K,, /nF =
(8.314 J/mol-K)(298.1 K) In (8.2 x 10°'’)/(96485 C/mol) = 0.951 V =
Fe ods
14.48
Gio as Gr
0. OONV aHenceA t=. 0.152, V.
The half-reactions are Hy
> 2H* + 2e and 2(AgBr +e — Ag +Br). The cell
reaction is H2 + 2AgBr hae
yee
2Ag.
4 =0.073 V. Equation (14.46)
gives £=£ — (RT/2F) In {a(H‘)a(Br
)*/Maddala
cea he
£ —(RTI/F) In [a(H*)a(Br)] = & = (RT/F) In (y+mj/m°)’. Hence 0.200 V =
0.073 V — [2(8.314 J/mol-K)(298.15 K)/(96485 C/mol)] In (0.100y+). We get
In (0.100y+) = -2.47 and ys = 0.84.
14.49
As in Prob. 14.40b, we find A,G°(Cl ) =~-131.2 kJ/mol. Since u°(HCl) =
.°(HCI) = n°(H") + p°(CI) [see Eq. (10.44)] and A ,G°(H") = 0, we have
A ,G°THCl(aq)] = -131.2 kJ/mol.
14.50
(a)
The half-reactions are Hy
> 2H* + 2e and 2[AgCl(c) +e > Ag+Ctr]J.
The cell reaction is Hy + 2AgCl(c)
+ 2H* + 2Cl + 2Ag. The solids’
activities can be taken as 1, and £= £ — (RT/2F) In {{a(H*)]*[a(Cl]*} =
& — (RT/F) In [a(H")a(Cl)]. Since we are writing H* rather than H30*,
198
we write the water ionization as H2O — H*+OH. Then Ki=
a(H*)a(OH_)/a(H20) and a(H*) = K°.a(H,0)/a(OH_). So a(H")a(CI) =
)m(Cl )/y(OH )m(OH_ ).
K°a(H,O)a(Cl )/a(OH ) = Kj,a(H,0)y(Cl
Substitution in the above equation for £ gives the desired result.
(b)
As Im — 0, the y’s go to 1 and a(H20O) goes to 1. Hence
£— F —(RTI/F) In [K,m(Cl )/m(OH_
)J.
Soln K\, =(FIRT){& — £-(RT/F) In [m(Cl )/m(OH )}* =
[(96485 C/mol)/(8.314 J/mol-K)(298.15 K)](—0.8279 V) = —32.22s and
Ken
14.51
(a)
1.010105
The half-reactions are the same as in Prob. 14.50a and the cell reaction is
H> + 2AgCl(c)
£
> 2H’ + 2Cl + 2Ag. As in Prob. 14.50a,
)]. For the ionization HX Lies xX ,we
=F —(RT/F) \n [a(H')a(CYI
have K; = a(H")a(X )/a(HX) and a(H') = K,a(HX)/a(X). Hence
a(H*)a(Cl) = K; a(HX)a(Cl lax) =
K, y(HX)m(HX)y(Cl )m(CY
yx
mC
ym”. Substitution in the above
equation for £ gives the desired result.
(b)
As Im — 0, the y’s — 1 and we get
InK? =(F/RT){& — &-(RTI/F) in [(m(Cl )m(HX)/m(X )m°}}* =
[(96485 C/mol)/(8.314 J/mol-K)(298.15 K)](—0.2814 V) =-10.953 and
K? =1.75x 10°.
14.52
Sn+Pb2* © Sn2*+Pb. & =-0.126
V+0.141
V=0.015 V. InK°=
nFEIRT= 2(96485 C/mol)(0.015 V)/(8.314 J/mol-K)(298 K) = 1.17 and
) =
K° = 3.>. The solids’ activities can be taken as 1, so 3.5 = a(Sn**)/a(Pb**
m(Sn**)/m(Pb**) = 2/(0.1 —z). We find z = 0.076. Hence m(Sn?*) =
so
0.076 mol/kg and m(Pb**) = (0.024 mol/kg. The solution is reasonably dilute,
we expect the activity coefficients to be determined mainly by the ionic
strength. Hence y(Pb™*) = y(Sn**) and the activity coefficients cancel in the
expressions for K°.
199
14.53
4, =612 mV and 4; = 741 mV. Equation (14.73) gives pH(X) = 6.86 +
[(612 — 741)10° V](96485 C/mol)/(8.314 J/mol-K)(298.1 K)2.3026 = 4.68.
14.54
In Eq. (14.78), we take ve = es , since the ionic strengths of the two
solutions are equal. Then oP — o* = [(8.314 J/mol-K)(298 K)/(96485 C/mol)] x
In (0.100/0.150) = -0.0104 V, where B is the NaNO3-KNO; solution.
14.55 w= Sdand § = pd = (3.57 x 10° C m)(1.30 x 107° m) = 2.75 x 10°" C.
Sle = (2.75 x 10°7° CV(1.60 x 10°'’ C) = 0.172.
14.56 (a)
>, O,x,
=(-0.5e)(-1.5 A) + (-1.5e)(1.0 A) = -0.75e A =
~0.75(1.6 x 10!° C)(10'° m) = -1.2 x 10°? C m. The direction is in the
negative x direction and the magnitude is 1.2 x 10° Cnn:
(b) x= D; Ox, =(-e)(l A) = -e A. py= X, Oy;
IN,
=(-e)(1 A) =
w= (2 +5)? =2'*¢ A=2.3 x10? Cm. The vector makes
an angle of 45° with the x and y axes and points from the negative
charges’ region toward the positive charge.
(c)
14.57
; O,x; =(-0.5e)(-2.5 A) + (2e)(-1.0 A) = -0.75e A.
Let a, b, and c be the x, y, and z coordinates of the new origin in the original
coordinate system. If x; 1s the x coordinate of charge i in the original system,
then p, in the new coordinate system equals >’; Q;(x; —-a) =; Q;x; -
ad; O; =X; O;x,; -a-0 =; Q,x;, whichis the same as in the original
coordinate system. Similarly for py and p,.
14.58
(a)
Wro- Wry = (1 — m/e
(r, = ry ny (7,a0
(b)
= (mM — r2)(1) + 12)/ryr2(r7) + 12) =
)e
Forr>>d, we have r) = rz =r, and I/ry — Wr, = (7? —r2)/2r°. The law
of cosines for triangle PAC gives ca = ee + d° — 2rd cos 0. (Because
r >> d, angle PAC is very nearly equal to angle PBC.) Then im a ie =
2rid cos 0 - d’ = d(2r, cos 0 — d) = 2rd cos 0, since d << rj).
(c)
= (Q/r2 — Olr;)/4m€o = O(2rd cos 0)/2r°*(4nep) = (u cos 0)/4it€gr”, since
p= Od.
200
14.59 (a)
w= J? Fdr=-(Q\Q./4ne) J? F? dr =—-~(O102/4n€9)(1/r; — Wr) =
(1.602 x 10°" C)?
1
4n(8.854 x 10°'* C?/N-m?)\10’m
(b)
Heint'yn
|
= 10°°m
)
=2.08x10!°J
From Eq. (14.89), we include a factor 1/€,, where (Sec. 10.8) €, = 78.40;
so w = (2.08 x 107’? J/78.40 = 2.65 x 1077" J.
14.60
w=0 for CCl, and Eq. (14.87) gives
1.24 153.8g/mol
6.022 10”* mol!
4.24 1.59g/cem*>
—3(8.854107'? C?/N-m’)
-
and o = 1.25 x 10°? cm?-C’/N-m? = 1.25 x 10°? C*-m/N.
a/4n€p =
(1.25 x 107°? C?-m/N)/41(8.854 x 1071? C?/N-m?) = 1.12 x 10°? m3 = 11.2 A?.
14.61
(a)
w=0 for CHy. Also, PV = (m/M)RT and M/p = RT/P. Equation (14.87)
gives
0.00094
-atm/mol-K)(273.1K)
(82.06 x 10% m?
_=
3.001
1.00 atm
6.022107? mol”!
3(8.854x107!? C?/N-m’)
a = 3.10 x 107° C? N! m. Also, o/4it€9 = (3.10 x 10° C?-m/NY
4n(8.854 x 107'? C?/N-m’) = 2.78 x 10°? m? = 2.78 A’.
(b)
€-+2=3.
M/p=RT/P.
w=0.
Equation (14.87) gives
€, — 1 = PN4O/RT€o = 4NPN4(O/4t€9)/RT =
4n(10.0 atm)(6.022 x 107° mol)(2.78 x 107° m’)/
(82.06 x 10° m?-atm/mol-K)(373.1 K) = 0.0068; and €, = 1.00687.
14.62
Equation (14.87) applies. We plot Mp '(€, — 1)/(€, + 2) vs. 1/T. Noting that
M/p = RT/P for a gas, we have
201
10°Mp'(e, — 1)/(e, + 2)
10°/(T/K)
5.73
2602)
5.35
5.00;
) 2:380NN2049eew?
4.67
0664
4.31
LOIG
[10°Mp '(e, — 1)/(e, + 2)\/(m?/mol)
2077.7x + age
0.3535
Ue. slahmeiieeaialss y = Sacha
0, ee
Se
5D)
5.0
4.8
4.6
4.4
4.2
0.0019)
SESS
50,0021,
Ssh
Sse
5.0.0023
eS Soave
0.00259"
sean
TEiKe
10.0027
where the units of the first line are m*/mol. The plot is reasonably linear with
slope 0.0208 m?-K/mol = Nayt/9€ok. So p? =
(0.0208 m?-K/mol)9(8.854 x 107!? C?/N-m7)(1.381 x 10°77? J/K)/
(6.022 x 1073 mol!) = 3.80 x 10°? C? m? and u = 6.17 x 10°° C m. The
intercept 1s 3.5 x 10° m*/mol = N,O/3€po, and
oe 3(8.854x10|?C*/N-m7)(3.5x10°° m?/mol)
6.022107 mol!
ot = 1.5 x 10° C*-m/N
[Using Eqs. (20.2) and (20.3), one finds that the dipole moment is 1.85 D.
Also, one find o/4m€ = 1.4 x 10°? m3 = 1.4 A?]
14.63
(a) Molecular,C m;
14.64
(a)
CSe, because of a greater a;
(b)
n-Ci9H22;
(c)
o-dichlorobenzene, because of a greater w.
(a)
Let f=(e,- 1), +2). dffde, = Me, + 2) —(€,—- 1)(e,
[ep +2) = (e, — Dent 2)? = 3/(e, #2)" = 0.
14.65
(b) molecular,
202
C m? V7;
(c) macroscopic, no units.
+2) =
(b)
We have 1 < €,<-o. Since dffde, > 0 for all €,, the minimum value offis
at the minimum value of €,; at €, = 1, fmin = 0. As €, 4 00, f approaches
its maximum possible value, which is 1.
14.66
ee
Then
oi 24 o* =
0.57440,
0.02569 V in P,P14100 ++ 0.0.0044P,,4649 0+
+ 0.5P; 540
Let P(K"*) =
= (0.02569 V) In (48.4/682) = -0.068 V =-68 mV
14.67
The half-reactions are Ag + Cl'(0.0100m°)
— AgCl+e and AgCl+e —
Ag + Cl (0.100m°). Equation (14.51) gives £= 4; + 0 (RT/F) In (0.100y_.r/0.0100y_7).
JIm.r = 0.100 mol/kg and the Davies equation
for Cl’ gives log y_.r = —0.1072 and y_z = 0.781. Similarly, y_., = 0.902. Then £
= -0.038 V — [(8.314 J/mol-K)(298.15 K)/(96485 C/mol)] In 8.66 = -0.093 V.
14.68
Im = 0.100 mol/kg. The Davies equation for H” gives log y, = —0.1072 and
Y, = 0.781. Then a(H’) = (0.781)(0.100) = 0.0781 and —log a(H") = 1.107.
14.69
K°/K°, = {[a(Ag*)]°/a(Cu*)}/{ fam(Ag*)}-/am(Cu’*)} = (0.997)7/0.997 =
0.997),
AG? =—-RI In Kh =—nFF- and
= (RI/nE) in K
so
EF.
=(RTnF\(In Kj, -In K?)=-(RT/nF) In K2/K>, =
—(RT/2F) |n 0.997 = 0.000039 V = 0.039 mV, which is insignificant.
14.70
(a) C; (b) m;
(c) N/C=V/m;
(g) dimensionless;
14.71
(d) V; (e) V; (f) Cm;
(h) J/mol.
If the left and right electrodes are at different temperatures, the emf is nonzero.
203
Chapter I5
15. 1 (a)yF i
(bars
15.2
(a) T;
(b) F;
15:33)
(a)
ee
(c) T;
(d) T.
3nRT = 1.5(1.00 mol)(8.314 J/mol-K)(298 K) = 3720 J = 889 cal.
(b) 3720J.
(c) (0.470 g)/(16.0 g/mol) = 0.0293 mol and 3nRT= 109 J.
15.4
(a)
(€,,) = 3kT = 1.5(1.381 x 10 J/K)(5S71.1 K) = 1.18 x 107°J.
G3) Sa
15.5
(€,,(T))/(€y(T;)) = (BKT2/2)/(3KT)/2) = T2/T; = (373.1 K)/(273.1 K) = 1.366.
15.6
Uims = (3RT/M)"”, SO Urms(Ne)/ Umms(He) = (Mrte/Mne)'”” = (4.0026/20.179)!? =
0.4454.
15.7
Ums =GRTy,/My,)” = (BRT, /Mo, )'” and Ty, = (My,/Mo, )To, =
(2.016/32.00)(293.1 K) = 18.5 K.
15.8 0Ey = 3nRT= 3 PV = 1.5(1.00 atm)(90 x 10° cm*)(8.314 J)/(82.06 cm? atm) =
1.37 x 10’ J. The answer is the same for 40°C.
15.9
(a) Oto;
(b) -coto : (ec) 0 (Fig. 15.5);
whose speed is in the range from vto v+ dv.
15.10
(a) T;
(b) F; (c) T;
(d) F.
204
(d) the number of molecules
15.11
(a)
=m/k=Nam/Nak = M/R. The interval is small enough to be considered
“infinitesimal” and Eq. (15.44) gives dN» =
4m.N (M/2MRT
)* eM" 28T1)?dy =
0.0320 kg/mol
ue
4n(6.02 x 107° |
27(8.314 J/mol-K)(300 K)
_ (0.0320 kg/mol)(500 m/s)? [som/s)* (0.001 m/s) =1.1«10!8
2(8.314 J/mol-K)(300 K)
(b)
Considering the interval as infinitesimal and using (15.42) with x
replaced by z, we have dN,
= N(M/2nRT)'exp (—Mv2 eRI a0s—
(6.02 x 107*)[0.0320 kg/mol/2n(8.314 J/mol-K)(300 K)]'” x
exp[—(0.0320 kg/mol)(150 m/s)*/2(8.314 J/mol-K)(300 K)] (0.001 m/s) =
TAS
10.
(c)
The fraction of molecules with x and z velocity components
simultaneously in the ranges v, to v, + dv, and UV, to VD, + dU; 1s
g(0,)2(V,) dv, dv, = (dN, /N )(dN,, /N) and the number of such
molecules is (dN,, /N)(dN, /N)N= dN, dN,_/N. From part (b), dN, |
= dN, =7.45x10'’ and dN, dN, /N = (7.45 x 10'7)/(6.022 x 10”) =
9.22 x 10''.
15.12 This probability equals dN/N = 4n(M/2nRT)*7
0 MY?87 9?dy=
4n[(0.0160 kg/mol)/27(8.314 J/mol-K)(300 K)}*” x
exp[—(0.0160 kg/mol)(400 m/s)*/2(8.314 J/mol-K)(300 K)] x
(400 m/s)*(0.001 m/s) = 1.24 x 10°.
15.13
(a)
(b)
15.147
(a)
Fig. 15.6 gives G(v) = 0.0018; s/m at v= 500 m/s, so dNJN =
G(v) dv = (0.0018; s/m)(0.0002 m/s) = 3.7 x 107’ and dN, =
(6.02 x 107°)(3.7 x 10°’) = 2.2 x 10"” molecules.
At 1000 K, Fig. 15.6 gives G(v) = 0.0008 s/m; dN,/N =
(0.0008 s/m)(0.0002 m/s) = 1.6 x 10°77; dNy=1x 10"".
Pr
ie G(v) dv= G(U;) re dv= G(v;)(U2 — V;) if G(v) is essentially
constant in the interval.
205
(b)
As in the example, M/2RT = 3.52 x 10° s*/m*. At v= 90.000 m/s,
G(v) = exp[-(90 m/s)°(3.52 x 10 s*/m’)] x
"(3.52 x 10° s2/m*)*742(90 m/s)” = 1.17328 x 10~ s/m. Replacing 90
by 90.002, we find that at 90.002 m/s, G(v) = 1.17333 x 107 s/m, and
the change is 0.004%.
15.15
(a)
Equation (15.44) gives (dN,,, IN)I(dN ,,IN
y2eMHAT
py? gM UAT — (42 /y?)exp[M (v7 —v3)/2RT ] =
5007 hs (0.0320 kg/mol)(15007 — 5007 )m7/s7
= 45000
2(8.314 J/mol-K)(298.1K)
15002.
(b)
From (15.44), (m/2nkT)’exp (-mv{/2kT) 410; =
(m/2nkT)*exp (—mv3/2kT) 4.05, and exp[m(v3 — v7;)/2kT] = 05/0;.
Taking logs, we have m(v3 — v;)/2kT = 2 In (v2/v,) and T=
M (v3 —v7)/[4R In (v2/01)] = (0.0320 kg/mol)(1500* — 5007)(m/s)*/
4(8.314 J/mol-K) In 3.00 = 1750 K.
15.16
(fiw) + foOv)) = Tm [fidw) +.Aw)lgQv) dw = J" fidw)gQv) dw +
es fw)g(w) dw = (fiw) + f,(w)).
(b)
15217
(=
le = cf(w) dw = arhe fdw= c(f).
The fraction of molecules with x and y velocity components simultaneously in
the ranges V, to v, + dv, and Vy to Vy + dvy is g(Vx)g(Vy) dv, dvy =
(m/2TtkT) ex p (—mv°/2kT) dv, dvy,5 where v” = v2x +02.
y Molecules with speeds
p
lying between v and v + dv have their velocity vectors lying within a thin
annulus (ring) of inner radius v and outer radius v + dv. The area of this
annulus is 1(v + dv)” — mv = 2mv dv, where the n(dv)’ term is negligible. The
probability that v lies in this annulus is the following sum over the annulus:
Y (m/2nkT eT
dy, dv, = (ml 2nkT ye”(2a dv) =
(m/kT )e-™”"*") dy, and this is the desired probability.
206
15.18
(a)
Since the integrand has the same value at —x as at +x, the areas on each
side of the origin are equal, and the integral from —° to 0 equals the
integral from 0 to °°.
(b)
For very small x, e- ax
~ 1 and the graph resembles that of y = x. For
large x, the exponential factor dominates the factor x and the function
goes to 0. The graph resembles the v = 1| graph in Fig. 18.18. The area on
the left side of the origin is negative and exactly cancels the area on the
right side of the origin.
15.19
(a)
(b)
Let z = ax’. Then dz = 2ax dx and J* xe"® dx = (1/2a) J? &* dz=
-(/2a)e*|" = -(1/2a)(0 — 1) = 1/2a.
(0/da) f= xe" dx = J" (/da) xe"™ dx = -J® x2 dx = (0/0a)(1/2a)
= —1/2a? and J xe°® dx = 1/2a”, Similarly, (8/0a) J" x3e°® dx =
- ie we™ dx= (d/da)(1/2a*) = —1/a® and Ikexe™ dx= Ia’.
15.20
15.21
(a)
Vims = (3RTIM)'? = [3(8.314 J/mol-K)(500 K)/(0.04401 kg/mol)]'” =
532 m/s.
(b)
(v) = (8/31)! Urms = 490 m/s.
(c)
‘ie See
ae Beiaiatly,
dG(v)ldv = (m/2nkT)?
[8mve "4? — Are? (mv/kT
ee” *" ]. The equation
dG(vy/dv = 0 gives 87Ump = 47unv;,,/kT and Ump = (2kT/m)'"" = (2RT/M)"”.
(v = 0 and v = ~ also satisfy G’ = 0 but these are minima.)
15.22
(v2) = J® v’G(v) dv = 4n(m/2nkTY” JF pte MAT chy =
4n(m/2nkTy?(2410"7/29-2)(2kT/m)” = 3kT/m = 3RT/M, where integral 3 (with
n = 2) of Table 15.1 was used.
207
15.23
(v3) = J? v'G(v) dv = 4m(m/2nkTY” Je vee” dv =
4n(m/2mkT)*?(2/2)(2kTim)y> = 2"?(kTimy?
i”? = 2’?(RTIM)>"/"*. We have
(v)(v7) = (8RT/M)'?(3RT/M) = 3(2°?)(RTIM)
tt? # (v3).
15.24
(a)
(v,) = S7, vye(v,) dv, = (m/2nkT)"?
J@,,ve" dv, = 0, where
integral 4 of Table 15.1 was used. From Fig. 15.5, v, is as likely to be
negative as positive, so (v,) =0.
(b)
Incalculating (v,), we average positive and negative values of v, and
get zero. In calculating (v~), all the (v2) values are nonnegative,
and we must get (v2) to be positive.
15.25
(Cc)
Verms = (02) 7 = (kT/m)'? = (RT/M)'”, where (15.37) was used.
(d)
Zeto7 see. Figs 155:
(vt) = J. v4 g(v,) dv, = 2(m/2mkT)"” J” vte7”**"
dv, .Integral 3 with
n=2in Table 15.1 gives (v2) = 2(m/2nkT)'?(242'7/2°-2)(2kT/m)>? =
3(kT/m)? = 3(RT/MY’.
15.26
Differentiation of the distribution function in Eq. (15.52) with respect to Ey
gives 2n(mkT) *?[5€,,/e S/T — (kT) eye €/*") = 0 and Exmp = ware
(5/3.
15.27
15.28
(a)
(2m)'71(u) = Jeeds = J¥[1 — 52/2 + (1/2!)54/4— (1/3!)s°/8 +---J]ds=
(b)
u—u'l6 + w/40 — u'/336 +---.
(2m)'"1(0.30) = 0.30 — (0.30)*/6 + (0.30)°/40 — (0.30)'/336 + --- =
0.2956 and 1(0.30) = 0.118.
(a) The probability is J” G(v) dv = (m/2mkT*? JY e-™”
AT amy? dy =
(Afr Vinp) Jo ew?
re
dv = (4/n"?
v3,)B,where Ump = (2kTIm)"?
2
and B = ips e ("¥m) »? dv. The integration-by-parts formula is J x ay
208
2
xy —
J y dx. Let x =v
and dy = eo (emp “ vdv.
SUC HeeR?
Dace SHU) as
ly
ph
tien Bi
Ao UR.)
Ui e
SU
Wye!
2
a wD
ee Oe
ee
Then y= =; mp e
OE OO)
cm
2
~(V/ Vip)”
CU SUT = 20) Dro,
MIO? WN ee Dorp€° = 52/2 "Ump2 Wel as re=
+ 5Uinp is
oy (OT
h
ee Dd Dg) and. |otG(0) a0.=
21(2"20' mp) — 20-"(0/rmple
(b)
The fraction with speed in the range from 0 to 4.243Ump is
212"? . 4.243) — 20 "(4.243)e ~(4243)" = 21(6.000) — (7.27 x 10°) =
2(0.4999999990) — (7.27 x 10°*) = 0.9999999253, since Fig. 15.10 gives
0.5 — 1(6) = 1.0 x 10°. The fraction with speeds exceeding 4.243Ump is
1 — 0.9999999253 = 7.5 x 10™.
15.29
(a)
The theory of relativity shows that speeds must be less than c, the speed
of light in vacuum, so the correct range is 0 toc.
(b)
The number of molecules whose speed according to the Maxwell
distribution lies between c and infinity is essentially zero at ordinary
temperatures.
15.30 From Eq. (15.58), 0.872 = (Mo,/M)' = [(32.0 g/mol)/M]'" and
M = 42.1 g/mol. The only hydrocarbon with this molar mass is C3H6.
15.31
(a)
Equation (15.58) applies. We have dN/dt = Na(dn/dt) = Na(An/At) and
P = (An/At)(20MRT)'7/A
ae (270(0.04496 kg/mol)(8.314 J/mol-K)(1690 K)]"”
7(0.001763/2)? m*
(10.510 (49.5x60)
kg)/(0.04496
kg/mol)
s
0 *m)°* atm 760 torr =0.0152 torr
P = (2.03 N/m’an ) 82.06 (1
ae
8.314 J
a
(b)
Equation (15.67) applies. The order of magnitude of d is a couple of
angstroms. The order of magnitude of A is
209
(82 cm? -atm/mol-K)(1690 K)
ee
2" 1(2x 10-8 cm)? (0.015/760)
atm (6x10? /mol)
Since dnote = 0.2 cm, the condition A >> dhole iS met.
15.32
The CO; partial pressure is 0.00033 atm and Eq. (15.56) gives dNw/dt =
[1.0x (10°?
m)}? (0.00033 atm)(6.02x107°/ mol) |8(8.314 J/mol-K)(298 i
4[82.06 (10°? m)? -atm/mol-K](298 K)
7(0.0440 kg/mol)
=F
7x10 es”.
The mass striking the leaf in 1 sec is [(7.7 x10'”)/(6.0 x 10” mol™')](44 g/mol)
= 5.6 mg.
T5333
P = (0.010/760) atm (10.1325 Pa/atm) = 1.33 N/m?. As discussed in the (ext
we can take the evaporation rate as essentially equal to the rate at which
molecules of vapor in equilibrium with the liquid strike the liquid. Equation
(15.58) gives
adN___(1.33 N/m*)(6.02x10**/mol)(10
*m7)
dt
[2m (0.3906 kg/mol)(8.3145 J/mol-K)(393 K)]!7_—
(8.9x10'’ molecules)
Lge
6.02x107* molecules
15.34
apie alee
ogUGe =0.58 mg
| mole
I< (dN/dt) x Pr(i), where dN/dt is the effusion rate and Pr(i) is the probability
of ionization. From Eq. (15.58), dN/dt « P/T'”. If x is the direction of motion
of the molecular beam, the statements in the problem give Pr(i)
I/(v,) «
1/T'””. Hence I « (P/T"?) x (A/T"”) and I~ PIT.
1535
(a) T; (b) F.
15.36
(a)
z(b) =(2 +2) collns./s =4 87,
(b)
Z,=[(2+242)s (1 x 10> cm’) =6 x 10° s7! cm.
Nszo(byV= 3(4 s'W(1 x 10° cm) = 12 x 10° s"! cm™ # Zp.
Noze(bYyV= 1(12 x 10° s' cm™) = 6 x 10° x"! cm™ = Zipp.
210
15.37
(a)
d = (v)(t) = (450 m/s)(4.0 x 107"? s) = 1.8 x 10°” m.
(b)
We have z(b) = 1/(5.0 x 107!° s) = 2 x 10” collisions per second and z;(c)
= 1/(8.0 x 10°'° s) = 1.25 x 10” collisions per second. z,(b) is greater than
z(c). From Eqs. (15.60) and (15.61), these rates are influenced by the
molecular diameters, the molecular concentrations, and the molecular
masses. Since no information is given about the concentrations (or the
molecular masses), the question is defective and we cannot say which has
the larger diameter. From (15.66), Ay = (385 m/s)/(3.25 x 10’ s') =
Pep at ON mt
15.38
(a)
zo(b) = 2"°td 3(SRT/MM;)'P,Na/RT =
2'2(3.7 x 107! m)’[8(8.3145 J/mol-K)(298 K)/n(0.0280 kg/mol)]'” x
(1.00 atm)(6.02 x 107°/mol)/(82.06 x 10° m?-atm/mol-K)(298 K) =
Ta
lO ssa
(b)
Zo» = =Nozo(bV = 5 zo(b)(PNa/RT) = 5(7.1 x 10° s~')(1.00 atm) x
(6.02 x 1073/mol)/(82.06 cm?-atm/mol-K)(298 K) = 8.7 x 107° s' cm™.
(c)
1.0x 10° torr=1.3x 10° atm.
z,(b) is proportional to P and Zp, is
proportional to P*, so 2.(b)i= Ole lex 10° s')(1.3 x 10° atm/1 atm) =
9.3 5! and Zp = (8.7 x 1078s"! cem)(1.3 x 10°7/1)? = 1.5 x 10's"! cm.
15:39
Let
b= CO, andc=N>.
P,=0.97(4.7 torr)(1 atm/760 torr) = 0.0060 atm and
P,.= 0.00019 atm.
(a)
From (15.61), z(b) = 4107? (RT/Mp)' PpNa/RT =
4n"(4.6 x 107!° m)*[(8.3145 J/mol-K)(220 K)/(0.044 kg/mol)]'” x
(b)
(0.0060 atm)(6.02 x 107*/mol)/(82.06 x 10° m’-atm/mol-K)(220 K) =
6.1x 10's".
z.(b) = (810)!(ry + re) [RT(M | + M7')]'? PANa/RT =
(c)
(810)"/7(4.15 x 107!° m)°(RT)'"[(0.044 kg/mol)" + (0.028 kg/mol) '}'” x
(0.0060 atm)Na/RT = 5.65 x 107s".
z.(c) = 4r'2(3.7 x 10°'° m)[RT/(0.028 kg/mol)]'*(0.00019 atm)Na/RT=
[oeetO. sss z-(c) + 2-(b) =o
(d)
10° s}+5.65;x10’s!'= 5.8) X Na ee
Zoe = Nozp(cV = Neze(b/V = 2(b)PeNa/RT =
(5.65 x 10” s')(0.00019 atm)Na/RT= 3.6 x 10° s' cm™,
211
15.40 (a)
(b)
(c)
15.41
A =RT/2'?nd’PN, = (82.06 cm’-atm/mol-K)(300 K)/
223.7 x 1078 cm)?(750/760) atm (6.02 x 107*/mol) = 6.8 x 10° cm.
2=(6.8 x 10° cm)(750/1) = 5.1 x 10° cm.
5.1 x 10° cm.
14100 ft = (14100 ft)(12 in./1 ft)(2.54 cm/I in.)(0.01 m/1 cm) = 4298 m.
P = Poe M887 — (760 torr) exp[—(0.029 kg/mol)(9.81 m/s*)(4298 m)/
(8.3145 J/mol-K)(290 K)] = 458 torr.
15.42 Mg2/RT= (0.029 kg/mol)(9.81 m/s*)(30 x 12 x 0.0254 m)/
(82 14/mol-K)(298 K) = 000105
ho ope lene
(1 atm)(1 — e °°!) = 0.00105 atm.
15.43 P/Py = e™"8"*" = 0.5 and z = -(RT/Mg) In 0.5 =
~[(8.3145 J/mol-K)(250 K)/(0.029 kg/mol)(9.81 m/s”)] In 0.5 = 5060 m.
15.44
M, =0.97(44) + 0.03(28) = 43.5. P= Pre eat =
(4.7 torr)exp[—(0.0435 kg/mol)(3.7 m/s”)(40000 m)/(8.314 J/mol-K)(180 K)] =
0.064 torr.
15.45
(a)
With T= To — az, (15.70) gives In(P’/P,) = —-(Mg/R) {2 (Ty — az)!
(Mg/aR)\n(Ty — az) Be (Mg/aR) \n[(To — az’)/Tp] = In{(Ty - age)
SOR)
15.46
(a)
dz =
Te lena
eel ean
ae (1h iz)
For Cy.m, the translation contribution is 3R/2, the rotational contribution
is 3R/2, and the predicted vibrational contribution is 2[3(5) — 6]R/2 = OR,
SO Cy.m is predicted to be 12R. Cpm=Cym+R
and Cpm 1s predicted to
be 13R.
(b)
No; the actual Cp is less than 13R at 400 K.
(c)
Cpm will reach 13R at high 7’s.
15.47 (a)
J*.g(w) dw=1= J®, Ae
dw = 2A(n!/2\(kTIc)"? and
A= (c/nkT)””, where integrals 1 and 2 in Table 15.1 were used.
212
(b)
(€,,.) =
1
2
15.48
ie Ewe(w) aw=A
ee cwe —cw2 /kT We
2Ac(2!10"7/271 N\(kTIc)*? _
T, where A = (c/mkT)'” and integrals 1 and 3 of Table 15.1 were used.
Lmy? = 5kT and Vims = (3kT/m)"" =
[3(1.38 x 10°? J/K)(298 K)/(1.0 x 10°"? kg)]'” = 0.00035 m/s.
15.49
(v2) = 3RTIM and (v)* = 8RT/MM, so (v0?) # (v)°*.
15.50
The possible values of s are 1, 2, 3, 4, 5, 6 and the probability of each value is
15.51
(a)
1/6. Hence (s) =X, sp(s) = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6)
= 21/6 = 3.5. Also, (s?) =, s’p(s) = 1°(1/6) + 27(1/6) + 3°(1/6) + 4°(1/6) +
5(1/6) + 6°(1/6) = 91/6 = 15.17. (s)? = (3.5) = 12.25 # (s”).
(b)
oy = (vz) -(v,)* =kTim—-Oand o, = (kT/m)"”.
(v,) =O and o = (kT/m)'”. The distribution function (15.42) is
Vv
and the desired fraction is
(21) /?7'e"**/?
Ones
oF
12
v,/26
ce e
2 dv;
=
2(2m)
2 dvx
z W265 a.1 ie e ane
v,/26
=
2020) 207 fie * ods =2I(1) = 0.68, where we made the
Sa)
2
substitution s = ,/o and used Eq. (15.51) and Fig. 15.10.
15.52
h(xy) = fx) + gy). We take (0/dx), of this equation. Let z = xy. The partial
derivative of the left side is 0h(xy)/dx = (dh/dz)(0z/0x) = h'(z)y. The partial
= y 'dfixldx.
derivative of the right side is df(x)/dx = h’(z)y; we have h’(z)
x 'dg(y)/dy. Hence
Similarly, taking (d/dy), of h(xy) = fix) + g(y) gives K@=
x! de(ydy = y! dfvo/dx and y[dg(y)/dy] = x{dflx\/dx] = k. By the argument
dx and f(x) =
that follows Eg. (15.33), k must be a constant. Hence df(x) = (k/x)
k In x +a, where a is an integration constant.
15.53
the plane
Molecules with v, = b have the tips of their velocity vectors lying in
axes and is a
v, = b. (This plane is parallel to the plane formed by the v, and v,
213
distance b from it.) The region corresponding to b < vx <c is the region
between the parallel planes at v, = b and v, = c.
15.54
(a)
1, since the probability density G(v) satisfies [e GD) dul
EquGlaezs):
(b)
A crude approximation to the area under the 300 K curve is found by
taking the area of a triangle with height 2 x 10° s/cm (the peak of the
curve) and base 10° cm/s (the width of the region for which G(v) has a
substantial value). This triangle’s area 1s 110° cm/s)(2 x 10~ s/cm) = 1.
15.55
(a)
Hp. The lighter H2 molecules move faster, so that their average kinetic
energy equals that of Oo.
15.56
(b)
(€,,) is the same, for both.
(c)
p= PM/RT is greater for O2 because of the greater mass of the O2
molecules.
(d)
From (15.67), A decreases as the molecular diameter increases. O> is
larger than Ho, so hu, = ho, -
(e)
The H2 molecules are moving faster and so collide more often with the
wall.
Substitution of 4; = w; (7) + RT In (P;/P°) into wp + Migz” = uP > Migz? gives
wh; (T) + RT In (P3/P°) + Migz” = 3 (T) + RT In (P?/P°) + Migz? or
RT In (P#/P?) = Mig(z’ — 2%). In (P2/P%) = —Mig(z — 2*)/RT and P® =
P® exp [-Mig(z — 2*)/RT].
1557)
(a) See Fig. 15.7a.
15.58
A, zp(b), and Lhe
15.59
(a) Statistical mechanics.
15.60
(a) False.
(b) True. See Fig. 15.5.
(f) True.
(g) False.
(b) See Fig. 15.6.
(h)
(b) Yes.
(c) True.
(d) False.
True. See Eq. (15.13). (i) False.
214
(e) False.
Chapter 16
16.1
(a) T;
(b) T;
(c) F.
16.2
(a) Equation (16.1) gives |g| = |K(AT/Ax)At| =
(0.80 J/K-cm-s)(24 cm’)(60 s)(50 K)/(200 cm) = 288 J.
(b)
AS;oq = 0 since the rod’s state is not changed. The temperature of each
end of the rod differs only slightly from the temperature of the reservoir
it is in contact with, so we can use dS = dg;ey/T = dq/T for each reservoir.
Theretore AS = AShorres + Aocoldres = (2008) (520s) + (288.J)/275 K)
= 0.161 J/K.
16.3.
Cym=3R/2 for this monatomic gas, and Eq. (16.12) gives
Les (a9 )
16\2
(8.314 J/mol-K)
(8.314 Jmol-K)(273 K) |
4)(6.02x1077/mol)(2.2 x107"° m)? | —_(0.00400 kg/mol)
k = 0.00142 J/K-cm-s. At 100°C, k = 0.166 J/K-m-s.
16.4
Cym+9R/4 = Cpm—R + 9R/4 = Cem + SR/4 =
(35.309 J/mol-K) + (5/4)(8.3145 J/mol-K) = 45.7 J/mol-K. Then
k = (5/16)(45.7 J/mol-K)[(8.314 J/mol-K)(298 K)/m(0.0160 kg/mol)]'"*/
[(6.02 x 107° mol')(4.1 x 107'° m)*] = 0.031 J/K-s-m =
31x 10°3 K's! cm", where (16.12) was used.
16.5
p =M/Vm= (18.015 g/mol)/(18.1 cm’/mol) = 0.995 g/cm? = 995 kg/m® and
k = 2.8(8.314 J/mol-K)[17.99/(17.72)(995 kg/m*)(4.46 x 107"? m?/N)]""7/
[(6.022 x 107 mol!)'2(18.1 cm*/mol)”*] = 6.05 x 10° JK! cm* ms" =
6.05 mJ K7! cm! $s".
(b) T;
(c) T;
(d) T.
16.6
(a) F;
16.7
The maximum (v,) for laminar flow makes Re = 2000, so (U,) max =
2000n/pd = 2000(0.0089 dyn s cm™)/(1.00 g/cm*)(1.00 cm) = 18 cm/s.
215
16.8
(a)
(32/760) atm (8.314 x 107 ergs)/(82.06 cm? atm) = 4.26. x 10° dyn/cm’.
Equation (16.17) gives n = (mr°/8V)(|AP|/|Ay\)t =
[70.100 cm)*/8(148 cm?)][(4.266 x 10* dyn/cm7)/(24.0 cm)](120 s) =
0.0566 dyn cm” s = 5.66 cP.
(b)
In time t, a volume V = mr’ (d) flows through the pipe of cross-sectional
area Tir, where (d) is the average distance traveled by the fluid. We
have (d) = (v,) rand (v,) = (d)/t= (V/nr?)/t = Vitr’t =
(7/8) |AP |/ |Ay |, where Eq. (16.17) was used. So (v,) =
[(0.100 cm)?/8(0.0566 dyn cm~ s)](4.266 x 10* dyn/cm’)/(24.0 cm) =
39.3 cm/s. Then Re = p (v,) d/n = (1.35 g/cm*)(39.3 cm/s)(0.200 cm)/
(0.0566 dyn cm” s) = 187 < 2000, so the flow is laminar.
16.9
(a)
From (16.17), (P2 — Pi)/2 — y1) = (89/nr*)(
Vit) =
~8[(0.04 dyn s cm~)/m(1.25 cm)*](5000 cm?)/(60 s) = -3.5 dyn/cem® =
—35 Pa/m.
(b)
(v,) = Vit = (5000 cm*)/n(1.25 cm)"(60 s) = 17 cm/s.
(c)
Re = p(v,) din = (1.0 g/em*)(17 cm/s)(2.5 cm)/(0.04 dyn s/cm?) =
1100 < 2000, so the flow is laminar and use of Eq. (16.17) 1s justified.
For a flow rate of 30 L/min, (Oy) and Re are 6 times as large and
Re = 6400 > 2000, so aortic flow is turbulent during vigorous activity.
16.10
P? — P? = (1.44 - 1.00) atm? = (0.44 atm’)(8.314 J)°/(82.06 x 10° m* atm)’ =
4.52 x 10° N/m’. Eq. (16.18) gives dn/dt = 1(0.000210 m)*(4.52 x 10° N*/m*)/
[16(0.0000192 N s/m*)(8.314 J/mol-K)(273 K)(2.20 m)] = 0.000018 mol/s,
which is 0.00058 g/s.
16.11
Ne.u,, = N,0(Pc,,,tccH,, /PH,0%H,0) = (1.002 cP)(0.659 g/cm*)(67.3 s)/
(0.998 g/cm*)(136.5 s) = 0.326 cP.
16.12
(a)
The pressures at the left and right ends of C exert forces Pms” and
—(P + dP)ms*, respectively, on C. The viscous force on C is given by
(16.13) as nA(dv,/ds) = n(2ns dy)(dv,/ds). So -(P + dP)ns? + Ps? +
n(21s dy)(dv,/ds) = 0 and dv,/ds = (s/2n)(dP/dy). Integration gives
216
Vy = (s?/4n)(dP/dy) + c, where c is a constant. Use of v, = 0 at s =r gives
c = -(r’/4n)(dP/dy). Therefore v, = (1/4n)(r" — s°)(-dP/dy).
(b)
The volume of fluid in the shell that passes a given location in time dt
equals the volume of a cylindrical shell with length v,(s) dt and inner and
outer radii s and s + ds; this volume is 1(s + ds) - Vy(s) dt — Ts Vy(S) d=
27sV,(s) ds dt, since (ds) is negligible compared with ds. Integration
over all shells from s = 0 to r gives dV = [J§ 2msv,(s) ds] dt. Substitution
of dV = dmi/p and vy = (1/4n)(r° — s*)(-dP/dy) gives dm/dt =
(np/2n)(-dPldy) J, (r’s — s*) ds = (r*p/8n)(-dP/dy). Separating P and y
and integrating from one end to the other, we have
= 4 dP = (dm/dt)(8r/nr'p) J>» dy, which becomes dm/dt =
(tr'p/8n)(P; — P2)/(y2 — yi). We have dm/dt = d(pV)/dt = p dV/dt = pVIt,
since the flow rate is constant with time. Hence (16.17) follows.
(c)
Substitution of p = PM/RT into (16.91), separation of P and y and
integration gives (RT/M)(dmldt)(8y/mr*) J}? dy = — Je PdP=
1 (p? — Py); use of (RT/M)(dm/dt) = RT d(m/M)/dt = RT dnidt gives
(16.18).
16.13
In time dt, the matter in the thin shell travels a distance Dy dt in the y direction.
The volume dV of matter in the thin shell that passes a fixed location in time dt
equals the length v, df times the shell’s cross-sectional area d4; dV = vy dt dA.
(The shell volume is the difference between volumes of cylinders of radii
s + ds and s, and the volume of a cylinder equals its length times its crosssectional area. Hence the shell volume equals the shell length times the shell
that
cross-sectional area.) We have p = dm/dV, where dm is the thin-shell mass
passes a given location in time dt, so dm = p dV = Pv, dt dA.
16.14
Equation (16.22) gives v = 2[(7.8 — 1.0) g cm (980.7 cm/s”)(0.050 cm)’/
9(0.0089 dyn s cm’) = 420 cm/s. For glycerol, v =
2(7.8 — 1.25) 107 kg(10 m) *(9.807 m/s”)(0.00050 m)'/9(0.954 N s m™) =
0.0037 m/s = 0.37 cm/s.
16.15
' =
From (16.25), @ = 5(MRT)'7/16nNan
0.1763(0.04401 kg/mol)!(8.314 J/mol-K)'7T'7/(6.022 x 10°/mol)n =
217
1.77 x 10°°° m? (7/K)!?/[y/(kg m7! s“')] = 1.77 x 10-74 m? (T/K)'7/(n/P),
where P stands for poise. At 0°C, d’ =1.77x 10% m’* (273)'7/(139 x 10°)
and d= 4.59 x 107°
16.16
m=4.59 A. At 490°C, d= 3.85 A; at 850°C, d= 3.69 A.
The diameters of H; and D> are the same and (16.25) shows that n is
proportional to M'”. So Np, /Nn, =(Mp, /Mu, )? and Np, =
(4.03/2.02)"(8.53 x 10° P) = 1.20 x 10° P.
16.17
(a)
M, =>) x)M; = 0.5(200 kg/mol) + 0.5(600 kg/mol) = 400 kg/mol.
My = (Xi xiM 7)/(i Mi) =
[0.5(200 kg/mol)” + 0.5(600 kg/mol)”]/(400 kg/mol) = 500 kg/mol.
(b)
16.18
Let us take 600 kg of each species. We then have 3.0 moles of the
molecular-weight 200000 species and 1.0 mole of the molecular-weight
600000 species. Hence M,, = >; x,M; =
0.75(200 kg/mol) + 0.25(600 kg/mol) = 300 kg/mol. M, =
[0.75(200 kg/mol)” + 0.25(600 kg/mol)*]/(300 kg/mol) = 400 kg/mol.
We plot (1, — 1)/pp vs. Pp. The points fit a straight line well, and extrapolation
to Pg = 0 gives [] = lim, _,o(n-— 1)/pp = 0.147 dm*/g = 147 cm’/g. Then
147 cm’/g = (0.034 cm?/g)(Mp/M°)°® and Mg = 390000 g/mol.
M,.p = 390000.
16.19
(a) sR (DSF
16.20
(a)
(Cc) Ty s(d)er (ey als a)
(Ax)ims = (2D1)'”, so t = (Ax)?_./2D = (1 cm?)/2(107! cm?/s) =
5 x 10% s=2~x 10" yr.
(b) ¢=(1cm’)/2(10-” cm’/s) = 5 x 10° s = 2 x 107 yr.
16.21
(a)
(Ax)ms = (2Dt)' = [2(0.52 x 107 cm?/s)(60 s)]!? = 0.025 cm.
(b)
t= 3600s
(c)
t= 86400 s and (Ax)ms = 0.95 cm.
and (Ax)ms = 0.19 cm.
218
16.22
Let the origin be at the t = 0 location of the particle and let the particle be at
point (x, y, Z) at time ¢. Then (r?) = (x? + y? +27) = (x7) + (a) 4 za BY
symmetry, Copa) = (2 acon ae
Fons = (12) 1? = (6D1)"”,
16.23
3(x*) = 3(2Dr) = 6Dt and
From (16.35), k= 3[(AX)mms} mnr/tT. For t= 30s, k= 3(7.1 x 10“ cm)n x
(0.011 dyn-s/em?)(2.1 x 107 cm)/(30 s)(290 K) = 1.26 x 107'° erg/K. Then Na
= R/k = (8.314 x 10’ ergs/mol-K)/(1.26 x 107'® erg/K) = 6.6 x 10°*/mol. For t =
60s, we getk = 1.41 x 107'° erg/K and Na = 5.9 x 10” mol. For 90s, k=
1.07 x 107'° erg/K and Na = 7.8 x 10°? mol.
16.24
(Ax)= 4 (-5.34+3.4-19-0.440.5+3.1-0.2-3.5+414+0.3-1.0+2.6)
um =—0.08 um. ((Ax)*) = (1/12)[(-5.3)? + 3.4)? +--+] um? = 6.28 pm’.
(Ax)ims = 2.5 um.
16.25
(a)
Equation (16.42) and N/V = PN,/RT give Dj =
(3/8r02)[(8.314 J/mol-K)(273 K)/(0.0320 kg/mol)]'”” x
(3.6 x 107! m)°[82.06 (10° m)’-atm/mol-K](273 K)/
(1.00 atm)(6.02 x 10”? mol!) = 0.000016 m7/s = 0.16 cm//s.
16.26
(b)
Since Dj is proportional to 1/P, we have Dj = 0.016 cm’/s.
(a)
The volume of each cell is 8r> , and the molar volume is Vnj = 8r>N
1/3
pel
,
Hence’g=s5 (Ving Na)
(b)
16.27
Dj ~[(6.02 x 10°/mol)/(18.1 cm?/mol)]'°(1.38 x 107° ergs/K)(298 K)/
27(0.00890 dyn-s/cm’) = 2.4 x 10° cm’/s.
The N> and HO molecules are not greatly different in size, so Eq. (16.38) is
appropriate. We have Dj, = (1.38 x 10°'° erg/K)(298 K)/
4n(0.0089 dyn-s/cm?)(1.85 x 10™ cm) = 2.0 x 10° cm’/s.
16.28
(a) Equations (16.25) and (16.42) give 6n/Sp = (61/32){v) A =
(31/16)(v)A = Dy.
219
(b) Dj = 6nRT/SPM = 6(0.000297 dyn-s/cm*)(82.06 cm*-atm/mol-K) x
(273.1 K)/5(1.00 atm)(20.18 g/mol) = 0.40 cm’/s.
16.29
Since a hemoglobin molecule is much larger than a water molecule, we use Eq.
(16.37). We have Vm/Na = 47/3 and r) = (3Vin/4Nan)'? =
[3(48000 cm?/mol)/4(6.02 x 107°/mol)m]'? = 2.67 x 10°’ cm. Then Dy =
(1.38 x 107° erg/K)(298 K)/6n(0.0089 dyn-s/cm”)(2.67 x 107’ cm) =
9.2 x 10°’ cm’/s.
16.30
(a)
Let z=x*. Then d(x’)/dt = dz/dt = (dz/dx)(dx/dt) = 2x(dx/dt). Also,
&(xdt? = (didt)[d()/dt] = (d/dt)[2x(dx/dt)] = 2(dx/dty’ + 2x(d’x/dt’).
Substitution of these two equations into the equation in the problem
transforms it to Eq. (16.33).
(b)
Averaging, we get 0- 1f (d(x? )/dt) = 4m(d*(x*)/dt*) - (m(dx/dt)’) .
We have & = 1mv{ = 1m(dx/dt)’, so 2(e,) = (m(dx/dt)*). Also,
€=€,+£,+€,and
(€) = (€,) + (€,) + (€,). By symmetry,
(€,) = (€,) = (€,); thus (€,) = (<)Bea B/DTB=
SAT. So
(rida die) = Oe) ikl
(c)
Substitution of (m(dx/dt)*) = kT into twice the first equation in part (b)
gives —f (d(x?)/dt) = m(d*(x’)/dt*) — 2kT. Let s = d(x*)/dt. Since the
derivative of a sum is the sum of the derivatives, we have s =
(d*(x?)/dt?) and ds/dt = (d7(x*)/dt”) . Thus —fs = m ds/dt — 2kT and
m ds/dt + fs = 2kT.
(d)
ds/dt = (2kT—fs)/m and m(2KkT — fs)' ds = dt. Integration gives
—(mif) In (2kT — fs) = t + c and 2kT—fs = ee". From part (e),
et" =~ 0, so s = d(x") /dT = 2kT/f. Integration gives (x) = 2kTt/f+ b.
The integration constant b is 0 if we take x = 0 at t = 0 for each particle.
Hence, (x*) = (2kT/f)t.
(e)
m=(3 g/cem*)[4n(10~ cm)*/3] = 10°'4 g and f= 6nnr =
67(0.01 dyn-s/em*)(10~ cm) = 2 x 10° dyn-s/em. Then e“”” =
exp [-(2 x 10° dyn-s/cm)(1 s)/(107'4 g)] = 27200000000 = 1 9-90000000
220
16.31
(a)
Substitution of dna and dng from (16.30) into 0 =dV= Vy dnat Vg dnp
gives 0 = Va [AD ap(dca/dx) dt] + Vz [ADga(dcp/dx) dt] and
Dap Va (dea/dx) + DgaVz (dcp/dx) = 0.
(b)
Division of (9.16) by V gives 1 =ca V, +cpVz,- Since there is no volume
change on mixing, V, and Vz are independent of the concentrations and
hence are independent of the x coordinate. Differentiation of the last
equation with respect to x gives 0 = Vy (dca/dx) + Vg (dcp/dx).
Substitution of V, (dca/dx) = —Vz (dcp/dx) into the result of (a) gives
Dap = Dba.
16.32
The number of moles of the diffusing species that lie between L and M equals
of
CLV, = CiA(AX)rms, Where V; is the volume between L and M. The number
ms.
moles moving left to right through plane M in time fis therefore Y2c;A(Ax)
Similarly, Ycr4(Ax)rms moles move right to left through M in time t. The net
rate of flow of the diffusing species through M is thus dn/dt = An/At =
c,
Y2(c, — CrA(AX)rms/t. Since dc/dx is constant, the average concentration
, the
equals the concentration midway between planes L and M. Likewise
average concentration cr between M and R equals the concentration midway
the
between M and R. The distance from the plane midway between L and M to
the
plane midway between M and R is %(Ax)mms + ¥2(AX)ms = (AX)ms; $0
ms. Substitution into
concentration gradient is dc/dx = Ac/Ax = (cr — C)/(AX)
(16.30) gives ¥2 (cr, — CRM(AX)rms/t = —DA(eL — Cr)/(AX)rms and (AX)rms= (Di:=
16.33
M; = RTs /D;,(\ — PV;) =
coy ES
(6.9x10-''
16.34
(a)
(b)
a3 a
ee a
= 63 kg/mol = 63000 g/mol
m*/s)[1—(0.998 g/cm )(0.749 cm’/g)]
The column chart shows a bell-shaped (Gaussian) curve.
locations after
Row 15 contains the numbers of molecules in the various
(Ax)
10 time intervals. The squares of the numbers in row 3 give the
this
values. The formula =(C15*(C3)42)/1000 is entered in cell C1020;
the
formula corresponds to the term n,s?/N in (s?) =X, n,s*/N . Using
method given in the problem, the contents of C1020 are copied to D1020,
GU1020 are
E1020,..., GU1020. In another cell, C1020, D1020,...,
221
summed to give ((Ax)*). The result is 6.667 for 10 time intervals. For
100 and 1000 time intervals, we use rows 105 and 1005 instead of row
15. The results are 66.67 and 664.8. The relation of proportionality to
time is well obeyed. (The deviation for the 1000 time-interval result can
be ascribed to the error introduced by using a finite length for the x axis;
the values in locations beyond 100 units from the origin were omitted,
thereby giving a ((Ax)”) value somewhat smaller than the true one.)
(d)
From (16.31), ((Ax)*) = 2Dt. The numbers —100,..., 100 in row 3
correspond to —100 x 10° cm,..., 100 x 10° cm, so the ((Ax)*) values
found in (b) should be multiplied by 107'* cm”. Thus ((Ax)”) =
664.8(107'? cm’) = 2D(1000 x 1 s) and D = 3.32 x 10°! cm’/s.
(e)
The formula in C6 becomes
=0.05*A5+0.25*B5+0.40*C5+0.25*D5+0.05*E5
This is copied to the appropriate cells and the same procedure is used as
in (b) and (d).
16.35
[= dQ/dt = Q/t since J is constant. Q = It = (1.0 A)(1.0 s) = 1.0 C. The number
of electrons is (1.0 C)(6.02 x 10” electrons)/(96485 C) = 6.2 x 10!°.
16.36 R=p2£/A= (1.67 x 10° Q cm)(250 cm)/(0.040 cm?) = 0.0104 Q.
16.37
|Ao| =/Rand/= |Ao|/R = (25 VV/(100 Q) = 0.25 A.
16.38 E = j/« = I/AK = (0.10 A)/(10 cm’)(0.010 Q™! cm!) = 1.0 V/em.
16.39 (a) T; (b) T; (c) F; (d) F; (e) T.
16.40
Q=It= (2.00 A)(30.0 x 60 s) = 3600 C.
(3600 C)
16.41
(a)
Imolee
ImoleCu
63.55g Cu
96485C
2molee
ImoleCu
=1.19g0f Cu
Kee = KR = [(0.012856/1.000495) Q*' cm'](411.82 Q) 2352917 cm
222
(b)
Kyo = Keew/R = (5.2917 cm™!)/(368000 2) = 0.00001438 Q7 cm”.
K = Kee/R = (5.2917 cm /(10875 Q) = 0.00048659 Q cm”! =
Ky,0 + Kmx, and Kyx, = (0.00048659 — 0.00001438) Caene =
(d)
0.00047221 Q7' cm".
Am = k/c = (0.00047221 Q7! em™)/(10 mol/em’) = 472.21 cm*/Q-mol.
Neg = Am/V,2, = Am/2 = 236.10 Q"! cm’ equiv”.
(a)
Am =K/c = (1.242 x 103 Q7 cm™)/[(5.000 x 10° mol)/(1000 cm*)] =
(c)
16.42
248.4 Q7' cm? mol”.
(b)
Neg = Am/Vs2+ = Am/1(2) = 124.2 Q7! cm? equiv”.
16.43
Equation (16.67) gives u(Na*) = (10.00 cm)(0.0023 13/82-cm)(0.
16.44
(a)
11 15 cm’)/
(0.00160 A)(3453 s) = 4.668 x 10 cm? V' s'. Equation (16.72) gives 1(Na’)
= 1(96485 C/mol)[(0.02000 mol)/(1000 cm*)](0.0004668 cm? V's!)
(0.002313 Q7 cm) = 0.3894.
Substitution of Eq. (16.67) (which applies to any ion) into (16.72) gives
tp = |¢p|Fcp(xkA/It)/K = |zp|FcpxA/Q, where Oat.
(b)
As in Fig. 16.23, the method gives the transport number of the cation.
The equation in (a) with xf =V gives 1(Gd>") = 3(96485 C/mol) x
[(0.03327 mol)/(1000 cm?*)](1.111 cm*)/(0.005594 A)(4406 s) = 0.434.
Then 7(CI) = 1 — 0.434 = 0.566.
16.45
The coulometer reaction is Ag’
+e = Ag. The 0.16024 g of Ag is 0.0014855
moles of Ag, so Q = (0.0014855 mol)(96485 C/mol) = 143.33 C. Letn=
0.0014855 mol. Then n moles of Cl enter the cathode compartment due to the
reduction reaction AgCl + e = Ag + CI. The total number of moles of charge
on the ions that cross the plane between the middle and cathode compartments
is n; this charge is composed of tn moles of CI” leaving and t,n moles of K*
Cl
entering the cathode compartment. The net change in number of moles of
the
in the cathode compartment is thus n—tn= (1 -t)n=tn, which is also
change in number of moles of K* in this compartment. The final composition
of KCI and
of the cathode compartment 1s (0.0019404)(120.99 g) = 0.23477 g
120.99 g - 0.235 g = 120.75s g of H20. The initial composition of this
223
compartment is 120.75; g of water plus x grams of KCI, where x/(120.755 + x)
= 0.0014941. We get x = 0.18069. The change in KC] mass in the cathode
compartment is 0.05408 g, which is 0.0007254 moles. Thus, t.n =
t,(0.0014855 mol) = 0.0007254 mol and t, = 0.4883. Then t = 1 —t, = 0.5117.
16.46
(a)
Using Eq. (16.84), we have A; (LiNO3) =A
> (Li?) +A
> (NO;) and the
following three equations:
(mee (LiCl =A (Li +A, (Gly);
(2) Aj, (KNO3) =A; (K") +47, (NO3);
GymA} (KGaA
a Ke ae (Ci)
We take (1) + (2) — (3) to get A>, (LiNO3) = A> (LiCl) + A> (KNO3) —
A® (KCI) = (90.9 + 114.5 -105.0) cm?/Q-mol = 100.4 Q7! cm? mol!.
(b)
(4) A; (HC) =A7, (H") +05 (CI);
(5) Aj, (NaCl) = A> (Na*)
+A (CI);
(6) Aj, (NaC2H302) = A 7 (Na) + A= (C,H303).
Taking (4) + (6) — (5), we get A; (H") + AZ,(C,H;,0;) = A* (HCI) +
A, (NaC2H302) — A® (NaCl) = (426 + 91 — 126) cm?/Q-mol =
391 cm’/Q-mol.
16.47 From (16.85) for NaCl, t? = 0.463 = (1)A@ ,(96.9 cm? Q* mol!) and
A, (Na*) = 44.9 cm7/Q-mol. Then A* (CI) = (96.9 — 44.9) cm2/Q-mol =
52.0 cm’/Q-mol. We now use (16.84) for the other electrolytes. For NaNOs;,
106.4 cm’/Q-mol = 44.9 cm’/Q-mol + 4= (NO5) and A® (NO;) =
61.5 cm?/Q-mol. For LiNO3, 100.2 cm2/Q-mol = X~ (Li*) + 61.5 cm?/Q-mol
and A (Li*) = 38.7 cm’/Q-mol. For NaCNS, 107.0 cm?/Q-mol =
44.9 cm?/Q-mol + d.*, (CNS) and A* (CNS’) = 62.1 cm?/Q-mol. For HCl,
192 cm*/Q-mol = 4 (H*) + 52.0 cm?/Q-mol and * (H*) = 140 cm’/Q-mol.
For Ca(CNS)2, 244 cm’/Q-mol = 4" (Ca*) + 2(62.1) cm?/Q-mol and
A (Ca**) = 120 cm?/Q-mol.
224
16.48
(a)
upg =A™ p/|zp|F = (67.2 Q! cm’ mol ')/1(96485 C/mol) =
Gog x10 cm Vv" Ss”.
(b) vg =uR E= (6.96 x 107% cm’/V-s)(24 V/cm) = 0.017 cm/s.
(c)
16.49
rp ~=|zele/6nnus = 1(1.6022 x 10°" C)/
6m(0.008904 x 107! Ns m™)[6.96 x 107 (10° m)*/V-s] =
1.37 x 107° m=1.37 A.
MST IN
Sia
ap
GS
(a)
A® =(73.5+71.4) cm’/Q-mol = 144.9 cm*/Q-mol.
(b)
A® =[2(73.5) + 159.6] cm?/Q-mol = 306.6 cm*/Q-mol.
A® =(106.1 + 159.6) cm?/Q-mol = 265.7 cm’/Q-mol.
(d) A =[118.0 + 2(199.2)] cm?/Q-mol = 516.4 cm*/Q-mol.
(c)
16.50
From (16.85), i (Mg"*) = 1(106.1)/[1(106.1) + 2(71.4)] = 0.426 and (NO; )=
1 — 0.426 = 0.574.
16.51
From (16.85), f2 = v,A%,/AZ = 2A%,, /(259.8 Q7! cm’ mol"') = 0.386 and
A® (Na*) = 50.1 Q7' cm? mol”!. From (16.84), AZ = 259.8 Q”' cm’ mol! =
2(50.1 Q7! cm? mol") + A%(SO2") and AS (SOZ) = 159.6 2"! cm? mol”.
16.52
We assume the electrodes are inert, so only Ag” and NO; carry the current in
the solution.
=t™ = v_A~_MviAR. t+V_An_) = 71.4/(62.1 + 71.4) =
0.535. If 10° mol of Ag’ is deposited according to Ag’ +e — Ag, then 105
Faradays of charge must have flowed through the circuit. The nitrate ions carry
a fraction t_ of the current in the solution, so the nitrate ions carried
L105, Faradays) of charge through the solution. Hence 10°t_ moles of NO,
crossed a plane in the solution, and this is (1.00 x 10~*)(0.535) mol =
0.535 mmol.
225
16.53
Because of its higher charge, Mg” is hydrated to a much greater extent than
Na’ and so its radius is much greater than that of Na’. Therefore u in (16.70)
has similar values for these two ions.
16.54
16.55
(a)
AW(CI), u(Cl);
(b)
no (interionic forces differ in the two solutions).
x.
16.56 (a)
Use of(16.70) in (16.78) gives A™ , =|zp|Fuy =|ze| eF/6mnrp. So
InA@ =In(zzeF/6mrg) — In yn and d In A®/dT = -(1/n) dn/dT.
(b)
(c)
dn/dT = AW/AT= (0.8705 — 0.9111)cP/(2 K) = -0.0203 cP/K and
(1/n)(dn/dT) = (1/0.8904 cP)(-0.0203 cP/K) = -0.023 K™, so
din) id= 0.023) Kes
Integration gives In (A* ,/A™ ,) = (0.023 K')(T7 — T;) and A™, =
A%1expl(0.023 K™')(T2 - T;)] = (71.4 Q* cm? mole?”
=
89.9 QT cm? mol.
16.57 (a)
A® =(73.5 + 71.4) cm/Q-mol = 144.9 cm?/Q-mol. From (16.87),
Am = 144.9 cm’/Q-mol — [60.6 + 0.230(144.9)](cm?/Q-mol)(0.00200)!”
= 140.7 cm’/Q-mol. K = A,c = (140.7 cm?/Q-mol)(0.00200 mol)/
(1000 cm*) = 0.000281 Q7! cm.
(b)
Neglecting the conductivity of the water, we have
R=pl/A= ¢/KA=
(10.0 cm)/(0.000281 Q7' cm™')(1.00 cm?) = 35500 Q.
16.58
K,.=[H'][OH
], since y+ = 1 at the extremely low J,, value in this solution.
Equation (16.90) with c, = 0 in the denominator gives the initial estimate c, ~
(5.47 x 10° Q7 cm'V/[(350 + 199) cm?/Q-mol] = 9.96 x 107!! mol/em? =
9.96 x 10° mol/dm*. We have S = 60.6 cm’/Q-mol + 0.230(549 cm?/Q-mol) =
186.9 cm’/Q-mol. Using the initial estimate of c, in the denominator of Eq.
(16.90), we have c, = (5.47 x 10° Q7! cm"!)/
[549 cm*/Q-mol — (186.9 cm?/Q-mol)(9.96 x 10°°)"”] = 9.96 x 107! mol/em?
226
= 9.96 x 10° mol/dm’, as before. Hence K; = (9.96 x 10° mol/dm’)* =
9.9, x 107° mol?/dm*.
16.59
(a)
Since z, =|z_|, Eq. (16.90) applies. We have A, +A
=
(118.0 + 159.6) cm?/Q-mol = 277.6 cm7/Q-mol and
S =8[a+ bm,
i,
Cy
+A%_)] = 996 cm/Q-mol. Therefore
221A
Ome@arcems.
~ 277.6 cm?/Q-mol — (996 cm?/Q-mol)[c, /(10~> mol/em?)]'/?
With c, = 0 in the denominator, we get the initial estimate c, =
7.96 x 10° mol/em>. With c, = 7.96 x 10° mol/cm’, we get the
improved estimate c, = 1.17 x 10° mol/cm?. Further repetitions yield the
successive estimates 1.30 x 10° mol/cm’, 1.35:x107 mol/cm’,
1.36 x 10° mol/cm’, and 1.37 x 10> mol/cm’. Hence c, =
1.37 x 10°? mol/dm’. The solution is dilute, so we can take the molality
as 0.0137 mol/kg. The Davies equation with J, = 0.0548 mol/kg then
gives log y. =—0.353 and y+ = 0.443. We can neglect the difference
between molality-scale and concentration-scale activity coefficients in
this dilute solution, so the concentration-scale Ksp 18 Ksp =
(0.443)°(0.0137 mol/dm*)? = 3.7 x 10° mol’/dm’*.
(b)
No.
Ksp= a [Ca**][SO re ]. We found the ionic concentrations and Ys.
The existence of the additional equilibrium Ca** + SO,” = CaSO,(aq)
does not invalidate our work.
16.60
We have A”, +A%,_ = (350 + 40.8) cm7/Q-mol = 391 cm*/Q-mol and
S = [60.6 + 0.230(391)] cm?/Q-mol = 150.5 cm’/Q-mol. Therefore
4.95x10° Q!' cm”!
CS
+
OO
TT
391cm?/Q-mol - (150 4.cm?/Q-mol)[c, /(10-* mol/em*)}"”
With c, = 0 in the denominator, we get the initial estimate c, =
1.27 x 1077 mol/cm’. Recalculation with this c, value in the denominator gives
c, = 1.27 x 1077 mol/cem? = 1.27 x 10“ mol/dm’, which is the H30*
concentration. For this dilute solution, we can take the ionic molalities as
1.27 x 107 mol/kg. The Davies equation then gives Y1 = 0.987. Neglecting the
slight differences between concentration-scale and molality-scale y’s, we have
794}
K. = (0.987)(0.000127 mol/dm?)7/[{(0.001028 — 0.000127) mol/dm*] =
1.74 x 10° mol/dm?.
16.61
We have A*,, +A*_ = (106 + 160) cm*/Q-mol = 266 cm*/Q-mol and S =
8[60.6 + 0.23(266)] cm?/Q-mol = 974 cm*/Q-mol. Then
3
6.156x10° Q' cm!
*
266 cm?/Q-mol — (974 cm?/Q-mol)[c, /(10°-3 mol/em?*)}'”
With c, = 0 in the denominator, we get the initial estimate c, =
2.31 x 107’ mol/cm*. With this value of c,, we get the improved estimate c, =
2.45 x 10°’ mol/cm®. Another repetition yields the final result c, =
Dida ax 10°’ mol/cm? = 2.46 x 10~* mol/dm*. For this dilute solution, we can
take the ion molalities as 2.46 x 10~ mol/kg, so J, = 9.8 x ir mol/kg. The
Davies equation gives logy+ = —0.061 and y: = 0.87. We can assume the
molality-scale and concentration-scale activity coefficients to be equal in this
dilute solution. We have [Mg”*] = [SO2 ]= 2.46 x 10~* mol/dm? and
[MgSO,(aq)] = 2.50 x 10“ mol/dm? — 2.45; x 10“ mol/dm? = 4.5 x 10%
mol/dm*. Hence, K, = (4.5 x 10° mol/dm?)/(0.87)°(2.46 x 107+ mol/dm*)” =
100 dm*/mol.
16.62
Omitting the S term and using Eq. (16.58), we have for (16.90):
CoN
16.63
eh
Nycl (Nae een
enema
cae =e (ee PA),
The Onsager equation predicts ac!” dependence of A,, in dilute solutions, so
we plot A, vs. c’”. The points give a rather good straight-line fit; the intercept
atc =\00seA —270:8icm /Qamal:
275
A p/(Q7 cm’/mol)
= -14.095x + 270.77
at
265
260
22
250
245
IN
O02
SV
hes
0.459
Ne
0,Gian 0) Cres lems
ee
ie
oe
[c/(mmol/L)]"”
228
ed
16.64
10 A=60.6:B=0.23
60 FOR J=1 TO 1000
15 INPUT “ZPLUS”;ZP
70 M=CP
20 INPUT “LAM+IN/(CM2/OM—
80 CP=K/(L-S*(CP/.001)10.5)
90 IF ABS(CP-M)/CP<0.0001
MOL)”;LP
THEN 200
25 INPUT “LAM-—IN”;LM
30 L=LP+LM
100 NEXTJ
go S=ArZP%34+L"B*ZP%3
150 PRINT “DIDN’T CONVERGE”
‘SoTOPR
40 INPUT “KAPPA/(1/OM-—CM)”
200 PRINT “C+=”";CP
;K
210 STOP
50 CP=K/L
16.65
From (16.32), (A%)ms= (Die
Substitution of the Stokes—Einstein equation
(16.37) forD gives (Ax)rms ~(KTt/37™y, 07492 Use of (16.70) for 7,,,2+
| 2452 eye ine se ore
gives (Ax)rms ~(2kTuy2+t/ t/|z
[2(1.38 x 10°? J/K)(298 K)(55 x 10° m*/V-s)(1 s)/2(1.60 x 10n Gl, a=
3.8 x 10> m = 380000 A = 0.0038 cm, where u” was taken from p. 513 of the
text. In 10 s, (AX)mms = (10/1)'7(38 wm) = 120 pm.
16.66
(a) Increases;
(b) increases;
(Cc) increases;
(d) increases;
(e) increases;
(f) increases.
16.67
Thermal conduction: g (units J), k (unitsJ Kom oe as T (units K), where the
orderis W, L, B. Viscous flow: py (kg m/s), 1 (N s/m? ), Vy (m/s). Diffusion: n;
(V).
(mol), Djx (m 1s), Cj (mol/m’). Electrical conduction: Q (C), K(Q"! m’), o
229
Chapter |7
17.1
(a) F; (b) F; (c) T; (d) F; (e) T; (f) F; (g) T; (h) F.
17.2
(a) s'; (b) Lmol's7!; (c) L’ mol™ s™.
17.3
(a).
17.4
0.002 mol L7! s7!.
Ulds)
(a)
na/V=Pa/RT = (0.10 atm)/(82.06 cm?-atm/mol-K)(298 K) =
(b)
r=[I/(-2)] d[N2Os]/dt = 7.1 x 10° mol/dm*-s and d[N2Os]/dt =
4.09 x 10™° mol/cm? = 4.09 x 10° mol/dm*. r= k[N2Os] =
(1.73 x 10° s“!)(4.09 x 10°? mol/dm’*) = 7.1 x 10° mol/dm’*-s.
J=rV=(7.1 x 10° mol/dm*-s)(12.0 dm*) = 8.5 x 107’ mol/s.
-1.4 x 10°’ mol/dm’-s.
(c)
(d)
(1.4 x 1077 mol/dm?-s)(6.02 x 107/mol)(1 s)(12 dm*) = 1.0 x 10"*.
N20; — 2NO2 + 40>. Here, J= —dny9,/dt as compared to J =
— = dny o,/dt for 2N20s > 4NO? + O2. So J = 2(8.5 x 10°’ mol/s) =
17x 107 mol/s.
r=J/V=
14x 10° mol/dm?*-s.
r= k[N2Os] and
k = r/[N2Os]; since r is doubled, so is k, and k = 3.46 x ie ee [In (c),
the approximation dn/dt = An/At was used. This approximation is
justifiable, as follows: We have (0.0041 mol/L)(12 L) = 0.049 mol,
which is 3 x 10°” molecules present; therefore a change of | x 10°
molecules will not significantly change the rate dn/dt over a period
of 1 s.]
17.6
From (17.2), J = (1/v;)(dn,/dt), where the stoichiometric coefficient v; is
negative for reactants. From (4.97), dn; = v; d&, so J = (1/v) v,(d&/dt) = dE/dt.
Ait
Cy =Ng/V =(Ng/Na)MV/N,) =nghVIN,)=[BIN,. r=—(1/b)d[B/dt.
ro = -(I/b)dC,/dt = -(/b)N ,d[B\/dt = N,r. So rc/r=Na. For simplicity,
230
let the rate law be r = k[B]”. We have r- =k-Cy = k-[B]" Nj. Hence re/r=
(kc /k)Nj . Also rc/r = Na. Hence (kc/k)NA =Naand k. =kNI.
17.8
(a)
PaV=naRT and Pa = [AJRT. The equation -(1/a)dP,4/dt = kpP.
becomes —(RT/a)d[A]/dt = kp[A]"(RT)" and r = —(1/a)d{A]/dt =
(RT)" 'kp[A]". Comparison with r = k[A]" gives kp = K(RT)'™.
(b)
Leo
Yes.
r, = k,[A] and rz = k2[B]. Since [B] might be much greater than [A], it is
possible for rz to exceed r; even though k, exceeds ko.
17.10
(a)
Since the first step produces one C molecule while the second step
consumes two C’s, the first step must occur twice for each occurrence of
the second step. The second step produces one F and the third step
consumes one F, so the second and third steps have the same
stoichiometric number. The stoichiometric numbers are thus 2 for the
first step, 1 for the second step, and 1 for the third step. Adding twice the
first step to the second and third steps, we get
2A +2B+2C+F+B—2C+2B+F+2A
is 3B > 2B +G.
(b)
17.11
+ Gand the overall reaction
Species A is consumed in the first step, regenerated in the last step, and
does not appear in the overall reaction. Hence A is a catalyst. B is a
reactant. C and F are reaction intermediates. B and G are products.
From Eq. (17.5), the units of k are (dm?/mol)"* st. Comparison with the value
of k shows thatn
—-1=1,n =2.
17.12
r, = k,[A]. Since [A] decreases exponentially with time (Fig. 17.3), so does rj.
r2 = k2[B].
[B] increases from 0 to a maximum and then decreases to zero (Fig.
17.3) and so does rp.
L713
(a) -k,[B], [BJ]; (b) -k:[B], Ai[B] —ko[E], k2[E]; (c) -k,[B] + k[E],
k,[B] — K.\[E]
— ko[E], ko[E]; (d) —(ki + k3)[B] + Ki[E] + ks[F],
k,[B] — k.i[E], k3[B] — k-s[F].
231
17.14
(a) F;
17.15
(a)
(b) F;
(c) T (assuming Vis constant).
[AV/[A]o = e7*4', so 0.65 = e 4955) and In 0.65 = —ka(325 s). We get ka
= 0.00133 s!. From Eq. (17.11), k = kala = ka/2 = 0.00066 s™.
(b) t = —(1/ka)In([A]/[AJo)= -[1/(0.00133 s”')] In (0.30 or 0.10) =
O0DiS Olas:
17.16
(a)
73 x 10°s')=
Kati = 0.693. From (17.11), ka = ak = 2(1.
(b)
10° s71(24%
46%
[A] = [Aloe “*’ = (0.010 mol/dm’) 73-
10° s"') = 2.00
3.46 x 10s. So typ = 0.693/ka = 0.693/(3.46 x
x 10° s.
36005) _
5.0 x 10“ mol/dm’.
17.17 d{AVdt=-kslAT,
J?LAT? d[A} =—J? ka dt, -1[AT*]i =-ka(o— 11),
W/[A]3 — WA]? = 2ka(t2 —t1) or L/LA]* — 1/[A]9 = 2kat, where we took t, = 0
17.18
and
=
(a)
Eqs. (17.22) and (17.19) with A = NO2, B = F2, a= 2, and b= | apply.
[A]o = (2.00 mol)/(400 dm?) = 0.00500 mol/dm? and [B]po =
0.00750 mol/dm?. (a[B]o — b[A]o)kt =
[2(0.00750) — 0.00500](mol/dm*)(38 dm?/mol-s)(10.0 s) = 3.80.
Using [B]/[B]o = 1 - ba'[A]o/[B]o + ba '[A]/[B]o, we have for Eq.
(17.22): 3.80 = In {(0.667 + [66.7 dm*/mol][A])/(200 dm*/mol)[A]} and
e°®° = 0.00333/(dm*/mol)[A] + 0.333. We find [A] = 7.5 x 10° mol/dm>
= [NO}]. Then na = (400 dm*)(7.5 x 10° mol/dm’*) = 0.0300 mol NO.
Let 2z moles of NO; react in 10 s. Then 2 — 2z = 0.0300 and z = 0.985.
Then ng, = 3-—z= 2.015 moles and nyo,p = 2z = 1.97 mol.
(b)
r = k[NO>][F2] = (38 dm*/mol-s)(0.00500 mol/dm*)(0.00750 mol/dm*) =
0.00142 mol/dm*-s initially. After 10 s, part (a) gives [NO>] = 7.5 x 10>
mol/dm? and [F2] = (2.01; mol)/(400 dm*) = 5.04 x 107° mol/dm’. So r=
(38 dm*/mol-s)(7.5 x 10° mol/dm*)(5.04 x 107 mol/dm’*) = 1.44 x 107
mol/dm?-s.
232
17.19
(a)
As noted in Section 1.8, (d/dx) J S(x) dx = fix). Using this identity, we
have dy/dx = e"(dwldx)(J efx) dx +c) + eM Te"MAR
X)]. Since w(x)
= J g(x) dx, we have dw/dx = g(x) and dy/dx = e"g(J efdx+c) +f, The
night side of the differential equation is f+ gy = f+ ge"(J e"fdx +c),
which is the same as dy/dx. Hence, y is the correct solution.
(b)
In 19539) iweitake y =(B], x =7,/f{x) = RiAlpes , (x) = —k>. Then
w=-] ky dt=-IKot and y=[B] = e "(Je k[A]oe™' dt+c) =
eo" {ki LA]oe2 | 1" Nk — ki) +c} = ki[Aloe ky — ky) + ce". To
evaluate c, we use the fact that [B] = 0 at r= 0; so0 = ki [A]o/(k2 — ki) +
and ¢ = —ky[A]o/(k2 — ki). Hence [B] = {kilA]o/(kz - ki) }(e™" — e"),
17.20
No. A kinetically reversible reaction is one where a significant amount of back
reaction occurs. A thermodynamically reversible process must go through
equilibrium states only, so a kinetically reversible reaction is a
thermodynamically irreversible process.
17.21
bromiEqn@d: ld) sr kK{A]? integrates to [A] = [A]o/(1 + akt[A]o), and
r = k[A]° becomes r= k[A]¢/(1 + akt{A]o)”.
17.22
r= k{A]® =k =—d[A]/dt, which
(A)
integrates to [A] =—kt+c.
At t=10, [A]o=.0 +.c. So [A}=
[Alo
[A]o — kt. The graph is linear.
ez3
For n = 1, the integrated rate law is (17.30), for which [A] reaches 0 only at t =
oo, For n # 1, the integrated rate law is (17.28). Setting [A] = 0 in (17.28), we
see there are two cases to consider. If n < 1, then the left side of (17.28) is zero
when [A] = 0, and the time 7” required for [A] to reach 0 satisfies 0 = 1 +
[A]g(n— Lkat* and f° = 1/[A]§
‘(1 —n), which is positive, since n < 1.
Hence [A] reaches 0 in a finite time when n < |. Forn> 1, the left side of
(17.28) becomes infinite when [A] equals 0 and an infinite amount of time is
required for [A] to reach 0.
233
17.24
a! d{A\/dt = -k{A]’[B]. Use of (17.19) gives
J?TAT? ([Blo — ba'[A]o + ba [A] [A] = - \? ak dt. A table of integrals
gives J [1/x*(p + sx)] dx =—l/px + (s/p’) In [(p + sx)/x]. We have
x=[A],
p=[Blo- ba'[A]o, and s = ba™', so
ba"!
Blo —ba™'[A]p LORIN i
([B]) —ba '[A]y)°
[A]
,
1
| (Bly —ba [A] LAI!
=—ak(t, —t,)
ae
a
soma
a[Bl)—b[Alp [Alo [A] [B]o-d[A]o
Me
[AVILA]
where we used (17.19) and took t; = 0 and f2 = 1.
17:25
17.26
(a)
(1.5)°(3) = 6.75.
(b)
(3[A]o)" = 27([A]o)” and n = 3.
In equation (17.28), let [A]/[A]o = a. Then Ce ale (Al ee (n — 1)Katg and ty
== (a! ~"—1)/[A] ae (n — 1)k,y. Taking the log of each side, we get the desired
result for n # 1. Forn = 1, we set [A]/[A]o = & in (17.30) to get
Q=e “kala and In a= —Katg.
17.27
For n # 1, Eq. (17.51) gives log 6 = log ((a! ~*_ 1)/(n — 1)]. For
a = 0.05 and n
= 0, log @ = log[(0.05 — 1)/(-1)] = -0.0223; for n = 1/2, 3/2, 2, and 3, one finds
log 6= 0.1911, 0.8416, 1.2788, and 2.2999, respectively. For n = 1, we have
log o = log (-In &) = log (-In 0.05) = 0.4765.
17.28
(a)
We use the fractional-life method, plotting log ta vs. log [A}o.
The data are
log ta
Did
2525
2.954
31057
log [A]o —2.090
—2.191
—2.509
—2.726
The slope of the straight line through these points 1s 0.44 = | —n, so
n= 1.44 = 1.5. The order is 3/2.
234
BQ. goneneene ge enn g ee YEH OMM2IA+ 1.8495
log (to/s)
B05
poe
Zui)
2.95
2.90
23
2,80)
ore
Tae hae To die
eee
Ce See ee eee es
f---7----4-- Pd
E---je-2- denote det
oe
ied=3Ee ed
el aia
een Mined Po |
alate)
SOLGf IM eyMEPeiay SOR
Oe).
aay) a ae)
log {[A]o/(mol/L) }
(b)
Putting [A]/[A]o = & = 0.69 and n = 1.5 in Eq. (17.28), we get ka =
0.408/1,[A] a . Substitution of the four pairs of tf and [A]o values gives
10°ka/(dm**/mol'*-s) as 7.67, 7.64, 8.14, and 8.25. Averaging, we get
ka = 7.9 x 10° dm*/mol’”-s.
17.29
(a)
Both plots give pretty good fits to a straight line. The log [A] vs. t plot
(corresponding to n = 1) gives a better fit, but considering the inaccuracy
of kinetics data, one could not absolutely rule out n = 2 from the plot.
n=1
Log({A]/c®)
SU
Rete 0.99989
Nes
1/(LA]/c?)
[
140
0
5
ORS
eiSige Ole eS
t/min
(b)
ie.
Saas
Lege
aaa
I
0
5
Oa
R? = 0.997
SE A
ae
ee?
Oe)
t/min
A good fit is obtained for n = 1 and for n = 3/2; the fit for n = 2 is not
good. To decide between n = | and n = 3/2, one would need data at later
times.
235
17.30
B = S20 - and A = C3H7Br.
Equation (17.22) with a = | and b= 1 applies. Let
Then a[B]o — b[A]o = 57.1 mmol/dm’. We use [A] — [A]o = [B] - [B]o to
calculate the [A] concentrations. A plot of In {({B]/[B]o)/[A]/[A]o)} =
In z vs. t is linear with slope (a[B]o — b[A]o)k. The data are
A
t/s
0
0.104
0.189
0.474
= 1.035
1110
2010
5052
ie
The slope is 9.29 x 10° s“!, so k = (9.29 x 10° s“')/(0.0571 mol/dm’) =
1.61 x 10°? dm? mol! st.
y = 9.21E-05x + 2.89E-03
1.2
ee poe
Le n
ana
asa
See
see
ay
Sal)
lS
alin
eile
Tie
iee
a
(lim
thick
sts
Valle
eI
TR fi!
0
eS
2000
SS
Fe
4000
0
a
|
6000
8000
IE es Ne
ea
10000
ale
eel
12000
t/s
17.31
(a)
The reaction has the form 2A — B. We have P = Pa + Pg = (ca + Cp)RT.
Let 2z mol/dm? of A react to form z mol/dm? of B. Then ca = Cao — 22
and cg = z= (cao — Ca). Hence P = (ca + Y2Ca0 — Y2Ca)RT =
Y2 (Cx +Cao)RT. Also, Po = Ca oRT, SO Ca = (2P — Po)/RT and & = Ca/cao
= 2P/Po — 1, where Po = 632 torr. We calculate the & values and plot o
vs. log t; comparison with the Powell-plot master curves shows the order
is 2. Alternatively, we can use the fractional-life method: The calculated
values of 100c,/(mol/dm*) are 1.0925 >a G 84560457403.
0.8929, 0.7349, 0.6252, and 0.4780. We plot ca vs. t and take a = 0.75.
From the graph, we find the times needed for
236
ca/(mol/L)
Gitte
Oo ON
a
a
natin
ale
oh
Sled
hark
ratant
gla
0.004
estes gy
0
ae
Pee grease eg
2000
mig
4000
ecto
6000
et
pe
8000
a
10000 12000
t/s
100ca/(mol/dm*) to undergo the following changes: 1.6 to 1.2, 1.4 to
1.05, 1.2 to 0.90, 1.0 to 0.75, 0.8 to 0.6. Values of log fo.75 are
log to.75
3.13805
ate 04
eee jae oon.
3.467
log [100ca/(mol/dm’)]
0.204
0.146
0.079
—0.097
log(to75/s)
= -1.1008x + 3.3693
3-50
S710
0
FY
1 PS a ET
E(TOF-0'05"
OE
0:00
0.059
a
0.108
OE
0:15
40200)
log[100c,4/(mol/L)]
where Ca is the initial
SOM
(b)
A concentration. The plot has slope —-1.1 = 1 —n,
25
Equation (17.16) applies. We plot I/ca = 1/[A] vs. t. The data are
(mol/dm*)/ca
t/s
59.11
64.29
69.66
74.34
85.15
0
367
731
1038
1751
237
ete.
(mol/dm*)/c,
0
2000
4000
6000
8000
10000
12000
t/s
The slope is 0.0141; dm’ mol s' = ka = ak = 2k, and
k = 0.0071 dm? mol"! s1.
17.32
(a)
The first two columns of data show that tripling [A]o at constant [B]o and
[C]o triples ro, so
@ = 1. The first and third columns show that tripling
[B]o at constant [A]o and [C]o multiplies the rate by 9, so B = 2. The
second and fourth columns show that y = 0.
(b)
ro=k[A],[B] ro and use of the first column of data gives
k = (0.0060 c? s'/(0.20 c°)(0.30 c°)’ = 0.33 dm® mol™ s™.
(c)
For the rate law (1) in Eq. (17.6), [HBr] is initially zero and initial-rate
data would yield the erroneous result r = k[H2][Brz] ue
17233
In the first run, we have [A]o >> [B]o, so [A]o 1s essentially constant. The B
concentrations are 0.400, 0.200, 0.100, and 0.050 mmol/dm? at 0, 120, 240,
and 360 s. The half-life is thus constant at 120 s; hence the order with respect
to B is 1. In the second run [B]o 1s constant and the A concentrations are 0.400,
0.200, 0.100, and 0.050 mmol/dm’ at 0, 69, 208, and 485 ks. The half-lives are
69, 139, and 277 ks. The half-life is doubled when the A concentration is cut in
half, so t1/2 1s proportional to 1/[A] and [see Eq. (17.29)] the order with respect
to A is 2. Hence r = k[A]’[B]. On the first run, r = —d[B]/dt = k{A]?[B] =
kp[B]. The reaction is pseudo first order and Eq. (17.14) gives [B] = [Blo eke.
we have kg = —f! In ([B]/[B]o) = (120 s)! In % = 0.00578 s™! =
k(0.400 mol/dm?*)’ and k = 0.036 dm°/mol?-s.
238
17.34
(a)
We plot a=[A]/[A]o vs. log t; comparison with the Powell-plot master
curves shows the order is 3/2. (Alternatively, the fractional-life method
can be used.)
(b)
From (17.28) with n = 1.5, a plot of ([A]o/[A])'” vs. tis linear with slope
1[A]o°k,. The data are
(AISA)
t/s
0
lems
100
1206)
200
294%
300
1399
400
1612)
600
201
1000
The slope is 0.00101 s' = 0.5(0.600 mol/dm’*)'7k, and ka =
0.0026, s! dm*? mol” = ak =k.
CN
y = 0.001013x + 0.997930
ae
oie oes oe a
DO ai
te
nad '
yp eee lt
co eae ,
1.8 £----- pone manne rern5y fo --- ro-= =~
Le ee ee Fe a
: Soe ante 8 Se
Oe ee
ieee
1.0
0
a
LA
200
400
cy
ee
600
800
eee
el
1000
1200
t/s
17.35
For the first run, [A]o >> [B]o, so [A] is essentially constant. A Powell plot of
a = [B]/[B]o vs. log t shows the order with respect to B is 2. (Alternatively, the
fractional-life method can be used.) For the second run, [A] is also essentially
constant. For runs | and 2, we have r; = —d[B]/dt = k[A] at [B]° =j,[B]’ and
ry = K{A] 0-2 (B} = jalB]’. Hence jy/j2 = kLA] 9, /KLA] 9.2 = ((A]o,1/[A]o.2)%. The
pseudo order is 2; a plot of 1/[B] vs. tis linear with slopej.For run 1, such a
plot has slope 0.0119 dm? mol s“'; for run 2, the slope is 0.0067,
dm? mol”! s7!. Then (0.800/0.600)* = 0.0119/0.0067; = 1.773 and
o = (log 1.773)/(log 1.333) = 1.99 = 2. Also, j) = k[A] 9, andk =
(0.0119 dm? mol! s~!)/(0.800 mol/dm?)” = 0.0186 dm” mol? s™'. (Data from
the second run give 0.0186 dm’ mol? s”'.)
239
Run 1
1/ {[B]/(mol/L)}
y = 0.00671x + 502.61523
1200 ee
200
ee ee
0
25000
50000
Run 2
1/{[B]/(mol/L)}
y =0.01192x + 502.53457
eee
Pepa eS reel
500
75000 100000
ee2 a ee
0
25000
50000
(a)
:
ee!
100000
t/s
t/s
17.36
bees
75000
3
The regression line has slope 1.289 x 107 L mmo!
a
es
s”
and ka = 0.1285
Lmo!! s”.
(b)
The Solver gives 1.289 x 10“ L mmol"! s”.
(c)
The 1/{A] vs. t regression line has slope ka = 1.3647 x 107 L mmot!!
min”! = (1.3647 x 107)(1000/60) L mol s"' = 2.2745 x 10° L mol! s™.
Also (Sec. 17.3) k = ka/3 = 7.58 x 107 L mol! s”'. For a least-squares fit
of the [A] vs. t data, the Solver gives ka = 1.2901 x 107 L mmol! min"!
= 2.150 x10~ L mol! s7!, and k = ka/3 = 7.17 x 10° L mol s"'
1s 4, (a) T;
17.38
(b) T.
For an elementary reaction, k;/k, = K-. We have —RT In Kp = AG®° =
[2(51.31) — 97.89] kJ/mol = 4730 J/mol. Then In K, =-1.908 and K;, =
0.148. From Eq. (6.25), K% = K® RT/(bar-dm*/mol) and K° =
0.148(bar-dm*/mol)/(0.08314 dm?-bar/mol-K)(298.1 K) = 0.00597; so K, =
0.00597 mol/dm?. Then ky = ky/K; = (4.8 x 10* s')/(0.00597 mol/dm*) =
8.0 x 10° dm? mol"! s™!.
17239
From (17.4), r= —d[A]/dt = ¥2 d[C]/dt, so d[A]/dt = —r and d[C]/dt = 2r. From
(17.5) for this elementary reaction, r = k[A][B]. Hence d[A]/dt = —k[A][B] and
a[(C]/dt = 2k[A][B].
240
17.40
(a) T;
17.41
The charge is not balanced.
17.42
k, and k> have different dimensions (units) and can’t be compared with each
(b) F.
other.
17.43
The rds composition is given by rule 1 as HyNO2Br. One possibility is
Hebe
wes
(rapid equilib.), HBr + HNO2 — ONBr + H20
(slow rds),
ONBr + Ce6HsNH2 > CeHsN3 +H20+Br.
17.44
(a)
For runs | and 3, the CIO” initial concentration is cut in half while the
other initial concentrations are kept constant; since the rate is cut in half,
we conclude that the order with respect to CIO is 1. Similarly, runs 2
and 3 tell us that the order with respect to I is 1. For runs 3 and 4, the
OH initial concentration is multiplied by 4 and the rate is quadrupled;
so the order with respect to OH is —1. Thus r= k{I-J[ClO )/[OH J. For
run 1, 0.00048 c°/s = k(0.00200 c°)(0.00400 c°)/(1.000 c°) and k = 60 can
(b)
We assume a mechanism with a rate-determining step (rds). From rules
1b and 2 in Sec. 17.6, the total composition of the rds reactants is
fe Olt
OH
oro
ClO
> =; ana inespecics Olims a product
in an equilibrium step that precedes the rds and OH’ does not appear in
the rds. To keep 2x — 1 nonnegative, x must be | or greater. The simplest
assumption is x = 1. This gives the rds composition as CIIOH .A
plausible mechanism that meets the preceding two requirements and that
gives the proper stoichiometry is OCI + H2O — OH +HOCI (rapid
equilib.); HOC] +I — HOI+ CT (rds); HOI+ OH — OT + H20
(rapid).
17.45
The rate-determining-step reactants’ composition is NO2CI. One plausible
> NO; + Cl (slow), Cl + NO2Cl + NO) + Cl (rapid).
NO); + Cl (slow), Cl + Cl > Cl (rapid), with
Another possibility is NOzCl
mechanism is NO,CI
stoichiometric numbers 2 and 1, respectively, for steps 1 and 2.
241
17.46
The rate-determining-steps reactants’ composition is CrT1°*. One possibility is
Ons Tl* S Gr Tl slow), Il Grams Gr + ble (rapid)sAnother
possibility is Cr°* + TP* 2 CrTI* (equilib.), CrTP°* 3 Cr°* + TI** (slow),
erapid),
Teer
One
17.47
The rate-determining-step reactants’ composition is NO2F. A possible
mechanism is NO> + F2 > NO2F + F (slow),
F + NOz
— NOszF (rapid).
Another possibility is NO2 + F2 = NOsE (equilib.), NO2F2
— NO2F + F
(slow), F + NO2 — NO%F (rapid).
17.48
XeF,
17.49
The rate-determining-step reactants have composition N2Os. A likely
mechanism is the slow step N20;
NO» + NO; followed by a series of rapid
+ NO
> XeF; + NOF.
steps that yield the correct stoichiometry. One of the many possibilities for this
series of rapid steps is NO2 + ClO
NO;3Cl
+O,
O+0O-—
+ NOC] + OCI,
OC] + NO3 —
Og. (The stoichiometric number of all steps but the last
is 2.)
17.50
The rate-determining-step reactants have overall composition HeTl**, and Hg*
is a product in an equilibrium that precedes the rate-determining step. Two
other mechanisms besides that given in Example 17.1 are: Hg3* + TI*
HgTI** + Hg” (rapid equilib.), HgTI°* > Hg** + TI* (slow); and Hg3* + TI**
— Hg* + TI* + Hg” (rapid equilib.), Hg* + TI’* > Hg”* + TI* (slow).
17.51
The reverse reaction would then proceed by the one-step mechanism
N2 + 3H2
occur.
17.52
> 2NH3. But a tetramolecular elementary step is far too unlikely to
(a)
d[O2}/dt = 2k,[O][O3] + k\[O3][M] — k-1[O2][O][M].
d[O3]/dt = —k,[O3][M] + k-1[O2][O][M] — k2[O][03].
(b)
d[O}/dt = 0 = k\[O3][M] — k,[02][O][M] — k2[O][O3], so
k,[O3][M] — k_,[O2][O][M] = k2{O][O3]. Substitution into the expressions
242
for d[O2)/dt and d[O3]/dt in (a) gives d[O2]/dt = 3k,[O][O3] and
d[O3]/dt = —2k2[O][O3].
(c)
From (b) we get [O] = k;[O3][M]/(A_1[O2][M] + k2[03]). We have
r =—Y2 d{O3]/dt = k2[O][O3] = kik2[O3]°[MJ/(K-1[O2][M] + k2[Os]) =
kik2[O3]}"/(k1[O2] + k2[O3])/[M]). Also, r = (1/3)d[O2)/dt = k2[O][Os] =
etc:
(d)
If step 1 is in equilibrium, then k\/k_; = [O2][O][M]/[O3][M] and [O] =
k,[O3]/k_,;[O2]. As noted in the problem statement, r= + d(O)/dt =
ks[O][O3] = kiks[O3]°/k1[02].
17.53
(e)
If k2[O3]/[M] << k_;[Oo] (i-e., if k_;[O2][M] >> k2[O3]), the second term
in the denominator of the steady-state expression can be neglected,
thereby giving the rate-determining-step expression.
(a)
d{NO]/dt = 0 = k,[NO2][NO3] — k-[NO][NO3] and k.[NO][NO3] =
ky[NO2][NO3]. d[NO3]/dt = 0 = ka[N20s] — k-a[NO2][NOs] kp[NO2][NO3] — k-[NO][NO3] = ka[N2Os] — kK-a[NO2][NO3] ky[NO2][NO3] — k,[NO2][NO3] and [NO3] = ka[N20s]/(k-a + 2k»)[NOz].
Then r= -$4[N20s)/dt = -3 (—ka[N2Os] + k_a[NO2][NO3]) =
5KalN20s] = 5 kakalN2Os)/ (kK_q + 2kp) = [kakp/(k-a + 2k») [N205] =
k[N2Os] and k = kgkp/(k_a + 2k»).
(b)
If step b is the rate-determining step and step a is in equilibrium, then
Kal k-q = [NO2][NO3]/[N20s]. The rate of the reaction equals the rate of
the rate-determining step b (since the stoichiometric number of step b
is 1), so r= kp[NO2][NO3] = (kpka/k_a)[N20s].
(c)
If k_4 >> 2k,, then the steady-state rate law of part (a) reduces to the rate-
determining-step rate law of (b). (Of course, this is a necessary condition
for the validity of the rate-determining-step approximation. )
(d)
From this problem, we see that k of the N»O5 decomposition is a function
of ka, ka, and kp. Hence, it is clear that the mechanism of the reaction in
Prob. 17.49 starts off with steps a and b of the mechanism in Eq. (17.8).
After step b, we need steps that give the correct stoichiometry. A possible
mechanism is N2Os5 i NO; + NO; (rapid equilib.), NOz
+ NO3 >
NO + O> + NO; (slow, rate determining), NOz + ClxO — NO CI + OCI,
OC] + NO > NO,Cl, NO2Cl + 02 — NO3Cl + O, O + O > Oj. (The
stoichiometric number of all steps but the last is 2.)
243
17.54
For (17.60), r= % d[NOz]/dt = k2[N202][O2]; the initial equilibnum gives
[N,O,}/[NO}? = K,.1 = ky/k; and [N02] = (ki/k.1)[NO]’, so r=
(koky/k_;)[NO]*[O2]. For (17.61), r= % d[NOo}/dt = k2[NO3][NO] and k)/kK., =
[NO3]/[NO][O9], so r= (koki/k-1)[NO][O2]. For (17.62), r = k[NO}°[O2].
17.55
The exactnecultaml.00 siisfAl= 1A loan = Gl.00 mol leven Gala amines —
0.860708 mol/L and is 0.637628 mol/L at 3.00 s. For this reaction g([{A]) in
(17.64) is g([A]) = —-k[A]" = -K[A] and [A]; = [A]o — k[A]o At. The first number
in column A (the time column) is 0 and the next entry in column A is the first
entry plus Ar. The first number in column B is | and the next entry contains the
Euler formula. The results for 1 s and 3 s are 0.85873 and 0.63325 mol/L for At
= 0.2 s and 0.85973 and 0.63546 for Ar = 0.1 s.
17.56
The exact value at 1.00 s is [Eq. (17.17)]
[A] =
(1.00 mol/L)/[1 + (0.15 L/mol-s)(1.00 s)(1.00 mol/L)] = 0.869565 and is
0.689655 at 3.00 s. If the Prob. 17.55 spreadsheet is properly set up, one need
only change n from | to 2 to get the Euler results for this problem, which turn
out to be 0.86627 and 0.68421 mol/L for At = 0.2 s and 0.86795 and 0.68697
mol/L for At = 0.1 s.
17.57
We have g[A] = -A[A] in the modified-Euler formulas. The top entries in the r
and [A], columns are 0 and | respectively, and the top entry in the [A]n41/2
column is [A]1/2 = [A]o — k[A]o At/2. The n = 1 modified-Euler results at 1 and
3 s are 0.860728 and 0.637672 mol/L for At = 0.2 s and 0.860713 and
0.637639 mol/L for At = 0.1 s. Comparison with the exact values in Prob.
17.55 shows the modified-Euler results are quite good and much better than the
Euler results.
17.59
(a) F; (b) F; (c) F.
17.60
k= Ae™*e!®T
k(T>)/k(T) = exp [(Eg/R\(T)! -T 5'))
k720 = (0.0012 dm*/mol-s) exp er
alK 7 K )
0.018 dm? mol! $7!
244
17.61
k= Ae *«'®T andink=In A — E,/RT. We plot In k vs. 1/T. The data are
i @eOay
ie
ay
a
KOE
Mast)
RSC
SOR
ee
kee.
The slope is -19500 K! = -E,/(8.314 J/mol-K) and E, = 162 kJ/mol. The
intercept at 1/T =0 is 25.9 = In(A c° s);
A=7 x 10'° dm’ mo! s'. (The
intercept is calculated from the slope and one point on the graph.
y = -19458x + 24.943
Inte. Ss)
. bee
0.0014
0.0015
a
0.0016
0.0017
1/(T/K)
17.62
k= Ae **!®T | ky [kp = exp [(Eg/R)(WUT) — W/T2)], In (hy, /kp,)=
(E,/R)(1/T; — 1/T>) = In (0.000030/0.0012) =
[E,/(1.987 cal/mol-K)][1/(700 K) — 1/(629 K)] and Ey = 45.5 kcal/mol.
ie
keealRT —
(0.0012 dm? mol"! s”') exp[(45500 cal/mol)/(1.987 cal/mol-K)(700 K)]
= [5-< 10!! dm? mol! s™.
17.63
k=
Ae &*'®"
As T- ©, E,/RT goes to 0 and k goes to A. As T 4 &, the
collision rate goes to infinity and the fraction of collisions having at least the
activation energy goes to 1, so the collision-theory picture leads one to expect
the rate to increase without limit as T — 9, rather than approaching an upper
limit as predicted by the Arrhenius equation.
245
17.64
(a)
Ink=InA-E,/RT. A plot of In N vs. 1/T (where N is the chirping rate)
has slope —E,/R. We have
IiVN ie Siisaemeed
83, ~ 4/60.
107T
03408
G5dme
44cm
ke.
The plot is linear with slope —6.48 x 10° K"! = -E,/(1.987 cal/mol-K) and
Exes) eto kcal/mol.
y = -6476.7x + 26.906
Sh
In N
alan
lee
sie
Tal
lima
ie
4.4
et
0.00335
0.00340
0.00345
0.00350
1/(T/K)
(b)
At 14.0°C, 10°/T = 3.4825 K”' and we read the value In N = 4.35 from the
graph. Thus N = 77 per minute. The rule gives the Fahrenheit
temperature as 40 + '4(77) = 59°F. Actually, 14.0°C is 57°F, so the
crickets are in error by 2°F.
17.65
(a)
A =2.05 x 10'* s! and E, = 24.65 kcal/mol.
(b)
k(0°C) = (2.05 x 10!? s!) x
exp [(—24650 cal/mol)/(1.987 cal/mol-K)(273.15 K)] = 3.87 x 10°’ sl.
(c)
The units of k and the rate law in Eq. (17.6) show that the order is 1.
Equations (17.15) and (17.11) give katy2 = 0.693 = akty/2 = 2kty2 and
t/2 = 0.693/2k. From the given k(T) expression, we find k(—50°C) =
1.47 x 107"! s“' and k(50°C) = 4.36 x 10~* s"!. We find 11/.(-50°C) =
2.36 x 10'° s; ty2(0°C) = 8.95 x 10° s; and t4/2(50°C) = 795 s.
246
17.66
This fraction of collisions equals e*"’ = exp(—Nae/RT) =
exp[—(80000 J/mol)/(8.3145 J/mol-K)T] = exp[-9622/(T/K)].
(a) exp(-9622/300) =e?" =1.2x 10°";
(b) exp(-9622/310) =e?! = 3.3 x 10°";
(c)
17.67
Cpe e887
alinti(Each 10°C increase nearly triples the fraction.)
From (17.43), the observed rate constant is k = kj + kz =
Aye @
+ A,e °**" From Eq. (17.68), Ea =RT d In dT =
RT°(d/dT) In (ky + ky)=RT? (ky + 2) \(dk\/dT + dk/dT)=
[RT M(ky + ko) |[(Ea/RT)A
6 08” + (Eq o/RT)Are 88] =
(Eaiky + Eg 2k2)i
(ki ko).
17.68
AH 593 = (-393.509 + 90.25 + 110.525 — 33.18) kJ/mol = —225.91 kJ/mol.
Since An, = 0, we have AH® = AU®, and use of Eay— Ea» = AU*® gives
116 kJ/mol — E,, = —226 kJ/mol. So Eq», = 342 kJ/mol.
17.69
The equilibrium condition for step 1 gives k,/k_; = [C][D]/[A][B]. We have
= d[G]/dt = k[C]’ = ko(i [A] [BV/K1[D])°=A kok, [A] [B]/[D]’, so k =
manke vand Ae ©eteen Ae 0) Ae
—]
aa
a2 Be Ae eae Then
= 2E,) + Eg2 —2Ea,-1 = (240 + 196 — 192) kJ/mol= 244 kJ/mol.
17.70
a7.71
= (Ea RXWT, -WT2)
yy (ko/ky) =
(a)
klk, = AeFa! RT | Ag Fa! RN
(b)
(E,/R)(A/T, — 1/T>) and In 6.50 = [Eq/(8.314 J/mol-K)] x
[(300.0 K)! — (310.0 K)'] and E, = 145 kJ/mol = 34.6 kcal/mol.
ko/ky = exp [(19000 J)/(8.314 J/mol-K)}[(300 Ky '— (310 K)'] = 1.28.
(a)
The linear regression slope is —12459 K"! =-E,/R, and E, = 24.76
kcal/mol. The intercept is 30.81, = In(A/s) and A = 2.42 x [Opec lhe
absolute percent errors range from | to 5% and the sum of the squares of
the deviations is 1.9 x 10’ s™.
(b)
Since the sum of the squares of the deviations is so small, it is helpful to
minimize 10!° times this sum, rather than the sum itself. One finds E, =
25424 cal/mol and A = 6.55 x 10'* s"! as the optimum values with a sum
247
of squares of deviations equal to 8.91 x 10°'°. The absolute percent error
is 0.1% at the highest T and about 10% at the three lowest T’s.
17.72
(a)
ForH2 +h — 2HI, AG‘4o) =—-23.6 kJ/mol = —-RT In Kp and Kp = 57.7 =
K,, since An = 0. Then KS = k/ky = 0.064/0.0012 = 53 = (58)'”, so
Sie
(b)
17.73
K's = K, = ky/ky = 0.0025/0.000030 = 83.
kikp— Kon
(a)
kr/kp = K-. Let
r= kpZ be the rate law of the reverse reaction. At
equilibrium, the forward and reverse rates are equal, so kpZeq =
KIBO. jeg SO.se leatealey and Zen BLO) sen SOuge eal
kig leq
ii Ke BT SO ge |i BiO. leis slew Duetetor
k,Z = ky[Br ][SO7 ]° [H*V[SO3"]°.
(b)
k/ky = K\?. From part (a), ke/kp = Zeq/[BrO 5JeqlSO 3” JeqlHJeq =
Ki? = [Br] ig,[SO4
Jeg IBrO31.¢ [SO ]2q, and k,Z =
ol Bilwe (SO al:
17.74
1He|(BrOs iSO:
aio
kplkp = K's and In ky — In ky = (1/s) In K,. Differentiation of this equation with
respect to T and use of Eq. (17.68) and the result of Prob. 6.19 gives
Eq/RT — Eqy/RT’ = (1/s)AU°IRT’ and Ey — Ea» = AU°s.
IWATE)
We reverse the designations of forward and back reactions and use primes to
denote the newly designated constants; i.e., ky =ikyand:k, = kealso. kK =
1/K,. Since the mechanism of the reverse reaction consists of the reverse of the
mechanism of the forward reaction and (as noted on p. 546 of the text) the
reverse-reaction’s rate-determining step is the reverse of that for the forward
reaction, we have s’ = s. Substitution into k/k, = K).” gives k,/ky = (1K; Nee
rye
and k,/k,
+ vis’
=(K.)°.Q.E.D.
248
17.76
Consider two chemical reactions
aA+bB > cC+dB
(), pP+qQ > rR+wW
(I)
Let us form a third reaction that is the sum of m times the first reaction and n
times the second reaction:
(Hl)
maA + mbB + npP + ngQ mcC +mdB+nrR+nwW
The concentration-scale equilibrium constants of these reactions are
Ki = (C} (DIAN IBY’, Ku = [RYEW) "(PYLON
Km = (C)"(DI"
[RI (W) "A BY PP"1Q)”
(where all concentrations are equilibrium concentrations) and we see that
Kw = K{' Kj,. A similar result holds for a reaction formed by multiplying more
than two reactions by integers and adding them. Since the elementary reactions
of a mechanism multiplied by their stoichiometric numbers and then added
yield the overall reaction, we have K, = K;!K3? ---K," = II; (K;)" , where
K, is the equilibrium constant of the overall reaction and Kj, K2,---, Km are
the equilibrium constants of the elementary reactions. But we know that for an
elementary reaction K; = kj/k_; [Eq. (17.53)], so K- = pike
1729
Step (b) has stoichiometric number s = 1, so k¢/ky = K i . Let r, = kpZ be the
rate law of the reverse reaction. We know that the forward reaction has rate
law rp= kf N2Os]. At equilibrium, k{N2Oseq = koZeq and Zeg/[N2Osleq = kp/ko =
K, = [NOo] 4,[O2] eq /IN2Os] 2q» 80 Po = keZ = ko[NO2]}"[O2}/[N205].
17.78
1/kuni = kK-1/kik2 + 1/[M]k1. Since PoV = NokT, we have [M] = Po/RT and
VWkuni = K\/kik2 + RT/Pok\. A plot of 1/kuni vs. 1/Po is linear with slope RT/k,
and intercept k_\/k\k2. The data are
Iki
10°/P)
10440
9.09
9620
4.74
9010
1.316
9260
2.58
s
tom 1
The slope is 1.83 x 10° s torr = 240 s atm = (0.08206 dm’-atm/mol-K) x
(743.1 K)/ky and k, = 0.253 dm? mol! s'. The intercept is
RTTO ES A Hl Kinip2 SO Kuni pee = 11-40 x 107 serand kay/k3 =
(8770 s)(0.253 dm? mol"! s“') = 2220 dm*/mol.
249
ne ee
__y= 182844x+8772.2
W(Kyni/S”')
10000
9500
9000
Read
liames
0e
laghy
Sahl
[pa
lieatgeeliont’
9599
Cees
0.000
0.002
tt
0.004
0.006
0.008
0.010
1/(P)/torr)
17.79
The B and C molecules are smaller than the A molecules and so undergo
collisions with A less often than A molecules do.
17.80
d{Br2]/dt = —k\{[Bro][M] + k,[Br]?[M] — k3[H] [Bro].
d{Br]/dt = 2k,[Br2][M] — 2k_,[Br]?[M] — k2[Br][H2] + k2[HBr][H]+4[H] [Bro].
17.81
(a)
Step (1) is the initiation step, (2) and (3) are the propagation steps, and
(4) is the termination step.
(b)
The chain-propagating steps (2) and (3) occur many times for each
occurrence of step (1) or step (4). The overall reaction is therefore the
sum of steps (2) and (3), namely, CH3CHO
(c)
— CH, + CO.
We use the steady-state approximation for the intermediates CH; and
CH3CO: d[CH3]/dt = 0 = ki[CH3CO] — k.[CH3][CH3CHO] +
k3[CH3CO] — 2k4[CH3]?. d[CH3CO}/dt = 0 = k2[CH3][CH3;CHO] —
k3[CH3CO]. Addition of these two equations gives 0 = ki[CH3;CHO] —
2k4[CH3]° and [CH3] = (k;/2k4)![CH3;CHO]"”. Then r = d[CHy]/dt =
ko[CH3][CH3CHO] = k2(ki/2k4)"”” [CH3CHO]*”. (Alternatively, we can
find [CH3CO] and use r = d[CO]/dt.)
17.82
The reverse of step I has a near-zero E,, so E, of step I equals AU 7;- From the
Appendix, AH; = 104 kcal/mol and AU; = 103 5 kcal/mol at 300 K. This
250
103+ kcal/mol E, is far higher than the 45 kcal/mol E, of Br;
+M
2Br + M [see the discussion after Eq. (17.95)], so the dissociation of Hz by M
can be neglected. Step II is the reverse of step 3 in (17.88). Appendix data give
AH, = AU; =-41 ; kcal/mol = E52
angie, 3 = Ea y=
421 kcal/mol (since E,.3 = | kcal/mol). Ea, is far higher than the
18 kcal/mol E, of Br + H,
~ HBr + H, so Br reacts preferentially with H2
rather than with HBr, and reaction II can be neglected. The atom combination
> HBr + M has E, 1 = O. Reaction 3 in (17.88), namely,
H + Br, > HBr + Br, has E,3 = | kcal/mol, not much different from Eq; thus
the rate constants k3 and ky are of the same order of magnitude. We have r3 =
reaction
H + Br+ M
k3{H][Br2] and ny = kin{H][Br][M]. The concentration of the reactant Bry is
high and is of the same order of magnitude as the M concentration. The very
low concentration of the reactive intermediate Br makes ry << 73, SO We can
neglect reaction III.
17.83
(a)
Initiation is step 1: propagation is steps 2 and 3; termination is step —1.
(b)
[Cl]*/[Cl2] = ki/k_1 and [COCI]/[C]][CO] = ky/k_z. So [Cl] =
[CO] [Clo]. For the
(ky/ky)![C19]"? and [COC]] = (k2/k-2)(ki/k-1)”
forward reaction, ry = d[COCI]2]/dt = k3[(COCI][Cl2]. (Since we want the
rate law for the forward reaction, we assume negligible amount of
product has formed and we do not consider the reverse of step 3.)
Substitution for [COC]] from (b) gives rf=
Uele
aiGlaaglC OL
Wek. kie
(c)
k1 [As
ICI," [COCI,}.
ry = —d[COCI|/dt = k3[COCh][Cl] = (k_3k}7/
a check, putting r= rp, we get [COC]s]}eq/[COJeqlCl2Jeq =
(ko/k_2)(k3/k_3) = K-, where the result of Prob. 17.76 was used. ]
17.84
(a)
=
/
k= ko(k/k.,)'? and Ae Fy (RT _ vA
= )\4?
Al; e
Eea,2
I SSR
ie
a,l
a,-1
7
2
2
so Eq = Ea2 + LE, a 5 Eat.
(b)
As noted after Eq. (17.95), Ea: = 0 and Ea,1 = AU, . Data in the
Appendix give AH; = 46.1 kcal/mol; then AU; = AH; — RT = 45.5
kcal/mol at 298 K. So Eq.2 = Eq — YE.) = (40.6 — 22.7) kcal/mol = 18
kcal/mol. The ratio k\/k_; is K, for the reaction Br2(g) e 2Br(g); data in
251
the Appendix give AG 39, = 38.64 kcal/mol; (6.14) and (6.25) give
Kei p= Aer and 'K}sf a0leo sol DaeSo m/e HOGI
mol/dm? at 298 K. Substitution in Ae~£« "7 = Aye*?/*" (k,/k_,)”? at
298 K gives A> = (1.6 x 10'! dm*?/mol"”?-s)(1.9 x 107° mol/dm*)'” x
exp {[(18000 — 40600) cal/mol]}/(1.987 cal/mol-K)(298 K)} =
3 x 10° dm? mol! s"!. Then kp = Ae
17.85
Ps2
(a)
[Root] = (fki/k) 7)" = (0.008 mol/dm*)'”7(0.5 x 5 x 10° s7')'”
/(2 x 10’ dm*/mol-s)"”* = 1.0 x 10°’ mol/dm’.
(DP) = kp[M\/(fkik,)'
[1] = (3 x 10° dm*/mol-s)(2 mol/dm°)/
(0.5 x 1000 dm?/mol:s”)'7(0.008 mol/dm?)'” = 3000. —d[M]/dt =
kp (fkilky) [MI] = kp[M] [Rio] = (3000 dm*/mol-s)(2 mol/dm°) x
(1.0 x 10°’ mol/dm?) = 0.0006 mol/dm?-s. d[Po]/dt = ki[Rtor]” =
(2 x 10’ dm?/mol-s)(1.0 x 10-7 mol/dm?)” = 2 x 107’ mol/dm*-s.
(b)
When termination is by disproportionation, two polymer molecules
(instead of one) are formed whenever two radicals combine. This doubles
d[PioJ/dt and hence cuts (DP) in half. Thus d[Pro/dt = 4 x 10°
mol/dm?-s and (DP) = 1500. The other quantities are unchanged.
17.86
The initiation reaction 2M — 2R- contributes a term 2k,[M]° to —d[M]/dt and
(17.97) is modified to —d[M]/dt = 2k,[M]° + k,[M][Rio?]. To apply the steadystate condition d[Rtot:]/dt = 0, we note that the initiation step has (d[Rtot:]/dt); =
2fki[M]°. So 0 = (d[RioV/dt); + (d[Riov Vdt), = 2fki[M} - 2k,[Rro]” and [Riot] =
(fki/k,)'""[M]. Substitution in the above —d[M]/dt equation gives —d[MJ/dt =
[2k; + kp(fkilk,)?)[M]’. Use of the above [Rio] expression in (17.105) gives
d[Prol/dt = ki{Rro]” = fki{MI]°. Equation (17.104) becomes (DP) =
—(d{MV/dt)/d[Pros\/dt = [2ki + kp(fki/
ki)! Vp.
17.87
d[AJ\/dt = -k{A] + kG
Let [A]eg and [C]eg be the equilibrium concentrations
under the new conditions, and let x = [A]eg — [A]. Then dx/dt = —d[A]/dt. Since
2 moles of C are formed when | mole of A reacts, we have [C]eg — [C] = -2x.
Then —dx/dt = —ky({A]eq — x) + kx((C] 2, + 4x1Ceq + 4x°) =
252
—kp[Aleq + Kol] a + xkp(kplkp + 4[Cleq + 4x). At equilibrium, d[A]/dt = 0 and
the first equation in this paragraph gives —k{Aleq + ko[C] Be = 0. Since the
perturbation is small, [C] is close to [C]eq and we can neglect 4x in comparison
with 4[C]eg. We then have dx/dt = —xkp(k/kp + 4[Cleq) = ~t~'x, where t =
(ky + 4kp[C]eq) .Integration gives x = xoe* or [A] — [Aleq = ([A]o - Nee
17.88
(a)
(b)
CH3;CH2CH3, CH3CH3, CH;CH»CH2CH;, and N2.
CH3CH2CH;3 and N> (cage effect).
17.89
x
Equation (17.112) gives kp = 27(6.02 x 107°/mol)(4 x 10° cm)
Sap!
(8.4 x TOs cm//s) = 1.3 x 10'? cm?/mol-s = 1.3 x 10'° dm? mol!
17.90
(a)
). Then
From Eggs. (17.114) and (17.1 15), In kp = In T— In n + In (const.
Eq. (17.68) gives Eq = RT? d\n kp/dT= RT’ [l/T — (1/n)dn/aT] =
RT — (RT’/n)dn/aT.
(b)
.023 K')
Eq = (1.987 cal/mol-K)(298 K) — (1.987 cal/mol-K)(298 K)*(-0
= 4.7 kcal/mol = 19 kJ/mol.
(Cc)
ar
ro
25
20.3
12.5
6.45
WiSIo
0.800
0.400
0.200
0.050
(bd) Ie
17.91)
(aie
17.92
The data give
dm’-s/mmol
dm*/mmol
y = 38.98x + 4.7321
1/{ro/(mmol/L-s)]
40
35
30
phs)
0
0.2
0.6
0.4
253
0.8
1
1/{[S]o/(mmolV/L) }
The slope is 39 s = Ky/k2[E]o and the intercept is 4.7 dm?-s/mmol = 1/k2[E]o.
So ky = (2.8 x 10° mol/dm?)"'(4700 dm?-s/mol)"! = 7.6 x 10° s' and
Ky = (39 s)/(4.7 dm?-s/mmol) = 8.3 x 107° mol/dm’.
17.93
Since [P] =~ 0 and since [ES] is assumed small, it follows that [S] = [S]o. From
(17.122) with [S] = [S]o = Ku = (kK: + k2)/k; and [P] = 0, we have [ES}/[E]o =
(ky + ko/(kKy + ko + ky + ky) = %. From (17.125), ro = ka[E]oKm/2Kmu =
k2[E]lo/2 = ro max/2, SINCE Tomax = Kal[E]o:
17.94
We plot log t1/2 vs. log Po. The data are
log ti2
0.88;
0.56,
0.239
log Po
2.423
2.114
M .763
log(t;/min)
Ow
y = 0.9858x - 1.5106
(CR
Lo
clFES RY) ES
LS
EES)
2
RAY SS
Ne
CE
VyWY a ER
eee
HU
VY
US TW
ees ae
ee
(SUE
0
|
ee)
log(Po/torr)
The slope is 0.99 = 1 —n andn=0.
17.95
The overall reaction is N2 + 3H2
2NH3. To produce this overall reaction
from the steps listed, we must multiply steps a, b, c, d, e and fby 1, 3, 2, 2, 2,
and 2, respectively, and then add them. These are the stoichiometric numbers
of steps a to f. The fact that the rate-determining step probably has
stoichiometric number s = | indicates that the rate-determining step is probably
step (a), Nz + 2*
17.96
— 2N*.
The rate of the bimolecular desorption reaction 2A(ads) > A2(g) is
proportional to (n,/A) (na/A) and so is proportional to 64 ee
254
kgs, . The
rate rag; of the adsorption reaction A2(g) — 2A(ads) is proportional to the
A»-surface collision rate, which is proportional to P. Since two adjacent vacant
sites are required for A2 to be adsorbed dissociatively, the adsorption rate is
also proportional to (1 — @,)(1 — 8,), the square of the fraction of vacant sites.
Hence, rags = Ka(1 — 9,)°P. SO rads = Faes and ka(1 — 9,)°P = kg, ork? (1 O4)P'7/k'!? = O,. Solving for 04, we get 0, = (kalka)?P'7/[1 + (kalka)'P'"),
which is (13.38) with b = ka/ka.
17.97
(a)
We assume that each CO molecule occupies one adsorption site. (This
isn’t always true.) The number of sites per cm’ is (2.3 x 10°’ mol) x
(6.02 x 107°/mol) = 1.33 x 10".
(b)
The total number of sites is (1.38 x 10!°/cm7)(5.00 cm’) = 6.9 x 10'°. The
number of occupied sites is (9.2 x 10-'° mol)(6.02 x 10°7/mol) =
5.5 x 10'*. So 0 = (5.5 x 10'*)/(6.9 x 10'*) = 0.080. The adsorption rate
per unit area is r; = (5.5 x 10'*)/t(5.00 cm’) = (1.1 x 102 cm~)t! and the
equation in Sec. 17.18 that defines s becomes s =
(1.1 x 10!“/em?)(2nMRT)'7/tPNa. We have tP = (0.43 x 10° torr - s) x
(1 atm/760 torr)(101325 Pa/1 atm) = 5.7 x 10° Pa-sands=
(1.1 x 10'4 10¢ m™)[2m(0.028 kg/mol)(8.314 J/mol-K)(300 K)}""7/
(5.7 x 10° s N/m’)(6.02 x 10°°) = 0.67. Since 8 is close to 0, this is
approximately So.
17.98 (a)
(b)
/RT
ty. = 0.693/kges = 0.693 08"
MAges =
5.3.%10 s:
17.99 d=(2D1)'? =(2Dye
1)? =
dis 100"” = 10 times d at 1 s, which is 3.7 x 10° cm.
17.100 (a)
With Egaas = 0, Eq. (17.71) gives Eades = —AU jag = ATjas |AH ays|/RT
IRT
|Ages =
/Ag., = 0.693¢
tp = 0.693/kges = 0.693e"**"'
0 60fe0e
J/mol)/(8.314 J/mol-K)(300 S08
255
sr) oo
107’ S.
(b)
180s.
(c) 4.6x 10!’ s.
17.101 Let the half-reaction be M* + ne” = M* or its reverse (where z and z’ are the
charges on the species M). Let B in Eq. (17.128) be the species e . Then
|val=nand r,='|val7'|dnp/dt| = (1/n)| dn, /dt|, where n_ (not to be
confused with n, the number of electrons in the half-reaction) is the number of
moles of electrons. Since the Faraday constant is the absolute value of the
charge per mol of electrons, we have
|Q| =Fn_ ,so |dn__ /dt| = F'|dQ/dt| = I/F and r, = (/AnF)I
=jinF.
17.102 N = (0.000420 g)(1 mol/347 g)(6.022 x 107*/mol) = 7.29 x 10!” atoms of *°U.
Then A = A/N = (9.88 x 10° s"')/(7.29 x 10'”) = 1.355 x 10° 71.
ty. = 0.693/A = 0.693/(1.355 x 10°! s') = 5.11 x 10'* s = 1.62 x 10° yr.
17.103 (a)
(20.0 g)(1 mol/63.0 g)0.00200(6.022 x 107°/mol) = 3.82 x 107° atoms of
7H. Then A = AN = 0.693N/t/2 = 0.693(3.82 x 10°°)/
(12.4 x 365.25 x 24 x 60 x 60 s) = 6.77 x 10!' 571.
(b) A= Ape =Apen
= (6.77 x10! /s)270 9936 2012-4) — 4.79 x 10!! s1.
17.104 The rates of the two decay modes are (dN/dt),
= —A,N and (dN/dt)> = -A2N.
Since dNior = (dN); + (dN)2, the total decay rate is the sum of the contributions
of the two modes: (dN/dt) = (dN/dt), + (dN/dt)2 = —(A, + A2)N = -AN. Then
Ati = 0.6931 and ty2 = 0.6931/A = 0.693 1/(A; + Ad).
17.105 (a)
(1.00 g)(1 mol/12.01 g)(6.022 x 107*/mol) = 5.01 x 107” atoms in one
gram of carbon. Then N = A/A = At)/2/0.693 =
(12.5 min™')(5730 x 365.25 x 24 x 60 min)/0.693 = 5.43 x 10!° atoms of
C. So, % '4C = 100(5.43 x 10!%)/(5.01 x 1077) = 1.08 x 107!° %.
(b)
A =
Age
CAre
eee
(c)
0.0296 min! (g Cy)".
7.0 = 12.5e°° 079%.
sat [12.5 min”!
(g Gye
ce
ee)
In (7.0/12.5) = -0.693/(5730 yr) and
t = 4800 yr.
256
sm
17.106 Let A = 2°U and B = 25U. We have Na = Naoe **" and Ng = Ngoe **’.
We want Nao = Ngo at t= 0. So Na/Np = exp[(Ap — Aa)t] =
exp[0.6931(1/ti/2.8 — Wth/2,)]. So In (99.28/0.72) =
0.693z[1/(7.0 x 10° yr) — 1/(4.51 x 10° yr)] and 1 = 5.9 x 10” yr.
17.107 A = Age™ and A = 0.693/ty/2 = (In 2)/ti2, 80 A = Age
= Ay(s)
17.108 (a)
t/ty)2
(In 2)(—1)t/t 1/2
=
Ao2 (-I)t/t 1/2
"”.
Let the steps be called I, II, Ill, IV. Step II requires two *He nuclei, but
only one >He is produced in step II. Hence the stoichiometric number of
step II is 2; this requires that the stoichiometric number of step I be 2.
The stoichiometric number of step IV is also 2, so as to get rid of the two
positrons formed when step I is multiplied by 2. Multiplication of steps I,
Il, IW, and IV by 2, 2, 1, and 2 gives as the overall reaction
4'H+2°e — jHet+ 2v+6y.
(b)
AUm is -2.6 x 10'? J/mol. There are (3.9 x 107° J)/(2.6 x 10° J/mol) =
1.5 x 10'* moles of “He produced each second.
(c)
The number of neutrinos produced each second is
AG
ayes 10'* mol)(6.02 x 107°/mol) = 1.8 x 10°°. The area of a sphere of
radius 1.5 x 10° km is 4n(1.5 x 10!! m)? = 2.8 x 10” m’. One cm? =
10~ m? and the number of neutrinos hitting a square centimeter of the
Beir ni
second isGksecl0 )1¢10s, ms )/(2 SO)
el Om
(which is a lot of neutrinos).
17.109 (a) 10;
17.110 (a)
(b) 5;
(c) 10;
(d) 6.
For '3CH4, the answers are 10;5; 10; 7.
=
For this elementary reaction, r = kmax{ B][C] = —d[B]/dt = —(d/dt)(np/V)
—(d/dt)(Np/NaV) = —(1/V)(dNp/dt)/Na = Zac/Na and kmax = Zpc/Na[B][C].
At each collision, one B molecule disappears, and Zgc is the collision rate
per unit volume, so Zgc = —(1/V)(dNp/dt).
(b)
=
We have [B] = Np/VN and [C] = Nc/VNa, so (15.62) gives Kmax
=
ZaclNa{B][C] = Na(rs + rc) (8RTm)'(M5| +My
(6.02 x 1073/mol)(4 x 107° m)?[8(8.314 J/mol-K)(300 K)m]"* x
257,
[(0.030 kg/mol)! + (0.050 kg/mol)']'” = 1.8 x 108 m* s! mol"! =
1.8 x 10'! dm? mol"! $7.
J = (1/v;) dnj/dt
(e) T. From (17.2) and (4.97),
= dé/dt. (f) F. See Eq. (17.77). (g) T. (h) F;e.g., see (17.60). (i) $i
(j) T. (k) F. (1) F(r increases with tin an explosion). (m) F. (n) F.
17.111 (a) T.
(b) T.
(c) T.
(d) T.
258
Chapter 18
ay"
18.1
(a)
ve
dR/idv =0= |
C2
So 3 = (AVmax/KT) eT
enan —1)
=xe/(e - 1),where x =
hVmax/kT. Then 3e* — 3 = xe‘ and multiplication by e“ gives x + 3e™ = 3.
(b)
For x = 0, 1, 2, 3, the function x + 3e” equals 3, 2.104, 2.406, and 3.149.
So the nonzero root lies between 2 and 3. We have (3 — 2.406)/
(3.149 — 2.406) = 0.80, so interpolation gives x = 2.80. For x = 2.80, 2.81,
2.82, 2.83, we find x + 3e~* = 2.98243, 2.99061, 2.99882, 3.00704. The
root lies between 2.82 and 2.83, and interpolation gives x = 2.8214. The
Excel Solver gives 2.82143937... . (Use Options to change the Precision
(c)
for Omme)
At 300 K, Vmax = kTx/h = (1.3807 x 10°*° J/K)(300 K)(2.821)/
(6.626 x 104 Js) =1.76 x 10’? s’. At 3000 K, Vmax is 10 times as large,
namely, 1.76 x 10'* s'. From Fig. 21.2, these frequencies lie in the
infrared.
(d)
T = Vmaxxk= (6.626 x 107 J s)(3.5 x 10" s1)/2.821(1.38 x 10-7 J/K)
= 6000 K.
18.2
(e)
Vmax = kTx/h = (1.38 x 107? J/K)(306 K)2.821/(6.626 x 10°" Js) =
1.80 x 10'° s'. Infrared.
(a)
The total emission per unit time and per unit area is Jo Rv) dv =
(2nhic?) J° [vi(e"*? — 1)] dv. Let z = Av/KT; then dz = (h/kT) dv. We
(b)
dz —
have J® R(v) dv = (2mh/c*)(kTIh)* Jy ize =
8/15 = 2K TSC.
(2mh/c2)(kTIh)
The emission rate is (21K T*/15c7h?)(4nr’)=
9
8° (1.38110 —23 ULSY(5800 K)* Oligo Te ays Saou
15(2.998 x10° m/s)? (6.626 x10 **Js)’
(similar to the value given in Prob. 72LGS),
(c)
In 1 year, AE = (3.95 x 10° J/s)(365.25 x 24 x 60 x 60 s) = 1.25 x MO Ip
SOAR = (1.2-< 10 412.998 x 10° m/s) = 14% 10 kp.
259
18.3
(a)
hv=@+
i mv’ and hvin, = @. For K we have Vin, = ®/h =
(2.2 eV)(1.60 x 107!? J/eV)/(6.626 x 1074 J s) = 5.3 x 10'* s! and
Aine = C/Vine = (3.0 x 10!° cm/s)/(5.3 x 10'4 5!) = 5.7 x 10° cm. For Ni we
find Vin, = 1.2 x 10'° s7! and Ag = 2.5 x 10° cm.
(b)
K-yes; Ni-no.
(c)
Lv’ = hv — © = (6.63 x 10°" J s)(3.00 x 10° m/s)/(4.00 x 1077 m) —
(2.2 eV)(1.60 x 107? eV) = 14x 10
J=0.9 eV.
18.4
E = hy = hcld = (6.626 x 1074 J s)(2.998 x 10° m/s)/(700 x 10” m) =
210
oo)
18.5
Ephoton = hv = held = (6.626 x10 J s)(2.998 x 10° m/s)/(590 x 10? m) =
3.37 x 10°!° J. Then 100 J/s = N(3.37 x 107"? J) and N = 2.97 x 10° photons/s.
18.6
hv=O+ Lmy* = ® + Kmax and Kmax = hv —®.
[0 aKa eresoes A le 256
10°\4v/s"!
91593,
el. OS
OS
GISi213. 18 7.408!
= 8490
where we used v = c/X. The slope is 6.53 x ioe erg-S=h. @®can be found
from the graph as the value of hv at Kmax = 0 or as the negative of the intercept
at v = 0. We find ® = 2.8; x 10°” erg = 1.8 eV.
y = 0.652x - 2.840
eo
10'*Kmay/ergs
VE ES
4
5
a
6
a
iy
Se
a
8
a
ee
ae
9
10°'4v/s"!
260
10
18.7
18.8
(a)
(b)
A=hA/mv= (6.626 x 10° J s\/(1.67 x 10-7’ kg)(6.0 x 10° m/s) =
6.6 x 10°? m.
2A=(6.626 x 10°* J s)/(0.050 kg)(1.20 m/s) = 1.1 x 10° m.
(a)
sinO=A/w anddA =h/m», so sin 9 = h/mvw =
6.26x10°** J -s
):
(9.11x1077! kg)(6.0x10° m/s)(2400x10-'° m) i
5.05 x 10 and 0 = 0.0289° = 5.05 x 10~ rad.
(b)
Let z be the width of the central maximum. Figure 18.4 gives tan 0 =
PE
(c)
18.9
DE
52/(40 cm) and z = 2(40 cm) tan 0.0289° = 0.040 cm.
Ap, = 2h/w = 2(6.63 x 10°" J s)/(2400 x 107° m) = 5.5 x 10°’ kg m/s.
AxAp, > hand Ap, > A/Ax = (6.6 x 104 J s)/(1 x 107"? m) =
~
Gx
ie
kgm s!. We have Ap, = A(mv,) = m Av, and Av, = Ap,/m 2
(6.6x 10g kgm sW(9.1 x 105) ke} = 9x 10° m/s, which is very large.
18.10
The time ¢ and the 9 spatial coordinates x,, yj), Z),X2» V2» Zz, 3, V3, 23 Of the
three particles.
ren(c) L, (dye
13:11
(a) Fb)
18.12
|z|? =z2* and |z| = (zz*)!”.
18.13
ie) ie (ty: 7.
(a)
Bl-2 l= 2
(by
B20
(c)
|cos 6 +7sin 0 | = [(cos 8 + i sin 8)(cos 8 —7 sin 9)!" =
(d)
(cos’0 + sin?@)! = 1'*=1.
[23274 | ua (ex tae Cer uO ee a (Oey? ay
1B=2nGa24So0'+4
Pe ise
Let Isis19and rsjg19denote the left side and right side of (18.10), respectively.
We have Is}g10= rSig.10- 10 see if c'¥ is a solution of (18.10), we replace ‘¥ in
(18.10) by c¥ and see if (18.10) is satisfied. With c'¥ as the proposed solution,
the left side of (18.10) becomes (—h /i)(d/dt)(c¥) = c(-h /i)(O'¥/dt) = c 1818.10.
261
The night side becomes
~(h?/2m)[0(CPVOx? +--+
J—-- + -(h7/2m,)[0?(CP)lox7 +++ - 1+ Ve =
CTS} 19. Since V is a solution of (18.10), we have Isig10= rSig.10. Then
c ISig.10 = C FSig19 and so c® satisfies (18.10).
18.14
f° [4 (1ANe(Oyn(o) dr] dO do = J? [4g@)A(O)LS, Ar) dr] dO do = Ji,f(r) dr x
J? [J e(@)h(o) dO] do = J‘ fir) dr J4(8) dO J? h(o) do
18.15
The result of Prob. 18.14 with r, 8, @ replaced by x, y, z, gives Wale
Iga (2 fc?"
eat
e728 1" (2imc2)"/2 e729 1° (2 Ic? )"2 @-2e° 10° dx dy dz =
(2 Imc?)!!2 fe, ee gilts dx -(2inc?)'!? J~., eV le? dy. (2/mc?)!? ha popes AES
Use of integrals 1 (with n = 0) and 2 in Table 15.1 gives
(2Imc2)'"? J, ernie Pe NO
OLE Be
ice kee
2(2htc? yn
11 2\(C 12)" 7 = |. By symmetry the y and z integrals also equal 1
and V is normalized.
18.16
The 9 spatial coordinates x,, ¥,, Z}, X95 Yo» Za» 3» Y3> 23 Of the three particles.
18.17
(a) I; (b) F; (c) T, (d) T.
18.18
The time-dependent Schrodinger equation is more general, since the timeindependent equation applies only to stationary states.
18.19
Let
f= fi + if) and g = g; + ig2, wheref) 1s the real part of f,andfo is the
coefficient of the imaginary part of f.Then (fg)* = [(fi + i/2)(g1 + ig2)]* =
[igi —frg2 + i(fogi + fig2))* =figi —frg2 — igi + fig2). Also, f*g* =
(fi + if2)*(e1 + ig2)* = (fi — tf2)(g1 — 182) =figi —fog2 — Wf2e1 + fig2) = (fe)*.
18.20
Let Isig.24 and rs;g.24 denote the left and right side of (18.24). To see if ky is a
solution of (18.24) we replace y in (18.24) by kw and see if (18.24) is satisfied.
With ky as the proposed solution, the left side of (18.24) becomes
(-h* /2m,)[0? (ky)/Ox) +---J—--- + Vik) =
262
k[ (—h*/2m, \(0°w/dx; +:--)----+Vy)]
= k |sig.24. The right side becomes
Eky = k(Ew) = k rsig.24. Equation (18.24) is Isig.24 = rsig.24 and multiplication
by k gives k |sig24 = k rsjg.24, so ky is a solution of (18.24).
18.21 (a) Yes. The integral |" e-?” dx with a >0 is finite, as shown by integral 1
with n = 0 and integral 2 in Table 15.1.
(b) No. The integral of the square of this function is infinite.
(c) No. J®, x? dx =), x? dx +JE x? dx =—x' 2, -x7' [P= -0-(0--)
(d)
No. For x <0, |x| =—x, and the integral from —
to o¢ of the square of
this function is J°,, [I1M(—x)!/7]
dx+Jp (1/x'’") dx =
- 2(—x)7|" i Dicatle = 0 — (-2)(00) + 2(ce) —0 =,
(e) False. Consider the function
f= em jlx|'’*, which is the product of the
functions in (a) and (d). The integrand is infinite at x = 0. Despite this, the
integral J*_| f |?dx is not infinite. For xbetween —1 and 1, the integrand
is less than |x |"t'?(exceptiat
i= 0)ps0 [vey
2
Ixlh gdxi<
Phy pxp'? de=22,|-xf "2 deMyla?de= -2-a9!7]2, +2272]! =
0
=
iz
be
2+2=4.S00<!J!,| f |?dx <4. Over the rest of the integration range we
deal with Jo) e728” |x[-!/?dx and f° e?® |x[-/?dx.Here, the I/|x|!/
factor makes the integrand less than e— 2ax" and, since integration of
2
.
.
.
eek gives a finite area under the curve (see part a), the integrals
ee
a
ie aeeandal, IES |x|'/* dx are finite. Hence,
\s if |?dx is finite and fis quadratically integrable even though it is
infinite at the origin.
18.22
E=n’h’/8ma’; |AE| =hv =hcla;
A =hcl|AE| = hc[8ma‘/h*|n5 —n?|] =
0x10°° kg)(6.0x10-'°
BN a=
8ma’clh|\n5 ean | = BUSY AU ee SED
(6.6310
1.45 x 10°’ m.
263
m)*(3.0x10° m/
temek
Te
-" J-s)(25—-16)
eS =
18.23 (a)
J3/*|w|? dx =(2/a) J6’* sin? (nmx/a) dx=
al4
;
rea]
(2/a)[ 4x — + (a/ntt) sin (2nttx/a)] here
‘
1
wa (1/2ntt) sin 5mm where we
used J sin* cx dx = ix Ca sin 2cx.
(b)
18.24
Forn=1, 2, 3, we get 0.091, 0.250, and 0.303, respectively.
The interval 0.0001 A is much, much smaller than the box length, so we can
consider this to be an “infinitesimal” interval. The probability is then
|w|? dx = (2/a) sin* (ntex/a) dx.
(a) Forn=1, we get [2/2 A)] sin? [m(1.6 A)/(2.0 A)](0.0001 A) =
3.45 x 10°.
(b) Forn =2, we get 9.05 x 10°.
18.25
Note that y* has zero slope at the nodes.
18.26
d’y/dx’ = 0. Integration with respect to x gives dw/dx = c and a second
integration gives y = cx + d, where c and d are integration constants. The
continuity condition at x = 0 requires that y = 0 at x = 0, so0 = c(0) + d and
264
d=0. Then y = cx. The continuity condition at x = a requires that
ainennl = 0,
Y= aaand.0 = coysoi=0
Wie Oat
18.27
The lowest frequency transition corresponds to n = | — 2. Use of (18.7) gives
2°h?/8ma? — 1°h?/8ma? = hv and a = (3h/8mv)'” =
[3(6.63 x 107°4 J s)}/7/[8(9.11 x 107°! kg)(2.0 x 10'* s')]* = 1.2 x 10° m.
18.28
hv = Eupper — Etower = (n2 —n rf)h?/8ma’, where u and / stand for upper and
lower. v = (n? —n?)h/8ma?. V3-54 = (4° — 3°)h/8ma? = Th/8ma’.
V639 = (97 — 6’)h/8ma? = 45h/8ma*. So V6-49/V3-34 = 45/7 and
V639 = (45/7)(4.00 x 10! s“') = 2.57 x 10s".
18.29
[ow*w, dx=2a" J§ sin(nrx/a) sin(njnx/a) dx, n; # nj. A table of integrals
gives J sin cx sin bx dx = [1/2(c — b)] sin [(c — b)x] — [1/2(c + b)] sin [(c + b)x],
provided c? # b*. So JG w*w; dx =
2| sin[(@;,—n; )mx/a}
a
2(n; —n,)M/a
sin[(n; +n, )ttx/a]}
2(n, +n,)tla
a
0
since sin[(n; — n,)7] = 0, sin[(1; + nj)1] = 0, and sin 0 = 0.
18.30
The left side of (18.28) becomes d’y/dx* = (2/a)'?(-1)(n’1/a’) sin(ntx/a).
With use of (18.34), the right side of (18.28) becomes —(2m/h? )(n7h*/8ma’) x
(haya sin(nttx/a) = (wn? /a*)(2/a)' sin(nttx/a), which equals the left side.
18.31
For (a), || is a maximum at (4a, +4) and at (eam = 4); where a is the box
length. For (b), || is a maximum at (4a, 7a), (4a, pany (ei +a), (Fa; 2a).
where the dashed lines are nodal lines and the x axis is horizontal.
18.32
(a)
2
E=(h/8ma’)(n2 + n> +n :). There are 17 states with n; +7 Airs ee a
16, namelWanwi neti lie oll toil
222 el25mleene 1523 leSlee
(b)
2al 222i 222 oll ISL. 113,
These states give a total of 6 different values for n- + i +n :, namely,
n> TA
p
+n° = 3,6, 9, 11, 12, 14, so there are 6 energy levels in the
given range.
(ears
(f)-Prite)o Ge
18.33
(ajerra(b) Fs (c) F; (d)akay
18.34
(a) Yes;
18.35
Since the wave function is an eigenfunction of the Hamiltonian (energy)
operator, we must get the eigenvalue 25h7/8ma’.
18.36
(a)
(b) no;
(c) yes; | (d) no.
AB (x) — BAf (x) = (d/dx”)Lxflx)] - x[(d"/dx")flx)] =
(d/dx) (xf (x) + fx) — xf (20) = xf (x) + £0) + £0) — fC) = FO).
(b)
(A+ B)(e* +cos 2x) = (d2/dx? + e* +008 2x] =
(Pidx)(e* + cos 2x) +x(e* +cos 2x) =d’(e* idx + d*(cos 2x)/dx +
2
Xe it x COS 2y= Je
18.37
2
2
+.4x7e)
2
2
ta
—-4icos
2x + xe,
tt COS 2X
(a)
(
ip and(_i+)* are nonlinear; the others are linear.
(b)
A (f+g)=[(-hA?/2m,)V? ----+V\(f+2)=
[(-h?/2m,)V? —--- (f+ 9) + Vf + g) = (-h?/2m,
Vi f-(h?/2m,
)V? g
—-+++ Vf+ Ve =[-(h?/2m)V? —---+ Vif+ [-(h?/2m,)V? ---- + Vg
= Hf + He , where the definition of the sum of operators and the fact
that (0°/dx? \(f+g) = 0°fldx; + 0°g/dx; were used. Similarly, one finds
H (cf) = cHf .So H is linear.
266
18.38
(a) When
B operates on g(x), it turns g into another function, which we shall
call f(x). When A operates on f(x), we get another function, so ABg(x) is a
function.
18.39
18.40
(b) Operator.
(c) Function.
(d) Operator.
(a)
p. = [(h/i)(d/dx)} = -(h7/i)(07/Ox’).
(b)
pt = [(A/i)(d/dz)}* = f4 04/02").
(e) Function.
(d*/dx*)(sin 3x) = (d/dx)(3 cos 3x) = —9(sin 3x), so sin 3x is an eigenfunction of
d°/dx? with eigenvalue —9.
(d*/dx’)(6 cos 4x) = -96 cos 4x = -16(6 cos 4x), so
6 cos 4x is an eigenfunction of d’/dx* with eigenvalue —16. (d’/dx’)(5x°) = 30x,
which does not equal a constant times 5x°; so 5x° is not an eigenfunction of
Pid.
(@ldx?)x' = 2x? # (const.)x'. (d’/dx*)(Be™) = 75e™* = 25(3e™),
and the eigenvalue is 25.
18.41
(a)
(d*/dx’) In 2x = —1/x? # (const.)In 2x.
(p,)=J. w*p,w deaJ®, y *(h/i)(0/dx)w dx= (Ali) 9. w *(w/dx) dx
+ (hli) Jo w*(Ow/dx) dx + (h/i) J? w *(dw/dx) dx =
(h/i) i? w*(dw/dx) dx = (h/i) (2/a)(nt/a) ip sin (nttx/a) cos (nttx/a) dx =
(2nnh/ia? (a/nt) > sin? (nmtx/a) |6 =O, since sin nt =O for
1,2,3,...
(b)
—
. (We used the fact that y* = 0 outside the box.)
(x) = [2 w*xy dx = J8 (2/a)'” sin (nmxla)x(2/a)"” sin (nttx/a) dx =
(2/a) ikex sin? (nmtx/a) dx. A table of integrals gives
J x sin’ cx dx = dg — (x/4c) sin 2cx - (1/8c’) COS ZEX SOY
=
(2/a)[ +.x°— (ax/4nm) sin (2nx/a) - (a?/8n?1) cos (2nnx/a)} |: ~
(2/a)(4.a? — a?/8n°1 + a’/8n71) = a/2, since sin 2nt = 0 and cos 2nt = 1
sal 2s
{OWE
(c)
ok
(x*) = iF wey dx = (2/a) Ik x’ sin? (nmtx/a) dx. A table of integrals
gives | x sin? cx dx = x°/6 — (x7/4c - 1/8c°) sin 2cx — (x/4c°) cos 2cx, so
(x?) = (2/a)7/6 — (ax’/4nt — a?/8n°n°) sin (2nmx/a) (a*x/4n°n’) cos (2nttx/a)] |3 = (2/a)(a°/6 — a /An?t) = a?/3 — a7/2n*n’.
267
18.42 The time-independent Schrodinger equation Hy = Ey for (18.64) is
Ey and (Ay + How +--+ Hw) = Ew (1).
w= )
(H, + A, +--+ +H
Taking y = fi(qi)f2(q2) - - - f(r), we have Aw = Ay titan flq2) -: f(g) =
(fa- = ay) Afi, since H, involves only q;. Equation (1) becomes (fp - - Af
file des Division by fiz -*
+ A fe a f=
+ (fifse Po bafets
gives (I/f;) Hifi + (/f) Hof t-:+:+ (Uf) H,f-=E (2). By the same kind of
argument used after Eq. (18.39), each term on the left side of equation (2) must
be a constant. Calling these constants E), E2,...,E,, we have (1/fi) Af S12)
or Af = Efi, etc., and equation (2) gives FE, +E.+-::+£,=E.
18.43 y = (2/a) sin (mymxj/a) sin (nymxx/a). E=n?h?/8mia? + n3h'/8moa°.
(b) vee, quantum number.
18.44
(a) nu, frequency;
18.45
Viight = (Eupper — Elower/h = [(Dupper We 5 )AVose — (Viower + 5 )AVosc\/h =
(Dupper — Viower)Vose = (8 — 7)Vose = Vosc = 6.0 x 10, cue
18.46
Squaring the curves in Fig. 18.18, we get the following curves (note the
unequal peak heights):
y?
18.47
(a)
From Fig. 18.18, Wo is a maximum at x = 0; likewise, Wo is
at x = 0 and this is the most probable value of x.
268
amaximum
(b)
dy? /dx = 0 = (402/m)'2(2xe" — 2ox°e
+), sox= t1/a!”? =
+(h /2tvm)"?. (x = 0 is a minimum.)
18.48
adyoldx*
=
(o/n)!*(d?/dx)e
(wn)'*(—oe pur
-ax2/2
=
(o/n)'*(d/dx)(-axe
-ax*/2
)=
2a axe -ox?/2 )= (02x? — a) Wo =
(L6n4v?m?x/h? — 42°’vm/h) yo. Equation (18.73) gives k = 4r’v'm, so Lkx*yo =
2n’v*mx
Wo. Then —(h 7/2m)(d’yo/dx’) + Lk Yo =
(21° v?mx" + Lhv)yo+ 2m vmxWo= LhvYo = EoWo.
18.49
Fyn de = (403/m)!? Jere dx = 2(408/m)'? JF
Pe" dx =
2(403/n)!/?(270/7/2302) = 1, where we used integrals 1 and 3 (with n = 1) in
Table 15.1.
18.50
(a)
(x) = J° w*xy dx = (O/T)
1/2 ie
co
Me
d= 0, where integral 4 (with n
=) in Table 15.1 was used. This result is obvious from Fig. 18.18.
(b)
(x2) = [Pee de = (ovm)!? [ee
dx = 2(a/n)!? JP Pe dx =
2(ov/n)? (2107/23) = 1/2 = h/8n°vm, where we used integrals 1 and 3
in Table 15.1.
()
(p,) = Pow* bv de= (on)? [ee2(Al@dNe*7dx=
(o/n)?( A 1i)(—a) J”, xe “ax" dy = 0) (from integral 4).
18.51
18.52
+ Lk? = Lim{(kim)'"A cos [(k/m)"7t + b]}* +
kA? sin? [(k/m)'7t + b] = 1 kA*{cos" ((kim)71 + b] + sin? [((k/m)'"t + b]}
(a)
by =K+V=1m(dddty
(b)
mad?xidt? = m(a’/dt){A sin [(k/m)!71 + b]} = —mA(K/m) sin [(k/m)'"t + B]
=—k{A sin [(k/m)'t + b]} = —kx
(a)
(b)
v = (1/2n)(k/m)!? and k = 41°v°m = 417°(2.4 s')°(0.045 kg) = 10.2 N/m.
E= 1kA* =0.5(10.2 N/m)(0.04 m) = 0.0082 J=(v+ 1)hv,sov+ + =
(0.0082 J)/(6.626 x 10°" J -s)(2.4s"') =5.2 x 10° =v.
269
18.53
(a)
The Hamiltonian is the sum of three one-dimensional harmonic-oscillator
Hamiltonians, one for each coordinate; the separation-of-variables
theorem [Egs. (18.65) and (18.66)] gives the energy as the sum of three
one-dimensional-harmonic-oscillator energies:
E=E,+ E, +E, = (vx + 2)hVy + (Vy + Ya)hvy + (v; + Y2)hv:., where
= OF EO
Dy
Dy
U7 =O) ee eee ani
On eters,
Vv, = (1/20)\(k/m)!2, vy = /2n)(kim)'?, v- = (1/20)(k/m)'”, where m is
the particle’s mass.
(b)
18.54
The lowest level has v, = Vy = Vv; = 0 and E = ShWvx + Vy + V:).
H = (1/2u)(uv? + pos + uv?) + V4 (1/2M\(M'vy + M’v? +M’v;)=
V+luwoi tv; +v2)+ 1M(vy +0} +7). Using Eq. (18.77), we have
v, = dxldt = dx,/dt — dx\/dt = Vx2 — Vx. Similarly, Vy = Vy2 — Vy,1 and
VD: = Vz2 — Vz. Since X = (mx; + m2x2)/M, we have vx = dX/dt =
[m(dx,/dt) + m2(dx2/dt)\/M = (my0x,1 + MyVx,2)/(m + m2); similar equations
hold for vy and vz. SoH = V+ 5[m,mz/(m, + m2)] X
9)
2
2
ergy BP oe (my +m )(m, + m2) ~ X
(On SP
2mymyW,1Vx2 +MZV_2
v2, + MM2Vx,1Vx,2
(m2
Wee,
FMV,
9 +7) =
aa
V+ $(m +m)
=
2
2
[(m, + m2z)mvy, + (mm, + m2)mqv., +°°° J =
V+ lomv%,2 + mv? 4 +md>, +mV>\, +mv2, +mW=,)=
2
2
V+ imyv; 2 + imy 3 =Vtm?
v7 /2m +m35v5/2m2
= p;2 /2m, + p5/2m
+ V.
(The dots indicate similar terms in y and z.)
18.55
(a)
(b)
(c)
w= mymp/(m, + m2) = [(12.0 g/mol)/Na][(16.0 g/mol)/Na]/
[(28.0 g/mol)/Na] = (6.86 g/mol)/Na = 1.14 x 10> g.
[= pd? =(1.14 x 10° kg)(1.13 x 10°'° m)’ = 1.45 x 10“ kg m’.
Ep = JJ + 1)h7/21. hh? 121 = (6.626 x 10°" J s)?/807(1.45 x 10° kg m?)
= 3.83 x 10°? J. For J=0, 1, 2, 3, we have Exot = 0, 7.66 x 10° J,
23.0 x 10°" J, 46.0 x 10°*° J, respectively. The levels are (2J + 1)-fold
degenerate, so the degeneracies are 1, 3, 5, 7.
(d)
ForJ=Oto 1, AE=7.66 x 107° J-0=7.66x 10° J=hv=
(6.626 x 10-4 J s)v and v = 1.16 x 10'' s"'. For J= 1 to 2, AE=
(23.0 — 7.66)10° J = Av and v = 2.32 x 10's.
270
18.56
(a)
Let Nx(a—x) be the normalized function. So Ie [Nx(a — x)]*Nx(a — x) dx
=1and|N| = 1/[J%x*(a—x)? dx]'”. From Example 18.8,
Jox°(a —x) dx = a°/30, so| N| = (30/a”)'”.
(Dem) = Gla) [xia xx ae de = 30/0") In (Gx ax ex) ax —
(30/a°)(a'/5 — a'/3 + a'/7) = 30a7/105 = 2a°/7 = 0.2857a’. The true value
is found by setting n = 1 in Prob. 18.41c to give (x?) =a7(1/3 - 1/2r’) =
0.2827a*. The error is 1.1%.
18.57
The value of k that minimizes the variational integral W satisfies OW/dk = 0.
We have 0W/dk = 0 = (h2/ma’)[(8k + L)/(2k — 1) - 2(4K + K)/(2k - 1)] =
(h*/ma?)(8k° — 8k — 1)/(2k — 1)* and 8k* — 8k — 1 = 0. The solutions are k =
1.112372 and —0.112372. The negative value of k makes } = e at x = 0 and so
is rejected. For k = 1.112372, W= (h7/ma’)(4k° + k)/(2k — 1) = 4.94949 h*/ma 2
= 4.94949h7/47ma* = 0.125372h?/ma’ compared with the true value h7/8ma* =
0.125h*/ma*. The percent error is only 0.30%.
18.58
(a)
I o*o dx = [6x°(a—x)°x"(a -— x) dx=
§2 (ax! — 4a°x? + 6a’x° — 4ax’ + x°) dx = a°/5 — 2a°/3 + 6a°/7 — 12 +
a’/9 = a°/630. We have H 6 = (h7/2m)(d’/dx*)(x’a’ — 2ax? + x*) =
~(h7/2m)(2a’ — 12ax + 12x’). So J? o* H 6 dx =
_(h7/m) lfx(a —x)"(a’ — 6ax + 6x’) dx =
—(h?/m) Ik (ax
Sax +19ax, — 18ax° + 6x°) dx =
~(h2/m)(a’/3 — 2a’ + 19a’/5 — 3a’ + 6a7) = h°a"/105m = h’a'/420n'm.
Then | o* Ho dx/J o*o dx = (h’a"/420n’m)(630/a’) = (3/20°)(h7/ma’) =
0.152h7/ma? ~ Egs. The true Egs is h?/8ma” = 0.125h"/ma’. The error is
22%.
(b)
18.59
Itis discontinuous at x = a, since o = 0 outside the box.
H = H°+ 4H’, where H ° is the particle-in-a-box Hamiltonian operator and
H’ =kx for0<x<a. We have E\) = J y* HW, dt=
(2/a) J¢ kx sin*(ntx/a) dx=
(2k/a)[1x° — 1(ax/nm) sin (2nmx/a) — (a°/8n°1) cos (2nmx/a)]§ = 4.ak, where
271
sin 2ntt = 0 and cos 2nm = | were used. E® = n°h?/8ma’, so FO+E
=
n’h?/8ma? + 3 ak.
18.60
(a) a1; .(b) E(cjal.
18.61
We are given that B and C are Hermitian operators, so from (18.92) we have
.(d) Faa(eyat:
[ f* Be dt=J e(Bf)*dt (1) and J f*Cg dt=J g(Cf)*dt (2). To prove
thatB+C
is Hermitian, we shall prove that it satisfies (18.92):
[ ¢*(B+ Og dt=/ g{(B+C)f]*dt (A). The left side of equation (A) is
J ¢*(B+O)g at= J f*(Be+Cg)dt =) f*Bg dt +) f*Cg dt =
f) at=
+(Cfyl*
J g(Bpytat + | g(Cf)*dt = J gl(Bf)*+(Cfy*ldt = J gl(B
J ol(B + Chit * dt, which completes the proof of Eq. (A). In the proof, we
used the definition (18.51) of the sum of operators, the integral identity (1.53),
the given equations (1) and (2), and the identity (z) + z2)* = z1* + 22”, which is
easily proved by writing z; and z; as a, +ib, and a, +ib,, respectively, where
the a’s and b’s are real.
18.62
With ¥ = f+ cg, Eq. (18.91) becomes J (f +cg)*M(f +cg) dt=
J (f +cg)[M(f +cg)]* dt. Using the identity (z; + z2)* = z1* + 22* (which is
easily proved by writing z, and zz as a, + ib, and a, +ib,, respectively, where
the a’s and b’s are real) and using the result of Prob. 18.19 and the fact that M
is a linear operator, we get
J f*Mf dtt+c*] 2 * Mf dt+c] f*Mg dtt+c*c] 2¢*Me at
J f(Mf)* dt+cJ e(Mf)* dt+c*) f(Mg)*dt+cc*) g(Mg)*dt
Equation (18.91) with ‘¥ replaced by either for by g shows that the first
integral on the left side of this equation equals the first integral on the right
side, and that the last integral on the left side equals the last integral on the
right side. Therefore we are left with
c*] g*Mf dt+cJ f*Mg dt=c] g(Mf)* dt+c*) f(Mg)* dt
Putting c = 1 we get
J o*Mf dt+J f*Mg dt=J g(Mf)* dt+J f(Mg)*dt
Putting c =1 and then dividing by 7, we get (since 1* = —1)
—~J g*Mf dt+J f *Mg dt=J g(Mf)* dt—J f(Mg)*dt
272
Adding the last two equations and dividing by 2, we get J f * Mg dus
J g(Mf)* de .
18.63
(a)
| ftse dt= |". fitxe dx=J|o_ ex* f * dx =!" e(xf)* dx, where
(ab)* = a*b* [the equation after (18.19)] and the fact that x is real
(x = x*) were used.
(b)
Jf*p,g dt=J", f *(h/i)Og/dx)dx. Let u =f
* and
dv = (dg/dx)
dx. Then v = g and the integration-by-parts formula
Judv=
w-Jvdu
gives
J f * pg dt=(hli)f * |". —(h/i)J@, g (of */dx) dx. Equation (18.92)
requires that fand g be well-behaved, which includes the requirement of
quadratic integrability. In order to be quadratically integrable, the
functions
f and g must go to zero as x goes to tee. So Jf * DEUS
lee g[(h/i)df /dx]* dx = ne g(p,f)* dx, which completes the proof.
18.64
We must prove that M(cf, + C) f,) = b(c,f,+c) f,). Using the linearity
equations given after Example 18.5, we have M (cf, +c,f,)= M(c,f,) +
M (cof) = ¢,Mf, + coMf> = cbf, + cobfy = b(c, f, + Coo), where the given
eigenvalue equations for M
18.65
Jgitg, dt= PGs
+kf,)dt = ee
Vet frat hhc
18.66
were used.
di+k] f* f,dt=
fered) Atha =0.
F =x’(1-x),
G=>D™,c,W,,) where w, = 2'’* sin(nnx) . From Eq. (18.98),
eam Jw*F atin
* iF (x* — x*)sin(ntx)
dx. A table of integrals (or use of
the website integrals.wolfram.com or a calculator that can do symbolic
integration) gives Jx? sin kxdx =k? (2—k?x”)coskx + 2k °xsin kx and
Jx? sin kxdx = k-3 (6x —k?x?)coskx +k“ (3k?x° —6)sin kx. We get
C, = —2"?
(nn) 3[4(-1)” + 2], where sinnm =0 and cosmm =(-1)" were used.
So G = D™,(-2)
(nn) °[4(-1)" + 2] sin nx. We set up a spreadsheet with x
values going from 0 | | in steps of 0.02 in column A, the values of F at these
273
the
points in column B, and the values of the first, second,..., fifth terms in
series G in columns C, D, E, F, and G. In column H we sum the first three
terms of the series and in column I we sum the first 5 terms of the series. The
data in columns B, H, and I are graphed versus x on the same plot. The fiveterm sum gives a more accurate representation of F’ than the three-term
function. For example, some values are
| ee
Se mnie
a
le
|
F
[ies [00153 |oos44 [loos [o.1as [0.126 | 0072
x
18.67
18.68
(a)
Since |y|’ dx isaprobability and probabilities are dimensionless, lw?
(b)
has units of length’ and w has units of length”. The SI units of y are
m!”* for a one-particle one-dimensional system.
lw? dx dy dz is dimensionless and y has units of length >”.
(c)
lw? dx, dy, dz, dx2 dyz dz2 is dimensionless and y has units of length”.
The blackbody function (18.2) depends on the combinations of constants
hic? and h/k. In 1900, c was known reasonably accurately, so by fitting the
observed blackbody curves Planck obtained values for both h and k. Use of
R= Nak then gave Na. Use of F = Nae then gave e.
18.69 (a)
(b)
(c)
18.70
E=n7h’/8ma’ and doubling a multiplies E by “4.
E=J(J+1)A7/21 = JJ + I)h?/2ud’ and doubling d multiplies E by 4.
E= hv =h(1/2n)(k/m)'” and doubling m multiplies E by 1/2.
(a) T. (b) T. The future state is predicted by integrating the time-dependent
Schrodinger equation. (c) T. (d) T. (e) F. “Sum” must be replaced by
“product” to make the statement true.
274
(f) T.
(g) F.
(h) F.
(i) F.
Chapter 19
19°
(a) bab)
19.2.
The equation shows that 4m€ has the same units as Q1Q)/rV,
which are
C?7/(m J=C?N"! m™”’, since
lJ=1N m.
19.3
c* = I/po€o, so 1/49 = ploc2/4n = (40. x 10-7 N sIC*)c7/41 = 10°C? N s? C?,
19.4
=
els (c)iT:
Q1Q2/4m€Eor.
(a)
V=(1.602 x 10°” C)*/4m(8.854 x 107!” C?/N-m’)(3.0 x 107! m) =
Gal X10 J = (7.7 x 100.3) (MeVil.602 x10 2.1) = 4. grey
(b)
Let the electrons be numbered | and 2, let the proton be p and let e
denote the proton charge. Then V = V;> + Vip + Vrp = (1/411€9) x
[(-e)7/(3.0 x 107'° m) — e7/(4.0 x 107! m) — €/(5.0 x 107! m)] =
[(1.602 x 107° C)7/4n(8.854 x 1071? C2/N-m?)](-1.167 x 10° m7!) =
E21 x 10 = (2, Fx 10° Te V/N602 < 10) Wyse ev.
19.5
Q/m decreases at high v due to the relativistic increase of mass with speed.
19.6
V=4nr°/3, 80 VauelVatom = Foy¢/P io, = (107? cm)3/(10° em)? = 1 x 1072.
19.7
Vm = M/p = (197 g/mol)/(19.3 g/cm’) = 10.2 cm*/mol.
Vatom = (10.2 cm?/mol)/(6.02 x 107*/mol) = 1.69 x 102 cm? = ¢3 and
€=2.57x 10° cm=2.6A.
19.8
(a) T; (b) T; (c) F, wis #0 atr=0.
(i) F.
19.9
(a) 8; (b) r
(Cc) o.
275
(d) T; (e) F; (f) F; (g) T:; (h) F:
19.10
(a) T; (b) T; () F-
19.11
(a)”
“OFT; 2h
(b)
-5, 4, -3, -2,-1, 0, 1, 2, 3, 4, 5.
(a)
The only n= | state is lso, so the degeneracy is | (..e., nondegenerate), if
19.12
spin is not considered.
(b)
The states 2s, 2p;, 2po, 2p_; have the same energy, so the degeneracy 1s 4.
(c)
The states 3s, 3p1, 3po, 3p-1, 3d2, 3d), 3do, 3d_, 3d_2 have the same
energy and the degeneracy is 9. (The general formula is n’.)
19.13
These are hydrogenlike species, so £ = —(Z7In*)(13.60 eV). The ionization
energy IE is —E for the ground state, n = 1.
19.14
(a)
JIBS 2°(13.60 eV) = 54.4 eV and the ionization potential IP is 54.4 V.
(b)
IE= 37(13.60 eV) = 122.4 eV and IP = 122.4 V.
|AE| =-(13.60 eV)(1.602 x 107! J/1 eV)(1/3” — 1/2”) = 3.026 x 10°”
J=hv
and v = (3.026 x 107! J)/(6.626 x 10°°4 J s) = 4.567 x 10'*/s.
d= clv = (2.9979 x 10° m/s)/(4.567 x 10'*/s) = 6.564 x 10°’ m = 656.4 nm.
ONS
For a hydrogen atom, ft = m.m,/(me + mp) =
(9.1095x1L0*! kg)(1.67265x10~’ kg)
=9.1045x107*! kg
(9.1095x10*!kg +1.67265x107’ kg)
We have a = 4m€oh 7/e* = 4m1(8.854 x 10°!” C*/N-m’)(6.6262 x 107°" J - s)”/
4n°(9.1045 x 10°! kg)(1.6022 x 10°” Cy’ = 5.295 x 107"! m = 0.5295 A.
19.16
This is a hydrogenlike species, so its energy levels are given by Eq. (19.14) as
E =—-(Z’In’)[e?/(4m€o)2a] = -(Z’/n’)[we*/2(4m€0)" h 7]. Let m, be the electron
mass; the positron has mass m,. SO positronium = 7 :/(me + m,) = m,/2, as
compared with pw = m, for an H atom. Since E 1s proportional to pt, each
positronium energy level is half the corresponding H-atom energy. The
positronium ionization potential 1s thus *(13.6 V)=6.8 V.
276
19.17
Using Table 19.1 and Eq. (19.24), (r) = Jw*ry dt=1'(Z/ay x
Jo" I SG etre ""r* sin © dr dO do = (Z’Ima’) Jp re?” dr J" sin 0 dO J5™ do.
A table of integrals gives J ce dz= e(2 /b —327/b* + 62/b* — 6/b*), sO
JS re?" dr = 6/(-2Z/a)* = 3a*/8Z', since e“”" vanishes at r =o. (This
result also follows from the definite integral JF z"e~’* dz = n!/b"*' found in
most tables.) Then (r) = (Z*/ma*)(3a*/8Z*)(2)(2m) = 3a/2Z.
19.18
e = 1 + ib + (i)°/2! + (i)°/3! + (10)*/4! + (iQY/5! +--+ =
1 + id — 07/2! — ig°/3! + 0°/4! + i0°/5! 4 ---.
Also, cos
6+ isin d=(1 — 07/2! + 94/4! -—---)+i(0- 67/314 0/5!----)=
1 + id — 67/2! — i9°/3! + 0°/41.4 9/5! +--- =e".
19.19
2p, = 2"'7(2p, + 2p-1) = 2? (1/80?)(Z/ayrre
"“r sin O (e® + e*). We have
e° +e =cos O +i sin d + cos (-d)
+ isin (0) = cos 6 +7 sind + cos oi sin @ = 2 cos 9, so 2p, = (217/23 n!”)(Zlayrre "rr sin 8 cos =
n'(Z/2a)?xe
"4, where Eq. (19.7) was used. Also, 2py = (2p; — 2p-:liv2 =
(27)77i)(1/820"”)(Zlayr?e
@" r sin 0 (e’® — e"*). We have
ee"
Ove
19.20
(a)
(b)
=cos 6 + isin b— (cos d —i sin o) = 2i sin , so 2Py =
izlayice
a
rsin @ sino = mo (Zi2aye
vest
a,+ib; = az + ib2. We must have a; = az and b, = bp.
(2ny ei" = (2m) tein + 27) (2m) Vein e2nmi 55 | = o2tuni
cos (27m) + i sin (27m), where (19.21) was used. Equating the real parts
and the imaginary parts of this last equation [as shown in part (a)], we
have cos (27m) = | and sin (2mm) = 0. The cosine function equals | only
for angles of 0, +27, +4, +67, ... and the sine vanishes at these angles.
Therefore 21071 = 0, +20, t4n,...
19.21
andm=0, 11,172.78.
Wo). has the form bze~”’; b and d are constants. Along the z axis, x = 0 = y and
ra(e+y +7)? = (27) =|z|. Thus W>,, = bze“""! along the z axis. Near z =
0, e4*! = 1 and Wop. = bz (a straight line through the origin). For large values
of |z|, the exponential causes y to fall to zero. Also, |W>,_ Pelelce |
along the z axis. lw? is parabolic near the origin and is positive for negative
values of z. The graphs are:
277
2
Ilr
19.22
We have 0.9 = (1/32na°) J (2 - rlay’r’e dr JF sin 0 dO J," db =
e98!4(_p? /2@? — ra — 1 —r3,/8a"] + 1, where a table of integrals was used.
Let v = r,/a. We must solve e(4v" +80+8+ v') = 0.8. Trial and error (or a
spreadsheet Solver) gives v = 9.125 and r2; = 9.125a = 4.83 A.
19.23
de 2"V0x = (de ""“/ar)\(Orlax). We have r = (x? + y’ + 2’)'” and dr/dx =
Gs ag BayObra die Se de" sox = (Zxlraye™".
Pe Ol0x" =
(d/dx)(-Zxe?""Ira) = -Ze*"/ra — x{(0/dr)(Ze""/ra)\(Or/dx) =
Ze" Ira + (PLP ayer" + (Zelarye
2". Similar equations hold for 0’/dy"
and 07/02 of e-~”". Then (07/dx* + 07/dy" + O°/0z)e~" =
Bo7em Gira
aye DL ira jen + (Ze + y+ 2varje" =
=376 "Ira + (Lila jew it (Zirayee haw Ze6 “Iran Z
janes =
~2ue*Ze~""/4nepr hh? + (Z’/a*)e“", where (19.14) was used. The Hamiltonian
operator is (19.5) and H wy;=(h7/2p)(0°/dx° + 0°/dy" + 7/dz*)[t (Layee) (Ze7/Aneor)[1 (Zia?"") = -(Ze*/4neor)
mt 7(Z/ay7e "" —
nt (Zlay?( h7/2p)[—2we?’Ze —"/4nerh? + (Zla)e"") =
(h 27a
19.24
ee (Zlaye
oh A ~[(Z267/2(4nep)a][e2(Z/a)ye2e2""] = EW.
(V) = (-Ze*/4neyr) =—(Ze7/4neo) J wt. r'wi; dt=
~(Z'e*l4repa’n) Jon [EIS ere?" r° sin 0 dr dO db =
~(Z'e7/4neoa’n) JF re?" dr J5 sin © dO J2" do. Using either the definite
integral JF r"e”” dr =n'/b""' or the indefinite integral J re?” dr =
—e’'(rlb + 1/b’), we get JF re?“"" dr = a /4Z°. Then (V) =
—(Z'e7/Aneoa’n)(a’/4Z’)2(2n) = -Z’e”/4neoa for the ground state.
19.25
(a)
Wehave: (r) = J WS, T Wop, dt=
(2°/32na°) J5" Ji S5 re?" cos @ r re""* cos 0 r’ sin 8 dr dO do =
278
(2°/32na?) Je re" dr §2" do Jp cos’ @ sin @ d@. A table of definite
integrals gives JF r"e”” dr= n\/b"*' for b > 0 and na
lr re~ dr=5!a°/Z. Let
positive integer. So
t= cos 0; then dt = -sin 0 dO and
J” cos’®@ sin 8 dO = -J;! 1 dt = 2/3. So
(r) = (Z/32ma’)(120a°/Z°)(2/3)(21) = SalZ.
(b)
The 2p- and 2p, orbitals have the same shape and the same size and differ
only in spatial orientation. Since r does not depend on spatial orientation,
(r) must be the same for the 2p, and 2p, states.
GS
Oy Saat
eee
lh Pe" dr Ji" cos’ dbx
if sin’ d@. The r integral was found in (a). A table of integrals gives
J cos’ do = to+ + sin 2 and J sin°@ d0=-= cos 0 - i cos 8 sin’0
and we find (r),, = (Z/32na°)(120a°/Z°)(1)(4/3) = SalZ.
19.26
Equations ee =) and (19.19) give the ground-state H-atom radial distribution
re". The maximum is found by setting the
function as Ryie r= 4(Z/ay
derivative equal to zero: 0 = 4(Z/a) ‘(Ore ma (22Zr’la)e°“"") and r = a/Z. (The
root
19.27
r=0 is a minimum.)
Letc = 2.00 A. To find the desired probability, we integrate y*w dt over the
volume of a sphere of radius c. The angles go over their full ranges and r goes
from 0 to c. Table 19.1 and Eq. (19.24) EN the probability as (1/ma*) x
jel ale en 7 sin
drdo dy =(1/na®) JSr°e?”* dr Jf sin 0 dO J5" do. The
radial integral has the same form as the radial nee in Prob. 19.22 except
that r2, is replaced by c. So J6 re?"
dr=
oe
S|
(+ ac’ + lac+ ta)t+ 4a.
1
The 0 and 9 integrals are given in Prob. 19.22, and the akira probability iS
_%!4(22}a? + 2cla + 1) + 1. We have cla = (2.00 A)/(0.5295 A) = 3.777, and
the probability is 1 — &7°7[2(3.777)° + 23.777) + 1] = 0.981.
19.28
[2"|@|? do = J3" @*@ do = (1/2) J" ein gin® dy = (1/2r0) J2" do= 1.
19290
Ra) el
Dye
rc) al
279
19.30
Let 0 be the angle between the positive z axis and an angular-momentum
vector. For m = +1 in Fig. 19.10, cos @ = L/|L| =
/./2h = 1/V2 =0.7071
and 0 = 45°. Form = 0, 0 = 90°. For m =-1, 8 = 180° — 45° = 135°.
19.31
(a)
(b)
L=rp sin B, where f is the angle between r and p. For circular motion,
the velocity vector v is perpendicular to the radius, and so is
p =mv; thus B = 90° and sin B = 1. Since p = mv, we get L = mur.
The L vector is perpendicular to both r and p and r and p lie in the plane
of the circular motion. Hence L is perpendicular to the plane of the
circular orbit.
19.32
19.33
(a)
From Sec. 19.4, |L| = [J+ )]'"h =0, since / = 0 for the Is state.
(b)
From Sec. 18.3, Bohr had |L| = mur = nh/2n = h/2r for the ground state.
The Bohr theory had the wrong value of |L].
+p)?A = (12)? = V2h = V2(6.626x10™ Js)/2n =
IL] =
TAS log On eis:
19.34
[S|= [s(s + 1]? = [0.5(1.5)]'7(6.626 x 10°" J s)/2n = 9.13 x 10° Js.
19735
Let 0 be the angle between the z axis and a spin vector. For m, = +i, we have
cos 0 = }nl.{3/4h = 1/V3 = 0.57735 and 0 = 54.7°. For m, = — 1
7 >
O'="180?
19.36
19.37
547 = 12553":
(a)
Electronic orbital angular momentum; |L| = [/(/ + Oa
(b)
z component of electronic orbital angular momentum; L; = mh.
(c)
Electronic spin angular momentum; |S| = [s(s + 1)]'7h.
(d)
z component of electronic spin angular momentum; S, = mh.
(a)
For s = 3/2, Eq. (19.29) gives m, = 3/2, 1/2, -1/2, and —3/2. The possible
z components of the spin are m,h . The length of the spin vector is
Sis
ln = sV15h . The possible orientations are
280
(b)
cos@= 1.5/5 15h =
0.7746 and 6 = 39.2°.
19.38
(a) SF: e(b) GF; (oc) 2; (d).T:
19.39
(a)
Neither, since f(2)g(1) # tf(1)g(2).
(b)
Symmetric, since g(2)g(1) = g(1)g(2).
(c)
Antisymmetric, since f{2)g(1) — g(2)fU) = —[f(1)2(2) - gC)f{2)].
(d)
Symmetric.
19.40
(e) Antisymmetric.
The true ground-state energy of He is -79.0 eV (Sec. 19.6). The variational
value —86.7 eV is less than the true E,,; this violates the variation theorem
(18.86), so there must be an error in the calculation.
19.41
There is one electron, soS =s =
anid 2Sue lee
SO
(a) 2S 2c(b). ¢Bew (0) 97D:
19.42
28 +1=4,so S = 3/2. The code letter F means =s
19.43
(a)
(b)
(c)
19.44
(a)
(b)
Total electronic orbital angular momentum; |L| = [L(L + Di
ne
Total electronic spin angular momentum, [S| =[SO+ 17h.
= Msh.
z component of total electronic spin angular momentum; S.
= 2, so\LI= IL Dla G te
D means
Foeteeends <1, so |S) =o), = Selie
281
19.45
(a)
Electrons in filled subshells can be ignored. The 3d electron has s = 2
and / = 2. With only one electron outside filled subshells, there is only one
term, namely *D, with L = 2 and S=%.
(b)
19.46
7P.
Let 0 be the angle between the z axis and a spin vector. For spin function a(1)
and vector S1, we have cos @ = 1/(3/4)""? = 1/3'” = 0.57735 and
@ = 54.7° (as in Prob. 19.35). Likewise, 8 = 54.7° for S2. For a(1)a(2), the
total spin vector S has magnitude 2'* fh and z component fh, so cos @ =
h/2'h = 1/2"? = 0.7071 and 0 = 45° for S. S lies closer to the z axis than do
S, and S5. S; and S; lie on the surface of a cone making angle 54.7° with the z
axis and §S lies within this cone:
v
Zz
19.47
(a)
/=1 fora 3p electron and / = 2 for a 3d electron. The maximum and
minimum L values are 2 + | and |2-1 | and the possible L values are 3,
2, and 1. A 3p electron has s = ; and so does a 3d electron, so the
maximum and minimum S values are 5 + ; and = = 4 ; the possible S$
values are | and 0. (Electrons in filled subshells were ignored.)
(b)
Combining each S value with each L value, we have as the terms:
3F°D,°P,'F,'D, and 'P, where 2S + 1 = 3 and 1 for S= 1 and 0, and P,
D, F denote
19.48
L = 1, 2, 3.
(a)
L=OandS=!,soJ= !12 and the only level is 75,2.
(b)
L=1 and S = 3/2. SoJ=5/2, 3/2, 1/2, and the levels are *Ps/>, *P 3/7, *Py).
(c)
L=3.and S =2, so the levels are 2 Fed aay Mie Ey yey
(d)
L=2 and S = 1s; the levels are Dh, tase Dyk
282
19.49
Let the electrons be numbered 1, 2, and 3. The nuclear charge is 3e. As was
done with He in Eq. (19.32), we use the electron mass m, in the Hamiltonian.
Then H =-(h7/2m,)V? -(h7/2m,)V3 = (h7/2m.)V 5— 3e7/4neor 3e7/Ameor2 — 3e7/A4n€or3 + e/AnEor\> + e’/Aneori3 + e/4n€orr, where r; is the
distance between electron | and the nucleus.
19.50
Let fand g denote the n = | and n = 2 spatial functions, i.e.,
f= (2/a)'” sin (mx/a) and g = (2/a)'” sin (2nx/a). With interelectronic repulsion
ignored, the spatial wave function is a product of one-electron spatial
functions. Analogous to Eqs. (19.42) and (19.43), we form the linear
combinations 27"7[f{1)g(2) + A2)g(1)] and 27 [f(1)g(2) — fl2)g(1)] that don’t
distinguish between the electrons. To satisfy the Pauli principle, the symmetric
spatial function must be combined with the antisymmetric two-electron spin
function (19.38) and the antisymmetric spatial function must be combined with
one of the symmetric spin functions. The approximate wave functions are
therefore
2-7(R1)g(2) + fl) 12" Lax) B) - B(L)a(2))
Ja 1)a(2)
2" [A 1)g(2) — fl2)g(1)
27" TA1)g(2) — fl2)g(D)IBO)B(2)
7” + BU) a(2)]
[ae1)B(2)
2-11) 9(2) — A2)g(L))]2
The first wave function has S = 0. The second, third and fourth have S = | and
have the same energy as one another (since they have the same spatial factor).
According to Hund’s rule, the S = | functions lie lower.
19.51
The ground-state configuration 1s 1s?2s°. To make the approximate wave
function antisymmetric, we use a Slater determinant. Analogous to Eq. (19.51),
we have
2s5(1)Bd)
2s5(1)a(1)
Is(DBC)
Is()ad)
_
fls(2)a(2)
1s(2)B(2)
2s(2)0(2)
— 2s(2)B(2)
Mee
1s(3)a(3)
1s(3)B(3)
25(3)a(3)
25(3)B(3)
1s(4)a(4)
1s(4)B(4)
25(4)a(4)
25(4)B(4)
where N is a normalization constant (equal to 1/24 ).
19.52
H (1s), Li (152s), B (1s?2s?2p), C (1s2s°2p*), N (1s72s?2p), O (1s72s2p'),
S #0
and F (1572s?2p”) all have one or more unpaired electrons and so have
283
and have paramagnetic ground states. He (157), Be (1572s°), and Ne (1572s?2p°)
have all electrons paired, have S = 0 and L = 0 and do not have paramagnetic
ground states. (Ne has two 2p electrons with m = +1, two 2p electrons with
m = 0, and two 2p electrons with m = —1, and so has total orbital angularmomentum quantum number L = 0.)
19.53
We want the energy needed for ;gAr'’* > Ar'**. The ion Ar'’* has one electron
and so is a hydrogenlike species. From Eq. (19.18) with n = 1, the ionization
potential is (18)°(13.6 V) = 4406 V.
19.54
€ = —(Z2,.eff In*)(13.6 eV).
(a)
In Li, the first ionization potential is for removal of a 2s electron,
so 5.4 eV = (Z>,/2”)(13.6 eV) and Zet, = 1.26.
(b)
93eV= (Ze /2°)(13.66 eV) and Zerr = 1.65. (The increase over Li is due
to the poor screening of one 2s electron by the other.)
19.55
(a)
If s = 3/2, the m, values are 3/2, 1/2, -1/2, -3/2. For s = 3/2, the electrons
are still fermions and the Pauli exclusion principle still holds. The four
values of m, mean that 4 electrons (instead of 2) can go in each orbital.
The Ls, 2s, and 2p subshells would therefore hold 4, 4, and 12 electrons
(double their capacities for s = ¥2). The ground-state configurations are
15°, 1s*2s*2p, and 1s‘25'2p”.
(b)
Fors = 1, the electrons would be bosons and there would be no
restriction on the number of electrons in a spin-orbital. The ground-state
configurations would therefore be Is*, 1s°, and Is'’.
19.56
(a)
The outer electron in K is further from the nucleus, so Na has the higher
ionization potential.
(b)
The ineffective screening of one 4s electron by the other makes Zegr
greater in Ca than in K, so Ca has the higher ionization potential.
(C))
Ch
(d)
Kr.
284
1957,
For Z = 10, the figure gives ./e/€,,
= 8, 2.2, and 1.¢. Since €4 = —13.6 eV, we
get €),=—-870 eV, &; =-66 eV, and €2, = —35 eV. Substitution in € =
—(Z24 In*)(13.6 eV) gives Zetts = 8, Zettas = 4-4, Zett.2p ~ 3.2.
19.58
(a) T;
19159
Nitrogen, with 3 unpaired electrons.
19.60
Ionization energy data in Sec. 19.8 show that AE = 5.1 eV for Na > Na’ +e,
(b) T.
so E(Na* +e) > E(Na) and E(Na’ + 2e) > E(Na+e ). Electron affinity data
give AE =-0.5 eV forNa+e — Na, so E(Na )< E(Na+ e ). The lowest-
energy (most stable) system is Na ; the highest-energy system 1s Na’ +2e.
19.61
(a) Sr;
(b) F;
(ce) K;
(d) C;
(e) CI.
Cl and Ar are isoelectronic and the
higher Z in Ar means a smaller size.
19.62
d= es bigx, where gx = Nyexp(—Cir/ao). Replacement of Z by & in the Is
orbital in Table 19.1 gives Ny = 1 '(Cy/ao)””. So o = bigi + b2g2 ++» + bsgs =
rao)
0.7681 ¥(1.417/a)*exp(-1.417 r/ao) + 0.23310-'7(2.377/ao)”exp(-2.377
+ 0.041107 '7(4.396/a9)*exp(—4.396r/ao) —
rao) + 0.0020 "”(7.943/a9)”exp(-7.943rlao).
0.0102 (6.527/ap)>exp(—6.527
19.63
e=ee ROO
1s(2)a(2)
eee
|
25(2)a(2)
2-771 s(1)a(1)25(2)an(2) — 15(2)0(2)25(1)0(1)] =
15(1)2s(2) — 1s(2)2s(1)}o(1)@(2), which is the S = 1, Ms = 1 function in
2-1
Fig. 19.13. Replacement of & by f in the preceding equations shows that D4
equals the S = 1, Ms =—1 function.
r=eae
Is(a(l) 2s(DBC)
1s(2)0(2) 2s(2)B(2)
2-11 s(1)2s(2)a(1)B(2) — 28(1)18(2)B(1)(2)]. Interchange of o and B in D>
gives D3 = 2-/7[1s(1)2s(2)B(1)a(2) — 2s(1)1s(2)ox(1)B(2)]. We have
[ax 1)B(2) + B(L)aX(2)] - 2s(1)1s(2)2"" x
2-2(D, + D3) = 27!?{ 1s(1)25(2)27
285
[B(1)a(2) + a(1)B(2)]}, which is the S = 1, M, = 0 function. Similarly,
ue)
— D3) is found to be the S = 0, Ms = 0 function.
19.64 [27 J" [2 P sin
@dr dO do = J2" do J% sin @ dO Jé r° dr = 2n(2)(a'/3)=
4 na’.3
3
19.65
19.66
(a)
Eis proportional to a7a so Ey > ENa :
(b)
The ionization energy of K”.
(c)
The energy-level spacing for these one-electron species is proportional to
Z°, so V is proportional to Z’. Thus Viet > Vu and Ay >A -
(d)
These quantities are equal.
(a)
Particle in a box; rigid rotor.
(b)
Harmonic oscillator.
(c)
Hydrogenlike atom.
19.67 (a)
dt= nr’ = 210.0010 Ay’ = 4.19 x 10” AY. lw|? = (/na*)e?"? =
[2(0.5295 Ay =2.14A~.
||? dt=9.0x 10”.
(b) fw)? = (/na%)e2 9 = 0.324. dt=4.19x 10° A’.
(c)
lw|? dt=1.4x 10°.
|w|? =1.35 x 10% and |y|? dt=5.7 x 107".
19.68 From (19.25) and (19.19), this probability is Pr=|R1,(r)|°r° dr =
(4a )e?"r
19.69
dr.
(a)
(b)
(c)
(d)
Pr =4(0.5295 A) 2620 109295)(g 100 A)*(0.001 A) = 0.000185.
4(0.5295 A)*exp[-2(0.500/0.5295)](0.500 A)?(0.001 A) = 0.00102.
0.00062.
4.2 x 10°’. (See also Fig. 19.8.)
(a)
dt=dxand-«o<x<o.,
286
(b)
dt=dxdydz.
x, y, and z each range from —° to ©, but since lw? =0
outside the box, we need integrate over only the region
yi b..0 <2
4,08
OR
(c)
19.70
dt=rsinOdrdOdb.
0<rs~, 0<O0<7, 0<o<2n.
Yes. The gravitational force is far smaller than the electrostatic force and so
can be neglected. |Fyray|/| Fal = (Gmemp/r)/(e?/4m€or?) = 4ne9Gm,m,/e? =
4n(8.85 x 10°!* C?/N-m?)(6.67 x 107!! m3/s?-kg)(9.1 x 10°! kg) x
(1.67 x 10°°’ kg)/(1.6 x 107? CP? = 4 x 10.
19.71
Wop. = 1 (21) 7a? re"
(a)
4
cos 8. The maximum value of cos 0 occurs at
6 = 0, where cos 0 = 1. Setting cos 8 = 1 in w and then taking dw/dr = 0,
we get YW, /dr=0= Hie
aie
~ (1/2a)re""]. Solving for r,
we get r = 2a. Setting r = 2a and cos 8 = 1 in Wop, > We get Wmax =
(b)
This problem is incompletely stated, in that it is intended that the
calculations be done for points in the yz plane, where x = 0. In the yz
plane, Wop. Rais 5(z|/aye exp {-5 [G/a)? + (z/ay’]""} =k.
A BASIC program is
15 INPUT "PSI/PSIMAX";K
65 Y=SQR(4*W*W*-Z*Z)
25 PRINT "PSI/PSIMAX=";K
75” PRINT "ZA="Z2 YUWA=
35 FOR )Z = 0.01 TON0 STEFF 0.01
Yormten=
45 W=LOG(Z/2)+1—-LOG(k)
85 NEXT Z
55 IF W<Z/2 THEN 85
95 SLOP
19.72
As ahint, elements were named for the physicists in (a) and (c).
19.73
(a) F;
(b) F.
(c) F; this is true only for identical fermions.
(d) F.
(f) F; (g) F; only well-behaved solutions are possible stationary states.
(h) T.
(i) F; this is true only if the states have the same energy.
287
(e) F.
Chapter 20
20.1.
The table of bond radii in Sec. 20.1 gives the following estimates.
(a)
(b)
20.2.
for CH, 0.77 A + 0.66 A = 1.43 A for CO, and
0.30A+0.77 A =1.07A
0.66 A + 0.30 A = 0.96 A for OH.
0.30A 40.77 A = 1.07 A for HC and 0.60 A+ 0.55 A = 1.15 A for CN.
Each BF bond has some double-bond character, as shown by the Lewis
E:
structure
20.3.
(a)
ee
see
ee
\ee
F=B
and two others.
I’:
The TeBr2 Lewis dot formula has four electron pairs around Te and two
lone pairs on Te. The geometry is bent with bond angle somewhat less
than 10912°.
(b)
Hg has electron configuration - - - 5d'°6s and has 2 valence electrons. So
Hg has two valence pairs in the Lewis structure, and HgCl) is linear.
20.4
20.5
(c)
With 3 pairs on Sn, SnCl2 is bent with angle a bit less than 120°.
(d)
With 5 pairs on Xe, XeF> 1s linear.
(e)
The dot formula has four pairs on Cl, so the ion is bent with angle
somewhat less than 10912°.
(a)
With 5 pairs on Br, BrF; is T-shaped (Fig. 20.2b).
(b)
Three pairs on Ga. Trigonal planar.
(c)
4 pairs on O. Trigonal pyramidal with angles a bit less than 10912°.
(d)
Four pairs on P. Trigonal pyramidal with angles a bit less than 10912°.
(a)
4 pairs on Sn. Tetrahedral.
(b)
5 pairs on Se. Seesaw shape.
(c)
6 pairs on Xe. Square planar.
(d)
4 pairs on B. Tetrahedral.
(e)
6 pairs on Br. Square planar.
288
20.6
20.7
20.8
(a)
5 pairs on As. Trigonal bipyramidal.
(b)
6 pairs on Br. Square-based pyramid.
(c)
6 pairs on Sn. Octahedral.
Each multiple bond is counted as one pair.
(a)
“3” pairs on O. Bent, with angle close to 120°.
(b)
3” pairs on N. Trigonal planar.
(c)
“3” pairs on S. Trigonal planar.
(d)
“3” pairs on S. Bent. Angle close to 120°.
(e)
“4” pairs on S. Approximately tetrahedral.
(f)
“4” pairs on S. Pyramidal with angles close to 109'2°.
(g)
“4” pairs on I. Pyramidal with angles close to 10912°.
(h)
“5” pairs on S. Trigonal bipyramidal.
(i)
“4” pairs on Xe. Trigonal pyramidal with angles close to 10912°.
(j)
“6” pairs on Xe. Square-based pyramid.
(a)
There are four electron pairs around the methyl carbon, so the HCH and
HCC bond angles will be close to 1092°. (Because the four groups
attached to the methyl carbon are not all the same, we cannot expect the
angles at this C to be exactly 10912°.) There are “2” pairs around the CN
carbon, so the CCN angle will be 180°.
(b)
There are “3” pairs around the CH? carbon and the bond angles at this
carbon will be close to 120°. Because of the greater repulsions exerted by
the double-bond’s electrons, the HCH angle will be a bit less than 120°
(c)
and the HCC angles a bit more than 120°. The bond angles at the CH
carbon will be close to 120°. The angles at the CH3 carbon will be close
to 10912°.
The HCH and HCN angles at the methyl carbon are close to 10912°;
There are 4 pairs around the N and the HNC angles are a bit less than
1092":
(d)
Near 1091'4° for the HCH and HCO angles. A bit less than 10912° for the
HOC angle.
(e)
A bit less than 10912° for each FOO angle.
289
20.9
The Lewis dot structure is -§_G=6:
<>»
:6=0-6:-
The bond angle is
close to 120°. The lone pair on the central atom makes the dipole moment
nonzero.
20.10
(a)
Aa 50g = Aatll 39¢,r¢ — Ault 298, pr =
(2(415) + 812 + 2(436)] kJ/mol — [344 + 6(415)] kJ/mol = -320 kJ/mol.
The true value is [-84.68 — 226.73 — 2(0)] kJ/mol = -311.4 kJ/mol.
AHSo, = [946 + 3(436) — 2(3)391] kJ/mol = -92 kJ/mol.
(b)
AH 59, = 2(-46.1) kJ/mol = —92.2 kJ/mol.
In truth,
20.11
We assume tetrahedral angles. As noted in Sec. 20.1, the vector sum of three
CH moments in a CH; group equals the moment of one CH bond. The H3CF
+
=
ee
H—C and C—F. We have
dipole moment is thus the sum of the moments
D=0.4D+ Hg and
1.85D=0.4D+ Ucp and Ucp = 1.45 D. Similarly, 1.62
Ue; = 1.2.D.
20.12
(a)
The net moment of the CH; group equals the CH moment and the dipole
+
moment is the sum of the moments
=
H—C
~
-
and C—-Cl.
Sou =0.4D+1.5D=1.9D.
(b)
w=04D+15D=1.9D.
(c)
wp=04D+1.5D=1.9
on 3 H
(d)
C=
Cl
Ua
N
D. (Here, agreement with experiment is poor.)
aes
Uy = 2c
cos 0+ 2Zucy cos O=
H
2(1.5 D) cos 60° + 2(0.4 D) cos 60° = 1.9 D.
20.13
My = 0. Hence p= 1.9D.
The moments listed in Sec. 20.1 give the H3CCN moment as the sum of the
+
H-C
=
+
=
and C=N
bond moments, namely, as p = 0.4 D + 3.5 D=3.9 D. If we
-
now assume the polarity
+
H—C, then the H3C moment is oppositely directed
290
from the CN moment, and we would have 3.9 D = -0.4D + Ucyn and Uciy =
4.3 D (instead of 3.5 D).
20.14
(a)
Acu/(kS/mol) = 415 — +(344 + 436) = 25.
|xc —xy |= 0.102(25)!? = 0.5
(compared with 0.3 in Table 20.2).
(b)
Aco/(kJ/mol) = 350 - 5 344 + 143)= 1065.
|xc - x0 | = 0.102(106)!”
= 1.0; (compared with 0.9 in Table 20.2).
(c)
Acev(kI/mol) = 328 — +(344 + 243) = 34... lxc-xei| = 0.102(34 +)"
= 0.6 (compared with 0.7 in Table 20.2).
20.15
(a)
For H, there is only one valence electron and (Ene) = 13.6 eV (Sec.
19.3), so xq = 0.169(13.6) = 2.30.
(b)
Li has one valence electron whose ionization energy is given by the table
in seg, 19. Sias 5°46 Vso,
(c)
= 0:169(5-4) =091-
Be has electron configuration 1s?2s° and each 2s electron has ionization
energy 9.3 eV (Sec. 19.8), so xpe = 0.169(9.3) = 1.57.
(d)
xna = 0.169(5.1) = 0.86.
20.16
xq = 1.66(1/0.667)'? + 0.37 = 2.27. x1; = 1.66(1/24.3)'” + 0.37 = 0.94.
Xpe = 1.66(2/5.60)'? + 0.37 = 1.55. xp = 1.66(3/3.03)'" + 0.37 = 2.05.
Ke SALOGUAlite)e ee Beene
PNG 3
O33 Op
4.
20.17
(a)
|xa—xp| +|xp—Xc| =xa — XB + XB — XC =Xa—XC =|Xa— XC.
Substitution of Eq. (20.3) into this equation gives
0.102[Aqp/(kJ/mol)]!”? + 0.102[Agc/(kJ/mol)]'* = 0.102[Aac/(kJ/mol)]"”,
SO
(b)
= Ne
Table 20.1 gives Aon/(kJ/mol) = 175 — =(143 + 159) = 24;
Anc/(kJ/mol) = 292 — 1(344 + 159) = 40.5; Aoc/(kJ/mol) =
350 — +(143 + 344) = 106.5. Ave
ANC = lle (Kiimoly
A\2 = 10.3 (kJ/mol)'”. The relation in (a) is obeyed fairly well.
291
:O:
20.18
H-G-S-9-H
(a)
:O:
(b)
In the dot formula of (a), the S has /2(8) = 4 valence electrons, as
compared with 6 valence electrons in a free S atom. The formal charge
on S is +2 for this dot formula. This is an unlikely value for a nonmetallic
element.
(c)
(d)
The SF, dot formula shows that S can share as many as 12 valence
electrons. (This is due to the presence of 3d orbitals on S.) A dot
zs
formula for H2SO, that gives S a zero formal charge is H -0 ziS-O =H:
:O:
Here S has 2(12) = 6 valence electrons, as in a free S atom.
A dot formula for SO7” that gives S a zero formal charge is
(e)
<O:
‘Ox S-O:
42>
. In addition, there are other resonance structures in which
the double bonds and single bonds are permuted. Each sulfur—oxygen
bond is intermediate between a single bond and a double bond.
20.19
:C=0: The carbon has 2 + /2(6) = 5 valence electrons, as compared with 4 in
the free C atom. The formal charge on C is —1. (This formal charge opposes the
greater electronegativity of O and produces a dipole moment with the polarity
-
+
COx)
20.20
(a)
Because of the electronegativity difference between H and Cl, we expect
the H — Cl bond energy to be larger than the average of the H — H and
Cl — Cl bond energies, so AH” is negative.
(b)
AH° <0, for reasons similar to those in (a).
292
20.21
(a) T (b) F; (c) F; (d) T; (e) T.
20.22
K, =-(h°/2m,)V{ -(h?/2m,)V3; Ky =—(h?/2m,)V2 -(h?/2m,)V2;
Va Ze Zne (ATERaw sonVu= Zhen Alene Zee
Ane re
A
4
Ze 2 MATE aya ne2 WATE Ty, ' Vee
JATear
20.23
(a)
1
KF > K°+F
2
+ K+F. According to the model, the energy needed to
dissociate KF to K* + F is AE, = e7/4meoR, = (1.602 x 107° C)’/
4n(8.854 x 107'* C?/N-m?)(2.17 x 107!° m) = 1.063 x 107!® J = 6.63 eV.
The energy change for step 2 is AE, = 4.34 eV + 3.40 eV =-0.94 eV.
The net AEF is 6.63 eV — 0.94 eV = 5.69 eV.
(b)
According to the model, u = eR, = (1.602 x 10°? @@xgeq107"° m)=
3.48 x 10° C m=
(c)
20.24 (a)
(b)
10.4 D, where (20.2) was used.
Both compounds are essentially completely ionic with charges of +1 and
—1 on the cation and anion. The larger size of Cl. as compared with Fmakes R, greater in KC] and gives KC] the greater dipole moment
(which is approximately eR-).
At R., OE./OR = 0 =-12B/R” + e7/4neoR? and B = e?R!/12(4n€)).
At equilibrium, the electronic energy is Eecq = B/R ug _ e’/4me oR. =
eR’ /12(4mep)R)” — e7/4meoR, = —11e7/12(4neo)Re =
~11(1.602 x 10°'? C)?/12(41)(8.854 x 107!? C?/N-m’)(2.36 x 107!° m) =
-8.96 x 10°’? J
=-5.59 eV. According to the model, it requires 5.59 eV
to dissociate the NaCl molecule to Na* + Cl’. AE for Na* + Cl
4 Na +
Cl is —5.14 eV + 3.61 eV =-—1.53 eV (where data was taken from
Example 20.1 in the text). The model gives D, of NaCl as 5.59 eV — 1.53
eV = 4.06 eV.
(c)
The Pauli repulsion decreases D,. Since 4.06 eV is less than the true D,,
the function B/R'* overestimates the Pauli repulsion. For E, =
AIR” — e/4meoR, we have at R,, 0E./0R = 0 =—mA/R™*! + e7/4meoR? and
A=eR™ |/4meom. So Eeeq= AIR™ — e/4neoR. = e°R™' /4egmR™ e*/Ate
oR, = —(1 - 1/m)e?/4meoR-. Then 4.25 eV = (1 — I/m)e’/4neoR, —
5.14 eV + 3.61 eV. We have e7/4megR, = (1.602 x 10°"? C)*/
293
4n(8.854 x 107!? C2/N-m2)(2.36 x 107!° m) = 9.77 x 10°’ J = 6.10 eV, so
1 — 1/m = 0.948 and m = 19.
20.25
(a) T; (b) F; (c) T; (d) T.
20.26
(a) Even;
20.27
The box size is small enough to be considered “infinitesimal.” The probability
(b) neither;
(d) even;
(c) odd;
(e) odd.
is |y|? dV =(2 + 2S) ‘(sq + Isp)(10 A’), where
S =e *!% (1 + Rlag+ RiGag) At the equilibrium separation of R = 1.06 N=
2.00ao, we find S = e7(1 + 2 + 4/3) = 0.586 and (2 + 2S) = 0.315.
(a) Atnucleus A, ra = 0 and rg = R = 2.00ap, so Isa = 1 (1/ao)*e =
1.466 A®? and sg = 10 '7(1/ao)2e* = 0.198 A>”. Then |y|* dV =
0.315(1.466 + 0.198)? A7(10~ A?) = 8.7 x 10°”.
(b)
At the midpoint of the internuclear axis, ra = rg = R/2 = 1.00ao,
Isa = Isp
= 11a)?! = 0.539 A? and |y|* dV =
(c)
20.28
0.315(0.539 + 0.539)°(10°) = 3.7 x 10°”.
ra =R/3 =2a3, rp = 2R/3 = 4apj/3, 1s, =0.753 AW”,
1sp =0.386 A>”, |w|? dV=4.1 x 107.
The MO electron configuration is (G,ls)(6 *Is)° . To achieve antisymmetry, we
use a Slater determinant. Analogous to Eqs. (19.51) and (20.20) and the Be
wave function in Prob. 19.51, we wnte
o, Is(1)a(l)
o, 1s(1)BU)
o* Isa)
o*ls()Bd)
G,Is(2)a(2)
oO, Is(2)62)
eomlsQ)0
(2 wae oF
0, lsG)aG)
oo, Is(S)pG)e
oO
ls(3)O(
see as (SL
0, 1s(4)a(4)
oO, 1s(4)B(4)
o * Is(4)a(4)
9 * 15(4)B(4)
where N is a normalization constant.
20.29
We use the homonuclear diatomic MOs in Fig. 20.14.
(a)
(G¢1s)°(o* Is).
(b)
(Ggls)(o* 1s)°(o,2s).
294
i252)
(c)
(dels) (6* 1s)'(6,2s)°(o*
2s)”.
(d)
(d¢1s)(6* 1s)°(6,2s)(6 * 25)°(11,2p)*.
(Ge1s)"(o* 1s)°(6,2s)° (6*2s)"(1,2p) (G¢2p)’. He5 has an unpaired
electron and so is paramagnetic. All the others have filled subshells and
are not paramagnetic.
(e) with (1* 2p)’ added.
20.30
(2-—1/2=%.
(b)7(4-—2)/2=1.
(c) 4=-4)/2=0.
(d) 8-—4)/2=2.
(10 — 4)/2 = 3 (in agreement with the dot formula :N=N:).
20.31
(f) 1.
The N2 MO configuration is given in Prob. 20.29e. The highest occupied
N>2 MO is a bonding MO, so N3 has one less bonding electron than No.
Therefore N> has the higher D,.
(b)
The O2 MO configuration is shown in Fig. 20.17. The highest-occupied
shell, 1% 2p, is half-filled and is antibonding. Therefore O; has one
fewer antibonding electron than O2 and has two fewer antibonding
electrons than O;, so O3 has the highest D,.
20.32
(H +c)y=
H wt+ev =Ey+cw =(E+0)y, where the definition of the sum
of operators was used.
20.33
(a)
We feed the valence electrons into the MOs of Fig. 20.21, where n = 2
and n’ = 2. NCI has 5 + 7 = 12 valence electrons and has the valence-
electron configuration (os)"(o*s)*(mp)*(op)’(m*p)’. NCI* and NCI have
11 and 13 electrons, respectively, and have the configurations
(o8)°(6*s)
(mp)"(op)"(n*p) and (os)"(o*s)"(mp)
(op) (rp),
respectively.
(b)
(8 — 4)/2 = 2 for NCI (which is similar to O2); (8 — 3)/2 = 2.5 for
NGEAiSi- 5)/2.= 1. 5dfor NCI.
(c)
Each of these species has a partly filled m* subshell, so each has
S #0 and each is paramagnetic.
295
20.34
(a)
Cls, C2s, C2p,, C2p,, C2p., Ols, O2s, O2p,, O2py, O2p:.
(b)
Cls, C2s, C2p., O1s, O2s, O2p, contribute to
o MOs. C2p;, C2py, O2p;,
O2p, contribute to 7 MOs.
307,
20.35
oe
=
2 oy
.
rr
—
|
|
|
|
‘
|
Be
;
|
|
|
|
|
20.36
(a) g,o.
(b) 2,5.
(c) g,d5.
(d) g,7.
20.37
(a) ©, since it has no nodal planes that contain the internuclear (z) axis.
(b) 1, since it has one nodal plane containing the internuclear axis.
(d) o.
(e) o (see Fig. 19.6.)
the internuclear axis.
20.38
(a)°
(C) 1.
(f) 5, since it has two nodal planes containing
(g) 5. (h) 7.
(i) 7.
(Halsiiigis. Gis; C2s, Gln, C2p G2n7 Olss O25. O27 O2p-0) 2p.
H
(b)
The dot formula
e = O; suggests the following description of the
H
occupied localized MOs. We use sp” hybrid AQOs on C to form the CH
bonds and the o bond of the double bond. These sp hybrids are formed
from C2s and the in-plane p orbitals C2p, and C2p-.. The bonding coMO
between Ha and C is a linear combination of Hals, C2s, C2p,, and C2p..
The bonding 6 MO between Hg and C 1s a linear combination of Hgls,
C2s, C2p,, and C2p.. As we did with carbon, we form in-plane sp”
hybrids on oxygen, using O2s, O2p,, and O2p.; these hybrids go to form
the o bond of the double bond and the lone-pair AOs on oxygen. Overlap
between C2p, and O2p, forms the 1 bond of the CO double bond. The o
bond of the CO double bond is formed by overlap of those sp’ hybrids on
C and O that point along the z (CO) axis; the C2p, and O2p, AOs do not
contribute to these sp” hybrids—the C2p, and O2p, AOs each have one
nodal plane containing the z axis and cannot contribute to the CO o MO.
Therefore the CO localized 6 bonding MO is a linear combination of
296
C2s, C2p,, 02s, and O2p-. The lone-pair localized MOs on O are formed
from the O2s, O2p,, and O2p- AOs. There is an inner-shell localized MO
on C that is essentially identical to the Cls AO and an inner-shell
localized MO on O that is identical to Ols.
20.39 (a)
3H2(g) + 6C(graphite) —“-» 6H(g) + 6C(g) —2> CoHo(g). Appendix
data give AH”59, = [6(217.96) + 6(716.68) — 3(0) — 6(0)] kJ/mol =
5607.8 kJ/mol. Viewing C¢H, as containing three CC single bonds and
three CC double bonds, we use the bond energies in Table 20.1 to get
AH j,59g = —[6(415) + 3(344) + 3(615)] kJ/mol = —5367 kJ/mol. Then
AAT593 = AH,393 + AH;993 = 241 kJ/mol. The Appendix gives the
experimental value as 83 kJ/mol, so benzene is far more stable than it
would be if it were composed of isolated single and double bonds.
(b)
5H2(g) + 6C(graphite) —*> 10H(g) + 6C(g) —-> CeHio(g).
Appendix data give AH’59, = [10(217.96) + 6(716.68) — 5(0) — 6(0)]
kJ/mol = 6479.7 kJ/mol. The bond-energy table gives AH;, 592 =
—[10(415) + 5(344) + 615] kJ/mol = -6485 kJ/mol. Then Ay 59. =
AH 4 90g + AH 4 29g =~ —5-3 kJ/mol. The Appendix gives the experimental
value as —5.4 kJ/mol.
20.40
(a)
The following sketches show a view from above the molecular plane.
The dashed lines denote nodal planes perpendicular to the molecular
plane. The molecular plane is a nodal plane for each MO. The lobes
below the molecular plane have signs opposite those of corresponding
lobes above the plane.
=
-
oS
———
ee
297
+
|
(b)
204le
The MO p, + p2+p3 + ps + Ps + Po has the fewest nodes and builds up
the most electron probability density between the nuclei; this MO is
lowest in energy.
et {= (2s. 2p.) V2 . Along the z axis we have x = 0, y=0, r=
Raley (22)"? =|z| and f= 22 (1/4)(2m)y"? x
(i+
[a27(\z|/a — 2)188 + ze t19] = (1/870?)e | (w + |w| — 2), where
w = da. We have 1/81'7a"” = 0.183 A>? and we find
f
Za
0.050
4
-0.082
—3
-0.135
—2
—-0.222
—|
0.366
0
0
-0.143
Ope
y
z/a
0.086
135
0.135
2
0.163
a2
0.149
4
0.120
5
0.091
6
0.066
7
The graph shows a sharp negative peak at z/a = 0 anda broad positive region
for za > 1.
O28,
20.42
(a) T; (b) T; (c) F.
20.43
(a) Yes.
(b) No, since the HOOH dihedral angle must also be specified. Changing this
dihedral angle without changing any bond distances or angles changes the
structure.
(c) Yes. Imagine adding the H atoms one a time. When H, is added to N, the
NH, bond distance specifies the structure. When H) is now added, the structure
is fully specified by the two bond distances and the HNH angle (as 1s true for
298
H2O). When the third H is added, the values of the N-H; bond distance, the
H,NH3 bond angle and the H2NH; bond angle give three conditions that
determine the three spatial coordinates of atom H3.
(d)
No. The CICCC] dihedral angle must be specified.
(e) No. Twisting one CH: group with respect to the other generates different
structures but does not change any bond distances or angles.
20.44
Let Tne Oy) = lay? sin(n,7x,/a) (for 0 < x, <a) denote a particle-in-a-box
wave function, where n, is the quantum number and x; is the coordinate. For
noninteracting particles, the wave function is the product of wave functions for
each particle, and one's first impulse might be to take y as ikea heel eey
However, this function is not symmetric when n, # n2, and spin-O particles are
bosons and require symmetric wave functions. Therefore, the normalized wave
function [analogous to (19.42)] is y= 27°, (x1) fu, (42) + {peer Call.
To get Pp, we integrate lw? over the coordinates of all particles but one and
multiply by the number of particles: p =
Aer Nol
Ca Ut AGS nate ialCNP ACA Nid @adnnul ciate diniein Vi, (anil.
= fn 1)Jo Sr (42) Oxo + 2F, (m1) Fp, 150 Fa, (82)Fa,(2) dey +
Fr (41) Jo Sn,2) ep
Use of the normalization condition for
fand the orthogonality property (18.36)
gives P(x,) = iti(x,)+0+ fi,(x,). Dropping the unnecessary subscript |, we
have p(x) = f, (x) + fy, (x) = (2/a)sin® (nyx/a) + (2/a) sin? (nynx/a) for
0 <x
<a, and p = 0 elsewhere. Also, jige P(x) dx = ik OC) xen
aelee
because of normalization.
20.45
(a) Yes;
20.46
(a)
(b) no;
(c) yes;
(d) yes.
E, in (20.8) depends in the coordinates of the nuclei; for fixed nuclear
positions, F, is a constant.
299
(b)
The probability density p is a function of the spatial coordinates x, y, z.
(In a molecule, it also depends parametrically on the nuclear locations.)
(c)
Exc in (20.44) is the sum of definite integrals (which are numbers) and E,,
is a constant for a fixed nuclear configuration. (Of course, it changes
when the nuclear positions change.)
(d)
Vxe in (20.49) is a function of x, y, z (and also depends parametrically on
the nuclear coordinates).
20.47
Multiplication of (20.7) by yw and integration over all space gives
dt=E, | wy, dt. Because of normalization, the integral on the
right side equals 1 and J ww, dt=E,. Use of (20.6) gives E, =
J VAL,
[WHR + Ve + Vee t+Vaw We at=JysKey, dtl wiVyeW, at +
Jwev,w, dt + JweVyyW, dt = (K,) + Vue) + Veo) + Vin)
20.48
(a) T;
(b) T; (c) F; (d) T; (€&) T.
20.49
The potential energy of interaction between the infinitesimal charge elements
dQ, and dQ; of the continuous distribution is given by Eq. (19.1) as
dQ, dQ>/4m€ori2, where rj is the distance between dQ, and dQ>. If p is the
electron probability density, then —ep is the charge density (the charge per unit
volume) and multiplication by the infinitesimal volume gives dQ, =
—ep(x1, y1, 21) dx, dy; dz, and dQ2 = —ep(x2, y2, Z2) dx2 dy2 dz. Integration of
dQ, dQ>/4m€or;2 over the coordinates x2, y2, 22 of dQ2 gives the energy of
interaction of dQ, with the entire continuous charge distribution. If we then
integrate over the coordinates x), yi, Z; Of dQ), we get the total energy of
interaction of all the infinitesimal elements of charge with one another, except
that we must divide by 2 to avoid counting each interaction twice. Thus the
interaction energy is given by Eq. (20.48).
20.50 We have g = -(3/4)(3/n)'?p** so 0g/dp=— (3/4)(3/m)'°(4/3)p" =
—(3/n)'3p'3 and dg/dp, =0. Then, since F= Ey?” , we have SEL?’ /Sp =
= Gino.
;
300
20.51
There are 2 pi electrons in each double bond, plus the lone pair on N, for a total
of 2k + 4 pi electrons. These fill the lowest k + 2 pi MO’s and the transition is
fromn=k+2ton=k +3. There are 2k + 2 conjugated single and double
bonds and addition of an extra bond length at each end gives a box length of
(2k + 4)(1.40 A) = a. Then |AE| = (h?/8ma?)[(k + 3)” — (k + 2)°] =
(2k + 5)(h7/8ma*).
|AE| =hv =hc/d andi = hcl |AE| = 8marcl(2k +5)h =
8m(2k + 4)°(1.40 A)c/(2k + 5)h = 8(9.11 x 107°! kg)(2k + 4)°(1.40)? x
(107'° m)?(3.00 x 10° m/s)/(2k + 5)(6.63 x 107° J s) =
(64.6 nm)(2k + 4)°/(2k +5). Fork = 1, A = (64.6 nm)(2 + 4)7/(2 + 5) = 332 nm.
20.52
For CHy(g), Apt 59g = Na(Ee — Da Naa) =
(6.022 x 10*/mol)[-40.19517 — 1(-37.88449) — 4(-0.57077)] hartrees x
(4.35974 x 107'® J/hartree) = -72.5 kJ/mol. For H»O2(g), Ee — Xa Nae =
[-150.76479 — 2(-0.57077) — 2(-74.78852)] hartrees = 0.04621 hartrees and
Ad593 = -121.3 kJ/mol.
20.53
The molecule is CH3CHO, with H4, H5, and H6 bonded to C1 and H3 and O
bonded to C2. The atoms Cl, C2, H3, and O lie in the same plane. Atom H4
eclipses the O atom and H3 is staggered between H5 and H6.
20.54
(a)
The OH hydrogen lies on the same side of the C—O bond as the carbonyl
oxygen in one conformer and on the opposite side in the other conformer.
For the first-mentioned conformer, a possible Z-matrix is
Gal
7a
Lad
Fee ee
Ha 2-096
Hoa
C7
3
1
2
20.0
105.0737 0.0
120. 0a su
where bond lengths were estimated from the bond radii in Sec. 20.1. The
Z-matrix of the second conformer is the same except that the dihedral
angles in row four and row five are changed from 0.0 and 180.0 to 180.0
and 0.0, respectively.
(b)
One possible answer is
301
Gl
Xo eelewleO
Ome L232
90:0
O4N lo 232890035
20.55
20.56
130.0
For the Windows version of Gaussian, the input in Fig. 20.37, appropriately
modified, can be used. For part (a), only the basis set needs to be changed. For
part (b), HF/3-21G in the Route Section should be changed to B3LYP/6-31G*.
In using Gaussian, you are asked to specify a filename for the output. Note that
filenames with an asterisk are forbidden. Results are (atoms as in Fig. 20.38):
B3LYP/6-31G*
HF/3-21G
_HF/6-31G*
HF/6-31G**
wD
ep
ley
1.83
1.69
R(COY/A
1.441
1.399
1.398
1.419
R(OH)/A
0.966
0.946
0.942
0.969
R(CH4)/A
1.085
1.088
1.088
1.101
R(CH5)/A
1.079
1.081
1.082
1.093
ZCOH/?
110:3
109.4
109.6
107.7
ZOCH4/°
2
112.0
11221
112.7
ZOCHS/°
106.3
1072
107.3
106.7
D(H6COHS3)
61.4°
Ole
Gle23
61:53
D(H5COHS3)
180°
180°
180°
180°
(a)
The Z-matrices of Prob. 20.54a are used. One finds —188.7623096
hartrees for the conformer with the OH hydrogen near the carbonyl O
and —188.7525454 hartrees for the other conformer. This is an energy
difference of 0.0097642 hartrees. From Eq. (20.1) and the equation after
(19.18), one hartree corresponds to 27.211(23.061 kcal/mol) = 627.51
kcal/mol, so the conformer with the OH hydrogen close to the CO
oxygen is predicted to be more stable by 6.1 kcal/mol.
(b)
The calculated energies are —189.7554562 hartrees for the conformer
with the OH hydrogen near the carbonyl O and —189.7471663 hartrees
for the other conformer. This is an energy difference of 0.0082899
hartrees or 5.2 kcal/mol.
302
20.57
CS Chem 3D Net Version 4.0 gives R(CO) = 1.427 A, R(OH) = 0.942 A, all
CH distances = 1.112 A, ZCOH = 107.2°, ZHCO = 108.6°.
20.58
CS Chem 3D Net Version 4.0 gives a steric energy of 2.174 kcal/mol for the
geometry-optimized trans (anti) conformer and 3.049 kcal/mol for the
optimized gauche conformer. The predicted energy difference is 0.88 kcal/mol,
with the trans being of lower energy.
20.59
CS Chem 3D Net Version 4.0 gives a steric energy of 1.311 kcal/mol for the
cis isomer and 2.509 kcal/mol for the trans isomer. The cis isomer is predicted
to be more stable than the trans by 1.2 kcal/mol, which is not what one would
expect from chemical intuition. [For references on this "cis effect, " see N. C.
Craig et al., J. Phys. Chem. A, 102, 6745 (1998).]
20.60
(a)
There are seven bonds and hence 7 terms in Ver;.
(b)
There are 6 bond angles at each carbon atom and therefore a total of 12
bond angles and 12 terms in Vpeng. [Each carbon is bonded to four atoms
and the bond angle at a carbon is described by specifying the two atoms
at the ends of the angle; the number of ways of selecting two objects
from four objects is /2(4)(3) = 6.]
(c)
Each Cl at one carbon has a 1,4 relation with three C] atoms on the
second carbon. There are thus a total of 3(3) = 9 terms in Vioys.
(d)
There are no 1,5 or higher interactions and as in (c), there are 9 pairs of
atoms that have a 1,4 relation, so there are 9 terms in Vygw.
(e)
20.61
9, as in (d).
(a) =400 kJ/mol;
(b) =(3/2)RT=4 kJ/mol;
(c) = 12 kJ/mol;
(d) ~600 kJ/mol;
(e) = 1300 kJ/mol, which corresponds to
13.6 eV/molecule. So (b) < (c) < (a) < (d) < (e).
20.62
(a)2ie(b)ere
(c) F.
(d) Fo
(e) Po
303
(f) BF. (g) 1.
th),
Gi) T-
Chapter 21
21.1
(a),
21.2
E = (1.00 eV)(1.602 x 107'? J/1 eV) = 1.602 x 10° J =hv, so
v = (1.602 x 107'? J)/(6.626 x 1074 J s) = 2.42 x 10" s™.
A =clv = (2.998 x 10'° cm/s)/(2.42 x 10'4/s) = 1.24 x 107 cm.
o = 1/A = (1.24 x 10% cm)! = 8.06 x 10° cm".
21.3
4(b) Ee (Cym
(ak
eae) aI
Cu,0 = c/n y,9 = (2.998 x 10 cm/s)/1.33 = 2.25 x 10'° cm/s. The frequency
is unchanged in water and Ay. =Cy,0/V = CH,0Avacl€ = Avac/N H,0 =
(589 nm)/1.33 = 443 nm. V = Vvac = C/Avac = (2.9979 x 10* m/s)/(589 x 10” m)
= 5.09 x 10'* Hz.
21.4
(a) F; (b) F; (c).T;. dd), T;, (eT.
2165
(a) sor Hz;
21.6
Using identities before Eq. (21.6), we get Umn =
(c) m/s;
(b) m!:
(e) m’/mol.
(d) no units;
(20/a) - J2 {x cos [(m — n)nx/a] — x cos [(m + n)nx/a]} dx =
O
a’
= Gee ae
a|(m-n)
(m—n)Tx
xa
ee
eee
Tt
ae
— —————
(m+n)? 1"
(ie
Dh)IX
aa
(m—n)Tt
a
a
: (m+n)Tx
xa
_ (m+n)tx
- ———__ sin ——_
cos ———a
(m+n)tt
a
||"
0
which reduces to Eq. (21.6), since sin jm = 0 forj an integer and cos 0 = 1.
2157
Viight = |AE| /h = (v2 + 5 hvviv = (Ui 5 hvviv\/h = (V2 — V1) Vib = vib.
21.8
(a)
{i = Ox, since this is a one-particle, one-dimensional system.
O Sway, dx = O(n)" (403/ny"4 Jee dx =
304
Qox(2/10)'"°2(2!)r0'"7/23 1103? = O/(2a)"”, where integrals 1 and 3 of Table
15.1 were used.
(b)
O(w/n)'*(ov/4n)!4 lee (20x° — xe ~ax! dx = 0, where integral 4 was used.
(c) O(orn)'*(cu9n)"4 J@, (2082x4 — 3a!7x7)e"" dx =
O(o/32)'?[2072(4!)07/252
105? — 3q!72(2 1/7/2311 !02/7] = 0. The
results (a)—(c) are consistent with Av = +1.
21.9
Ec-—Exy=helkac.
Ec-Ep=hclApc.
Eg — Ea, =hec/Aap =
(Ec — Ea) — (Ec — Ep) = hclAac — hclApc, 80 W/Aap = UAac— Age =
(485 nm)! — (884 nm)! = 0.000931 nm‘! and Aap = 1075 nm.
21.10
Since E increases as n* + n, the spacing between levels increases as n
increases. Therefore the lowest-frequency absorption is due to the transition
from n= 1 ton =3. We have Viowes: = 80 GHz = |AE |/h = [b(3)(5) — b(1)B)\/h
= 12b/h and b = (80 GHz)h/12. The next-lowest absorption frequency is that
from n = 2 ton = 4 and its frequency is v = |AE |/h = [4(6)b — 2(4)b]/h =
16b/h = 16(80 GHz)h(12)
'/h = (16/12)(80 GHz) = 107 GHz.
21.11
(a)
|AE |ie (h7/8ma*)(n 3 —n ;Vis (h/8ma’)(n 3 —n a =
(6.626 x 10°" Js)(n3 —n})/8(9.109 x 107°! kg)(2.00 x 107!° m)? =
(2.273 x 10'°/s)(n5. —n;). The selection rule is that An is odd, so the
lowest frequencies result from n = 1 > 2,n=1—4,andn=1-— 6. The
frequencies are (2.273 x 10'°/s)(4 — 1) = 6.82 x 10'° Hz, 3.41 x 10!° Hz,
and 7.96 x 10'° Hz.
(b)
The smallest values of ns - n> with nz — n, odd are forn = 1 3 2,
n=2—3,andn=3
— 4. We get 6.82 x 10s Hz, 1.14 x 10!° Hzaand
1.59 x 10'° Hz.
21.12
v=
(Eupper = Eiewern)/h = [ANupper(upper + 4) — Attiower(Miower + 4)\/h =
a[(Niower + 3)(Niower + 3 + 4) — Mower(Mower + 4)]/h = (ONtower + 21)a/h, where
Tike ees:
21.13
3 De
Toe
Using u and @ for upper and lower, we have
v=h'A[K,(K, + 3)—K (K, + 3)) = An UK, + (K, +4)-K (K, +3)] =
305
2Ah"(K, +2), where K, = 1, 2,3,.-.. Then 60 GHz = 2Ah”'(4);
A=7.5h GHz: v=(15 GHz)(K, +2). 135 GHz = (15 GHz)(K, +2) and
Keats
21.14
T = 10%. For A = 0.1, T= 10°! = 0.79 and 21% is absorbed. ForA = 1,
T = 0.10 and 90% is absorbed. For A = 10, T= 10°'° and 99.99999999% is
absorbed.
21.15
aan ©
COnIV = PIREand T= B/ino =a ele
(a) Pp, ¢/RT = (10° dm*/mol-cm)[{(10/760) atm](1.0 cm)/
(0.08206 dm?3-atm/mol-K)(298 K) = 5.35. T= 10°°* = 4.2 x 10°.
(b)
T=10>°%=16x10™.
21.16
T = Dl/h.o = 10-8".
(a) ecg’ = (150 dm*/mol-cm)(10~ mol/dm’)(1.0 cm) = 0.15 and
T=10, °=0.71,
(b) T=10'° =0.033.
21.17
A = log (Ip/I) = log (T') = log (1/0.083) = 1.08.
c = (0.080 g/cm3)(1 mol/14600 g)(10° cm*/1 dm*) = 0.0055 mol/dm*. T= I/Ip
= 10! and € = -(1/cl) log T = -(0.0055 mol/dm*) (0.010 cm)! log 0.083 =
2.0 x 10 dm? mol” cm".
21.18
A2/A, = [log (h.o/b)2\/[og Ch.o/D)1) = €2¢B.2!2/€.¢B
1h = Cp.2/Ccp.1 = 2.
So log (h.o/I))2 = 2 log (1/0.60) = 0.444 and Kh o/I, = 2.78. We have
T = 1/2.78 = 0.36 and 36% of the light is transmitted.
21.19
Let the subscripts 3 and 4 denote Fe(CN) re and Fe(CN) ar respectively. Use
of A = (€3c3 + €4C4)l and c3 + c4 = 1.00 x 10°° mol/dm’ gives 0.701 =
[(800 dm?/mol-cm)c3 + (320 dm*/mol-cm)(0.00100 mol/dm? — c3)](1.00 cm).
We find c3 = 7.94 x 10“ mol/dm’. Then c3/c3.9 = 0.000794/0.00100 = 0.794.
The % reacted is 20.6%.
306
21.20
(a) T; (b) T; (c) F; (d) F; () T; (f) T; (g) F; (hb) T.
21:21
(a) F:
“(b) T;°
(c) F.
21.22 Division of Eq. (21.25) by he gives Do/he = D,/he - 19, +45 ,x.
(a)
Do/he = [79890 — +(2359) + =(14)] cm! = 78714 cm;
Do = (78714 cm™')(100 cm/m)(6.6261 x 10~** J s)(2.9979 x 10° m/s) =
1.5636 x 10°'® J = 9.759 eV.
(b)
Do/he = [90544 - +(2170) + =(13)] cm”! = 89462 cm”;
Do = 1.7771 x 107'8 J = 11.092 eV.
21.23
(a)
D, is the depth of the electronic energy curve E, and k, equals E, (R,). In
the Born—Oppenheimer approximation, E,(R) is found by solving the
electronic Schrodinger equation (20.7) in which the nuclei are fixed; the
nuclear masses do not occur in (20.7) and (20.6). Hence, E,(R) is the
same for 7H*°Cl and 'H™Cl, which have the same nuclear charges. From
Eq. (21.25), Do differs from D, by the zero-point vibrational energy. The
vibrational frequency v, equals (1/2m)(k/)"” * The reduced mass u differs
substantially for 7H*°Cl and 'H*°Cl, so their v,’s differ and their Do’s
differ.
(b)
For 'H*°Cl, Do/hc = D-/hc - -V, + -V,x, =
[37240 — +(2990.9) + 7(52:8)] em" =35758 em);
Do = 7.103; x 107? J = 4.4334 eV. From Eqs. (21.23) and (18.79):
VepcvVeuc! = (MucYMpc)7; Hci = 1.00782(34.969)(g/mol)/35.977Na =
(0.97959 g/mol)/Na; Mpc = (1.9044 g/mol)/Na. So V, pc =
(2990.9 cm™!)(0.97959/1.9044)'* = 2145.1 cm”. Also, De.pci = Dene.
For “HCl we then have Dolhe = [37240 — 5(2145) + oa)Lcms =
36181 cm™! and Dp = 7.187 x 10°!” J = 4.486 eV (where we neglected the
change 1n Vx).
21.24 VYhv, = YekA*. But v, = (1/2m)(k/p)'’, sok = 417pv2; “hv, = 2n’uv? A’ and
A = (h/4ruv,)!””. For H°°Cl, p = (1.0)(35.0)g/(36.0)(6.02 x 107°) =
307
1.6 x 104 g and v, = (2.998 x 10!° cm/s)(2991 cm”) = 9.0 x 10" s', so
A =[(6.63 x 1074 J s\/4n2(1.6 x 1077 kg)(9.0 x 10'/s)]'? = 1.1 x 107" m=
0.11 A. For '*Ny, p= 14(14)g/28Na = 1.16 x 10 g, v.=7.1 x 10° s™ and
A=0.045 A.
21.25
Use of (21.31) gives A = c/v = c/2(J + 1)B, so 2.00 cm = c/2(3)B and B =
c/(12.0 cm). We then have A = c/2(J + 1)c(12.0 cm) |= (6.0 cm)/(J + 1).
hoes? = (6.0 cm)/7 =.0.86 cm.
21.26
From Eqs. (21.31), (21.33), and (21.15): Vyos+1 = 2V + 1)Bo = 2Bo =
2h/8tIp= h/4r? wR? and Ro = (h/pv)'"/21.
(a)
Use of the table of atomic masses gives Hh = mym)/(m, + m2) = -
(b)
(1.0078250)(78.91834)g/(79.926165)(6.02214 x 10°°) =
1.652431 x 10-4 ¢ for 'H’°Br and p = 1.652945 x 10 g for 'H®'Br.
For 'H’’Br, Ro =
[(6.62607 x 107° J s)/(1.652431 x 1077 kg)(500.7216 x 107/s)]'"/2n =
1.424257 A. For 'H®'Br, we get Ro = 1.424257 A.
ForJ=1 to 2, v = 2( + 1)Bo = 4Bo = 2Vs-0-41 = 2(500.7216 GHz) =
1001.4432 GHz with centrifugal distortion neglected.
(c)
For the J=0
to | transition, Vps/Vusr = 2Bo,pp/2Bo.HBr = LHBR/Mpsr, Since
Ro is essentially unchanged on isotopic substitution.
Upp; = (2.014102)(78.91834)g/(80.93244)(6.02214 x 10°*) =
3.261264 x 10°" g. UnBr/Mpsr = 1.652431/3.261264 = 0.506684 and
Vpsr = 0.506684(500.7216 GHz) = 253.708 GHz.
21.27
From Eq. (21.31), Vsss41 = 2BoV + 1), so Bo?’ K*’Cl) = (22410 MHz)/2(3) =
3735 MHz.
(a)
v= 2(3735 MHz)! = 7470 MHz.
(b)
The reasoning in Prob. 21.23a shows that R, for the two isotopic species
is the same; further, Ro should differ only very slightly for the two
species (see, for example, Prob. 21.26). Equations (21.33) and (21.15)
then give Bo?’ K*°C1)/Bo(?K*’Cl) = In??K* C/o? KCl) =
w??K7Cl/uCPK
Cl) = 18.9693/18.4292 = 1.02931, where the reduced
masses were found from mamp/(ma + mp). Then BoC KRG) =
308
1.02931(3735 MHz) = 3844+MHz and vj-9..; = 2(3844+ MHz) =
7689 MHz for *°K*°Cl.
21.28
The equation preceding (21.37) is ¥ = V origin + BU’ + 1)- BS’ + 1).
For P. branch lines, J =J” —-1 and/(V +1)-J
4g =e a
21.29
cae TG
ne 1) = a) Ie so
Vp =v
J
+ 1)=
origin — 2B ae
Eq. (21.39) multiplied out gives Vp = V origin + 2B, J” + 1) Gv’ (J? + 3S” +2) + &, 0" + I") — &, (J + 1). With centrifugal
distortion neglected, Eq. (21.26) gives Vp = (Eysiy — Ey
Ve +2
a, (v’ o
)— Vixt0 +o)
ys" cs Due
. 2\s
Gi. (WH EF #1) = 910
eB Poe
he =
2)
Vv,(0” - ‘) ‘2 Voxe(0" +
0) = V_2(0° £0 =
Dios BJS’
+ 1) fs
A=) +
B,(2J” +2)+ &, 0°77 +I") — & v(I + 3” + 2) - &, Ul” + 1), which Eq.
(21.34) shows to be the same as the above multiplied-out form of Eq. (21.39),
verifying (21.39). When multiplied out (21.40) is
Vp =Voigin — 2B” + GU" - I) + GW" +I) + &, J". Then Fp =
(Ey. — Eyyhe
and use of (21.26) leads to the multiplied-out form of
(21.40).
21.30
ve = (1/2m)(ke/)!? and ke = 41°v 2p = 41°? V2p = 400°(2.998 x 10'° cm/s)” x
(1580 cm™)°(15.99491) g/2(6.0221 x 107) = 1.176 x 10° dyn/cm =
11.76 mdyn/A = 1176 N/m, where we used LL = mama/(ma + ma) = ma/2.
21.31
(a)
From Eq. (21.34), Vorigin(O > LI = V, —2V,%e and Vorigin (0 — 2) =
DUNE GY fxHence BV one, (0 => 1) Vote tO > aaa Vee
3(2886.0 cm”) — 5668.0 cm™ = 2990.0 cm'. Then V, xe =
LV, = Vorigin (0 > 1} = 5(2990.0 - 2886.0)em™ = 52.0 cm”.
(b)
From (21.34), Voigin (0 > 6) = 6(2990.0 cm) — 42(52.0 cm) =
15756 cm |.
309
21.32
(a)
The distance of a line from the band origin is the change in rotational
energy for the line's transition. The selection rule is AJ = +1 and the
spacings between rotational levels increase as J increases. Hence the two
lines closest to the band origin involve the J = 0 and 1 levels. The 2139
cm’ line is lower in frequency and energy than the band origin and so
must be the J = 1 — 0 line, where the rotational energy is decreasing.
The 2147 cm”! line has J = 0 > 1. The 2151 cm" line has J = 1 > 2.
The 2135.5 cm” line has J = 2 > 1.
(b)
Let a, b, c, d be the four given wavenumbers in order of increasing
se
epumber. From (21.40) and (21.39), we have
Be
(2B,
20)
en40 er
4B,
p= Vor (2B, — 26, )1-&, =Voigin — 2B, + &,
erat (2B 20g NaC
d= Vonen
ea ee a 30.
(2B, —20,)2-46, = Vorigin + 4B, — 80,
Three spreadsheet cells are designated for Voisin» B,, and &,. The initial
guess for Vici, Can be taken as the average of b and c, namely 2143.25
cm’'.The initial guesses for B, and @, can be taken as zero. The four
formulas for a, b, c, and d are entered into four cells and the squares of
the deviations of the a, b, c, and d formula values from the observed
values are calculated, and summed. The Solver is set up to minimize the
sum of the squares of the deviations by varying V origin ? B, yagd.
subject to the constraints that B, and @, be positive. An excellent fit to
the observed lines is found with V,,,.;,
ongin = 2143.2695 cm"!, B, = 1.93023
m',and @, = 0.016411.
(c)
From (21.16) and (21.15), B, =h/8n* UR? and R, = (h/8n7pB,)'”?.
1 = [(12)(15.994915)/27.9949 15]9/(6.02214 x 107°) = 1.138500 x 10 g.
B, = B,c = (1.93023 cm ')(2.997925 x 10'° cm/s) = 5.78668 x 10!°s7!,
R. = [(6.62607 x10 J s)/8107(1.138500 x 10° kg)( 5.78668 x 10!° sy]
= 1.12863 x 107'® m = 1.12863 A, which is smaller than Ro in Example
Ba:
310
21.33
With centrifugal distortion neglected, the v = 0 vibration-rotation levels are
given by Eq. (21.26) as Evip-rot = 5 Ve — Fhvexe + hB IJ + 1) - shad foi
For
03 Evi-rot(O) =
5 Ve =
Nee
We
have Evib-rot\) a E\ib-rot(O) =
h(Be- 5Oe)J(J + 1) =(B, - 5&, acJ(J +1). (B, — 4G, helkT=
[10.594 — (0.31)] cm” (100 cm/m)(6.6261 x 10°" J s)(2.9979 x 10° m/s)/
CE S50 77% 1Oe2 J/K)(300 K) = 0.050064. The degeneracy of each level is
2J + 1, so the Boltzmann distribution law gives the relative populations as
NjINo = (2J + Ve"? = (2 + 1) exp [-(B, - + &, \acJJ + L/KT] =
(Oy ll \camaeneds ae Werrind
al
N,/No
21.34
0
1
1
2
3
2.) 14 yr3a7 03 ae) 328395
4
5
hp3150d SeeUY
6
1.588
B,—rotational constant; o©,—vibration-rotation interaction; D—centrifugal
distortion; v.x.—anharmonicity.
21.35
(a)
From Fig. 20.17 O2 has 4 more bonding electrons than antibonding
electrons, O53 has 5 net bonding electrons, and O; has 3 net bonding
electrons. Therefore O 5 has the strongest bond and the largest k, and O 5
has the smallest ke.
(b)
Use of Fig. 20.14 shows that N> has 6 net bonding electrons and N35 has
5 net bonding electrons, so N2 has the stronger bond, the larger k, and the
larger Vv, since V, = (1/2)(ke/p)”?.
(c)
N>zhas a triple bond and O; a double bond. The N2 bond is stronger and
Nz has the larger ke.
(d)
Ex = J(J + 1)A7/21, where
I= wR2. An Na atom is heavier and larger
than an Li atom, so Naz has the larger pt and the larger R,. So Exot jai 1S
greater for Lip.
21.36
(a) T;
21.37
(a)
(b) T;
(c) T;
(d) F.
A C2 axis and two symmetry planes.
311
(b)
A C3 axis (through the CCI bond) and three symmetry planes (each one
containing C, Cl, and one F).
(c)
The molecule is square planar. A C4 axis perpendicular to the molecular
plane; an Sq axis and a C) axis, each coincident with the C4 axis; a
symmetry plane coincident with the molecular plane; four symmetry
planes perpendicular to the molecular plane; four C2 axes in the
molecular plane (two pass through pairs of opposite F’s and two lie
between the F’s); a center of symmetry.
(d)
The structure is trigonal bipyramidal. A C3 axis; an S3 axis coincident
with the C3 axis; a (horizontal) plane of symmetry containing the
equatorial Cl’s; three planes of symmetry that each contain the two axial
Cl’s; three C2 axes, each lying in the horizontal symmetry plane.
21.38
(e)
The VSEPR method shows the structure is a square-based pyramid. A C4
axis and four symmetry planes.
(f)
A C? axis perpendicular to the molecular plane; a center of symmetry;
two C2 axes in the molecular plane; three symmetry planes—one
coincident with the molecular plane and two perpendicular to it. See
ProbsZ1.3 9c.
(g)
A C.. axis through the nuclei and an infinite number of symmetry planes
that each contain the C... axis.
(h)
A C.. axis (which is also an S.. axis), an infinite number of symmetry
planes through this axis, a symmetry plane perpendicular to this axis, a
center of symmetry, an infinite number of C2 axes perpendicular to the
molecular axis.
(a)
The symmetry elements (Prob. 21.37a) are a C2 axis and two symmetry
planes, which we call 6, and 6». The symmetry operations are E ‘ eo :
(obey Oy,0:
(b)
21.39
(a)
Moves a nucleus at x, y, z to —x, —y, z.
(b)
From x, y, Z, (0.x) Va—c:
312
(c)
The S, rotation about the z axis consists of a C > rotation about z
followed by reflection in the xy plane. From the answers to (a) and (b)
this moves a point at x, y, z to —x, —y, -z. We see that Ss =].
21.40
(a)
The three principal axes intersect at the B nucleus (which is the center of
mass). One principal axis is perpendicular to the molecular plane (and
coincides with the C3 axis). The other two principal axes lie in the
molecular plane; one of these can be taken to coincide with a BF bond,
and the other is perpendicular to this one. (As in XeF4, the orientation of
the principal axes 1s not unique.)
(b)
The three principal axes intersect at the center of mass, which lies on the
C2 axis. One principal axis coincides with the C2 axis. The second lies in
the molecular plane and is perpendicular to the C2 axis. The third is
perpendicular to the molecular plane.
(c)
The three principal axes intersect at the C nucleus. One principal axis
coincides with the molecular axis (which is a C.. axis); the other two can
be taken as any two axes through the C that are perpendicular to the
molecular axis and perpendicular to each other.
21.41
21.42
(a)
SF, has more than one noncoincident C4 axis and is a spherical top.
(b)
IF; (which is a square-based pyramid) has one C4 axis and is a symmetric
top.
(c)
One C2 axis. Asymmetric top.
(d)
One C3 axis. Symmetric top.
(e)
One Cz axis. Symmetric top.
(f)
One C.. axis. Symmetric top.
(g)
One C; axis. Symmetric top.
(h)
Symmetric.
The principal axes intersect at the center of mass, which is the B nucleus. One
principal axis is the C; axis. For this axis, J. = 2 mir; =
3(18.998 amu)(1.313 A)” = 98.3 amu A’. The other two principal axes lie in
the molecular plane and we can take one of them to coincide with a B-F bond.
313
Hence, In = Xj mir2,, = 2(18.998 amu){(1.313 A)(sin 60°)]° = 49.1 amu A®.
With one C3 axis, BF; is a symmetric top, so J, = J, = 49.1 amu A’. I, can also
be calculated by taking distances from the F atoms to an in-plane line through
B and perpendicular to a B-F bond: J, =
(19.0 amu)(1.313 A)? + 2[(cos 60°)(1.313 A)]?(19.0 amu) = 49.1 amu A’.
21.43
One principal axis of this symmetric top is the C3 axis through the axial bonds.
For this axis, J = 3(34.97 amu)(2.02 iP = 428 amu A’. Another principal axis
can be taken to coincide with an equatorial P—Cl bond, and for this axis J =
2(34.97 amu)(2.12 A)’ + 2(34.97 amu)[(2.02 A)(sin 60°)]* = 528 amu A’.
Since this is a symmetric top, the moments of inertia about the principal axes
that are perpendicular to the C3 axis are equal, and the third principal moment
is 528 amu A’.
21.44
The molecule is a symmetric top with J, 4 I, = I.. From Eq. (21.45), Ero/h =
BU(J +1) +(A—B)K’ =[BU(J +1)+(A-B)K’Ic.
(a)me
FOr, Oth — 0 and Eas
On bore
el ke
— =| On Per or,y
—leand
K =0, Eph = 2Bc = 2(0.05081 cm')(2.9979 x 10!° cm/s) =
3.046 x 10° s'. For J=1 and |K| =1, E,./h = (2B +(A- B)]c =
(B+ A)c = (0.2418 cm')c = 7.249 x 10° 71.
(b)
From Eq. (21.47), v = 2B(J + 1) = 2Bc (J+ 1) = 2Bc, 4Bc,... =
3.046 x 10’ s!, 6.093 x 10’ s!,... = 3046 MHz, 6093 MHz.
21.45
(a)
(b)
(ce)
Let the molecule lie on the positive half of the z axis with the origin at the
oxygen nucleus. Then Zcom = [12(1.160 A) + 31.972071(2.720 A)]/
59.966986 = 1.682 A.
Io =D; mir? = [(15.994915)(1.682)° + 12(1.682 — 1.160)" +
31.972071(2.720 — 1.682)7](g A)*/(6.02214 x 1074) =
1.3777 x 10° g A? = 1.3777 x 10° kg m*.
Vyon1 = 2(J + 1)Bo. Bo = h/817Ip = (6.62608 x 10°*4 J s)/
87°(1.3777 x 10 kg m’) = 6.091, x 10’ s7!. vo_5; = 2By = 12.18 GHz:
V1 52 = 4Bo = 24.36 GHz; V2_,3 = 6Bo = 36.55 GHz.
314
21.46
Bo = Boc = (0.39021 cm™!)(2.99792 x 10!° cm/s) = 1.16982 x 10's").
Ip =
h/81By= (6.62607 x 10-** J s)/817(1.16982 x 10'° s"!) = 7.1738 x 10° kg m?.
The center of mass is at the carbon atom and the principal axes pass through
this atom. If dis the CO bond length, then Jp = >; mir? = mod’ + mod’ =
2mod’, so d = (Ig/2mo)'”” =
[(7.1738 x 10° kg m’)(6.02214 x 107°)/2(15.994915 x 107 )kg]!” =
1.162 x 107° m= 1.62 A.
21.47
21.48
(a)
From the VSEPR method, SO; is nonlinear; 3V— 6 =9 —-6 =3.
(b)
Linear. 3V-5=7.
(c) 3V-6=9.
We look for sets of integers 7, 7, k such that 36577 + 15957 + 3756k is slightly
greater than 7252. Systematic trial and error (best done by first setting j = 0
and looking for i and k values that fit, then setting 7= 1 and looking for i and k,
then settingj= 2, etc.) gives the v;v3v4 possibilities for the 7252 cm! band as
(calculated frequencies in parentheses)
21.49
(a)
5 i hv; = she eS
200 (7314),
101
(7413),
002
(7512).
+hc(1340 + 667 + 667 + 2349) cm” =
4.99 x 10°°° J=0.311 eV.
21.50
21.51
21.52
(b)
+he(3657 + 1595 + 3756) cm! = 8.95 x 10°? J = 0.558 eV.
(a)
Inactive, since the dipole moment remains unchanged.
(b)
Active, since the dipole moment changes.
Ve
(L/2I) ak)
Ue
(a)
The C=C bond is stronger and has the greater & and the greater v.
(b)
C—Hhas
(c)
Bending vibrations are generally lower-frequency than stretching, so
C—H stretching has the greater v.
the smaller yp and the greater v.
Vea) C= (1/2n)(k/)" "Ic. Isotopic substitution does not affect the electrons
and hence doesn’t affect k. We have pt = mym2/(m, + m2) = mym/m2 = m,,
315
where m = my (Or mp) and mz is the mass of the rest of the molecule, and we
used m7 >> m). Therefore cp = 2Ucu and Vep = Voq/2"”” = (2900 cm ')/2""
= 2050 cm”!.
21.53 v =(1/2n)(k/p)” and k = 4r°v2p = 407 9 2C7. cn = 12(1) g/13(6.02 x 10°°)
= 1.53 x 1074 g, and uco = 12(16) 2/28(6.02 x 10”) = 1.14 x 10° g.
So kcu = 410°(3000 cm™!)*(100 cm/m)?(3.00 x 10° m/s)?(1.53 x 107°’ kg) =
489 N/m. Also, kco = 41°(1750 cm™')*(100 cm/m)*(3.00 x 10° m/s)* x
(1.14 x 10°°° kg) = 1240 N/m.
21.54
21.55
(a) T;
(b) T;
(Lele
(2 er.
(a)
(c) F;
(d) T (provided it is not a spherical top);
(e) T.
The rotational levels of a linear molecule are E,., = BhJ(J + 1)
[Eq. (21.45) with K = 0] and the pure rotational Raman selection rule is
AJ = +2. So Vo — Vscar = AE/h = t(Bh/h)[(VJ + 2)(VJ + 3) —-J(J + 1] =
+(4J + 6)B, where J=0, 1, 2,... . The spacing between adjacent lines is
[4(/ + 1) + 6]B- (4/ + 6)B = 4B.
(b)
4B =7.99 cm! and B = 1.998 cm! = 199.8 m!. We have B = Bia=
h/8n°Ic = h/8r?cuR? and R = (h/8r°cu B )". Also, 1 = mym2/(m, + mo) =
m;/2m, = mj/2 and R =
2(6.626x107*4 J s)(6.022x1023 mol!)
1/2
81° (2.998x10* m/s)(0.01401 kg/mol)(199.8 m7!)
1.098 x 10°'° m = 1.098 A (as in Table 21.1).
(c)
The lowest J is 0 and Vo — Vscat = [4(0) + 6]B = 6B.
(d)
vo =c/Ag = (2.9979 x 10° m/s)/(540.8 x 10° m) = 5.543; x 10!4 57.
Vo — Vscat = +6B = +6 Bc = +6(199.8 m™!)(2.998 x 10° m/s) =
+0.00359 x 10'* s7'. Vecat = 5.5435 x 10'4 5! + 0.00359 x 10'4 57! =
5.547, x 10'* s! and 5.5395 x 10'4 s7!. Then Ascat = C/Vscat = 540.45 nm
and 541.15 nm.
21.56
(a) F; (b) T;
(c) F.
316
21.57
For the Balmer series, n, = 2 in Eq. (21.50); for the series limit, ng = ce and 1/A
= R/4 = (109678 cm !)/4. Then A = 3.647 x 10° cm = 364.7 nm.
21.58
For the Paschen series, n, = 3 in Eq. (21.50); the first three lines have ng =
4,5, and 6. So 1/A = R(1/9 — 1/16), R(1/9 — 1/25), R(1/9 — 1/36). We get A =
1.8756 x 10~* cm, 1.2822 x 107 cm, 1.0941 x 107 cm.
21.59
Li** is a hydrogenlike atom and Eq. (19.14) gives the energy levels. The Li
nucleus is substantially heavier than the H nucleus, so we can take pt equal to
the electron mass. So E = —9[21’me*/(4m€9)’n°h’] and 1/A = v/c = |AE| /he =
9[21?me*/(4m€o)*ch*|(1/n? — 1/n2) = 9(109736 cm ')(1/1 — 1/4) =
7.4072 x 10° cm”. Then A = 1.3500 x 10° cm.
21.60
Using the notation of Fig. 21.38 and Eq. (21.52), we have Do of the B state as
Dj, = hVcont — hVoo = (6.6261 x 10 J s)(2.998 x 10° m/s) x
[1750.5 x 107?° m)! — (2026.0 x 107"? m)"] = 1.543 x 10°” J = 0.963 eV.
Do of the ground state is given by (21.52) as
Dy, = (6.6261 x 10-** J s)(2.9979 x 10° m/s)(1750.5 x 10°'° my! x
(1 eV)/(1.60218 x 107'? J) - 1.970 eV = 5.11 eV.
21.61
Ny-\/Nj-9 = 0.181 = G/L) and Ae/kT = -In(0.181/3) = 2.81. So T=
0.356Ae/k. From (21.15), (21.33), and (21.32), Ae = 1(2)h7/21-0 = h’/4rI=
2 Byhe = 2(B, - ; &, )hc = 2(1.931 — 0.009) cm (1 cm/0.01 m) x
(6.626 x 10° J s)(2.9979 x 10° m/s) = 7.636 x 10° J. T= 0.356 Ae/k =
0.356(7.63¢ X 10°? J)/(1.3807 x 10°°° J/K) = 1.97 K.
21.62, (a) 1
e(D) elec) cP se(d)ul pe(e)ah a(0) F) (2) ste
ee Ueh ss.() a0;
(kK) Pee:
21.63 F =BQvsin 0 = (1.5 T)(1.60 x 10°? C)(3.0 x 10° m/s) sin 6 =
(7.2 x 10°? N) sin 8.
(a) F=0. (b) F=(7.2x 10° N) sin 45° =5.1 x LOpaea
(c) 7.2x10°3N. (d) 0.
317
21.64
From the paragraph before (21.61), |m| = Qur/2 =
21.65
(a)
(2.0 x 107! €)(2.0 x 10° m/s)(25 x 107! m)/2 = 5.0 x 10° J/T.
e/2m, = (1.602176 x 10"! C)/2(1.672622 x 10° kg) = 4.78942 x 10’
Hz/T, where (21.54) was used.
21.66
21.67
(b)
y/2n = (4.78942 x 10’ Hz/T)5.58569/2n = 42.5775 x 10° Hz/T.
(a)
Equations (21.65) and (21.63) give E = —yh BM, =
(b)
15000 G = 1.50 T; energies are the same as in (a).
(a)
From (21.67) and (21.63), v = |y|B/27 =
(b)
27.32 MHz.
~1.792(4.7894 x 10’ s'/T)1.792(6.626 x 10°" J s)(2m) '(1.50 T)M7 =
~(1.35g x 10°26 J)M,. Since I = 3/2, M, = 3/2, 1/2, -1/2, and -3/2. The
levels are —2.04 x 10° J, -0.679 x 107° J, 0.679 x 10° J, 2.04 x 10°° J.
(4.7894 x 10’ s'/T)1.792(1.50 T)/2m = 2.049 x 10’/s = 20.49 MHz.
21.68
From (21.67), V,/Vg =|Yal/|¥g | and v(°C) = (10.708/42.577)(600 MHz) =
151 MHz.
21.69
(a)
Equations (21.67) and (21.63) give B = 2nv/|y| =
2n(60 x 10° s!)/[5.5857(4.7894 x 10’ s'/T)] = 1.41 T.
21.70
(b)
(300/60)(1.41 T) = 7.05 T.
(a)
M, = +2 and —'2. The energy separation is given by (21.65) as AE =
|y| 2B = (4.7894 x 10’ s"'/T)5.5857(6.626 x 10°" J s)(2)'(1.41 T) =
3.98 x 10° J. The levels are nondegenerate and the population ratio is
eoMEAT = exp [-(3.98 x 10°°° JV/(1.381 x 10°? J/K)(298 K)] =
exp (-0.00000967) = 0.9999903.
(b)
An increase in B increases the separation between energy levels, thereby
producing a greater population difference between the initial and final
states. Hence the absorption intensity increases.
318
21.71
From (21.72), vi — vj = 10°(60 x 10° Hz)(1.0) = 60 Hz.
21.72
From the tables in the text, 6 is 2 to 3 for the CH; protons and is 9 to 11 for the
CHO proton; J is 1 to 3 Hz. The CH; doublet has three times the total intensity
of the CHO quartet. Thus (all splittings are about 2 Hz)
(a)
i
21.73
eae 300
Hz
(a)
Hz
ee
(b)
00 Hz12 cee
1500 Hz a=
From (21.70) and (21.68), increasing 5; means decreasing 6; and
increasing V;, SO V, increases to the left.
(b) To the left.
(c) To the nght.
(d) From (21.70) and (21.69), increasing 5; means decreasing 6; and
decreasing By;,S0 By; increases to the nght.
21.74
(a)
One singlet peak.
(b)
One proton NMR peak that is split into a doublet by the F.
(c)
One singlet.
(d)
The four methylene protons give a quartet of relative intensity 2; the six
methyl protons give a triplet of relative intensity 3.
(e)
The (CH3)2 protons give a doublet of relative intensity 6; the CH proton
gives a septet of relative intensity 1.
(f)
Two equal-intensity singlet peaks. The CH groups are not equivalent
and don’t split each other.
(g)
Three quartet peaks of equal intensity. In each of the three quartets, the 4
lines have equal intensity but only two of the 3 spacings are equal.
(h)
The CH; protons give a triplet of relative intensity 3, the CHO proton
gives a triplet of relative intensity 1; the CH2 protons give an octet of
relative intensity 2.
319
21.75
(a)
(b)
21.76
nts: one for
This molecule has two different proton—'’F coupling consta
ent coupling
the 'H and '°F nuclei that are cis to each other, and a differ
constant for 'H and !*F nuclei that are trans to each other.
r to the
In this molecule, the F atoms lie in a plane that is perpendicula
coupling
plane containing the CH? group, and there is only one HF
constant.
All peaks except in (b) are singlets.
(a)
One peak.
(b)
One peak that is split into a doublet by the Ee
(c)
One peak.
(d)
(e)
Two peaks of equal intensity.
Two peaks with 2:1 intensity ratio.
(f)
Three equal-intensity peaks.
(g)
Two peaks of equal intensity.
(h)
Three equal-intensity peaks.
(Db), 59 (c)r3-
21.77
4 kinds of carbons);
(a) 4 (there are
21.78
In Fig. 21.44, 100 Hz corresponds to a length of 28 mm and J corresponds to a
length of 22 mm, so J = (2 -/28)(100 Hz) = 8 Hz.
21.79
(a)
(c)
Unchanged; see Eq. (21.70) and the following paragraph.
Each v and the difference between the v’s is multiplied by 10; see Eqs.
(21.68) and (21.72).
(e) Unchanged, as noted after (21.73).
21.80 From Eq. (21.69), 6; = 1 — 27Vspec/|¥i|Bo,i. $0 8; = 10°(Grer — 51) =
10°(—21Vspec!|'¥i|)(U/Bo,ret — 1/Bo.i) = 10°Bo(Bo,et — Bo,i)/Boset Bo,i. Since 0; << 1,
we have Bo; ~ 21Vspec/|'7i| = Bo, and ; = 10°(Boet— Bo.i)/Bo.ret-
320
21.81
(a)
For a60-MHz spectrometer, coalescence occurs at 120°C and the
formula gives k, at 120°C as ke =
|v2 — v1 |/2"”. From (21.72), v2 — V1 =
10-°(60 MHz)(2.94 — 2.79) = 9.0 Hz and k. = 1(9.0 s'y/2"" = 20s".
(b)
Eq. (21.72) and the fact that 6; is independent of Vspec show that an
INCTEASE 1N Vepec INCreases V; — V;. Coalescence occurs when
Vexch >>
lv, —V2 |, and the increase in lv, - v2| and the fact that Vexch
increases as T increases mean that the coalescence temperature will
increase WhEN Vgpec 18 Increased.
21.82 v = eB. Bolh = 2.0026(9.274 x 104 J/T)(2.50 T)/(6.6261 x 10° Js) =
7.01 x 10'° 5’ = 70.1 GHz.
21.83
There is one set of 4 equivalent protons and a second set of 4 equivalent
protons, so there are 5(5) = 25 lines.
(b) 3(3)=9;
(c) 10;
(d) 3(2)=6.
21.84
(a) 2(7)=14;
21.85
[a] = W[pp/(g/cm*)](J/dm) = 1.90°/(0.0650)(2.00) = 14.6°.
21.86
The observed o of the mixture is the sum of the @’s of the a and B isomers:
=
Og + Ap = [OJaPal(em*/dm g) + [x.]spgl(cm*/dm g) (1). The total solute
mass m is m = mq, + mp and division by the solution’s volume gives Pp = Pa +
Pp. Division of Eq. (1) by pl(cm?/dm g) gives [O] = [Q]owat+ [A]awp (2),
where Wo = Po/(Pa + PB) = Ma/ (ma + mB) and we are the mass fractions (and
also the mole fractions) of the a and B isomers. Equation (2) gives 52.7° =
Wo112.2° + (1 — Wa)17.5° and Wo = 0.372, or 37.2% a-D-glucose.
21.87
(a)
The photon energy is hv = hc/X =
(6.626 x 1074 J s)(2.998 x 10° m/s)/(30.4 x 10” m)] x
(1 eV/1.602 x 10°'? J) = 40.8 eV. Hence the energy used to ionize the
molecule is 40.8 eV — 31.5 eV = 9.3 eV.
(b)
Replacement of 30.4 nm by 58.4 nm in the calculation in (a) gives a
photon energy of 21.2 eV; subtraction of the 9.3 eV ionization energy
gives 11.9 eV as the photoelectron energy.
321
21.88
As discussed in Sec. 21.15, the 17 eV band results from loss of a 2p electron.
Since this is a bonding electron, the N3_ ion produced is more weakly bound
than N> and has a smaller force constant k and hence has a smaller vibration
frequency Vv = (1/2m)(k/p)'”.
21.89
(a)
Two carbon peaks of equal intensity. One oxygen peak.
(b)
One C peak. One O peak.
(c)
Three C peaks of equal intensity. Two O peaks of equal intensity.
Three C peaks of relative intensities 2:2:1. One O peak.
(d)
21.90 E=hv =hc/d. The first entry in (21.79) is E = (6.626 x 10°" J s) x
(2.998 x 108 m/s)/(200 x 10-° m) = 9.93 x 107'” J = 6.20 eV; etc. The first
entry in the following line is Nahv = Nahc/A = (6.022 x 10°*/mol) x
(6.626 x 1074 J s)(2.998 x 10° m/s)/(200 x 10° m) = 5.98 x 10° J/mol; etc.
21.91
(a)
1 is the energy per unit time that falls on unit cross-sectional area
perpendicular to the beam. In time f, the beam travels a distance c’t and
the photons that pass through cross-sectional area A in time f are
contained in a volume Ac’t. Each photon has energy hv. If N is the
number of photons in the volume Ac’t, the energy of the photons in this
volume is Nhv. The intensity equals this energy divided by the time rt and
the cross-sectional area A: so J = Nhv/tA. So N = ItA/hv. Then the number
of photons per unit volume is N/V = N/Ac’t = (ItA/hv)/Ac’t = I/hvc’.
(b)
c’ = c/1.34 = 2.24 x 10° mis. v =cl(488 x 10° m) = 6.14 x 10s".
NIV = Whvc! = (10"° W/m’)/
[(6.63 x 10°** J s)(6.1 x 10'* s“!)(2.2 x 10° m/s)] = 1 x 10” photons/m? =
ie 10 photons/cm’. 18 g of water contains 6 x 10°° molecules in a
volume of 18 cm’, and the number of molecules per cm’ is 3 x 107”.
21.92
The energy absorbed is 0.744(0.00155 J/s)(110 s) = 0.1269 J. The number of
moles of photons absorbed 1s (0.1269 J\(Nahc/A) = 4.92 x 10°’ mole. So ® =
(6.80 x 10° mole)/(4.92 x 107’ mole) = 13.8.
322
21.93
(a)
d[HI)/dt = —)$, — k2[H][HI]. The steady-state approximation for H gives
d[H]/dt = 0 = 0S, — k2{H][HI] and k2[H)[HI] = 05,. So d[HI)/dt =
—20S8q = —25q, since 6 = 1.
(b)
The number N of photons absorbed satisfies 4184 J = Nhv = Nhc/Ad and
N = (4184 J)(250 x 10°? m)/(6.626 x 10°" J s)(2.998 x 10° m/s) =
5.27 x 10°'. The number of HI molecules decomposed is 2(5.27 x 107') =
IO Scala
21.94
With the inclusion of reaction (5), Eq. (21.84) becomes r = k[A*][A] —
k4[A>] + ksfAy. Reaction (5) does not involve A*, so (21.85) still holds and we
have r= ko[A]Sq/(ko[A] + k3) — ka[A2] + ks[A]’. For the photostationary state, r
= 0 and we get [A2] = ks[A]-/ka + k[A]Sq/(kak2[A] + k3kq). In the absence of
radiation, $5, = 0 and the last equation becomes [Az] = ks[A]*/ka. The
concentration of A> is greater in the presence of radiation.
21.95
(a) Yes;
(b) no; (c) no. (1 + 1 is not a member of the set.)
21.96
(a) Yes; (b) no;
21.97
(a)
(c) yes;
(d) no.
We have AJ = JA =A, since / is the identity element. Since each group
element appears exactly once in each row of the multiplication table, we
must have AA = J. The multiplication table is thus
I
A
(b)
Using the property of the identity element, we start with
323
where w, x, y, and z are to be determined. w cannot equal A because this
would put two A's in row 2. So w must either equal J or B. If we put w =
I, then the theorem that each group element appears exactly once in each
row means that x = B. But with x = B, we have B appearing twice in
column three, which is not allowed. Hence w = B. Filling in the rest of
the table, we get
This table shows that each element has an inverse, and using the table,
one finds that associativity is satisfied, so this is the multiplication table
of a group.
21.98
(a) F; (b) T.
21.99
As in Fig. 21.55, the coordinate axes do not move when a symmetry operation
is applied.
(a) They do not commute.
(b) They do not commute.
(c) They commute.
21.100 (a) E; (b) 6; (c) 7; (d) C2; (e) C2; (f) $3. (Since S$? involves two
reflections, which amounts to no reflections, it is not the inverse of S;.)
21.101 (a) T.
(b) T.
(c) T. (If we call the rotation axis the z axis, then the 180°
rotation part of Ss converts the x and y coordinates to their negatives and the
reflection converts the z coordinate to its negative.)
21.102 (a)
(d) T.
The structure is trigonal pyramidal. A C3 axis and three planes of
symmetry (each of which contains one of the bonds).
(DyemCnLC?. E.G
a?
Ommce
324
21.103 Symmetry elements: A Cg axis that is also an S¢ axis, an S3 axis and a C> axis.
A center of symmetry. Seven symmetry planes, six perpendicular to the
molecular plane and one coincident with it. Six C2 axes lying in the molecular
plane. Symmetry operations: E , Cx Ca Ge Ce Ge S6> Se) Se vad Pasix
C, rotations, and seven 6 reflections, for a total of 24 operations. (The
operation o is the same as 7; see the answer to Prob. 21.101c.)
21.104 Four C3 axes (one along each bond). Three $4 axes (which are also C2 axes),
one through each pair of opposite faces of the cube in Fig. 21.20. Six
symmetry planes, each of which contains two bonds.
21.105 (a) Tz; (b) Cav;
(i) D3,;_ G) Danas
DIGG
21.107 (a)
(€) Den; (d) D3n; (e) On; (1); Cav (8) Dans
(K) C3, ( Car; (m) Ci; (n) C;.
ACyO
eh. Orderme
Yes, since the product C,,6, is a symmetry operation.
(b) When n is even, since Cr? = C, 1s then a symmetry operation.
(C)i
21.108
D3;
Con ee
Cop.
Dog.
LORE
aOR a2
9 12
325
(hh) Cp;
spiipene
21.111
0.2(4)+4(5)
+ 4(8) z
0.2(1)
20: Sas 22
— 0.20819
Ba -(
—1(4)+3(5)
+3(8)
—1(1)
[inen23
s/n
744
[ All Pesedics! iss
=
=
3. AG
3(2) + 4(6)
30
21.112 Let k be the number of columns in R and the number of rows in S. (These
quantities must be equal or the matrix product is not defined.) The element fm
k
is calculated using row m of R and column n of S. SO thn = Diy ni Sin +
2
A OMe =4
21.113 MIN =(5,
alg
(
MN,
=(7)(6) =(42)
9
1
0
21.114 For example, the product of the G(xz) and G(yz) matrices (in either order) is
he
re
ON pb
the
cot tew AO
a)
O
-1
Of
1
O}=}
O
-—-1
OJ}, which is the matrix that corresponds
OS
Odiyels
eel
On
OPE
1
O
0 dr).O
to CZ), in agreement with the product in Table 21.2.
21.115 (a) C,(z) moves the point at (x, y, z) to (—y, x, z) and Eq. (21.93) becomes
Xs
0
YS
Za
(b)
-!l
OYx
rod)
0.
GOs
moles
7 moves the point at (x, y, z) to (—x, —y, —z) and the matrix that does this
=
lO
ee
1S) | 90s ee—
ee
326
(c)
(xy) moves the point at (x, y, z) to (x, y, -z) and the matrix is
15*0;
470
0
1
O
OM O8R=
1
21.116 (a)
Using associativity, we have (P-'AP)(P"'CP) = (P"'A)(PP™')(CP) =
(P"'A)(1)(CP) = (P"'A)(CP) = P"'ACP = PEP, since multiplication by
a unit matrix has no effect.
(b)
The result of part (a) of this problem shows that the transformed matrices
P'AP, P'BP.... multiply the same way as the original matrices A, B....;
since the original matrices multiply the same way as the group members,
so do the transformed matrices.
21.117 As noted a couple of paragraphs before (21.96), each element of a
commutative group is in a class by itself, so the number of classes c equals the
number of elements / in a commutative group. There are thus A terms on the
left side of (21.96). The smallest possible value of each term on the left of
(21.96) is 1, and if any representation had a dimension greater than 1, then the
value of the left side of (21.96) would exceed h and (21.96) would not hold. So
all irreducible representations of a commutative group are one-dimensional.
21.118 For example, the products of the matrices corresponding to C,(z) and 6(xz)
are (1)(—1) = (-1) and (—1)(1) = (-1), and (—1) is the matrix corresponding to
G6( yz), which according to Table 21.2 is the correct product. The other
products are verified similarly.
21.119 To find the elements in the same class as E, we form the products A~'EA,
B'EA,.... ButA'EA=A_'A = E for every element A, so the identity element
E is in a class by itself.
21.120 (a) 2p,;
(b) —2p,, since the positive and negative lobes are interchanged;
(c) 2p, (d) -2p,; (e) 2py (f) -2py; (g) -2py; (h) —2py.
327
21.121 (a)
The orbitals O2s, O2p., and H1s + H21s are each unchanged by each of
the four symmetry operations and the C2, character table (Table 21.3)
gives the symmetry species as a}.
(c) The O2p, orbital is unchanged by E and by 6(yz) and is transformed to
—O2p, by C,(z) and by 6(xz) and H;1s — H21s shows the same behavior.
The character table gives the symmetry species as b2.
21.122 (a)
We examine the effects of each symmetry operation on the vectors
showing the normal-mode motions of the atoms. The point group is Czy
and the z axis coincides with the C2 axis through the C and O atoms. The
G 5(Z) operation reverses the direction of a vector pointing in the +x or
—x direction (and also interchanges the vectors on the H atoms), so for
this normal mode the C,() character is —1. For G( yz), the direction of a
vector pointing in the +x or —x direction is reversed and the character is
—1. The 6(xz) operation interchanges the vectors in the H atoms but
leaves the directions of vectors pointing in the +x direction unchanged
and the character is +1. The character table (Table 21.2) gives the
symmetry species as by).
(b)
The point group is C3. The N-atom vibration vector points along the z
axis and is unaffected by each of the six symmetry operations listed in
the character table (Table 21.3). The H-atom vibration vectors point
along the bonds and although the symmetry operations may interchange
or permute these vectors, the direction of each vector on each H is
unchanged by each symmetry operation. Thus the characters are all +1
for this mode and the mode has species a.
21.123 The representation I consists of the matrices in (21.94), and the point group is
C2» with order h = 4. The irreducible-representation characters x; are taken
from Table 21.3. The characters xr of I are found by taking the traces (the
sums of the diagonal elements) of the matrices in (21.94); these characters are
3,-1,1,and 1 for E, Gala): 6(xz), and 6( yz), respectively. Eq. (21.98)
BIVES Ay(21.94) = (1/ 4)[1(3) + (-I) +10) +10) = 1,
A, 21.94) = (L/4)(1(3)
F1C-D + (-D@) + CDW] =9
328
Ap(21.94) = A/ 403) + (-I(-1) +10) + -D(D)) = 1
p,(21.94) = A/ 403) + (-1)(-D) + (-DG) + 10)] = 1 so Tai.94) = A; ® B, ® By.
21.124 (a)
E is represented by a unit matrix whose order equals the dimension of
the representation. The trace of a unit matrix of order n is n. Hence this
representation has dimension 9.
(b)
We use Eq. (21.98) with the x; values taken from the character table in
Table 21.3. We have a, = (1/4)1(9) + (-D + 10) + 13)] =3,
Gael
AU)
(De (a) |a
apr —~U/ 410) + -DEDT1O+EDG)I—2
Gp DIO) DE) + DDI
)ies
sol = 3A, @ A, © 2B, ©3B,.
21.125 We use Eq. (21.98) with the x; values taken from the character table in Table
21.3. The sum in (21.98) is over the h symmetry operations. The characters in
the C3, character table are listed for each class, and if we take the sum in
(21.98) to be over the classes, we must multiply each term in the sum by the
number of operations in that class, so as to include each of the h operations in
the sum. We have a, = (1/6)[1(293) + 20.)(-118) + 3)9] = 14,
Ag = (1/6)[1(293) + 20)(-118) + 3(-1)9] = 5
Gp
16)[(2(293) + ZeCE 11s)30)9] = 137
so P14
AeA
37
21.126 The trace of a matrix such as M in (21.91) that is in block-diagonal form is
clearly equal to the sum of the traces of each block. (For example, the trace of
M is—1 +8 +7 = 14 and the traces of M,; and M2 are 8 — 1 = 7 and 7.) Hence
the trace of a matrix of a reducible representation equals the sums of the traces
of all the irreducible-representation matrices that it is the direct sum of. So
% (E) = 4(1) +14 6(2) =17, x7 (C3) = 40) +14 6-1) = -1,
%r (6) = 4(1) + (-1) + 6(0) = 3.
21.127 (a)
(b)
HBr, HS, CH;Cl, which have nonzero dipole moments.
HBr, CO>, HS, CHs, CH3Cl, CoH, which have vibrations that change
the dipole moment.
329
(c)
21.128 (a)
N>, HBr, CO2, H2S, CH3Cl, and CH, which are not spherical tops.
k= joes (R.). The function E,(R) is found by solving the electronic
Schrodinger equation, which is the same for H2 and D2, so E.(R) is the
same for H> and D> and this makes k,, D., and R, the same for Hz and D>.
(This answer neglects very slight deviations from the Born—Oppenheimer
approximation that make E,, k., Re, and D, differ extremely slightly for
H> and D>.)
(b)
Va= (1/27)(ke/p)'”. Since pis greater for D2 and k; 1s essentially the
same, Vv, is smaller for Dp.
(c)
Ie= UR? . Since pt is substantially greater for D2 and R,; 1s essentially the
same for the two species, /, is greater for Do.
(d)
B= h/81 Ie. Since I, is greater for D2, B, is greater for Ho.
(e)
The same, since E,(R) is the same for the two.
(f)
Do = D, — Yehve. Since Ve is smaller for D2 and D, is essentially the same
for the two, Do is greater for Dp.
(g)
Since v, is smaller for D2 and D, is the same, D2 has more bound
vibrational levels.
(h)
Since v, is smaller for D2, the separation between vibrational levels is
smaller for D2, and more D2 molecules are in excited vibrational levels at
a given T, so the fraction in v= 0 is greater for Hp.
(i)
Since I, is greater for D2, the separation between D> rotational levels is
smaller for D2 and the fraction in J = 0 is greater for Ho.
21.129 (a)
joule, newton, watt, pascal, hertz, coulomb, volt, tesla, ohm, kelvin,
ampere, siemens;
(b)
21.130 (a)
(b)
poise, debye, gauss, angstrom, svedberg, dalton, torr, bohr, hartree.
aL,
F. It cannot change its rotational state by absorption or emission of
radiation, but can change its rotational state during collisions.
(c)
F. Counterexamples are CHy and BF3.
(d)
F. The energy might be transferred to another molecule in a collision.
330
(e)
F. A counterexample is H2O.
(f)
ly
(g)
(h)
(i)
F. This formula is only for linear and spherical-top molecules.
(j)
(k)
18.
le
Ei
This question is too silly to answer.
331
Chapter 22
22.1
(a) T, V, and mole numbers.
(b)
he V, I
UC oor
(c) Over states.
E ,/kT
22.2
Z=;e
"
is dimensionless.
22.3.
The Helmholtz energy A is extensive, so Az is 25/10 = 2.5 times A. Hence
Ad/A, = 2.5 = (-kT In Z2)/(-AT In Z;) = (In Z2)/(n Z;).
22.4
G=A+PV=-KT InZ + VET(0 In Z/OV)7.y, =KTV[(V™ In ZV] y, -
22.5
Let a subscript o denote the partition function and thermodynamic properties
before the constant b is added to the levels. Then
IE
J rss pee ah
en ale yi obit = eZ, and In Z = —b/kT + In Zo.
(a)
P=kT(0 In Z/V);y, =kTO In ZJOV)7y, = Po:
(b)
U=kT'(0 In ZT);y, =kT [b/kT’ + 0 In ZJOT) yy, ] = b + Uo.
(c) S=U/T+k\n Z= DIT+ UST -kb/kT + k ln Zo = UST + k In Zo = So.
(d) A=-kT In Z=-KkT(-b/kT + In Z,.) = b — kT In Z, = b + Ao.
22.6
From p; = exp(-BE,)/Z [Eq. (22.15)], we have In pj = -BE; — In Z.
So -Y
p; In pj = B Dj pjEj + In Z Dj pj = (kT)
'U + In Z = S/k, where (22.3),
(22.33), and 2; pj = 1 were used.
Erk
©
ie
:
.
221
ee
As dimensionless:
22.8
From the equation before (22.53), the number of available translational states
is roughly 60V(mkT/h?)*” = [60(10 cm*)(1 m?/10° cm*)/(6.6 x 1074 J s)*] x
[(0.020 kg/mol)(1.38 x 10°” J/K)(300 K)/(6.0 x 10°°/mol)]}*? = 3 x 10°”.
332
22.9
(a)
Each dot with n,, n,, and n, being positive integers corresponds to a
stationary state with quantum numbers nx, ny, n,. The states with
Er S Emax Satisfy Eq. (22.52). The distance of a dot from the origin is
r= (n* +n ; sete :)"”?| so the the positive square root of (22.52) is
The region defined by this inequality is a sphere of
PaASmV- hea
radius rmax = (8mV"?h Emax)” and the number of dots in 1/8th of this
sphere equals the number of quantum states with energy €; S Emax} We
take 1/8th of the sphere because n,, ny, and n, must each be positive,
which is true only in 1/8th of the sphere.
(b)
Fig. 24.8 shows the 4 cubes at the same altitude that share a dot, and 4
more cubes above these 4 also share this dot. The number of translational
states with €, < max is then +(Sar;
) = (n/6)[(8mV"Fh? - 3kT)'?PF=
(1/6)(24mVe
nh?RT)”.
22.10
(a) T; (b) F.
22.11
(a)
(b)
We get 3628810.
9.332621569 x 10'°’, where we used 100!” = (107)! = 1077!
(c)
Using (LO.
eee
=
10-1024
and py
= (ene yl? =
(3.72007597601 x 10~4)!'° = 507595.88975 x 10°", we get
4.0238726006 x 10°°°’,
22.12
In (300!) =In 1 +1In2+1n3+---+1n 300. A BASIC program is
10 S=0
40 NEXT |
20 FOR !=1 170300
50 PRINT S
30 S=S+LOG(I)
60 END
One finds In (300!) = 1414.905850. Also, N In N — N = 300 In 300 — 300 =
1411.134742.
22.13
From (22.51) and (22.56), In Z = Nc In zc — In Nc! + Np In zg — In Ng! +: -Ne In zc —In Nc! + Ng In zp — Np In Ng + Ng +--: and (0 In ZIONB) TV Neo =
In zg — In Ng — 1 + 1 = In (zp/Ng). Substitution in Eq. (22.41) gives pp =
—RT |n (zp/Np).
333
22.14
(a) F; (b) F; (c) T; (d) T.
22.15
€, =nzh’/8ma’ and Ae, = (h’/8ma’)[(n, + 1)? — n2] = (h7/8ma*)(2n, + 1).
From €, = n2h’/8ma* = kT we get ny = a(8mkT)'"/h =
[(0.02 m)/(6.6 x 107*4 J s)][8(0.028 kg/6.0 x 107°)(1.38 x 10773 J/K)(273 K)}"”
and n, = 1.1 x 10°. Then Ag,J/kT = (h?/8ma’)(2n, + IWkT = 1.8 x 10°”.
22.16
At room T, kT = (1.38 x 10-7? J/K)(298 K)(1 eV/1.6022 x 107'? J) = 0.026 eV;
=|.
kT/he = (1.38 x 10°? J/K)(298 K)/(6.63 x 10-** J s)(3 x 10° cm/s) = 207 cm";
RT= (1.987 cal/mol-K)(298 K) = 0.59 kcal/mol = 2.5 kJ/mol.
22.17
With anharmonicity neglected, (N,)/(No) =e plain AT joa tol JkT
—hv,/kT
Cae eh:
(a)
Avo/kT= (6.626 x 107*4 J s)(6.98 x 10!'7/s)/(1.381 x 10°? J/K)(298 K) =
11.24 and (N,)/(Ny) =e7''** =.0.000013.
(b) (N,)/(Ng) =e "7! = 0.044.
(c) Ye 7 =046.
22.18
Use Eq. (22.76). Let x = AV/KT = hcV/kT = (6.6261 x 10°" J s) x
(2.9979 x 10® m/s)(2329.8 cm™!)(100 cm/m)/(1.38065 x 10°*° J/K)T =
Gs5 20K) Tawa
eee! ch cnt) / (lee
(a) x = (3352 K)/(298.15 K) = 11.243. zyip = W/(1 —e |79) = 1.000013.
No = (6.0221 x 107°)1/1.000013 = 6.0220 x 107°.
N, = Nae ''743/1.000013 = 7.888 x 10'°.
(b) x = (3352 K)/(1073.15 K) = 3.1235. zyip = 1.0460. No = Na(1)/1.046 =
BS TpalO- aeN= Naer
4 a)104.65 932 xl Oaee
(c) x=1.0241. zp = 1.5604. No =3.859x 107. N, = 1.386 x 107°.
22.19
With centrifugal distortion neglected, the rotational levels are BohJ(J + 1). The
degeneracy of each level is 2J + 1, and (N,)/(No) =3¢770"/*7 fe® =
3 ?BohlkT =O Boh/k = 2(5.96 x 10!°/s)(6.626 x 104 J s)/(1.381 x 10°23 J/K) =
5.72 K.
334
(a)
(N,)/(No)
(b)
aaa a2
= 3076.72 K)/(200 K) _ 2.92.
K)/(600 K) o
2.97.
(Cc)
22.20
(a)
The vibrational levels are nondegenerate and (N 1)/(No) ayes
e'", We have (N>)/(No) = e2""T = (Ni/No)’. We find (0.528)? =
0.279, as is observed. Therefore there is an equilibrium distribution in
these levels.
(b)
In (N{/No) = -AV/KT and T = —hv/k In (N,/No) =
(c)
(6.626 x 10° J s)(6.39 x 10!” s)/(1.381 x 10°73 J/K)(In 0.528) = 480 K.
N3/No = e°""" = (Ni/No)’ = (0.340) = 0.0393. This is an approximation
since anharmonicity has been neglected.
22.21
With the +1 neglected, (22.77) becomes (N,),- =e ®/*” /eM". Use of the
Prob. 22.13 result gives —w/RT = In(z/N), so eY*" = el) = 2/N. Hence
(N,)., =e °"" Niz, which is (22.69).
22.22
In Zpp =—BUN/Ng + D, (41) In [1 + e PH -€) ), Where Eqs. (1.67) and (1.68)
were used. Since | >> eP(#/"a-®) | we can use Eq. (8.36) to expand the log and
we need include only the first term in the sum; so In Zhe ~
—BUNIN, + >, (£1 [te P/N =") | = -BuN/N, +X, eH NA-EOT
Ror
/kT
kT
4-6,
(N,) << 1, the +1 in (22.77) can be neglected to give (N,) we gW/Na
So In Zip ~—BUN/Na + ©, (N,) =—WN/NakT + N. Use of (22.77) with the
+I neglected gives >). (N,)
=N= ne.
8 el
N(x, e*'*" ) and p/NakT= In N
tun sock
eh Bie
-In(X, e ©"/*" ). Hence In Zin =~-NInN+
Nind,e *!’ +N=NInz—In N! and Zpp ~z'/N! for (N,) << 1.
22.23
We have exp(€,/k7) = exp(m vs /2kT)exp(mv*, /2kT)exp(m v2 /2kT). However,
because of the +1 term in (22.77), the population (N,) ce (which is related to
the probability of occupation of state r) is not equal to the product of separate
factors for v,, V,, and v,; hence, v,, V,, and vV, are not statistically independent.
335
(N,) << 1 does hold, the +1 can be neglected and (N,) becomes the
When
product of factors for the three velocity components. Here, the components are
statistically independent.
22.24
(a) All; (b) Zu.
22.25
(a) m;
22.26
Zel < Zrot < Zvib < Ztr-
22.27
As explained at the end of the problem, we omit the 1/N! from Z.
(aye
(b)
(c)
(b) Jando;
(c) v.
in =
eer ey
a
0 eS) We
he or
Nin(i tem). U=kT(O In ZloT)yy = kT [Ne (alkT VL + 0) =
S=U/T+k1nZ
kT (e""" +1).
Nal(e™" +1). Cy = (QU/0T)y = Ne
= NalT(e*™ +1) + NkIn(1 +e").
a/kT=1.81 andz=1+e!*=1.163. U=(6.0x 10%)(1.0 x 107° J)/
(e!®! 41) =844J =202 cal. S = U/T+ Nk In z= (844 J)/(400 K) +
(6.0 x 107°)(1.38 x 10° J/K) In 1.163 = 3.36 J/K = 0.80 cal/K.
As T—4 09, we have eV*T
_, 1 and U 3 Na/2. This is because in the
T = © limit, the populations of the two levels become equal and U(T = °)
equals the average energy of the levels multiplied by the number of
molecules. As T > 0, Cy > Na?/4kT’
> 0. At very high T, the
populations of the two levels have become essentially equal, so we get no
further increase in U as T increases further; hence, Cy = (QU/0T)y
becomes 0.
(d)
AsT> ~, S 40+ Nk In 2 = Nk In 2.
(e) Let
t= kT/a. From the.expression in (a), Cy = Nke'“/t?(e!" + 1)° =
nRe\/t?(e! + 1)? and Cy.m/R = e'/t?(e' + 1). (The first printing of the
textbook erroneously requested a plot of Cy.n/RT.) The plot is
336
Ge
Sy
0.4 |
08 +
Oe
Oy
!
(f)
0
‘
0.5
7
|
1.5
cuekia
2
O has only translational and electronic contributions to Cy. O has a very
low-lying electronic state whose energy a above the ground electronic
state is such that kT/a > 0.42 (the maximum on the above graph) for T =
300 K and Cy decreases for temperatures above 300 K.
22.28
Equations (22.76) and (22.90) give (N,)/N =e 8? fein =e zyip = UW evi =
d - eeuey.
22.29
(a)
(d)
Wein = 1 -—e7'° = 0.999955.
1—e'=0.632. (e) 0.095.
(a)
(N,)/N =e oweulKt Feri
(b) 1—e° =0.950.
Ce ues
eee
(c)
1—e? = 0.865.
ye
e POvn/T (7 _ @Ovn/T) where (22.75), (22.90), and (22.88) were used.
(b) From (22.75) and (22.76) with
v=0: (No)/N = Uzviy and zyiy = N/(No).
A similar formula holds for Zrot.
(c)
Ovm=Avi/k = hVclk = (hclk)(2143 cm’) = 3084 K, as in the first
example in Sec. 22.6. At 25°C: Oyi,/T= 10.344;
(Ny)
(G022 6107 i= 2 ns 41 )i=.6,022
1058.
(Ne View em(leten es) = 1.94 210%,
At 1000°C: @yit/T= 2.422. (Ny) = 5.49 x 10°. (N,) = 4.87 x 107.
(d) For ©yi, = 3352 K and v = 1, we have (N,)/N = €©? OT] — ¢C352 KT),
We find
TIK
0
(N,)/N 0
2000 4000 6000 8000 10000 12000 15000
0.152. 0.245 0.245 0.225 0.204 0.184 0.160
337
N> is not a harmonic oscillator, so anharmonicity and the finite number of
vibrational levels make the high-T values inaccurate.
(N,N
O03"
7
O22
Osa
O
0
1
|
=
4000
8000
12000
16000
T/K
2230 na Na IN\al = 0 =(didine =)
(v + 1)(OQy/Te PPT
latesme aye (Ge Een
vib /T
at
= 0 and v = (v + Le?” , so Qvi/T =
—In[v/(v + 1)] and T = Oyjp/In[(o + 1/0] = Ovin/InC. + 1/v). For Oyii/T=
-In[v/(o
iy 1)), we
get (N,) ING
e In{v/(v+ 1)] as ee
UinluKos)) ae
exp{In[v/(v + 1)]°} —exp{In[v/(v + L]’*!} = v'/(v + LP - v7
(SP ae
ee Sains be
22.31
+ Y=
Yes. Despite the fact that €,
> €,in Eq. (22.72), N, might exceed N, if the
degeneracy g, exceeds g,. (For example, see Prob. 22.19.)
22,325 (A) ez» cn(levels\ SinCale bewkekl Gav Gahpeek sme Meme
(b)
= 3.055
Equation (22.71) gives (N(e,)) = Ng,e°9/*" Jz,
Sov N)) = (G02 sql Ossie
is Qac e 1 oo aa.
(N>) = (6.02)x 10ay3a uy” 9/3.93p ye met OF
(N;) = (6:02 x 10ssbe@Be pane 12-095. =[ieTlel Og:
(c)
AsT—3
~~, we havee o"*
5 ¢°%=1 andz514+34+5=9.
So (N,) — (6.02 x 10/9 = 0.669 x 10°’, (N,) — (6.02 x 1077)3/9 =
2.01 x 10°, and (N3) — (6.02 x 107°)5/9 = 3.35 x 10°.
22.33 (a) N!=1-2-3---NandInN!=In1+In2+In3+---+mnN=>D%, Inx.
338
(b)
For large N, the main contributions to the sum come from the later terms
(where x is reasonably close to N); the later terms don’t vary greatly as
we go from x to x + 1. (For example, In 50 = 3.912 and In 51 = 3.932.) So
Eq. (22.79) can be used.
22.34
(c)
Dea Inx~ JY Inxdv=(xInx-x|* =NInN-N+1=NINN EN.
(a)
Let 0 = Ojo. Then Oz = DF (20 + Le M*"",
a = 0, and f(/) = (2/ + le so0
J corresponds to n,
Dy So f(O) = 1. Differentiation gives
flee O(2i I)iTleat? 1)" ony
ONT
f(D) = [602 + DIT + 0°27 + 1)}/PJe8077
FED =I T 12072)
Te 0 Os LITlen a
fa) ——120/ T1205 20 17s
f°) = (602) + DOT’ — 20027 + 18+ (27 + 1P04/TYe 88 tT
Te
[12005/ Teal SOQ 411)-07/T 2 3027
10
Te OF)
at
e IT
¢Q) = 1200/7? — 1806°/T + 300°/T* — 65/T°.
Noting that J* (27 + 1)e°Y "+" ay = J" e"" dw = TIO, we have
G2
WO
1422 0/1/12 + Gl20/T + 120711 6 1 20
(12007/T° — 1800°/T* + 3004/T* — 6°/T°)/30240 +--+ and
Zrot = (T/6@ro)(1 + Oro
3T + 7, /15T? + 40? /315T? +---).
rot
(b)
O,o/T = 85.3/273.15 = 0.3123. Equation (22.85) gives Zor = 1/2(0.3123) =
1.601. Equation (22.86) gives Zror =
1.601[1 + 0.3123/3 + (0.3123)7/15 + 4(0.3123)7/315 +--+
]=1.779 and
the error is -10%.
(c)
22.35)
Oro/T= 2.862/273.15 = 0.01048. Then Z,o: = 1/2(0.01048) = 47.72 and
Zot = 47.72[1 + 0.01048/3 + (0.01048)7/15 + 4(0.01048)°7/315] =
47.72(1.0035) and the error is -0.35%.
Urop = NRT*(d In Zo)/dT) = nRT?(d/dT){In T — In 6©,o:) = nRT.
Uviy = MRT? d In zyp/dT = —nRT*[d In (1 = dT] = RT" (-v/kT?)/
(1 ~ eT)
= (nRhv/k)i(e"™"" — 1). Uy = nRT (d/dT) In ge1o = 0.
339
22.36
Hence,
As T > ©, the Taylor series for e* shows that eovr/T _» 1 4 Oyi/T.
22.37
Si: = UlT +Nk In 2 — Nk(In N - 1) =
/ TY = nR.
TY (Ovi/
Cyviy 9 NR@viv
3nR + NK{3 In (2nmkih’) + > In T+ In V]—Nk InN + nR = snk +
nR In [(2nmk/h?y?T??(NKTIP)(AI/N)] = >nR +nR In [(2nm)7(kT)Y/hP.
Srot =
T + NK In Zot = NR + nk In [T/OQ,zot].
Urol
nRQwiT(eo* =1)—nRin( —e 9").
0+nR
Svib = Uvir/T+ Nk In Zvib =
Sa=Uesl T+ NE In 21 =
In 8e1,0 = nR \n £el0-
22.38
Siem@/R = 2.5 + 2.5 In (T/K) — In (P/bar) + 1.5 In M, +
In{(2n x 10° kg/mol)*” (1.38065 x 10° J/KY"?K” x
(6.02214 x 107/mol)*7(6.62607 x 10° Js) *(10° N/m*)'] =
2.5 + 2.5 In (T/K) — In(P/bar) + 1.5 In M, — 3.65171 =
2.5 In (T/K) — In(P/bar) + 1.5 In M, — 1.1517.
22.59
For these gases of closed-shell monatomic molecules, S = Sy.
Equation (22.108) gives Sj,793 (He) =
(8.3145 J/mol-K)(1.5 In 4.0026 + 2.5 In 298.15 —In 1 — 1.1517) =
293 (Ar) = 154.85
126.15 J/mol-K. Similarly, S*,59g(Ne) = 146.33 J/mol-K, Sj,
Jmol-K, $2,593 (Kr) = 164.09 J/mol-K.
22.40
For H, geio = 2 and Eqs. (22.107) and (22.108) give S= late beim els
(8.3145 J/mol-K)(1.5 In 1.0079 + 2.5 In 298.15 — In 1 — 1.1517 + In 2)=
114.72 J/mol-K.
Cp 203 = Cy m29g + R = 20.79 J/mol-K.
22.41
Cy 29g = (3/2)R = 12.47 Jimol-K.
22.42
Equations (22.94) and (22.95) give Uj,
593 — Umo = Utem = (3/2)RT =
1.5(8.3145 J/mol-K)(298.15 K) = 3718 J/mol for each gas.
340
22.43 S=(U-Up/T+kInZ; InZ=[S—(U-Up\/TVk=
[154.8 J/K — (3718 J)/(298.1 K)](1.3807 x 10-7? J/K)! = 1.03 x 10” =
2.303 log Z. So log Z = 4.48 x 10" and Z = 10**8*!" for 1 mole. Z=*/N!
and In Z=NInz—InN!=NlInz-NInN+N.
Solnz=(InZ+NInN-N)/N
= (1.03 x 10°°)/(6.02 x 107) + In (6.02 x 1078) — 1 = 70.9 and z=6 x 10°”.
22.44
(a)
V5 = VP 2 V Xa23586'em OC 4ercms) — 2530.0'enn
Be=iBs 5 &, = 1.998 cm™ — 10.017 cm”) = 1.9895 em”.
hclk = (6.62607 x 107*4 J s)(2.99792 x 10° m/s)/(1.38065 x 10°*° J/K) =
0.0143877 m K = 1.43877 cm K.
Ovi» = Vo he/k = (2330.0 cm ')(1.43877 cm K) = 3352.3 K.
(b)
(c)
Orot = (1.9895 cm™')(1.43877 cm K) = 2.862 K.
2 = (2MMKTINa)?V/h?. V=nRTIP = 12310 cm>. 2, = (0.012310 m*) x
[27(0.0280 kg/mol)(1.3807 x 10°? J/K)(300 K)/(6.022 x 107*/mol)]*7/
(6.626 x 10°47 Js)? = 1.78 x 10°”.
Zrot = T/O@ro = (300 K)/2(2.862 K) = 52.4.
eee ee
(lee
le) OOU0 tare gale
V = nRTIP = 1.026 x 10° cm? and zy, = 3.57 X 10°73 Zot = 437;
pa
22.45
(a)
hte
WAIsh
ores yan
Ovip = Vy hclk = 5696 K and Ovit/T = 19.104. Orr = By hc/k = 29.58 K.
Equations (22.105)—(22.108) give at P = | bar:
S°..m, tr = (8.3145 J/mol-K)(1.5 In 20.0063 + 2.5 In 298.15 — 1.1517) =
146.22 J/mol-K.
S* sot = (8.3145 I/mol-K){ 1 + In[(298.15 K)/1(29.58 K)]} = 27.53 J/mol-K.
5°.iy= (8.3145 J/mol-K)[19.104/(e171 — 1) = Ind = e*)] = 8 x 107
(b)
J/mol-K.
S°,, = 0 (since all electrons are paired).
SO
ee Saito mmet om)
mee | ae
aOR
Equations (22.100)—(22.102) give: Cy um = 1-5(8.3145 J/mol-K) =
12.47 mol-K.
C2com = 8.3145 S/mol-K. Cy vin =
1? %/(e!? 1?— 1)° = 1.5 x 10° J/mol-K.
(8.3145 J/mol-K)(19.104)°e
So Cy m293 = 20.79 J/mol-K.
341
(c)s
22.46
29410 S/mol ks
(Cp = Cet
p= m’/2m = m/2 = +(126.90 g)/(6.022 x 10 yea 0530 108 cee he
(1.0536 x 10°? g)(2.67 x 10% em) =7.51x 10° gem’. On = h?/810Ik=
0.0536 K. Qyp = hvo/k = 306.9 K and ©,i,/T = 0.6138.
(a)
From Egs. (22.94)-(22.98) at 500 K:
Uj. = (3/2)RT=
1.5(8.3145 J/mol-K)(500 K) = 6236 J/mol. U},.,o¢ = RT =4157 J/mol.
U*,sy= (8.3145 J/mol-K)(306.9 K)/(e?*'** — 1) = 3011 S/mol.
U?,4=0. U%,509 — Uino = 13404 J/mol.
OO)
(c)
Pea
ee Ons
13404 J/mol + (8.3145 J/mol-K)(500 K) = 17561 J/mol.
From (22.105)-(22.108): S°,,, =
R(L.5 In 253.8 + 2.5 In 500 — 1.1517) = 188.65 J/mol-K.
S°,so = R{1 + In[500/2(0.0536)}} = 78.55 J/mol-K.
S? ap = R(0.6138/(e°°'** — 1) — In(1 — €°*!**)] = 12.50 J/mol-K.
S220,
(dq)
Uae
SOS)
hy SV
Oia,
Giisoo -Umo =4mso0 — TS Ey a
any
17561 J/mol — (500 K)(279.70 J/mol-K) = —122.29 kJ/mol.
22.47
From Fig. 21.13 the pure-rotational lines are at 2Bo, 4Bo, 6Bo, . . . , so the
separation is 2B, = 20.9 cm’! and By = 10.45.cm™’. Also, Vp = 2885 em So
Orot = Byhclk = 15.04 K; vip = Vy hc/k = 4151 K and Oyip/T = 13.92. We
shall assume the spectral data are for the predominant species HCl and shall
calculate S° 593forH Cl. At 25°C:
S*. = R(1.5 In 36.0 + 2.5 In 298.1 — In 1 — 1.1517) = 153.5 J/mol-K.
S° rot = R{1 + In[298.15/1(15.04)]} = 33.1 J/mol-K.
See R[13.92/(€i ae 1) = Intl ee 2) |= 0.0001 1/moleKeens
S0.5 jos = 1 80,0,)/mol Ke
342
y= 0)
22.48
(a)
For this relatively light diatomic molecule, only translation and rotation
contribute to Cy and Cy,,,= (3/2)R + R = 2.5R = 20.79 J/mol-K.
Cpom= Cy m +R =3.5R = 29.10 J/mol-K. The true value is 29.13 J/mol-
K.
(b)
All gases of relatively light diatomic molecules have Cp,,= 3.5R =
29.1 J/mol-K.
22.49
€.,,/k = (0.0149 eV)(1.6022 x 107! J/eV)/(1.38065 x 1077? J/K) = 172.9 K.
Zeal = elo He Baines Sin
ward + Ze 2 TAL 30 Kozy ai 2iei2e aise a= 2.006.
At 150 K, z1 = 2.63. At 300 K, z,) = 3.12.
22.50
Using the z.; expression found in Prob. 22.49, we have Ue, = NkT°(d Ineze al) =
(nRT"/z-1)(dzeVdT) = (nRT7/z1)(2(173 K)T 761? 87) = nRB46 Kye 72,
and Ueim = (2875 J/molje
1? 6"y[2 + 26177 8) = (1438 Jmol/e"? 7 + 1).
Cveim = dela = -(1438 J/mol)[-(173 KVP Jen (eB OF + 17 =
(2.49 x MO
K/mol)T7e"'? ST Kens we
ie We find (note the maximum in
Cv cim)
Cvetm/(J/mol-K)0.86
T/K
30
2.52 3.47 3.42 2.67 2.02 1.30 0.88 0.64
45
60
90
120 150° 200 5250, 300
C\e.n/(J/mol-K)
4x
ntxe
txt -(x4+0
29-51. (a), saxselex
(b)
—»).
Sos=
+°--)=1.Wd
The Taylor series for 1/(1 — x) about x = 0 is given by Eq. (8.8) as
x + store x Weel
te
ee
343
22.52
(a)
Zvib = ss 2 =eanay hal = » Seeseres
= Bol tds aE ie, BORAT —
eVAKTI( — e""1) where Eq. (22.89) was used.
(b)
Uyy = nRT-(0 In 2viv/OT)y =
In vip = —SAV/AT—In (1 - eT
nRT[L hvlkT? + &(avikTYL — =
=Niv +
nRhv/k(e""" — 1), which differs from Eq. (22.97) by M(hv/2); this agrees
with the Prob. 22.5b result.
22.53
From Eq. (22.100), Cvirm = 3R/2. The translational levels are so closely spaced
that this result holds at all temperatures not extremely close to 0 K. From p.
846, O,o1 is typically of order of magnitude | K (except for hydrides). Figure
22.11 shows that Cy,;otm = R for T above 1.30,. From p. 846, Ovi is typically
10° K. Figure 22.10 shows that Cy,vib.m is negligible below 0.10,» and
increases to R as T increases to | or 2 times ©yjp. So:
CHeya/
Cym
2.5R
1.5R
TIK
1
22.54
10
100
1000
Zot = [(298 K)/1(2.77 K)][1 + (1/3)(2.77/298) + - - +] = 107.5.
Fron et
(a)
Le */2] = J(J + 1)k©yot, where (22.83) was used.
Use of Eq. (22.71) gives
[eyo (NZ) VN = Zi
Dye (QJ + Lexpl-JU + @ro/T) =
Zin Jy (2S + expl-(J? + N@ro/T) dd = 25), 17*7 exp[-(@,o/T)w] dw =
Zo (—T/OQyo:)[EXp(—306@yo/T) — 1] = (107.9) '(298/2.77)[1 — 2002 77/298) —
0.939, which is 94%; we took w = J’ + J and dw = 2J + 1. The sum has 17
terms, so 17 was used as the upper limit of the integral.
344
(b)
The required sum is (107.9)!
D'6, (20 + Le ~ +2728). Direct
evaluation gives 0.932 = 93.2%. The individual percentages are 0.927,
2.73, 4.38, 5.80, 6.93, 7.71, 8.15, 8.26, 8.07, 7.63, 7.00, 6.25, 5.44, 4.61,
3.82, 3.09, and 2.44 for J/=0, 1, 2,..., respectively.
A BASIC program
iS
10 S=:0
70
20 A=2.77/298.1
80 NEXT J
30 FORJ=0TO
16
S=S+B
90 PRINT "SUM=";S
40 B= (2*J+1)*EXP(-A*(J*J+J))/107.9
95 END
50° P = 100*B
SOPRRINT d=40, POP=—e
22.55
v =(k/p)'/2m. N> and F; have roughly the same p values, but N> has a triple
bond and F; has a single bond, so ky, >> kp, (cf. Table 21.1). The high ky,
makes vy, and ©,;,, high, so the vibrational contribution to Seeneeis
negligible at room temperature. The light mass of H makes Uy << Hr, , SO
VHF >> Vp, and S Vin m298,HE << 5 vib.m,298.F) °
22.56 (a) wr= Soe? dv=(TIOwpe |, = T/Ovin.
(b)
For high T the Taylor series for e* gives PEE
SO Zyib
22.57
Ne IESE.
2 L/(AV/kKT) = T/Oyin.
Equations (22.104) and (22.82) are not applicable at extremely low T since the
translational levels are then no longer closely spaced compared with kT and the
sum in 2, [Eq. (22.78)] can’t be replaced by an integral. Another reason is that
at extremely low T, the condition (N,) << 1 does not hold, so Bose-Einstein
or Fermi—Dirac statistics must be used.
22.58
(a)
Zt,N, * Ztr,COs since My, =McoOsince one
Zrot,.Ny and Zrot.co differ substantially
2 but Gco = 1. We have Un, =Hco and ky, = kco for these
isoelectronic triply bonded molecules, so VN = Veo (sce Labier 1.1):
therefore Z,;, nj, ~ Zvib,cO-
Zel for each species 1s essentially 1.
345
(b)
From (22.64) and (22.104)-(22.107) and the results of part (a),
S m,298,CO(g)
S m,298,N > (g) =5 298.m,rot,CO(g)
N 2(g) a
— Ssm,298,rot,
R In (Gy, /Ocq ) = R In 2 = 5.76 J/mol-K, since we expect Oot to be
about the same for the two species. The actual difference is 6.06 J/mol-K.
ppanh)
(a)
Br> is the heaviest molecule and has the largest moment of inertia and the
smallest ©,o:. N> is the lightest molecule and also has the shortest bond
(b)
length (since it has a triple bond), so N> has the smallest / and the largest
One
v= (1/2n)(k/p)'”. The large p: of Br2 makes this singly bonded species
have the smallest v and hence the smallest Oyj». The high & and small pt
of N==N makes vy, and ©,,,n, the largest.
(c)
The low value of ©, for Brz2 makes more of its excited rotational levels
populated and makes Z,,; p,, largest.
(d)
The small v,,, and O,i,.5,, make excited vibrational levels populated
and Z,i,p,, 18 largest.
(e)
All these species have essentially the same Cy,rot.m at room temperature
since they all have essentially attained the high-T limit Cyrot.m = R at
room temperature.
(f)
22.60
Bry, for which excited vibrational levels are accessible at room 7.
Zr is very roughly equal to the number of translational energy levels that have
significant populations at temperature T. The particle-in-a-box translational
levels have €, proportional to 1/m (Chap. 18), so an increase in molecular mass
m lowers the translational levels and allows more of them to be populated at T,
thereby increasing Z,, and increasing St.
22.61
(a)
At 20 K, Zot = T/O@yor= (20 K)/(29.577 K) = 0.676. At 30 K, Zrot = 1.014.
At 40 K, Zot = 1.352.
(b)
Zrot = 0.6762[1 + 1/3(0.6762) + 1/15(0.6762)° + 4/315(0.6762)°] = 1.136
at 20 K. At 30 K, Zoe = 1.426. At 40 K, Zor = 1.742.
346
(c)
A BASIC program is
1
Ont =1293577.
60 IF A< 1E-12 THEN 90
fq ooreyeasiid)
70
S=S+A
SUMINEU Teta rsT
80 NEXT J
40 FOR J =0 TO 10000
SOIRRINT ZROT=2S
50 A= (2*J + 1)*EXP(-J*(J + 1)*TH/T)
95 GOTO 20
99 END
he exact values are 7.4; =1.1 565 at 20 Koo — 14I at 30 Kee
1.7439 at 40 K.
22.62 S=U/T+kInZ=kT@ In Z/0T)yy, +k In (“/N!) =
kT(0 In (2“/NV/OT)
yy, +NkInz—kIn N! = NKT(Q In /OT)y y, + Nk In zk In N! = (NKTIz)(0Z/0T)
yy, +NkInzDse°""" /z—kIn N!. Use of dv/0T=
(O/0T) Xe 88/" = (UT?) &, ee 8"
(Nk/z) Inz
gives S = (NizT) Ds ee 8"
+
se!" —k InN! =NKY, [(e/kT + In z\(e *8""" /z)] — k In NI.
From (22.69), x; =e ~€s/KT 7 and In x; =—€,/kT —1n z, so
S =-Nk ¥, x5 In x;—k In N!. The other condition is (N,) << 1, as in (22.49).
22.63 In Zot = In(T/GO,o) + IN(1 + Oyo/3T +O2,/1ST? +--+) = In T= In(GO yor) +
(OLY ec ae eyeee 1(Qro/3T + ©; NG Esdayne =
Tot
rot
In T — In (GO,q) + Oro
3T + (1/15 — 1/18)(O2,/T?) +--+ = In T= In(GO x) +
Oro3T + 02, /90T? +--+. dln Gold = WT —Oro3T - OF, /45T? +--+.
Uror = nRT?(d In Zo/dT) = NRT— NRO; o/3 — NRO 7,/45T +--+. As T > &, all
terms after the second go to zero and U,. > n(RT— RO,o/3).
22.64
(a)
We have the original orientation plus the following: 120° and 240°
rotations about the C3 axis give 2 indistinguishable orientations; rotation
about a C> axis followed by rotations about the C3 axis give 3 more
indistinguishable orientations. So o = 6.
)
%
(c)
1 (the same as for a heteronuclear diatomic).
347
(d)
120° and 240° rotations about each of the four C3 axes produce 2(4) = 8
indistinguishable orientations; 180° rotations about each of the three C2
axes (these coincide with the S4 axes—see Fig. 21.22) produce 3 more
indistinguishable orientations. Adding in the original orientation, we get
eel”.
(e)
2 (the same as for a homonuclear diatomic).
(f) There are three C2 axes and o = 4.
(Cay er
22.65
S*_, = R(L.5 In 34.08 + 2.5 In 298.15 — 1.1517) = 152.87 J/mol-K.
Ay =hI8r7clao, By = h/8n°clyo, Cy = etc. Equation (22.109) becomes Zrot =
(!7/6)(kTIncy?(AI Ay By Cy)”. Using o = 2, we get Zr = 125.9.
So $2, = RU.5 + In 125.9) = 52.68 I/mol-K. @vin,i/T=(2615 em™ Jhc/kT =
12.62. Ovip2/T =5.709. Ovin.3/T= 12.68. Sip m= RU12.62/(e'7* — 1) In (1 — e712) + 5.709/(e7? — 1) — In (1 — >) + 12.68/(e'7 — 1) -
In (1 —e7!?*)] = 0.186 J/mol-K. S%,, =0. Adding, we get So, =
205.74 J/mol-K.
22.66
Siem = R15 In 44.01 + 2.5 In 298.15 — In 1 — 1.1517) = 156.05 J/mol-K.
O.S
By hc/k = 0.561 K. For this linear molecule, Eq. (22.105) gives
Srom =R{1 + In [(298.15 K)/2(0.561 K)]} = 54.73 J/mol-K.
Ovin./T=
(1340 cm! )hc/kT = 6.466; Qyin.2/T = (667 cm )hc/kT = 3.219;
Ovin3/T =
3.219;
Oyin.4/T= 11.34. Equations (22.118) and (22.106) give S\,
R[6-466/(e OS 1) Sn Cle)
eS 21D Ve
wes) oe
=
nl er) +
11.34/(e!!34 — 1) — In (1 — e!!™4)] = 0.3616R = 3.01 J/mol-K. Si,_ =0.
S°,x93= (156.05 + 54.73 + 3.01 + 0) J/mol-K = 213.79 J/mol-K.
22.67
Cy tnm = (3/2)R.
Cy vipsm — ¥ at high T for each vibrational mode s.
Cyrtm — Kat high T for linear molecules; Cy om
nonlinear molecules.
348
— (3/2)R at high T for
(a)
Linear. 3(3) - 5 = 4 normal modes.
54.04 J/mol-K.
(b)
Cy,, > RU.5+4+1)=6.5R=
Cp,, =Cy , +R = 62.36 J/mol-K.
Nonlinear. 3(3) - 6 =3 normal modes.
Cy,, > R(1.5+3 + 1.5) =6R=
49.89 Jimol-K. C% _—> 58.20 J/mol-K.
(c)
12 normal modes.
Gre)
22.68
35.03)
Cy , > R(1.5 + 12 + 1.5) = 15R = 124.72 J/mol-K.
/moles
Since z, is the same as for a gas of diatomic molecules, U, is given by (22.95).
For a gas of linear molecules, Z,o 1s the same as for a gas of diatomics, so
(22.96) gives Zo. Use of Eq. (22.109) gives for a gas of nonlinear molecules:
Urot = NRT d\n ZjodT = nRT d(= In T + const)/dT =
=
3nRT. Equation (22.110) gives Uvin = SrRinidiabimi le (iesen
*nRT(didT) Yin (1 =e ©:
mR
Ou es
1)
22.69
(a)
yi= =nRT Valet
|
COVE
= etat
)) =
AtOK, there is no translational, rotational, or vibrational energy (above
the zero-point energy) for the H2 molecules; AU 4 is determined by the
change in electronic energy and AU 9 = (4.4780 eV)Na = (4.4780 eV) <
(1.602176 x 107! J/eV)(6.02214 x 10°*/mol) = 432.06 kJ/mol.
(b)
At 25°C, we need consider only translational and rotational contributions
to Um of H2(g); the translational contribution is =RT and the rotational is
RT, for a total of > Rip The translational contribution to Um of 2H(g) is
2( >RT) = 3RT. Hence AU9, — AUg = 3RT— 5 RT = 5 RT. Then AH 59g
°
°
5
1
°
= AU 59g + AngRT/mol = AUg + Asie 4+ RT =AU> + =RT = 432.06
kJ/mol + 1.5(0.0083145 kJ/mol-K)(298.15 K) = 435.78 kJ/mol. The
Appendix gives 2(217.965 kJ/mol) = 435.93 kJ/mol.
22.70 (a) Hoy —Hijg = Usp +RT-Unny =U inte — Uno + RT =GI2)RT+ RT
= 2.5(8.3145 J/mol-K)(298.15 K) = 6.1974 kJ/mol.
(b) 2.5RT = 20.786 kJ/mol.
349
(C)i
Gp
hao
har oO San
ge aT
io ean rae YC
take H;,, —H},9 from (a) or (b) and use (22.108) for S 7, [Alternatively, (22.123) can be used.] Gj,59g — Hi, = 6.1974 kJ/mol —
(298.15 K)(0.0083145 kJ/mol-K)[1.5 In 39.948 + 2.5 In 298.15 — 1.1517]
= —39.970 kJ/mol.
(d)
22.71
Gi jo00 — Amo = 20-786 kJ/mol — 180.00 kJ/mol = —159.22 kJ/mol.
T _
From (22.123) with 7,9 =U ;,o, we have z = Ne —(Gmr-Hmo)/R
(6.022 x 107*)exp[(257.7 J/mol-K)/(8.314 J/mol-K)] = 1.739 x 10°”.
22.72
A-—Up=-kT In Zand In Z=—-(A — UpWkT.
A-Up = U- Up -TS=
H — Ho) -TS, since there is negligible difference between H — Ho and U — Up
for a solid. So A — Up = [523 J/mol — (298.15 K)(2.377 J/mol-K)](1 mol) =
~185.4J.
22.73
In Z= (185.7 JKT =4.51 x 10”.
Substitution of (22.124) into )); vu; =0 gives
0= ¥, VUmo, — RTE; v,In(z,/N,;) = AUS — RT In], (z,/N,)”']. So
AUCIRT = In{T]; [(z,/VN.,)/(N,/VN, 1" } = In{T, [(z,/VN..)/c, J}. Then
eXpAUG/RI)Y=l1(GZIVNS)/
cc) Nl, GivN) “Wiklete) lsc
Ket
P(e
= exp AU Rd oleae VN)
22.74 S=kIn Wand AS =k In (Wena Winitiat), SO Wrinad Winitiat = Co.
AS mix= —R(1.00 mol In 0.5 + 1.00 mol In 0.5) = 11.53 J/K.
So Wena Winitiat = exp[(1 1.53 J/K)/(1.381 x 10-7? J/K)] = exp(8.35 x 107°). Let
10” = e83*! Taking base-10 logs gives y = 8.35 x 107 log e = 3.63 x 10,
SO Wena Winitial = 102
22.75
ee
S=kIn Wand In W= S/k = (191.61 J/K)/(1.3807 x 10°73 J/K) = 1.388 x 1029,
where the Appendix was used.
350
22.76
Stom = SNan =kIn
Wrom —kIln
Wan
=k In [eo
Wnan] —kln
Wan
=
kine") +k In Wyan — k In Wyan = K(10"2) = (1.4 x 107 I/K)10" = 107" I/K,
which is utterly negligible.
22.77
(a) Yes;
22.78
(a)
Ne.
(b)
Each has 2 electrons. The single +2-charged nucleus in He holds the 2
electrons more tightly than the two +1-charged nuclei in H2, so H2 is
more polarizable.
(c)
Ch.
(d)
Each has 18 electrons. H2S is more polarizable.
(e)
CrHe.
22.79
(b) no;
(c) no.
[In the first printing, Eq. (22.133) has an error; the 47t€, should be squared. ]
Var’ = —2[(1.60 D)(3.336 x 10°°° C m/D)]*/
{3(1.38 x 10°? J/K)(298 K)[4n(8.854 x 107!” C?/N-m7)]?} =-106 x 10°” J m®.
Vaigr® = -2[(1.60 D)(3.336 x 10°°° C m/D)]*(6.48 x 10°? m*)/
[4(8.854 x 107!? C?/N-m7] = -33 x 10°” J m°.
Vaisp?” = -[3(11.3 eV)7/2(22.6 eV)](1.602 x 107"? J/eV)(6.48 x 10° m*)’ =
=5/0 10; Im.
22.80
22.81
F =—dVidr = 4e(120'/r? — 60°/r’) = (24¢/r)[2(a/r)"” — (G/r)°).
€ = 0.013 eV =2.1 x 107! Jando =3.8 A.
10> (Selo mileG.e/8), — (4.8/8) |= =a
(ane eto
10> NO (Cc) 30. < 107 N,
Ne (byl
(a)
(b)
(ce)
(d)
lOn
v=Alr'?- Bir’. 0 =A/o'? — Bio®, so Bo’ = A and v = Bo'/r'? — Bir’.
dv/dr =-12Bo°/r'? + 6Bir’ = 0. So r8,, = 26° and rmin = 2'°0.
€ = (ce) — V(rmin) = 0 — [BO*(2!a)!? - Bi(2'*a)*] = B/40° and B = 4o°e.
Substitution of 6 = rmin2-”” in (22.136) gives v =
Ae(r!2 /4r'? 76 /2r°) = e[(rmin/7)'? — 2(rmin/ 7].
22.82
22.83
(b)
Tmin = 2! = 2'°(3.5 A) = 3.9 A.
min = 2!°(8.6 A) = 9.65 A.
(a)
As noted on p. 868 of the text, for a nonpolar molecule € = 1.3k7) =
(a)
1.3(1.38 x 10°23 J/K)(27.1 K) = 4.86 x 10°” J.
22.84
(b)
From Sec. 8.3, T, = 1.67», 80 € = 1.3k(Te/1.6) = 3.42 x 10-7" J.
(a)
We approximate the intermolecular potential by the Lennard-Jones
potential. F(r) = -dV/dr = -4e (-126'7/r?’ + 60°/r’) =
24e(20'4/r'? ~ o°/r’) = c/r'? — d/r'. The F (r) curve has the same general
appearance as the v(r) curve, being 0 at r = 9, negative (i.e., attractive)
for large r, positive for small r, and o at r= 0.
(b)
0 = 24e(20'7/b"3 — o/b’), so 20° = b® and b = 2"o. [Cf, Prob. 22.81b;
F(r) = 0 at the minimum in v.]
22.85
22.86
(a)
Xe (higher M);
22.87
(a)
v/e = 4[(6/2'%o)'* — (6/2'%o)"] = 4(~ — +) = =I (this is the minimum),
(b)
4[(o/1.50)!? — (o/1.50)°] = -0.320;
(c)
0.062;
(a)
For an ideal gas, /= 0; for linear molecules Eq. (22.145) becomes
22.88
(b) C2H5OH (H bonding);
(d) —0.016;
(c) H20.
(e) —0.005.
Zeon = (40) J§ sin 0, dO, --- [5 sin O@v dOy J5” do, --- J2" doy x
Jo dx J dy: [6 dz -- + S5 dxy J8 dyy JS dey=(4m) %2%(20)% (abc) =
352
V™. For an ideal gas of nonlinear molecules, (47) is replaced by
(81°) = (2n) “(411)” and we have the additional integral factors
Ro aaee
(b)
Jo" dx = (2m), so we still get Zoon = VY.
P=kT(0 In V*/OV)rN = NKT(O In V/OV) = NKTIV.
22.89 B=-2nNal? (eT — 1) dr =-2nNalJ§ (CO — Dr drt [5 (€ - Vr ar]
= —2nNa(-J¢r° dr + 0] = 2mNad’/3 = 4Nal =n(d/2)"].
22.90
At 100 K, log(kT/e) = log[(1.381 x 10° J/K)(100 K)/(1.31 x 107" J)} =
0.0229. For this log(kT/e), Fig. 22.22 gives Big Na]—9 4nd
B= 5.4
(3.74 x 10-8 cm)*(6.022 x 1077/mol) = -170 cm*/mol. At 300 K, log(kT/e) =
0.500; Fig. 22.22 gives Bio’Na =-0.1 and B= -3 cm*/mol. At 500 K,
log(kT/e) = 0.722; B/o*N, = 0.5 and B = 16 cm*/mol.
22.91 (a)
B=-2mNalo(e — lr dr—2nNa Jo (Ee — Vr? dr -2nNa Jz (e - Dr? ar
= 2nN,0°7/3 — 2nNa(e™" — 1)(a? — 0/3 + 0 = 20Na(O? — a )e""/3 +
2nNaa’/3.
(b)
Substitution of numerical values gives B(100 K) =-163 cm?/mol,
B(300 K) = —4 cm?/mol, B(500 K) = 17 cm*/mol.
(c)
22.92
From (8.5) and (8.6), Z= 1+ B'P=1+BP/RT =
1 — (163 cm°/mol)(3.0 atm)/(82.06 cm?>-atm/mol-K)(100 K) = 0.94.
U=kT°(A In Z/OT)yn = kT In Ziel )v.n + kT In Zeon/OT)v.n = Via +
kT°(-N)[(1/Vm)dB/dT+ (1/2V 7 )dClaT +--+ =
Uig — nRT’[(1/Vm)dBlaT+ (1/2V>)dClaT +--+). S=U/T+klnZ=
UsalT —nRT [(1/Vm)dBIdT + (1/2V2 )dClaT +--+) + k An Zig + KN In V—
NE[(/Vm)B + (1/2V2,)C +--+] -Nk In V=
Sig —nR{(A/Vm)(B + T dBIdT) + (1/2V2, (C + T dCldT) + - - «J, since Sia =
) =
UsalT +k in Zig. G=A+ PV =-KT In Z + KTV In Zeon/OVrN
_kT In Zig— KIN In V + KTN(B/Vm + C/2V5,+- +>) +kIN In V+
kTVN(A/V + B/VVm + C/VV2, +--+) = Gia + NRT(2B/Vin + 3C/2V2, +---),
since Gig =
Aia + (PV)ia = -KT In Zig + NAT.
353
22.93
A=-kT
In Zand Afalse = —kT In Zfatse, SO Afalse — A = -kT In (Ztatse/Z) =
(1.38 x 1073 J/K)(298 K) In 1019"? = -(4.11 x 107?! J)2.30 log 101° =
(9.5 x 10°77! J(£10'5) = F 10° J, which is negligible.
)IN= 6
22.94 (N,)IN=e §""" Iz, s0 (Ng,
22.95
(a)
z= Iz = 1/154.1 = 0.00649.
Since A is proportional to n and A = -4T In Z, we see that In Z is
proportional to n. From (22.81), Zr is proportional to V, which is
proportional to n; sO Z is proportional to n. Since Zr is proportional to n,
z iS proportional to n.
22.96
(b)
Zot, Zvib» aNd Ze) are independent of n.
(c)
None.
(d)
Zrot, Zvibs and Zel.
Every thermodynamic system in equilibrium.
354
Chapter 23
23.1
Az=keFa!®l = ke Fw!RT ol? — Nand? e'?(4RT/1Mg)'”, where Eqs. (17.69),
(23.5), and (23.4) were used.
23.2
(a)
Equation (23.6) gives A =
(6.0 x 107°/mol)n(3.4 x 107'° m)*[8(8.314 J/mol-K)(500 K)/r]'”” x
[mol/(0.030 kg) + mol/(0.048 kg)]'"(2.72)'" =
3 x 108m? s7! mol! =3 x 10'* cm? mol! st.
(b)
8x 10!! cm? mol! s™ = p(3 x 10'* mol! s“') and p = 0.003.
23.3
391 nm.
23.4
(V,c) = IF Vreg(Vre) Bre = Womyl2MKT)"” Jy Vie EXP(-MrcV eekly dvs =
" = (2kT Item)".
(2KTIm,-)/2.
2(myel 2UKT)
23:5
The energy per molecule is (9.6 kcal/mol)/Na =
[(9600 cal/mol)/(6.02 x 107°/mol)](4.184 J/1 cal)(10’ erg/1 J) =
6.7 x 10°" erg.
23.6
(a)
p1H{H, d=0.930 A (Example 23.2). Let the origin be at D and let
the molecular axis be the x axis. The center of mass 1s at Xem = > miX/ Mor
= [2.014(0) + 1.008d + 1.008(2d)]/4.030 = 0.698 A. The principal axes
pass through the center of mass (which is on the molecular axis and
0.698 A away from D); one principal axis is the molecular axis; the other
two are perpendicular to this axis.
I, pn, = Li mir, =Nq' [(2.014)(0.698)° + 1.008(0.930 — 0.698)" +
1.008(1.860 — 0.698)?](g/mol) A? = 2.397N, (g/mol) A’.
(b) For Hp, fy, = PR? = (mj,/2my)R° = ‘mpR® =
N; *(1.008)(0.741)°(g/mol) A? = 0.2767N| (g/mol) A*. Then
Iy.pn,/ly, = 2-397/0.2767 = 8.66. Since 04, =2 and py, =I, the
rotational-partition-function ratio is 2(8.00) =e
355
23.7
(a)
Zvinn, =[l-exp-h¥c/kT)]; hVc/kT =(6.626 x 10> sis)
(4400 cm™!)(2.998 x 10!° cm/s)/(1.3807 x 10°°° J/K)(450 K) =
14.07; Zyipn, =(1-e7'*°”)' = 1.000. The activated complex has 3
ordinary vibrations, with wave numbers
1764, 870, and 870 cm”! and
zi is the product of factors for each of these vibrations. We have
AV guclkT = h(1764 cm” |)c/kT = 5.640, AV penac/kT = h(870 cm” )c/kT =
2.78). Then z?,=(l-er ss )) (lee) = 1.141 bom (22.81) we
find: (2° /VV[(zu.p/V)(Z un, /V)) = (pn, fp my, P(r /2nkTY” =
[4.030/(2.014)2.016]°(mol/10~ kg)*”(6.022 x 10°°/mol)*” x
(6.626 x 10° J s)>[27(1.3807 x 10°73 J/K)(450 K)]}°? =
5.51 x 107! m?=5.51 x 10° cm’.
NakTh ‘exp (—Ae§/kT) =
(6.022 x 107°/mol)(1.381 x 10-7? J/K)(450 K)(6.626 x 10°" Js)! x
exp{(—5.79 x 10°? J)/[(450 K)(1.381 x 10°? J/K)]} = 5.07 x 10°*/mol-s.
Then k, = (5.07 x 10°7/mol-s)(5.51 x 10°? em®)(17.3)(1.141)(1) =
iy
(b)
5.5 x 10’ cm? mol! $7.
23.8
Replacing 450 K by 600 K in Prob. 23.7b, we find NakTh ‘exp (Ne
6.94 x 10°*/mol-s. Replacing 450 K by 600 K in Prob. 23.7b, we find
t
(25 /V@eD/V)@
wn, (V)] =3.58 x 10 m°, Zinn, = 1.000, zii,
= 1.323.
So k, = (6.94 x 10°*/mol-s)(3.58 X 10° cm*)(17.3)(1.323)(1) =
5.7 x 10!° cm?/mol-s. The tunneling correction 1s 7.5/5.7 = 1.3, which is less
than that at 450 K; tunneling becomes less important as T increases.
23.9
VY =vie=(1/2nc)(Kp)'”.
Uh,
kis the same for H2 and D2 so Vp,/Vu, =
ye (<my/ mp)" = (1.0078/2.0141)'” = 0.7074 and
Vp, = 3112 cm’. The ZPE of the activated complex is
+he(1762 + 694 + 694) cm”! = 3.13 x 10°’ erg. The ZPE of H is 0 and that of
Dz is ~heVp, = 3.09 x 10° erg. A(ZPE) = 0.04 x 10° erg. The classical
barrier height 1s the same as that for D + H2, namely, 6.68 x 10.3 erg.
So Ags = 6.77 X10)—13 ere sAlso, 2442 oznel
356
iea 20
Pop Mp, 0° = 2F Ip, . Ip, = WR? = mpR’/2 = Nj! 1(2.014)(0.741)(g/mol)
A? =0.553N;! (g/mol) A?. H“D£D, d=0.930 A (Example 23.2). The
procedure of Prob. 23.6a gives (with the origin at H) xm=1.116A,
[=
2.440 (g/mol) A’, and z*./z,01,.p, = 8-82. Proceeding as in Prob. 23.7a, we find
Zvib,p, = 1-001, ae = 1.545, (25,/View/V)(Z up, )) = 5.00 x 10° cm’.
NakTh ‘exp (-Aeé /kT) = 2.26 x 10°*/mol-s. So k, = (2.26 x 10°?/mol-s) x
(5.00 x 10°7° cm*)(8.82)(1.544)(1) = 1.5 x 10'° cm? mols.
23.10
The Taylor series for e* gives at high 7, e IVSIAT ~ 1 — pv kT and Zins *
L/(hv /kT) = kT/hv,. There are fyi» factors in Zyip in Eq. (22.110), so Zyin is
proportional to T ey high T.
23.11
Let lin, nonlin, lin*, and nonlin? denote linear and nonlinear reactant molecules
and linear and nonlinear transition states.
(a)
First consider the reaction between an atom and a molecule. Let the
reaction be atom, + ling ~ lin*. The ratio ZI LeAzeB is proportional to
PPC”?
= T°” for any bimolecular reaction. Also, Ze iap ec) =
T°. Let ling have w atoms. Then ling has 3w — 5 normal modes. Further,
lin* has w + 1 atoms and 3(w + 1) —5 — 1 = 3w —3 “ordinary” normal
modes and one anomalous mode (the “vibration” along the reaction
coordinate). So lin? has 2 more ordinary normal modes than does ling.
Also, the vibrational frequencies of lin* and ling should be of the same
order of magnitude in most cases, so we expect that 2+» /ZvibsB iS
proportional to T/, where 0 <j < =(Fviv — fviv.p). Hence, pda karin Cola
where 0 <c < 1. Equation (23.23) then gives A « T'T °1°T' = T°” =
T”, where -> <m< >: Now consider atom, + ling
nonlin*. We have
velzkiyic T°’/T = T'”’. Reasoning similar to the above shows that the
species nonlin* has | more ordinary normal mode than ling, so
3/2-pl/2
é
Cea T = T",
A
z¥.,/zvibp % T°, where 0 <b < +. Hence, A
where
T'T
OS ms + For atom, + nonling ~ nonlin’, we find similarly that
ane sewherm0iaas |. So-5 <ms< +
ole ie
AMocbhiv
357
(b)
Now let the reaction be between two molecules. First consider
nonling + nonling > nonlin*. Let A and B have wa and wz atoms. Then
A and B have a total of 3wa + 3wg — 12 normal modes and nonlin* has
3(wa + wp) — 6 — 1 = 3wa + 3Wp — 7 Ordinary normal modes, which is 5
more than the total of A and B. So Zula
Hence, A
WI?
For ling + nonling
a = T", wheree
Terk al
ais x TJ’, where 0 <d<2 7
2:5 7S
*
nonlin’, we find similarly that nonlin* has 4 more
ordinary normal modes than A + B, so A x Pte Wig
where 0 < e < 2 and =
ms
eaeCe STs,
* For ling + ling 3 nonlin’, nonlin* has
3 more ordinary normal modes than A and B together, so
0S = $ and-l<m<_1!.
Aare where
aT
27 7
Mea
tN
23.12
From the Example 23.2 in Sec. 23.4, Ae}, = 5.79 x 10° erg = 5.79 x 105)
So AE® = (5.79 x 10°? J)(6.02 x 10°°/mol) = (3.49 x 10° J/mol) x
(1 cal/4.184 J) = 8.33 kcal/mol. m is the exponent of T in k,. The example in
Sec. 23.4 shows the z, ratio and the Z,. ratio to be proportional to T>”? and Ta
respectively. At 300 K, the z,ip ratio is nearly equal to | and its temperature
dependence is essentially negligible. ze; is independent of T at 300 K. So Eq.
(23.19) gives kc T'T 71° = T'” and m = —¥2. Then E, = AE}, + mRT =
8330 cal/mol — ’2R(300 K) = 8.0 kcal/mol.
23.13
The activated complex A? is an A molecule in the hole or within a distance 8
beyond the hole. We have Aci = 0. Also, Zyip, Zrot, and Ze; are the same for A
and A*. The volume factor in z*_ is 4 8, where A is the hole area; also, z*
contains an extra factor of % since A* is moving only forward along the
reaction coordinate (which is the direction perpendicular to the hole).
Equations (23.16), (23.18), and (22.81) give r= 5'(2kT/mum)!?(z* /zu.a)[A] =
51 (2RT/nm)" L(2nmkT/h?)?A S{(2amkT/h?)??
Vy [A] = (RT/20M)"? PA/VRT,
since [A] = P/RT. So J = rV = —dna/dt = PA/(2NMRT)"”.
23.14
From Sec. 21.9, Voy = 3000 cm! and (see p. 903) exp[-A(Ae})/kT] =
exp[-0.146(6.63 x 10° J s)(300000 m™')(3.0 x 108 m/s)/(1.38 x 10-7? J/K) x
358
(300 K)] = 0.122 = 1/8.2. As T
increases, the exponential comes closer to 1 and
the isotope effect decreases.
23.15
v = (1/2m)(kK/a)"”. Ver/Ven = (Mew! er)? = (my)? = (1/3)'” and ver =
vcy/ V3. The ZPE of the reactant is lowered by “2h(Vcu — Ver) =
Vehvcu(1 — 1/3) = 0.21 Lhvcy and the ZPE of the activated complex is
unaffected. Replacement of 0.146 with 0.211 in Prob. 23.14 shows k, is
multiplied by 0.048 = 1/21.
23.16
The rate-determining step doesn’t involve breaking the Ar—H bond, so (a) and
(c) are ruled out.
23.17
(a)
AH‘ =AH* = E,—nRT = 2500 cal/mol — 2(1.987 cal/mol-K)(270 K) =
1.4; kcal/mol. From Eg. (23.35), AS°* = R{In[Ah(c°)"~ '/kT] — n} =
R{In[(6 x 10° dm?/mol-s)(6.6 x 107 J s)(mol/dm’*)/
(1.38 x 10°73 J/K)(270 K)] — 2} = -22.1 cal/mol-K.
AG** = (1.4; kcal/mol) — (270 K)(-0.0221 kcal/mol-K) = 7.4 kcal/mol.
(b)
The equations used in (a) give AH** = 40.3 kcal/mol, AS** =
~23.2 cal/mol-K, and AG?? = 103.0 kcal/mol.
23.18
(a)
Equation (23.44) gives log (k/k® ) = 1.02(-6)('7/(1 + 1’) — 0.301].
For 1 = 10°, log (k/k* ) = 0.1858 and k/k* = 0.65.
For1= 107, k/k* =0.29. For I= 101, k/k* = 0.052.
(b)
Replacement of the factor -6 by +6 gives k/k> = 1.53, 3.45, and 19.4 for
A= 10p 110, 107%
23.19
(a)
Equation (23.44) shows that for an elementary reaction in very dilute
solutions, k, increases as / increases if zpzc > 0 and decreases as [
increases if zgzc < 0. (i) Little change. (ii) Increases. (iii) Decreases.
(b)
As /J increases from zero, the function C7
+ lun — 0.30] increases to a
maximum and then decreases. Hence (23.44) predicts that k, will go
through a maximum if zpzc > 0 and will go through a minimum if
359
zpzc < 0 [provided (23.44) holds at least qualitatively at moderate ionic
strengths].
23.20
The slope of a plot of log (k,/k°) vs. [7/1 + 1'*) - 0.301 is 1.02zczs.
The data are
log(k/k°)
P41 +1?) 0.301
0.979
0.0464
-0.928
0.0724
0.951
0.0559
-0.900
0.0817
-0.854
0.0966
The straight-line fit is at best only fair; the slope is 2.37 = 1.02zczp and
AGG is Oe)
D2
-0.84
log(k/k°)
020e
0.86
-0.88
-0.90
E(n02
-0.94
-0.96
-0.98
+@fiiisstss ttt
0:04
23.21
Box
8.0.057.0106R
(0.078
0:05an0.09 me Ot
PPGCATSe) 2030!
Equation (23.49) gives AS°/R = In (Ahc°/kTe) = In 0.000113 = —9.0s. So
AS°* = R(-9.05) = —18.9 cal/mol-K. From (23.48), AH°? = E, — RT =
15700 cal/mol — R(300 K) = 15.1 kcal/mol. AG°*t = AH* — T AS° =
(15.1 kcal/mol) — (300 K)(-0.0180 kcal/mol-K) = 20.5 kcal/mol.
23.22
(a)
We get [Chreonar = [C][1 — kehem/(Kaitr + Kchem)]. For diffusion-controlled
reactions, kehem >> kaite and [C]p_,, ,./[C] = 1 — WC. + Kaitt/kenem) =
1 — (1 — kgitt/Kenem) = Kaite/kKchem, Where Eq. (8.8) was used. For chemically
controlled reactions, Kchem << Kai and (Chea
(b)
ae = (Gh
This ratio is given by (a) as 1 — Kchem/(Kaitt + Kchem) = 1 — 1/(Kaitt/Kchem + 1).
We find 0.091, 0.5, and 0.91 for kchem/Kaire = 10, 1, and 0.1, respectively.
23:23
1. A unimolecular reaction in the falloff region. 2. A diffusion controlled
reaction, where the depletion of C molecules around a given B molecule 1s a
360
departure from the Boltzmann distribution. 3. A photochemical reaction. 4. A
laser-illuminated reaction. 5. A reaction with a very low E, (see Sec. 23.1).
23.24
In the unimolecular decomposition of C)HsCl, the transition state of Fig.
23.20b has two reactant bonds partially weakened and an equal number of new
bonds partially formed; moreover, the transition state and the reactant have
similar sizes. Thus the degree of disorder in the activated complex is similar to
that in the reactant molecule, and we expect AS ef to be close to zero for this
and similar decompositions. In the C2H¢ decomposition (Sec. 23.6) and similar
decompositions, the bond elongation in the transition state produces a
“disordered” transition state, and we expect AS°* >> 0 here.
361
Chapter 24
24.1
(a) Metallic;
(f) ionic;
24.2
(a)
(b) molecular;
(c) covalent;
(d) ionic;
(e) molecular;
(g) metallic.
For C(graphite) — C(g), Ec,208 = AH sygp= 716.7
0) kJ/mol =
716.7 kJ/mol.
(b)
For SiC(c)
> Si(g) + C(g), Ec.298 = (455.6 + 716.7 + 65.3) kJ/mol =
1237.6 kJ/mol.
(c)
For SiO2(c) > Si(g) + 2 O(g), Ec.298 =
[455.6 + 2(249.2) — (-910.9)] kJ/mol = 1864.9 kJ/mol.
24.3
E, = (411.15 + 107.32 + 121.68) kJ/mol + (6.022 x 107°/mol) x
(5.139 — 3.614)eV(1.6022 x 107"? J/1 eV)(1 kJ/1000 J) = 787.3 kJ/mol.
24.4
(a) I, (greater M);
(b) NH3(H bonding);
(c) SiO> (covalent solid);
(d) MgO (greater ionic charges).
For H20(£) > H20(g), Ec.208 = AH 39, = 44.0 kJ/mol.
Ecou = —(1.6022 x 107!? C)?(1.74756)(6.0221 x 10°°/mol)/
4n(8.8542 x 107}? C?/N-m7)(2.798 x 107'° m) = -867.8 kJ/mol.
24.7
n= 1+72n(8.854 x 107!? C?/N-m*)(2.798 x 107'° m)*/
(3.7 x 107! (cm?/dyn)(m2/10* cm?)(dyn/10~ N)}(1.602 x 107! C)°1.74756 =
8.4.
24.8
(a)
From (24.11) with Ro at O K approximated by Ro at 25°C, we have
n= 1+72n(8.854 x 107'? C?/N-m7)(3.299 x 10°'° m)* (10° Pa)/
(5.5 x 10°)(1.602 x 10°’? C)71.74756 = 10.6. From (24.9),
E. = (1.602 x 107!? C)?1.74756(6.022 x 107°/mol)(1 — 1/10.6)/
4n(8.854 x 107!? C?/N-m*)(3.299 x 107'° m) = 666 kJ/mol.
362
(b)
From Fig. 24.17, the nearest-neighbor distance Ro is one-half the length d
of the diagonal of the unit cell. As in Fig. 15.1 and Eq. (15.1),
d’ =a’ +a’ +a’ and d = 3"a, where a is the edge length. So Ro =
¥2(3")a = ¥2(3")(4.123 x 107! m) = 3.571 x 107° m. Fig 24.17 shows
that there is one CsCl ion pair per unit cell, so the molar volume is
Vino = Nag’ = Na(2Ro/3"”)’ = 8R4,.Na/3>” and Eq. (24.10) becomes n =
1 + 32me9(3'”)RG/Ke°M= 1 + 32n(8.854 x 10°? C?/N-m2)(3") x
(3.571 x 107"? m)*(10° Pa)/(6 x 10°°)(1.602 x 107!? C)?1.762675 = 10.9.
Eq. (24.9) gives Ex = (1.602 x 107'? C)*1.762675(6.022 x 107/mol) x
(1 — 1/10.2)/4m(8.854 x 107'* C*/N-m’*)(3.571 x 107'° m) = 620 kJ/mol.
In both (a) and (b), the theoretical value is less than the experimental
value due to neglect of the dispersion energy.
24.9
Vin = 2NaR’, SO (OVm/OR)r.p = 6NaR?. Taking P as 1 atm = 101325 N/m? and
putting R equal to the NaCl equilibrium value 2.80 A, we have
P(0V m/OR)r p = (101325 N/m”)6(6.02 x 107*/mol)(2.80 x 107!° m)* =
2.9 x 10'° N/mol. Taking Um = Ep, we have (0U,/OR)r.p = (E,/OR)rp =
(e7/4m€) MNAR 9 (—Ro/R° + R2/R"*') = (e7/4€o) MNa(RO'/R"*! — 1/R?). At the
equilibrium value R = Ro, QU/OR is zero, but for a value R = 2.81 A that
deviates by 0.01 A from Ro, we have, using n = 8.4 and
e*/Amteg = (1.6 x 107!” C)7/40(8.85 x 10°12 C? N! m”) = 2.3 x 10°78 N m’,
0U,/OR = (2.3 x 10°78 N m’)(1.75)(6.02 x 107°*/mol) x
(2. 8054107,.m)) #8 loelOaeiny ce t/2:81 4109 mA —
-8.0 x 10'° N/mol, and P(0Vp/9R) is negligible compared with 0U,/OR.
24.10
(a)
(0’Up/OV2,)r = T(0°P/OVm OT) — (OP/AVm)r. AS T — 0, the first term on
the right side of this equation goes to zero and (0°U,»/0V
5,)r>
~(OP/OVm)r = 1/KVm, Where K = —V_! (OVn/0P)r was used.
(b)
Use of Vee = c!?R and Vet — eR A in (24.8) gives the desired equation
for —E,; differentiation of this equation gives 201k
jd
=
(e7/4ne.MNaRo ((-1)(-3)VinoVm "SHS —- DV oVm I
|pan, = (€/4MEOMNAR9|Vino X
Setting Vm = Vmo. We get -J°EplOV
(1 —n)/9. Substitution of this result into the O-K equations
363
| =O Un/OVin|nny = 1/KVmo gives (e7/4me)MNaRo! Xx
PE /OV>pony
Vz2)(n— 1/9 = 1/KVmo; solving for n, we get (24.10).
24.11
A given positive ion has two negative ions at a distance R, two positive ions at
2R, two negative ions at 3R, two positive at 4R, etc. The potential energy of
interaction between one positive ion and all the other ions is
(e2/4me€o)(-2/R + 2/2R — 2/3R + 2/4R — + - +) = —-(e*/4rt€o MIR, where M =
2(1 — 1/2 + 1/3 - 1/4 + 1/5---- ). By symmetry, the potential energy of
interaction between a given negative ion and all the other ions is
MR by Ng and division by 2 (to
—(e7/4m€9M4/R. Multiplication of ~2(e7/4mte9
avoid counting each interionic interaction twice) gives Ecoui =
—(e7/4m&) MNA/R, as in Eq. (24.4). Equation (8.36) with x = 2 gives In 2 =
fea
24.12
4 =2 In 2 =i38622"
so 4e
1/3 —-1/
(a)
—E, = (6.022 x 107*/mol)(118 K)(1.381 x 10° J/mol-K) x
[24.264(3.50 A/3.75 A)!? — 28.908(3.50 A/3.75 A)°] and E, = 8.3 kJ/mol.
(b)
At equilibrium, 0E,/0R = -0E,/0R = 0 =
Nat{—12(24.264)o'7/R" + 6(28.908)6°/R’};
so RS = [12(24.264)/6(28.908)]o° and Ro/o = 1.09. The experimental
value is Ro/o = (3.75 AV/(3.50 A) = 1.07.
24.13
E, = -24(6.022 x 107°/mol)(0.0101 eV)(1.602 x 10°” J/eV) x
[(3.50/3.75)!* — (3.50/3.75)°] = 5240 J/mol = 1.25 kcal/mol.
24.14
(a)
Each of the 8 points at the corners is shared with a total of 8 unit cells.
The point within the unit cell is not shared. So each unit cell has
8/8 + 1 = 2 lattice points.
24.15
(b)
8/8 +2/2=2
(a)
The unit cell has 8/8 + 6/2 = 4 lattice points. There is one basis group at
each lattice point, so each unit cell has 4 basis groups.
(b)
Each unit cell has 8/8 = | lattice point and therefore has | basis group.
364
24.16
An orthorhombic lattice has 90° angles and Eq. (24.12) gives
Z = pabcNa/M = (2.93 g/cm’)(4.94 x 10° cm)(7.94 x 10-8 cm) x
(5.72 x 10° cm)(6.022 x 107*/mol)/(100.09 g/mol) = 3.96 = 4. There are 4
formula units and hence 4 Ca”* ions per unit cell.
24.17
A tetragonal lattice has 90° angles and has a = b. Equation (24.12) gives
p = MZ/abcNg = (79.899 g/mol)2/(4.594 x 10° cm)?(2.959 x 107° cm) x
(6.022 x 107*/mol) = 4.249 g/cm’.
24.18
The c intercepts are all at o0, so the Miller / index is 0 in each case. With origin
at the leftmost dot in the third row from the bottom, the leftmost p2 plane
intercepts the a axis at | and the b axis at —’. The reciprocals of these
intercepts give the Miller indices as (120). With origin at the sixth dot in the
bottom row, the s3 surface intersects the a axis at | and is parallel to the b axis
(intercept at oo). The reciprocals give (100) as the Miller indices. [The p>
planes can also be called (i200)
24.19
(11)
(10)
e
ie
24.20
o_o
(02)
_@
(12)
|
e
o_o
e
(11)
e
ry
—e—_e—_e—
@
e
A body-centered lattice has 8/8 + 1 = 2 lattice points per unit cell. There is one
basis group per lattice point, so there are 2 basis groups per unit cell. The basis
therefore has 16/2 = 8 molecules of COCI.
24.21
For a spherical atom of radius r inscribed in a cubic unit cell of edge length 2r,
the atom’s volume is snr and the unit cell’s volume is (2r)° = 87°. The
percentage of occupied space 1s (£17°/8r°) 100% = (1/6)100% = 52.4% and
there is 47.6% empty space.
365
24.22
The shaded atom touches the six atoms that have a single dot; all seven of
these atoms lie in the (111) plane. The shaded atom also touches the three
atoms with two dots [which lie below the (111) plane of the shaded atom] and
touches the three atoms with three dots [which lie above the (111) plane of the
shaded atom]. (The two atoms drawn with broken circles lie on the back faces
of the unit cells.)
24.23
A cubic unit cell has right angles and has a = b =c, so Eq. (24.12) gives
0 = MZ/abcNa = (58.10 g/mol) 4/a*(6.022 x 10”*/mol) = 2.48 g/cm’. We get
a = 5.38 x 10° cm = 5.38 A. As is clear from Fig. 24.16b, the nearest-neighbor
distance is Yaa = 2.69 A.
24.24
The CsCl space lattice is simple cubic with Z = 1; Eq. (24.12) gives a =
MZ/ONa = (212.8 g/mol) 1/(4.44 g/cm*)(6.022 x 107*/mol) = 7.96 x 10°? cm’,
so a = 4.30 A. From Fig. 24.17a, the nearest-neighbor distance is half the
length of the diagonal of the cubic unit cell, which is 113 a =3.72 A.
24.25
From Fig. 24.18 and the associated discussion, the lattice is face-centered
cubic with a = b = c; there are 8 F ions and 8/8 + 6/2 = 4 Ca’* ions per unit
cell, so Z = 4. Equation (24.12) gives a= MZ/pPNa = (78.08 g/mol)4/
(3.18 g/cm?)(6.022 x 107*/mol) = 1.63 x 10°** cm? and a = 5.46 A.
24.26
The lattice is face-centered cubic with 90° angles and a = b = c. Equation
(24.12) gives a = (MZ/pNa)'” =
[(12.011 g/mol)8/(3.51 g/cm*)(6.022 x 1077/mol)]'? = 3.569 A. Nearest-
366
neighbor atoms are at points 0 0 0 and <a 74 7a. The distance between the
:
ped
2
;
point (x, y, z) and the origin (x* + y +2z°)'? so the distance
between nearest-
neighbor atoms is (a°/16 + a°/16 + a°/16)'"? = 1 J3a= 1.545 A,
24.27
There is one atom at each lattice point. In the face-centered cubic unit cell in
Fig. 24.7, the closest distance between points is the distance between the point
at a center of a face and a point at a corner of that face. (This also equals the
distance between two points on adjacent faces.) The nearest-neighbor distance
is thus one-half the length of the diagonal of a unit-cell face, namely, 1)2a=
2
12 (5.311 A) = 3.755 A.
24.28
(a)
From Fig. 24.10b, the (100) planes are spaced by a = 4.70 A; Eg. (24.13)
gives sin 8 = nA/2dhy = n(1.54 A)/2(4.70 A) = 0.1638n = 0.1638, 0.3276,
O49147
(b)
= - We set O = 9.42" 191°) 20.4940 :02° 55 02. and 79.4%
Planes s and u in Fig. 24.10a are (110) planes. We see that the distance
between these planes is half the length of the diagonal of the bottom face
of the cubic unit cell, namely, d;19 = 12a = Gino [This also follows
from the formula dp: = a/(h* + + 7)!” in Sec. 24.9.]
So sin 8 = n(1.54 A)/2(3.323 A) = 0.2317n and 0 = 13.4°, 27.6°, 44.0°,
and 67.9°.
24.29
(a).
The sin’ @ values are 0.1069, 0.1424, 0.2849, 0.3916, 0.4273, 0.5696, and
0.6767. The ratios of these sin? 6 values are
1: 1.33 : 2.67 : 3.66 : 4.00 : 5.33 : 6.33. So the lattice is face-centered
(F).
(b)
From Sec. 24.9, these are the 111, 200, 220, 311, 222, 400, and 331
reflections.
(c)
a=Ath? +h +1)'7/2 sin @ = (1.542 A)(1? + 17 + 17)'7/2 sin 19.08° =
4.085 A. Similarly, the other angles give 4.086, 4.086, 4.086, 4.086,
4.086, and 4.085 A.
367
24.30
The band extends from 392 A to 422 A, which (using v = c/A) is from
24.31
(a)
7.65 x 10'° Hz to 7.10 x 10!° Hz. So the band width is AE =h Av =
(6.626 x 10°** J s)(0.55 x 10’ s"')(1 eV/1.60 x {07 J) =2.3 eV
(b)
Equations (24.21) and (22.38) give U= kT°(0 In Z/OTv.n =
| oie
2 Vl—e
KTIUHEE eONCOnTe)
Uy + 3Nk@gM(e 28"? — 1).
88"? Ke Pt"T — 1)’.
Cy = (QUIAT)y = 3NK(Og/T)e
TAI
EOI TOG
24.32
IAT
24.33
U = Uo + 3NkOz/[exp(Oz/T) — 1) (Eq. (24.22)].
(a)
In the high-7 limit, O,/T goes to 0 and we can use the e’ Taylor series to
write exp(O,/T) — 1=(1+ O-/T+---)-1= O,/T.
Then U — Up + 3NKT.
(b)
In the low-T limit, exp(©;,/T) is very large and the —1 in the denominator
can be neglected to give
U > Up + 3NkO ple 28/7 , which becomes Up at
=a),
24.34
(a)
(b)
T + 3NKO;,/T(e°8'" — 1)- UT S=U/T+kinZ= Up
e — 1) -3NkIn (1—e7°£"7).
3Nk In (1 —e78£/7) = 3NK(@z/TI(8/7
For Al, N/n = Na, Nk/n = R, Sm = 3(1.987 cal/mol-K)(240 K/298 K)/
(¢74098 _ 1) — 3(1.987 cal/mol-K) In (1 — e"""*) = 7.41 cal/mol-K. For
diamond, replacement of 240 by 1220 gives Sm = 0.514 cal/mol-K.
Agreement with experiment is fair.
24.35
(a)
@,/T = (240 K)/(50 K) = 4.8. Dividing Eq. (24.23) by n and using
Nkin = Nak = R, we have Cy m= 3(1.987 cal/mol-K)(4.8)°e**/(e** — 1)° =
1.15 cal/mol-K = 4.81 J/mol-K.
(b)
O,/T= (240 K)/(100 K) = 2.40, and we get Cy.m = 3.77 cal/mol-K.
(c)
O-/T = 1.00 and Cy m = 5.49 cal/mol-K.
(d)
O,/T = 0.600 and Cy m = 5.79 cal/mol-K.
368
24.36
The Solver is set up to minimize the sum of the squares of the deviations of
Einstein values from experimental values. The Solver converges for any
reasonable initial guess for O¢ to give Og = 229.1 K and b = 0.000729
J/mol-K’. The Einstein curve is pretty accurate except that it shows a
significant deviation from the 50 K value.
24.37
(a)
Using the equation in Problem 15.19, we have U = kT°(0 In Z/0T)yy=
Kip Uig/k Tas CONV WEl pleas
UGrECONIUY
(b)
24.38
RNR Tal see
vt yard
ai ont (veut =) av,
Cy =(0U/0T)vn = ONAIV3.) Jo™ [(AvUKT ee"? — 1)°] dv.
Let x = Av/kT. Then dv = (kT/h) dx and
Cy = (QNWIKT-v >) SPY" [(kTxth)*e'e* — 1)\(kTIN) dx =
ONK(T/Op) S027 [x*e*"/(e* — 1)°] dx, where Op = hVm/k.
One mole of a metallic element has Na molecules, and the Einstein and Debye
theories show that the limiting high-temperature Cy m for a solid metallic
element is 3Nak = 3R. The Debye temperature for most metals is not high, so
Cy. m Of most metals is reasonably close to 3R for temperatures near room
temperature. Thus Cy,m = Cpm = 3R = 6 cal/mol-K. For a metallic element,
Cp.m = cp(A; g/mol), where cp is the specific heat (capacity) and A, is the
(dimensionless) atomic weight. Hence, cpA; = Cpm mol/g = 6 cal/(g K).
24.39
(a)
Téis quite low; Eq. (24.31) gives O}, = T°(12"Nak/5Cym) =
(10 K)°1274(6.022 x 107*/mol)(1.381 x 107° J/K)/
5(0.96 x 4.184 J/mol-K) = 4.84 x 10° K* and Op = 78.5 K.
(b)
Equation (24.31) shows that Cy is proportional to T° at low temperatures,
so at 12 K we have Cy m= (12/10)°(0.96 cal/mol-K) = 1.66 cal/mol-K.
24.40
(a)
T/Op = 298 K/320 K = 0.931. Figure 24.28 gives Cy/3Nk = 0.94 at
T/©p = 0.93. NaCl has 2Na particles per mole and Nk/n = 2Nak = 2R, so
Cv.m,ebye = 3(2R)(0.94) = 47 J/mol-K. Cy.m and Cp.m don’t differ greatly
for solids at room temperature (Sec. 4.5) and the Appendix Cp. value is
50.5 J/mol-K.
369
(b)
7/Op = 298/2230 = 0.134. Figure 24.28 gives Cy/3Nk = 0.17 at
T/Op = 0.134, so Cym = 3R(0.17) = 4.2 J/mol-K compared with the
experimental Cp m = 6.1 J/mol-K.
24.41
(a)
Adding Eg. (24.32) to (24.31), we have at low T for a metal: Cy.m =
(12n*Nak/5)(T/Op) + bT and Cym/T = b + 12n°R/SO >,)T-. A plot of
Cy m/T vs. T? should be linear with intercept b and slope 121°R/5Oj,.
(b)
The data are:
10°(Cym/T)/(cal/mol-K’)
Tek?
0.188 * OB 1st "0523
4 OO pee 002
1\S2eenn
80057
ml 6-00
The intercept is 1.4 x 107 cal mol"! K~* = b; the slope is 3.95 x 10°
cal/mol-K‘. So @3, = 12n4(1.987 cal/mol-K)/5(3.95 x 107 cal/mol-K") =
1.176 x 10’ K? and @p = 227 K.
(Cy.m/T)/(cal/mol-K’)
0.0008
0.0007
0.0006
0.0005
0.0004
0.0003
0.0002
0.0001
24.42
(a)
Scotia — Ssotiao = Ssoia = 19 (Cr/T’) aT’ = Jq (CyIT’) aT’ =
(12n'*NK/5@3,)I. 1? aT’ = 4n°NKT'/5O},.
(b)
Sygas — Ssotia = Str + Set — Ssotia = 2R In Selo + 2.9nR +
nR In{(21m)?(kTY/hP| — 4°NkT°/5©3,= AcupH/T. Solving this
equation for P, we get P = eio(27m)>*(kTe)”” *h x
exp[—4n°NkT°/5nRO
>,J exp (—AgypH/nRT). At very low T (T << Op), the
first exponential is accurately approximated as 1, and P = geio(2mm)>” x
370
(kTey
he Aue RT
hich has the form aT’
since Asupff varies
only slowly with T. As T > 0, both T’” and e~” go to 0, so P > O.
24.43
(a)
Fora
gas of hard spheres of diameter d, there is zero probability for a pair
of molecules to be closer than d, and (since there are no intermolecular
forces for distances greater than d) there is equal probability for all
distances greater than d to occur.
(b)
Fora
solid, the atoms, molecules, or ions vibrate about fixed equilibrium
locations and g(r) shows narrow peaks at the various equilibrium
separations.
g(r)
g(r)
(a)
(b)
r
24.44
r
U of an ideal gas is independent of pressure, sO Um,liq— Um gas = —AvapU 59g =
—(AH 59 — P°AV°) = -AH 59g + P°V >),
4, =-AH393 + RT =44012 J/mol + RT
m,gas
= —41.53 kJ/mol. S of an ideal gas depends on P. The 25°C molar volume of
liquid water (density 0.99704 g/cm?) is 18.07 cm*. We use this 25°C path for
one mole:
liq(P®, 18.07 cm’) as ideal gas(P°) & ideal gas(18.07 cm’). The
Appendix and Eq. (3.29) give Stiq — Sgas = (AS; + AS2) =
—118.92 J/mol-K — R In [(18.07 cm*)P°/RT] = —(118.92 — 60.06) J/mol-K =
—58.86 J/mol-K = —14.07 cal/mol-K.
24.45
(a)
Each molecule has 6 bonds and V., has 6(300) = 1800 terms.
(b)
Each methyl hydrogen atom has two 1,4 van der Waals interactions with
other atoms in the same molecule and so there are 3(2) = 6 van der Waals
interactions in each molecule. Multiplication by the number of molecules
gives 6(300) = 1800 terms for intramolecular van der Waals interactions.
Each atom in a given molecule has an intermolecular van der Waals
interaction with 7(299) = 2093 atoms in other molecules. The number of
intermolecular van der Waals interactions involving the atoms of a
particular molecule is then 7(2093) = 14651. Multiplication by the total
number of molecules and division by 2 to avoid counting each interaction
371
der Waals
twice gives 14651(300)/2 = 2197650 intermolecular van
, we have 2197650
interactions. Adding in the intramolecular interactions
+ 1800 = 2199450 van der Waals terms.
24.46
brium, so one
The initial configuration of the molecules may be far from equili
one uses the
allows the system to reach a near-equilibrium configuration before
MD data.
24.47
(a)
d by
Special theory of relativity, Brownian motion theory, distance travele
quantum
diffusing molecules, photon explanation of photoelectric effect,
theory of Cy of solids, Bose-Einstein statistics.
(b)
ics
Probability interpretation of wave function, work on matrix mechan
Born—
form of quantum mechanics, calculation of E, of ionic solids,
Oppenheimer approximation.
(c)
24.48
(a)
Debye—Hiickel theory, Debye theory of Cy of solids, Debye—Langevin
equation, Debye equation for ionic diffusion-controlled reactions, work
on electrical conductivity of solutions.
One gram of solid Ar has a volume of (1.00 g)/(1.59 g/cm’) =
0.629 cm>. The hard-sphere atoms occupy (100 — 26)% = 74% of this
volume, which is 0.465 cm?>. One gram of liquid occupies (1.00 g)/
(1.42 g/cm’) = 0.704 cm*. The empty-space volume in the liquid is
(0.704 — 0.465) cm? = 0.239 cm’, so the percent empty space is
(0.239/0.704)100% = 34%.
(b)
24.49
The volume of 1 g of the gas is V = (12/39.9 g mol ')RT/P = 179 cm’.
The percent of filled space in the gas is (0.46;/179)100% = 0.26% and
the empty space is 99.74%.
(a) T.
(b) T.
(c) T.
(d) T.
372
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0-07-2393L0-2
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