Solution Manual for Physical Chemistry 4th Edition
PI:il~~AL ヘ竺 C H E M I S T R Y I , - - - - P.W. ATKINS---....... . ._ Solutions Manual for Physical Chemistry Fou
Views 220 Downloads 7 File size 57MB
Recommend stories
- Author / Uploaded
- darko58
Citation preview
PI:il~~AL
ヘ竺
C H E M I S T R Y
I
, - - - - P.W. ATKINS---....... . ._
Solutions Manual for
Physical Chemistry Fourth Edition
P. W. ATKINS Solutions Manual for
Physical Chemistry FOURTH EDITION
©
P W. Atkins, 1990
All rights reserved. No part of this publication may be reproduced, stor ed in a retrieval system, o r transmitted, in any form or by any means, electronic, mechan ical, photocopying, r ecording, or other wise, w itho ut the prior permission of Oxford U niversity Press. 'Reprinted by arra ngement w ith Oxford Publishing Limited' or 'This reprint has been authori zed by Oxford Publishing Limited for sale in J apan only and not for export ther efr om' together w ith the fu ll copyright line as printed in the Work.
Preface to the fourth edition I have reworked all the solutions in this edition from scratch and in the light of comments received on the earlier editions. I have also adopted, within the constraints of space to which a Solutions manual is subject, a slightly more generous style , with more words, more details, a more open layout, and more guidance. The solutions have been examined in detail by Michael Fuson, of Denison University, Granville, Ohio and by Charles Trapp, of the University of Louisville, Louisville, Kentucky. I am greatly indebted to them both for their good advice, which I have tried to follow, and their detailed comments. If errors remain, they are probably at locations where I ignored what they advised.
Oxford, April1990
P.W.A.
Contents PA RT 1: EQU ILI B RI UM I . Th e prope rti es o f gases 2. The first law: th e co ncepts 3. The first law: the mac hinery 4. Th e second law: the concepts 5. Th e second law: the mac hinery 6. Cha nges o f sta te : ph ysica l tra nsfo rm ati o ns of pure substa nces 7. Cha nges of state: physica l tra nsfo rm ati o ns of sim ple mixtures 8. Cha nges of sta te: the ph ase rule 9. C ha nges of state : che mi ca l reactio ns 10. Equilibrium electroch e mi st ry
1 18 40 59 78 93 108 132 147 170
PA RT 2: ST R UCTUR E II . Qu a ntum th eory: introduct io n a nd principles 12. Q ua ntum th eory: techniq ues a nd applicati o ns 13. A to mic structure a nd ato mic spectra 14 . Mo lec ul a r structure 15 . Symme try: its de te rmin a ti o n a nd co nseque nces 16. Ro ta ti o na l a nd vibratio na l spectra 17 . E lectro nic tra nsiti o ns 18. Magnetic reso na nce 19. Statist ica l the rmod ynamics: the concepts 20 . Stati stica l the rmod yna mics: the machin e ry 2 1. Diffractio n me thods 22. T he e lectric a nd mag ne tic pro pe rties of mo lecules 23. Macromo lecul es
198 216 234 247 266 28 1 304 321 334 349 372 390 409
PA RT 3: CHA NGE 24. The kin e ti c th eo ry of gases 25. Mo lecules in mo tio n 26. T he ra tes of che mica l reactio ns 27. T he kineti cs of complex reactio ns 28 . Mo lecul ar reactio n d yna mics 29 . Processes at solid surfaces 30. D yna mic e lectroche mist ry
432 454 472 497 516 539 563
APPENDI X: Linea r regressio n
577
PART 1: EQUILIBRIUM 1. The properties of gases Exercises
1.1
V; Pr= Vr X p; (3]
V; = 1.0 L= 1000 cm 3 , Vr= 100 cm 3 ,p;= 1.00 atm Pr=
1000cm 3 x 1.00 atm = 10 x 1.00 atm = 10 atm ---100 em·1
1.2 (a) Find what pressure a perfect gas exerts from pV=nRT. Since the molar mass of Xe is 131 g mol - I, the sample has n = 1.00 mol X e. Therefore, with p = nRTIV,
p=
1.00 mol x 0.0821 L atm K -I mol - I x 298.15 K J.OL
24atm
That is, the sample hasp= 24 atm, not 20 atm. (b) The van der Waals equation is (1 1]:
nRT an 2 p= V-nb- V 2 For xenon, Table 1.4 gives a= 4. l 94 L2 atm mol - I and
b = 5.105 x 10- 2 L mol- 1• Since n = 1.00 mol and V= 1.0 L,
nRT V-nb an 2 - 2 V
l.OOmol x 0.0821LatmK - 1 mol - 1 x 298.15K _ (1.0-0.05I)L =25.8atm 4. 194 L2 atm mol- 1 x (1.00 mol) 2 (1.0 L) 2
Therefore,
p = 25 .8 atm- 4.194 atm = 22 atm
4.194atm
The properties of gases
2
Vr 1.3 p;= V Xpr [3] I
Vr = 4.65 L, V; = 4.65 L + 2.20 L = 6.85 L p 1 =3.78 x l0 3 Torr Therefore 4.65 L (a) p ; = _ LX 3.78 X !OJ Torr= 2.57 X !OJ Torr 6 85 (b) Since l atm = 760 Torr exact ly, p; = 2.57 x JOJ Torr x
1.4
1 atm Torr 760
3.38 atm
Vr Tr= V X T; [5] I
V; = l.O L, V 1 = 100 em·\ T; = 298 K 100 cm 3 Tr= lOOO em' x 298 K = 30 K Tr 1.5 Pr= T Xp; [5] I
Internal pressure= quoted pressure+ atmospheric pressure p ; = 24lb in - 2 + 14.7 lb in - 2 = 38.7lb in - 2 T; = 268 K (- 5 oq, Tr= 308 K (35 oq
Pr=
308 K _ _ K X 38.7lb in - 2 = 44.5 lb in - 2 268
Therefore p(internal) = 44 .Slb in - 2 - 14.7 lb in - 2 = 30 lb in - 2 Complications include the change in volume of the tyre, the change in rigidity of the material from which it is made , and loss of pressure by leaks and diffusion. 1.6
v,
T1= V
X
T; [5]
I
Vr= 1. 14 V; (a 14 per cent increase) , T; = 340 K
The properties of gases
Therefore, !.14 V; Tr = -V--- X 340 K = !. 14 x 340 K = 388 K I
V; = 2.0 m' , p; = 755 Torr , p 1 =(a) 100 Torr, (b) 10 Torr Therefore: (a) V1 =
755 Torr x 2.0m 1 =15m1 IOOTorr --
(b) V1 =
755 Torr 1 1 Torr x 2.0 m = !. 5 x 102 m 10
I.s 11
=
nRT p=vliJ 0.255 g g mol _ 1 = 1. 26 x 10 - 2 mol , T= 122 K. V = 3.00 L 20 18 .
Therefore, 1.26 x 10 - 2 mol x 0.0821 L atm K - I mol - 1 x 122 K
p=
1.9
3.00 L
4.22 x 10 - 2 atm
n 1 RT (a)V=- - (7] PJ 0.225 g g mol _ 1 = 1.11 5 x 10 - 2 mol, p(Ne) = 66.5 Torr. T= 300 K 20 18
n(Ne) =
.
Therefore, since there is on ly o ne vo lume, ! . liS x 10- 2 mol x 62.36 L Torr K - I mol - 1 x 300 K V=
(b) p =
66.5 Torr
nRT
V
n( Cl-1)4 -
[I], n = n(Cl-1 4 )
0.320 g 16.04 g mol- 1
+ n(Ar) + n(Ne)
1.995 x 10 - 2 mol
=3. 14L
3
4
The properties of gases
n(Ar) =
0.175g g mol _ 1 =4.380 x 10 - 2 mol 39 95 .
n = (1.995 + 4.380 + l.llS)
10- 2 mol= 7.490 X 10- 2 mol
X
Therefore 7.490 X 10- 2 mol X 62 .36 L Torr K - l mol - l X 300 K p=
3.137 L
447Torr
mRT RT Therefore, M = - V = p p p p= 1.23 g L - 1, T=330 K , p = 150 Torr Hence 1.23 g L - l X 62.36 L Torr K - l mol - 1 x 330 K 150Torr = 169 g mol - l
M=
1.11
p=
RT p
M = p - [Exercise 1.10]
33.5 mg _ mL =0.1340 g L - 1, p = 152 Torr, T=298 K 250
0.1340 g L - l x 62.36 L Torr K - l mol - 1 x 298 K M= 152 Torr = 16.4 g mol - l
1.12
nRT (a) p=v[1J
n = 1.0 mol , T= 273.15 K (i) or 100 K (ii) V = 22.414 L (i) or 100 cm 3 (ii)
1.0 mol
X
8.206 X
(i) p = (ii) p =
w- 2 L atm K -l mol - l X 273.15 K 22.414 L
1.0 mol
X
8.206 X
w-~
L atm K - l mol - l X 1000 K
0.100 L
1.0 atm 8.2 x 102 atm
The properties of gases
From Table 1.4, a = 5.489 L2 atm mol - 2 and b = 6.380 x 10- 2 L mol - 1• T herefore , 1.0 mol X 8.206 X 10- 2 L atm K - I mol - l X 273.15 K
nRT (i) - - = V- nb
(22.414 -1.0 X 6.380 X 10- 2) L
= 1.003 atm
5.489 U atm mol- 2x (1.0 mol) 2 _ V2 = (22.414 L)2 = 1.09 x 10-2 atm
an 2
and p = 1.003 atm - 1.09 x 10- 2 atm = 0.992 atm = 1.0 atm
nRT 1.0 mol X 8.206 X 10- 2 L atm K - I mol- l X 1000 K 103 (ii) V- nb = (0.100- 0.06380)L = 2 ·27 x atm 5.489 L 2 atm mol- 1 x (1.0 mol) 2 V2 = (0. 100 L) 2
an 2
_
2
5.49 x 10 atm
and p = 2.2'7 x 103 atm - 5.49 x 102 atm = 1.7 x 10 3 atm
Vc= 3b(12a) = 3 X 0.0226 L mol - 1 =6.78 x 10- 2 L mol - 1 a 0. 751 L 2 atm mol - 1 2 Pc= 27b 2 (l b] = 27 X (0.0226 L mol- 1) 2 = 54 ·5 atm 1.13
8a 8 x 0.751 U atm mol - 1 T = - - (12c] = - - ------=---,- ----,--- - -----,c 27Rb 27 X 8.206 X 10- 2 L atm K -I mol-l X 0.0226 L mol - l = 120K
1.14
Z=
~; (9]; for a perfect gas V~ = RTIp . Since the molar volume is 12
per cent smaller than that of a perfect gas,
RT Vm=0.88 V~, = 0 .88 p
Therefore, p RT (a) Z=RTx0.88 p =0.88
ZRT 0.88 X 8.206 X 10- 2 L atm K - I mol - l X 300 K p 20 atm
(b) V~= - =
1.1 L
5
6
The properties of gases
Since V"' < 1.15
v::, attractive forces dominate
pV"'
Z = RT [9], implying that Vm=
ZRT
p
Since Z = 0.86 , T= 300 K, p = 20 atm, 0.86 x 8.206 x l0 - 2 LatmK - 1 mol - 1 x 300K Vm=
20 atm
_ =1.059Lmo l- 1
(a) V = n V"' = 8.2 X 10- 3 mol X 1.059 L mol - 1 =8.7mL (b) B=Vm(':;'-1) [lOb]=Vrn(Z - 1) = 1.059 L mol - 1 x (0.86-1) = -0.15 L mol - 1 1.16
n = n(H 2) + n(N 2) = 2.0 mol+ 1.0 mol= 3.0 mol
2.0mol (a) x(H 2) = . mol= 0.67 30
RT 8.206 x 10- 2 LatmK- 1 moi - 1 X 273.15K
V
22
.4 L
= 1.00 atm mol - 1
p(H 2) = 2.0 mol x 1.00 atm mol - 1 = 2.0 atm p(N 2) = 1.0 mol x 1.00 atm mol - 1 = 1.0 atm
(c) p = p(H 2) + p(Nz) [7] = 2.0 atm + 1.0 atm = 3.0 atm 1.17
b= ! Vc [12a , Vc=98.7cm 3 mol - 1] = 1 x 98.7 cm3 mol - 1 = 32.9 cm3 mol- 1
a= 27b 2pc= 3 V ~ pc [12b, Pc= 45.6 atm] = 3 X (98.7 X 10- 3 L mo l- 1) 2 X 45.6 atm = 1.33 L 2 atm mol - 2
The properties of gases
7
As b is approximate ly the volume occupied per mole of particles
b u
1-
32.9 x
w - ~> m -' mol -
-
NA- 6.022x 1W' mol - l
""' -
1
=5.46 X 10 - "9 m 1
Then, with Vrnol = }:rrr -',
3 ) 9 ( 4:rr x 5.46 x l0 - " m-'
r=
11.1
=0 .24nm
From Table 1. 4, a= 6.493 L" atm mol - 2 , b = 5.622 x 10 - 1 L mol- 1• There fore , 6.493 U atm mo l- 2 , , = 1.4 X JO-' K 5.622 x 10 - Lmol - 1 x8.206xlo - - La tmK - 1 mo l- 1
Tn =
b 5.622 X J0 - 5 m-' mol - 1 _ (b) As in Exa mpl e 1.1 7, V111., 1 = N A= X _. mol _ 1 6 022 102 r=
3 ) 29 ( 4:rr X 9.3 X J0 - m-'
11.1
9.3X J0 - "9 m-'
= 0.28 nm
1.19 At 25 oc and 10 atm, the reduced tempe rature and pressure [Sect ion 1.5] of hyd roge n are
T,=
p,=
298 K _ _ K = 8 .968 33 23 1.0 atm 12 .8 at m
_ =0 .078 1
[Tc= 33 .23 K, Tabl e 1.3] [pc= 12.8at m, Table 1.3]
Hence , the gases named will be in corresponding states at T= 8 .968 x Tc and at p = 0.0781 x p, . (a) For ammo ni a , Tc= 405 .5 K and Pc= 111 .3 atm [Table 1.3], so
T= 8.968 X 405.5 K = 3.64 X 10-' K p = 0.078T x 111. 3 at m = 8.7 atm
(b) For xe non, Tc= 289.75 K and Pc= 58.0 atm, so
T= 8.968 X 289.75 K = 2.60 X JO-' K
8
The properties of gases
p = 0.0781 x 58.0 atm = 4.5 atm (c) For helium, Tc = 5.21 K and Pc = 2.26 atm, so T = 8.968 x 5.21 K = 46.7 K
p = 0.0781 x 2.26 atm = 0.18 atm Problems 1.1
p Vr = ---'x V; [3] andp = pgh [Example 1.2] Pr
Total pressure: p; = 1.0 atm pr= 1.0 atm + pgh pgh = 1.025 x 103 kg m- 3 x 9.81 m s- 2 x 50 m = 5.03 x 105 Pa Hence, Pr= 1.01 x 105 Pa + 5.03 x 105 Pa = 6.04 x 105Pa _ _ l.Oix 105Pax 3 3 Vr- 6.04 x 105 Pa 3m - 0.5 m
1.2 External pressure is p; and pressure at foot of column is Pr+ pgh . At equilibrium the two pressures are the same, so Pr-p;=pgh =l.Ox10 3 kgm- 3 x9.81ms- 2 x0.15m = 1.5 X 103 Pa ( = 1.5 X 10- 2 atm) 1.3 p V = nRT [1] implies that, with n constant, PrVr p;V; --Tr T; or V Vr
Tr
(r·)3 X-
Pr =~ X - Xp;= ~
T;
rr
Tr Xp; T;
1.0 mv 253 K 2 = ( 3.0 m} X 293 Kx 1.0 atm = 3.2 X 10- atm
The properties of gases RT
p
9
RT
Thatis p = p - or- = , M' 'p M For a real gas
nRT RT p= (1 + B'p+ · · ·)=p-(1+B'p+ · · ·) V
M
which rearranges to
p RT RTB' - = - + - -p + .. . p M M Therefore, plot pip against p and expect a straight li ne with intercept RTI M at p = 0. Draw up the fo llowing table:
piTorr
91.74
188.93
277.3
452.8
639.3
760.0
pl(kg m- 3) (plp)1(105 m2 s- 3)
0.225 0.544
0.456 0.552
0.664 0.557
1.062 0.568
1.468 0.581
1.734 0.584
T he points are plotted in Fig. 1.1, and the limiting behaviour is confirmed 0.59
"'"'
_&,
I
"'E
057
V)
~
~
~
0.55
~
.8-
0.53
~
0
.. "
0.2
0.1.
VIJ
0.6
0.8
The intercept at p = 0 is at
~I (105 m 2 s- 2) = 0.540, or pip= 0.540 x 105 m 2 s- 2 Therefore,
RT M=
o
Fig 1.1
/
o
0.540 x 105 m- s --
1.0
5
p/10 Po
10
The properties of gases 8.314J K- 1 mo l- 1 x298.15 K 0.540 x 105 m 2 s- 2 =4.59
X
10- 2 kg mol - 1 = 45.9 g mo l- 1
p = 1.0 atm , T= 298 K
l.Oatm x JJ 3xl03 L (a) n = 8.206 x 10 2 L atm K 1 mo l 1 x 298 K
4 ·62 x IO' mol
(b) m(H 2) = nM(H 2) = 4.62 X 103 mo l X 2.02 g mol - 1 = 9.33 x W g Mass of displaced air= 113m 3 x 1.22 kg m - >= 1. 38 x I0 2 kg Therefore, th e payload is 138 kg- 9.33 kg= 129 kg (c) For helium, m = nM(H e) = 4.62 x 101 mo l x 4.00 g mo l- 1 = 1R kg The payload is now 138 kg- 18 kg= 120 kg 1.6 The mass of displaced gas is pV, whe re V is the vo lum e of the bulb and p is the density of the gas. The balance conditio n for the two gases is m(bulb) = pV(bulb ), m(bulb) = p' V(bulb ) which implies that p = p'. however , because [Prob lem 1. 4] pM p= RT
the bal a nce condition is pM=p'M'
which impli es th at
This relation is va lid in the limit of zero pressure (for a gas be having perfectly). In expe rim e nt l , p=423 .22To rr . p '=327. 10To rr ; hence M' =
423 .22 Torr lT x 70.014 g mo l- 1 = 90.51) g mol - 1 37_7. 1( orr
The properties of gases
11
In experiment 2, p = 427.22 Torr , p' = 293.22 Torr; hence M' =
427 .22 Torr _ Torr x 70.014 g mol - 1= 102.0 g mol - 1 293 22
In a proper series of experiments one should reduce the pressure (e.g. by adjusting the balanced weight). Experiment 2 is closer to zero pressure than experiment 1, it may be safe to conclude that M = 102 g mol - 1. The molecule CH 2FCF3 has M = 102gmol - 1. T I. 7 At constant volume , p = T
x p 1 where T1 and p 1 are the temperature and
3
pressure of the triple point. Therefore, 274.16 K ) _ K -1 PJ (a) P274.1oK-P m. lr. K= ( 273 16 =
1 1 _ x 50.2 Torr= 0.184 Torr _ x p3 = 273 16 273 16
(b) For 100 oc (373 K) p=
373 K _ K X 50.2 Torr= 68 .6 Torr 273 16
374 K ) 68.6 Torr (c) PJ74K-Pm K= ( K-I Pm K= =0 . 184Torr 373 373 1.8 Draw up the fo llowing table, which is based on the reaction N 2+ 3H 2---7 2NH 1
Initial amounts Final amounts Specifically Mo le fractions
N2
H2
NH1
Total
n
n 0
0
n+n'
Jn'
0 0
1.33 mol 0.80
n + .\ n' 1.66mol
n- *n ' 0.33 mol
0.20
nRT p=--y= 1.66 mol x
8.206 X
1.00
w- 2 L atm K - I mol - l X 273.15 K 22
.4 L
12
The properties of gases
= 1.66 atm p(H2) = x(H2)P = Q p(N 2) = x(N 2 )p = 0.20 x 1.66 atm = 0.33 atm p(NH 3) = x(NH 3)p = 0.80 x 1.66 atm = 1.33 atm 1.9
RT 8.206 X 10- 2 L atm K - I mol -l X 350 K (a) V = - = -- - - - - -- - -"' p 2.30 atm = 12.5Lmol- 1
RT a RT (b) From p = V m_ b- V~ [11 b], we obtain V m= --a-+ b [rearrange llb] p+
v2 m
Then , with a and b from Table 1.4, 8.206 X
w- 2 L atm K - I mol - l X 350 K
L' I , 6 .493 - atm mo 2.30 atm + (12.5 L mol 1)2
Von=
+5.622 x 10- 2 L mol - 1
28.72Lmol - 1 - - - - - + 5.622 X 10 - 2 L mol - 1 2.34
= 12.3 L mol -
1
Substitution of 12.3 L mol- 1 into the denominator of the first expression results in V"' = 12.3 L mol - 1, so the cycle of approximation may be terminated .
8 3
=-X
40atmx160 x 10 - 3 Lmol - 1 8.206 X 10 - 2 L atm K - I mol I
1 Vc 160 X 10 - 6 m3 mol - 1 Umol = NA = 3 N A = 3 X 6.022 X 1023 mol I b
4n - 3 r3 Umol-
210K
8.86 X 10- 29 m3
The properties of gases
Hence, with V, and T, from Table 1.3, r=
3 ) 29 3 ( 4.nx8.86x10- m
1/3
=0.28nm
a
V,=2b, T,= bR (Table 1.5] 4
1.11 Hence
b = tV,=t x 118.8 cm 3 mol - 1 =59.4cm 3 mol - 1
a = 4bRT, = 2RT, V, = 2 x 8.206 x 10- 2 L atm K - I mol - 1 x 289.75 K x 118.8 x 10- 3 L mol- 1 = 5.649 U atm mol- 2 Hence
nRT V-nb
= - - e-na! R'TV
1.0 mol x 8.206 x 10- 2 L atm K - I mol - 1 x 298 K l.OL-l.Omol x59.4x 10
3
L mol
1
) 1.0 mol x 5.649 L2 atm mol- 2 xexp ( - 8._2_0_6 _x_1_0__~ x_l_ . O_L_2~a-t_m_m __o_l-~1 2 x--2-98__
=26.0 atm x e- 0·231 =21 atm
RT V m( 1 -
a
~) - V ~'
13
14
The properties of gases 2
a ] -1 +b-, + .. .) =RT( - 1+ [ b - vm RT VITI V;;, Compare this expansion with B+C p = -RT( 1+ , + . .. ) [lOb] Vm V"' V;;,
and hence find a B=b-- and C = b 2 RT -___:_:__:__
__
a = RT(b- B) = 8.206 X 10- 2 X 273 L atm mol - 1 X (34.6 + 21.7) cm 3 mol - 1 = 22.40 L atm mol - l X 56 .3 X 1.13
w-} L
mol - l= 1.26 L 2 atm mol - 2
RT p = - - e - ai RTV.,. [Table 1.5] =
vlll-b
RT
vm
(1 -vmb) -
e - ai R TV.,.
1 Now use the expansions--= 1 +x +x 2 + · · · and e -.r = 1- x + h 2 + · · · 1- x -
and obtain 2
RT{ 1+-+-, b b +. . . } p=V"' Vm V~,
{
1 - -a- + -1 ( -aRTV111 2 RTV"'
)2+ ... }
2
=
RT{
vm
(
a ) 1 ( 2 ab a ) 1 } l+ b- RT VITI+ b - RT+2R 2 T 2 v ~ + ...
Comparing with the virial expansion [lOb] gives a ab a2 B= b -RT and C= b 2 -Rf+2Rifi To find a and b we form C - t B 2 = ! b2 , implying that b=(2C -8 2) 112 and then use a = RT(b-8) . From the data:
The properties of gases
b= (2 x 1200 cm 6 mol - 2 -471 cm 6 mol - 2) 112 = 43.9 cmJ mol - 1 a=22.40 L atm mol - l X (43.9+21.7)
= 1.47 U atm mol -
X
w-.1 L mol - l
2
1.14 For critical behaviour, show th at there is a point of inflexion with zero slope, and identify the critical constants.
RT
B
C
p=y-- v" +v.1 m
m
m
That is, -
RTcV~ =2BVc -3C:O RTcV c 3BVc+6C-O
l
which solve to
Now use the equation of state to find Pc: RTC
B
c
vc
v~
v~
Pc=-----:;+]
It follows that
1.15
pVm
RT = 1 + B'p+ C'p 2 + ··· (lOa]
pV"' B C - = 1 + - + -, +···(lOb] RT
Vm
V~,
15
16
The properties of gases
whence B 1 p+C 1p 2
• • •
B
C
vm
v~,
=-+-+ · · ·
Now multiply through by Vm, replace pVm by RT{1 + (B/Vm) +···},and equate coefficients of powers of 1/V"':
8 RT+ 1
BB 1 RT+ C 1 R 2 T 2
vm
C
+ · · ·=B+-+ · · ·
von
B
Hence , 8 1 RT=B, implying that BI=RT 2 , , , , C-8 2 Also , BB RT+ CIR T - = C , orB-+ C'R-T-= C , implying that C' =R2Tz -1
Therefore, the limiting slope of a plot of pip against p is B RTf M. From Fig. 1.1, the slope is I
B 1 RT
~=
(0.584- 0.544) x 105 m2 s- 2 (1.013- 0.122) X 105 Pa
4.5 X 10- 2 kg - 1m3
Therefore, since the intercept lies at RTf M = 0.540 x 105 m2 s- 2 [Problem 1.4] BI
=
4.5 X 10- 2 kg - 1 m 3 0.540 X 105m2s-2 = 8.3 X 10 -7 Pa - 1 [1 Pa = 1 kg m - I s-2]
Hence, B I = 8.3 X 10- 7 Pa -I
X
1.0133 X 105 Pa atm - I = 8.4 X 10 - 2atm - 1
Since B = RTB [Problem 1.15] 1
B = 8.206 X 10- 2 L atm K - I mol - l X 298 K X 8.4 X 10 - 2 atm - 1
=2.1 Lmol- 1 1.17 Hydrostatic pressure is given by p = pgh [example 1.2]; therefore dp = - pg dh [p decreases ash increases]
pM
Since p= RT [Problem 1.4]
The properties of gases dp = -
17
pMgdh . . dp Mgdh RT , 1mplymg that =-Iff
p
This relation integrates to
p =Poe - M~h ! IIT For air, M = 29 g mol - 1and at 298 K Mg 29 X 10-·1 kg mol - 1X 9.81 m s- 2 _ RT = 2.48 x 10-' Jmol -l =1.15x10 - 4m - l [1J=1kgm2s-2] (a) h = 15 em
p =Po X e-O.I) mx 1.1:\x 111-"m · l = 0.99998po (b) h = 1350 ft, which is equivalent to 412 m [1 inch= 2.54 em]
2. The first law: the concepts Exercises 2.1
w = -mgh[3]
(a) w=-l.Okg x 9.81ms - 2 x 10m=-98J (b) w =- 1.0 kg x 1.60 m s- 2 x 10m= -16 J 2.2
w= -mgh (3] =- 65 kg x 9.81 m s- 2 x 4.0 m =- 2.6 k~
2.3
w=
-pcx ~V
(5]
Pcx = 1.0 atm X 1.013 X 105 Pa atm - I= l.Ol X 10 5 Pa ~ V=
100 em 2 x 10 em= 1.0 x 10 3 em 3 = 1.0 x
w-> m 3
w = -l.Ol X 105 Pa X 1.0 X 10- 3 m 3 =- 1.0 X 102 ]
as 1 Pa m3 = 1 J . 2.4
(a)
w = -p 0 ~V[5]
Pcx=200Torr X 133.3 Pa Torr - 1 = 2.666 X 104 Pa ~V=3.3L = 3.3x 10- 3 m 3
Therefore, w =- 2.666 X 104 Pa X 3.3 X 10- 3 m3 =- 88 J Vr
(b) w= -nRTin V (7] I
4.50 g ~ n = 16.04g mol-l= 0.280) mol RT=2.577kJmol - 1, V;=l2.7L. Vr=16.0L _
w =- 0.2805 mol x 2.577 kJ mol - 1 x In
2.5
Vr w= -nRTin V [7] I
16.0L . L = -167 J 12 7
The first law: the concepts
nRT= 52.0 X w-] mol X 8.314 J K- 1 mol - l X 260 K = 1. 124 X 10 2 J 2
w =- 1.124 X 10 J X In
t = + 124 J
w= -pc,t-.V[5]
2.6
Pcx =95bar=95X 10 5 Pa !-. V =- 0.450 LX 0.67 =- 0.302 L =- 0.302 x 10- 1 m '
Therefore,
w= + 95 X 105 PaX 0.302 X 10- ' m1 = + 2.9 kJ
w=-pcxl-.V[5]
2.7
Mg(s) + 2HCI(aq)--> H 2(g) + MgCI 2(aq) , M(Mg) = 24 .31 g mol - 1
nRT V;=O , Vr= - -, pi=Pcx Pr
nRT w=- pc.x(Vr- V;) =- PcxX--=- nRT Pcx n=
15 g 1 . g mol _ 1 = 0.617 mol, RT= 2.479 kJ mol 24 31
Hence , w = - 0.617 mol x 2.479 kJ mol - 1 = - 1.5 kJ
t-.H /f,, =2.60 kJ mol - 1 [Table 2.2].
2.8
750 x 10' g n
22 .99 g mol - 1
q=nl-.H /f,, =
750x 101 g
_ 1 X 2.60 kJ mol - 1 =8.5 x 104 kJ
22. 99 g mo 1
q . 229 J C=- [Sect1on 2 SJ = -- =89 8J K- 1 !-. T ·2.55 K .
29 .
The molar heat capacity (at constant pressure) is the refore C1 = '
89 .8JK - 1 3.0mol
= 30 J K- 1 mol - 1
19
20
The first law: the concepts
For a pefect gas, C" - Cv= R [16, molar quantities]
Hence Cv= C"- R = (30- 8.3) J K- 1mol - 1=22 J K- 1mol - 1 2.10
q= C!).T, C = nC" , V=75 m3
pV l.Oatm x 75xl03 L n=-= 2 RT 8.206 X L atm K- 1mol -l
w-
_ 1 = 3.07 x lO"mol
X 298 K 1 1 q=3.07 x 10 mol x 21 J K- mol - x 10 K=6.4 x 10 2 kJ 1
Since q = P x t, where Pis the power of the heater and tis the time for which it operates, 6.4 X 105 J , . t=-p = l.O x 103J s _1= 6.4 X 10- s (about 11 mm) q
In practice, the walls and furniture of a room are also heated. q = -1.2 kJ [energy leaves the sample]
2.11
!).H= -1.2 kJ [!).H=q at constant pressure] q 1.2 kJ C = !).T= 15 K =80J K- 1 2.12
q=C!).T=nC,,!).T
=3.0 mol x 29.4J K- 1mol - 1x 25 K = +2.2 kJ !).H=q [lOa] = +2.2 kJ !). U = !).H- !).(p V) (9] = !).H- !).(nRT) = !).H- nR!).T = 2.2 kJ- 3.0 mol x 8.314 J K- 1mol - 1x 25 K
= 2.2 kJ - 0.62 kJ = + 1.6 kJ 2.13
q=0.50 mol x 26.0 kJ mol - 1= + 13 kJ
(5] = - Pcx V(g) (V(g) ~ V(/)] nRT = pc xX - - = -nRT Pcx Therefore, w = -0.50 mol x 8.3141 K - I mol - 1x250 K= -1.0 kJ !).H = q [lOa] = + 13 kJ W
= - Pcx/). V
!).U=q+ w= +13 kJ -1.0 kJ = +12 kJ
The first law: the concepts
21
CfiHsCzHs(l) + ¥ 0 2(g) ~ 8C0 2(g) + 5H 20(l)
2.14
11H'! = 811H f((C0 2 , g)~ 5!1Hf((H 20 , l)- !1Hf (eb, l) [eb = ethylbenzene]
= 8(- 393.51) + 5(- 285.83)- ( -12 .5) kJ mol - 1 =-4564.7kJmol - 1 2.15
11H'! =- 4003 kJ mol - 1
C6 Hdl) + 90 2 (g)~6C0 2 (g) + 6H 20(l)
11H'! =- 4163 kJ mol - 1
C6 H 14(l) + Jf0 2(g) ~6C0 2 (g) + 7H 20(l) The difference of these two reactions is C6H1 2(l) + H 2 0(l)~ C1,H 14(l) + f0 2(g)
!1H 9 = + 160 kJ mol- 1
To replace the H 20 by H 2 we subtract H 2 0(l)~
H 2(g) +
!1H 9 =- !1H 9 (H 20, /) = + 285 .83 kJ mol- 1
~ 0 2 (g)
Giving CfiH12(/) + H2(g)~CoH 14 (/)
!1H 9 = -126 kJ mol - 1
2.16
3C(s) + 3H 2(g) + 0 2 (g)~CH 3 COOCH 3 (/)
/!,. U =
!1H- !1nMRT (11),
!1H f( =- 442 kJ mol - 1
!1nM= - 4 mol
!1nMRT=- 4 mol x 2.479 kJ mol - 1= - 9.916 kJ
Therefore !1U f = -442 kJ mol - 1+9.9 kJ mol - 1= -432 kJ mol - 1
The reverse reaction is 10C0 2(g) + 4H 2 0(l)~ C 10Hx(s) + 1202(g)
!1H 9 = + 5157 kJ mol - 1
The C0 2 and H 20 can be replaced by adding the following two reactions [and using !!,.Hf((C0 2) and !!,.H f((HzO), Tab le 2.10]: 10C(s) + 100 2 (g)~ 10C02(g) !1H 9 =- 3935 kJ mol - 1 4H 2(g) +20 2 (g)~4Hz0(l) !1H 9 = -1143 kJ mol - 1 Thus overall: lOC(s) + 4H 2 (g)~ CwHx(s) !!,.H 9
=
+ 5157-3935-1143 kJ mol - 1= + 79 kJ mol - 1
22
The first law: the concepts
2.18
C= ~qTand q = /Vt [section 2.5]
Hence 3.20 A X 12.0 V X 27.0 s C = 1.617 K
641JK - 1
because 1 A V s = 1 J. 2.19
q=n~H ~, ~H ~ =
-5157 kJ mol - 1 [Table 2.9]
Therefore , 120 x w-Jg lql= _ X5157Jmol - 1 =4.83kJ 128 .1 8 g mo 1 1 4.83 kJ q C = - = - - = 1.58 kJ K -·l ~T 3.05 K When phenol is used, since ~H ~ = -3054 kJ mol - 1 [Table 2.9], 100 x w-3 g _ lql _1 x 3054 kJ mol - 1 = 3.245 kJ 94 .1 2 gmo 1 Therefore, q
3.24SkJ 58 kJ K_ 1 =2 .05 K
~T=c= 1.
2.20
q
C~T
MC~T
n
n
m
q=C~T, I~Hcl=-=--=--[m: massof sample]
Therefore, since M = 180.16 g mol - 1, 180.16 g mol - 1 X 641 J K- 1 X 7.793 K _ I~Hcl= _ g =2802kJmol - 1 0 3212 Therefore, since the combustion is exothermic,
~He=
-2.80 MJ mol - 1
The combustion reaction is C~H 1 2 0~(s)
Hence
+ 60 2(g)--76C0 2(g) + 6H 20(l)
~Uc = ~He;
therefore
~Uc =-
~n~ =
0
2.80 MJ mol - 1
For the enthalpy of formation we combine 6C0 2(g) + 6H 20(l)--7 C~H 1 2 06 (s) + 60 2(g) 6C(s) + 60 2(g)--76C0 2(g) 6H 2(g) + 30 2(g)--7 6H 20(l)
~H=
+2.8 MJ mol - 1
~H =~H=
2.36 MJ mol - 1
-1.72 MJ mol - 1
The first law: the concepts The sum of the three is 6C(s) + 6H 2(g) + 30"(g)---.,> ChH I20h(s)
!1H1 = 2.80- 2.36- 1. 72 MJ mol - 1 = - 1.28 MJ mol - 1 2.21
AgCI(s)-,> Ag+(aq) + Cl - (aq)
!1H 6 = !:l.H f'(Ag +, aq) + !:l.Hf'(CI - , aq)- !:l.Hf'(AgCI, s) = 105.58 + ( -167.16)- ( -127.07) kJ mol - 1 =+65.49kJmol - 1
NH >+ S0 2-,> NH 3S0 2
!1H 6
=-
40 kJ mol - 1
!1Hf'(NH 3S0 2, s) = !1H )"(NH 3 , g)+ !1H f'(S0 2, g)- 40 kJ mol - 1 =- 46 .11-296.83-40 kJ mol - 1 =- 383 kJ mol - 1 2.23
C(gr) + 0 2(g)-,>C0 2(g) C( d) + 0 2(g)---.,> C02(g)
!1H 6 =- 393.51 kJ mol - 1 !1H 6
=-
395.41 kJ mol - 1
The difference is C(gr)---.,> C( d) 2.24
!:l.H f}. =- 393.51- ( - 395.41) kJ mol - 1 =- 1.90 kJ mol - 1
q = n!:l.H 'ji 1.5g 1 . g mol _ 1 x ( - 5645 kJ mol - ) = -25 kJ 342 3
Effective work available =25 kJ x 0.25 = 6.25 kJ Since w = mgh, with m = 65 kg h=
6.25 X W J
, 65kg x 9.81ms-
2.25
9.8m t;.H'j
Ml f'(8)
C>Hx(l) + 50"(g)~C> H x(g) + 50"(g)~3CO"(g) + 4H"O(I)
(a) !:l.H 'ji(l) = !:l.H?.,P+ !:l.H'ji(g) = 15 kJ mol - 1 - 2220 kJ mol - 1 =- 2205 kJ mol - 1 (b) !:l.n g= - 2 [50 2 replaced by 3C0 2]
23
24
The first law: the concepts
Therefore D. U f((l) = D.H f((l)- (- 2)Rl' =- 2205 kJ mol - 1 + 2 x 2.479 kJ mol - 1 =- 2200 kJ mol - 1 2.26 D.H 9 > 0 indicates an endothermic reaction and D.H 9 < 0 an exothermic reaction. Therefore, (a) is exothermic, (b) and (c) are endothermic. 2.27 0 =
L vJSJ; hence J
(a) 0=C0 2 +2H 20-CH4-20 2 v(C0 2) = + 1, v(H 20) = + 2, v(CH 4) = -1, v(0 2) = - 2 (b)
0=~H 2 -2C - H 2
v(~H 2 )
= + 1, v(C) = - 2, v(H 2) = -1
(c) 0= Na +(aq) + Cl-(aq)- NaCI(s) v(Na+) = + 1, v(Cl-) = + 1, v(NaCl) = -1 (a) D.H 9 = D.H f3(N 2 0 4 , g)- 2D.H f3(N0 2 , g)
2.28
= 9.16-2 x 33.18 kJ mol- 1 = - 57.20 kJ mol - 1 (b) D.H 9 = D.Hf3(NH 4Cl, s)- D.Hf3(NH 3 , g)- D.H f3(HC1 , g) =- 314.43- (- 46.11)- (92.31) kJ mol - 1 = -176.01 kJ mol- 1 (c) D.H 9 = D.H f3(propane , g)- D.Hf3(cyclopropane , g) =20.42-53 .30 kJ mol - 1 = -32.88 kJ mol - 1 (d) The net ionic reaction is obtained from
and is H +(aq)
+ OH - (aq)~ H 20(/)
D.H 9 = D.H f3(H 20, l)- D.H f3(H+, aq)- D.Hf3(0H- , aq) .
= - 285.83-0 - (- 229.99) kJ mol - 1 =- 55 .84 kJ mol - 1
The first law: the concepts
The sum of the three reactions is
2.29
1:1H 9 /( k1 mol - 1) 2NO(g) + 0 2(g) ~ 2N0ig)
-114.1
! 0 2(g) + 2N02(g) ~ N20 5(g)
t(- 110.2)
N2(g) +
0 2 (g)~
2NO(g)
180.5
N2(g)+ t0ig)~N20 s(g)
+11.3
Hence, 1:1Hf (N 20 5 , g)=+ 11.3 k1 mol - 1 1:1H 9 /(k1mol - 1)
2.30 (a)
K(s) + 1 Cllg)~ KCl(s)
-436.75
KCl(s) + tOz(g)~ KClOls)
·H89.4)
-392.1 Hence , 1:1H f (KC10 3 , s) =- 392.1 k1 mol - 1 (b)
Na(s) + t 0 2(q) + !H 2 (g)~ NaOH(s)
-425.61
NaOH(s) + COig)~ NaHC0 3(s) C(s) +
02(g)~C02(g)
-393.51
Na(s) + C(s) + t Hig) + t0 2 (g) ~ NaHCOJ(s) Hence, 1:1H f (NaHC0 3 , s) =- 946.6 k1 mol (c)
tN 2 (g) + t0 2 (g)~NO(g)
- !(75.5)
! N2(g) + iO(g) + tC12(g) ~ NOCl(g)
2.31
-946.6
1
+90 .25
NO(g) + tCllg)~ NOCl(g)
Hence, 1:1Hf (NOCl , g)=52.5 k1 mol -
-127.5
+ 52.5 1
1:1H 9 (T2) = 1:1H 9 (T1) + 1:1C,I:1 T (Example 2.12]
1:1C,, = C"(N 20 4 , g)- 2C,(N0 2, g) =77.28-2 x 37.201 K- 1 mol - 1 = +2.881 K- 1 mol - 1 1:1H 9 (373 K) = 1:1H 9 (298 K) + 1:1CPI:1 T = -57 .20 k1 mol - 1 +2.88 1 K- 1 x 75 K = -57 .20+0.22 k1 mol - 1 = -56.98 k1 mol - 1
25
26
The first law: the concepts Mg 2+(g)+ 2 Cl(g)+ 2e-
2
X
24 1.6 Mg 2+(g)+ Cl 2(g)+ 2e-
15.035 eV = 1450.7
3.78 eV = 729.4 Mg 2+(g) + 2 Cl - (g)
2
X
383.7 = 767.4
Mg•(g)+CI 2(g)+ e-
7.646 eV = 737.7 Mg(g) + Cl2(g)
Mg 2 •(g) + 2 Cl - (aq)
167.2 D.Hhvd (Mg 2+) =
Mg(s) + Cl 2 (g)
-x
641.32 MgCI2(s)
X
I 150.5
MgCI2 (aq)
2.32 Distance up on left= distance down on right. Therefore, 150.5 + 641.32 + 167.2 + 737.7 + 1450.7 + 241.6 = X+ 767.4 + 729.4
Solving to x = 1822.2, implying that t..H ~.Hf(CH3CHO, g)
= 2 x ( -484 .5)- 2 x ( -166.19) k1 mol- 1 = - 636.62 k1 mol - 1 f..G 9 = -636.62 k1 mol - 1 -298.15 K x ( - 386.11 K- 1 mol - ') = -521.5 k1 mol- 1 (b) f..H 9 = 2t>.Hf (AgBr, s) - 2t>.Hf( AgCI, s) = 2 x ( -100.37)- 2 x ( -127.07) k1 mol- 1 = +53.40 k1 mol-' f..G 9 = +53.40 k1 mol - 1 - 298.15 K x 92 .61 K- 1 mol- 1 = +25.8 k1 mol- 1 (c) f..H 9 = t>.Hf (HgC1 2 , s) = -224.3 k1 mol- 1 f..G 9 = - 224.3 k1 mol - 1 - 298.15 K x ( -153.11 K- 1 mol - ') = -178.7 k1 mol - 1 (d) f..H 9
=
t>.Hf (Zn 2 +, aq)- t>.Hf (Cu 2 + , aq)
= +153 .89- 64.77 k1 mol - 1 = -218.66 k1 mol- '
62
The second law: the concepts
!:l.G 6 = -218.66 kJ mol - 1 - 298.15 K x ( -21.0 J K - I mol - 1) = -212.40 kJ mol- 1 (e) !:l.H 6 = !:l.Hf( = -5645 kJ mol- 1 !:l.G 6 = -5645 kJ mol- 1 -298.15 K x 512 .0 J K- 1 mol - 1 = -5798 kJ mol- 1 (a) !:l.G 6 = 2!:l.G f (CH 3COOH, l)- 2!:l.Gf (CH 3CHO , g)
4.10
= 2 x ( -389.9) - 2 x ( -128.86) kJ mol - 1 = -522.1 kJ mol - 1 (b) !:l.G 6 = 2/:iG f (AgBr, s)- 2/:iG f (AgCl , s) =2 X ( -96.90)- 2 x ( -109.79) kJ mol - 1 = + 25.78 kJ mol - 1 (c) !:l.G 6 = !:l.G f (HgC1 2 , s) = -178.6 kJ mol - 1 (d) !:l.G 6 = !:l.Gf (Zn 2 +, aq)- !:l.Gf (Cu 2 +) = -147 .06-65.49 kJ mol- 1 = -212.55 kJ mol - 1 (e) !:l.G 6 = 12!:l.G f (C0 2, g)+ ll!:l.G f (H20 , l)- !:l.G f(CI 2 H 22 0 1 ~> s) = 12 x ( -394.36) + 11 x ( - 237.13)- ( -1543) kJ mol - 1 = -5798 kJ mol - 1 6C(s) + 3H2(g) + f 02(g)~ C6HsOH(s)
4.11
!:l.S 6
= S 6 (C6H 50H, s)- 6S 6 (C, s)- 3S 6 (H2 , q)- t S 6 (0 2, g) = 144.0-6 X 5.740-3 X 130.68 - 1 X 205.14= -385.05 J K- 1 mol - 1
C6H 50H(s) + 70ig)~6C0 2 (g) + 3H 20(l) !:l.H f( = 6!:l.Hf (C0 2 , g)+ 3!:l.Hf (H 20, l)- !:l.Hf (C6H 50H , s)
Hence , !:l.Hf (C6H 50H , s) = 6!:l.Hf (C0 2, g)+ 3!:l.Hf (H 20 , l ) - !:l.H f( = 6 x ( -393.51) + 3( -285 .83)- ( - 305.4) kJ mol - 1 = -164.55 kJ mol - 1 Hence !:l.G f = -164.55 kJ mol - 1 - 298.15 K x ( -385.05 J K - I mol - 1) = -49.SkJ mol- 1
The second law: the concepts 4.12
63
Vr (a) dS(gas) = nR In V I
14 g -::-::---:-:-----,----;1
28.02 g mol
x 8.314 J K - I mol - 1 x In 2
= +2.9 J K - 1 AS( surroundings) = -2. 9 J K - I [overall zero entropy production] dS( total) = Q (b) dS(gas) = +2. 9 J K - I [Sa state function]
dS(surroundings) = Q[Section 4.4] dS(total) = +2.9 J K -
I
(c) dS(gas)=O [qrcv= OJ dS(surroundings) = 0 and dS (total)= 0 4.13 The same final state is attained if the change takes place in two stages , one isothermal compression : Vr dS 1 =nRln v=nR In t = -nRin2 I
and the second, heating at constant volume: Tr dS2 =n Cvln T= nCv In 2 I
the overall entropy change is therefore
dS = -nR In 2+nCv In 2 =n(Cv- R) In 2 4.14 2.9]
CH 4 (g) +20 2 (g)~C0 2 (g) + 2H 20(/), dG 9 = -817.90 kJ mol - 1 [Table
Therefore, the maximum non-expansion work is 817 .90 kJ mol - 1 [since lwei= ldGIJ . 4.15
Tc t: = 1 - - [11] Th
64
The second law: the concepts
(a) t:= 1 -
333K K = 0.11 (11 per cent efficiency) 373
(b) t: = 1 -
353 K K = 0.38 (38 per cent efficiency) 573
4.16
!::.H,,. + 1.9 kJ mol - 1 1 1 !::.S,,. = T,,. = 000K =+0.95JK- mol 2
4.17
(a) No work need be done because the cooling is spontaneous.
Tc 295 K (b) t:=Th-Tc[ 15] = 303K-295K
36 ·9
= 20kJ
qh Th qc c
4.18 - = T, therefore
w=~ = (Th;cTc) Xq (13 , 15)
4.19
200K - 80K 80K
----,--- X
4.20
Tc c = Th _ Tc [15]
(a) c =
273K K = 14 20
(b) c=
263K K = 8.8 30
2.10 kJ =3.15 kJ
The second law: the concepts 4 _21
w = lf. = (Th-Tc) x!:J.H=(293K - 273K) c Tc 273 K
X
250g 18.02 g mol- 1
x 6.01 kJ mol- 1 = 6.11 kJ This amount of work can be done in
Problems T
4.1
(a) !:J.S",(t~s, T) = !:J.S",(l~s, Tr) - !:J.CP!n Tr
with !:J.CP = Cp(l ) - CP(s) = + 37.3 J K- 1 mol- 1 Therefore ,
- 6.01 x 103 J mol- 1 268 -37.3 J K- 1 mol- 1 x In K 273 273 = - 21.3 J K- 1 mol- 1 !:J.S( surroundings)
!:J.Hrus(T) T
!:J.Hrus(Tr) (T- Tr) T +!:J.CP T
6.01kJmol- 1 268 - 273 268K . + 37.3JK -I mol- l x 268 = + 21. 7 J K- 1 mol- 1 !:J.S(total) = 21.7-21.3 J K - I mol- 1 = +0.4 J K -I mol- 1 Since !:J.S(total) > 0, the transition
z~s
is spontaneous at - 5 ac.
T (b) !:J.S",(l~g, T) = !:J.S,rs(l~g, Tb)+!:J.CP!n Tb
65
66
The second law: the concepts T
I:J.Hvap
=~+ I:J.C, In Tb' I:J.C,= -41.9 J. K
I:J.S",(l~g, T)=
-1
mol
-1
40.7 kJ mol - 1 368 K -41.9JK- 1 moi - 1 X In 373 373
= +109.7 J K - 1 mol - 1 !:J.S(surroundings) =
- I:J.Hvap(T) T
- I:J.Hvap(Tb)
!:J.CP(T- Tb)
T
T
-40.7 kJ mol - 1 378 K
( -41.9 J K -I mol-l) X
368-373 368
= -111.21 K - 1 mol - 1 !:J.S(total) = 109.7-111 .2 J K- 1 mol - 1 = -1.5 J K - 1 mol - 1 Since !:J.S(total) < 0, the reverse tra nsition,
4.2 !:J.S=
r r C,
d; =
g~ l ,
is spontaneous at 95 oc.
(a+:T) dT
=91.47 J K - 1 moi - 1 Jn
300K K+0 .075J K - 1 mol - 1 X27K 273
=10.7JK - 1 mol - 1 Therefore, for l.OOmol , I:J.S=+llJK - 1
At 298 K, S 6 (NH 3 , g)= 192.45 J K - I mol - 1. Therefore ,
The second law: the concepts
=200.7 1 K - I mol- 1 (b) S 9 (773 K) = 192.451 K - I mol- 1 + 29 .751 K - I mol - 1 1n +25.10x
773
298
w-) 1 K - 2 mol- l X475 K
= 232.01 K - 1 mol - 1
4.4
!::iS depends on only the initial and final states, so we can use
Tr !::iS = nCP In T I
Since q = nC/Tr- Ti),
/ 2Rt
q
Tr= Tj+-= Tj+- [q = ltV = l 2Rt] nCP nCP
That is, 2
I Rt) !::iS = nCP In ( 1 + nCpTi
Since n =
500 g _ g mol 63 6
1
7.86 mol ,
!::iS=7.86 mol x 24.41 K- 1 mol - 1 x ln
(1
(1.00 A) 2 X 1000 Q X 15.0 s) + 7.86x24.41K- 1 x293K
= 1921 K - 1 x In 1.27 = + 45.41 K - 1 For the second experiment,
67
68
The second law: the concepts
dqrev(net) = 0; therefore dS = 0 and .!lS = 0. However, for the water LlS =
Jd~ev
=
q;v= f
:t
2
(1.00 A) 2 X 1000 Q
X
15.0 s
293K
+51.2JK- 1
[1J=1A Vs=1A 2 Q s] 4.5
C(s) + t0 2 (g)+2H 2 (g)~CH3 0H(l), Lln 8 = -2.5mol
M e = .!lG 9
il(p V) = .!lG 9
-
-
iln(RT) = .!lG 9 + 2.5RT
= - 166.27 + 2.5 X 2.479 kJ mol - 1 = -160.07 kJ mol- 1 4.6
Calculate the final temperature as in Exercise 4.6:
Tr
Tr
.!lS=n 1CPin T; +n 2Cpln T;z 1
n
=n 1CP In T; T;z [n 1 =n2] 1
200 g 3182 18.02 g mol-l x 75.5 J K-1 mol- l x In 273 x 363
+17.01 K - 1
(b) Heat required for melting= n 1 ilHrus = 11.1 mol X 6.01 kJ mol - 1 = 66. 8 kJ The decrease in temperature of the hot water as a result of its causing the . . q . 66.8 kJ meltmg IS ilT= nCP= 11.1 mol x 75 .5 J K-1 mol-l =79.6 K At this stage the system consists of 200 g water at 0 oc and 200 g water at 90 oc -79.6 oc = 10 oc (283 K). The entropy change so far is therefore
n ilHrus 283 K .!lS =----;:;;- + nCP In 363 K
The second law: the concepts 11.1 mol x 6.01 k1 mol- 1 273
K
+ 11.1 mol x 75.51 K - I mol - 1 ln
283 K 363
K
= 244 1 K- 1 - 208.6 1 K- 1 = +35.7 1 K - 1 The final temperature is Tr= t(273 K + 283 K) = 278 K, and the entropy change in this step is
n
~S = nCP In T;JTi2 = 11.1 X 75 .51 K- ln 1
278 2 X 273 283
+0.27 1 K- 1
Therefore, overall, ~S = 35.71 K - I+ 0.271 K -I = +36 1 K - I 4.7 We need ~G and ~A under the stated conditions, and begin by calculating ~G for the transition l--"'s : ~G(T)
= ~H(T) - T ~S(T)
=
~H(Tr)- ~Cp(T- Tr) -
=
~H(Tr)- ~ ~H(Tr) - ~cP{ T- Tr- Tin~}
=(
T{
~S(Tr)- ~CP In ~}
~ -1) ~Hrus(Tr) - ~cP{ T- Tr- Tin~}
T=268 K, Tr=273 K, ~Hrus =6.01 k1 mol - 1, ~CP = +37.3 1 K - 1 mol - 1: ~G(268
K) = 268 -1 ) x 6.01 k1 mol - 1 ( 273 -37.31mol - 1 x 268-273 - 268ln 268} { 273 = - 0.11 k1 mol - 1
For
~A
we use
~A = ~G-~(pV)=~G-p ~V =~G - pM ~(lip)
= -0.11 k1 mol - l- 1.013 X 10 5 Pa X 18.02 X w-) kg mol - l
69
70
The second law: the concepts
x (917
k~ m -
3
999
k~ m -
3
)
= - 0.11 k1 mol - 1 Therefore: (a) Maximum work is 0.11 k1 mol - 1 (b) Maximum non-expansion work is also 0.11 k1 mol - 1
From the data , draw up the following table : T IK
TIK
10
15
20
25
30
50
0.284
0.47
0.540
0.564
0.550
0.428
70
100
150
200
250
298
0.333
0.245
0.169
0.129
0.105
0.089
Plot CPIT against T (Fig. 4.1). This has been done on two scales. The region 0 to 10 K has been constructed using CP= aT 3 fitted to the point at T= 10 K , at which CP=2.8 1 K - 1 mol - 1, so a = 2.8 x 10- 3 1 K - 4 mol- 1 • The area can be determined (primitively) by counting squares, which gives area A= 38.281 K - 1 mol - \ area B (up to 0 °C)=25 .601K - 1 mol - 1, area B (up to 25 oq = 27.801 K - 1 mol - 1• Hence: Sm(273 K) =Sm(O) +63.88 1 K - 1 mol - 1 Sm(298 K) = Sm(O) + 66.081 K - I mol - 1 Th- Tc 1200 K
4.9
e = ~=
2273 K=0.53
w=mgh, w = eq , q=nflHc
The second law: the concepts
-
oo
Fig4.1
I 1'\. I
05
'
~ ~
~
I'.
A
~
r-.... ........, 01
0
'
l8 0
20
1.0
~ t-..,
8
Etra cia on 60
80
100
200
300
TjK
Eq
Hence h = - = 0.53 x mg
3 X 103 g 114.2 g mol
1
5512 X 103 J mol - 1 x -:-:,.-::--c----::-:-----::;2 1000 kg x 9.81 m s
=7.8km 4.10
t.G 9 = t.H 9
-
T t.S 9 = 26.120 kJ mol - 1
t.H 9 = +55.000 kJ mol- 1
Hence t.S 9 =
(55.000-26.120) kJ mol - 1 _ K = +96.864 J K - I mol - 1 298 15
t.S 9 = 4S 9 (K + , aq) + S 9 ([Fe(CN) 6] 4 -, aq) + 3S 9 (H 20, /) - S 9 (K 4 [Fe(CN) 6] • 3H 20, s) Therefore,
r-, aq) = t.S
S 9 ([Fe(CN) 6
9 -
4S 9 (K +, aq)- 3S 9 (H 20, /)
+ S 9 (K 4 [Fe(CN) 6]· 3H 20, s) = 96.864-4 X 102.5-3 X 69.9 + 599.7] K - I mol - 1 =+76.9JK - 1 mol - 1 4.11 Draw up the following table:
71
The second law: the concepts
72
TIK
10
20
30
40
50
60
(Cp!T)! (J K - 1 mol- 1)
0.209
0.722
1.215
1.564
1.741
1.850
TIK
70
80
90
100
110
120
(CPIT)I (J K - 1 mol - 1)
1.877
1.868
1.837
1.796
1.753
1.708
TIK
130
140
150
160
170
180
1.665
1.624
1.584
1.546
1.508
1.473
TIK
190
200
(Cp! T)! (J K- 1 mol- 1)
1.437
1.402
(C/T)/ (J K - 1 mol - 1)
Plot CPIT against T (Fig. 4.2a). Extrapolate to T= 0 using CP = aT 3 fitted to the point at T = 10 K, which gives a = 2.09 mJ K - 2 mol- 1• Determine the area under the graph up to each T and plot Sm against T (Fig. 4.2b). (a)
2 ·0
-
ie-.
J
I
300 (b) Fig4.2
-...
_E 200
~ ~
/
~
'0
l
~
/
100
/
0
/
\!2.
I l1 0
/
I
~
~ ;;::::
IT
0
/
I
0
v
0
50
200
0
50
/
The second law: the concepts
73
TIK
25
50
75
100
125
150
175
200
{Sm- Sm(O)}/ (J K- 1 mol- 1)
9.25
43.50
88.50
135.00
178.25
219.0
257.3
293.5
4.12 Draw up the following table and proceed as in Problem 4.11:
TIK
14.14
16.33
20.03 31.15 44.08 64.81
(Cp!T)I (JK- 2 mol- 1)
0.671
0.778
0.908 1.045
1.063
TIK
183.59
225.10
262.99
298.06
(C/T)/ (J K- 2 mol- 1)
0.787
0.727
0.685
0.659
100.90 140.86
1.024 0.942
0.861
Plot CP against T (Fig. 4.3a) and CP/T against T (Fig. 4.3b), extrapolating to
~
Fig4 .3 (b) 1·0
Ia)
200
"'~ ...... ,., ........
__..,....
I 150 0
E
/
\.:: 100
,.
v-
,-
.......... li
II
/
:i,
(j
50
0
0'2
f ~ 0
0 100
T/K
200
300
0
100
T/K
200
300
T= 0 with CP = aT 3 fitted at T= 14.14 K, which gives a= 3.36 mJ K - 3 mol - 1 It then follows that
The second law: the concepts
74
J
298 K
CPdT=34.4 kJ mol - 1, so H m(298 K) = Hm(O) + 34 .4 kJ mol - 1
0
J
298 K
C dT . --T- = 243 J K - 1 mol- 1, so Sm(298 K) =Sm(O) +243 J K - 1 mol - 1
0
c=
4.13
Tc r
_
1 h
q
nCP!'!.T
c
c
T , w = -= c
Th = 1.20 K, (Tc)mcan= t (l.lO K + 0.10 K) = 0.60 K 0.60K
c = 1.20 K- 0.60 K
l.OO
1.0 g 3.9 X 10 - 5 J K -I mol - 1 X 1.00 K 1.00 =0. 6 l.uJ w = 63 .54gmol_ 1 x For the more realistic calculation ,
= -n
=n =
r(AT~ +B) ( Th-Tc)dTc
r
(AT~ - AThT~ + BTC-
BTh) dTc
n{tA(T1 - T;) - j-A (Ti - Tf)Th +
ts(n - n)- B(Tr-1i).Th}
We evaluate this expression with n = 0.016 mol , A = 4.82 x w-s J K - 4 mol - 1, B = 6.88 x 10- 4 JK - 2 mol - 1, Th =1. 20K , Ti=l.lOK , Tr=0.10K: W
= 0.016 mol.
4.21
X
X
10- 4 J mol - 1 = 6.7 ttl
4.14 The four episodes of heat transfer are
qh VB -=nRlnTh VA (b) 0 [adiabatic] Vo
(c) qc= nRTc In Vc
The second law: the concepts
75
(d) 0 [adiabatic] Therefore
f
Therefore
rJ.dqT = 0
dq qh qc VaVo - =- +- = nR In - T Th Tc VAVC
If the first stage is replaced by isothermal , irreversible expansion,
w= -pc.(V 8 - VA), implying that q=pc.(V8 - VA) [as t..U=O] Therefore,
Va However , Pcx(V 8 - VA) < nRTh InVA because less work is done in the irreversible expansion, so
rJ. Tdq ,u(l), the gas tends to condense into a liquid.
Therefore, !'iT =
Tpgh !'1 V !'lH 234.3 K X 13.6 X 103kg m- 3X 9.81 m S- 2 X 10m X 0.517 X 10 - 6 m 3 mol - 1 2.292 x 103J mol - 1
=0.070K Therefore, the freezing point changes to 234.4 K 6 .5 AteqUit Tb num,n . p 1V H vap, 1 =RT ' q = -nl!'l
!'l
T =-c q
"
Changes of state: physical transformations of pure substances
99
Therefore ,
tiT=
-piV tiHvap RTC p
-23.8 Torr x 50.0 L x 44.0 x 103 J mol - 1 62.364 L Torr K -I mol- 1 x 298.15 K x 75.5 J K -I mol - 1 X
250g g mol - l 18 02 .
= -2.7 K
The final temperature will be about 22 oc d lnp
6.6
d"T=
tiHvap
RT 2 [Sa], tiHvap
In p =constant- RT Therefore, plot lnp against liT and identify -tiHva/R as its slope. Construct the following table:
ere
0
20
40
50
70
80
90
100
TIK
273
293
313
323
343
353
363
373
1000 KIT
3.66
3.41
3.19
3.10
2.92
2.83
2.75
2.68
In p/Torr
2.67
3.87
4.89
5.34
6.15
6.51
6.84
7.16
The points are plotted in Fig. 6.2. The slope is -4546, so 7·5
~
...
Fig 6.2
......
~
.......
......, ~
""• 2 ·5
~6
'r-...
....... ~8
30
32 (103/ TiK
3~
3~
3~
100
Changes of state: physical transformations of pure substances
- !"J.Hvap R
_ -1 - 4546 , or !"J.Hvap - +37 .8 kJ mol
the normal boiling point is reached at p = 760 Torr, which occurs at 1000 KIT = 2.80, so Tb = 357 K (84 °C). [Alternatively, do a least-squares fit of In p to 11 Tusing the procedure outlined in the Appendix.] 6. 7 Adapt the procedure in Problem 6.6, but note that Tb = 227 .5 oc is obvious from the raw data. Draw up the following table:
ere
57.4
100.4
133.0
157.3
203 .5
227.5
TIK
330.6
373.6
• 406.2
430.5
476.7
500.7
1000 KIT
3.02
2.68
2.46
2.32
2.10
2.00
lnpiTorr
0.00
2.30
3.69
4.61
5.99
6.63
The points are plotted in Fig. 6.3. The slope is -6.6 x 103 , so - 6.6 X 103 K, implying that !"J.Hvap= +55 kJ mol- 1
Fig 6.3
6 1'.. ..........
'
~ ~ .........
0 2·0
6.8
!'-.. 2·2
2·4 2·6 (1rftTJK
(a) Solid-liquid boundary:
!"J.Hrus T p=p * + !"J.V · In T * (6a) fus
(b) liquid- vapor boundary:
2·8
3·0
- !"J.Hvap
R
Changes of state: physical transformations of pure substances
(c) solid-vapor boundary
We need !).Hsuh= !).Hrus + !).Hvap=41.4 kJ mol - 1 1
1 )
!).Vrus=M ( p(l)- p(s)
=
78.11 g mol gcm- 3
1 (
X
1 1 ) 0.879-0.891
=+1.197cm 3 mol- 1 10.6xl03 Jmol - 1 T (a) p=p*+l.197xl0- 6 m3 mol _ 1 lnT*
-
T
= p* + 8.855 x 109 PaIn T * T
= p* + 6.64 x 107 Torr In T* [1 Torr= 133.322 Pa] This line is plotted as a in Fig. 6.4, starting at (p * , T * ) = (36 Torr, 5.50 oc [278.65 K]).
100
Fig 6.4
a 80
SA /_..;A
_.d Vb
.s 40 20
~ .-
(Fe' +. aq) =2 x 77 . 1 - ( -78 .90) kJ mo l- 1 = +223. 1 kJ mol - 1 = 2!'!.G 8 (Ag +, aq)- !'!.G ((Fe ~ + . aq)
!'!.H 9
=2 a!'!.G 9 ) -( aT
x 105.58- ( -89.1) kJ mol - 1 = +300.3 kJ mol - 1 =-!'!. 5 ° =
!'!.G
01
-
!'!.H
8
[!'!.G=!'!.H-T!'!. S ]
T
"(223.1- 300.3) kJ mol 298 . 15 K
1
- 0 .259 kJ mol - 1 K
Therefore, !'!.G 9 (333 K) ""'223. 1 + 5 K
- I
x ( -0 .259 K - 1) kJ mol - 1
""' +22UskJmo l- 1 10.22
C u,(P0 4 Hs) :;:::= 3Cu' +(aq) + 2P01- (aq)
K,P = a(Cu' -.- ) 1 a(Po~ - )~ ""' m(Cu 2 + )'m(P01- )2 (a) S= m(Cu,(P0 4) 2) = \m(C u 2 +) Howeve r . m(Po ~ - ) = ~ m(Cu ~+ ) Therefore . ,.1es t h at ·'(' = K 'I' = 4,11n (C u-, +)'·, w h.1c h 1mp
I
1
115 x ( 'I• K 'I' )
Hence , S= .\
xn
X 1.3 X lo- ' 7t'= 1.6 X Io- ~ [mol kg -
1 ]
(b) The cell react ion is R:
C u 2 +(aq)+2e - ___.,.Cu(s)
L:
2H+ (aq) +2e - -->H~(g )
Overall: Cu 2 +(aq)
+0.34 ()
+ H 2 (g)--> Cu(s) + 2H +(aq)
From the Nernst eq uation ,
RT E=E e - - ln Q vF
25.693 x IW ' V = 0 .34V -
2
v
a(H +) 2 ln a(Cu +)
+ 0 .34
v
177
178
Equilibrium electrochemistry 25 .693 x = 0.34 Y-
w- 3 Y In
2
3
1 X _ x 16
w- x[m(Cu 2 +) = 35]
=0.34 y -0.22 y = +0.12 y 10.23 R:
L:
(a) Sn(s) +
Sn 4 + (aq)~2Sn 2 +(aq)
Sn 4 ++2e--?Sn 2 +(aq) 2
Sn +(aq)+2e - -?Sn(s)
+0.15Y} -0.14 Y
£ 9 = +0.29 y
vF£ 9
2 X 0.29 Y _ In K =Iff= 25.693 my= 22.6, K = 6.5 x 109 (b) Sn(s) + 2AgCI(s)~SnCI 2 (aq) + 2Ag(s)
R: L:
2AgCI(s)+2e - -72Ag(s)+2CI - (aq)
+0 .22Y}
2
Sn +(aq)+2e - -7Sn(s)
InK=
-0.14Y
2X 0.36 Y _ _ mY= +28.0, K = 1.5 X 10 12 25 693
(c) 2Ag(s) + Cu(N01Maq) ~Cu(s) + 2AgN01(aq) R:
Cu 2+(aq)+2e - -?Cu(s)
+0.34Yl
L:
2Ag+(aq)+2e - -?2Ag(s)
+0.80Y
InK=
-0.46 y
2 X ( -0.46 Y) _ _ mY =-35.8,K=2. 8x i0 - 1h 25 693
(d) Sn(s) + CuSO.(aq)~Cu(s) + SnSO.(aq)
R: Cu 2 +(aq)+2e - -?Cu(s) L:
2
Sn +(aq)+2e - -?Sn (s)
+0.34 Y} +0.14
+0.48 y
2 X 0.48 Y _ InK= 25.693 mY= +37.4, K = l.7 x IOih
(e) Cu 2+(aq) + Cu(s) ~2C u +(aq) R:
Cu 2 +(aq)+e - -7cu +(aq)
+0. 16Y}
L:
Cu +(aq)+e--?Cu(s)
+0.52Y
InK=
-0.36 y _ my 25 693
- 14.0, K = 8.2 x 10- 7
-0.36 y
+0.36 y
Equilibrium electrochemistry S(AgCI) = m(Ag +)
10.24
AgCI(s)~Ag+(aq)
+ Cqaq)
K,r = m(Ag +)m(Cl - )/m 6
2
Since m(Ag +) = m(cn
K,P = m(Ag+) 2/m 62 = S 2/m 62 = (1.34 X 10- 5) 2 = 1.80 x
w-w
S(BaS0 4) = m(Ba 2 +) BaS0 4 (s)~Ba2+ (aq)
+ so~ - (aq)
As above, K,r=S / m e' =(9.51 x 10 - 4) 2 = 9.04 x 10 - 7 2
We can estimate the activity coefficients usi ng lg Y± = -A lz +z- 11 112 , A= 0.509 For AgCI, I=S, lz +z-1= l , and so lg Y± = -0.509 X (1.34 X 10- 5) 112 = - 1. 86 X 10- ' , Y± = 0.9957 Hence , K,r=y~
X K~, [K ~,
calculated previously] = 0 . 991K~,
For BaS0 4 , /=45, lz +z- 1=4, and so lg Y± = -0.509 X 4 X (4 X 9.51 X 10 - 4 ) 112 = -0. 126, Y±= 0.75 Hence K,r= y~K~,= 0.75 2 K~r=0 . 56K~r Thus, the neglect of activity coefficients is significant for BaS0 4 . 10.25 The half-reaction is Cr 2 0~- (aq)
+ 14H +(aq) + 6e----?2Cr'+ (aq) +7H 20(/)
The reaction quotient is a(Cr1 +) 2 Q=
a(Cr 2 0~
)a(H +) 14
v=6
Hence,
10.26
R: L:
2AgCI(s)+2e - ---?2Ag(s)+2CI-(aq) 0.22 v 2H +(aq)+2e - ---?H 2(g) 0
+0.22V}
179
Equilibrium electrochemistry
180
Overall: 2AgCI(s) +
H 2 (g)~ 2Ag(s)
+ 2H +(aq) + 2Cqaq)
Q = a(H +)2a(Cn 2 v = 2 =a(H +) 4
[a(H +) = accnJ
Therefo re, fro m the Nernst eq uatio n,
RT
= £ 9 +2 X2.303 F
pH
Hence F
pH= 2 X 2.303 R TX (£- £
E-0.22 V
9 )
= 0.1 183 V
0.322 v- 0.22 v - - -- - =0.86 0.1183 v 10.27
R: AgB r(s) + e - ~ Ag(s) + Br - (aq) L:
Ove rall :
Ag + (aq) + e - ~Ag(s)
Ag Br (s) ~
Ag+(aq) + Br - (aq)
Therefore, sin ce the cell reaction is the so lu bili ty equi li brium , fo r a sat urated solution there is no further te ndency to dissolve and so E = 0.
+0.80 L:
Agl(s)+e - ~Ag(s)+I - (aq)
Overa ll (R-L) : ln K=
0.9509V . V 25 693
v} - v
- 0.15V
Ag + (aq)+! - (aq)~Ag!(s)
+ 0.9509
v= 1
_ 37 .010, K=l.l84 x 10 1fi
However, K ,r = K - I since the so lu bi li ty eq uilibrium is writte n as the reverse of the cell reactio n. T herefore, K,P= 8.45 X 10- 17 The solubility is obtain ed fro m m(Ag +) = m(I - ) and S= m(Ag +) , so K,r = m(Ag +)2 , im plying that S = (K,p) 112 = (8.45 X 10- 17 ) 11 2 = 9. 19 X 10- 9 (mo l kg- 1]
Equilibrium electrochemistry
Problems 10.1
R:
Hg~ SO.(s)+2e - ~2Hg(l)+SO~ - (aq)
+0.62V
L:
PbSOb)+2e - ~Pb(s)+SO~ - (aq)
-0 .36V
R- L:
Pb(s) +
Hg~SO.(s)~
PbS0 4(s) + 2Hg(/ )
+0 .9R V
He nce, a suitab le ce ll wo uld be
o r, a lte rn ati ve ly, Pb(s) IPbSOis) IH,S04(aq)ll H ~S04(aq) IHg~ SO.(s) IHg(/) Fo r the ce ll Pb(s) IPbSOis) IPbS04(aq )IIHg~S04(aq) IHg~SO.(s) IHg(l ) T he electrode pote ntials ~t re
beca use K,r =a +a_ = (a _)' T herefo re,
RT K, 11 (Hg~S0 4 ) 1 ;~ E= 0.98 V- 2F In K,"(PbS04)1 '~ RT K ,"(Hg~S0 4 ) =0. 98 V- 4F ln K (PbSO) 'P
25.693 =0.98V= 0.98 10.2
X
4
.t
1
IW V
6.6 X 10- 7 lnl. ox lO - x [Tab le l0.6]
v- 0 .024 v = 0.96 v
(a) H~(g)
+ lO,(g)~ H ~O(/)
t.G 0 = t.G ~'(H 2 0. /) = -237. 13 kJ mo l- 1 [Tab le 2. 10] t.G +237. 13 kJ mol- 1 Ee = ---= =+ 1. 23V vF 2 x 96.485 kC mo l- 1
181
182
Equilibrium electrochemistry
(b) C4 H 111(g) + lf0lg)~4C0 2 (g) + 5H 20(/) !l.G 9 = 4!l.G }"l(C0 2 , g)+ 5!l.G }"l(HP, /)- !l.G}"l(C4HIIl, g) =4 x ( -394.36) +5 x ( -237.13)- ( -17.03) kJ mol - 1 (Tables 2.9 , 2.10] = -2746.06 kJ mol- 1 To find the number of electrons transferred , note that the cathode halfreaction is the reduction of oxygen to produce 5H 20: t Oz(g) + lOe - + 10H + (aq)~5H 2 0(!)
v= 10
Therefore , 9
£ 10.3
-!l.G 9
+2746 .06 kJ mol - 1 =~= 10 x 96.485 kC mol - 1 = + 2 ·85 y Ptl H 2(g) IHCI(aq )J Hg 2Clz(s) IHg(l)
H2 (g) +
Hg 2 CI 2 (s)~ 2Hg(l)
+ 2HCI(aq)
v=2
a(H +) 2a(Cl - ) 2 y ~ m 2 Q= [(H2)/p 9 = flp 9 RT RT RT f E=£ 9 --ln Q=£ 9 --lny m+-ln9 2F
f= yp with In y=
F
z;
J:: (
±
1 ) dp
2F
p
(Chapter 5]
Therefore, fl
In y=
J (5.37
X
10- 4 atm - 1 +3.5 X 10 -xp atm- 2 ) dp
II
= 5.37 X J0- 4 (p/atm) + 1
X
3.5 X 10-x(p/atm?
Hence , at 500 atm , lny=0.268S+0.0044=0.2729 , y=1. 314 At 0.10 mol kg- 1, y ± = 0. 798 (Table 10.4], so RT RT 500 atm x 1.314 E= £ 9 - - l n 0.798 x 0.10+-ln - -- -F 2F 1 bar _ _ 25 .693 mY In 1.314 X 500 X 1.0133 = 0.27 Y- 25.693 mY In 0.0798 + 2
183
Equilibrium electrochemistry
= 0.27 y + 0.0650 y + 0.0835 y = +0.42 y H 2(g)JHCl(aq)JCl 2(g)
10.4
RT E=E e -2FlnQ ,
Q=
H 2(g)+Cl2(g)->2HCl(aq)
v =2
a(H +)2a(cl -}" f(Cl2)/p e , E e =+l.36Y,f/p e =l
Form= 0.010 mol kg - 1 , Y±= 0.905 [Table 10.4], a(H+)a(cn = y~ m 2 E = 1.36 Y -25 .693 mY In 0.905 2 X 0.010 2+25.693 mY In ( ;e)
-
f
= 1.602 Y + 1 X 25.693 mY In e p Therefore ,
f E-l.602Y In p e = 0.0 1285 y
withp 9 = 1 bar
Hence , we can draw up the following table: p/bar
1.000
50 .00
100.0
ElY f!p e y
1.5962 0.637 0.637t
1.6419 22.3 0.446
1.6451 28.6 0.286
t This seems abnormally low at this pressure .
10.5
H 2(g)JHCl(aq)JHg 2Clls)JHg(/)
E=E9
RT
-F In a(H +)a(cn
2RT · 2RT =E 9 - - l n m - - l n y F F ±
= E 9 - 0.1183 Y lg m + 0.1183 Y x Ax m 112 [Debye-Hiickel and 2 x 2.303RTI F= 0.1183 Y) E + 0.1183 Y lg m = E 9
+ constant x m 112
Therefore, plot E + 0.1183 Y lg m against m 112 , and the intercept at m = 0 is E 9 /Y. Draw up the following table:
184
Equilibrium electrochemistry
(m/m e )ll2 EIY+0 .1183lgm
1.6077
3.0769
5.0403
7.6938
10.9474
0.04010 0.27029
0.05547 0.27109
0.07100 0.27186
0.08771 0.27260
0.1046 0.27337
The points are plotted in Fig. 10.1. The intercept is at 0.26835, so E e = +0.26835 V. The least-squares best fit (Appendix) gives E e = +0.26838 V and a coefficient of determination equal to 0.99895. For the activity coefficients we form: E e -E m In Y±= 2RT/F -In me m 0.26838- ElY ------ln0.05139 me
and draw up the follow ing table: 0-274
Fig 10. 1
/
~
l
/
0·272
/
.Ql
~
.I
~
/
0
+
:0,.
.
~
/
027
Lil
/ L 02fJJ 0
(}02
004
006
008
0 ·10
(ml me) " 2
m/(mmol kg- 1)
1.6077
3.0769
5.0403
7.6938
10.9474
In Y± Y±
-0.03562 0.9650
-0.05135 0.9500
-0.06639 0.9358
-0.08090 0.9223
-0.09597 0.9085
Equilibrium electrochemistry 10.6
PtiH 2(g)iNaOH(aq), NaCl(aq)IAgCI(s)iAg(s)
H 2(g) + 2AgCl(s)---c> 2Ag(s) + 2Cl - (aq) + 2H +(aq)
E = E9
-
RT F In Q, Q=a(H +)"a(cn" 2
[J!p 8 = 1]
RT
RT K,a(cn £ 8 - F In a(OH - ) RT
= E8 - -
=
v= 2
-F In a(H +)a(CI - )
= E8
=
185
F
In
Kwy +m(Cl - ) = E8 Y±m(OH - )
-
RT K".m(Cn - In----,--,------,F m(OH - )
RT RT m(Cl - ) 6 E - F In K"'-F In m(OH - )
= E
RT RT m(Ci- ) +2.303Fx pK"-F In m(OH - )
8
He nce,
pK"' =
E-£ 8 lnm(Cl- )/m(OH - ) 2.303RTI F + 2.303
E-E 8 0 051 14 = 2.303RTI F + · We then draw up the follow ing table with E 9 = +0.2223 Y:
ere
20 .0
25.0
30.0
ElY (2.303RTI F)!Y pKw
1.04774 0.058 19 14.23
1. 04864 0.059 18 14 .01
1.04942 0.06018 13.79
d In Kw
-
6.H 8 RT
- = --2
dT
He nce,
186
Equilibrium electrochemistry =2 .303 X 8.3141 K- 1 mol - 1 X (298.15 K) 2 X
13.79-14.23 10 K
= +74.9 kJ mol - 1
t::..G 9 = -RT in Kw = 2.303RTx pKw= +80.0 kJ mo l- 1 t::..H 9 -t::..G 9 t::..S 9 = -17 .1 J K - t mol - 1 T
See the origin al reference for a careful analysis of the precise data. Ag(s)jAgX(s)jMX(m 1)jM,Hg(s)
10.7
Hg
R: M+(m 1) + e- -----'?M,Hg(s) ama lga m] L:
[Red ucti on of M + and for mation of
AgX(s)+e - ~Ag(s)+X - (m 1 ) J-Ig
Ag(s) + M+(m 1) +X - (m 1 )~MxHg(s) + AgX(s)
R- L:
v=1
a(M,Hg)
Q = a(M +)a(X -) RT £=£9 - F JnQ For a pair of such cells back to back, Ag(s)jAgX(s)jMX(m 1)jM,Hg(s)jMX(m 2 )jAgX(s)jAg(s) RT
ER =
£9
-F In QR
EL =
£9
--
RT
RT
F
In QL
QL
RT
(a(M +)a(X - ))t.
F = - - In - = - In -:--:----,-.,.....-,--:-:. F QR F (a(M +)a(X ))R
2RT m 1 2RT Y±( l) =-ln-+--ln-F m2 F Y±(2) Take m 2 =0 .09141 mol kg - 1 (t he refe rence va lue), and write m =m 1/m 9 2RT ( In m E=-+In Y+ - ) F 0.0914 1 Y±(ref.)
Equilibrium electrochemistry
187
Form= 0.09141, the extended Debye-Hiickellaw gives - 1.461 X (0.09141) 112 _ lg Y± = 1 + 1.70 X (0.0 9141 ) 112 + 0.20 X 0.09141 = -0.2735 Y± =0.5328
and
m Y± ) E=0.05139V x ( ln0.09141 +ln0.5328
E
m
In Y± = 0.05139 V -In 0.09141
X
0.05328
We then draw up the following table:
ElY y
0.0555
0.09141
0.1652
0.2171
1.040
1.350
-0.0220 0.572
0.0000 0.533
0.0263 0.492
0.0379 0.469
0.1156 0.444
0.1336 0.486
A more precise procedure is described in the original references for the temperature dependence of E 8 (Ag, AgCl, Cl - ), see Problem 10.10. 10.8
H 2(g)IHCl(m)IAgCl(s)IAg(s)
t H 2(g) + AgCl(s)- HCI(aq) + Ag(s) E= E 8
RT
2RT
2RT
-F In a(H +)a(cn = E 8 -F In m-F In Y±
2RT RT = E 8 - F In m-2 x 2.303 F lg Y±
Therefore, with 2RTI F x 2.303 = 0.1183 V, ElY+ 0.1183lg m- 0.0602m 112 = E 8 !Y- O.J183km
hence, withy= ElY+ 0.1183lg m - 0.0602m 1' \ y = E 8 !Y - 0.1183km
[m ;; mfm 8 ]
188
Equilibrium electrochemistry
We now draw up the following table:
y
123.8
25.63
9.138
5.619
3.215
0.2135
0.2204
0.2216
0.2218
0.2221
(a) The last three points are plotted in Fig. 10.2, and extrapolate to 0.2223 V, 0·2224
~ ~a
Fig 10.2
\
0·2222
1\
aI
e: .l2\ 0 ·2220
\
\
-.. / / / v/ / ~ / / / / / / v /. >--i/ / / / / / / / / / v / 1/ / / 1/ / / / /
Fig 10.4
/ / / / / / / / / 7
0
0 ·01
002
0·03
m/mol kg-1
0 ·04
005
infinity as m~o, but we are confident about the validity of the Debye-Hi.ickel limiting law in this region, and evaluate the integral analyt ically up to m = 0.010m9[See Problem 10.18 for details]:
f
m (
II
¢)
1 - - dm=_j_A' 3 m
I"' -dm= 1.A'm J/1
II
m -
3
112
,
A'=2.303A
Therefore, up tom= 0.010m 9 the value of the integral is 0.0781. Above that molality, evaluate the integral numerically. In the range from 0.010m 9 to 0.050m 9 its value is 0.106. Therefore, In Y± =0.926 - 1 - (0.106+0.0781) = -0.258
192
Equilibrium electrochemistry
Hence Y± = 0.77.
10.12 K '= a
K. =
a(H+)a(A -) a(HA)
y~m(H+)m(A -)
m(HA)
y~K~
m(H+)m(A -) a 2m = -m(HA) 1- a
Hence, lg K~ = lg K., - 2lg Y± = lg K. + 2A(IIm e) 112 = lgK. + 2A(am) 112
[! = am]
We therefore construct the following table:
1000(am/ m 9 ) 112 105 X K~ lgK~
0.0280
0.114
0.2184
1.0283
2.414
5.9115
3.89 1.768 -4.753
6.04 1.779 -4.750
7.36 1.781 -4.749
11.3 1.799 - 4.745
14.1 1.809 -4.743
17.9 1.822 - 4.739
lg K~ is plotted against (am/m 9 ) 1' 2 in Fig. 10.5, and we see that a good
-4·74
-4·75
-4·76
0
~/
0 04
l.-"
v
0 08
./
.,.
012
/"
0 16
v
0 20
.V( 10- 23
201
El(kJ mol - 1)
p/(kgms - 1)
199 218 299 598 7.98 X 105 0.012
1.10 X 10 -27 1.20 x w- 27 1.66 x w- 27 3.31 x w- 27 4.42 X 10- 24 6.63 X 10- 32
. h 6.626 X 10- 34 J s 11.12 p = )..= , ; the entries are in the table above.
11.13 If a photon is absorbed, the atom acquires its momentum p. It therefore reaches a speed u such that p = mu. We use mH
= 1.008 U = 1.008 X 1.6605 X 10- 27 kg= 1.674 X 10- 27 kg
and draw up the following table using the information in the table above: ..:Unm 600 550 400 200 150pm 1.00 em
ul(m s- 1)
0.66 0.72
0.99 1.98 2640 3.96 X 10-s
11.14 The energy of a photon of 650 nm light is E =he/ A with A= 650 nm. The total number of photons emitted in an interval r is
Pr PrA =E he
N=-
with P = 0.10 Wand r = 10 y. The total momentum emitted is
Nh PrA h Pr =X-=A he A c
p= -
202
Quantum theory: introduction and principles
The momentum is acquired by the source, and since its mass is m, and
p = mv, its final speed is Pr
0.10Js - 1 X3.16X10 8 s
em
2.998x10 8 ms- 1 x5 .0xl0 - 3 kg
v=-=
= 21ms- 1 Note that the answer is independent of the wavelength of the radiation emitted: the greater the wavelength ther smaller the photon momentum, but the greater the number of photons emitted. 11.15
p N= hv
p).
[P=power in J s- 1]= he
6.626 X 10- 34 J Hz - 1 X 2.998 X lORm s- 1 =
(P!W) x (J../nm) s- 1 1. x _ 16 =5 .03xl0 15 (P/W)x(A/nm)s - 1 99 10
(a) N = 5.03 X 10 15 X 1.0 X 550 s- 1 = 2.8 X 10 18 s- 1 (b) N=5.03xl0 15 X l00 x 550s - 1 =2.8 x l0 20 s- 1 11.16 From Wien's law,
n.,.,x= tcz= 1.44 em K [8] Therefore , 1.44 em K T= 5 X 480 X 10- 7 em 11.17
°
600
K
he +mv 2 = hv- = - - J..
=2. 14 eV ~ 2.14 X 1.602 X 10- 19 J = 3.43 X 10- 19 J
(a) hc!J..=
6.626 X 10- 34 J S X 2. 998 X lQR Ill 700x 10 9 m
= 2.84 X 10 - 1'1 J < ,
SO
S- 1
no ejectio n occurs
19
(b) hc/J..=6 .62 x !0- J Hence
~ mv 2
= (6.62- 3.43) X 10 - 19 J = 3. 19 X 10- 1'1 J
Quantum theory: introduction and principles
2 X 3.19 X 10- 19 J) 112 u= ( 9.109 x 10 31kg =837kms - l 11.18
oA. = A.,(1- cos 8)
[16] =A., when 8 = 90°
(a) A.c = 2.43 pm [16], so oA. = 2.43 pm h
(b) Ac= mrc
[18 , mc--7 mp]
1.673 X
6.626 x w- 34 1 s 1.32 X 10- ISm 10 27 kg X 2.998 X 10Km S- 1
Therefore, OA = 1.32 fm h
!!.E = hv=T [T= period]
11.19
(a) !!.E=
6.626 x w- 341 s w-1 5s
corresponding to N A (b) !!. E =
X
7X
7 x 10- 19J
w-19 J = 400 kJ mol - l
6.626 x w-34 1 s 1 = 7 x 10- 20 J4 , -"O::. . :k!::.J__,_,m"-'o'-'--1-_ w - 14 s
h
h
A= - = p mu
11 .20
6.626 x 10 - 34 1 s 6.6 X 10 - 29 m ( a) A- 1.0 X 10- 3 kg X 1.0 X 10 - 2m S - I ( b) A-
- 1.0 X
(c) A.-4.003 11.21
6.626 x w-341 s 6.6 X 10 - 36 m 10 3 kg X 1.00 X 10 5 m S 1
X
~-mu 2 =
6.626 x w- 34J s ] .6605 X 10- 27 kg X 1000 m S
1
99.7 pm
2e !!.¢) 112
e !!.¢, implying that u = ( ----;;;-
and
203
Quantum theory: introduction and principles
204
that p = mv = (2mef'...¢ ) 112 . Therefore , h A. =- = p
h ..,...-----,--= 112
(2mef'...¢)
6.626 X 10- 34J S (2x9.109 x 10 31kg x 1.602 x 10 19C x f'...¢) 1'2 1.226 nm = (f'...¢/V) I/2 [1 J = 1 c V] (a) /1¢ = 100 V, A.=
1.226 nm 10.0 = 123 pm
(b) f'...¢=l.OkV , A. =
1.226 nm 1. 3 6
39pm
(c) f'...¢=100kV , A.=
1.226 nm . 316 2
3.88pm
11.22
f'...pf'...q~ fh , f'...p=mf'...v
1.055 X 10- 34 ] S f'...vmin=2m f'...q = 2 X 0.500 kg X 1.0 X 10 h
6m
1.1 x 10 - 28 m s- 1
h 1.055 X 10- 34 J S 1 x 10- 27 m f'...qmin =2m f'...u = 2 X 5.0 X 10- 3 kg X 1 X 10 - 5 m s- 1
1.055 X 10- 34 ] S f'...pmin = 2f'...q = 2 X 100 X 10- 12 m = 5 X 10- 25 kg m S- 1 h
11.23
c 2 =hv-l v= 11.24 +mu '" , A. 6.626 x 10- 34 J s x 2.998 x 108 m s- 1 l=T - ! mv = 150 X 10- 12 m
he
2
-f
X
9.109 X 10- 31kg X (2.14 X 10 7 m s- 1? = 1.12 X 10- 15J
Quantum theory: introduction and principles
205
Problems 11.1
1
8nhc (
p = ---;:s- e'"m r -1 1.439 X 10- 2 m K
he
8nhc
[12], ~au=p~A,A=652 . 5nm
2.205 X 104 K
A
Ak -
)
8n X 6.626 X 10- 34 1 S X 2. 998 X 10Hm S-
5
A
(652 .5 X 10
~au= 4.221 X 10 1 m7
4
X (
9
m)
e 22 osx
II~
5
KI T _
) 1
X
1
4.221 x 107 1 m - 4
5 X 10 - 9 m
0.211 1m - 3 (a) T = 298 K, ~au= e220Sx lo-'/29R-1 = 1.6 X 10-33 1 m -3
0.2111 m - 3 (b) T = 3273 K, ~au= e2.2US xlo-'/3273_1 = 2.5
11.2 AmaxT = hc/5k he Amax= 5k
X
10- 4 1 m - 3
[8, and c2 = hc!k]. Therefore
1
xT
and if we plot Amax against liT we can obtain h from the slope. We draw up the following table:
eloc T/K 104 /(T/K)
Amaxfnm
1000
1500
2000
2500
3000
3500
1273 7.86 2181
1773 5.64 1600
2273 4.40 1240
2773 3.61 1035
3273 3.06 878
3773 2.65 763
The points are plotted in Fig. 11.1. From the graph, the slope is 2.83 x 10~, so
he nm = 2 .83 X 10- 3 m K -5k = 2 .83 X 1061/K and 5 X 1.38066 X 10- 23 1 K - 1 X 2.83 X 10- 3 m K h= 2.99792 X 1QH m S I = 6.52 X 10- 341
S
206
Quantum theory: introduction and principles 2500 2000
v.
..
E 1500
~
~
v
I/
Fig 11 . 1
/v
r
cU = 8nhc
11.3
au =
he
1.439 X 10- 2 m K
AkT
AT
~
1
e"c/Ak T
dA
[12]
[Problem 11.1]
8nhc = 4. 992 X 10- 24 J m
For A1 = 350 nm, A2 = 600 nm (a) OU(100°C)=7.47 x 10- 29 Jm - 3 (b) OU(500 °C) = 4.59 x 10- 14 J m- 3 (c) OU(700oC)=3.49xl0- 11 Jm - 3 [Using standard integration programs.] For the classical calculation , use p = 8nkT/ A4 [10]; then OU=
2 J2cU 8nkT ( 1 1) J pcU=8nkT y =- M-M 1
= 8; X 1.381 X
1
3
-
w-n J X (T!K) X { (3 .50 X
= 2.16 x 10- 3 J m - 3 x (T!K) Then , (a) OU(100 oC)=0.807 J m- 3 (b) OU(500 °C)=l.67Jm- 3 (c) OU(700 °C)=2.10Jm- 3
\o
-7 mr- ( 6.00 X 110 -7 mr}
Quantum theory: introduction and principles
207
All three classical values are very much larger than the quantum values. Js x s- 1
hv 11.4
8E=k, (8E] = J K - 1
K
The Einstein formul a [14] reverts to the classical expression when kT ~ hv, or T ~ hv/ k = eE . The criterion for classical behavior is therefore that T ~ e E 6.626 X 10 -> 4 J Hz - \ X v 11 8E=k = 1. 381 x 10-l> JK _1 =4.798 x 10- (v/Hz)K
hv
(a) For v=4.65 X 10 13 Hz, 8E=4 .798x 10 - 11 X 4.65 X 10\J K=223l K (b) For v=7.15 X 10 12 Hz , 8E=4 .798 X 10 - \\ X 7 .15 X 10 12 K=343 K
(jE)2{ e (T e
1
V 2T 11 1.,.
Cv = 3R-
E
}2
(14],classicalvalue=3R
-1
2231 K) 2 { e22.1 1 12 x2~x } 2 (a) Cvi3R= ( 298 K X em ll29x_l =0 .031 4
343 K) 2 { e> J12x2.:~
'\.
A
=
1218 pm 312
cs
&6'.6'
a
'?'!
~ ---L..C.....------==.8
from B. Therefore, e -22.4152.9 + e -86.6152.9
tjJ =
Fig 14.8
200pm
86-0pm
0.65 +0.19 1218 pm 312
6.97 X 10- 4 pm- 312
tjJ 2 = 4.9 X
w-
7
pm- 3 , so P = 4.9 X
w-
7
Molecular structure
257
For the antibonding orbital , we proceed similarly: (a) tjJ2_(z = 0) = 19.6 X 10 - 7 pm - 3 (Problem 14.5) , soP= 2.0 X 10- 6 (b) By symmetry, P=2.0x 10- 6 (c) lfJ2_(t R)=O, so P=Q. (d) We evaluate '1/J- at the point specified in Fig. 14.8: 0.65-0.19 ''' - 7 40 X 10- 4 pm -J/ 2 -r - - 622 pm312 - ·
ljJ2_ = 5.47 X
w-
7
pm - 3, soP= 5.5 X
w-
7
14.7 Draw up the following table using the data in the question and using a0 e2 au e2 e2 --=--x-= x2 4nt:oR 4nt:o£lo R 4nt: 0 X (4nt:cNimce ) R
so that
Rla0 e2
4nt:oR (V1
0
I
RH
+ V2)/RH
(E- EH)/RH
1
00
2.000 00
1.465 0.212
2
3
4
00
0.500
0.333
0.250
0
0.879 -0.054
0.529 -0 .059
0.342 - 0.038
0 0
The points are plotted in Fig. 14.9. The minimum occurs at R=2.5a0 , so R = 130 pm . At that bond length E- EH = - 0.07RH = - 1.91 eV Hence, the dissociation energy is predicted to be about 1.9 eV and the equilibrium bond length about 130 pm.
14.8 We proceed as in Problem 14.7, and draw up the following table:
258
Molecular structure 0 ·4
Fig14.9
\
0·3
\
\
'
1\
\
1
0
r--.. '-""'
' -..._
uo3
2
'
-D·1
1'.. 4
5
...........
Rla0
0
1
2
3
4
co
- ez- ; -R 4nE 0R H
co
1
0.500
0.333
0.250
0
0
-0.007 1.049
0.067 0.338
0.131 0.132
0.158 0.055
0 0
(VI - Vz)IRH (E- EH)/RH
00
The points are also plotted in Fig. 14.9. T he contribution V2 decreases rapidly because it depends o n th e overlap of the two orbitals.
nmU' h n =l ,2, . . . and 1/J,. = (2)112 L sin (nnx) L 2 2
14.9 E,.=
8
Two e lectrons occupy each level (by the Pauli principle) , and so butadiene (in which there are four n electrons) has two e lectrons in 1/J 1and two electrons in 1/Jz:
1/1 1 =
(nx)
2)1/2 ( L sin L
_ (~) 112 .
1/Jz-
L
sm
(2nx) L
These o rbitals are sketched in Fig. 14.10a. The minimum excitation energy is
Molecular structure
259
Fig 14.10
Compared with
Compared with
(a)
CH 2=CH-CH=CH-CH=CH-CH=CH 2 there are eight n electrons to accommodate, so the HOMO will be 1/J4 and the LUMO 1/J5 . From the particle-in-a-box solutions (Chapter 12), h2 9h 2 6.£=£, -£.=(25-16)--,=--, 8mcL- 8mcL-
9 X (6 .626 X J0 - :14 J s) 2 -:---::--:-::-::----c---;-;---,-----:;---:-:; = 8x9.109x 10 ' 1 kg x( 1.12 x 10 " m) 2
4 3 X J0 - 19 J •
which corresponds to 2.7 eV. It follows that he
6.626 X J0 - :1< J S X 2.998 X 10' 4.3 x w - IY J
111 S- 1
,t = - = - -- - -- - - - - , - - - - - - - -
6.£
=
4.6 X l0 - 7 m, or 460 nm
The wavelength 460 nm corresponds to blue light ; so the molecule is likely to appear orange in white light [since blue is subtracted]. The HOMO and LUMO--are-
260
Molecular structure
1/J, =
(L2)
112
sin (n:rrx) L with n = 4, 5 respectively and the two wavefunctions
are sketched in Fig. 14.10b.
14.10 Since1fJ2r = R20 Y00 =
2
3'2(2 v1 2 (2) ~
p)e -P' 2 X ( 1:rr)1n [Table 13.1] 4
1)1 /2 (2)3/2 =41 ( 2:rr ~ (2-p)e -pt2 2
1/J2p, = R21Y10=
1( 1 ) 4 2:rr
=- -
=-1
2 ~6 (~r pe-P12 X(4~r
112(2) 3/2pe -P12 cose a0
(2)3' 2 pe- P12 (- 3 ) 112sin8cos ¢ -
Y6 a0
8:rr
1(1) 1/2(2) 3/2pe -P' sin8cos¢
=4 ~
=
~
2
1(1) 1/2(2)3/2
4
~
~
12
pe-P sin8sin¢
2
cos 8
[Table 13.1]
Molecular structure
261
Therefore,
=41 ( 1n ) 6
11z(z)3' 2 {2-p-psin8cos¢+Y3psin8sin¢}e -P12 ~
1( 1 ) =4 n 6
112(z) ~
312
1( 1 ) n 4 6
112(z) ~
312
=
12
{2-p(1+sin8cos¢-Y3sin8sin¢)}e- P {2-p(1+[cos¢-Y3sin¢]sin8)}e- P'2
The maximum value of 1/J occurs when sine has its maximum value ( + 1) and the term multiplying p has its maximum negative value, which is -1 when ¢ = 120°. 14.11 2
{(V 1 - V2)(1 + S) + (1- S)(V 1 + V2)} _:__ __:::....:.__._:____:_ __:_:_.:___:_ + -2e- + 2£H (1- S)(1 + S)
4nE 0 R
2(SV2 -V 1) 2e 2 2 1- S + 4nEof?. + 2£H The nuclear repulsion term is always positive , and always tends to raise the mean energy of the orbitals above EH. The contribution of the first term is difficult to assess. Where S = O, SV2 = 0 and V1 = 0, and its contribution is dominated by the nuclear repulsion term. Where S = 1, SV2 = V1 and once again the nuclear repulsion term is dominant. At intermediate values of S , the first term is negative, but of smaller magnitude than the nuclear repulsion term. 14.12 N2
J
1fJ 2 dr= 1, 1/J = N(A +B) in a simplified notation .
J(A+B?dr=N J(A +B +2AB)dr=N (1+1+2S) 2
2
2
2
Molecular structure
262
1 2 Therefore, N = ( + S) 2 1 h2
e2
H= --\7 2 - - -
2m
e2
1
• ---- •
4nt:0 rA
e2
1
1
-+-- ·-
4nt:0 r6
4nt: 0 R
H'ljJ = E'ljJ implies that
e2
h2
e2
1
e2 1
4nt:0 r6
4nt:0 R
1
- - V ' z'ljJ --·-'ljJ--·-'ljJ+--'ljJ=E'ljJ
4nt:0 rA
2m
Multiply through by 'ljJ *( = 'ljJ) and integrate using
h2 e2 1 --2 V' 2'1/JA--4- • -'ljJA=£H'lfJA m nt:o rA
Then:
Whence 2
EH
J
1 'ljJ 2 dr+ -e- · 4nt:0 R
J
ez 'ljJ 2 dr---N 4nt: 0
J
'ljJ ) dr=£ 'ljJ ('1/J ~+~
rA
r8
and so
Then use
... J
'lfJA
~'1/JA dr = rs
J
[by symmetry] = V 1/(e 2/4nt:0)
'ljJ 6 !_'ljJ 8 dr
rA
which gives EH = 4::
0
•
~- 1 ~ S) (V, + V (
2)
=E
Molecular structure or
v, + v2 e 2 4nt:u 1+5
263
1
E=E ~-~----+ -·-R
14.13 The Walsh diagram is shown in Fig. 14.11. The steep rise in energy of
Fig 14.11
2e
HAH angle
the 3a 111a;' orbita l arises from its loss of s character as the molecule becomes planar (120°). (a) In NH 3 there are 5 + 3 = 8 va lence electrons to accommodate. This demands occupancy of the 3a 1/la2 orbital , and the lowest energy is obtained when the molecule in nonpl anar with the configuration 2a~1 e 4 3aT. (b) C H j has only 4 + 3- 1 = 6 electrons. The 3a 11la2 orbital is not occupied, and the lowest energy is attai ned with a planar molecule with 2 configuration 2a ; 1e ''.
264
Molecular structure
J~~~
dr=
J~ r e-
2 k'
J~ d¢ = : 2
dr J>in e de
J~V 2~dr = J~~dd:2 (re-k')dr= J~(k2 - :)~dr 2
:rr k
2:rr k
:rr k
Therefore, h2
J
:rr
e2
:rr
~H~dr =- x-- - x -2
2Jl
k
4:rrE 0
k
and h2:rr
E=
e2:rr
2Ji.k- 4:rrEok 2 h 2k 2 :rr/ k
e2k
= ---
3
2Jl
4:rrE0
The optimum energy is therefore e4Jl E =- 2:rr 2E6h 2 3
-hC!JtH, the exact value.
(b) ~=e-k' , Has before. 2
J~~~dr= J~ J~ V 2~
2 2 re- k' dr
dr = - 2
= _2
J:
sine de
J~(3k - 2k 2r 2)~
I:
I~ d¢=~
dr
2 24 2 2 (3kr _ 2k r ) e - k' dr
I:
(3k) (.!!..._) 1/2- (.!!..._) 1/2} 8 2k 16 2
= -
{
B:rr
sin e de
3
3k
2k 5
I~ d¢
Molecular structure
265
Therefore, 3hzk
ezk 112
E =----~
2!1
co(2.n) 1/2
and the optimum energy is therefore e 4f1 8 E = - 12.n3cYz 2 - h x he8l.H Since 8/3.n < 1, the energy in (a) is lower than in (b), and so the exponential wavefunction is better than the gaussian.
15. Symmetry: its determination and consequences Examples 15.1 Since the number of symmetry species of irreducible representations is equal to the number of classes [end of Section 15.6], there are four classes of -operation in the group. 15.2 The elements, other than the identity E , are a C 3 axis and three vertical mirror planes a•. The symmetry axis passes through the C-CI nuclei. The mirror planes are defined by the three CICH planes. 15.3 A D group and a cubic group cannot possess an electric dipole moment [Section 15.3], so of the molecules listed only (a) pyridine, (b) nitroethane, and (c) chloromethane may be polar. 15.4 We use the procedure illustrated in Example 15 .8, and draw up the following table of characters and their products:
J3=p, !2=z J1=Px JJJ3
E
2C4
C2
2a.
2ad
1 1 2 2
1 1 0 0
1 1 -2 -2
1 1 0 0
1 1 0 0
The number of times that A 1 appears is 0 [since 2 0 -2 0 0 are the characters of E itself], and so the in tegral is necessarily zero. 15.5 We proceed as in Example 15.8, considering all three components of the electric dipole moment operator:
Component: AI r(.u) A2 A 1f(.u)A 2
1 2 1 2
1 -1 1 -1
1 0 -1 0
'--or---'
E
z
y
X
1 2 1 2
1 -1 1 -1
1 0 -1
0
'--or---'
E
1 1 1 1
1 1 1
1 1 -1
]
-]
'-------v------
A2
Symmetry: its determination and consequences
267
Since A 1 is not present in any product, the transition dipole moment must be zero. 15.6 We can determine the irreducible representations that contribute to the characters using the technique adopted in Example 15.8 and expressed formally in footnote 3 on p. 448 of the text. Thus, in this group of order 8, the numbers of appearances of each irreducible representation is AI: t(5+2+1+6+2)=2
A 2 : t(5 +2+1-6-2)=0
8 1: t(5 -2+1 -6+2)= 1 8 2: t(5-2+1-6+2)=0 E: t(10+0-2+0+0)= 1
That is, the orbitals span 2A 1 + 8 1 +E. One selection of atomic orbitals is therefore
and the composition of the hybrids is p 1d 4 • 15.7 Under each operation the function transforms as follows: E
c2
c.
av
X
X
y xy X
y xy 1
-x -y xy 1
y -x -xy - 1
-y -xy -1
X
ad -y -x xy 1
From the C4 v character table, we see that this set of characters belongs to 8 2 . 15.8 In each case we must identify an improper rotation axis, perhaps in a disguised form (S 1 =a, S2 = i) (Section 15 .3). Thus, D 211 contains i, C 111 contains a 11 , T11 contains i, Td contains S4 [Refer to more extensive sets of character tables than those provided in the text: see Further reading .) 15.9 By inspection of the outcome of successive operations we can construct the following table:
268
Symmetry : its determination and consequences
First operation :
Second operation
E
r
Cz
C2
q
E
Cz
C2
c;
C2
Cz
E
c;
C2
C2
E
C2 Cz
C'{
C'{
C'{ C2
Cz
E
15.10 List the symmetry elements of the objects (the principal ones, not necessarily all the implied ones) , then use the remarks in Section 15.2, and Fig. 15.1
(a) Sphere: an infinite number of symmetry axes; therefore R3. (b) Isosceles triangle: E , C2 , av, and (c) Equilateral triangle: E , C3, C2 ,
a ~;
therefore C2v
ah
D3 '----v---'
D 3h
(d) Cylinder: E , C. , C2 ,
therefore
ah ;
D ., h
(e) Sharpened pencil: E , C.,, av; therefore C.,v· (f) Propellor: E , C3, 3C2 ; therefore D 3. (g) Table: E, C4 , av; therefore C4 v. (h) Person: E, av (approximately) ; therefore C, . 15.11 (a) N02 : E , Cz , av, a~ ; C2v
(b) N20: E, C.,, Cz, av ; C.,v (c) CHCJ3: E , C3, 3av; C3v (d) CH 2=CH2 : E , C2 , 2C:!,
ah;
D 2h
(e) cis-CHCI=CHCJ; E , C2 , av, (f) trans-CHCI=CHCI; E, C2 ,
a ~;
ah,
15.12 (a) Naphthalene: E , C2 , C',
(b) Anthracene: E , C2 , (c) Dichlorobenzenes:
C2,
Czv
i; C2h ah ; D 2h
ah; D zh
Symmetry: its determination and consequences (a)
(b)
I~
(c)
I~
Fig 15. 1
~ C;,
I
c,
m
(e)
(h)
(i) 1,2-dichlorobenzene: E, C2 , av,
a~;
C2v
(ii) 1,3-dichlorobenzene: E, C2 , a,
a~;
C2v
(iii) 1,4-dichlorobenzene: E, C2 ,
c;, ah;
a,l
D 211
15.13 (a) No D or cubic point group molecule may be polar; so the only polar molecules are N0 2 , N20, CHCI 3 , 1,2-dichlorobenzene, and 1,3-
269
270
Symmetry: its determination and consequences
dichlorobenzene . The trans-dichloroethene molecule is also ruled out by its a 11 plane and its inversion center. (b) All the molecules have at least one mirror plane (a=S 1) and so none is chiral. 15.14 Refer to the C2v character table. The s orbital spans A 1 and the p orbitals of the central N atom span A 1(p, ), B 1(p;), and B 2(py)· Therefore, no orbitals span A 2 , and hence Px(A)- Px(B) is a non bonding combination. If d orbitals are available, we could form a molecular orbital with d xy > which is a basis for A 2. 15.15 The electric dipole moment operator transforms as x (B 1), y(B 2), and z(A 1) [ C2v character table] . Transitions are allowed iff 'ljJ tftl/J; dr is non-zero [Example 15.10], and hence are forbidden unless rr x r(ft) x r ; contains A 1• Since C=A I, this requires rrxr(fi.)=AI. Since Bl X BI=AI and BzX Bz=AI, and A 1 x A 1 =A~> x-polarized light may cause a transition to a 8 1 term, y-polarized light to a 8 2 term , and z-polarized light to an A 1 term. 15.16 The product rrx r(fl.) X r; must contain AI [Example 15.10] . Then , since r; = B I, r(fl.) = r(y) = Bz [C 2v character table], we can draw up the following table of characters:
Bz Bl BIB2
E
Cz
Ov
a'v
1 1 1
-1 -1 1
-1 1 -1
1 -1 -1
=A 2
Hence, the upper state is A 2 , because A 2 x A 2 = A 1• 15.17 (a) The point group of benzene is D 611 , but we can draw conclusions by considering the smaller group C6v because all the elements of C6v are present in D 611 • In this group the components of fl. transform as E 1(x , y) and A 1(z). The ground term is A 1• We note that E 1 x A 1 = E 1 and A 1 x A 1 = A 1• Therefore , the upper term must be E 1 (since E 1 x E 1 contains A 1) or A 1 (since A 1 X A 1 = A 1). In D 611 itself, fl. spans E 1u(x , y) and A 2u(z), and the ground term is A 1g· Then, using A 2uX A;g= A 2u, E 1u X A 1g= E 1u, A 2u X A 2" = A 1g, and E 1ux E 1u= A 1g+ A 2g+ E 2g, we conclude that the upper term is either E 1u or A 2u. (b) Naphthalene belongs to D 211 , but we can consider the simpler subgroup C2•. The ground term is A 1 so we can use the conclusions in Exercise 15.15 for the same group. The upper terms are B 1 (x-polarized) , B2 (y-polarized) and A 1 (z-polarized). In D 211 itself, the components span B 3u(x), B 2u(y) , and B 1u(z)
Symmetry: its determination and consequences
271
and the ground term is A g. Hence, since A g x f =fin this group, the upper terms are B 3" (x-polarized), B2u (y-polarized) , and B 1u (z-polarized).
15.18 We examine how the operations of the C3v group affect l, = xp,.- yp, when applied to it. The transformation of x , y, and z , and by analogy p, , P_v , and p, are set out in Section 15.7:
El, =xp_v -YPx=l, a), = -xpv+ yp,= - 1,
[(x , y , z )~(-x,y , z)]
c;t== ( - t x + t Y3y)(- t V3p, -
tP_v)- (- -}V3x- ~ y)(- t p, + t V3p,.) [(x ,y , z)~( - ~x+ t Y3y, - t Y3x- ty, z)]
= ~ (V3xp,
+ XP_v - 3yp, - V3yp-" - V3xp, + 3xp,- YPx + V3yp-" )
=xp_v -yp, =l, The representatives of E, a" and c; are therefore all one-dimensional matrices with characters 1, -1, 1 respectively. It follows that I, is a basis for A 2 [see the C3v character table].
15.19 We consider the integral
I=
f~.td2 dO= r~a sin 0 cos 0 dO
and hence draw up the following table for the effect of operations in the group C,:
/ 1 / 2
=sin 0 =cos 0
E
ah
sin 0 cos 0
-sin 0 cos 0
In terms of characters: E
!I !2 fd2
ah
-1
1 -1
A" A' A"
272
Symmetry: its determination and consequences
Since fJ2 does not span A', the integral must vanish. If the range of integration is not symmetrical, the reflection ah is not a symmetry element and the group becomes C 1, in which [ 1 and [ 2 both span A , and A X A= A; so the integral does not necessarily vanish.
Problems 15.1 (a) Staggered CH3CH 3 : E , C3 , C2 , 3act ; D 3ct
[see Fig. 15.4 of the text]
(b) Chair C6H 12 : E , C3 , C2 , 3act; D 3ct Boat C6H 12 : E , Cz, av , (c) B 2H 6 : E, C2 , 2C2, ah;
a~;
Czv
D 2h
(d) [Co(en) 3j3+: E, 2C3 , 3C2 ; D 3 (e) Crown S8 : E , C4 , C2 , 4C 2, 4ad, 2S8 ; D4d Only boat C6H 12 may be polar, since all the others are D point groups. Only [Co(en) 3j3+ belongs to a group without an improper rotation axis (S 1 x a) , and hence is chiral. 15.2 The operations are illustrated in Fig. 15.2. Note that R 2 = E for all the Fig 15.2
operations of the group, that ER = RE = R always, and that RR' = R 'R for this group. Since C2ah = i, ahi = C2 , and iC2 = ah we can draw up the following group multiplication table: E
Cz
ah
E
c2
C2
ah
c2
E
i
ah
ah
ah i
E
C2
Cz
E
E
ah
Symmetry: its determination and consequences
273
The trans-CHCI=CHCI molecule belongs to the group C 2".
15.3 Consider Fig. 15.3. The effect of a 11 on a point Pis to generate a"P, and p
the effect of C2 on a11 P is to generate the point C2a1,P. The same point is generated from P by the inversion i , so Cp"P= iP for all points P. Hence, C2a" = i, and i must be a member of the group.
15.4 Refer to Fig. 15.3 of the text. Place orbitals h 1 and h 2 on the H atoms and s, p,, p,, and p , on the 0 atom. The z-axis is the C~ axis; x lies perpendicuiar to a:., y lies perpendicular to av. Then draw up the following table of the effect of the operations on the basis: E
Cz
av
a'v
hi hz s Px P. .
hi hz s P.•. P. .
hz hi s
hz hi s Px
hi hz s -p..
p,
p,
- pv p,
p,, p,
- p .. - p.. p,
Express the columns headed by each operation R in the form (new)= (originai)D(R) where D(R) is the 6 x 6 representative of the operation R. We use the rules of matrix multiplication set out in the Further information section of Chapter 15.
Symmetry: its determination and consequences
274
is reproduced by the 6 X 6 unit matri x. (ii) C2: (hz, h,,
S,
-p-" -py, p,)~(h" hz,
S,
Pn Py, p, )
is reproduced by 0 1 0 0 0 0
1 0 0 0 0 0
0 0 0 0 0 0 0 0 1 0 0 0 D(C2 ) = 0 - 1 0 0 0 -1 0 0 0 0 0 1 (iii) av: (hz, h, , s , Pn -pY' p,)~(h 1 , h2 , s, Pn pY' p,) is reproduced by 0 1 0 0 0 0 1 0 0 0 0 0 D(av) = 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 -1 0 0 0 0 0 0 1 (iv) a~ : (h 1, h2 , s, -p-" pY, p,)~(h 1 , h2 , s, Pn py, p,) is reproduced by 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 D(a~) = 0 0 0 - 1 0 0 0 0 0 0 1 0 0 0 0 0 0 (a) To confirm the correct representation of C2av = a~ we write
D(C2)D(av) = 0 1 0 1 0 0 0 0
0 0 0 0 0
0 1 0 0 0
0 0 0 - 1 0 0
1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 0 -1 0 0 0 0 - 1 0 0
0 0 0 0 0 1 0 0 0 0 1 0
0 1 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0
=D(a ~)
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 -1 0
0 0 0 0 0 1
Symmetry: its determination and consequences (b) Similarly, to confirm the correct representation of 0 1 0 0 0 0
1 0 0 0 0 0
0 0 1 0 0 0 0 1 0 0 0 0
0 0 0
0 0 0 0 -1 0
1 0 0
1 0 0 0 0 0
0 0 1 0 0 0
0 0 0 -1 0 0
1 0 0 0 0 0
0 0 0 0 0 0 0 0 0
-1
0 1 0 0 0 0
0 0 0 0 0
0 0
0 0 0
1
-1
0 0 0
0 0
0 0 0 0 1 0
av o~= C~.
275
we write
0 0 0 0 0
=D(C2)
0
The characters of the representatives are the sums of their diagona l elements:
6
0
2
4
(a) The characters are not those of any one irreducible representation, so the represe ntation is reducibl e. (b) The sum of the characters of the specified sum is
E 3A 1 81 282 3AI + 81
+28~
a~
c2
Ov
3
3
3
1 2
-1 -2
1 -2
3 - 1 2
6
0
2
4
wh ich is the same as the original. Therefore the representation is 3A 1 + 8 1 + 282. 15.5 Representat io n I: D(C,)D(CJ = 1 X 1 = 1 =D(Ch)
and from the character table is either A 1 or + I or -1 respective ly.
A~.
Hence, either D(av) = D(ad) =
Symmetry: its determination and consequences
276
Representation 2: D(C3 )D(C2 ) = 1 X ( -1) = -1 =D(C6) and from the character table is either B 1 or B2 . Hence, either D(av) = -D(ad) = 1 or D(av) = -D(ad) = -1 respectively. 15.6 Use the technique specified in the Comment of Example 15.4. £: All four orbitals are left unchanged, hence
x= 4
x= 1 C No orbitals are left unchanged , hence x= 0 5 No orbitals are left unchanged, hence x= 0 ad: Two orbitals are left unchanged, hence x= 2 C3 : One orbital is left unchanged, hence 2:
4:
The character set 4, 1, 0, 0, 2 spans A 1+ T 2 . Inspection of the character table of the group Td shows that s spans A 1 and that the three p orbitals on the C atom span T 2 • Hence, the s and p orbitals of the C atom may form molecular orbitals with the four H1s orbitals. In Td, the d orbitals of the central atom span E+T2 [Character table , final column], and so only the T 2 set (d,r, dY" d,x ) may contribute to molecular orbital formation with the H orbitals. 15.7 (a) In C3v symmetry the H1s orbitals span the same irreducible representations as in NH 3 , which is A 1+ A 1+E. There is an additional A 1 orbital because a fourth H atom lies on the C3 axis. In C3v. the d orbitals span A 1+ E + E [see the final column of the C3v character table]. Therefore , all five d orbitals may contribute to the bonding. (b) In Czv symmetry the H1s orbitals span the same irreducible representations as in H 20 , but one 'H 20 ' fragment is rotated .bY 90° with respect to the other. Therefore , whereas in H 20 the Hls orbitals span A 1+ Bz [H 1+Hz, H 1- Hz], in the distorted CH 4 molecule they span A 1+ 8 2 + A 1+ B 1 [H 1 + H 2 , H1- H 2 , H 3 + H4, H 3 - H4]. In C2v the d orbitals span 2A 1+ B 1+ B2 + A z [Czv character table] ; therefore , all except A 2(dxy) may participate in bonding. 15.8 (a) C2v . The functions x 2 , y 2 , and z z are invariant under all operations of the group, and so z(5z 2 - 3rz) transforms as z (A 1), y(5y 2 - 3r 2 ) as y(B 2) , x(5x 2 - 3r 2 ) as x (B 1), and likewise for z(x z- y z), y(x 2 - z 2) , and x (z z- y z). The function xyz transfers is 8 1x Bz x A 1= A 2 .
Therefore, in the group C2v. t~ 2A 1+ A 2 + 28 1+ 28 2 . (b) C3v . In C3v. z transforms as A ~> and hence so does z 3 . From the C3v character table , (x 2 - y 2 ,xy) is a basis"for E , and so (xyz, z(x 2 -y 2 )) is a basis
Symmetry: its determination and consequences
277
for A 1x E =E. The linear combinations y(S/ - 3r 2) + 5y(x 2 - z 2) o: y and x(5x 2 - 3r 2) + 5x(z 2 - y 2) o::x are a basis for E. Likewise, the two linear combinations orthogonal to these are another basis for E . Hence, in the group c )v , J-7 A1 + 3E. (c) Td. Make the inspired guess that the f orbitals are a basis of dimension 3 + 3 + 1, suggesting the decomposition T + T +A. Is the A representation A 1 or A 2? We see from the character table that the effect of S4 discriminates between AI and A2. 'Under s4 ' X-7 y ' Y-7- X ' Z-7- z' and so xyz-7 xyz. The character is x= 1, and so xyz spans A 1. Likewise, (x 3 , y ', z 3)-7 (y ", -x\ -z 3) and x= 0 + 0- 1 = -1. Hence, this trio spans T 1. Finally, {x(z 2- y 2), y(z 2- x 2), z(x 2 - /)}-7 {y(z 2- x 2 ), - x(z 2 - / ) , - z(/- z 2)}. resulting in
x= 1, indicating T 2. Therefore , in
Td , f-7 AI+ Tl + T 2.
(d) Oh. Anticipate an A+T+T decomposition as in the other cubic group. Since x, y, and z all have odd parity, all the irreducible representatives will be u. Under S4 , xyz-?xyz (as in (c)) , and so the representation is A 2" [see the character table]. Under S4 , (x 3 ,y 3 , z 3)-?(y 3 , -x 3 , -z 3 ), as before , and x= -1 , indicating T 1u. In the same way, the remaining three functions span T 2u. Hence, in Oh,j-?A2u+T1u+T2u· (The shapes of the orbitals are shown in Inorganic chemistly, D. F. Shriver, P. W. Atkins, and C. H . Langford, Oxford University Press and W . H . Freeman & Co (1990).] The f orbitals will cluster into sets according to their irreducible representations. Thus (a) f-7 A 1+ T 1+ T 2 in Td symmetry, and there is one nondegenerate orbital and two sets of triply degenerate orbitals. (b) f-7 A 2u + T 1u+ T 2u, and the pattern of splitting (but not the order of energies) is the same .
15.9 (a) In Td, the dipole moment transforms as T 2 [see the character table], and we require fr x T 2x f; to contain A 1 if the transition is to be allowed. (i) f(d,,) = E, r(dn,) = T2 , fr X T 2 X f; = T 2 X T 2 x E Then, since T 2 x E = (6, 0, -2, 0, 0), it follows that
T 2 X T 2 X E = (18, 0, 2, 0, 0) The number of times that A 1 appears in this set is determined using the recipe in Example 15.8 (and footnote 3 on p. 448 of the text), and is 1. Therefore the transition dxy-7 d,, is not forbidden. However, closer analysis (dealing with the representations rather than the characters) shows that the transition is not in fact allowed. (ii) For the transition dxy-7 J: yz we use
278
Symmetry: its determination and consequences
f;=T 2 , Tr=A 1 [Problem 15.8] fr XTzX r , = A, X T zX T z=TzX Tz=A , + E+T, +Tz Since the product contains A" the transition is allowed . (b) In Oh the electric dipole moment operator transforms as T 1u (i) f(d,2 ) = Eg, f(dxy ) = T 2g But g X u X g = u; therefore the product f; x r," x fr cannot contain A 1g and so the transition is forbidden. (ii) [;(dxy ) = T 2g, r r(fxyz ) = A 2u [Problem 15.8] fr X T 1u X f; = A2u X Ttu X T2g = A2u X (A zu+ Eu + T,u +Tzu) = A 1g + Eg+T2g + T,s The product contains A 1g, so the transition is allowed. 15.10 (a) xyz changes sign under the inversion operation (one of the symmetry elements of a cube) ; hence it does not span A 1g and its integral must be zero. (b) xyz spans A 1 in Td [Problem 15.8] and so its integral need not be zero. (c) xyz~ -xyz under z~ - z (the ah operation in D 6h) , and so its integral must be zero. 15.11 Refer to Fig. 15.4, and draw up the following table:
n,
1Cz
1[3
1[4
ns
1[6
1[1
1Cg
1[9
nw
X 10 0 0 2
n,
1Cz
1[3
1[4
ns
1[6
1[1
1Cg
1[9
nw
ns
1[6
1[1
1Cg
n,
1Cz
1[3
1[4
1[9
av
1[4
1[3
1Cz
n,
1Cg
1[1
1[6
ns
nw nw
a~
1Cg
1[1
1[6
ns
1[4
1[3
1Cz
n,
1[9
1Cw
E Cz
1[9
[xis obtained from the number of unchanged orbitals.] The character set CT ' v
n:,
Fig15.4
n;
rr,
~
"TT3 1T5"
a-:_
rr,._
(10, 0, 0, 2) decomposes into 3A 1 + 2A 2 + 2B 1 + 3B 2 • Now form symmetry adapted linear combinations as explained in Section 15.9:
Symmetry: its determination and consequences
279
n(AI)=.n1+.n4+ns+nx [from column I] .n(AI) = Jt2+n.1 +.n6+n1 [column 2] .n(A1)=.nq+Jt111
[column 9)
n(Az) = Jt1 + .7ts- Jt4- .7tx
[column 1)
n(Az) =n2 +.n6- .7tJ- Jt7 [column 2] n(81)=.nl-ns +n4-nx [column l] .n(81)=nz-n6+.nJ -.n7 [column 2] n(8z) = Jt1- Jts- Jt4+ .7tx [column 1] n(8z)=nz-.7t6-.nJ+.n7 (column 2) n(8z) = Jtq- Jt111
[column 9]
[The other columns yield the same orbitals.] 15.12 We proceed as in Problem 15.11, and begin by drawing up the following table:
N2s
N2p,
N2p v N2p ,
02p,
02pr 02p ,
0'2p, 0'2p ,,
E N2s
N2p,
N2p, N2p,
02p,
02p,. 02p,
0'2p,
0'2p , X
0'2pr 0'2p , 10
C2 N2s -N2p, -N2p)' N2p , -0'2p, - 0'2pv 0'2p, -02p, -02p,. 02p , av N2s N2px -N2p.v N2p, 0'2p, -0'2p, 0'2p, 02p... -02p. 02p , Uv·
N2s -N2p...
N2py N2p , -02p,
02py 02p , -0'2p,
0 2
0'2p, 0'2p , 4
The character set (10, 0, 2, 4) decomposes into 4A 1+ 28 1+ 38 2 + A 2• We then form symmetry adapted linear combin ations as described in Section 15.9: 1/J(AI)=N2s
[column 1]
1/J(AI) = N2p,
(column 4]
1/J(AI) = 02p, + 0 '2p , [column 7] 1/J(AI) = 02pv- 0'2p,
[column 9]
1fJ(8 1)=N2p,
(column 2]
1/J(BI) = 02p, + 0'2py [column 5] 1/J(B 2) = N2pv
[column 3]
1/J(82) = 02pv+ 0'2py [column 6]
Symmetry: its determination and consequences
280
._
1/J(B 2)=02pz-0'2pz 1fJ(A2) =02p,-0'2p,
[column 7] [columnS]
[The other columns yield the same combinations.] 15.13 We shall adapt the simpler subgroup C6v of the full D 611 point group . The six .n-orbitals span A 1 + B 1 + E 1 + E 2, and are 1
a1 =
v 6 (.nl + .n2 + .n3 + .n4,p- .ns + .n6)
b1 =
v 6 (.nl- .n2+ .n3- .n4 + .7ts- .n6)
el =
.) 1v
e =
1
(2.nl- .n2- .n3 + 2.n4- .ns- .n6) 12 1 (.n2- .n3 + .ns ~ .n6) 2
1~12
2
(2n, + n,- n,- :"',- n, + n,)
v2 (.n2 + .n3- .7ts- .n6)
The hamiltonian transforms as A 1; therefore all integrals of the form f 1/J 'H'f/J dr vanish unless 1/J' and 1/J belong to the same symmetry species. It follows that the secular determinant factorizes into four determinants:
f =~ f (.n~-.n2+
A I: Ha,a, = B1: Hb,b,
~
E1: H , ,(a )e,(a)
Hence: E2:
(.nl + ... + .n6)H(.nl + ... + .n6) dr =a+ 2/3
= a- /3,
··
·)H(.n~-n2 + · · ·) dr=a -
H , , (b)e,(b)
=a- /3,
,a-~-t: a-~-t:~ =0
Hence:
la+~-t: a+~-t:~ =0
0,
solves to t:=a -/3 (twice)
P _ it is necessary for (3 < 0. For a negative temperature to describe a three-level system, the populations are specifically inverted as T ~- T only if the separations E2 - E1 and E1 - Eo are equal.
20. Statistical thermodynamics: the machinery Exercises Cv=t(3+v~+2vt)R
20.1
[18)
with a mode active if T > eM· (a) v~ = 2, vv=O; hence Cv=t(3+2)R=tR
[Experimental: 3.4RJ
(b) v~ = 3, vv=O; hence Cv=t(3+3)R=3R
[Experimental: 3.2R]
(c)
v~=3,
vv=O; hence Cv=H3+3)R=3R
[Experimental: 8.8R)
Some of benzene's 30 vibrational modes must be at least partly active. 20.2
0.6950 TIK q=--x ( ') a Blcm 0.6950 X (TIK) 10.59
[Table 20.2]
[a= 1] = 0.06563(T/K)
(a) q = 0.06563 x 298 = 19.6 (b) q=0.06563x523=34.3 20.3 Look for the rotational subgroup of the molecule (the group of the molecule composed only of the identity and the rotational elements, and assess its order). (a) CO. Full group Coov; subgroup C 1; hence a= 1 (b) 0 2 • Full group D oo h; subgroup C2 ; hence a=2 (c) H 2S. Full group C2v; subgroup C2 ; hence a=2 (d) SiH4 . Full group Td; subgroup T; hence a= 12 (e) CHC1 3 . Full group C3v; subgroup C3 ; hence a = 3 1.0270 (T/K) 312 20.4 q = -a-(ABC/cm 3 ) 112
2X
[Table20.2]
1.0270 X 298 312 (27.878 X 14.509 X 9.287) 112
[a=2)=43.1
350
Statistical thermodynamics:the machinery
The high temperature approximation is valid if T > (}R, where hcB
(}R
=k
= 1.4388 em K x B
= 1.4388 K x 27 .878
[inside front cover]
[choose the 'worst case']
= 40K Therefore, the approximation is valid so long as Tis substantially greater than 40K. 1
(a) q = :Le-P£'=;:2:C21+1)2e-"cBPJ(J+I)
20.5
JMK
1.4388 K hc8{3 =
q = _1 12
X
T
2:
J
5.28
7.597 = TIK, a =
( 2 J + 1)2
12
-
e -7.5971(1+ 1)/(T/ K)
J
1 = 12 (1.0000 + 8.5526 + 21.4543 + 36.0863 + . .. )
=
1 X 439.27 = 36 .61 at 298 K 12
Similarly, at 500 K 1 q = 12 (1.0000 + 8. 7306 + 22.8218 + 40.8335 + ... )
=
1
12
X
950 .06=79.17
[Note that the results are still approximate because the symmetry number is a valid corrector only at high temperatures. To get exact values of q we should do a detailed analysis of the rotational states allowed by the Pauli principle.] 1.0270 (T/K) 312 (b) q = -a- x (B/cm-1)3'2 [Table 20.2, A= B = C] 1.0270 =-u
(T/K) 312 _ ) = 7.054 X 10- 3X (T/K) 312 5 28 312
X (
Statistical thermodynamics:the machinery
351
At 298 K, q = 7.054 X 10- 3 X 298112 = 36.3 At 500 K, q =7.054 X 10- 3 X 500312 =78.9
For 0
2,
11 = tm(O) =
-t X 16.00 u = 8.00 u
and a= 2; therefore
q= 8n 2 X 1.381 X 10 - 23 J K - I X 300 K X 8.00 X 1.6605 X 10 - 27 kg X (1.20 X 10 - 111 m) 2 2 X (6.626 X 10- 14 J s) 2 = 71.2 1.0270
20 .7
(T!K) 312
q = - a - ·(ABC/em
3) 112
[Table 20.2, a= I]
1.0270 X (T!K) 312 49 (3.1752 X 0.3951 X 0.3505) 112 = 1.5 (a) q = 1.549 X 298 312 = 7.97
X
312 X
(T/K)
101
(b) q = 1.549 X 373 312 = 1.12 X 104
20.8
Cv!R = f,f=
fJv) ( (T 1 _ e
e -JoviT ) Ovi T
[17]
We write x = flvl T ; then
This function is plotted in Fig. 20.1. For the acetylene (ethyne) calculation, use the expression above for each mode. We draw up the following table using kT/ he= 207 cm - 1 at 298 K and 348 em - I at 500 K , and fJviT=hcv/kT.
352
Statistical thermodynamics:the machinery 1·0
"\
0-8
0·6
Fig20. 1
'\
~ (}T,
(}R]
Hence S~ =
Urn- Um(O) { q~ } T +R lnNA+1
6
(Table20.1]
= i R + R{ln 3.02 x 108 + 1} = 23.03R =191.4JK- 1 mol - 1 The difference between the experimental and calculated values is negligible , indicating that the residual entropy is zero.
Statistical thermodynamics:the machinery
N A= 2.561
X
10 -"(TIK) 512 (M!g mol - 1) 312 [Table 20.2]
q T(I2)/ N A= 2.561
X
10 - 2 X 1000 512 + 253.8 312 = 3.27 X 109
qT(I)/ N A= 2.561
X
10- 2 X 1000512 + 126.9 312 = 1.16 X 109
20.18
qT/
357
TIK B/cm -
0.6950 a
qR(J, ) = - - X - -1
-
1000
-
= t X 0.6950 X 0.0373 = 9316 qv(I 2) =
1
1 v/cm - 1 [Table 20.2] _ e - "' a= 1.4388 TIK 1
= 1 - e - -' 14 .36/695 = 3. 77
qE(l) = 4 qE(l 2) = 1 Hence K =
(q e(I)!NS q 6 (1z)INA
P
=
e-D/ RT
[22]
(1.16 X 109 X 4) 2 e- 179 3 3.27 X 10 9 X 9316 X 3.77 = 3 ·2 X 10 -
20.19 We need to calculate q 6 ( 9 Br2)q 6 (8 1Br2) K = e -1!. £, / RT P q 6 ( 9Brx 1Br? The ratio of the translational partition functions is virtually 1 [because the masses nearly cancel; explicit calculation gives 0.999]. The same is true of the vibrational partition functions. Although the moments of inertia cancel in the rotational partition functions, the two homonuclear species each have a = 2 , so
qR(9Br2)qR(RIBr2) qRCYBrxiBr)2 = 0.25 The value of !)..£0 is also very small compared with RT, so
KP=0 .25
Problems 20.1
qv =
1
1 _e
hnVJ
[Table 20.2]
358
Statistical thermodynamics:the machinery
which rearranges to ii = -kTln
he
{_I } I I- -
q
Therefore, if q = l.OOI , -500 K
ii= 1.4388 em K ln
{
I } __ I_ = 2.4 x I03 cm-1
I
20.2
l.OOI
q = 2 + 2e-P•
x 2 e-..Therefore, Cvl R = (I + e x)2' X= {JE. We then draw up the following table:
TIK
50
298
500
(kT/hc)/cm- 1 X Cv!R Cvi(J K -I mol - 1)
34.8 3.46 0.354 2.94
207 0.585 0.079 0.654
348 0.348 0.029 0.244
Note that the double degeneracies do not affect the results because the two factors of 2 in q cancel when U is formed. In the range of temperatures specified, the electronic contribution to the heat capacity decreases with increasing temperature.
,
20.3
x- e-x Cvl R = (I+ x) 2
~
~
[Problem 20.2], x = 211 8 B{J
Therefore , if B=5.0T, 2 X 9.274 X 10- 24 J T- 1 X 5.0 T
X=-------=------:---1.38I X 10- 23J K -I X T
6.72
TIK
(a) T=50 K, x=O.I34, Cv=4.47 x 10- 3 R , implying that Cv= 3. 7 x 10- 2 J K -I mol- 1. Since the equipartition value is about 3R [v~ = 3, vt =O], the field brings about a change of about O.I per cent.
Statistical thermodynamics:the machinery
(b) T=298 K, x=2.26x 10- 2, Cv=l.3x 10- 4 R, implying that Cv= 1.1 mJ K - I mol - 1, a ch~,mge of about 4 x 10 - 3 per cent.
n1 =- 'X
1
f oo
a
-
=-
1 (2/kT) 112 e -m~!Jl/ 2/kT dm=- - oo a h2
f oo
e -x:! dx
-00
= ~ (2nf~!:\ 112 a h- } U- U(O) =-
Cv=(a~
aT} v
Sm =
Naq
N
q a(3 = 2(3 = i_-NkT
[or get by equipartition]
= i.-R = 4.2JK - 1mol - 1
Um- Um(O) T + R In q
[Table 20.1]
1 (2nlkT) 112 =iR+Rln- - 2
-
a
h
47 kgm 2X 1.381 X 10- 23 J K -I X 298)112 - (1.055x10 34 Jsf
~(2nX 5.341 X 10-
_ J.
- 2R+Rln3
= -! R+1.31R = l.81R, or 15JK - 1mol- 1 20.5
q = 1 + 5e-/lc
t: = E(J = 2) - E(J = 0) = 6hcB U- U(O) N
5t: e -/Jc 1 +5e-/J
' + a,c, b,b, + b,by+ b,b, bxcx + bycv + b,c,
c.,a, + c,aY+ c,a,
Cxb, + Cyby + c,b,
az
a·b
b·a
c·a
az
C,Cr + CyCy + C,C,
b2
a·c b·c
ab cosy
ab cosy b2
c·b
c-'
ac cos {3
be cos a
= a 2b 2c 2(1- cos 2 a- cos 2 {3- cos 2 y + 2 cos
ac cos {3 be cos a cz
a cos {3 cos y) 112
Hence,
V = abc(l- cos 2 a- cos 2 {3- cos 2 y + 2 cos a cos f3 cosy) 112 For a monoclinic cell a= y = 90°, and
V = abc( I - cos 2 {3) 112 =abc sin f3 For an orthorhombic cell , a=f3= y =90°, and
V=abc
388
Diffraction methods
21.14
J=
NV T ,where N is the number of atoms in each unit cell , V,, their c
individual volumes , and Vc the volume of the unit cell itself. Refer to Fig. 21.8.
(b) N=2 , V .. = t nR
3 ,
Vc= (
~r
[body diagonal of a unit cube is V 3]
2 x-j-nR 3
nV3
!= (4R/V3)3 =-8-=0.6802 (c) N=4 , V,, = t nR 3 , Vc=(2 V 2R) 3
4 X t nR 3 JT != (2 V 2R) 3 3V 2 = 0 ·7405
For each A atom use t ! A[each A atom shared by eight cells] but use fs for the central atom [since it contributes solely to the cell]. Fltkl = t / A{l + e 2mlt + e LTik + e LTil + e 2lri(lt+k) + e h i(lt +l) + e LTi(k+ l ) + e 2"i(lt +k+ ll}
+ f s e " i(lt+k+ l ) =fA+ ( -1 )" +k+'Js
[ h , k, I are all integers , e;" = -1]
Diffraction methods
389
(a) !A=f,Js=O; F,k,=f [no systematic absences]
(b) fs=+JA; F,k,=fA{l+t(-l)"+k +l} Therefore, when h + k +I is odd, Fhk 1 = fA{l- t} = tfA, and when h + k +I is even, F,k, = ffA· That is , there is an alternation of intensity (I rx F 2) according to whether h + k +I is odd or even . (c) fA= fs = f; F,+k +f= f{l + ( -l)" +k+l} = 0 if h + k +I+ is odd. Thus, all h + k +I odd lines are missing.
22. The electric and magnetic properties of molecules Exercises C= t:,C0 [Example 22.1] = 35.5 x 6.2 pF = 220 pF
22.1
2
f1.
3t:oPm
a+ 3kT= NA
22.2
2
p. Therefore, k
(
3
1 1) TT'
[6a]
3t:0 = NA (P-P')
[Pat T, P' at T']
and hence
(9t:okl NA)(P - P')
p.2 = - - - - - -
1
1
T
T'
9 X 8.854 X 10- 12 J - I C 2 m- 1 X 1.381 X 10- 23 J K -I X (70.62- 62.47) X 10- 6 m 3 mol- 1 6.022 X 1023 mol - l X
(
351~0 K - 423\ K)
= 3.064 X 10- 59 C 2 m2 and hence
f1.
= 5.5 X 10- 30 c m, or 1.7 D.
3t:oPm
Then
p.
2
a=---,:;-;- __, 3kT 3 X 8.854 X 10- 12 r
1
C 2 m- 1 X 70.62 X 10- 6 m 3 mol - 1
6.022 x 1023 mol- 1 3.064 X 10- 59 C 2 m2 3 X 1.381 X 10- 23 J K -I X 351.0 K = 1.01 X 10- 39 J - I C 2 m2
The electric and magnetic properties of molecules
a corresponding to a'=--= 9.1 x 10- -' 4 em 3 4nEu
E,-1 22.3 - E,+2
1.89 g cm - 3 x 27.18 cm 3 mol - 1 92.45 g mol
pPm M
[6b]- ---'----:-::---,-::--...,--;--1
--
=0.556 Hence,
E,
=
1 +2 X 0.556 _ 0. 1 556
4.8
22.4 A D 3h (trigonal planar) molecule is nonpolar; hence the second structure (with symmetry group C2v) is more likely . 22.5 Follow Example 22.4 Rml(cm 3 mol - 1) = 10 X 1.65 + 2 X 1.20 + 2 X 1.41 = 21.72
74.12gmol- 1 _ 1 3 Vm= 0.7l gcm _3 =103.6cm mol 5
_{V"' + 2R"'}"2_ {103.6+2 x 21.72}"2_
n,-
Vm-Rm
-
103 .6-21.72
-1. 34
The experimental value is 1.354.
22.6
3EoRm
M{n~-1}
a = - - with R"'=- - , NA p n~+2
Therefore, a = 3c0M {n: -1} pNA n~+2
and a'= 1.28 x 10- 23 cm 3.
[9]
391
392
The electric and magnetic properties of molecules
22.7 Refer to Fig. 22.2 of the text , and add moments vectorially [see diagram 2 of the text , Section 22.1]. ·
(a) p-xylene: the resultant is zero, so fl = Q (b) a-xylene: fl=0.4 D cos 30° +0.4 D cos 30° =0.7 D (c) m-xylene : 11-=0A D cos 30°+0.4 D cos 60° =0 .5 D The p-xylene molecule belongs to the group D2h , and so it is necessarily nonpolar. 22.8 fl = (fli+fl ~ +2fllfl2 Cost9) 1 12
[Section 22.1]
= (i.5 2+ 0.80 2+ 2 X 1.5 X 0. 80 X cos 109.5°) 112 D = 1.4D 22.9 fl* = a"&
[la] = 4.7te 0a '"&
[3]
= 4.n X 8. 854 X 10- 12J - I C 2 m2 X 1.48 X 10- 30 m3 X 1.0 X 105 V m- 1 = 1.6 x 10 - 35 Cm
[IJ=lCV]
which corresponds to 4.9flD.
22.10
V m
n, =
R:
v +2R { ;m _
}1'2
[10]
M 18.02gmol- 1 -18.0cm3 mol - 1 p 1.00 g cm - 3
Hence, -{18 .0+2 x 3.8}1'2 _ n,18.0-3.8 -1. 34
The electric and magnetic properties of molecules Discrepancies may be due to a more complicated local field correction than has been assumed, and vibrational contributions.
2
Hence,
E,
1 + 2x . pN A ( fl ) _ x with x = EoM a + kT 1 3 3
=
1.173 x 10~ g m - 3 x 6.022 x 1023 mol - 1 x= 3 X 8.854 X 10- 12 J - I C 2 m 2 X 112.6 g mol - 1 X { 4.n X
8.854 X 10- 12 ]
- I
C 2 m 2 X 1.23 X J0 - 29 m 3
(1.57 X 3.336 X 10-Jo C m? } + 3 X 1.381 X 10- 23 ] K - I X 298.15 K =0 .848 Therefore ,
E,
=
1 + 2 X 0.848 _ _ = 18 1 0 848
22.12 (a) Rm(CaC1 2 ) = 1.19 + 2 X 9.30 cm 3 mol - 1= 19.79 cm 3 mol - 1 Vm(CaC1 2)=
111.0 g mol- 1 _ =51.6cm 3 mol - 1 2 .1 5 gem 3
_ {Vm + 2Rm}l /2_ {51.6+ 2 X 19.79}1/2_ Vm-Rm 51.6-19.79 -1. 69
n,-
(b) Rm(NaCl) =0.46+ 9.30 cm 3 mol - 1=9.76cm 3 mol - 1 Vm(NaCl)=
58.4 g mol - 1 =27 .0cm 3 mol - 1, . gem 3 216
which gives [as above] n, = 1.64
393
394
The electric and magnetic properties of molecules
which gives n, = 1.23 22.13
{(0)12 -;. - (a)6 -;. } 12 6a 6} 24t:{ (a)l3 12a
F = - dV dr with V = 4t:
Therefore, F= 4t: { ~ -7
=---;;-
2 -;.
- (a)7} -;.
The force is zero when 116 a)l3 = (a) -;. 7, or r = 2 a 2 ( -;.
22.14
m = gc{S(S+ 1)} 112,U 8
[Section 22.11 , ge=2]
Therefore, since m = 3.8l.u 8 S(S + 1) = t x 3.81 2 = 3.63 , implying that S = 1.47 Since S = f, there must be three unpaired spins. 22.15
xM xm=xVm [24]= p -7.2 X 10- 7X 78.11 g mol- 1 0.879 gem 3
22.16
_ Xm- 6.3001
X
-6.4 X 10- 5 cm 3 mol - 1
_1 S(S+1) [Example 22.7] TIK em 3 mol
Since Cu(II) is a d 9 species , it has one unpaired spin , and so S = s = t . Therefore, Xm =
6.3001 xt x t cm 3 mol - 1 = --'+0 cm 3 mol - 1 298 --"-'"--"016 "'--"-'"'-'-'--'-=~
Problems 22.1 The positive (H) end of the dipole will lie closer to the (negative) anion . The potential arising from a point dipole is
-.u
¢ = -2 [11 , V=q 2¢] 4nt:ur
The electric and magnetic properties of molecules and since the electric field is the negative gradient of the potential , · hence C = KR = cA. mR
Therefore, from the data C = 0.0200 mol dm - 3x 138.3 S cm 2 mol - 1X 74.58 Q = 206.3 cm 2dm - 3= 0.2063 cm - 1 25 1 ·
K(CH3COOH)
R(KCI)
33 .21 Q
K(KCI)
R(CH 3COOH)
300.0 Q
Therefore, 33.21 1C(CH3COOH) = 300.0 X 1.1639 X 10 - 2sem -I = 1.288 X 10 - 3sem - I But this value includes a contribution of 7.6 X 10- 4 sem - I from the water; hence the conductivity of the acetic acid itself is (1.288 - 0.76) x 10- 3S em - 1= 5.3 X 10- 4 s cm - 1. Therefore , 5.3 X 10- 4 S cm - 1 Am = 0.100moldm 3 5.3Scm2mol - l
Therefore, we draw up the following table: elM
0.0005
0.001
0.005
0.010
0.020
0.050
(c/M)l/2 R/Q Am/(S cm 2 mol - 1)
0.0224 3314 124.5
0.032 1668 123.7
0.071 342.1 120.6
0.100 174.1 11 8.5
0.141 89.08 115 .8
0.224 37.14 111 .1
The value of Am are plotted against c 112 in Fig. 25.1. The limiting value is A~ = 126 cm 2 mol - 1. The slope is -76.5; hence ':1{ =76.5 S cm 2 mol - 1M- 112 (a) Am = (50.1 + 76.8) S cm 2 mol - 1-76.5 S cm 2 mol - 1X (0.010) 112 = 119.2 S cm 2 mol - 1 (b) 'K =cAm= 0.010 mol dm - 3X 119.2 S cm 2 mol - 1 = 1.192 S cm 2dm - 3 = 1.192 mS cm - 1 C 0.2063 cm - 1 (c) R = ;=1.192 x 10-3 Scm - 1 173.1 Q
Molecules in motion 126
1\.
124
0
"'E 122
Fig25.1
'
~ ~
§
'•
~
E12
~
118
11 6
1(
'
-
'\
0
003
006 009 .V(c/M)
1"'-
012
~. 075
1(
25.3 c = - = i\~
i\m
[c small, conductivity of water allowed for in the data.]
1.887 X 10- 6 S cm - 1 c= 138 .3 S cm 2 mol_ 1 [Exercise 25.7]
= 1.36 X 10-Kmol em - 3 = 1.36 X 10- 5 M K ,P
= (1.36 X 10- 5) 2 = 1.86 X 10- 10
We can correct for activities using Y±= w - AVc= 0.996; hence 1.86 X 10- 10 = 1.85 X 10- IO
K,P=y~ X
25.4
i\~(NaCH,C0 2 )
= A.(Na+) + A.(CH 3C02) = 91.0 S cm 2 mol - 1
i\~(HCl). =A.(H + )
A~(NaCI) i\~(CH,COOH)
+A.(Cl-) =425.0 S cm 2 mol- 1
=A.(Na +) +A.(cn = 128.1 S cm 2 mol- 1 =A.(H+) +A.(CH 3C02) = i\~(HCl) + i\~(NaCH 3 C0.1 )- i\~(NaCl) = (425 .0 + 91.0 -128.1) S cm 2 mol - 1 = 387.9 S cm 2 mol - 1
Am
'K
C
1\~
c~
cRi\~
a=-=-=--
0.2063 cm - 1 -0.020 X 10- 3 mol cm- 3 X 888 Q X 387.9 S cm 2 mol - 1 =0.030
459
460
Molecules in motion
with A~ = A.(H + )+A.(CH 3 C02) = 390 . 5Scm 2 mol - 1 • We draw up the following table using Am=Kic = CicR and C = 0.2063cm - 1:
Am/ (S cm 2 mol- 1) 105cAm/(S cm- 1) 100/ (Am/S cm 2 mol- 1)
0.49
0.99
1.98
15.81
63 .23
252.9
68 .5 3.36 1.46
49.5 4.90 2.02
35.6 7.05 2.81
13.0 20.6 7.69
6.56 41.5 15.2
3.22 81.4 31.1
We now plot 100/ Am against 105 cAm (Fig. 25.2). A least-squares fit of the data
40 Fig 25.2
.::-1
~
'E ll
)
$ (§ a
30
/ /v
20
10
~
0
~
,.,
v
v
./fti'
0
100
gives an intercept at 0.352 and a slope of 0.01559. Since we are actually plotting 100 S cm 2 mol- 1 _ 100 S cm2 mol - 1 102 S cm 2 mol - 1 (105cAm) S cm - 1 Am Nm + K a (A om) 2 S em - l x 105 the slope of the plot is 102 S 2 em mol - l Slope = 105 Ka(A~r
10- 3 S2 em mol-l
Ka (A~f
= 0.352
Molecules in motion Hence, 10- 3S2 em mol - l K. = 0.352 x (390.5 S cm 2 mol - 1?
= 1.86 X 10-s mol em - 3= 1.86 X 10 -s M Therefore, pK. = -lg 1.86 x 10- 5 =4.73 25.6 s = uE
lOY with E = - =10Vcm- 1 1. 00 em
[7]
10- 4 cm 2 S- 1 X 10 V cm- 1 = 4.0 X 10- 3em s- 1 s(Na+) = 5.19 X 10- 4 cm 2 S- 1 X 10 V cm - 1 = 5.2 X 10- 3em s- 1 s(Li+) = 4.01
X
s(K+) = 7.62 X 10- 4 cm 2 s- 1 X 10 V cm - 1 = 7.6 X 10- 3em s- 1 d t=- with d=l.Ocm: s
t(Li +) =
l.Ocm 4.0 x 10 3ems
1
-250s ---
t(Na+) = 190 s, t(K +) = 130 s likewise. For the distance moved during a half-cycle , write d=
f
lnv s dt = fl nv uE dt = u£ fl nv sin(2.n:vt) dt 0
0
0
0
uE
uxlOVcm -
.n:v
.n:X 10 . X 103S
1
_1 =3.18x10- 3 uVscm- 1
That is, diem= 3.18 X 10- 3 X (u/cm 2 v
- I
s- 1)
Hence, d(Li +) = 3.18 X 10- 3 X 4.0 X 10- 4 em= 1.3 X 10- 6 em d(Na+) = 1.7 X 10- 6 em , d(K +) = 2.4 X 10- 6 em These correspond to about 43, 55, and 81 solvent molecule diameters respectively. 25.7
t(H+) =
u(H +) u(H +)+u(cn 3.623 3.623 + 0.791
[llb]
°· 82
461
462
Molecules in motion
When NaCI is added , c(H +)u(H +) t(H +) - ----,----,--,----,---,---:-:--:-:---::-::----:----:-=:-:-----:-::~ - c(H +)u(H+ + c(Na +)u(Na +) + c(CI )u(CI )
( 1OJ
1.0 X 10- 3 X 3.623 1.0 X 10
3
X 3.623 + 1.0 X 0.519 + 1.001 X 0.791
=0.0028 25.8 t =
zcVF zcAFx !!it (Section 25 .2] = J""M 1
21 mol m- 3 x n x (2.073 x 10- 3 m) 2 x 9.6485 x 104 C mol - 1 x 18.2 X 10- 3 A XM
=
x/mm) 1.50 ( M/s
Then we draw up the following table: M ls
200
400
600
800
1000
64 0.48 0.52
128 0.48 0.52
192 0.48 0.52
254 0.48 0.52
318 0.48 0.52
Hence, we conclude that
t+ =
0.48 and
t_
= 0.52.
For the mobility of K + we use t+ =
. 1. + N
(12]
m
to obtain t + i\~
0.48 x 149.9 S cm 2 mol- 1 u+ = F = 9.6485 X 104 Cmol - 1 =7.5x 10- 4 cm 2 s- l y-l ..1.+ = t+ i\~
(12]
= 0.48 x 149.9Scm2 mol - 1
= 72 S cm 2 mol- 1
Molecules in motion zcA F
x
t =-- x + I !:it
25.9
463
[Problem 25.8]
Since the density of the solution is about 0.682 g cm- 3 , the concentration c is related to the molality m by 1 elM= _ x m/(mol kg- 1) = 14.7m/ (mol kg- 1) 0 682
(2.073 X 10- 3 m) 2= 1.350 X 10- 5 m2 czAF 1.350 X 10- 5 m 2 X 9.6485 X 104 C mol - 1 I !:it = 5.000 X 10- 3 A X 2500 s Xc
A = nr 2 = n
X
=0.1042 m 2 mol - 1 x c =0.1042/mm x elM =0.153/mm x m/(mol kg- 1) and so t+
= 0.153 x (x/mm) x m/(mol kg - 1)
(a)
t+
= 0.153 X 286.9 X 0.013 65 = 0.60
(b)
t+
= 0.153 X 92.03 X 0.042 55= 0.60
Therefore, t(H+) = 0.60 and the mobility is not as abnormal as in water where t(H +) = 0.82. C
25.10
0.2063 cm - 1
R=-=
o-s s em _ =3.75MQ
x 55 . x1
A~ =A.(H +) +A.(OH - ) =
1
349.8 + 197.6 S cm 2mol - 1 = 547.4 S cm 2 mol - 1
Am x 5.5 x 10-x Scm- 1 and so a = - = - = - - - - - - = - - - - --=---:A~ cA~ 55 .5 mol dm - 3 x 547.4 S cm 2 mol - 1 = 1.8 x 10- 9 We then write Kw = a(H +)a(OH- ) = c(H+)c(OH - )/M 2
= a 2c(H20nM 2 = a 2 X (55 .5? = (1.8 X 10- 9) 2 X (55.5) 2 = 1.0 X 10 - 14 pKw = -Jg Kw= 14.0 pH= -lg a(H +) = -lg K ~ 2 = - -!Ig Kw=7.0
464
Molecules in motion
'!J' = -
25 .11
RT -
c
de dx
X -
(15)
de (0.05 - O.lO)M dx = O.lOm = -0.50Mm-l RT= 2.48 X 103 J mol- 1 =2.48 X 103 N m mol - 1
- 2.48 kN m mol - 1 1 O.lOM x (-0.50Mm - )
(a) '!F =
= 12 kN mol - 1, 2.1 X (b) '!F =
w--20 N molecule- 1
- 2.48 kN m mol - 1 x (-0.50Mm - 1) _ M 0 075
= 17 kN mol - l' 2.8 X 10- 20 N molecule- 1 (c) '!F =
- 2.48 kN m mol - 1 x (- 0.50Mm - 1) _ M 0 05
= 25 kN mol-l ' 4.1 25.12
D s = kT'!F
X
w-20 N molecule - 1
[Section 25.4]
1.381 X 10- 23 J K -l X 298.15 K = 1.26 x 10" m s- 1('!1'/N) ['!F is the force per molecule.]
(a) s = l.26x10"ms - 1 x 2.1 x l0- 20 = 2.7nms- 1 (b)
S=
1.26 X 10" m S- 1 X 2.8 X 10- 20 = 3.5 nm s- 1
(c)
S=
1.26 X 10" m s- 1 X 4.1
X
10- 20 = 5.2 nm S- 1
We could monitor the concentration by refractive index, optical rotation , infrared spectroscopy . The initial flux through a region is the same at every point because dc/clx is a constant except at the left boundary and at the right , open side (Fig. 25 .3a). The initial change is then as shown in Fig. 25.3b. This initial distortion is then magnified as time goes on , and as dc/clx is no longer the same everywhere, dcldt changes everywhere (Fig. 25.3c) . After a long time, the concentration becomes virtually uniform and sinks toward 0.075 M (Fig. 25.3d).
Molecules in motion
25 .13
uRT
D=-
[16]
zF
ze and a=-6nryu
[Example 25.7]
8.3141 K - I mol- 1 X 298.15 K X u D = - - - - - - . . . , . . .4 - - --...,....---9.6485 x 10 C mol - 1 =
so
2.569 X 10- 2 V
D/(cm 2 s- 1) =
X u
2.569 X 10- 2 X u/(cm 2 s- 1 v-I)
1.602 x w- 19 c a= - - - - - - , 3- --,--..,......1 1 6n X 1.00 X 10- kg m- s-
8.50 X
w
- IH
c kg -
I
X u
ms
u
8.50 X 10-IRy u
- l
m3 S- l
[1J = lCV, lJ =l kgm 2 s- 2]
and so a/m = 8.50 X l0- 14/(u/cm 2 s- 1 V and therefore a/pm = 8.50 X l0 - 2/(u/cm 2 s- 1 V
- I)
- I)
We can now draw up the following table using data from Table 25 .2:
104 u/(cm 2 s- 1 V - 1) 10 5 D/cm 2 a/pm
4.01 1.03 212
Na+
K+
5.19 1.33 164
7.62 1. 96 112
7.92 2.04 107
465
Molecules in motion
466
The ionic radii themselves (i .e. , their crystallographic radii) are Li +
Na+
K+
Rb +
59
102
138
149
and it would seem that K + and Rb + have effective hydrodynamic radii that are smaller than their ionic radii . The effective hydrodynamic and ionic volumes of Li + and Na + are
4n
--gna
3
and
4n
3 nr ~
respectively , and so the
volumes occupied by hydrating water molecules are
4n
(a) Li +: ~V=3 X (212 3 -59 3)
X
10- 36 m3 = 3.9 X 10- 29 m 3
The volume occupied by a single H 20 molecule is approximately (4n/3) X (150 pm) 3 = 1.4 X 10- 29 m 3 . Therefore, Li + has about three firmly attached H 2 0 molecules whereas Na + has only one (according to this analysis). 25.14 If diffusion is an activated process , we expect D rx e -E,I RT
Therefore , if the diffusion constant is D at T and D' at T' ,
R In (D'/D) E = - _ __:__ ___:_ a
(;,-~) 8.314 J K - l mol - 1 x In 2.89) _ ( 2 05 --------~-.!..._ = 9 3 kJ mol - 1 1 1 . ----298 K 273 K
That is , the activation energy for diffusion is 9.3 kJ mol - 1• 25.15
c = A(nDt) 112
and we know that n 0 =
[24]
342
10 g g mol _ 1 x 6.022 x 1023 mol- 1 = 1.76 x 1022
Molecules in motion A= .nR 2 = 19.6 em\ D = 5.21 x 10- 6 cm 2 s- 1 [Table 25.4]
A(nDt) 112 = 19.6 cm 2 x (n x 5.21 x 10- 6 cm 2 s- 1 x t) 112
=7.93 x 10- 2 cm 3 x (tl s) 112 x2
4Dt
25 cm 2 4x5.2lx10- 6 cm 2 s- 1 xt
1.20 X 106 (tis)
Therefore ,
e -1.20x 10°/(1/s)}
= 2.22 x 1023 em - 3 x
_
{e
(tis) 112
{
- 1.20 x wot( t/s)}
-369M X
(tis) 112
(a) t=10s, e - 1.2 x I(}'
c=369MX~=0
(b) t=1yr=3.16x10 7 s, e -o.o3R
25.16
(x 2) = 2Dt
[26b]
'
kT D= - 6nary
[Example 25.7]
kT kTt Hence, r y = D-a = 3na (x-') n 6
1.381 X 10- 23 1 K -I X 298.15 K X t 3n x 2.12 x 10
7
m x (x 2) t
= 2.06 X
10- 15 1 m- 1 X (x 2)
467
Molecules in motion
468
and therefore
We draw up the following table:
ti s
30
60
90
120
10s(x2)/cm2 1037]/(kg m- 1 s- 1)
88.2 0.701
113.5
128 1.45
144
1.09
1.72
Hence , the mean value is 1.2 X 10- 3 kgm - 1 s- 1• 25.17 The current /i carried by an ion j is proportional to its concentration ci, mobility ui , and charge number lzJ Therefore, /i = Aciui lzil where A is a constant. The total current passing through a solution is
The transport number of the ion j is therefore
t = ~ = Aciuilzd = ciuilzil '
I A
L ciui lzd L ciui lzil
If there are two cations in the mixture ,
t'
c'u 'z'
c'u '
-= - - = - i f z'= z" t" c"u"z" c"u" 25.18 Consider the consequence of the passage of 1 mol of electrons through the cell Ag IAgCl lHCl( c1) IHCl( c2) IAgCII Ag
Molecules in motion
469
Right compartment: 1 mol Cl - are formed, butt_ mol migrate out across the junction, giving a net change of (1- L) mol= t + mol. Left compartment: 1 mol Cl- is lost (by formation of solid AgCI), but t_ mol flows in across the junction , giving a net change of ( -1 + t+) mol= - t + mol. The reaction Gibbs function is therefore
Therefore, since !'!..G= -FE,
For the same cell without transfer, the Nernst equation gives
-RT a 2 E=--lnF a1 Therefore, E, = t+E For electrodes reversible with respect to the cations, 1 mol M + is generated butt+ mol migrates out , giving a net change of (1- t + ) mol= t_ mol. By the same argument,
25.19
ac a 2c at= D ax 2
no
e -x2t4DI
[21] with c= A(.nDt)"2 [24]
' then When c =a- e- bx-tl t"2
,
& = _ I (aft 312 ) e-bx 2il +a X_ ~2 ' e-bx-11
-
at
2
t" 2
c bx 2 =--+-c 2 2t
t
t2
Molecules in motion
470
ac
-= -
ax
t
a112 (-2bx) X
__
'
e - bx-tr
t
2
a c = - (2b) ax2 t (atl / 2) =_
=D
e -bx 2t r
+ ( tla/2) (2bx) - t- 2 e -bx2t r
2 2bt c+ (2bx) 2c = _ (-1)c+ (bx \c t 2Dt Dt 2 }
ac at as required .
Initially the material is concentrated at x = 0. Note that c = 0 for x > 0 when t=O on account of the very strong exponential factor [e-bx 21 '~0 more strongly than 1/t 1 ' 2 ~ oo ]. When x .= 0, e -x 2140' = 1. We confirm the correct behavior by noting that (x) = 0 and (x 2) = 0 at t = 0 [26], and so all the material must be at x=Oatt = O. 25.20
N! P(x) = t(N+s)!t(N-s)! 2N [A2]
, s=xld
N! P( 6d)= t(N+6)!t(N-6)!2N (a) N=4 , P(6d)=Q
[m!=oo form < O]
6! 1 1 (b) N = 6, P(6d) = !0! 26= 2_6 = = 0.016 64 6 12! 12 X 11 X 10 (c) N=12, P(6d) = , ,212 = x x 212 = 0.054 93 3 2 [NB 0!=1] 25.21 Draw up the following table based on eqns A2 and A3: N
4
6
8
10
20
P(6A.hxact P( 6A.) Approx.
0 0.004
0.016 0.162
0.313 0.0297
0.0439 0.0417
0.0739 0.0725
Molecules in motion
N
30
40
60
100
P(6A.)Exact P( 6,1,) Approx
0.0806 0.0799
0.0807 0.0804
0.0763 0.0763
0.0666 0.0666
471
The points are plotted in Fig. 25.4. The discrepancy is less than 0.1 per cent
010
Fig25.4
E act p
I
r
If-.
~sy
0
when N > 60.
0
20
r!nPt
fie
40
- -
60 n 80
100
26. The rates of chemical reactions Exercises 26.1
1 d[J] dt
(1],
u=--
SO
d[J] -dt =V1V
VJ
The reaction has the form 0=3C+D-A-2B Rate of formation of C = 3v = 3.0 M s- 1 Rate of formation of D = v = 1.0 M s- 1 Rate of consumption of~= v = 1.0 M s- 1 Rate of consumption of B =2v=2.0 M s- 1 26.2
1 d[J] v=-V1 dt
[1]
For the reaction 2A + B__,.. 2C +3D , 1
v = tx 1.0 M s- =0.50 M s-
Vc
= + 2; hence
1
Rate of formation of D = 3v = 1.5 M s- 1 Rate of consumption of A= 2v = 1.0 M s- 1 Rate of consumption of B = v = 0.50 M s- 1 26.3 The rate is expressed in
1 M s- ;
therefore
1
MS- =(k) X M X M
requires the units of k to be
M-
1
s -I
(a) Rate of formation of A= v = k[AJ[BJ (b) Rate of consumption of C = 3v = 3k[AJ[BJ 26.4
d~~J = k[AJ[BJ[CJ
1 d[JJ . v = - - w1th v1 = vc = 2 V1 dt
The rates of chemical reactions 1 d[C] Therefore v=2dt=!k[A1[B1[C]
The units of k must satisfy M S-
1
= (k1X
M X M X M
= (k 1M 3
which requires k to have the units 26.5
2N 2 0 5 ~ 4N0 2 + 0 2 ,
M-
2
s- 1
v = k[N 20 51
Therefore, rate of consumption of N20 5 = 2v = 2k[N 20 51 d[Nz0 s1 dt
[Nz0 s1 = [NzOs1o e -Zkt which implies that 1 [NzOs1o t = 2k In [N 20 51
and therefore that
Since the partial pressure of N20 5 is proportional to its concentration , p(NzOs) = Po(NzOs) e- Zkt (a) p(N 20
5)
= 500 Torr X e- 2.76 x 111 - 'x 102 = 499 Torr
(b) p(N 20 5) = 500 Torr X e -z. 76 x 111 - 'x 6000 = 424 Torr 26.6 (a) For a second-order reaction, denoting the units of k by [k1: M
s- 1 = [k1 x
2 M ,
implying that [k1 =
M-
1
s- 1
For a third-order reaction , M
s- 1 = [k1 x
M\
implying that [k1 =
M
2
s- 1
(b) For a second-order reaction atm s- 1 = [k1 x atm 2 , implying that [k1 = atm - 1 s- 1 For a third-order reaction atm s- 1 = [k1 x atm 3 , implying that [k1 = atm- 2 s- 1
473
474
The rates of chemical reactions
In 2 k= 28 .1 y =0.0247 y- l Hence, with [Sr] replaced by its mass , m = 1.00 flg x e- 0.0247(tly) (a) m = 1.00 flg x e- oo247xIB = 0.64 flg (b) m = 1.00 flg X e-0 0247 x7o = 0.18 flg
26.9
1 [A ]0 ([B] 0 - x) kt= [B]o- [A]o In ([A]o- x)[B]o [7b]
which rearranges to [A]o[B ]o{ ek =-----:,.----:, kr+k[A] If k[A] ~ kr, 4> = 2(k/kr)[A], and the efficiency is determined by the availability of A molecules in the vicinity of A *. If k[A] ;;> kr , 4> = 2, and the rate is determined by the excitation step, because there is now plenty of A to react to form A 2 .
27.16 Write the differential equations for [X] and [Y]: (i)
d~] = k.(A][X] -
(ii)
dt = kb[X][Y] -
d[Y]
kb[X][Y]
kc[Y]
and express them as finite-difference equations: (i) X(ti +I)= X(ti) + k. [A ]X(ti) !:J.t- kbX(ti)Y(ti) !:J.t (ii) Y(ti +l)=Y(ti) - kcY(ti) !:J.t+kbX(ti) !:J.t and iterate for different values of [A], X(O) , and Y(O). For the steady state, (i)
d~~] = k.[ A] [X] -
kb[X] [Y] = 0
The kinetics of complex reactions (ii)
d[YJ
dt = kb[XJ[Y]- kc[YJ = 0
which solve to
Hence, k k [A] [X] = _:, [Y]= - · kb kb
27.17
(i)
d~~J = k.(AJ + kb[Xj2[Y]- kc[BJ[X]- kd[XJ
(ii)
d~~J = - kb[Xj2[Y] + kc(BJ[X]
Express these equations as finite-difference equations: (i) X(t;+I) = X(t;) + {k.[A] + kbX 2(t;)Y(t;) - kc[B]X(t;)- kd[X]} M
(ii) Y(t;+ I)= Y(t;) + {kc[B]X(t;)- kbX 2(l;)Y(t;)} !:J.t and iterate. See Figs . 27.7 to 27.9 of the text. 27.18
(i)
d~~J = k.[AJ[Y]- kb(XJ[Y] + kc[BJ[X]- 2kd[X]
(ii)
dt= -
d(Y]
2
k.[AJ[Y]- kb(X](Y] + ke[Z]
Express these differential equations as finite-difference equations:
(i) X(t;+ I) = X(t;) + {k.(A]Y(t;) - kbX(l;)Y(t;) + kc[B]X(l;)- 2kdX 2 (t;)} M (ii) Y(l; +I)= Y(t;) + {ke[Z]- k.[A ]Y(t;)- kbX(t;)Y(t;)} !:J.t
Solve these equations by iteration. More sophisticated procedures are available programmed in the Library of Physical Chemistry Software that is available to accompany the text.
515
28. Molecular reaction dynamics Exercises 21'2acp
28. 1 z=~
[eqn 9 of Chapter 24]
_ (8kT) 112 and c= [eqn 7b of Chapter 24] nm 4ap . , Therefore, z = (nmkT) 112 w1th a=nd 2 = 4nR Similarly,
Z AA
4kT) =a ( nm
112('!\ 2 V} [eqn lOb of Chapter 24] [NIV=plkT]
We express these equations in the form 16nR 2 x 1.0133 x 105 Pa {n X (Mig mol - 1) X 1.6605 X 10- 27 kg X 1.381 X 10 - 23 J K -
z=~~------~--------~~----------=---~------~
1.10 X 10311 m- 2 s- 1 X R 2 (Mig mol 1) 112 ZA A
=
1.10 X 106 X (Ripm) 2 s- 1 (Mig mol - 1) 112
, ( 4 x 1.381 x 10- 23 1 K -I x 298.15 K) 4nR , n x (Migmo l- 1)xl.6605 x 10-- 7 kg
112
5
X
=
)2 1.0133 X 10 Pa ( 1.381 X 10 - 23 J K -I X 298.15 K
1.35 X 1055 m -J s- 1 X R 2 (Mig mol - 1) 112
1.35 X 1031 (Ripm) 2 m- 3 s- 1 (Mi g mol - 1) 112
(a) NH1 ; R=190pm, M = l7gmol - 1 1.10 X 106 X 1902 S- 1 z= , =9 .6 x l0 9 s- 1 171 2
I X
298.15 K} 112
Molecular reaction dynamics (b) CO; R=180pm, M=28gmol - 1 1.10 X 106 X 1802 Z = S- I = 6 7 X 109 S- I 281/2 .:::..:. :.·--'-'-o..:::_.::..._
For the percentage increase at constant volume, use 1 dz 1 de 1 1 dZ 1 ~ dT = e dT= 2T' Z dT= 2T Therefore,
oz oT oz oT - = - and-=z 2T Z 2T and since oTIT= 10 K/298 K = 0.034, both z and Z increase by about 1.7 per cent. 28.2 In each case use f = e- E,.tRT [Section 28.1]: E. 10 x 103 J mol- 1 401 (a) RT=8.314JK - 1mol - 1x300K 4 ·01 'f=e - = 0 ·018 E. 10 x 10 3 J mol- 1 RT- 8.314 J K -I mol- 1x 1000 K
(b)
100 x 103 J mol- 1 RT = 8.314J K 1mol 1x 300 K E.
100x103 Jmol- 1 RT= 8.314 J K 1mol 1x 1000 K E.
1.20, f= e-uo = 0.30 40.1, f= e - 40.I = 3.9 X 10-IR 12.0, f= e- 120 = 6.0 X 10- 6
28.3 The percentage increase is
of 1 ( df) 100E. 100x - =100x- xoT=--oT f f dT RT 2 (a) Ea=10kJmol- 1,oT=10K 100
of 100x10x103 Jmol- 1x10K y= 8.314JK- 1moi- 1XT 2 1.20 x 106 - (T/K) 2
{13 per cent at 300 K 1.2 per cent at 1000 K
517
518
Molecular reaction dynamics
(b) Ea = 100 kJ mol- 1, oT= 10 K
of 1.20 X 107 100- = ----,--.,..,..! (TIK?
{
130 per cent at 300 K 12 per cent at 1000 K 8X8.314JK- 1 moi- 1 X298K
8RT
28.4
kd =
"""3-i
[7] =
3TJ
6.61 x 103J mol- 1 TJ
6.61 x 103 m3 mol - 1 s- 1
6.61 x 103 kg m 2 s- 2 mol- 1 (TJ ikgm
1
s 1) xkgm
1
1
s
(TJikgm- 1 s- 1)
6.61 X 106 M- 1 S- 1 6.61 X 109 M- 1 S- 1 1
(TJ f kg m
S
1 )
(YJ/cP)
(a) Water, TJ = 1.00 cP, kd = 6.61 x 109 M- 1 s- 1 (b) Pentane,rJ = 0.22cP,kd =
6.61 X 109 _ M- 1 s- 1 =3.0x10 10 M- 1 s- 1 0 22
(c) Decylbenzene ,rJ =3.36cP,kd=
6.61 X 109 _ M- 1 s- 1 = 2.0x109 M- 1 s- 1 3 36
2
28.5
k2 = ac:fJ-~JI NAe-EiRT
(1]
= (3.72 X 10 12 M-l min -I) X e -E.tRT 3.72x10 12 - - - - X 10-3m3 mol-l S-1 60
Xe-E.fRT
Therefore, we must evaluate 9
3. 72 X 10 3 1-1 -I a= 60 m mo s
X ( TC!J- ) 1/2
8kT
1 NA
16 X 100 with fJ- = a=
16
+ 100 u = 13.79 u
3.72x109 m3 mol- 1 s- 1 ( nx13.79xl.6605x10- 27 kg )112 x 60 X 6.022 X 1023 mol- 1 8 X 1.381 X 10- 23 J K -I X 298 K
= 1.52 X 10- 19 m 2 ,
Of
0.152 nm 2
Molecular reaction dynamics
519
a* P =-
28.6
a
For the mean collision cross section, write a A= ndi, as= nd~, and a = nd 2 , with d = f(dA + do): a = -!-n(dA + ds) 2 = tn(di + d~ + 2dAds) =t(aA + as+ 2a~ 2 a~ 2 ) =-!-{0 .95 + 0.65 + 2 X (0 .95 X 0.65yt 2} nm2 = 1.03 nm 2 Therefore, P
9.2 X 10- 22 m 2 = 1.03 X 10-IS m 2
28.7
dt = k2[A][B]
8.9 X 10- 4
d[P]
Therefore, the initial rate is
28.8 k
z
!::J.Ht = E.- RT [20b] kT RT x [19] h pe
= Be 6511 Re-tJ.HitRT B= -
,
Molecular reaction dynamics
520
Therefore, A= e B e 65 11 R, implying that Therefore, since
~H * =
~5 1 = R (In~- 1)
8681 K X R ,
E. = ~H * + RT= (8681 K + 303 K)R = 8984 K X 8.314 J K - I mol - 1=74.7 kJ mol - 1 1.381 X 10- 23 J K -I X 303 K 8.3141 K -I mol - 1X 303 K B= x-----= ----6.626 X 10- 34J s 105 Pa = 1.59 x 10 11 m3 mol- 1s- 1= 1.59 x 10 14M - 1s- 1
and hence 13 1 2.05 X 10 M - l s- ) } ~S * = R { In ( 1.59 x 1014 M Is I -1
= 8.314J K - I mol - 1x (-3 .05) = -25 J K - I mol - 1
28.9
~H * =
E. - RT [20b], ~H 1 =9134 K x R=75.9 kJ mol - 1
~S* =R (In ~-1)
[Exercise 28.9]
kT RT with B = - X(19] = 1.59 X 10 14 M - 1s- 1 at 30 °C h pe Therefore, 1
=
1
14 7.78 x 10 )
~S * =8.314JK - mol - X { In ( 1. x l0 14 -1 59
}
+4.9 J K - I mol - 1
Hence , ~G 1
=
~H * -
T
~s + =
75 .9-303 X 4.9 X 10- 3 kJ mol - 1
=74.4 kJ mol - 1 28.10
~H * =E" -2RT
[20a]
= 56.8-2 x 8.314 x 10- 3 x 338 kJ mol - 1=51.2 kJ mol - 1
Molecular reaction dynamics k 2 =A e - E,tiiT implies that A = kz e E) RT =7 .84 X 10- 3 kPa - 1S- 1X e56.Rx iO'IR.3 14 x33R = 4. 705 x 10 6 kPa - Is - I= 4. 705 x 10 3 Pa - Is - I
In terms of molar concentrations V = kzPAPB = k z(RTn A][B) and instead of
dp A
dt = -k2PAPB
and hence use
A= 4.705 x 103 Pa- 1s - 1x 8.314 J K - I mol - 1x 338 K = 1.322 x 10 7 m 3 mol - 1s - 1 Then
kT RT Xh pe
B= -
1.381 X 10- 23 X 338 K 6.626 X 10- 34 J s
8.314 J K - I mol - 1X 338 K 10 5 Pa
--------~~- x -------~-------
and
and hence ~:t.G +
28.11
= ~:t.H + - T ~:t.S + = 51.2-338 X ( -96.6 X 10 - 3) kJ mol - 1 = +83.9 kJ mol - 1 k2 = N A a
8k!:\ 112 *
(
lrfl-}
e -f>Eui RT,
which implies that
521
522
Molecular reaction dynamics
For identical particles, Jl = A
4a*hp 9
B
(:n:mk3T3)1 tz
- {:n; X
tm,
so
4 X 0.4 X 10- 18 m 2 X 6.626 X 10- 34 J S X 10 5 Pa 50 X 1.6605 X 10- 27 kg X (1.381 X 10- 23 J K -I X 300 K) 3} 112
=7.78 x 10- 4 and hence ilS t = R {In (
~)- 2} = 8.314 J K
-I
mol- 1{ln 7.78 X 10- 4 - 2}
= -76 J K - I mol- 1 28.12
kT
RT
h
pB
B = -X-
1.381 x 10- 23 J K -I X 298.15 K
8.314 J K -I mol - 1 x 298.15 K
--~~~~~~---x --------~---------34
6.626 x 10
Js
10 5 Pa
= 1.540 x 10 11 m 3 mol- 1 s- 1 = 1.540 X 10 14 M- 1 S- 1 Therefore, 12
4.6 X 10 ) } (a) !lS *= R { In ( l.S 0x10 14 -2 =-45. 8JK- 1 mol- 1 4 (b) ilH *=E. -2RT=10.0- 2x2.48kJmol- 1 = +5.0 kJ mol- 1 (c) ilG *= ilH*- T ilS*= 5.0-298.15 K x ( -45.8 x 10- 3) kJ mol - 1 = +18.7 kJ mol- 1
Molecular reaction dynamics
523
28.13 If cleavage of a C-D or C-H bond is involved in the rate-determining step, use 2 k 2(D) (hk/' ) { 1 1 } [16] k2 (H) = e'' A= 2kT 1-lgl-1-lgA 2X 12 !"-(CD)= + u=l.71 u 2 12 1 X 12 !"-(CH)=--u=0.92u 1+ 12 A
1.054xl0- 34 Jsx(450Nm- 1) 112 { 1 1 } 1 x - -112- - x - - -- 27 1.71 0.92 112 (1.6605 X 10- kg) 112 2 X 1.381 X 10- 23 J K -I X 298 K
= -1.85 k 2(D) = e-185 = 0.156 Hence, k (H) 2 That is, k 2(H) = 6.4 X k 2(0), in reasonable accord with the data.
28.14
2 .! _ (hk/' ) {-1 _ _ 1 } k 2(T) (a) k 2(H) = e • A- 2kT 1"-g¥ 1-lgA
12X 3 flcr=12+3 u=2.40U,flcH =0.92u 1.054 X 10- 34 J S X k/' 2 A= _2_x_1_.3_8_1_x_10__-=2-=-3J-K-_...,.1-x T 1 1 } X { (flcrlu)lt2 - (!"-cH/u)l'2 x
1 --
. ·-
1 112 93.65 X (krfN m- ) { 1 1 } (T/K) X (/A-cr/u)lt2- (!"-CH/u)l/2 93.65 X 450 298
112
{ X
1 1 } 2.401/2 -0.921/2 =-2.65
k 2(T) Therefore, k2(H) = e - 265 = 0.071, so kz(H) = 14k2(T)
524
Molecula r reaction dynamics
93.65 (b) A=
1750 298 X
112
{ X
1 1 } 7.20112- 6.861'2 = -0.12
12 X 16 _ since .uC 2C 160) = - - - u =6 86 u 12+ 16 . 12 X 18 _ =.u C2C 1x0)=--u=7 20u 12 + 18 . kC 2CIRO) Therefore , kC 2C 160)
e-
0
12 = 0.89
and k( 12C 160) = 1.1 X kC 2C 180) Increasing the temperature reduces the magnitude of A, so the isotope effect is likewise reduced. 28.15
1gk2 =1gk2' +2AzAzsf 112 (24)
Hence , lg k2= lg k 2 - 2Az Az 8 ! 112 = lg 12.2-2 X 0.509 X 1 X ( -1) and k2=20.9M - 2min - 1
X
0.0525 112= 1.32
28.16 Fig. 28.1 shows that lg k, is proportional to the ionic strength for neutral
0. 19 Fig 28. 1
u
.;:::--
'c
.J
~0. 17
............
:>-... ............
o--..
"
~.!2> I
-
..__ ............ \..1....
0. 15
0
05
1.0
1.5
I
molecules. From the graph , the intercept at 1=0 is -0.18 , so k~= 0.66M - 1min- 1. -
Molecular reaction dynamics [H +J[A -]
28.17
Ka= (HA]YHA y ~ =
525
[H +J[A - Jy ~ (HAJ
[HA]Ka
Therefore , [H +j = [ _ 2 A 1Y ±
[HA] and lg[H +j = lg Ka+ Ig [A _ - 2lg Y±
1
= lg K.
+ lg [HAJ_ 2A/ 112
a
[A-]
Write u = k2[H +J[B] , then lg u = lg(k 2[B]) + lg[H +j = lg(k2[B]) + lg
-1
- gu
o
[[~~i + 2Al
112
[BJ[HA] + 2Aflt2 , uo_k - 2 [A _
1
That is, the logarithm of the rate should be proportional to the square root of the ionic strength , lg u ex: / 112 .
Problems 28.1
A=NAa * C:f..l~
1 12
[Exercise 28 .1l ; f..l= t m(CH 3)J
K
K)
s x 1.381 x 10- 23 1 - l x 298 112 =a * X 6.022 X 1023 mo l-l X ( .n: X f X 15 .03 X 1. 6605 X 10 27 kg = 5.52 x 1026 x a * mol - 1m s- 1 2.4 x 10 10 mol - 1dm 3s- 1 2.4 x 107 mol - 1m 3s- 1 (a) a *= 5.52 x 1026 mol - 1m s- 1 = 5.52 x 1026 mol - 1m s- 1 = 4.4 X 10-20 m2 (b) Take a=.n:d 2 = .n: X (154 X 2 X 10- 12 m) 2 =3 .0 X 10- 19 m 2
a * 4.4 X 10- 20 0.15 Hence P=-;;= . x 3 0 10 19
526
Molecular reaction dynamics
28.2 Draw up the following table as the basis of an Arrhenius plot: TIK
600
700
800
1000
103 KI T k/(cm 3 mol- 1s- 1) ln(k/cm 3 mol- 1s- 1)
1.67 4.6 X 1()2 6.13
1.43 9.7 X 103 9.18
1.25 1.3x10S 11.8
1.00 3.1x106 15.0
The points are plotted in Fig. 28.2. The least-squares intercept is at 28.3,
30
Fig 28.2
i" V) ..... 0
r-....
"'-.,
E: 20 E:
I"'-
"'~
.......
2!:."' ~
"'
10
0
0
1·0
' ' "'
10~(T/K}
2·0
which implies that A/(cm 3 mol- 1s- 1) = e 28 ·3 = 2.0 x 10 12 As in Problem 28.1,
6 3 1 27 = Aexptl(:n;m) l/2 = 2.0x 10 m mol-l s- X ( :n;X46X 1.6605x 10- kg )1/2 1 23 23 4N A kT 4 x 6.022 x 10 mol 1.381 x 10 J K -lx 750 K
= 4.0 X 10- 21 m2, or 4.0 X 10- 3 nm 2 4.0 X 10- 3 nm 2 P =-= 0 60 a . nm 2 =0.007 a*
Molecular reaction dynamics
527
28.3 For radical recombination, Ea = 0. The maximum rate of recombination is obtained when P = 1 (or more), and then
k2=A
=a*NAC:~t~
112
=4a*NA(::)
112
[~t= tmJ
a* =nd 2 = 7C X (308 X 10- 12 m) 2= 3.0 X 10- 19 m2 Hence, 1.381 x w- 23 1 K -l x 298 K) 112 k2 = 4 X 3.0X 10-19m2X 6.022 X l023mol-l X ( :JCX 15.03 X 1.6605 X 10 27 kg = 1.7 x 108 m 3mol- 1s- 1= 1.7 x lOu M- 1s- 1 This rate constant is for the rate law
v = k 2[CH3]2 d[CH 3] ----ctt= - 2k 2[CH3]2
Therefore,
and its solution is 1 1 [CHJ]- [CHJ]o = 2k2t
For 90 per cent recombination, [CH 3] = 0.10 X [CH3]0 , which occurs when
9
9
2k2t = [CHJ]o' or t= 2k2[CHJ]o The concentration of CH3 radicals in a mixture in which the mole fraction is 2 X 0.10/(1 +0.10) =0.18 and the total pressure is pis 0.20p, expressed as a molar concentration. That is
Therefore,
9RT t = - - -k2 X 0.40p
9x8.314JK- 1mol- 1x298K 1.7 X 108 m3 mol- 1s- 1X 0.36 X 1.013 X 105 Pa
=3.6 ns 28.4 Draw up t.h e following table for an Arrhenius plot:
528
Molecular reaction dynamics
ere
-24.82
-20 .73
-17.02
-13.00
-8.95
TIK 103/(T/K) ln(k/s - 1)
248.33 4.027 -9.01
252.42 3.962 -8.37
256.13 3.904 -7.73
260.15 3.844 -7.07
264.20 3.785 -6.55
The points are plotted in Fig. 28.3. The intercept at liT= 0 is +34.8 and the
9
Fig 28.3
/ ;:::-
I
8
v
'"' g-
"
.s I
I J.
7
_;
v
I
6
37
38
39
t.O
1.1
703/(T/ K}
slope is -10.91 x 103 • The former implies that ln(A/s - 1)=34.8 , and hence that A = 1.3 x 10 15 s- 1• The slope implies that E) R = 10.91 x 103 K, and hence that Ea = 90.7 kJ mol- 1• In solution
!lH ~ =
11H ~ =90.7
E .. - RT, so at -20 °C,
kJ mol- 1 -8.3141 K - I x 253 K= +88.6 kJ mol - 1
For a first-order reaction we write
and hence identify kT kl = - etJ.SI ! R e h
llS ~
E) RT
by writing
e =A
e-E) RT
Molecular reaction dynamics
529
and hence obtaining
~S 1 =R{ln (:~) -1} 15 1 } 6.626 x l0 - 34 Jsxl.3xl0 s- ) =8.314JK - ImoJ - I x { In ( 1.381 x l0 2J JK lx 253K -1 = +37.5 J K - I mol - 1
Therefore, T ~5 1 = 88.6 kJ mol - 1-253 K x 37.5 J K -I mol - 1 = +79.1 kJ mol- 1
~G 1 = ~H 1 -
28.5
lg k = lg k o + 2AzAz 81 112 with A= 0.509
M-
112.
This expression suggests that we should plot lg k against 1 112 and determine z 8 from the slope, since we know that lzA I = 1. We draw up the following table: ffM
0.0025
0.0037
0.0045
0.0065
0.0085
(I/M)' '2 lg(k/M - 1 s - 1)
0.050 0.021
0.061 0.049
0.0067 0.064
0.081 0.072
0.092 0. 100
These points are plotted in Fig. 28 .4. The slope of the limiting line is 2.4. 0·10
/
Fig 28.4
j
J v
/
0'
.?
J
/__ 0 ODS
006
OD7 OD8 1'12
009
0·10
Since this slope is equal to 2A zAz 8 x M 112 = 1.018zAz 8 , we have zAzn = 2.4. But lzAI = 1, and so lzsl = 2. Furthermore , zA and z 8 have the sa me sign because ZAZB > 0. (The data refer to I - and s2 o~- . )
530
Molecular reaction dynamics
28.6 The work w needed to bring two ions from infinity to a separation a medium of relative permittivity E, is w=
z'z"e 2 4.7tE0E,
R '~
R '~
in
[Coulomb potential x charge]
The electrical work is a contribution to the Gibbs function, so ~G '~ = ~G '~ +
z'z"N e 2 A
4.nt: 0 t:,R '~
Since kerr