Unformatted text preview: Part 1: Equilibrium
1
The properties of gases
Solutions to exercises
Discussion questions
E1.1(b) The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it occupied alone the same container as the mixture at the same temperature. It is a limiting law because it holds exactly only under conditions where the gases have no effect upon each other. This can only be true in the limit of zero pressure where the molecules of the gas are very far apart. Hence, Dalton's law holds exactly only for a mixture of perfect gases; for real gases, the law is only an approximation. The critical constants represent the state of a system at which the distinction between the liquid and vapour phases disappears. We usually describe this situation by saying that above the critical temperature the liquid phase cannot be produced by the application of pressure alone. The liquid and vapour phases can no longer coexist, though fluids in the socalled supercritical region have both liquid and vapour characteristics. (See Box 6.1 for a more thorough discussion of the supercritical state.) The van der Waals equation is a cubic equation in the volume, V . Any cubic equation has certain properties, one of which is that there are some values of the coefficients of the variable where the number of real roots passes from three to one. In fact, any equation of state of odd degree higher than 1 can in principle account for critical behavior because for equations of odd degree in V there are necessarily some values of temperature and pressure for which the number of real roots of V passes from n(odd) to 1. That is, the multiple values of V converge from n to 1 as T Tc . This mathematical result is consistent with passing from a two phase region (more than one volume for a given T and p) to a one phase region (only one V for a given T and p and this corresponds to the observed experimental result as the critical point is reached.
E1.2(b)
E1.3(b)
Numerical exercises
E1.4(b) Boyle's law applies. pV = constant pf = E1.5(b) so pf Vf = pi Vi pi Vi (104 kPa) (2000 cm3 ) = 832 kPa = Vf (250 cm3 )
(a) The perfect gas law is pV = nRT implying that the pressure would be nRT V All quantities on the right are given to us except n, which can be computed from the given mass of Ar. 25 g = 0.626 mol n= 39.95 g mol1 p= (0.626 mol) (8.31 102 L bar K1 mol1 ) (30 + 273 K) = 10.5 bar 1.5 L not 2.0 bar. so p =
4
INSTRUCTOR'S MANUAL
(b) The van der Waals equation is p= so p = RT a  2 V m  b Vm (8.31 102 L bar K1 mol1 ) (30 + 273) K (1.5 L/0.626 mol)  3.20 102 L mol1  E1.6(b) (1.337 L2 atm mol2 ) (1.013 bar atm1 ) = 10.4 bar (1.5 L/0.626 mol)2 so pf Vf = pi Vi
(a) Boyle's law applies. pV = constant and pi =
pf Vf (1.48 103 Torr) (2.14 dm3 ) = = 8.04 102 Torr Vi (2.14 + 1.80) dm3 (b) The original pressure in bar is pi = (8.04 102 Torr) E1.7(b) Charles's law applies. V T and Tf = so Vi Vf = Ti Tf 1 atm 760 Torr 1.013 bar 1 atm = 1.07 bar
E1.8(b)
Vf Ti (150 cm3 ) (35 + 273) K = = 92.4 K Vi 500 cm3 The relation between pressure and temperature at constant volume can be derived from the perfect gas law pV = nRT so pT and pi pf = Ti Tf
The final pressure, then, ought to be pf = E1.9(b) pi Tf (125 kPa) (11 + 273) K = 120 kPa = Ti (23 + 273) K
According to the perfect gas law, one can compute the amount of gas from pressure, temperature, and volume. Once this is done, the mass of the gas can be computed from the amount and the molar mass using pV = nRT so n = pV (1.00 atm) (1.013 105 Pa atm1 ) (4.00 103 m3 ) = 1.66 105 mol = RT (8.3145 J K1 mol1 ) (20 + 273) K
and m = (1.66 105 mol) (16.04 g mol1 ) = 2.67 106 g = 2.67 103 kg E1.10(b) All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm /T will give the best value of R.
THE PROPERTIES OF GASES
5
m RT M m RT RT which upon rearrangement gives M = = V p p The best value of M is obtained from an extrapolation of /p versus p to p = 0; the intercept is M/RT . The molar mass is obtained from pV = nRT = Draw up the following table
p/atm 0.750 000 0.500 000 0.250 000 (pVm /T )/(L atm K1 mol1 ) 0.082 0014 0.082 0227 0.082 0414 (/p)/(g L1 atm1 ) 1.428 59 1.428 22 1.427 90
From Fig. 1.1(a), From Fig. 1.1(b),
pVm = 0.082 061 5 L atm K1 mol1 T p=0 = 1.42755 g L1 atm1 p p=0
8.206
8.20615
8.204
8.202
m
8.200 0 0.25 0.50 0.75 1.0
Figure 1.1(a)
1.4288 1.4286 1.4284 1.4282 1.4280 1.4278 1.4276 1.4274 0 0.25 0.50 0.75 1.0 1.42755
Figure 1.1(b)
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INSTRUCTOR'S MANUAL
M = RT
= (0.082 061 5 L atm mol1 K1 ) (273.15 K) (1.42755 g L1 atm1 ) p p=0 = 31.9987 g mol1
The value obtained for R deviates from the accepted value by 0.005 per cent. The error results from the fact that only three data points are available and that a linear extrapolation was employed. The molar mass, however, agrees exactly with the accepted value, probably because of compensating plotting errors. E1.11(b) The mass density is related to the molar volume Vm by Vm = M pM = RT
where M is the molar mass. Putting this relation into the perfect gas law yields pVm = RT so
Rearranging this result gives an expression for M; once we know the molar mass, we can divide by the molar mass of phosphorus atoms to determine the number of atoms per gas molecule M= RT (62.364 L Torr K1 mol1 ) [(100 + 273) K] (0.6388 g L1 ) = = 124 g mol1 . p 120 Torr
The number of atoms per molecule is 124 g mol1 31.0 g mol1 = 4.00
suggesting a formula of P4 E1.12(b) Use the perfect gas equation to compute the amount; then convert to mass. pV RT We need the partial pressure of water, which is 53 per cent of the equilibrium vapour pressure at the given temperature and standard pressure. pV = nRT so n= p = (0.53) (2.69 103 Pa) = 1.43 103 Pa so n = (1.43 103 Pa) (250 m3 ) (8.3145 J K1 mol1 ) (23 + 273) K = 1.45 102 mol
or m = (1.45 102 mol) (18.0 g mol1 ) = 2.61 103 g = 2.61 kg E1.13(b) (a) The volume occupied by each gas is the same, since each completely fills the container. Thus solving for V from eqn 14 we have (assuming a perfect gas) V = nJ RT pJ nNe = 0.225 g 20.18 g mol1 pNe = 66.5 Torr, T = 300 K
= 1.115 102 mol, V =
(1.115 102 mol) (62.36 L Torr K1 mol1 ) (300 K) = 3.137 L = 3.14 L 66.5 Torr
THE PROPERTIES OF GASES
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(b) The total pressure is determined from the total amount of gas, n = nCH4 + nAr + nNe . nCH4 = 0.320 g 16.04 g mol1 = 1.995 102 mol nAr = 0.175 g 39.95 g mol1 = 4.38 103 mol
n = (1.995 + 0.438 + 1.115) 102 mol = 3.548 102 mol p= nRT (3.548 102 mol) (62.36 L Torr K1 mol1 ) (300 K) [1] = V 3.137 L = 212 Torr E1.14(b) This is similar to Exercise 1.14(a) with the exception that the density is first calculated. M= RT [Exercise 1.11(a)] p 33.5 mg = = 0.1340 g L1 , 250 mL
p = 152 Torr,
T = 298 K
M= E1.15(b)
(0.1340 g L1 ) (62.36 L Torr K1 mol1 ) (298 K) = 16.4 g mol1 152 Torr
This exercise is similar to Exercise 1.15(a) in that it uses the definition of absolute zero as that temperature at which the volume of a sample of gas would become zero if the substance remained a gas at low temperatures. The solution uses the experimental fact that the volume is a linear function of the Celsius temperature. Thus V = V0 + V0 = V0 + b, b = V0 At absolute zero, V = 0, or 0 = 20.00 L + 0.0741 L C1 (abs. zero) (abs. zero) =  20.00 L 0.0741 L C1 = 270 C
E1.16(b)
which is close to the accepted value of 273 C. nRT (a) p= V n = 1.0 mol T = (i) 273.15 K; (ii) 500 K V = (i) 22.414 L; (ii) 150 cm3 (i) p = (1.0 mol) (8.206 102 L atm K1 mol1 ) (273.15 K) 22.414 L = 1.0 atm
(ii) p =
(1.0 mol) (8.206 102 L atm K1 mol1 ) (500 K) 0.150 L = 270 atm (2 significant figures) b = 4.34 102 L mol1
(b) From Table (1.6) for H2 S a = 4.484 L2 atm mol1 nRT an2 p=  2 V  nb V
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INSTRUCTOR'S MANUAL
(i) p =
(1.0 mol) (8.206 102 L atm K1 mol1 ) (273.15 K)
22.414 L  (1.0 mol) (4.34 102 L mol1 ) (4.484 L2 atm mol1 ) (1.0 mol)2  (22.414 L)2 = 0.99 atm (1.0 mol) (8.206 102 L atm K1 mol1 ) (500 K) 0.150 L  (1.0 mol) (4.34 102 L mol1 ) (4.484 L2 atm mol1 ) (1.0 mol)2  (0.150 L)2 = 185.6 atm 190 atm (2 significant figures).
(ii) p =
E1.17(b)
The critical constants of a van der Waals gas are Vc = 3b = 3(0.0436 L mol1 ) = 0.131 L mol1 pc = and Tc = a 1.32 atm L2 mol2 = = 25.7 atm 27b2 27(0.0436 L mol1 )2
E1.18(b)
8(1.32 atm L2 mol2 ) 8a = 109 K = 27Rb 27(0.08206 L atm K1 mol1 ) (0.0436 L mol1 ) The compression factor is Z= pVm Vm = RT Vm,perfect
(a) Because Vm = Vm,perfect + 0.12 Vm,perfect = (1.12)Vm,perfect , we have Z = 1.12 Repulsive forces dominate. (b) The molar volume is V = (1.12)Vm,perfect = (1.12) V = (1.12) RT p = 2.7 L mol1
(0.08206 L atm K1 mol1 ) (350 K) 12 atm
E1.19(b)
(a)
o Vm =
RT (8.314 J K1 mol1 ) (298.15 K) = p (200 bar) (105 Pa bar1 ) = 1.24 104 m3 mol1 = 0.124 L mol1
(b) The van der Waals equation is a cubic equation in Vm . The most direct way of obtaining the molar volume would be to solve the cubic analytically. However, this approach is cumbersome, so we proceed as in Example 1.6. The van der Waals equation is rearranged to the cubic form
3 Vm  b +
RT p
2 Vm +
a ab RT Vm  = 0 or x 3  b + p p p
x2 +
ab a x =0 p p
with x = Vm /(L mol1 ).
THE PROPERTIES OF GASES
9
The coefficients in the equation are evaluated as b+ (8.206 102 L atm K1 mol1 ) (298.15 K) RT = (3.183 102 L mol1 ) + p (200 bar) (1.013 atm bar1 ) = (3.183 102 + 0.1208) L mol1 = 0.1526 L mol1 1.360 L2 atm mol2 a = 6.71 103 (L mol1 )2 = p (200 bar) (1.013 atm bar1 ) (1.360 L2 atm mol2 ) (3.183 102 L mol1 ) ab = 2.137 104 (L mol1 )3 = p (200 bar) (1.013 atm bar1 ) Thus, the equation to be solved is x 3  0.1526x 2 + (6.71 103 )x  (2.137 104 ) = 0. Calculators and computer software for the solution of polynomials are readily available. In this case we find x = 0.112 or Vm = 0.112 L mol1
The difference is about 15 per cent. E1.20(b) (a) Vm = Z= 18.015 g mol1 M = 31.728 L mol1 = 0.5678 g L1
pVm (1.00 bar) (31.728 L mol1 ) = 0.9963 = RT (0.083 145 L bar K1 mol1 ) (383 K) a RT  2 and substituting into the expression for Z above we get Vm  b V m
(b) Using p = Z= =
a Vm  Vm  b Vm RT 31.728 L mol1 31.728 L mol1  0.030 49 L mol1  5.464 L2 atm mol2 (31.728 L mol1 ) (0.082 06 L atm K1 mol1 ) (383 K)
= 0.9954 Comment. Both values of Z are very close to the perfect gas value of 1.000, indicating that water vapour is essentially perfect at 1.00 bar pressure. pVm The molar volume is obtained by solving Z = [1.20b], for Vm , which yields RT Vm = (0.86) (0.08206 L atm K1 mol1 ) (300 K) ZRT = = 1.059 L mol1 p 20 atm
E1.21(b)
(a) Then, V = nVm = (8.2 103 mol) (1.059 L mol1 ) = 8.7 103 L = 8.7 mL
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INSTRUCTOR'S MANUAL
(b) An approximate value of B can be obtained from eqn 1.22 by truncation of the series expansion after the second term, B/Vm , in the series. Then, B = Vm pVm  1 = Vm (Z  1) RT
= (1.059 L mol1 ) (0.86  1) = 0.15 L mol1 E1.22(b) (a) Mole fractions are xN = nN 2.5 mol = 0.63 = (2.5 + 1.5) mol ntotal
Similarly, xH = 0.37 (c) According to the perfect gas law ptotal V = ntotal RT ntotal RT V (4.0 mol) (0.08206 L atm mol1 K1 ) (273.15 K) = = 4.0 atm 22.4 L (b) The partial pressures are so ptotal = pN = xN ptot = (0.63) (4.0 atm) = 2.5 atm and pH = (0.37) (4.0 atm) = 1.5 atm E1.23(b) The critical volume of a van der Waals gas is Vc = 3b so b = 1 Vc = 1 (148 cm3 mol1 ) = 49.3 cm3 mol1 = 0.0493 L mol1 3 3 By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e., twice their radius); that volume times the Avogadro constant is the molar excluded volume b b = NA 1 2 4(2r)3 3 so r = 1 2
1/3 3b 4 NA 1/3
r=
3(49.3 cm3 mol1 ) 4(6.022 1023 mol1 )
= 1.94 108 cm = 1.94 1010 m
The critical pressure is pc = a 27b2
so a = 27pc b2 = 27(48.20 atm) (0.0493 L mol1 )2 = 3.16 L2 atm mol2
THE PROPERTIES OF GASES
11
But this problem is overdetermined. We have another piece of information Tc = 8a 27Rb
According to the constants we have already determined, Tc should be Tc = 8(3.16 L2 atm mol2 ) 27(0.08206 L atm K1 mol1 ) (0.0493 L mol1 ) = 231 K
E1.24(b)
However, the reported Tc is 305.4 K, suggesting our computed a/b is about 25 per cent lower than it should be. dZ vanishes. According to the (a) The Boyle temperature is the temperature at which lim Vm d(1/Vm ) van der Waals equation pVm Z= = RT so dZ = d(1/Vm )
RT Vm b
 Va2 Vm
m
RT dZ dVm dZ dVm
=
Vm a  Vm  b Vm RT
dVm d(1/Vm )
2 = Vm
2 = Vm
Vm a 1 + 2 + Vm  b Vm RT (Vm  b)2
2 Vm b a  2 RT (Vm  b) In the limit of large molar volume, we have
=
Vm
lim
dZ a =b =0 d(1/Vm ) RT
so
a =b RT
a (4.484 L2 atm mol2 ) = 1259 K = Rb (0.08206 L atm K1 mol1 ) (0.0434 L mol1 ) (b) By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e. twice their radius); the Avogadro constant times the volume is the molar excluded volume b and T = b = NA 1 2 4(2r 3 ) 3 so r= 1 2
1/3 3b 4 NA 1/3
r= E1.25(b)
3(0.0434 dm3 mol1 ) 4(6.022 1023 mol1 )
= 1.286 109 dm = 1.29 1010 m = 0.129 nm
States that have the same reduced pressure, temperature, and volume are said to correspond. The reduced pressure and temperature for N2 at 1.0 atm and 25 C are pr = p 1.0 atm = = 0.030 pc 33.54 atm and Tr = T (25 + 273) K = = 2.36 Tc 126.3 K
12
INSTRUCTOR'S MANUAL
The corresponding states are (a) For H2 S p = pr pc = (0.030) (88.3 atm) = 2.6 atm T = Tr Tc = (2.36) (373.2 K) = 881 K (Critical constants of H2 S obtained from Handbook of Chemistry and Physics.) (b) For CO2 p = pr pc = (0.030) (72.85 atm) = 2.2 atm T = Tr Tc = (2.36) (304.2 K) = 718 K (c) For Ar p = pr pc = (0.030) (48.00 atm) = 1.4 atm T = Tr Tc = (2.36) (150.72 K) = 356 K E1.26(b) The van der Waals equation is p= RT a  2 Vm  b V m RT (8.3145 J K1 mol1 ) (288 K) 4 3 m mol1  a = 4.00 10 0.76 m6 Pa mol2 p + V2 4.0 106 Pa + 1 2 4 3 m
(4.0010 m mol )
which can be solved for b b = Vm 
= 1.3 104 m3 mol1 The compression factor is Z= pVm (4.0 106 Pa) (4.00 104 m3 mol1 ) = 0.67 = RT (8.3145 J K1 mol1 ) (288 K)
Solutions to problems
Solutions to numerical problems
P1.2 Identifying pex in the equation p = pex + gh [1.4] as the pressure at the top of the straw and p as the atmospheric pressure on the liquid, the pressure difference is p  pex = gh = (1.0 103 kg m3 ) (9.81 m s2 ) (0.15 m) = 1.5 103 Pa (= 1.5 102 atm) P1.4 pV = nRT [1.12] implies that, with n constant, p f Vf pi Vi = Tf Ti Vi Tf pi Vf Ti
Solving for pf , the pressure at its maximum altitude, yields pf =
THE PROPERTIES OF GASES
13
Substituting Vi = 4 ri3 and Vf = 4 rf3 3 3 pf = (4/3) ri3 (4/3) rf3 Tf pi = Ti = P1.6 ri 3 T f pi rf Ti 1.0 m 3 3.0 m 253 K 293 K (1.0 atm) = 3.2 102 atm
The value of absolute zero can be expressed in terms of by using the requirement that the volume of a perfect gas becomes zero at the absolute zero of temperature. Hence 0 = V0 [1 + (abs. zero)] 1 All gases become perfect in the limit of zero pressure, so the best value of and, hence, (abs. zero) is obtained by extrapolating to zero pressure. This is done in Fig. 1.2. Using the extrapolated value, = 3.6637 103 C1 , or Then (abs. zero) =  (abs. zero) =  1 = 272.95 C 3.6637 103 C1
which is close to the accepted value of 273.15 C.
3.672 3.670 3.668 3.666
3.664 3.662 0 200 400 p / Torr 600 800
Figure 1.2
P1.7
The mass of displaced gas is V , where V is the volume of the bulb and is the density of the gas. The balance condition for the two gases is m(bulb) = V (bulb), m(bulb) = V (bulb) which implies that = . Because [Problem 1.5] = the balance condition is pM = p M p which implies that M = M p This relation is valid in the limit of zero pressure (for a gas behaving perfectly). pM RT
14
INSTRUCTOR'S MANUAL
In experiment 1, p = 423.22 Torr, p = 327.10 Torr; hence M = 423.22 Torr 70.014 g mol1 = 90.59 g mol1 327.10 Torr
In experiment 2, p = 427.22 Torr, p = 293.22 Torr; hence M = 427.22 Torr 70.014 g mol1 = 102.0 g mol1 293.22 Torr
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 mol1 . The molecules CH2 FCF3 or CHF2 CHF2 have M 102 g mol1 . P1.9 We assume that no H2 remains after the reaction has gone to completion. The balanced equation is N2 + 3H2 2NH3 We can draw up the following table
N2 Initial amount Final amount Specifically Mole fraction n 1 n  3n 0.33 mol 0.20 H2 n 0 0 0 NH3 0
2 n 3
Total n+n n + 1n 3 1.66 mol 1.00
1.33 mol 0.80
p=
nRT = (1.66 mol) V
(8.206 102 L atm K1 mol1 ) (273.15 K) 22.4 L
= 1.66 atm
p(H2 ) = x(H2 )p = 0 p(N2 ) = x(N2 )p = (0.20 (1.66 atm)) = 0.33 atm p(NH3 ) = x(NH3 )p = (0.80) (1.66 atm) = 1.33 atm P1.10 (8.206 102 L atm K1 mol1 ) (350 K) RT = = 12.5 L mol1 p 2.30 atm RT RT a + b [rearrange 1.25b] (b) From p =  2 [1.25b], we obtain Vm = Vm  b Vm p+ a (a) Vm = Then, with a and b from Table 1.6 Vm (8.206 102 L atm K1 mol1 ) (350 K) (2.30 atm) +
6.260 L2 atm mol2 (12.5 L mol1 )2
2 Vm
+ (5.42 102 L mol1 )
28.72 L mol1 + (5.42 102 L mol1 ) 12.3 L mol1 . 2.34
Substitution of 12.3 L mol1 into the denominator of the first expression again results in Vm = 12.3 L mol1 , so the cycle of approximation may be terminated.
THE PROPERTIES OF GASES
15
P1.13
(a)
Since B (TB ) = 0 at the Boyle temperature (section 1.3b): Solving for TB : TB = c ln
a b
B (TB ) = a + b ec/TB2 = 0 = 501.0 K
=
(1131 K 2 )
1 ln (0.1993 bar ) 1
(0.2002 bar
)
(b)
Perfect Gas Equation:
Vm (p, T ) =
RT p
Vm (50 bar, 298.15 K) = Vm (50 bar, 373.15 K) =
0.083145 L bar K1 mol1 (298.15 K) = 0.496 L mol1 50 bar 0.083145 L bar K1 mol1 (373.15 K) = 0.621 L mol1 50 bar RT (1+B (T ) p) = Vperfect (1+B (T ) p) p
Virial Equation (eqn 1.21 to first order): Vm (p, T ) = B (T ) = a + b e

c 2 TB
B (298.15 K) = 0.1993 bar 1 + 0.2002 bar 1 e B (373.15 K) = 0.1993 bar 1 + 0.2002 bar 1 e
 
1131 K2 (298.15 K)2 1131 K2 (373.15 K)2
= 0.00163 bar 1 = 0.000720 bar 1
Vm (50 bar, 298.15 K) = 0.496 L mol1 1  0.00163 bar 1 50 bar = 0.456 L mol1 Vm (50 bar, 373.15 K) = 0.621 L mol1 1  0.000720 bar 1 50 bar = 0.599 L mol1 The perfect gas law predicts a molar volume that is 9% too large at 298 K and 4% too large at 373 K. The negative value of the second virial coefficient at both temperatures indicates the dominance of very weak intermolecular attractive forces over repulsive forces. P1.15 From Table 1.6 Tc = 2 3 2a 1/2 , pc = 3bR 1 12 2aR 1/2 3b3 12bpc . Thus R
2a 1/2 may be solved for from the expression for pc and yields 3bR Tc = 2 3 12pc b R = = vmol = vmol = r= b = NA 4 3 r 3 1 3 Vc NA 8 3 8 3 = p c Vc R
(40 atm) (160 103 L mol1 ) 8.206 102 L atm K1 mol1
= 210 K
160 106 m3 mol1 = 8.86 1029 m3 (3) (6.022 1023 mol1 )
1/3 3 (8.86 1029 m3 ) = 0.28 nm 4
16
INSTRUCTOR'S MANUAL
P1.16
Vc = 2b,
Tc =
a [Table 1.6] 4bR
1 1 Hence, with Vc and Tc from Table 1.5, b = 2 Vc = 2 (118.8 cm3 mol1 ) = 59.4 cm3 mol1
a = 4bRTc = 2RTc Vc = (2) (8.206 102 L atm K1 mol1 ) (289.75 K) (118.8 103 L mol1 ) = 5.649 L2 atm mol2 Hence p = = RT nRT na/RT V ea/RT Vm = e Vm  b V  nb (1.0 mol) (8.206 102 L atm K1 mol1 ) (298 K) (1.0 L)  (1.0 mol) (59.4 103 L mol1 ) exp (1.0 mol) (5.649 L2 atm mol2 ) (8.206 102 L atm K1 mol1 ) (298 K) (1.0 L2 atm mol1 )
= 26.0 atm e0.231 = 21 atm
Solutions to theoretical problems
P1.18 This expansion has already been given in the solutions to Exercise 1.24(a) and Problem 1.17; the result is p= RT Vm 1+ b a RT b2 1 + 2 + Vm Vm RT Vm 1+ B C + + Vm Vm 2 [1.22]
Compare this expansion with p = and hence find B = b 
a and C = b2 RT
Since C = 1200 cm6 mol2 , b = C 1/2 = 34.6 cm3 mol1 a = RT (b  B) = (8.206 102 ) (273 L atm mol1 ) (34.6 + 21.7) cm3 mol1 = (22.40 L atm mol1 ) (56.3 103 L mol1 ) = 1.26 L2 atm mol2 P1.22 For a real gas we may use the virial expansion in terms of p [1.21] p= nRT RT (1 + B p + ) = (1 + B p + ) V M p RT RT B = + p + M M
which rearranges to
THE PROPERTIES OF GASES
17
Therefore, the limiting slope of a plot of Solutions Manual, the limiting slope is
B RT p against p is . From Fig. 1.2 in the Student's M
B RT (4.41  5.27) 104 m2 s2 = 9.7 102 kg1 m3 = M (10.132  1.223) 104 Pa RT From Fig. 1.2, = 5.39 104 m2 s2 ; hence M B = 9.7 102 kg1 m3 = 1.80 106 Pa1 5.39 104 m2 s2
B = (1.80 106 Pa1 ) (1.0133 105 Pa atm1 ) = 0.182 atm1 B = RT B [Problem 1.21] = (8.206 102 L atm K1 mol1 ) (298 K) (0.182 atm1 ) = 4.4 L mol1 P1.23 Write Vm = f (T , p); then dVm = Vm Vm dT + dp T p p T Restricting the variations of T and p to those which leave Vm constant, that is dVm = 0, we obtain Vm Vm = T p p T From the equation of state RT p 3 =  2  2(a + bT )Vm Vm T Vm Substituting
R b Vm + Vm2 Vm = T P  RT  2(a+bT ) 2 3 Vm Vm
p p 1 = T Vm Vm T
p  T p = p T Vm V
Vm
m T
p R b = + 2 T Vm Vm Vm R + Vb m
RT Vm
=+
+ 2(a+bT ) V2
m
RT (a + bT ) =p From the equation of state 2 Vm Vm Then P1.25 Z= R + Vb Vm m = RT RT T p Vm + 2 p  Vm = R + Vb m 2p 
RT Vm
=
RVm + b 2pVm  RT
Vm o o , where Vm = the molar volume of a perfect gas Vm From the given equation of state Vm = b + RT o = b + Vm p then Z=
o b + Vm b =1+ o o Vm Vm
o o For Vm = 10b, 10b = b + Vm or Vm = 9b 10b 10 then Z = = = 1.11 9b 9
18
INSTRUCTOR'S MANUAL
P1.27
The two masses represent the same volume of gas under identical conditions, and therefore, the same number of molecules (Avogadro's principle) and moles, n. Thus, the masses can be expressed as nMN = 2.2990 g for `chemical nitrogen' and nAr MAr + nN MN = n[xAr MAr + (1  xAr )MN ] = 2.3102 g for `atmospheric nitrogen'. Dividing the latter expression by the former yields xAr MAr 2.3102 + (1  xAr ) = MN 2.2990 so xAr MAr 2.3102 1 = 1 MN 2.2990
2.3102 2.3102 1 2.2990  1 and xAr = 2.2990 = = 0.011 MAr 39.95 g mol1 1 1 MN 28.013 g mol1
Comment. This value for the mole fraction of argon in air is close to the modern value. P1.29 pVm p c Vm pr Vr Tc p = = [1.20b, 1.28] RT T pc RTc Tr V 3 8Tr = r [1.29]  Tr 3Vr  1 Vr2 8 pc V V RTc pc V 8 But Vr = = = [1.27] = Vr Vc p c Vc RTc 3 RTc 3 3 8Tr V  Therefore Z = r Tr 3 8Vr  1 8Vr 2 3 Z=
3
V = r Tr = Vr Z=
Tr 27  Vr  1/8 64(Vr )2 27 1  Vr  1/8 64Tr (Vr )2 (2)
Vr 27  Vr  1/8 64Tr Vr
To derive the alternative form, solve eqn 1 for Vr , substitute the result into eqn 2, and simplify into polynomial form. Vr = ZTr pr
pr ZTr /pr 27 Z = ZT  1 r 64Tr ZTr pr  8 8ZTr 27pr =  8ZTr  pr 64ZTr2 = 512Tr3 Z 2  27pr (8Tr Z  pr ) 64Tr2 (8ZTr  pr )Z
2 64Tr2 (8ZTr  pr )Z 2 = 512Tr3 Z 2  216Tr pr Z + 27pr 2 512Tr3 Z 3  64Tr2 pr + 512Tr3 Z 2 + 216Tr pr Z  27pr = 0
THE PROPERTIES OF GASES
19
Z3 
2 27pr pr 27pr =0 Z + 1 Z2 + 8Tr 64Tr2 512Tr3
(3)
At Tr = 1.2 and pr = 3 eqn 3 predicts that Z is the root of Z3  27(3) 27(3)2 3 Z =0 + 1 Z3 + 8(1.2) 64(1.2)2 512(1.2)3
Z 3  1.3125Z 2 + 0.8789Z  0.2747 = 0 The real root is Z = 0.611 and this prediction is independent of the specific gas. Figure 1.27 indicates that the experimental result for the listed gases is closer to 0.55.
Solutions to applications
P1.31 Refer to Fig. 1.3.
h
Air (environment)
Ground
Figure 1.3 The buoyant force on the cylinder is Fbuoy = Fbottom  Ftop = A(pbottom  ptop ) according to the barometric formula. ptop = pbottom eMgh/RT where M is the molar mass of the environment (air). Since h is small, the exponential can be expanded 1 in a Taylor series around h = 0 ex = 1  x + x 2 + . Keeping the firstorder term only 2! yields ptop = pbottom 1  Mgh RT
20
INSTRUCTOR'S MANUAL
The buoyant force becomes Fbuoy = Apbottom 1  1 + = pbottom V M RT Mgh RT = Ah n= pbottom M RT pbottom V RT g
g = nMg
n is the number of moles of the environment (air) displaced by the balloon, and nM = m, the mass of the displaced environment. Thus Fbuoy = mg. The net force is the difference between the buoyant force and the weight of the balloon. Thus Fnet = mg  mballoon g = (m  mballoon )g This is Archimedes' principle.
2
The First Law: the concepts
Solutions to exercises
Discussion questions
E2.1(b) Work is a transfer of energy that results in orderly motion of the atoms and molecules in a system; heat is a transfer of energy that results in disorderly motion. See Molecular Interpretation 2.1 for a more detailed discussion. Rewrite the two expressions as follows: (1) adiabatic p 1/V (2) isothermal p 1/V The physical reason for the difference is that, in the isothermal expansion, energy flows into the system as heat and maintains the temperature despite the fact that energy is lost as work, whereas in the adiabatic case, where no heat flows into the system, the temperature must fall as the system does work. Therefore, the pressure must fall faster in the adiabatic process than in the isothermal case. Mathematically this corresponds to > 1. E2.3(b) Standard reaction enthalpies can be calculated from a knowledge of the standard enthalpies of formation of all the substances (reactants and products) participating in the reaction. This is an exact method which involves no approximations. The only disadvantage is that standard enthalpies of formation are not known for all substances. Approximate values can be obtained from mean bond enthalpies. See almost any general chemistry text, for example, Chemical Principles, by Atkins and Jones, Section 6.21, for an illustration of the method of calculation. This method is often quite inaccurate, though, because the average values of the bond enthalpies used may not be close to the actual values in the compounds of interest. Another somewhat more reliable approximate method is based on thermochemical groups which mimic more closely the bonding situations in the compounds of interest. See Example 2.6 for an illustration of this kind of calculation. Though better, this method suffers from the same kind of defects as the average bond enthalpy approach, since the group values used are also averages. Computer aided molecular modeling is now the method of choice for estimating standard reaction enthalpies, especially for large molecules with complex threedimensional structures, but accurate numerical values are still difficult to obtain.
E2.2(b)
Numerical exercises
E2.4(b) Work done against a uniform gravitational field is w = mgh (a) (b) E2.5(b) w = (5.0 kg) (100 m) (9.81 m s2 ) = 4.9 103 J w = (5.0 kg) (100 m) (3.73 m s2 ) = 1.9 103 J
Work done against a uniform gravitational field is w = mgh = (120 103 kg) (50 m) (9.81 m s2 ) = 59 J
E2.6(b)
Work done by a system expanding against a constant external pressure is w = pex V = (121 103 Pa) (15 cm) (50 cm2 ) (100 cm m1 )3 = 91 J
22
INSTRUCTOR'S MANUAL
E2.7(b)
For a perfect gas at constant temperature U= 0 so q = w H is also zero
For a perfect gas at constant temperature, dH = d(U + pV )
we have already noted that U does not change at constant temperature; nor does pV if the gas obeys Boyle's law. These apply to all three cases below. (a) Isothermal reversible expansion w = nRT ln Vf Vi
= (2.00 mol) (8.3145 J K 1 mol1 ) (22 + 273) K ln q = w = 1.62 103 J (b) Expansion against a constant external pressure w = pex V where pex in this case can be computed from the perfect gas law pV = nRT so p =
31.7 L = 1.62 103 J 22.8 L
(2.00 mol) (8.3145 J K1 mol1 ) (22 + 273) K (1000 L m3 ) = 1.55 105 Pa 31.7 L (1.55 105 Pa) (31.7  22.8) L and w = = 1.38 103 J 1000 L m3 q = w = 1.38 103 J
(c) Free expansion is expansion against no force, so w = 0 , and q = w = 0 as well. E2.8(b) The perfect gas law leads to p1 V nRT1 = p2 V nRT2 or p2 = p1 T2 (111 kPa) (356 K) = 143 kPa = T1 277 K
There is no change in volume, so w = 0 . The heat flow is q= CV dT CV T = (2.5) (8.3145 J K 1 mol1 ) (2.00 mol) (356  277) K
= 3.28 103 J U = q + w = 3.28 103 J
THE FIRST LAW: THE CONCEPTS
23
E2.9(b)
(a) w = pex V = (b) w = nRT ln
Vf Vi 6.56 g = 39.95 g mol1 = 52.8 J
(7.7 103 Pa) (2.5 L) = 19 J 1000 L m3
(8.3145 J K 1 mol1 ) (305 K) ln
(2.5 + 18.5) L 18.5 L
E2.10(b)
Isothermal reversible work is w = nRT ln = +6.01 J Vf = (1.77 103 mol) (8.3145 J K 1 mol1 ) (273 K) ln 0.224 Vi
E2.11(b)
 H = n( vap H  ) = (2.00 mol) (35.3 kJ mol1 ) = 70.6 kJ Because the condensation also occurs at constant pressure, the work is
q=
w=
pex dV = p V
The change in volume from a gas to a condensed phase is approximately equal in magnitude to the volume of the gas w p(Vvapor ) = nRT = (2.00 mol) (8.3145 kJ K 1 mol1 ) (64 + 273) K = 5.60 103 J U = q + w = (70.6 + 5.60) kJ = 65.0 kJ E2.12(b) The reaction is Zn + 2H+ Zn2+ + H2 so it liberates 1 mol H2 (g) for every 1 mol Zn used. Work at constant pressure is w = p V = pVgas = nRT =  5.0 g 65.4 g mol1
(8.3145 J K1 mol1 ) (23 + 273) K = 188 J E2.13(b) E2.14(b) 500 kg 39.1 103 kg mol1 (a) At constant pressure
 q = n fus H  =
(2.35 kJ mol1 ) = 3.01 104 kJ
q=
Cp dT =
100+273 K 0+273 K
[20.17 + (0.4001)T /K] dT J K 1
373 K 273 K
1 = [(20.17)T + 2 (0.4001) (T 2 /K)] J K1
= [(20.17) (373  273) +
2 1 2 (0.4001) (373
 2732 )] J = 14.9 103 J =
H
24
INSTRUCTOR'S MANUAL
w = p V = nR T = (1.00 mol) (8.3145 J K 1 mol1 ) (100 K) = 831 J U = q + w = (14.9  0.831) kJ = 14.1 kJ (b) E2.15(b) U and H depend only on temperature in perfect gases. Thus, H = 14.9 kJ and 14.1 kJ as above. At constant volume, w = 0 and U = q, so q = +14.1 kJ Vi 1/c Vf U =
For reversible adiabatic expansion Vf Tfc = Vi Tic where c = so Tf = Ti
Cp,m  R (37.11  8.3145) J K 1 mol1 CV ,m = = = 3.463 R R 8.3145 J K1 mol1 So the final temperature is Tf = (298.15 K) 500 103 L 2.00 L
1/3.463
= 200 K
E2.16(b)
Reversible adiabatic work is w = CV T = n(Cp,m  R) (Tf  Ti ) where the temperatures are related by [solution to Exercise 2.15b] Tf = Ti where c = Vi 1/c Vf Cp,m  R (29.125  8.3145) J K 1 mol1 CV ,m = = 2.503 = R R 8.3145 J K1 mol1 400 103 L 2.00 L
1/2.503
So Tf = [(23.0 + 273.15) K] and w =
= 156 K
E2.17(b)
3.12 g (29.125  8.3145) J K 1 mol1 (156  296) K = 325 J 28.0 g mol1 For reversible adiabatic expansion p f V f = pi V i pf = pi
so 500 103 L 3.0 L
1.3
Vi = (87.3 Torr) Vf = 8.5 Torr
E2.18(b)
For reversible adiabatic expansion p f V f = pi V i
so
pf = pi
Vi Vf
We need pi , which we can obtain from the perfect gas law pV = nRT so p= nRT V
THE FIRST LAW: THE CONCEPTS
25
pi =
1.4 g 18 g mol1
(0.08206 L atm K 1 mol1 ) (300 K) 1.0 L 1.0 L 1.3 = 0.46 atm 3.0 L
= 1.9 atm
pf = (1.9 atm) E2.19(b) The reaction is
nC6 H14 + 19 O2 6CO2 + 7H2 O 2
 H = fH fH fH cH       19 f H (nC6 H14 )  2     f H (H2 O)  c H fH  19 2  (O  f 2)  H  (O   = 6 f H  (CO2 ) + 7 f H  (H2 O) 
so
(nC6 H14 ) = 6 f H
 
(CO2 ) + 7
2)
   
(nC6 H14 ) = [6 (393.51) + 7 (285.83) + 4163  (0)] kJ mol1 (nC6 H14 ) = 199 kJ mol1
E2.20(b)
qp = nCp,m T Cp,m = qp 178 J = = 53 J K1 mol1 n T 1.9 mol 1.78 K
CV ,m = Cp,m  R = (53  8.3) J K 1 mol1 = 45 J K1 mol1 E2.21(b) H = qp = 2.3 kJ , the energy extracted from the sample. qp = C T E2.22(b) so C= qp 2.3 kJ = = 0.18 kJ K1 T (275  288) K
H = qp = Cp T = nCp,m T = (2.0 mol) (37.11 J K 1 mol1 ) (277  250) K = 2.0 103 J mol1 H = U+ (pV ) = U + nR T so U= H  nR T U = 2.0 103 J mol1  (2.0 mol) (8.3145 J K 1 mol1 ) (277  250) K = 1.6 103 J mol1
E2.23(b)
In an adiabatic process, q = 0 . Work against a constant external pressure is w = pex V = (78.5 103 Pa) (4 15  15) L = 3.5 103 J 1000 L m3 w n(Cp,m  R)
U = q + w = 3.5 103 J w = CV T = n(Cp,m  R) T T = so T =
3.5 103 J = 24 K (5.0 mol) (37.11  8.3145) J K 1 mol1 H = U + (pV ) = U + nR T , = 3.5 103 J + (5.0 mol) (8.3145 J K 1 mol1 ) (24 K) = 4.5 103 J
26
INSTRUCTOR'S MANUAL
E2.24(b)
For adiabatic compression, q = 0 and w = CV T = (2.5 mol) (27.6 J K 1 mol1 ) (255  220) K = 2.4 103 J U = q + w = 2.4 103 J H = U+ (pV ) = U + nR T = 2.4 103 J + (2.5 mol) (8.3145 J K 1 mol1 ) (255  220) K = 3.1 103 J The initial and final states are related by Vf Tfc = Vi Tic where c = Vi = so Vf = Vi Ti c Tf
27.6 J K1 mol1 CV ,m = = 3.32 R 8.314 J K1 mol1
nRTi 2.5 mol 8.3145 J K 1 mol1 220 K = 0.0229 m3 = pi 200 103 Pa 220 K 3.32 = 0.014 m3 = 14 L 255 K
Vf = (0.0229 m3 ) pf = E2.25(b)
nRTf 2.5 mol 8.3145 J K 1 mol1 255 K = = 3.8 105 Pa Vf 0.014 m3
For reversible adiabatic expansion p f V f = pi V i where = and Vi =
so
V f = Vi
pi 1/ pf
Cp,m 20.8 J K 1 mol1 = 1.67 = Cp,m  R (20.8  8.3145) J K 1 mol1
nRTi (1.5 mol) (8.3145 J K 1 mol1 ) (315 K) = 0.0171 m3 = pi 230 103 Pa pi 1/ = (0.0171 m3 ) pf 230 kPa 1/1.67 = 0.0201 m3 170 kPa
so Vf = Vi Tf =
pf Vf (170 103 Pa) (0.0201 m3 ) = 275 K = nR (1.5 mol) (8.3145 J K 1 mol1 )
w = CV T = (1.5 mol) (20.8  8.3145) J K 1 mol1 (275  315 K) = 7.5 102 J E2.26(b) The expansion coefficient is defined as = 1 V V = T p ln V T p
so for a small change in temperature (see Exercise 2.26a), V = Vi T = (5.0 cm3 ) (0.354 104 K 1 ) (10.0 K) = 1.8 103 cm3
THE FIRST LAW: THE CONCEPTS
27
E2.27(b)
In an adiabatic process, q = 0 . Work against a constant external pressure is w = pex V = (110 103 Pa) U = q + w = 36 J w = CV T = n(Cp,m  R) T w T = n(Cp,m  R) = so (15 cm) (22 cm2 ) = 36 J (100 cm m1 )3
36 J = 0.57 K (3.0 mol) (29.355  8.3145) J K 1 mol1 H = U + (pV ) = U + nR T = 36 J + (3.0 mol) (8.3145 J K 1 mol1 ) (0.57 K) = 50 J E2.28(b) The amount of N2 in the sample is n= 15.0 g = 0.535 mol 28.013 g mol1 pi 1/ pf
(a) For reversible adiabatic expansion pf Vf = pi Vi where =
so
Vf = Vi
Cp,m where CV ,m = (29.125  8.3145) J K 1 mol1 = 20.811 J K1 mol1 CV ,m 29.125 J K1 mol1 so = = 1.3995 20.811 J K1 mol1 nRTi (0.535 mol) (8.3145 J K 1 mol1 ) (200 K) and Vi = = 4.04 103 m3 = pi 220 103 Pa so Vf = Vi Tf = pi 1/ = (4.04 103 m3 ) pf 220 103 Pa 110 103 Pa
1/1.3995
= 6.63 103 m3 .
(110 103 Pa) (6.63 103 m3 ) pf Vf = = 164 K nR (0.535 mol) (8.3145 J K 1 mol1 ) so pex (Vf  Vi ) = CV (Tf  Ti )
(b) For adiabatic expansion against a constant external pressure w = pex V = CV T pf Vf = nRTf Solve the latter for Tf in terms of Vf , and insert into the previous relationship to solve for Vf Tf = pf Vf nR so pex (Vf  Vi ) = CV p f Vf  Ti nR CV Ti + pex Vi pex +
CV ,m pf R
In addition, the perfect gas law holds
Collecting terms gives CV Ti + pex Vi = Vf pex + C V pf nR so Vf =
28
INSTRUCTOR'S MANUAL
Vf =
(20.811 J K1 mol1 ) (0.535 mol) (200 K) + (110 103 Pa) (4.04 103 m3 )
K 1 mol1 )(110103 Pa) 110 103 Pa + (20.811 J8.3145 J K1 mol1
Vf = 6.93 103 m3 Finally, the temperature is Tf = E2.29(b) (110 103 Pa) (6.93 103 m3 ) pf Vf = = 171 K nR (0.535 mol) (8.3145 J K 1 mol1 )
At constant pressure q=
 H = n vap H  = (0.75 mol) (32.0 kJ mol1 ) = 24.0 kJ
and w = p V pVvapor = nRT = (0.75 mol) (8.3145 J K 1 mol1 ) (260 K) = 1.6 103 J = 1.6 kJ U = w + q = 24.0  1.6 kJ = 22.4 kJ Comment. Because the vapor is here treated as a perfect gas, the specific value of the external pressure provided in the statement of the exercise does not affect the numerical value of the answer. E2.30(b) The reaction is C6 H5 OH + 7O2 6CO2 + 3H2 O
cH     = 6 f H  (CO2 ) + 3 f H  (H2 O)  fH    (C6 H5 OH)  7 f H  (O2 )
= [6(393.51) + 3(285.83)  (165.0)  7(0)] kJ mol1 = 3053.6 kJ mol1 E2.31(b) The hydrogenation reaction is C4 H8 + H2 C4 H10
hyd H        
=
fH
(C4 H10 ) 
fH
(C4 H8 ) 
fH
(H2 )
The enthalpies of formation of all of these compounds are available in Table 2.5. Therefore
hyd H  
= [126.15  (0.13)] kJ mol1 = 126.02 kJ mol1
fH  
If we had to, we could find
 
(C4 H8 ) from information about another of its reactions
fH    (C4 H8 )  6 f H  (O2 )
C4 H8 + 6O2 4CO2 + 4H2 O,
cH   = 4 f H  (CO2 ) + 4 f H  (H2 O)  fH       (C4 H8 ) = 4 f H  (CO2 ) + 4 f H  (H2 O)  6 f H  (O2 )  c H  1 = [4(393.51) + 4(285.83)  6(0)  (2717)] kJ mol
so
= 0. kJ mol1
hyd H  
= 126.15  (0.)  (0) kJ mol1 = 126 kJ mol1
This value compares favourably to that calculated above.
THE FIRST LAW: THE CONCEPTS
29
E2.32(b)
We need
fH
 
for the reaction
(4) 2B(s) + 3H2 (g) B2 H6 (g) reaction (4) = reaction (2) + 3 reaction (3)  reaction (1) Thus,
fH        
=
rH
{reaction (2)} + 3
rH
{reaction (3)} 
rH
{reaction (1)}
= {2368 + 3 (241.8)  (1941)} kJ mol1 = 1152 kJ mol1 E2.33(b) The formation reaction is
1 C + 2H2 (g) + 2 O2 (g) + N2 (g) CO(NH2 )2 (s)
H =
fU  
U+
(pV )
U + RT
ngas
so
fU
 
=
fH
 
 RT
ngas
= 333.51 kJ mol1  (8.3145 103 kJ K1 mol1 ) (298.15 K) (7/2) = 324.83 kJ mol1
E2.34(b)
The energy supplied to the calorimeter equals C T , where C is the calorimeter constant. That energy is E = (2.86 A) (22.5 s) (12.0 V) = 772 J So C = 772 J E = = 451 J K1 T 1.712 K For anthracene the reaction is C14 H10 (s) +
cU cH cU      
E2.35(b)
=
33 O2 (g) 14CO2 (g) + 5H2 O(l) 2 5    ng RT [26] ng =  mol cH 2
= 7163 kJ mol1 (Handbook of Chemistry and Physics)
5 = 7163 kJ mol1  ( 2 8.3 103 kJ K1 mol1 298 K) (assume T = 298 K)
= 7157 kJ mol1
 q = qV  = n c U   =
2.25 103 g 172.23 g mol1
(7157 kJ mol1 )
= 0.0935 kJ C= q 0.0935 kJ = = 0.0693 kJ K 1 = 69.3 J K1 T 1.35 K
When phenol is used the reaction is C6 H5 OH(s) + 15 O2 (g) 6CO2 (g) + 3H2 O(l) 2
cH cU    
= 3054 kJ mol1 (Table 2.5) =
cH  

ng RT ,
3 ng =  2 mol
3 = (3054 kJ mol1 ) + 2 (8.314 103 kJ K1 mol1 ) (298 K)
= 3050 kJ mol1
30
INSTRUCTOR'S MANUAL
q = T =
135 103 g 94.12 g mol1
(3050 kJ mol1 ) = 4.375 kJ
q 4.375 kJ = = +63.1 K C 0.0693 kJ K1
  Comment. In this case c U  and c H  differed by 0.1 per cent. Thus, to within 3 significant   figures, it would not have mattered if we had used c H  instead of c U  , but for very precise work it would.
E2.36(b)
The reaction is AgBr(s) Ag+ (aq) + Br  (aq)
sol H    
=
fH
(Ag+ ) +
fH
 
(Br  ) 
fH
 
(AgBr)
= [105.58 + (121.55)  (100.37)] kJ mol1 = +84.40 kJ mol1 E2.37(b) The difference of the equations is C(gr) C(d)
trans H  
= [393.51  (395.41)] kJ mol1 = +1.90 kJ mol1
E2.38(b)
Combustion of liquid butane can be considered as a twostep process: vaporization of the liquid followed by combustion of the butane gas. Hess's law states that the enthalpy of the overall process is the sum of the enthalpies of the steps (a) (b)
cH cH    
= [21.0 + (2878)] kJ mol1 = 2857 kJ mol1 (pV ) =
cU  
 = cU  + The reaction is
+ RT
ng
so
cU
 
=
cH
 
 RT
ng
C4 H10 (l) + 13 O2 (g) 4CO2 (g) + 5H2 O(l) 2 so ng = 2.5 and
cU  
= 2857 kJ mol1  (8.3145 103 kJ K1 mol1 ) (298 K) (2.5) = 2851 kJ mol1
E2.39(b)
(a)
rH
 
=
fH
 
(propene, g) 
fH
 
(cyclopropane, g) = [(20.42)  (53.30)] kJ mol1
= 32.88 kJ mol1 (b) The net ionic reaction is obtained from H+ (aq) + Cl (aq) + Na+ (aq) + OH (aq) Na+ (aq) + Cl (aq) + H2 O(l) and is H+ (aq) + OH (aq) H2 O(l)
rH  
=
fH
 
(H2 O, l) 
fH
 
(H+ , aq) 
= [(285.83)  (0)  (229.99)] kJ mol = 55.84 kJ mol1
fH 1
 
(OH , aq)
THE FIRST LAW: THE CONCEPTS
31
E2.40(b)
reaction (3) = reaction (2)  2(reaction (1)) (a)
rH     (3) = r H  (2)  2( r H  (1)) 1 = 483.64 kJ mol  2(52.96 kJ mol1 )
= 589.56 kJ mol1
rU  
=
rH
 

ng RT
= 589.56 kJ mol1  (3) (8.314 J K 1 mol1 ) (298 K) = 589.56 kJ mol1 + 7.43 kJ mol1 = 582.13 kJ mol1 (b)
fH fH     1 (HI) = 2 (52.96 kJ mol1 ) = 26.48 kJ mol1 1 (H2 O) =  2 (483.64 kJ mol1 ) = 241.82 kJ mol1 rU  
E2.41(b)
rH
 
=
+
(pV ) =
rU
 
+ RT
ng
= 772.7 kJ mol1 + (8.3145 103 kJ K1 mol1 ) (298 K) (5) = 760.3 kJ mol1 E2.42(b)
  1 1 1 (1) 2 N2 (g) + 2 O2 (g) + 2 Cl2 (g) NOCl(g) =? fH   (2) 2NOCl(g) 2NO(g) + Cl2 (g) = +75.5 kJ mol1 rH 1 1 (3) 2 N2 (g) + 2 O2 (g) NO(g) 1 (1) = (3)  2 (2) fH   fH  
= 90.25 kJ mol1
1 (NOCl, g) = 90.25 kJ mol1  2 (75.5 kJ mol1 )
= 52.5 kJ mol1 E2.43(b)
rH  
(100 C) 
rH
 
(25 C) =
100 C 25 C
 rH  T
dT =
100 C 25 C
r Cp,m dT
Because Cp,m can frequently be parametrized as Cp,m = a + bT + c/T 2 the indefinite integral of Cp,m has the form
1 Cp,m dT = aT + 2 bT 2  c/T
Combining this expression with our original integral, we have
rH  
(100 C) =
rH
 
1 (25 C) + (T r a + 2 T 2 r b 
r c/T )
373 K 298 K
Now for the pieces
rH ra rb rc  
(25 C) = 2(285.83 kJ mol1 )  2(0)  0 = 571.66 kJ mol1
= [2(75.29)  2(27.28)  (29.96)] J K 1 mol1 = 0.06606 kJ K1 mol1 = [2(0)  2(3.29)  (4.18)] 103 J K2 mol1 = 10.76 106 kJ K2 mol1 = [2(0)  2(0.50) (1.67)] 105 J K mol1 = 67 kJ K mol1
32
INSTRUCTOR'S MANUAL
rH
 
1 (100 C) = 571.66 + (373  298) (0.06606) + 2 (3732  2982 )
(10.76 106 )  (67) = 566.93 kJ mol1 E2.44(b) The hydrogenation reaction is (1) C2 H2 (g) + H2 (g) C2 H4 (g)
rH  C  (T)
1 1  373 298
kJ mol1
=?
The reactions and accompanying data which are to be combined in order to yield reaction (1) and  C  r H (T) are
1 (2) H2 (g) + 2 O2 (g) H2 O(l) cH  
(2) = 285.83 kJ mol1
cH   c H (3)  
(3) C2 H4 (g) + 3O2 (g) 2H2 O(l) + 2CO2 (g)
5 (4) C2 H2 (g) + 2 O2 (g) H2 O(l) + 2CO2 (g)
= 1411 kJ mol1
(4) = 1300 kJ mol1
reaction (1) = reaction (2)  reaction (3) + reaction (4) Hence, (a)
rH  C  (T)
=
cH
 
(2) 
cH
 
(3) +
cH
 
(4)
= {(285.83)  (1411) + (1300)} kJ mol1 = 175 kJ mol1
rU  C  (T)
=
rH
 C  (T) 
ng RT [26]
ng = 1
= (175 kJ mol1 + 2.48 kJ mol1 ) = 173 kJ mol1 (b)
rH  
(348 K) =
J
rH
 
(298 K) +
r Cp (348 K
 298 K)
[Example 2.7]
r Cp =
J Cp,m (J) [47] = Cp,m (C2 H4 , g)  Cp,m (C2 H2 , g)  Cp,m (H2 , g)
rH
 
= (43.56  43.93  28.82) 103 kJ K1 mol1 = 29.19 103 kJ K1 mol1 (348 K) = (175 kJ mol1 )  (29.19 103 kJ K1 mol1 ) (50 K) = 176 kJ mol1
E2.45(b)
The cycle is shown in Fig. 2.1.
   hyd H  (Ca2+ ) =  soln H  (CaBr2 )   + vap H  (Br2 ) + fH  
(CaBr2 , s) +
ion H
sub H  
 
(Ca)
diss H
 
(Br2 ) +
(Ca)
   + ion H  (Ca+ ) + 2 eg H  (Br) + 2 hyd H  (Br  )
= [(103.1)  (682.8) + 178.2 + 30.91 + 192.9 +589.7 + 1145 + 2(331.0) + 2(337)] kJ mol1 = 1587 kJ mol1 and
hyd H  
(Ca2+ ) = 1587 kJ mol1
THE FIRST LAW: THE CONCEPTS
33
Ionization
Dissociation Vaporization Br Sublimation Ca Formation
Electron gain Br
Hydration Br
Hydration Ca Solution
Figure 2.1
E2.46
(a) 2,2,4trimethylpentane has five C(H)3 (C) groups, one C(H)2 (C)2 group, one C(H)(C)3 group, and one C(C)4 group. (b) 2,2dimethylpropane has four C(H3 )(C) groups and one C(C)4 group. Using data from Table 2.7 (a) [5 (42.17) + 1 (20.7) + 1 (6.91) + 1 8.16] kJ mol1 = 230.3 kJ mol1 (b) [4 (42.17) + 1 8.16] kJ mol1 = 160.5 kJ mol1
Solutions to problems
Assume all gases are perfect unless stated otherwise. Unless otherwise stated, thermochemical data are for 298 K.
Solutions to numerical problems
P2.4 We assume that the solid carbon dioxide has already evaporated and is contained within a closed vessel of 100 cm3 which is its initial volume. It then expands to a final volume which is determined by the perfect gas equation. (a) w = pex V Vi = 100 cm3 = 1.00 104 m3 , Vf = nRT = p 5.0 g 44.01 g mol1 p = 1.0 atm = 1.013 105 Pa (8.206 102 L atm K1 mol1 ) (298 K) 1.0 atm = 2.78 L
= 2.78 103 m3 Therefore, w = (1.013 105 Pa) [(2.78 103 )  (1.00 104 )] m3 = 272 Pa m3 = 0.27 kJ
34
INSTRUCTOR'S MANUAL
(b)
w = nRT ln =
Vf [2.13] Vi (8.314 J K 1 mol1 ) (298 K) ln 2.78 103 m3 1.00 104 m3
5.0 g 44.01 g mol1
= (282) (ln 27.8) = 0.94 kJ P2.5 w = pex V [2.10] Hence w (pex ) Vf = nRT pex nRT pex Vi ; so V Vf
= nRT (1.0 mol) (8.314 J K 1 mol1 ) (1073 K) 8.9 kJ
Even if there is no physical piston, the gas drives back the atmosphere, so the work is also 8.9 kJ P2.7 The virial expression for pressure up to the second coefficient is p = w = RT Vm
f i
1+
B Vm
[1.22]
f i
p dV = n
RT Vm
1+
B Vm
dVm = nRT ln
Vf Vi
+ nBRT
1 1  Vmf Vmi
From the data, nRT = (70 103 mol) (8.314 J K 1 mol1 ) (373 K) = 217 J 5.25 cm3 6.29 cm3 = 75.0 cm3 mol1 , Vmf = = 89.9 cm3 mol1 70 mmol 70 mmol 1 1 1 1 = (28.7 cm3 mol1 )   and so B 3 mol1 Vmi Vmf 89.9 cm 75.0 cm3 mol1 Vmi = = 6.34 102 Therefore, w = (217 J) ln Since U = q + w and U+ 6.29 + (217 J) (6.34 102 ) = (39.2 J) + (13.8 J) = 25 J 5.25 q= U  w = (83.5 J) + (25 J) = +109 J B Vm , as T =0
U = +83.5 J,
H =
(pV ) with pV = nRT 1 Vm = nRTB
1+
(pV ) = nRTB
1 1  Vmf Vmi
= (217 J) (6.34 102 ) = 13.8 J Therefore, P2.8 qp = H = (83.5 J) + (13.8 J) = +97 J H = n vap H = +22.2 kJ
vap H
=
qp = n
18.02 g mol1 10 g
(22.2 kJ)
= +40 kJ mol1
THE FIRST LAW: THE CONCEPTS
35
U=
H
ng RT ,
ng =
10 g 18.02 g mol1
= 0.555 mol
Hence U = (22.2 kJ)  (0.555 mol) (8.314 J K1 mol1 ) (373 K) = (22.2 kJ)  (1.72 kJ) = +20.5 kJ w= P2.11 U  q [as U = q + w] = (20.5 kJ  22.2 kJ) = 1.7 kJ qp (methane) = n vap H = 32.5 kJ = 8.18 kJ mol1 (Table 2.3) nRT ; therefore p This is constantpressure process; hence qp (object) + qp (methane) = 0. qp (object) = 32.5 kJ n= qp (methane) vap H
vap H
The volume occupied by the methane gas at a pressure p is V = V = qRT p vap H =
(32.5 kJ) (8.314 J K 1 mol1 ) (112 K) (1.013 105 Pa) (8.18 kJ mol1 )
= 3.65 102 m3 = 36.5 L P2.14 Cr(C6 H6 )2 (s) Cr(s) + 2C6 H6 (g)
rH  
ng = +2 mol
=
rU
 
+ 2RT , from [26]
= (8.0 kJ mol1 ) + (2) (8.314 J K 1 mol1 ) (583 K) = +17.7 kJ mol1 In terms of enthalpies of formation
rH    
= (2)
fH
 
(benzene, 583 K) 
 
fH
 
(metallocene, 583 K)
or
rH
(metallocene, 583 K) = 2 f H
 
(C6 H6 , g, 583 K)  17.7 kJ mol1
The enthalpy of formation of benzene gas at 583 K is related to its value at 298 K by
fH  
(benzene, 583 K) =
fH
(benzene, 298 K) + (Tb  298 K)Cp (l) + (583 K  Tb )Cp (g)
 + vap H   6 (583 K  298 K)Cp (graphite)
3 (583 K  298 K)Cp (H2 , g) where Tb is the boiling temperature of benzene (353 K). We shall assume that the heat capacities of graphite and hydrogen are approximately constant in the range of interest, and use their values from Table 2.6
rH  
(benzene, 583 K) = (49.0 kJ mol1 ) + (353  298) K (136.1 J K 1 mol1 ) + (583  353) K (81.67 J K 1 mol1 ) + (30.8 kJ mol1 )  (6) (583  298) K (8.53 J K 1 mol1 )  (3) (583  298) K (28.82 J K 1 mol1 ) = {(49.0) + (7.49) + (18.78) + (30.8)  (14.59)  (24.64)} kJ mol1 = +66.8 kJ mol1
fH  
Therefore, for the metallocene,
(583 K) = (2 66.8  17.7) kJ mol1 = +116.0 kJ mol1
36
INSTRUCTOR'S MANUAL
P2.17
We must relate the formation of DyCl3 to the three reactions for which we have information Dy(s) + 1.5 Cl2 (g) DyCl3 (s) This reaction can be seen as a sequence of reaction (2), three times reaction (3), and the reverse of reaction (1), so
fH fH  
(DyCl3 , s) =
rH
 
 (2) + 3 r H  (3) 
rH
 
(1),
 
(DyCl3 , s) = [699.43 + 3(158.31)  (180.06)] kJ mol1 = 994.30 kJ mol1
       1 (SiH4 )  2 f H  (O2 )
P2.19
(a)
rH
=
fH
(SiH3 OH) 
fH
1 = [67.5  34.3  2 (0)] kJ mol1 = 101.8 kJ mol1
(b)
rH
 
= =
fH
 
(SiH2 O)  (SiH2 O) 
fH
 
(H2 O) 
fH
 
(SiH4 ) 
 
fH
 
(O2 )
= [23.5 + (285.83)  34.3  0] kJ mol (c) P2.21
rH   fH   fH  
1
= 344.2 kJ mol1 (H2 )
(SiH3 OH) 
1
fH
= [23.5  (67.5)  0] kJ mol
= 44.0 kJ mol1
When necessary we assume perfect gas behaviour, also, the symbols w, V , q, U , etc. will represent molar quantities in all cases. w= p dV =  C dV = C Vn dV Vn
For n = 1, this becomes (we treat the case n = 1 later) (1) w= = = w=
final state,Vf C C = V n+1 n1 n1 initial state,Vi
1 Vfn1

1 Vin1
pi Vin n1
1 Vfn1
 1
1 Vin1  1 Vin1
[because pV n = C]
pi Vi Vin1 n1 pi Vi n1
(2)
Vi Vf
Vfn1 n1
1 =
RTi n1
Vi n1 1 Vf
But pV n = C or V = (3)
C 1/n for n = 0 (we treat n = 0 as a special case below). So, p
n1
Vi n1 pf n = n=0 Vf pi Substitution of eqn 3 into eqn 2 and using `1' and `2' to represent the initial and final states, respectively, yields p2 p1
n1 n
RT1 (4) w = n1
1
for n = 0 and n = 1
THE FIRST LAW: THE CONCEPTS
37
In the case for which n = 0, eqn 1 gives w= C 01 1 Vf1  1 Vi1 = C(Vf  Vi )
= (pV 0 )any state (Vf  Vi ) = (p)any state (Vf  Vi ) (5) w = p V for n = 0, isobaric case In the case for which n = 1 w= p dV =  C dV =  Vn C Vf dV = C ln V Vi Vi Vf
w = (pV n )any state ln
Vi Vf Vi = (RT )any state ln Vf V1 V2 = RT ln
= (pV )any state ln
(6) w = RT ln
p2 p1
for n = 1, isothermal case U = q + w = CV (Tf  Ti ). So
(7)
To derive the equation for heat, note that, for a perfect gas, Tf p f Vf  1 = CV Ti 1 q + w = CV Ti Ti p i Vi Vin1 = CV Ti  1 [because pV n = C] Vfn1 = CV Ti q = CV Ti pf pi pf pi
n1 n n1 n
1
[using eqn 3 (n = 0)] pf pi 1 1
n1 n n1 n
RTi 1  n1 pf pi pf pi
n1 n
1
[using eqn 4 (n = 0, n = 1)]
= CV Ti  = =
RTi n1
CV 1  RTi R n1
n1 n
CV pf 1  RTi Cp  C V n1 pi CV 1 =  RTi Cp n1 C 1
V CV
1
n1 n
[2.31] 1
pf pi 1 1
n1 n
=
1 1  RTi 1 n1
pf pi pf pi
n1 n
[2.37]
(n  1)  (  1) = RTi (n  1) (  1) = n RTi (n  1) (  1)
n1 n
pf pi
1
38
INSTRUCTOR'S MANUAL
Using the symbols `1' and `2' this becomes (8) q= n RT1 (n  1) (  1) p2 p1
n1 n
1
for n = 0, n = 1
In the case for which n = 1, (the isothermal case) eqns 7 and 6 yield (9) q = w = RT ln V2 p1 = RT ln for n = 1, isothermal case V1 p2
In the case for which n = 0 (the isobaric case) eqns 7 and 5 yield q = U  w = CV (Tf  Ti ) + p(Vf  Vi )
= CV (Tf  Ti ) + R(Tf  Ti ) = (CV + R) (Tf  Ti ) = Cp (Tf  Ti ) (10) q = Cp T for n = 0, isobaric case A summary of the equations for the process pV n = C is given below
n 0 1 Any n except n = 0 and n = 1
w p V p2 RT ln p1 CV T 0 RT1 n1 p2 p1
n1 n
q Cp T p1 RT ln p2 0 CV T 1 n RT1 (n  1)(  1) p2 p1
Process type Isobaric [2.29] Isothermal [2.13]
n1 n
Adiabatic [2.33] Isochoric [2.22]
1
Equation 8 gives this result when n = p  2 RT1 q= p1 (  1) (  1) Therefore, w = U q = U = CV T.
1
1
=0
Equation 8 gives this result in the limit as n lim q = = = 1 p2 RT1 1 1 p1 CV Cp  CV CV Cp  CV RT1 p2  p1 p1
n
V1 (p2  p1 )
However, lim V = lim
n
n
C C = 0 = C. So in this limit an isochoric process is being discussed and p1/n p
V2 = V1 = V and lim q = CV CV (p2 V2  p1 V1 ) = (RT2  RT1 ) = CV Cp  CV R T
n
THE FIRST LAW: THE CONCEPTS
39
Solutions to theoretical problems
P2.23 dw = F (x) dx [2.6], with z = x Hence to move the mass from x1 to x2 w=
x2 x1
F (x) dx x a [F = constant] x1 Fa x2  cos cos a a 2F a Fa (cos  cos 0) = Fa (cos 2  cos 0) = 0
Inserting F (x) = F sin w = F
x2 x1
sin
x a
dx =
(a) x2 = a, (b) x2 = 2a,
x1 = 0, x1 = 0,
w=
w=
The work done by the machine in the first part of the cycle is regained by the machine in the second part of the cycle, and hence no net work is done by the machine. P2.25 (a) The amount is a constant; therefore, it can be calculated from the data for any state. In state A, VA = 10 L, pA = 1 atm, TA = 313 K. Hence n= (1.0 atm) (10 L) pA VA = 0.389 mol = RTA (0.0821 L atm K1 mol1 ) (313 K)
Since T is a constant along the isotherm, Boyle's law applies p A V A = pB V B ; VB = pA VA = pB 1.0 atm 20 atm (10 L) = 0.50 L VC = VB = 0.50 L
(b) Along ACB, there is work only from A C; hence w = pext V [10] = (1.0 105 Pa) (0.50  10) L (103 m3 L1 ) = 9.5 102 J Along ADB, there is work only from D B; hence w = pext V [10] = (20 105 Pa) (0.50  10) L (103 m3 L1 ) = 1.9 104 J (c) w = nRT ln VB 0.5 [13] = (0.389) (8.314 J K 1 mol1 ) (313 K) ln VA 10 = +3.0 103 J The work along each of these three paths is different, illustrating the fact that work is not a state property. (d) Since the initial and final states of all three paths are the same, U for all three paths is the same. Path AB is isothermal; hence U = 0 , since the gas is assumed to be perfect. Therefore, 3 U = 0 for paths ACB and ADB as well and the fact that CV ,m = 2 R is not needed for the solution.
40
INSTRUCTOR'S MANUAL
In each case, q =
U  w = w, thus for path ADB, q = 1.9 104 J ;
path ACB, q = 9.5 102 J ; path AB, q = 3.0 103 J
The heat is different for all three paths; heat is not a state property. P2.27 Since (a) U is independent of path U (A B) = q(ACB) + w(ACB) = 80 J  30 J = 50 J
U = 50 J = q(ADB) + w(ADB) q(ADB) = 50 J  (10 J) = +60 J
(b)
q(B A) = U (ADB) =
U (B A)  w(B A) = 50 J  (+20 J) = 70 J U (A D) + U (D B); 50 J = 40 J + U (D B)
The system liberates heat. (c) U (D B) = 10 J = q(D B) + w(D B); q(ADB) = 60 J[part a] = q(A D) + q(D B) 60 J = q(A D) + 10 J; P2.29 w=
V2 V1
w(D B) = 0, hence q(D B) = +10 J
q(A D) = +50 J
V2 V2 dV dV + n2 a 2 V  nb V1 V
p dV = nRT V2  nb V1  nb
V1
= nRT ln
 n2 a
1 1  V2 V1
By multiplying and dividing the value of each variable by its critical value we obtain w = nR Tr = T , Tc
V2 nb n2 a Vc Vc V  Vc  Tc ln Vc  nb 1 Vc V2 V1 Vc  Vc V 8a Vr = , Tc = , Vc = 3nb [Table 1.6] Vc 27Rb
T Tc
w=
8na 27b
(Tr ) ln
Vr,2  1 3 Vr,1 
1 3

na 3b
1 1  Vr,2 Vr,1 awr 3bw , then w = and a 3b
The van der Waals constants a and b can be eliminated by defining wr = 8 Vr,2  1/3 wr =  nTr ln 9 Vr,1  1/3 1 1  Vr,2 Vr,1
n
Along the critical isotherm, Tr = 1 and Vr,1 = 1, Vr,2 = x. Hence wr 8 3x  1 =  ln n 9 2  1 +1 x
THE FIRST LAW: THE CONCEPTS
41
Solutions to applications
P2.30 1.5 g (5645 kJ mol1 ) = 25 kJ 342.3 g mol (b) Effective work available is 25 kJ 0.25 = 6.2 kJ Because w = mgh, and m 65 kg 6.2 103 J = 9.7 m h 65 kg 9.81 m s2 (c) The energy released as heat is (a)
 q = n cH  =  q =  r H = n c H  = 
2.5 g 180 g mol1
(2808 kJ mol1 ) = 39 kJ
(d) If onequarter of this energy were available as work a 65 kg person could climb to a height h given by 1/4q = w = mgh P2.35 so h= q 39 103 J = = 15 m 4mg 4(65 kJ) (9.8 m s2 )
(a) and (b). The table displays computed enthalpies of formation (semiempirical, PM3 level, PC Spartan ProTM), enthalpies of combustion based on them (and on experimental enthalpies of formation of H2 O(l) and CO2 (g), 285.83 and 393.51 kJ mol1 respectively), experimental enthalpies of combustion from Table 2.5, and the relative error in enthalpy of combustion.
Compound CH4 (g) C2 H6 (g) C3 H8 (g) C4 H10 (g) C5 H12 (g)
1  f H /kJ mol 1  c H /kJ mol
(calc.)
54.45 75.88 98.84 121.60 142.11
910.72 1568.63 2225.01 2881.59 3540.42
1  c H /kJ mol
(expt.)
890 1560 2220 2878 3537
% error 2.33 0.55 0.23 0.12 0.10
The combustion reactions can be expressed as: Cn H2n+2 (g) + 3n + 1 2 O2 (g) n CO2 (g) + (n + 1) H2 O(1).
The enthalpy of combustion, in terms of enthalpies of reaction, is
cH     = n f H  (CO2 ) + (n + 1) f H  (H2 O)  fH   fH  
(Cn H2n+2 ),
where we have left out % error =
cH
(O2 ) = 0. The % error is defined as:
fH  (expt.) 
 (calc.)   fH
 (expt.) 
100%
The agreement is quite good. (c) If the enthalpy of combustion is related to the molar mass by
cH  
= k[M/(g mol1 )]n
then one can take the natural log of both sides to obtain:
 ln  c H   = ln k + n ln M/(g mol1 ).
42
INSTRUCTOR'S MANUAL
 Thus, if one plots ln  c H   vs. ln [M(g mol1 )], then one ought to obtain a straight line with slope n and yintercept ln k. Draw up the following table:
Compound CH4 (g) C2 H6 (g) C3 H8 (g) C4 H10 (g) C5 H12 (g)
M/(g mol1 ) 16.04 30.07 44.10 58.12 72.15
1  c H /kJ mol
890 1560 2220 2878 3537
ln M(g mol1 ) 2.775 3.404 3.786 4.063 4.279
ln 
1  c H /kJ mol 
6.81 7.358 7.708 7.966 8.172
The plot is shown below in Fig 2.2.
9 ln  cH / kJ mol1
8
7
6
2
3 ln M/(g mol1)
4
5
Figure 2.2
The linear leastsquares fit equation is:
 ln  c H  /kJ mol1  = 4.30 + 0.093 ln M/(g mol1 ) r 2 = 1.00
These compounds support the proposed relationships, with n = 0.903 and k = e4.30 kJ mol1 = 73.7 kJ mol1 . The aggreement of these theoretical values of k and n with the experimental values obtained in P2.34 is rather good. P2.37 In general, the reaction RH R + H has a standard enthalpy (the bond dissociation enthalpy) of
 H  (R H) = fH      
(R) +
fH fH
(H) 
fH fH
(RH)
so
fH
 
(R) =
 H  (R H) 
 
(H) +
 
(RH)
Since we are provided with bond dissociation energies, we need H = So and
rH  
U+
(pV ) =
rU    
U + RT
ng
(R H) = (R) =
rH
(R H) + RT
fH  
rH
 
(R H) + RT 
(H) +
fH
 
(RH)
THE FIRST LAW: THE CONCEPTS
43
Inserting the bond dissociation energies and enthalpies of formation from Tables 2.5 and 2.6, we obtain
fH fH    
(C2 H5 ) = (420.5 + 2.48  217.97  84.68) kJ mol1 = 120.3 kJ mol1 (secC4 H9 ) = (410.5 + 2.48  217.97  126.15) kJ mol1 = 68.9 kJ mol1
fH
 
(tertC4 H9 ) = (398.3 + 2.48  217.97  134.2) kJ mol1 = 48.1 kJ mol1
3
The First Law: the machinery
Solutions to exercises
Discussion questions
E3.1(b) The following list includes only those state functions that we have encountered in the first three chapters. More will be encountered in later chapters. Temperature, pressure, volume, amount, energy, enthalpy, heat capacity, expansion coefficient, isothermal compressibility, and JouleThomson coefficient. E3.2(b) One can use the general expression for T given in Justification 3.3 to derive its specific form for 2 a van der Waals gas as given in Exercise 3.14(a), that is, T = a/Vm . (The derivation is carried out in Example 5.1.) For an isothermal expansion in a van der Waals gas dUm = (a/Vm )2 . Hence Um = a(1/Vm,2  1/Vm,1 ). See this derivation in the solution to Exercise 3.14(a). This formula corresponds to what one would expect for a real gas. As the molecules get closer and closer the molar volume gets smaller and smaller and the energy of attraction gets larger and larger. The solution to Problem 3.23 shows that the JouleThomson coefficient can be expressed in terms of the parameters representing the attractive and repulsive interactions in a real gas. If the attractive forces predominate then expanding the gas will reduce its energy and hence its temperature. This reduction in temperature could continue until the temperature of the gas falls below its condensation point. This is the principle underlying the liquefaction of gases with the Linde Refrigerator which utilizes the JouleThomson effect. See Section 3.4 for a more complete discussion.
E3.3(b)
Numerical exercises
E3.4(b) A function has an exact differential if its mixed partial derivatives are equal. That is, f (x, y) has an exact differential if x (a) f y f x f y f s f t = y f x and f = 6x 2 y x f = 6x 2 y y f = es s f = es t
(b)
y = 2x 3 y and x = tes + 1 and t = 2t + es and s = 3x 2 y 2
Therefore, exact.
Therefore, exact.
E3.5(b) E3.6(b)
dz = (a) (b)
z z dx 2x dy dx + dy =  2 x y (1 + y) (1 + y)3 z z dx + dy = (3x 2  2y 2 ) dx  4xy dy x y
dz =
2z = (3x 2  2y 2 ) = 4y yx y 2z and = (4xy) = 4y xy x
THE FIRST LAW: THE MACHINERY
45
E3.7(b)
dz =
z z dx + dy = (2xy + y 2 ) dx + (x 2 + 2xy) dy x y
2z = (2xy + y 2 ) = 2x + 2y yx y and E3.8(b) 2z 2 = (x + 2xy) = 2x + 2y xy x Cp = p T Because p 2H H = T p pT T T H p T p
H = 0 for a perfect gas, its temperature derivative also equals zero; thus p T
Cp = 0. p T E3.9(b) E3.10(b)
V (H /V )p (U/V )p + p H p p = = = = 1+ U p (U/V )p (U/V )p (U/V )p (U/V )p (U +pV )
dp = d ln p = We express T = 
p dV + V T dp 1 = p p
p dT T V p dT T V
1 p dV + V T p
p in terms of the isothermal compressibility T V T
1 p p 1 V = V so = p T V T V T T V 1 p in terms of T and the expansion coefficient = T V V
1 V
We express p T V
V T p
T V p
V = 1 p T
so
(V /T )p p = = T V (V /p)T T dT  dV V
so d ln p =  E3.11(b) U=
1 1 + = pV T pT pT so U = 0 p T
3 nRT 2
by direct differentiation
3 5 H = U + pV = 2 nRT + nRT = 2 nRT ,
so E3.12(b)
H = 0 by direct differentiation p T 1 V nRT = V = V T p p = 1 V V T = 1 T
V V nR = = T p p T
46
INSTRUCTOR'S MANUAL
T =  T =  E3.13(b)
1 V 1 V
V p T  nRT p2
V nRT = 2 p T p = 1 p
The JouleThomson coefficient is the ratio of temperature change to pressure change under conditions of isenthalpic expansion. So = T p H T 10 K = 0.48 K atm1 = (1.00  22) atm p dUm = dUm dT + T Vm Um Vm dVm
E3.14(b)
Um = Um (T , Vm )
dT = 0 in an isothermal process, so dUm = Um = a Um dVm = 2 dVm Vm T Vm
Vm2
22.1 L mol dV a a 22.1 L mol m dVm = a = 2 2 Vm 1.00 L mol1 Vm1 Vm1 Vm 1.00 L mol1 Vm a a 21.1a = + = = 0.95475a L1 mol 1 1 22.1 L mol 1.00 L mol 22.1 L mol1
dUm =
Vm2
1
1
a = 1.337 atm L2 mol2 Um = (0.95475 mol L1 ) (1.337 atm L2 mol2 ) = (1.2765 atm L mol1 ) (1.01325 105 Pa atm1 ) = 129 Pa m3 mol1 = 129 J mol1 w= so w =  Thus q=+
22.1 L mol1 1.00 L mol1
1 m3 103 L
pex dVm RT Vm  b
and p = dVm +
RT a  2 Vm  b V m
for a van der Waals gas Um
a dVm = q + 2 Vm
RT Vm  b
dVm = +RT ln(Vm  b)
22.1 L mol1 1.00 L mol1
= +(8.314 J K1 mol1 ) (298 K) ln 22.1  3.20 102 1.00  3.20 102
= +7.7465 kJ mol1 w = 7747 J mol1 + 129 J mol1 = 7618 J mol1 = 7.62 kJ mol1
THE FIRST LAW: THE MACHINERY
47
E3.15(b)
The expansion coefficient is = = = 1 V V (3.7 104 K 1 + 2 1.52 106 T K2 ) V = T p V
V [3.7 104 + 2 1.52 106 (T /K)] K 1 V [0.77 + 3.7 104 (T /K) + 1.52 106 (T /K)2 ]
E3.16(b)
[3.7 104 + 2 1.52 106 (310)] K 1 = 1.27 103 K 1 0.77 + 3.7 104 (310) + 1.52 106 (310)2 Isothermal compressibility is V 1 V V so p=  T =  V p T V p V T A density increase 0.08 per cent means V /V = 0.0008. So the additional pressure that must be applied is p= 0.0008 = 3.6 102 atm 2.21 106 atm1
E3.17(b)
The isothermal JouleThomson coefficient is H = Cp = (1.11 K atm1 ) (37.11 J K 1 mol1 ) = 41.2 J atm1 mol1 p T If this coefficient is constant in an isothermal JouleThomson experiment, then the heat which must be supplied to maintain constant temperature is H in the following relationship H /n = 41.2 J atm1 mol1 p so H = (41.2 J atm1 mol1 )n p
H = (41.2 J atm1 mol1 ) (12.0 mol) (55 atm) = 27.2 103 J E3.18(b) The JouleThomson coefficient is = T p H T p so p= T 4.5 K = 3.4 102 kPa = 13.3 103 K kPa1
Solutions to problems
Assume that all gases are perfect and that all data refer to 298 K unless stated otherwise.
Solutions to numerical problems
P3.1 1 atm 1.013 105 Pa For the change of volume with pressure, we use T = (2.21 106 atm1 ) dV = = 2.18 1011 Pa1
V 1 dp[constant temperature] = T V dp T =  p T V V = T V p [If change in V is small compared to V ]
V p T
p = (1.03 103 kg m3 ) (9.81 m s2 ) (1000 m) = 1.010 107 Pa.
48
INSTRUCTOR'S MANUAL
Consequently, since V = 1000 cm3 = 1.0 103 m3 , V (2.18 1011 Pa1 ) (1.0 103 m3 ) (1.010 107 Pa) = 2.2 107 m3 , or 0.220 cm3 . V T p
For the change of volume with temperature, we use dV = 1 V dT [constant pressure] = V dT = T p V V = V T [if change in V is small compared to V ] (8.61 105 K 1 ) (1.0 103 m3 ) (30 K) 2.6 106 m3 , Overall, or  2.6 cm3 V 2.8 cm3 V = 997.2 cm3
Comment. A more exact calculation of the change of volume as a result of simultaneous pressure and temperature changes would be based on the relationship dV = V dp + p T V dT = T V dp + V dT p p
This would require information not given in the problem statement. P3.5 Use the formula derived in Problem 3.25. Cp,m  CV ,m = R which gives = 1 (3Vr  1)2 =1 4Vr3 Tr
Cp,m CV ,m + R R = =1+ CV ,m CV ,m CV ,m
3 In conjunction with CV ,m = 2 R for a monatomic, perfect gas, this gives = 1 + 2 3
For a van der Waals gas Vr =
Vm Vm T 27RbT , Tr = (Table 1.6) with a = = = Vc 3b Tc 8a 2 2 2 1 4.137 L atm mol and b = 5.16 10 L mol (Table 1.6). Hence, at 100 C and 1.00 atm, RT where Vm = 30.6 L mol1 p Vr Tr Hence [(3) (198)  (1)]2 1 =1 = 1  0.0088 = 0.9912, (4) (198)3 (1.29) (1) + 2 (1.009) = 1.67 3 = 1.009 30.6 L mol1 = 198 (3) (5.16 102 L mol1 ) (27) (8.206 102 L atm K1 mol1 ) (5.16 102 L mol1 ) (373 K) 1.29 (8) (4.317 L2 atm mol2 )
THE FIRST LAW: THE MACHINERY
49
Comment. At 100 C and 1.00 atm xenon is expected to be close to perfect, so it is not surprising that differs only slightly from the perfect gas value of 5 . 3 P3.7 See the solution to Problem 3.6. It does not matter whether the piston between chambers 2 and 3 is diathermic or adiabatic as long as the piston between chambers 1 and 2 is adiabatic. The answers are the same as for Problem 3.6. However, if both pistons are diathermic, the result is different. The solution for both pistons being diathermic follows. See Fig. 3.1.
Diathermic piston
Diathermic piston
Figure 3.1
Initial equilibrium state. n = 1.00 mol diatomic gas in each section pi = 1.00 bar Ti = 298 K For each section nRTi (1 mol) (0.083 145 L bar K1 mol1 ) (298 K) Vi = = pi 1.00 bar = 24.8 L Vtotal = 3Vi = 74.3 L = constant Final equilibrium state. The diathermic walls allow the passage of heat. Consequently, at equlibrium all chambers will have the same temperature T1 = T2 = T3 = 348 K. The chambers will also be at mechanical equlibrium so p1 = p2 = p3 = = (n1 + n2 + n3 )RT1 Vtotal
(3 mol) (0.083 145 L bar K1 mol1 ) (348 K) 74.3 L = 1.17 bar = p2 = p3
The chambers will have equal volume.
50
INSTRUCTOR'S MANUAL
V1 =
Vtotal = Vi = 24.8 L = V1 = V2 = V3 3 T1
5 = (1 mol) 2 (8.314 51 J K 1 mol1 ) (348 K  298 K)
5 U1 = n1 CV T1 = n1 2 R
U1 = 1.04 kJ =
U2 =
U3 Utotal
Utotal = 3 U1 = 3.12 kJ =
Solutions to theoretical problems
P3.11 dw = w dx + x y,z w dy + y x,z w dz z x,y
dw = (y + z) dx + (x + z) dy + (x + y) dz This is the total differential of the function w, and a total differential is necessarily exact, but here we will demonstrate its exactness showing that its integral is independent of path. Path a dw = 2x dx + 2y dy + 2z dz = 6x dx
(1,1,1) (0,0,0) 1 0
dw =
6x dx = 3
Path b dw = 2x 2 dx + (y 1/2 + y) dy + (z1/2 + z) dz = (2x 2 + 2x + 2x 1/2 ) dx
(1,1,1) (0,0,0) 1 0
dw =
(2x 2 + 2x + 2x 1/2 ) dx =
2 4 +1+ =3 3 3
Therefore, dw is exact. P3.12 U = U (T , V ) U U dU = dT + dV = CV dT + T V V T For U = constant, dU = 0, and CV dT =  U dV V T or CV = 
U dV V T dV U = dT U V T V T U
U V T
This relationship is essentially the permuter [Relation 3, Further information 1.7]. P3.13 H = H (T , p) H dH = dT + T p
H dp = Cp dT + p T For H = constant, dH = 0, and H dp = Cp dT p T H dT = Cp = Cp p T dp H
H dp p T
T = Cp = Cp p H
This relationship is essentially the permuter [Relation 3, Further information 1.7].
THE FIRST LAW: THE MACHINERY
51
P3.16
The reasoning here is that an exact differential is always exact. If the differential of heat can be shown to be inexact in one instance, then its differential is in general inexact, and heat is not a state function. Consider the cycle shown in Fig. 3.2.
1 A
4
Isotherm at B 2 3 Isotherm at
Isotherm at
Figure 3.2 The following perfect gas relations apply at points labelled 1, 2, 3 and 4 in Fig. 3.2. (1) p1 V1 = p2 V2 = nRT , Define T =T T , Subtract (2) from (1) nRT + nRT = p2 V1 + p1 V1 giving V1 (p1  p2 ) RT Subtracting (1) from (3) we obtain T = T = V2 (p1  p2 ) RT T = T (2) p2 V1 = nRT , (3) p1 V2 = nRT T =T T
Since V1 = V2 ,
qA = Cp T  CV T = (Cp  CV ) T qB = CV T + Cp T = (Cp  CV ) T giving qA = qB and q(cycle) = qA  qB = 0. Therefore P3.18 dq = 0 and dq is not exact. a RT  2 p = p(T , V ) = Vm  b V m p dT + T V p dV V T
dp =
52
INSTRUCTOR'S MANUAL
In what follows adopt the notation Vm = V . p R ; = T V V b then, dp = R V b dT + p RT 2a = + 3 V T (V  b)2 V RT 2a  V3 (V  b)2 dV V is more readily evaluated with the use T p
Because the van der Waals equation is a cubic in V , of the permuter.
R T V V V b = = p 2a T p  (VRT 2 + V 3 V T b) p
=
RV 3 (V  b) RT V 3  2a(V  b)2
For path a
T2 ,V2 T1 ,V1
dp = =
T2 T1
V2 R RT2 2a  + 3 dT + V1  b (V  b)2 V V1
dV 1
2 V2
R RT2 RT2 (T2  T1 ) +  a V1  b (V2  b) (V1  b) RT1 RT2 + a V1  b V 2  b
V2 V1

1
2 V1
= For path b
T2 ,V2 T1 ,V1
1
2 V2

1
2 V1
dp = =

T2 RT1 R 2a + 3 dV + dT V2  b (V  b)2 V T1
RT1 RT1  a V 2  b V1  b RT1 RT2 + a V1  b V 2  b
1
2 V2

1
2 V1
+
R (T2  T1 ) V2  b
=
1
2 V2

1
2 V1
Thus, they are the same and dp satisfies the condition of an exact differential, namely, that its integral between limits is independent of path. P3.20 p = p(V , T ) Therefore, dp = p dV + V T p dT T V with p = n2 a nRT  2 [Table 1.6] V  nb V n2 a V3 V  2nb V  nb
p nRT 2n2 a p + = + = V T V  nb (V  nb)2 V3 p p n2 a nR = + = T V V  nb T TV2
THE FIRST LAW: THE MACHINERY
53
Therefore, upon substitution dp = = p dV V  nb + n2 a V3 (V  2nb) dV + dV V  nb + p dT T dT + n2 a V2 dT T
(n2 a) (V  nb)/V 3  p V  nb
3 a(Vm  b)/Vm  p Vm  b
p + n2 a/V 2 T dT
=
dVm +
2 p + a/Vm T
Comment. This result may be compared to the expression for dp obtained in Problem 3.18. P3.21 p= nRT n2 a  2 (Table 1.6) V  nb V p (V  nb) + nR na RV 2 1
p T V
Hence T =
(V  nb)
T V  nb Vm  b = = = nR p V R
For Euler's chain relation, we need to show that Hence, in addition to T and p V
T p V
p V T
V = 1 T p V = T p (V  nb) 1
T V p
p [Problem 3.20] we need V T na RV 2   2na RV 3 2na RV 3
which can be found from
T p = + V p nR = T V  nb
(V  nb)
Therefore, T p V p V T V = T p
p T p V V T T V p V nb nR T V nb nRT (V nb)2 + 2n 3a V
2
=
2na  RV 3 (V  nb)
=
T V nb T V nb
2na + RV 3 (V  nb) 2na  RV 3 (V  nb)
= 1 P3.23 Cp = T V V = T p T
T V p
 V [Relation 2, Further information 1.7]
T 2na T  = (V  nb) [Problem 3.21] V p V  nb RV 3
54
INSTRUCTOR'S MANUAL
Introduction of this expression followed by rearrangement leads to Cp = (2na) (V  nb)2  nbRT V 2 V RT V 3  2na(V  nb)2 RT V 3 to simplify the appearance of the expression 2na(V  nb)2 V =
b 1  Vm
Then, introducing = 1  nb V 1
Cp =
1
V
For xenon, Vm = 24.6 L mol1 , T = 298 K, a = 4.137 L2 atm mol2 , b = 5.16 102 L mol1 , b 5.16 102 L mol1 nb = = = 2.09 103 V Vm 24.6 L mol1 = (8.206 102 L atm K1 mol1 ) (298 K) (24.6 L mol1 )3 = 73.0 (2) (4.137 L2 atm mol2 ) (24.6 L mol1  5.16 102 L mol1 )2 1  (73.0) (2.09 103 ) (24.6 L mol1 ) = 0.290 L mol1 72.0
Therefore, Cp =
Cp = 20.79 J K1 mol1 [Table 2.6], so = 0.290 103 m3 mol1 0.290 L mol1 = 20.79 J K1 mol1 20.79 J K1 mol1
= 1.393 105 K m3 J1 = 1.393 105 K Pa1 = (1.393 105 ) (1.013 105 K atm1 ) = 1.41 K atm1 nb changes sign (  1 The value of changes at T = T1 and when the sign of the numerator 1  V is positive). Hence b = 1 at T = T1 Vm that is, T1 = 2a Rb or RT1 bV 3 =1 2na(V  nb)2 Vm implying that T1 = 2a(Vm  b)2 2 RbVm
1
b 2 b 2 27 Tc 1  = Vm 4 Vm
For xenon,
2a (2) (4.137 L2 atm mol2 ) = = 1954 K 2 L atm K 1 mol1 ) (5.16 102 L mol1 ) Rb (8.206 10 5.16 102 24.6
2
and so T1 = (1954 K) 1 
= 1946 K
Question. An approximate relationship for of a van der Waals gas was obtained in Problem 3.17. Use it to obtain an expression for the inversion temperature, calculate it for xenon, and compare to the result above.
THE FIRST LAW: THE MACHINERY
55
P3.25
Cp,m  CV ,m =
2 T V [3.21] = T V T
p [Justification 3.3] T V
p nR = [Problem 3.20] T V V  nb V 1 V = = T T p
V p
Substituting, T
p T V T V p
Cp,m  CV ,m =
so
2na T T  = (V  nb) [Problem 3.21] V p V  nb RV 3
Substituting, Cp,m  CV ,m = For molar quantities, Cp,m  CV ,m = R with 2a(Vm  b)2 1 =1 3 RT Vm 8a , Vc = 3b. 27Rb
nRT (V nb) T (V nb)

2na RV 3
(V  nb)
= nR
with =
1 1
2na RT V 3
(V  nb)2
Now introduce the reduced variables and use Tc = After rearrangement, 1 (3Vr  1)2 =1 4Tr Vr3
For xenon, Vc = 118.1 cm3 mol1 , Tc = 289.8 K. The perfect gas value for Vm may be used as any 1 error introduced by this approximation occurs only in the correction term for . Hence, Vm 2.45 L mol1 , Vc = 118.8 cm3 mol1 , Tc = 289.8 K, and Vr = 20.6 and Tr = 1.03; therefore 1 (61.8  1)2 = 0.90, =1 (4) (1.03) (20.6)3 and Cp,m  CV ,m 1.1R = 9.2 J K1 mol1 P3.27 (a) = Vm = 1 Cp H 1 = p T Cp T Vm  Vm T p [Justification 3.1 and Problem 3.24] giving 1.1
RT + aT 2 p R Vm = + 2aT T p p
56
INSTRUCTOR'S MANUAL
= = (b)
1 Cp aT 2 Cp
RT RT + 2aT 2   aT 2 p p
CV = Cp  T Vm = Cp  T But, p =
RT Vm  aT 2 p R RT (2aT ) =  2 T V (Vm  aT 2 )2 Vm  aT R 2aRT 2 = + (RT /p) (RT /p)2 2ap 2 p = + T R Therefore R p 2ap 2 CV = Cp  T + 2aT + p T R = Cp  RT p 1+ 2apT R 1+ 2apT R p T
p T V p Vm T p T V
CV = Cp  R 1 +
2apT 2 R
Solutions to additional problems
P3.29 (a) The JouleThomson coefficient is related to the given data by = (1/Cp )(H /p)T = (3.29 103 J mol1 MPa1 )/(110.0 J K1 mol1 ) = 29.9 K MPa1 (b) The JouleThomson coefficient is defined as = (T /p)H ( T / p)H Assuming that the expansion is a JouleThomson constantenthalpy process, we have T = p = (29.9 K MPa1 ) [(0.5  1.5) 101 MPa] = 2.99 K
4
The Second Law: the concepts
Solutions to exercises
Discussion questions
E4.1(b) Trouton's rule is that the ratio of the enthalpy of vaporization of a liquid to its boiling point is a constant. Energy in the form of heat (enthalpy) supplied to a liquid manifests itself as turbulent motion (kinetic energy) of the molecules. When the kinetic energy of the molecules is sufficient to overcome the attractive energy that holds them together the liquid vaporizes. The enthalpy of vaporization is the heat energy (enthalpy) required to accomplish this at constant pressure. It seems reasonable that the greater the enthalpy of vaporization, the greater the kinetic energy required, and the greater the temperature needed to achieve this kinetic energy. Hence, we expect that vap H Tb , which implies that their ratio is a constant. The device proposed uses geothermal heat (energy) and appears to be similar to devices currently in existence for heating and lighting homes. As long as the amount of heat extracted from the hot source (the ground) is not less than the sum of the amount of heat discarded to the surroundings (by heating the home and operating the steam engine) and of the amount of work done by the engine to operate the heat pump, this device is possible; at least, it does not violate the first law of thermodynamics. However, the feasability of the device needs to be tested from the point of view of the second law as well. There are various equivalent versions of the second law, some are more directly useful in this case than others. Upon first analysis, it might seem that the net result of the operation of this device is the complete conversion of heat into the work done by the heat pump. This work is the difference between the heat absorbed from the surroundings and the heat discharged to the surroundings, and all of that difference has been converted to work. We might, then, conclude that this device violates the second law in the form stated in the introduction to Chapter 4; and therefore, that it cannot operate as described. However, we must carefully examine the exact wording of the second law. The key words are "sole result." Another slightly different, though equivalent, wording of Kelvin's statement is the following: "It is impossible by a cyclic process to take heat from a reservoir and convert it into work without at the same time transferring heat from a hot to a cold reservoir." So as long as some heat is discharged to surroundings colder than the geothermal source during its operation, there is no reason why this device should not work. A detailed analysis of the entropy changes associated with this device follows.
E4.2(b)
Environment at Tc Pump
Flow Flow
"ground" water at Th
Figure 4.1
CV and Cp are the temperature dependent heat capacities of water
58
INSTRUCTOR'S MANUAL
Three things must be considered in an analysis of the geothermal heat pump: Is it forbidden by the first law? Is it forbidden by the second law? Is it efficient? Etot = Ewater + Eground + Eenvironment Ewater = 0 Eground = CV (Th ){Th  Tc } Eenvironment = CV (Th ){Th  Tc } adding terms, we find that and Tc . Stot = Swater + Swater = 0 Sground = qground Cp (Th ){Th  Tc } = Th Th Cp (Tc ){Th  Tc } qenvironment Senvironment = = Tc Tc Cp (Tc ) = Cp , we find that 1 1  Tc Th Etot = 0 which means that the first law is satisfied for any value of Th Senvironment
Sground +
adding terms and estimating that Cp (Th ) Stot = Cp {Th  Tc }
This expression satisfies the second law ( Stot > 0) only when Th > Tc . We can conclude that, if the proposal involves collecting heat from environmentally cool ground water and using the energy to heat a home or to perform work, the proposal cannot succeed no matter what level of sophisticated technology is applied. Should the "ground" water be collected from deep within the Earth so that Th > Tc , the resultant geothermal pump is feasible. However, the efficiency, given by eqn 4.11, must be high to compete with fossil fuels because high installation costs must be recovered during the lifetime of the apparatus. Erev = 1  Tc Th
with Tc 273 K and Th = 373 K (the highest value possible at 1 bar), Erev = 0.268. At most, about 27% of the extracted heat is available to do work, including driving the heat pump. The concept works especially well in Iceland where geothermal springs bring boiling water to the surface. E4.3(b) See the solution to exercises 4.3 (a).
Numerical exercises
E4.4(b) (a) (b) E4.5(b) S= q dqrev = T T S= 50 103 J = 1.8 102 J K1 273 K 50 103 J = 1.5 102 J K1 S= (70 + 273) K S where dqrev = T CV ,m dT Tf = CV ,m ln T Ti
At 250 K, the entropy is equal to its entropy at 298 K plus S=
THE SECOND LAW: THE CONCEPTS
59
so S = 154.84 J K1 mol1 + [(20.786  8.3145) J K 1 mol1 ] ln S = 152.65 J K1 mol1 E4.6(b)
250 K 298 K
E4.7(b)
Cp,m dT Tf = Cp,m ln T Ti 5 (100 + 273) K = 9.08 J K1 S = (1.00 mol) + 1 (8.3145 J K 1 mol1 ) ln 273 K 2 However the change occurred, S has the same value as if the change happened by reversible heating at constant pressure (step 1) followed by reversible isothermal compression (step 2) S= dqrev = T S= S1 + S2
For the first step S1 = Cp,m dT Tf = Cp,m ln T Ti (135 + 273) K 7 = 18.3 J K 1 (8.3145 J K 1 mol1 ) ln S1 = (2.00 mol) (25 + 273) K 2 dqrev = T
and for the second S2 = dqrev qrev = T T p dV = nRT ln Vf pi = nRT ln Vi pf
where qrev = w = so S2 = nR ln
pi 1.50 atm = 25.6 J K 1 = (2.00 mol) (8.3145 J K 1 mol1 ) ln pf 7.00 atm
S = (18.3  25.6) J K 1 = 7.3 J K1 The heat lost in step 2 was more than the heat gained in step 1, resulting in a net loss of entropy. Or the ordering represented by confining the sample to a smaller volume in step 2 overcame the disordering represented by the temperature rise in step 1. A negative entropy change is allowed for a system as long as an increase in entropy elsewhere results in Stotal > 0. E4.8(b) q = qrev = 0 (adiabatic reversible process) dqrev = 0 T i U = nCV ,m T = (2.00 mol) (27.5 J K 1 mol1 ) (300  250) K S= = 2750 J = +2.75 kJ w= U  q = 2.75 kJ  0 = 2.75 kJ H = nCp,m T Cp,m = CV ,m + R = (27.5 J K1 mol1 + 8.314 J K 1 mol1 ) = 35.814 J K1 mol1 So H = (2.00 mol) (35.814 J K1 mol1 ) (+50 K) = 3581.4 J = 3.58 kJ
f
60
INSTRUCTOR'S MANUAL
E4.9(b)
However the change occurred, S has the same value as if the change happened by reversible heating at constant volume (step 1) followed by reversible isothermal expansion (step 2) S= S1 + S2
For the first step S1 = Tf CV ,m dT CV ,m = Cp,m  R = CV ,m ln T Ti 3 700 K = (3.50 mol) (8.3145 J K 1 mol1 ) ln = 44.9 J K 1 2 250 K dqrev = T
and for the second S2 = qrev dqrev = T T p dV = nRT ln Vf , Vi
where qrev = w = so S2 = nR ln
pi 60.0 L = 32.0 J K1 = (3.50 mol) (8.3145 J K 1 mol1 ) ln pf 20.0 L
E4.10(b)
S = 44.9 + 32.0 J K 1 = 76.9 J K1 qrev S= If reversible q = qrev T qrev = T S = (5.51 J K1 ) (350 K) = 1928.5 J q = 1.50 kJ = 19.3 kJ = qrev q = qrev ; therefore the process is not reversible (a) The heat flow is q = Cp T = nCp,m T = 2.75 kg 63.54 103 kg mol1 (24.44 J K 1 mol1 ) (275  330) K
E4.11(b)
= 58.2 103 J (b) S= Cp dT dqrev Tf = = nCp,m ln T T Ti 2.75 kg 275 K (24.44 J K 1 mol1 ) ln = = 193 J K1 3 kg mol1 330 K 63.54 10 dqrev Vf pi qrev = nRT ln = where qrev = w = nRT ln T T Vi pf pi = pf 35 g 28.013 g mol1 (8.3145 J K 1 mol1 ) ln 21.1 atm = 17 J K1 4.3 atm
E4.12(b) so
S=
S = nR ln
THE SECOND LAW: THE CONCEPTS
61
E4.13(b)
qrev dqrev Vf = where qrev = w = nRT ln T T Vi Vf S so S = nR ln and Vf = Vi exp Vi nR We need to compute the amount of gas from the perfect gas law S= pV = nRT so n = pV (1.20 atm) (11.0 L) = 0.596 mol = RT (0.08206 L atm K1 mol1 ) (270 K) = 6.00 L
So Vf = (11.0 L) exp E4.14(b)
3.0 J K 1 (0.596 mol) (8.3145 J K 1 mol1 )
Find the final temperature by equating the heat lost by the hot sample to the heat gained by the cold sample. n1 Cp,m (Tf  Ti1 ) = n2 Cp,m (Tf  Ti2 ) Tf =
1 (m1 Ti1 + m2 Ti2 ) n1 Ti1 + n2 Ti2 = M 1 n1 + n 2 M (m1 + m2 ) m1 Ti1 + m2 Ti2 = m1 + m 2 (25 g) (323 K) + (70 g) (293 K) = = 300.9 K 25 g + 70 g
S=
S1 +
S2 = n1 Cp,m ln =
Tf Ti1
+ n2 Cp,m ln
Tf Ti2 Cp,m
25 g 300.9 300.9 70 g ln ln + 1 1 323 293 46.07 g mol 46.07 g mol
= 3.846 102 + 4.043 102 Cp,m = (0.196 102 mol) (111.5 J K 1 mol1 ) = 0.2 J K1 E4.15(b) Htotal = 0 in an isolated container. Since the masses are equal and the heat capacity is assumed constant, the final temperature will be the average of the two initial temperatures
1 Tf = 2 (200 C + 25 C) = 112.5 C nCm = mCs where Cs is the specific heat capacity
S = mCs ln
Tf Ti
200 C = 473.2 K; 25 C = 298.2 K; 112.5 C = 385.7 K 385.7 298.2 385.7 473.2 = 115.5 J K1 = 91.802 J K1
S1 = (1.00 103 g) (0.449 J K 1 g1 ) ln S2 = (1.00 103 g) (0.449 J K 1 g1 ) ln Stotal = S1 + S2 = 24 J K1
62
INSTRUCTOR'S MANUAL
E4.16(b)
(a) (b)
q = 0 [adiabatic] w = pex V = (1.5 atm) 1.01 105 Pa atm (100.0 cm2 ) (15 cm) 1 m3 106 cm3
= 227.2 J = 230 J (c) (d) U = q + w = 0  230 J = 230 J U = nCV ,m T T = U 227.2 J = nCV ,m (1.5 mol) (28.8 J K 1 mol1 ) = 5.3 K (e) S = nCV ,m ln Tf Vi Vf Tf + nR ln Ti Vi = 288.15 K  5.26 K = 282.9 K nRT (1.5 mol) (8.206 102 L atm K1 mol1 ) (288.2 K) = = pi 9.0 atm = 3.942 L Vf = 3.942 L + (100 cm2 ) (15 cm) = 3.942 L + 1.5 L = 5.44 L S = (1.5 mol) (28.8 J K1 mol1 ) ln 282.9 288.2 5.44 3.942 1L 1000 cm3
+ (8.314 J K1 mol1 ) ln
= 1.5 mol(0.5346 J K1 mol1 + 2.678 J K1 mol1 ) = 3.2 J K1 E4.17(b) 35.27 103 J mol1 = + 104.58 J K1 = 104.6 J K1 Tb (64.1 + 273.15) K (b) If vaporization occurs reversibly, as is generally assumed (a)
vap S  
=
vap H
 
=
Ssys + E4.18(b) (a)
rS  
Ssur = 0
so
Ssur = 104.6 J K1
        = Sm (Zn2+ , aq) + Sm (Cu, s)  Sm (Zn, s)  Sm (Cu2+ , aq)
= [112.1 + 33.15  41.63 + 99.6] J K 1 mol1 = 21.0 J K1 mol1 (b)
rS           = 12Sm (CO2 , g) + 11Sm (H2 O, l)  Sm (C12 H22 O11 , s)  12Sm (O2 , g)
= [(12 213.74) + (11 69.91)  360.2  (12 205.14)] J K 1 mol1 = + 512.0 J K1 mol1 E4.19(b) (a)
rH       2+ f H (Cu , aq) 1
=
fH
(Zn2+ , aq) 
= 153.89  64.77 kJ mol
rG  
= 218.66 kJ mol1
= 218.66 kJ mol1  (298.15 K) (21.0 J K 1 mol1 ) = 212.40 kJ mol1
THE SECOND LAW: THE CONCEPTS
63
(b) E4.20(b)
rH rG
     
= =
cH
 
= 5645 kJ mol1
  2+ f G (Cu , aq) 1
= 5645 kJ mol1  (298.15 K) (512.0 J K 1 mol1 ) = 5798 kJ mol1
fG  
(a) (b)
rG
(Zn2+ , aq) 
= 147.06  65.49 kJ mol
rG  
= 212.55 kJ mol1
fG  
  = 12 f G (CO2 , g) + 11 f G (H2 O, l) 
(C12 H22 O11 , s)
= [12 (394.36) + 11 (237.13)  (1543)] kJ mol1 = 5798 kJ mol1 Comment. In each case these values of 4.19(b). E4.21(b) CO(g) + CH3 OH(l) CH3 COOH(l)
rH   rG  
agree closely with the calculated values in Exercise
=
 J f H  (J)
= 484.5 kJ mol1  (238.66 kJ mol1 )  (110.53 kJ mol1 ) = 135.31 kJ mol1
rS  
=
 J S  (J)
= 159.8 J K1 mol1  126.8 J K 1 mol1  197.67 J K 1 mol1 = 164.67 J K1 mol1
rG  
=
rH
 
  T rS
= 135.31 kJ mol1  (298 K) (164.67 J K1 mol1 ) = 135.31 kJ mol1 + 49.072 kJ mol1 = 86.2 kJ mol1 E4.22(b) The formation reaction of urea is
1 C(gr) + 2 O2 (g) + N2 (g) + 2H2 (g) CO(NH2 )2 (s)
The combustion reaction is
3 CO(NH2 )2 (s) + 2 O2 (g) CO2 (g) + 2H2 O(l) + N2 (g) cH fH
=
 
fH
 
 (CO2 , g) + 2 f H  (H2 O, l)  fH  
(CO(NH2 )2 , s) =
(CO2 , g) + 2
= 393.51 kJ mol1 + (2) (285.83 kJ mol = 333.17 kJ mol1
fS  
  f H (CO(NH2 )2 , s)   f H (H2 O, l)  c H (CO(NH2 )2 , s) 1
)  (632 kJ mol1 )
         1  = Sm (CO(NH2 )2 , s)  Sm (C, gr)  2 Sm (O2 , g)  Sm (N2 , g)  2Sm (H2 , g) 1 = 104.60 J K1 mol1  5.740 J K 1 mol1  2 (205.138 J K1 mol1 )
 191.61 J K1 mol1  2(130.684 J K 1 mol1 ) = 456.687 J K1 mol1
fG  
=
fH
 
  T f S
= 333.17 kJ mol1  (298 K) (456.687 J K1 mol1 ) = 333.17 kJ mol1 + 136.093 kJ mol1 = 197 kJ mol1
64
INSTRUCTOR'S MANUAL
E4.23(b)
(a)
S(gas) = nR ln
Vf Vi
=
21 g 39.95 g mol1
(8.314 J K 1 mol1 ) ln 2
= 3.029 J K1 = 3.0 J K1 S(surroundings) =  S(gas) = 3.0 J K1 [reversible] S(total) = 0 (b) (Free expansion) S(gas) = +3.0 J K1 S(surroundings) = 0 S(total) = +3.0 J K1 (c) qrev = 0 so S(gas) = 0 [No heat is transfered to the surroundings] [S is a state function] [no change in surroundings]
S(surroundings) = 0 S(total) = 0 E4.24(b)
Because entropy is a state function, we can choose any convenient path between the initial and final states. Choose isothermal compression followed by constantvolume heating S = nR ln Vf Vi + nCV ,m ln Tf Ti
= nR ln 3 + nCV ,m ln 3 = n(CV ,m  R) ln 3
3 S = 2 nR ln 3 5 CV ,m = 2 R for a diatomic perfect gas
E4.25(b)
C3 H8 (g) + 5O2 (g) 3CO2 (g) + 4H2 O(l)
rG      = 3 f G (CO2 , g) + 4 f G (H2 O, l)  f G (C3 H8 , g)  0 = 3(394.36 kJ mol1 ) + 4(237.13 kJ mol1 )  1(23.49 kJ mol1 )
= 2108.11 kJ mol1 The maximum nonexpansion work is 2108.11 kJ mol1 since we  =  r G Tc (a) =1 Th 500 K =1 = 0.500 1000 K (b) (c) Maximum work = qh  = (0.500) (1.0 kJ) = 0.50 kJ max = rev and wmax  = qh   qc,min  qc,min  = qh   wmax  = 1.0 kJ  0.50 kJ = 0.5 kJ
E4.26(b)
THE SECOND LAW: THE CONCEPTS
65
Solutions to problems
Assume that all gases are perfect and that data refer to 298 K unless otherwise stated.
Solutions to numerical problems
P4.1 (a) Because entropy is a state function following cycle H2 O(1, 0 C) S1        
S(1s,5 C)
trs S(1s,0 C)
trs S(l
s, 5 C) may be determined indirectly from the
H2 O(s, 0 C)
Ss
H2 O(1, 5 C)       H2 O(s, 5 C)     
trs
s, 5 C) = Sl + trs S(l s, 0 C) + Ss Tf [f = 0 C, = 5 C] Sl = Cp,m (l) ln T T Ss = Cp,m (s) ln Tf T with Cp = Cp,m (l)  Cp,m (s) = +37.3 J K1 mol1 Sl + Ss =  Cp ln Tf  fus H trs S(l s, Tf ) = Tf
trs S(l
Thus,
trs S(l
s, T ) =
T  fus H  Cp ln Tf Tf 6.01 103 J mol1 268 =  (37.3 J K 1 mol1 ) ln 273 273 K = 21.3 J K1 mol1
Ssur =
fus H (T )
fus H (T )
T =  Hl + =
fus H (Tf ) 
Hs Cp (Tf  T )
Hl +
Hs = Cp,m (l)(Tf  T ) + Cp,m (s)(T  Tf ) =
fus H (Tf )  fus H (T )
fus H (T )
Cp (Tf  T )
fus H (Tf )
Thus,
Ssur =
(T  Tf ) T T T 268  273 6.01 kJ mol1 + (37.3 J K 1 mol1 ) = 268 K 268 = + Cp = +21.7 J K1 mol1
Stotal = (21.7  21.3) J K 1 mol1 = +0.4 J K1 mol1 Since Stotal > 0, the transition l s is spontaneous at 5 C (b) A similar cycle and analysis can be set up for the transition liquid vapour at 95 C. However, since the transformation here is to the high temperature state (vapour) from the low temperature
66
INSTRUCTOR'S MANUAL
state (liquid), which is the opposite of part (a), we can expect that the analogous equations will occur with a change of sign.
trs S(l
g, T ) = =
trs S(l vap H
g, Tb ) + + Cp ln T , Tb
Cp ln
T Tb
Tb
Cp = 41.9 J K1 mol1
trs S(l
g, T ) =
40.7 kJ mol1 368  (41.9 J K 1 mol1 ) ln 373K 373 Cp (T  Tb ) T 368  373 368
= +109.7 J K1 mol1 Ssur = =  vap H (T ) vap H (Tb ) =  T T 40.7 kJ mol1 368 K
 (41.9 J K 1 mol1 )
= 111.2 J K1 mol1 Stotal = (109.7  111.2) J K 1 mol1 = 1.5 J K1 mol1 Since P4.2 Sm = Stotal < 0, the reverse transition, g l, is spontaneous at 95 C.
T2 T2 a + bT Cp,m dT T2 + b(T2  T1 ) [19] = dT = a ln T T T1 T1 T1 a = 91.47 J K1 mol1 , b = 7.5 102 J K2 mol1 300 K + (0.075 J K 2 mol1 ) (27 K) Sm = (91.47 J K1 mol1 ) ln 273 K
= 10.7 J K1 mol1 Therefore, for 1.00 mol, P4.8
Process (a) Process (b) Process (c) S +5.8 J K +5.8 J K 1 +3.9 J K 1
1
S = +11 J K1
Ssur 5.8 J K 1.7 J K 1 0
1
H 0 0 8.4 102 J
T 0 0 41 K
A 1.7 kJ 1.7 kJ ?
G 1.7 kJ 1.7 kJ ?
Process (a) H = T =0 S+ Vf Vi [isothermal process in a perfect gas] Ssurr 20 L 10 L = +5.8 J K1 Stot = 0 = S = nR ln
[4.17] = (1.00 mol) (8.314 J K 1 mol1 ) ln
Ssurr =  S = 5.8 J K1 A= U  T S [36] U = 0 [isothermal process in perfect gas] A = 0  (298 K) (5.76 J K1 ) = 1.7 103 J G= H  T S = 0  T S = 1.7 103 J
THE SECOND LAW: THE CONCEPTS
67
Process (b) H = T =0 [isothermal process in perfect gas]
S = +5.8 J K1 [Same as process (a); S is a state function] qsurr Ssurr = qsurr = q = (w) = w [First Law with U = 0] Tsurr w = pex V = (0.50 atm) (1.01 105 Pa atm1 ) (20 L  10 L) = qsurr Ssurr = 5.05 102 J = 1.7 J K1 298 K 103 m3 L = 5.05 102 J
A = 1.7 103 J G = 1.7 103 J Process (c) U =w [adiabatic process] [same as process (b)] = 5.05 102 J
3 (1.00 mol) 2 8.314 J K 1 mol1
[same as process (a); A and G are state functions]
w = pex V = 5.05 102 J U = nCV ,m T T = U nCV ,m
= 40.6 K
Tf = Ti  40.6 K = 298 K  40.6 K = 257 K S = nCV ,m ln Tf Ti [20] + nR ln 3 2 Vf Vi [17] 257 K 298 K + (1.00 mol)
= (1.00 mol)
(8.314 J K 1 mol1 ) ln 20 L 10 L
(8.314 J K1 mol1 ) ln Ssurr = 0 [adiabatic process]
= +3.9 J K1
A and G cannot be determined from the information provided without use of additional relations developed in Chapters 5 and 19. H = nCp,m T
5 Cp,m = CV ,m + R = 2 R
5 = (1.00 mol) 2 (8.314 J K 1 mol1 ) (40.6 K) = 8.4 102 J
P4.9
    Sm (T ) = Sm (298 K) +
S dT = a ln 1 1 T2 1 + b(T2  T1 )  c  2 2 T1 2 T2 T1
S=
T2 T1
Cp,m
T2 a dT c = +b+ 3 T T T T1
68
INSTRUCTOR'S MANUAL
(a)
  Sm (373 K) = (192.45 J K1 mol1 ) + (29.75 J K 1 mol1 ) ln
373 298
+ (25.10 103 J K2 mol1 ) (75.0 K)
1 1 1 + 2 (1.55 105 J K1 mol1 ) (373.15)2  (298.15)2
= 200.7 J K1 mol1 (b)
  Sm (773 K) = (192.45 J K1 mol1 ) + (29.75 J K 1 mol1 ) ln
773 298
+ (25.10 103 J K2 mol1 ) (475 K)
1 1 1 + 2 (1.55 105 J K1 mol1 ) 7732  2982
= 232.0 J K1 mol1 P4.10 S depends on only the initial and final states, so we can use Since q = nCp,m (Tf  Ti ), Tf = Ti + That is, S = nCp,m ln 1 + I 2 Rt nCp,m Ti S = nCp,m ln Tf [4.20] Ti
q I 2 Rt = Ti + (q = I tV = I 2 Rt) nCp,m nCp,m
Since n =
500 g = 7.87 mol 63.5 g mol1 (1.00 A)2 (1000 ) (15.0 s) (7.87) (24.4 J K 1 ) (293 K)
S = (7.87 mol) (24.4 J K 1 mol1 ) ln 1 + = (192 J K1 ) (ln 1.27) = +45.4 J K1
For the second experiment, no change in state occurs for the copper; hence, However, for the water, considered as a large heat sink S(water) = q I 2 Rt (1.00 A)2 (1000 ) (15.0 s) = = = +51.2 J K1 T T 293 K
S(copper) = 0.
[1 J = 1A V s = 1A2 s] P4.12 (a) Calculate the final temperature as in Exercise 4.14(a) Tf = n1 Ti1 + n2 Ti2 1 = (Ti1 + Ti2 ) = 318 K n1 + n 2 2 [n1 = n2 ] [n1 = n2 ] = +17.0 J K1
S = n1 Cp,m ln =
T2 Tf Tf + n2 Cp,m ln = n1 Cp,m ln f Ti1 Ti2 Ti1 Ti2 (75.3 J K 1 mol1 ) ln
200 g 18.02 g mol1
3182 273 363
THE SECOND LAW: THE CONCEPTS
69
(b) Heat required for melting is n1 fus H = (11.1 mol) (6.01 kJ mol1 ) = 66.7 kJ The decrease in temperature of the hot water as a result of heat transfer to the ice is T = q 66.7 kJ = 79.8 K = nCp,m (11.1 mol) (75.3 J K 1 mol1 )
At this stage the system consists of 200 g water at 0 C and 200 g water at (90 C  79.8 C) = 10 C (283 K). The entropy change so far is therefore S = = n Hfus 283 K + nCp,m ln Tf 363 K (11.1 mol) (6.01 kJ mol1 ) 273 K + (11.1 mol) (75.3 J K 1 mol1 ) ln 283 K 363 K
= 244 J K1  208.1 J K1 = +35.3 J K1
1 The final temperature is Tf = 2 (273 K + 283 K) = 278 K, and the entropy change in this step is
S = nCp,m ln
Tf2 2782 = (11.1) (75.3 J K 1 ) ln 273 283 Ti1 Ti2 S = 35.3 J K1 + 0.27 J K1 = +36 J K1
= +0.27 J K1
Therefore, overall, P4.15
rH rH    
=
J
 J f H  (J) [2.41] fH  
(298 K) = 1
(CO, g) + 1
fH
 
(H2 O, g)  1
fH
 
(CO2 , g)
= {110.53  241.82  (393.51)} kJ mol1 = +41.16 kJ mol1
rS rS    
=
J
  J Sm (J) [4.22]
        (298 K) = 1 Sm (CO, g) + 1 Sm (H2 O, g)  1 Sm (CO2 , g)  1 Sm (H2 , g)
= (197.67 + 188.83  213.74  130.684) kJ mol1 = +42.08 J K1 mol1
rH  
(398 K) = =
rH rH
   
(298 K) + (298 K) +
398 K 298 K r Cp
r Cp
dT [2.44]
T
[heat capacities constant]
r Cp
= 1 Cp,m (CO, g) + 1 Cp,m (H2 O, g)  1 Cp,m (CO2 , g)  1 Cp,m (H2 , g) = (29.14 + 33.58  37.11  28.824) J K 1 mol1 = 3.21 J K1 mol1 (398 K) = (41.16 kJ mol1 ) + (3.21 J K 1 mol1 ) (100 K) = +40.84 kJ mol1 Tf Ti
rH
 
For each substance in the reaction S = Cp,m ln = Cp,m ln 398 K 298 K [4.20]
70
INSTRUCTOR'S MANUAL
Thus
rS  
(398 K) = =
rS
 
(298 K) +
J
J Cp,m (J) ln
r Cp
Tf Ti
rS
 
(298 K) +
ln
398 K 298 K
= (42.01 J K1 mol1 ) + (3.21 J K 1 mol1 ) = (42.01  0.93) J K 1 mol1 = +41.08 J K1 mol1
  Comment. Both r H  and r S  changed little over 100 K for this reaction. This is not an uncommon result. T C p,m dT Sm (T ) = Sm (0) + [4.19] T 0 Perform a graphical integration by plotting Cp,m /T against T and determining the area under the curve.
P4.17
Draw up the following table
T /K (Cp,m /T )/(J K1 mol1 ) T /K (Cp,m /T )/(J K1 mol1 ) T /K (Cp,m /T )/(J K1 mol1 ) 10 0.209 90 1.837 170 1.508 20 0.722 100 1.796 180 1.473 30 1.215 110 1.753 190 1.437 40 1.564 120 1.708 200 1.403 50 1.741 130 1.665 60 1.850 140 1.624 70 1.877 150 1.584 80 1.868 160 1.546
Plot Cp,m /T against T (Fig. 4.2(a)). Extrapolate to T = 0 using Cp,m = aT 3 fitted to the point at T = 10 K, which gives a = 2.09 mJ K2 mol1 . Determine the area under the graph up to each T and plot Sm against T (Fig. 4.2(b)).
T /K S  S(0) / m m (J K 1 mol1 ) 25 9.25 50 43.50 75 88.50 100 135.00 125 178.25 150 219.0 175 257.3 200 293.5
The molar enthalpy is determined in a similar manner from a plot of Cp,m against T by determining the area under the curve (Fig. 4.3)
    Hm (200 K)  Hm (0) = 200 K 0
Cp,m dT = 32.00 kJ mol1
Solutions to theoretical problems
P4.20 Refer to Fig. 4.5 of the text for a description of the Carnot cycle and the heat terms accompanying each step of the cycle. Labelling the steps (a), (b), (c), and (d) going clockwise around the cycle starting from state A, the four episodes of heat transfer are
THE SECOND LAW: THE CONCEPTS
71
(a) 2.0
(b) 300
1.5
200
1.0 100 0.5
0 0 50 100 150 200
0 0 50 100 150 200
Figure 4.2
300 250 200 150 100
50 0 0 40 80 120 160 200
Figure 4.3 (a) (b) (c) (d) qh = nRTh ln 0 [adiabatic] qc = nRTc ln 0 [adiabatic] VD VC qc VD = nR ln Tc VC VB VA qh VB = nR ln Th VA
dq qc VB VD qh + = nR ln = T Th Tc VA VC V B VD VB VD Tc c Th c However, = = [2.34 of Section 2.6] = 1 VC V A Th Tc VA V C dq =0 Therefore T If the first stage is replaced by isothermal, irreversible expansion against a constant external pressure, q = w = pex (VB  VA ) ( U = 0, since this is an isothermal process in a perfect gas) Therefore
72
INSTRUCTOR'S MANUAL
Therefore,
qh = Th
pex Th
(VB  VA ) VB because less work is done in the irreversible expansion, so VA That is, dq <0 T
However, pex (VB  VA ) < nRTh ln
VD VB dq + nR ln = 0. < nR ln VC T VA
Comment. Whenever an irreversible step is included in the cycle the above result will be obtained. Question. Can you provide a general proof of this result? P4.22 The isotherms correspond to T = constant, and the reversibly traversed adiabats correspond to S = constant. Thus we can represent the cycle as in Fig. 4.4.
1 Temperature
4
2
3
Entropy
Figure 4.4
In this figure, paths 1, 2, 3, and 4 correspond to the four stages of the Carnot cycle listed in the text following eqn 4.7 The area within the rectangle is Area = T dS = (Th  Tc ) (S2  S1 ) = (Th  Tc ) S = (Th  Tc )nR ln VB VA
(isothermal expansion from VA to VB , stage 1) Th  Tc VB VB nRTh ln [Fig. 4.5] = nR(Th  Tc ) ln Th VA VA Therefore, the area is equal to the net work done in the cycle. But, w(cycle) = qh = P4.23 S = nCp,m ln Tf Tf + nCp,m ln [4.20] Th Tc
1 [Tf is the final temperature, Tf = 2 (Th + Tc )]
1 In the present case, Tf = 2 (500 K + 250 K) = 375 K
S = nCp,m ln
Tf2 (Th + Tc )2 = nCp,m ln = Th Tc 4Th Tc
500 g 63.54 g mL1 ln 3752 500 250
(24.4 J K 1 mol1 ) = +22.6 J K1
P4.26
g = f + yz dg = df + y dz + z dy = a dx  z dy + y dz + z dy = a dx + y dz
THE SECOND LAW: THE CONCEPTS
73
Comment. This procedure is referred to as a Legendre transformation and is essentially the method used in Chapter 5 to express the differentials of H , G, and A in terms of the differential of U . P4.27 (a) According to eqns 2.43, 4.19, and 4.39:
30 K 0K
Hm (T ) = Hm (T )  Hm (0) = Sm (T ) =
30 K 0K
Cp (T ) dT
where Cp (T ) = aT 3 1  e/T
2
Cp (T ) dT T
and
Gm (T ) = Gm (T )  Gm (0) =
Hm (T )  T Sm (T )
The integral computations are easily performed with the builtin numerical integration capabilities of a scientific calculator or computer software spreadsheet. Computations at ten or more equally spaced temperatures between 0 K and 30 K will produce smoothlooking plots.
150 Sm / JK1 mol1 Hm / J mol1 6
100
4
50
2
0
0
10 T/K
20
30
0
0
10 T/K
20
30
0 Gm / J mol1
20
40
0
10 T/K
20
30
(b) According to the law of Dulong and Petit the constant pressure heat capacity of Ce2 Si2 O7 (11 moles of atoms per mole of compound) is approximately equal to 11 3 R = 274 J K1 mol1 . The experimental value at 900 K equals 287J K 1 mol1 . The law of Dulong and Petit gives a reasonable estimate of the heat capacity at very high temperature.
Solutions to applications
P4.29 (a)
rG rH    
= =
rH fH
   
  T rS
(secC4 H9 ) 
1
fH
 
(tertC4 H9 )
= (67.5  51.3) kJ mol
= 16.2 kJ mol1
74
INSTRUCTOR'S MANUAL
rS
   
rG
= (336.6  314.6) J K 1 mol1 = 22.0 J K1 mol1 = 16.2 kJ mol1  (700 K) (22.0 103 kJ K1 mol1 ) = 0.8 kJ mol1
    = Sm (secC4 H9 )  Sm (tertC4 H9 )
(b)
   = f H  (C3 H6 ) + f H  (CH3 )  f H  (tertC4 H9 ) 1 = (20.42 + 145.49  51.3) kJ mol = 114.6 kJ mol1 rH   = (267.05 + 194.2  314.6) J K 1 mol1 = 146.7 J K1 mol1 rS   = 114.6 kJ mol1  (700 K) (0.1467 kJ K 1 mol1 ) rG rH  
 
= 11.9 kJ mol1 (c)
   = f H  (C2 H4 ) + f H  (C2 H5 )  f H  (tertC4 H9 )   = (52.26 + 121.0  51.3) kJ mol1 = 122.0 kJ mol1 rH   = (219.56 + 247.8  314.6) J K 1 mol1 = 152.8 J K1 mol1 rS   = 122.0 kJ mol1  (700 K) (0.1528 kJ K 1 mol1 ) rG rH  
= 15.0 kJ mol1 P4.32 The minimum power output that is needed to maintain the temperature difference Th  Tc occurs when dp/dTc = 0 p = dw d = (qh   qc ) [911] dt dt d d qh  Th = 1 = 1 qc  qc  dt qc  dt Tc = Th dqc  1 = Tc dt Th  1 kATc4 Tc
At constant Th dp Th =  2 dTc Tc kATc4 + 4kATc3 Th 1 Tc
This is a minimum when equal to zero. Simplifying yields  Th Th +4 1 =0 Tc Tc
Th 4 = Tc 3
5
The Second Law: the machinery
Solutions to exercises
Discussion questions
E5.1(b)
2 See the solution to Exercise 3.14(a) and Example 5.1, where it is demonstrated that T = a/Vm for a van der Waals gas. Therefore, there is no dependence on b for a van der Waals gas. The internal pressure results from attractive interactions alone. For van der Waals gases and liquids with strong attractive forces (large a) at small volumes, the internal pressure can be very large.
E5.2(b)
The relation (G/T )p = S shows that the Gibbs function of a system decreases with T at constant p in proportion to the magnitude of its entropy. This makes good sense when one considers the definition of G, which is G = U + pV  T S. Hence, G is expected to decrease with T in proportion to S when p is constant. Furthermore, an increase in temperature causes entropy to increase according to S=
f i
dqrev /T
The corresponding increase in molecular disorder causes a decline in th Gibbs energy. (Entropy is always positive.) E5.3(b) The fugacity coefficient, , can be expressed in terms of an integral involving the compression factor, specifically an integral of Z  1 (see eqn 5.20). Therefore, we expect that the variation with pressure of the fugacity coefficient should be similar, in a very qualitative sense, to the variation with pressure of the compression factor itself. Comparison of figures 1.27 and 5.8 of the text shows this to be roughly the case, though the detailed shapes of the curves are necessarily different because is an integral function of Z  1 over a range of pressures. So we expect no simple proportionality between and Z. But we find < 1 in pressure regions where attractive forces are expected to predominate and > 1 when repulsive forces predominate, which in behavior is similar to that of Z. See Section 5.5(b) for a more complete discussion.
Numerical exercises
E5.4(b) = 1 V V T p T =  1 V V p T
E5.5(b)
S V = = V p T T p pf pf Vi at constant temperature, G = nRT ln = pi pi Vf = nRT ln Vi Vf 72 100
= (2.5 103 mol) (8.314 J K 1 mol1 ) (298 K) ln = 2.035 = 2.0 J
76
INSTRUCTOR'S MANUAL
E5.6(b)
G = S T p
Gf Gi = Sf and = Si T p T p Gf Gi S = Sf  Si =  + T p T p (Gf  Gi ) G = = T T p p {73.1 + 42.8 T /K} J = T = 42.8 J K1
E5.7(b)
See the solution to Exercise 5.7(a). Without knowledge of the compressibility of methanol we can only assume that V = V1 (1  T p) V1 . Then G=V p m m 25 g = = 31.61 cm3 so V = = V 0.791 g cm3 G = (31.61 cm3 ) = +3.2 kJ 1 m3 106 cm3 (99.9 106 Pa)
E5.8(b)
(a)
pi Vf = nR ln Vi pf Taking inverse logarithms S = nR ln
[Boyle's Law]
pf = pi e S/nR = (150 kPa) exp  = 274 kPa (b) G = nRT ln pf pi = T S
(15.0 J K1 ) (3.00 mol) (8.314 J K 1 mol1 )
[ H = 0, constant temperature, perfect gas]
= (230 K) (15.0 J K1 ) = +3450 J = 3.45 kJ E5.9(b) = f  i = RT ln pf pi 252.0 92.0
= (8.314 J K1 mol1 ) (323 K) ln = 2.71 kJ mol1 E5.10(b)
 0 =  + RT ln  =  + RT ln
p  p f  p
THE SECOND LAW: THE MACHINERY
77
 0 = RT ln  0 = RT ln
f p
f p
= (8.314 J K1 mol1 ) (290 K) ln(0.68) = 929.8 J mol1 = 930 J mol1 E5.11(b) or 0.93 kJ mol1
3
1 (160.0 cm3 mol1 ) 106m 3 B cm B = = RT (8.314 J K1 mol1 ) (100 K)
= 1.924 107 Pa1 = eB p+
7 1 6 e 1.92410 Pa 6210 Pa
e11.93 = 7 106 E5.12(b) or of the order of 106
G = nVm p = V p = (1.0 L) 1 m3 103 L (200 103 Pa)
= 200 Pa m3 = 200 J E5.13(b) Gm = RT ln pf pi 100.0 kPa 50.0 kPa
= (8.314 J K1 mol1 ) (500 K) ln = +2.88 kJ mol1 E5.14(b)
E5.15(b)
G RT + B + C p + D p2 [5.10] = p T p which is the virial equation of state. S p = V T T V V = For a Dieterici gas p= RT ea/RT Vm Vm  b
a R 1 + RVm T ea/RVm T p = Vm  b T Vm
dS =
S dVm = Vm T
p dVm T Vm
78
INSTRUCTOR'S MANUAL
S=
Vm,f Vm,i
dS = R 1 +
a RVm T
ea/RVm T
Vm,f  b Vm,i  b
For a perfect gas
Vm,f Vm,i S for a Dieterici gas may be greater or lesser than S for a perfect gas depending on T and the magnitudes of a and b. At very high T , S is greater. At very low T , S is less. S = R ln
Solutions to problems
Solutions to numerical problem
P5.2 For the reaction N2 (g) + 3H2 (g) 2NH3 (g) (a)
rG   rG    = 2 f G (NH3 , g)
  (500 K) = r G (Tc ) + (1  ) r H  (Tc )
Problem 5.1, =
T Tc
=
500 K (2) (16.45 kJ mol1 ) 298.15 K 500 K + 1 (2) (46.11 kJ mol1 ) 298.15 K
= 55.17 + 62.43 kJ mol1 = +7 kJ mol1 (b)
rG  
(1000 K) =
1000 K (2) (16.45 kJ mol1 ) 298.15 K 1000 K + 1 (2) (46.11 kJ mol1 ) 298.15 K
= (110.35 + 217.09) kJ mol1 = +107 kJ mol1
Solutions to theoretical problems
P5.5 We start from the fundamental relation dU = T dS  p dV [2] But, since U = U (S, V ), we may also write dU = U dS + S V U dV V S U = p V S
Comparing the two expressions, we see that U =T S V and
These relations are true in general and hence hold for the perfect gas. We can demonstrate this more explicitly for the perfect gas as follows. For the perfect gas at constant volume dU = CV dT
THE SECOND LAW: THE MACHINERY
79
and dS = Then CV dT dqrev = T T CV dT U = CV dT S V T =T
dU = dS V
For a reversible adiabatic (constantentropy) change in a perfect gas dU = dw = p dV Therefore, P5.8 U = p V S p T = [Maxwell relation] S V V S = 1
S T
V
V S
[chain relation] =
T
S V S T
T V
[inversion]
V T U T
=
p T S U
V
V
U T
p
[Maxwell relation] =
V
p  V S U
T
p
[chain relation]
V
V
= P5.10 H = p T

V T V p
T
U S U T
V
V
[inversion twice] =
T T CV
U =T S V
H [Relation 1, Further information 1.7] p S dH = T dS + V dp [Problem 5.6] H H dS + dp (H = H (p, S)) compare dH = S p p S Thus, H = T, S p H = V [dH exact] p S H =T p T S + V = T p T V + V [Maxwell relation] T p
H S p
S + p T
Substitution yields, (a) For pV = nRT
nR V , = T p p (b) For p = T =
hence
H nRT +V = 0 = p T p
an2 nRT  2 [Table 1.6] V  nb V p(V  nb) na(V  nb) + nR RV 2
80
INSTRUCTOR'S MANUAL
T p 2na(V  nb) na =  + 2 V p nR RV RV 3 T + V [inversion] = p +V 2na(V nb) na nR + RV 2  RV 3 p which yields after algebraic manipulation Therefore, nb  H = p T 1 When b Vm
2na RT 2na RT V
H T = T p T V
2 2
,
=1
nb V
1, 1 and
2na 1 2na p 2pa 2na = 2 2 = RT V RT nRT RT V R T Therefore, nb  2na H RT 2pa p T 1  R2 T 2
For argon, a = 1.337 L2 atm mol2 , b = 3.20 102 L mol1 , 2na (2) (1.0 mol) (1.337 L2 atm mol2 ) = 0.11 L = RT (8.206 102 L atm K1 mol1 ) (298 K) 2pa (2) (10.0 atm) (1.337 L2 atm mol2 ) = = 0.045 2 R2 T 2 (8.206 102 L atm K1 mol1 ) (298 K) Hence, H {(3.20 102 )  (0.11)} L = 0.0817 L = 8.3 J atm1 p T 1  0.045 H p T p (8.3 J atm1 ) (1 atm) = 8 J
H
P5.12
T = T
p  p [5.8] T V RT BRT p= + [The virial expansion, Table 1.6, truncated after the term in B] 2 Vm Vm BR RT B p RT + 2 = + 2 2 T Vm Vm T V Vm 2 B B RT 2 T V Vm T B T V
p R = + T V Vm RT 2 Hence, T = 2 Vm
Since T represents a (usually) small deviation from perfect gas behaviour, we may approximate Vm . Vm RT p T p2 R B T
THE SECOND LAW: THE MACHINERY
81
From the data Hence, (a) (b) T =
B = ((15.6)  (28.0)) cm3 mol1 = +12.4 cm3 mol1
(1.0 atm)2 (12.4 103 L mol1 ) = 3.0 103 atm (8.206 102 L atm K1 mol1 ) (50 K) so at p = 10.0 atm, T = 0.30 atm
T p 2 ;
Comment. In (a) T is 0.3 per cent of p; in (b) it is 3 per cent. Hence at these pressures the approximation for Vm is justified. At 100 atm it would not be. P5.13 Question. How would you obtain a reliable estimate of T for argon at 100 atm? H U and Cp = CV = T p T V (a) 2U 2U CV = = = V T T V V T CV 2U 2U = = = pT T p p T V U = T V T p Cp Since Cp = CV + R, = x T T T
T V
U =0 V T V U p T V =0 (T = 0)
[T = 0]
CV for x = p or V x T d2 U dCV dCV CV and Cp may depend on temperature. Since = is nonzero if U depends on , dT dT 2 dT T through a nonlinear relation. See Chapter 20 for further discussion of this point. However, for a perfect monatomic gas, U is a linear function of T ; hence CV is independent of T . A similar argument applies to Cp . (b) This equation of state is the same as that of Problem 5.12. 2U CV = = V T T V = = T [Part (a)] T V RT 2 2 T Vm 2RT 2 Vm RT 2 Vm B [Problem 5.12] T V V 2B T 2
B RT 2 + 2 T V Vm 2 (BT ) T 2
V
=
V
P5.15
T = T
p  p [5.8] T V nRT ean/RT V [Table 1.6] p= V  nb nRT nRT na nap p ean/RT V + ean/RT V = p + = T T V V  nb RT V V  nb RT V
82
INSTRUCTOR'S MANUAL
Hence, T =
nap RT V
P5.17
T 0 as p 0, V , a 0, and T . The fact that T > 0 (because a > 0) U is consistent with a representing attractive contributions, since it implies that > 0 and the V T internal energy rises as the gas expands (so decreasing the average attractive interactions). G dG = dp = V dp p T G(pf )  G(pi ) =
pf pi
V dp
In order to complete the integration, V as a function of p is required. V = T V (given), p T so d ln V =  dp
Hence, the volume varies with pressure as
V V0
d ln V = T
p pi
dp
or V = V0 eT (ppi ) (V = V0 when p = pi )
pf
Hence,
pi
dG =
V dp = V0
pf
pi
eT (ppi ) dp 1  eT p T
G(pf ) = G(pi ) + (V0 ) If T p
1  eT (pf pi ) T
= G(pi ) + (V0 )
1 2 1 2 1, 1  eT p 1  1  T p + 2 T p 2 = T p  2 T p 2
1 Hence, G = G + V0 p 1  T p 2 For the compression of copper, the change in molar Gibbs function is 1 G m = V m p 1  T p = 2 = 63.54 g mol1 8.93 106 g m3 M p 1 1  T p 2
1 (500) (1.013 105 Pa) 1  T p 2
1 = (360.4 J) 1  2 T p
If we take T = 0 (incompressible),
1 2 T
Gm = +360 J. For its actual value
1 p = 2 (0.8 106 atm1 ) (500 atm) = 2 104
1 1  2 T p = 0.9998
Hence,
Gm differs from the simpler version by only 2 parts in 104 (0.02 per cent)
THE SECOND LAW: THE MACHINERY
83
P5.19
S = 
1 V
1 V = p p S V V
S
The only constantentropy changes of state for a perfect gas are reversible adiabatic changes, for which pV = const Then, const =  V V S +1 1 Therefore, S = = p p V V S = S(T , p) S dS = dT + T p T dS = T Use p = V S const V +1 = p V
Hence, p S = +1 P5.21 S dp p T S S dT + T dp T p p T S H p H 1 = Cp T p T H = T , Problem 5.6 S p
S = T p
S V = [Maxwell relation] p T T p V dp = Cp dT  T V dp T p For reversible, isothermal compression, T dS = dqrev , dT = 0; hence Hence, T dS = Cp dT  T dqrev = T V dp qrev =
pf pi
T V dp = T V p
[ and V assumed constant]
For mercury qrev = (1.82 104 K 1 ) (273 K) (1.00 104 m3 ) (1.0 108 Pa) = 0.50 kJ P5.25 When we neglect b in the van der Waals equation we have p= RT a  2 Vm Vm a RT Vm
p o p a Z1 dp = dp RT Vm p o p
and hence Z =1
Then substituting into eqn 5.20 we get ln =
84
INSTRUCTOR'S MANUAL
In order to perform this integration we must eliminate the variable Vm by solving for it in terms of p. Rewriting the expression for p in the form of a quadratic we have RT a 2 Vm  Vm + = 0 p p The solution is 1 1 Vm = (RT )2  4ap RT /p 2 p applying the approximation (RT )2 Vm = 1 2 RT RT p p 4ap we obtain
Choosing the + sign we get RT Vm = which is the perfect volume p Then p a ap ln =  dp =  RT 2 (RT )2 0 For ammonia a = 4.169 atm L2 mol2 ln =  4.169 atm L2 mol2 10.00 atm (0.08206 L atm K1 mol1 298.15 K)2 = 0.06965 f = 0.9237 = p
f = p = 0.9237 10.00 atm = 9.237 atm P5.27 The equation of state qT pVm =1+ is solved for Vm = RT Vm
1/2
RT 2p
1+ 1+
4pq 1/2 so R
pVm 2q 1 Z1 qT R = RT = = p p pVm 1 + 1 + 4pq R
ln =
p 0
Z1 p
dp[24] =
dp 2q p R 0 1 + 1 + 4pq R
1/2
Defining, a 1 + 1 + ln =
a 2
4pq 1/2 R(a  1) da, gives , dp = R 2q da [a = 2, when p = 0]
a1 a
1 4pq 1/2 4pq 1/2 1 1 = a  2  ln a = 1 + 1+  1  ln + 2 R 2 R 2 Hence, =
1/2 2e{(1+4pq/R) 1}
1 + 1 + 4pq R
1/2
THE SECOND LAW: THE MACHINERY
85
This function is plotted in Fig. 5.1(a) when
1.2 1.0 0.1 0.01 1.0 0.01 0.8 0.1 1.0
4pq R
1, and using the approximations
Figure 5.1(a) 1 ex 1 + x, (1 + x)1/2 1 + x, and (1 + x)1 1  x [x 1] 2 pq 1+ R When is plotted against x = 4pq/R on a linear rather than exponential scale, the apparent curvature seen in Fig. 5.1(a) is diminished and the curve seems almost linear. See Fig. 5.1(b).
2
2
2
Figure 5.1(b)
Solution to applications
P5.28 wadd,max =
rG   rG
[4.38] Problem 5.1, = 310 K 298.15 K T Tc
  (37 C) = r G (Tc ) + (1  ) r H  (Tc )
=
310 K 298.15 K
(6333 kJ mol1 ) + 1 
(5797 kJ mol1 )
= 6354 kJ mol1
  The difference is r G (37 C)  r G (Tc ) = {6354  (6333)} kJ mol1 = 21 kJ mol1 1 Therefore, an additional 21 kJ mol of nonexpansion work may be done at the higher temperature.
86
INSTRUCTOR'S MANUAL
Comment. As shown by Problem 5.1, increasing the temperature does not necessarily increase the   maximum nonexpansion work. The relative magnitude of r G and r H  is the determining factor. P5.31 The GibbsHelmholtz equation is T G T
rG  
=
H T2
rH T2     r G2   r G1   rH T2 T rH  
so for a small temperature change T so d
rG  
= =
T
 dT 
and and
 
T2
= =
T1
 +
rH
  r G190
  r G220
T =
T2 +
T190 1
T220
1 1  T190 T220
  r G190
 T190  r G220 T220
rH
T190 T220 + (127 kJ mol1 ) 1  190 K , 220 K
For the monohydrate
  r G190   r G190
= (46.2 kJ mol1 ) = 57.2 kJ mol1 = (69.4 kJ mol1 ) = 85.6 kJ mol1 = (93.2 kJ mol1 ) = 112.8 kJ mol1
190 K 220 K
For the dihydrate
  r G190   r G190
190 K 220 K
+ (188 kJ mol1 ) 1 
190 K , 220 K
For the monohydrate
  r G190   r G190
190 K 220 K
+ (237 kJ mol1 ) 1 
190 K , 220 K
P5.32
The change in the Helmholtz energy equals the maximum work associated with stretching the polymer. Then dwmax = dA = f dl For stretching at constant T A U S = +T l T l T l T assuming that (U/l)T = 0 (valid for rubbers) f = f = T S =T l T 3kB l N a2 l T  3kB l 2 +C 2N a 2 l
=T 
=
3kB T N a2
This tensile force has the Hooke's law form f = kH l with kH = 3kB T /N a 2 .
6
Physical transformations of pure substances
Solutions to exercises
Discussion questions
E6.1(b) Refer to Fig. 6.8. The white lines represent the regions of superheating and supercooling. The chemical potentials along these lines are higher than the chemical potentials of the stable phases represented by the colored lines. Though thermodynamically unstable, these socalled metastable phases may persist for a long time if the system remains undisturbed, but will eventually transform into the thermodynamically stable phase having the lower chemical potential. Transformation to the condensed phases usually requires nucleation centers. In the absence of such centers, the metastable regions are said to be kinetically stable. At 298 K and 1.0 atm, the sample of carbon dioxide is a gas. (a) After heating to 320 K at constant pressure, the system is still gaseous. (b) Isothermal compression at 320 K to 100 atm pressure brings the sample into the supercritical region. The sample is now not much different in appearance from ordinary carbon dioxide, but some of its properties are (see Box 6.1). (c) After cooling the sample to 210 K at constant pressure, the carbon dioxide sample solidifies. (d) Upon reducing the pressure to 1.0 atm at 210 K, the sample vapourizes (sublimes); and finally (e) upon heating to 298 K at 1.0 atm, the system has resumed its initial conditions in the gaseous state. Note the lack of a sharp gas to liquid transition in steps (b) and (c). This process illustrates the continuity of the gaseous and liquid states. Firstorder phase transitions show discontinuities in the first derivative of the Gibbs energy with respect to temperature. They are recognized by finite discontinuities in plots of H , U , S, and V against temperature and by an infinite discontinuity in Cp . Secondorder phase transitions show discontinuities in the second derivatives of the Gibbs energy with respect to temperature, but the first derivatives are continuous. The secondorder transitions are recognized by kinks in plots of H , U , S, and V against temperature, but most easily by a finite discontinuity in a plot of Cp against temperature. A transition shows characteristics of both first and secondorder transitions and, hence, is difficult to classify by the Ehrenfest scheme. It resembles a firstorder transition in a plot of Cp against T , but appears to be a higherorder transition with respect to other properties. See the book by H. E. Stanley listed under Further reading for more details.
E6.2(b)
E6.3(b)
Numerical exercises
E6.4(b) Assume vapour is a perfect gas and ln p vap H =+ p R R
vap H vap H
is independent of temperature
1 1  T T ln p p
1 1 = + T T =
1 8.314 J K1 mol1 58.0 + ln 293.2 K 32.7 103 J mol1 66.0
= 3.378 103 K 1 T = 1 3.378 103 K 1 = 296 K = 23 C
88
INSTRUCTOR'S MANUAL
E6.5(b)
dp = dT
fus S
Sm Vm = Vm
fus S
dp dT
Vm
p T (1.2 106 Pa)  (1.01 105 Pa) 429.26 K  427.15 K
assuming
and
Vm independent of temperature.
3 1  142.0 cm3 mol1 ) fus S = (152.6 cm mol
= (10.6 cm3 mol1 )
1 m3 106 cm3
(5.21 105 Pa K1 )
= 5.52 Pa m3 K 1 mol1 = 5.5 J K1 mol1
fus H
= Tf S = (427.15 K) (5.52 J K1 mol1 ) = 2.4 kJ mol1
vap H RT 2
E6.6(b)
Use
d ln p =
dT
ln p = constant  Terms with
vap H RT
1 dependence must be equal, so T 3036.8 K vap H  = T /K RT
vap H
= (3036.8 K)R = (8.314 J K1 mol1 ) (3036.8 K) = 25.25 kJ mol1
E6.7(b)
(a)
log p = constant  Thus
vap H
vap H
RT (2.303)
= (1625 K) (8.314 J K 1 mol1 ) (2.303) = 31.11 kJ mol1
(b) Normal boiling point corresponds to p = 1.000 atm = 760 Torr log(760) = 8.750  1625 T /K
1625 = 8.750  log(760) T /K 1625 = 276.87 T /K = 8.750  log(760) Tb = 276.9 K
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
89
E6.8(b)
Tf fus V Tf pM p= fus S fus H fus H [Tf = 3.65 + 273.15 = 269.50 K] T =
fus V
p=
1
T =
(269.50 K) (99.9 MPa)M 8.68 kJ mol1
1 1  0.789 g cm3 0.801 g cm3 m3 106 cm3
= (3.1017 106 K Pa J1 mol) (M) (+ .01899 cm3 /g)
= (+ 5.889 102 K Pa m3 J1 g1 mol)M = (+ 5.889 102 K g1 mol)M T = (46.07 g mol1 ) (+ 5.889 102 K g1 mol) = + 2.71 K Tf = 269.50 K + 2.71 K = 272 K E6.9(b) dn dm = MH2 O dt dt dn dq/dt = = dt vap H where n = q
vap H
(0.87 103 W m2 ) (104 m2 ) 44.0 103 J mol1
= 197.7 J s1 J1 mol = 200 mol s1 dm = (197.7 mol s1 ) (18.02 g mol1 ) dt = 3.6 kg s1 E6.10(b) E6.11(b) The vapour pressure of ice at 5 C is 3.9 103 atm, or 3 Torr. Therefore, the frost will sublime. A partial pressure of 3 Torr or more will ensure that the frost remains. (a) According to Trouton's rule (Section 4.3, eqn 4.16)
vap H
= (85 J K1 mol1 ) Tb = (85 J K1 mol1 ) (342.2 K) = 29.1 kJ mol1
Solid c Pressure
Liquid b Critical point
Start d Gas Temperature a
Figure 6.1
90
INSTRUCTOR'S MANUAL
(b) Use the ClausiusClapeyron equation [Exercise 6.11(a)] ln p2 p1 =
vap H
R
1 1  T1 T2
At T2 = 342.2 K, p2 = 1.000 atm; thus at 25 C ln p1 =  29.1 103 J mol1 8.314 J K1 mol1 1 1  298.2 K 342.2 K = 1.509
p1 = 0.22 atm = 168 Torr At 60 C, ln p1 =  29.1 103 J mol1 8.314 J K1 mol1 1 1  333.2 K 342.2 K = 0.276
p1 = 0.76 atm = 576 Torr E6.12(b) T = Tf (10 MPa)  Tf (0.1 MPa) =
fus H
Tf pM fus H
1
= 6.01 kJ mol1 (273.15 K) (9.9 106 Pa) (18 103 kg mol1 ) 6.01 103 J mol1
T =
1 1  9.98 102 kg m3 9.15 102 kg m3 = 0.74 K Tf (10 MPa) = 273.15 K  0.74 K = 272.41 K E6.13(b)
vap H vap H
=
vap U
+
vap (pV ) 1
= 43.5 kJ mol
vap (pV ) vap (pV )
= p vap V = p(Vgas  Vliq ) = pVgas = RT [per mole, perfect gas] = (8.314 J K1 mol1 ) (352 K) = 2927 J mol1
vap (pV ) vap H
Fraction =
=
2.927 kJ mol1 43.5 kJ mol1
= 6.73 102 = 6.73 per cent E6.14(b) Vm = M 18.02 g mol1 = 1.803 105 m3 mol1 = 999.4 103 g m3
2 Vm 2(7.275 102 N m1 ) (1.803 105 m3 mol1 ) = rRT (20.0 109 m) (8.314 J K 1 mol1 ) (308.2 K) = 5.119 102 p = (5.623 kPa)e0.05119 = 5.92 kPa
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
91
E6.15(b)
1 1 = 2 ghr = 2 (0.9956 g cm3 ) (9.807 m s2 ) (9.11 102 m)
(0.16 103 m) = 7.12 102 N m1 E6.16(b) pin  pout =
1000 kg m3 g cm3
2 2(22.39 103 N m1 ) = r (220 109 m) = 2.04 105 N m2 = 2.04 105 Pa
Solutions to problems
Solutions to numerical problems
P6.3 (a) dp = dT
vap H [6.6, Clapeyron equation] Tb vap V 14.4 103 J mol1 = + 5.56 kPa K1 = (180 K) (14.5 103  1.15 104 ) m3 mol1 vap V vap S
=
(b)
dp dp vap H p 11, with d ln p = = p dT RT 2 3 J mol1 ) (1.013 105 Pa) (14.4 10 = = + 5.42 kPa K 1 (8.314 J K1 mol1 ) (180 K)2 The percentage error is 2.5 per cent (l)  p T 1 (s) = Vm (l)  Vm (s)[6.13] = M p T 1 1  = (18.02 g mol1 ) 3 1.000 g cm 0.917 g cm3 = 1.63 cm3 mol1
P6.5
(a)
(b)
(g)  p T
(l) = Vm (g)  Vm (l) p T = (18.02 g mol1 ) = + 30.1 L mol1
1 1  1 0.598 g L 0.958 103 g L1 Vvap p =
At 1.0 atm and 100 C , (l) = (g); therefore, at 1.2 atm and 100 C (g)(l) (as in Problem 6.4) (30.1 103 m3 mol1 ) (0.2) (1.013 105 Pa) + 0.6 kJ mol1 Since (g) > (l), the gas tends to condense into a liquid. pH2 O V The amount (moles) of water evaporated is ng = RT The heat leaving the water is q = n vap H The temperature change of the water is T = q , n = amount of liquid water nCp,m
P6.7
92
INSTRUCTOR'S MANUAL
Therefore,
T = =
pH2 O V vap H RT nCp,m (23.8 Torr) (50.0 L) (44.0 103 J mol1 )
250 g (62.364 L Torr K1 mol1 ) (298.15 K) (75.5 J K 1 mol1 ) 18.02 g mol1 = 2.7 K
The final temperature will be about 22 C P6.9 (a) Follow the procedure in Problem 6.8, but note that Tb = 227.5 C is obvious from the data. (b) Draw up the following table
/ C T /K 1000 K/T ln p/Torr
57.4 330.6 3.02 0.00
100.4 373.6 2.68 2.30
133.0 406.2 2.46 3.69
157.3 430.5 2.32 4.61
203.5 476.7 2.10 5.99
227.5 500.7 2.00 6.63
The points are plotted in Fig. 6.2. The slope is 6.4 103 K, so implying that
vap H
= +53 kJ mol1
 vap H = 6.4 103 K, R
6
4
2
0 2.0 2.2 2.4 2.6 2.8 3.0
Figure 6.2 P6.11 (a) The phase diagram is shown in Fig. 6.3.
2 0 2 4 6 8 0 100 200 300 400 500 600 Liquid
Solid Vapour
LiquidVapour SolidLiquid
Figure 6.3
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
93
(b) The standard melting point is the temperature at which solid and liquid are in equilibrium at 1 bar. That temperature can be found by solving the equation of the solidliquid coexistence curve for the temperature 1 = p3 /bar + 1000(5.60 + 11.727x)x, So 11 727x 2 + 5600x + (4.362 107  1) = 0 The quadratic formula yields 1 1 + 4(11 727) 5600 {(5600)2  4(11 727) (1)}1/2 56002 x= = 2(11 727) 2 11727 5600
1/2
The square root is rewritten to make it clear that the square root is of the form {1 + a}1/2 , with 1 1 a 1; thus the numerator is approximately 1 + (1 + 2 a) = 2 a, and the whole expression reduces to x 1/5600 = 1.79 104 Thus, the melting point is T = (1 + x)T3 = (1.000179) (178.15 K) = 178.18 K (c) The standard boiling point is the temperature at which the liquid and vapour are in equilibrium at 1 bar. That temperature can be found by solving the equation of the liquidvapour coexistence curve for the temperature. This equation is too complicated to solve analytically, but not difficult to solve numerically with a spreadsheet. The calculated answer is T = 383.6 K (d) The slope of the liquidvapour coexistence curve is given by dp vap H =  dT T vap V  so
vap H    = T vap V 
dp dT
The slope can be obtained by differentiating the equation for the coexistence curve. dp d ln p d ln p dy =p =p dT dT dy dT 10.418  15.996 + 2(14.015)y  3(5.0120)y 2  (1.70) (4.7224) (1  y)0.70 y2 p Tc At the boiling point, y = 0.6458, so dp = 2.851 102 bar K1 = 2.851 kPa K1 dT and P6.12
vap H  
dp = dT
= (383.6 K)
(30.3  0.12) L mol1 (2.851 kPa K 1 ) = 33.0 kJ mol1 1000 L m3
The slope of the solidvapour coexistence curve is given by
  dp sub H =  dT T sub V 
so
sub H
 
 = T sub V 
dp dT
The slope can be obtained by differentiating the coexistence curve graphically (Fig. 6.4).
94
INSTRUCTOR'S MANUAL
60 50 40 30 20 10 144 146 148 150 152 154 156
Figure 6.4
dp = 4.41 Pa K1 dT according to the exponential best fit of the data. The change in volume is the volume of the vapour Vm = So
sub H  
RT (8.3145 J K1 mol1 ) (150 K) = = 47.8 m3 p 26.1 Pa = (150 K) (47.8 m3 ) (4.41 Pa K 1 ) = 3.16 104 J mol1 = 31.6 kJ mol1
Solutions to theoretical problems
P6.14 G G G =  = V  V p T p T p T Therefore, if V = V , G is independent of pressure. In general, V = V , so that though small, since V  V is small. pV Amount of gas bubbled through liquid = RT (p = initial pressure of gas and emerging gaseous mixture) m Amount of vapour carried away = M Mole fraction of vapour in gaseous mixture =
m M m M pV + RT mRT p PVM mRT PVM m M
G is nonzero,
P6.16
Partial pressure of vapour = p =
m M
+
pV RT
p =
+1
=
mP A , mA + 1
A=
RT PVM
For geraniol, M = 154.2 g mol1 , T = 383 K, V = 5.00 L, p = 1.00 atm, and m = 0.32 g, so A= (8.206 102 L atm K1 mol1 ) (383 K) = 40.76 kg1 (1.00 atm) (5.00 L) (154.2 103 kg mol1 ) (0.32 103 kg) (760 Torr) (40.76 kg1 ) = 9.8 Torr (0.32 103 kg) (40.76 kg1 ) + 1
Therefore p=
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
95
P6.17
p = p0 eMgh/RT [Box 1.1] vap H p = p e = R
1 1  T T
[6.12]
Let T = Tb the normal boiling point; then p = 1 atm. Let T = Th , the boiling point at the altitude h. Take p0 = 1 atm. Boiling occurs when the vapour (p) is equal to the ambient pressure, that is, when p(T ) = p(h), and when this is so, T = Th . Therefore, since p0 = p , p(T ) = p(h) implies that eMgh/RT = exp 
vap H
R
1 1  Th Tb
It follows that
1 1 Mgh = + Th Tb T vap H
where T is the ambient temperature and M the molar mass of the air. For water at 3000 m, using M = 29 g mol1 1 1 (29 103 kg mol1 ) (9.81 m s2 ) (3.000 103 m) = + Th 373 K (293 K) (40.7 103 J mol1 ) 1 1 = + 373 K 1.397 104 K Hence, Th = 363 K (90 C) P6.20 Sm = Sm (T , p) Sm dT + dSm = T p Sm dp p T Vm Sm = [Maxwell relation] p T T p
Cp,m Sm = [Problem 5.7] T p T dqrev = T dSm = Cp,m dT  T
P6.22
Vm dp T p q p Hm = Cp,m  T Vm = Cp,m  Vm [6.7] CS = T s T s Vm   C C(graphite) C(diamond) = 2.8678 kJ mol1 at T. rG We want the pressure at which r G = 0; above that pressure the reaction will be spontaneous. Equation 5.10 determines the rate of change of r G with p at constant T . (1) rG = r V = (VD  VG )M p T where M is the molar mas of carbon; VD and VG are the specific volumes of diamond and graphite, respectively.   C C = 100 kPa at T. r G(T, p) may be expanded in a Taylor series around the pressure p
C r G(T,
(2)
p) = +
rG
 C  (T,
1 2
 C  r G (T, p  )  (p  p  ) p T  C  2 r G (T, p  )   (p  p  )2 + (p  p  )3 p 2  p ) + T
96
INSTRUCTOR'S MANUAL
We will neglect the third and higherorder terms; the derivative of the firstorder term can be calculated with eqn 1. An expression for the derivative of the secondorder term can be derived with eqn 1. (3) VG VD M = {VG T (G)  VD T (D)}M [3.13]  p T p T T  C Calculating the derivatives of eqns 1 and 2 at T and p  2 rG p 2 =
 C r G(T, p  ) = (0.284  0.444) p T  C 2 r G(T, p  ) 2 p
(4)
cm3 g
12.01 g mol
= 1.92 cm3 mol1
(5)
= {0.444(3.04 108 )  0.284(0.187 108 )}
T
cm3 kPa1 g
12.01 g mol
= 1.56 107 cm3 (kPa)1 mol1  It is convenient to convert the value of r G to the units cm3 kPa mol1 8.315 102 L bar K1 mol1 103 cm3   = 2.8678 kJ mol1 rG L 8.315 J K1 mol1 (6)
rG  
105 Pa bar
= 2.8678 106 cm3 kPa mol1  Setting = p  p  , eqns 2 and 36 give 6 3 2.8678 10 cm kPa mol1  (1.92 cm3 mol1 ) + (7.80 108 cm3 kPa1 mol1 ) 2 = 0 C when r G(T, p) = 0. One real root of this equation is
 = 1.60 106 kPa = p  p  or
p = 1.60 106 kPa  102 kPa = 1.60 106 kPa = 1.60 104 bar Above this pressure the reaction is spontaneous. The other real root is much higher: 2.3107 kPa. Question. What interpretation might you give to the other real root?
7
Simple mixtures
Solutions to exercises
Discussion questions
E7.1(b) E7.2(b) For a component in an ideal solution, Raoult's law is: p = xp . For real solutions, the activity, a, replaces the mole fraction, x, and Raoult's law becomes p = ap . All the colligative properties are a result of the lowering of the chemical potential of the solvent due to the presence of the solute. This reduction takes the form A = A + RT ln xA or A = A + RT ln aA , depending on whether or not the solution can be considered ideal. The lowering of the chemical potential results in a freezing point depression and a boiling point elevation as illustrated in Fig. 7.20 of the text. Both of these effects can be explained by the lowering of the vapour pressure of the solvent in solution due to the presence of the solute. The solute molecules get in the way of the solvent molecules, reducing their escaping tendency. The activity of a solute is that property which determines how the chemical potential of the solute  varies from its value in a specified reference state. This is seen from the relation =  + RT ln a,   where is the value of the chemical potential in the reference state. The reference state is either the hypothetical state where the pure solute obeys Henry's law (if the solute is volatile) or the hypothetical state where the solute at unit molality obeys Henry's law (if the solute is involatile). The activity of the solute can then be defined as that physical property which makes the above relation true. It can be interpreted as an effective concentration.
E7.3(b)
Numerical exercises
E7.4(b) Total volume V = nA VA + nB VB = n(xA VA + xB VB ) Total mass m = nA MA + nB MB = n(xA MA + (1  xA )MB ) m =n xA MA + (1  xA )MB n= where n = nA + nB
1.000 kg(103 g/kg) = 4.6701 mol (0.3713) (241.1 g/mol) + (1  0.3713) (198.2 g/mol) V = n(xA VA + xB VB ) = (4.6701 mol) [(0.3713) (188.2 cm3 mol1 ) + (1  0.3713) (176.14 cm3 mol1 )] = 843.5 cm3 E7.5(b) Let A denote water and B ethanol. The total volume of the solution is V = nA VA + nB VB We know VB ; we need to determine nA and nB in order to solve for VA . Assume we have 100 cm3 of solution; then the mass is m = V = (0.9687 g cm3 ) (100 cm3 ) = 96.87 g of which (0.20) (96.87 g) = 19.374 g is ethanol and (0.80) (96.87 g) = 77.496 g is water. 77.496 g = 4.30 mol H2 O 18.02 g mol1 19.374 g nB = = 0.4205 mol ethanol 46.07 g mol1 nA =
98
INSTRUCTOR'S MANUAL
100 cm3  (0.4205 mol) (52.2 cm3 mol1 ) V  n B VB = 18.15 cm3 = VA = nA 4.30 mol = 18 cm3 E7.6(b) Check that pB /xB = a constant (KB ) xB (pB /xB )/kPa 0.010 8.2 103 0.015 8.1 103 0.020 8.3 103
KB = p/x, average value is 8.2 103 kPa E7.7(b) In exercise 7.6(b), the Henry's law constant was determined for concentrations expressed in mole fractions. Thus the concentration in molality must be converted to mole fraction. m(A) = 1000 g, corresponding to 1000 g n(A) = = 13.50 mol 74.1 g mol1 Therefore, xB = 0.25 mol = 0.0182 0.25 mol + 13.50 mol
n(B) = 0.25 mol
using KB = 8.2 103 kPa [exercise 7.6(b)] p = 0.0182 8.2 103 kPa = 1.5 102 kPa RT 2 M 8.314 J K1 mol1 (354 K)2 0.12818 kg mol1 = 18.80 103 J mol1 fus H = 7.1 K kg mol1 Kb = RT 2 M 8.314 J K1 mol1 (490.9 K)2 0.12818 kg mol1 = 51.51 103 J mol1 vap H = 4.99 K kg mol1 E7.9(b) We assume that the solvent, 2propanol, is ideal and obeys Raoult's law. xA (solvent) = p/p = 49.62 = 0.9924 50.00
E7.8(b)
Kf =
MA (C3 H8 O) = 60.096 g mol1 250 g = 4.1600 mol 60.096 g mol1 nA nA xA = nA + n B = nA + n B xA nA =
SIMPLE MIXTURES
99
nB = nA
1 1 xA 1  1 = 3.186 102 mol 0.9924
= 4.1600 mol MB = E7.10(b)
8.69 g = 273 g mol1 = 270 g mol1 3.186 102 mol
Kf = 6.94 for naphthalene MB = mass of B nB T Kf (mass of B) Kf (mass of naphthalene)
nB = mass of naphthalene bB bB = MB = E7.11(b) so MB = T
(5.00 g) (6.94 K kg mol1 ) = 178 g mol1 (0.250 kg) (0.780 K) nB nB = T = Kf bB and bB = mass of water V (density of solution density of water) T = Kf RT Kf = 1.86 K mol1 kg V RT
= 103 kg m3 nB = T =
(1.86 K kg mol1 ) (99 103 Pa) = 7.7 102 K (8.314 J K1 mol1 ) (288 K) (103 kg m3 ) = nRT (xA ln xA + xB ln xB ) xAr = xNe = 0.5, n = nAr + nNe = pV RT
Tf = 0.077 C E7.12(b)
mix G
nAr = nNe ,
mix G
1 1 1 1 = pV 2 ln 2 + 2 ln 2 = pV ln 2
= (100 103 Pa) (0.250 L) = 17.3 Pa m3 = 17.3 J E7.13(b)
mix G
1 m3 ln 2 103 L 17.3 J  mix G = = 6.34 102 J K1 T 273 K  mix G xJ ln xJ [7.19] = mix S = nR T J
mix S
=
= nRT
J
xJ ln xJ [7.18]
n = 1.00 mol + 1.00 mol = 2.00 mol x(Hex) = x(Hep) = 0.500 Therefore,
mix G
= (2.00 mol) (8.314 J K 1 mol1 ) (298 K) (0.500 ln 0.500 + 0.500 ln 0.500) = 3.43 kJ
100
INSTRUCTOR'S MANUAL
+3.43 kJ = +11.5 J K1 298 K mix H for an ideal solution is zero as it is for a solution of perfect gases [7.20]. It can be demonstrated from
mix S
=
mix H
=
mix G + T
mix S
= (3.43 103 J) + (298 K) (11.5 J K 1 ) = 0
E7.14(b)
Benzene and ethylbenzene form nearly ideal solutions, so
mix S
= nR(xA ln xA + xB ln xB )
mix S, differentiate with respect to xA
To find maximum is zero.
mix S
and find value of xA at which the derivative
Note that xB = 1  xA so = nR(xA ln xA + (1  xA ) ln(1  xA )) 1 d ln x = x dx xA d ( mix S) = nR(ln xA + 1  ln(1  xA )  1) = nR ln dx 1  xA =0
1 when xA = 2
use
Thus the maximum entropy of mixing is attained by mixing equal molar amounts of two components. mB /MB nB = 1 = nE mE /ME mE ME 106.169 = 1.3591 = = 78.115 mB MB
mB = 0.7358 mE E7.15(b) Assume Henry's law [7.26] applies; therefore, with K(N2 ) = 6.51 107 Torr and K(O2 ) = 3.30 107 Torr, as in Exercise 7.14, the amount of dissolved gas in 1 kg of water is n(N2 ) = 103 g 18.02 g mol1 p(N2 ) 6.51 107 Torr = (8.52 107 mol) (p/Torr)
For p(N2 ) = xp and p = 760 Torr n(N2 ) = (8.52 107 mol) (x) (760) = x(6.48 104 mol) and, with x = 0.78 n(N2 ) = (0.78) (6.48 104 mol) = 5.1 104 mol = 0.51 mmol The molality of the solution is therefore approximately 0.51 mmol kg1 in N2 . Similarly, for oxygen, n(O2 ) = 103 g 18.02 g mol1 p(O2 ) 3.30 107 Torr = (1.68 106 mol) (p/Torr)
For p(O2 ) = xp and p = 760 Torr n(O2 ) = (1.68 106 mol) (x) (760) = x(1.28 mmol) and when x = 0.21, n(O2 ) 0.27 mmol. Hence the solution will be 0.27 mmol kg1 in O2 .
SIMPLE MIXTURES
101
E7.16(b)
Use n(CO2 ) = (4.4 105 mol) (p/Torr), p = 2.0(760 Torr) = 1520 Torr n(CO2 ) = (4.4 105 mol) (1520) = 0.067 mol The molality will be about 0.067 mol kg1 and, since molalities and molar concentration for dilute aqueous solutions are approximately equal, the molar concentration is about 0.067 mol L1
E7.17(b)
M(glucose) = 180.16 g mol1 T = Kf bB Kf = 1.86 K kg mol1 10 g/180.16 g mol1 0.200 kg = 0.52 K
T = (1.86 K kg mol1 )
Freezing point will be 0 C  0.52 C = 0.52 C E7.18(b) The procedure here is identical to Exercise 7.18(a). ln xB = =
fus H
R
1 1  T T
[7.39; B, the solute, is lead] 1 1  600 K 553 K xB n(Bi) 1  xB
5.2 103 J mol1 8.314 J K1 mol1
= 0.0886, xB =
implying that xB = 0.92 implying that n(Pb) =
n(Pb) , n(Pb) + n(Bi)
For 1 kg of bismuth, n(Bi) =
1000 g = 4.785 mol 208.98 g mol1 Hence, the amount of lead that dissolves in 1 kg of bismuth is n(Pb) = (0.92) (4.785 mol) = 55 mol, 1  0.92 or 11 kg
Comment. It is highly unlikely that a solution of 11 kg of lead and 1 kg of bismuth could in any sense be considered ideal. The assumptions upon which eqn 7.39 is based are not likely to apply. The answer above must then be considered an order of magnitude result only. E7.19(b) Proceed as in Exercise 7.19(a). The data are plotted in Fig. 7.1, and the slope of the line is 1.78 cm/ (mg cm3 ) = 1.78 cm/(g L1 ) = 1.78 102 m4 kg1 .
12
10
8
6 3 4 5 6 7
Figure 7.1
102
INSTRUCTOR'S MANUAL
Therefore, M= E7.20(b) (8.314 J K1 mol1 ) (293.15 K) = 14.0 kg mol1 (1.000 103 kg m3 ) (9.81 m s2 ) (1.78 102 m4 kg1 )
Let A = water and B = solute. aA = pA 0.02239 atm [42] = 0.02308 atm = 0.9701 pA nA aA and xA = A = xA nA + n B 0.122 kg 0.920 kg = 0.506 mol nA = = 51.05 mol nB = 1 0.01802 kg mol 0.241 kg mol1 51.05 0.9701 xA = = 0.990 A = = 0.980 51.05 + 0.506 0.990 B (l) = (l) + RT ln xB [7.50, ideal solution] B
E7.21(b)
B = Benzene
RT ln xB = (8.314 J K1 mol1 ) (353.3 K) (ln 0.30) = 3536 J mol1 Thus, its chemical potential is lowered by this amount.
pB = aB pB [42] = B xB pB = (0.93) (0.30) (760 Torr) = 212 Torr
E7.22(b)
Question. What is the lowering of the chemical potential in the nonideal solution with = 0.93? pA pA = yA = = 0.314 pA + p B 760 Torr pA = (760 Torr) (0.314) = 238.64 Torr pB = 760 Torr  238.64 Torr = 521.36 Torr pA 238.64 Torr aA = = 1 atm pA 3 Pa) (73.0 10 760 Torr
101 325 Pa atm
= 0.436
aB = A =
pB 521.36 Torr = 1 atm pB (92.1 103 Pa) 101 325 Pa 760 Torr atm
= 0.755
aA 0.436 = = 1.98 xA 0.220 aB 0.755 B = = 0.968 = xB 0.780
Solutions to problems
Solutions to numerical problems
P7.3 Vsalt = V mol1 [Problem 7.2] b H2 O with b b/(mol kg1 )
= 69.38(b  0.070) cm3 mol1
Therefore, at b = 0.050 mol kg1 , Vsalt = 1.4 cm3 mol1
SIMPLE MIXTURES
103
The total volume at this molality is V = (1001.21) + (34.69) (0.02)2 cm3 = 1001.22 cm3 Hence, as in Problem 7.2, V (H2 O) = (1001.22 cm3 )  (0.050 mol) (1.4 cm3 mol1 ) = 18.04 cm3 mol1 55.49 mol
Question. What meaning can be ascribed to a negative partial molar volume? P7.5 Let E denote ethanol and W denote water; then V = nE VE + nW VW [7.3] For a 50 per cent mixture by mass, mE = mW , implying that nE ME = nW MW , Hence, V = nE VE + which solves to nE = Furthermore, xE = or nW = nE ME MW
n E M E VW MW V VE +
ME VW MW
, nW =
ME V VE M W + M E V W
nE 1 = ME nE + n W 1 + MW ME = 2.557. Therefore MW
Since ME = 46.07 g mol1 and MW = 18.02 g mol1 , xE = 0.2811, At this composition VE = 56.0 cm3 mol1 Therefore, nE = xW = 1  xE = 0.7189
VW = 17.5 cm3 mol1 [Fig.7.1 of the text]
100 cm3 = 0.993 mol (56.0 cm3 mol1 ) + (2.557) (17.5 cm3 mol1 )
nW = (2.557) (0.993 mol) = 2.54 mol The fact that these amounts correspond to a mixture containing 50 per cent by mass of both components is easily checked as follows mE = nE ME = (0.993 mol) (46.07 g mol1 ) = 45.7 g ethanol mW = nW MW = (2.54 mol) (18.02 g mol1 ) = 45.7 g water At 20 C the densities of ethanol and water are, E = 0.789 g cm3 , W = 0.997 g cm3 . Hence, VE = VW = mE 45.7 g = = 57.9 cm3 of ethanol E 0.789 g cm3 mW 45.7 g = = 45.8 cm3 of water W 0.997 g cm3
104
INSTRUCTOR'S MANUAL
The change in volume upon adding a small amount of ethanol can be approximated by V = dV VE dnE VE nE
where we have assumed that both VE and VW are constant over this small range of nE . Hence V (56.0 cm3 mol1 ) (1.00 cm3 ) (0.789 g cm3 ) (46.07 g mol1 ) = +0.96 cm3
P7.7
mB =
T 0.0703 K = 0.0378 mol kg1 = 1 ) Kf 1.86 K/(mol kg
P7.9
Since the solution molality is nominally 0.0096 mol kg1 in Th(NO3 )4 , each formula unit supplies 0.0378 4 ions. (More careful data, as described in the original reference gives 5 to 6.) 0.0096 The data are plotted in Figure 7.2. The regions where the vapor pressure curves show approximate straight lines are denoted R for Raoult and H for Henry. A and B denote acetic acid and benzene respectively.
300 Extrapolate for KB
R 200 B p / Torr Raoult
Henry
100 Henry
H
A Raoult R H 0.8 1.0
0
0
0.2
0.4 xA
0.6
Figure 7.2
pA pB As in Problem 7.8, we need to form A = and B = x p for the Raoult's law activity xA p A B B pB coefficients and B = for the activity coefficient of benzene on a Henry's law basis, with K xB K determined by extrapolation. We use pA = 55 Torr, pB = 264 Torr and KB = 600 Torr to draw up
SIMPLE MIXTURES
105
the following table:
xA pA /Torr pB /Torr aA (R) aB (R) A (R) B (R) aB (H) B (H) 0 0 264 0 1.00  1.00 0.44 0.44 0.2 20 228 0.36 0.86 1.82 1.08 0.38 0.48 0.4 30 190 0.55 0.72 1.36 1.20 0.32 0.53 0.6 38 150 0.69 0.57 1.15 1.42 0.25 0.63 0.8 50 93 0.91 0.35 1.14 1.76 0.16 0.78 1.0 55 0 1.00[pA /pA ] 0[pB /pB ] 1.00[pA /xA pA ] [pB /xB pB ] 0[pB /KB ] 1.00[pB /xB KB ]
GE is defined as [Section 7.4]: GE =
mix G(actual)  mix G(ideal)
= nRT (xA ln aA + xB ln aB )  nRT (xA ln xA + xB ln xB )
and with a = x GE = nRT (xA ln A + xA ln B ). For n = 1, we can draw up the following table from the information above and RT = 2.69 kJ mol1 :
xA xA ln A xB ln B GE /(kJ mol1 ) 0 0 0 0 0.2 0.12 0.06 0.48 0.4 0.12 0.11 0.62 0.6 0.08 0.14 0.59 0.8 0.10 0.11 0.56 1.0 0 0 0
P7.11
(a) The volume of an ideal mixture is Videal = n1 Vm,1 + n2 Vm,2 so the volume of a real mixture is V = Videal + V E We have an expression for excess molar volume in terms of mole fractions. To compute partial molar volumes, we need an expression for the actual excess volume as a function of moles
E V E = (n1 + n2 )Vm =
n 1 n2 n1 + n 2
a0 +
a1 (n1  n2 ) n1 + n 2
so V = n1 Vm,1 + n2 Vm,2 +
n1 n2 a1 (n1  n2 ) a0 + n1 + n 2 n1 + n 2 The partial molar volume of propionic acid is V1 = a 0 n2 a1 (3n1  n2 )n2 V 2 2 = Vm,1 + + n1 p,T ,n2 (n1 + n2 )2 (n1 + n2 )3
2 2 V1 = Vm,1 + a0 x2 + a1 (3x1  x2 )x2
That of oxane is
2 2 V2 = Vm,2 + a0 x1 + a1 (x1  3x2 )x1
106
INSTRUCTOR'S MANUAL
(b) We need the molar volumes of the pure liquids Vm,1 = and Vm,2 = M1 74.08 g mol1 = = 76.23 cm3 mol1 1 0.97174 g cm3
86.13 g mol1 = 99.69 cm3 mol1 0.86398 g cm3 In an equimolar mixture, the partial molar volume of propionic acid is V1 = 76.23 + (2.4697) (0.500)2 + (0.0608) [3(0.5)  0.5] (0.5)2 cm3 mol1 = 75.63 cm3 mol1 and that of oxane is V2 = 99.69 + (2.4697) (0.500)2 + (0.0608) [0.5  3(0.5)] (0.5)2 cm3 mol1 = 99.06 cm3 mol1 P7.13 Henry's law constant is the slope of a plot of pB versus xB in the limit of zero xB (Fig. 7.3). The partial pressures of CO2 are almost but not quite equal to the total pressures reported above pCO2 = pyCO2 = p(1  ycyc ) Linear regression of the lowpressure points gives KH = 371 bar
80
60
40
20
0 0.0 0.1 0.2 0.3
Figure 7.3 The activity of a solute is aB = pB = xB B KH
so the activity coefficient is B = pB yB p = xB K H xB K H
SIMPLE MIXTURES
107
where the last equality applies Dalton's law of partial pressures to the vapour phase. A spreadsheet applied this equation to the above data to yield
p/bar 10.0 20.0 30.0 40.0 60.0 80.0 ycyc 0.0267 0.0149 0.0112 0.009 47 0.008 35 0.009 21 xcyc 0.9741 0.9464 0.9204 0.892 0.836 0.773 CO2 1.01 0.99 1.00 0.99 0.98 0.94
P7.16
GE = RT x(1  x){0.4857  0.1077(2x  1) + 0.0191(2x  1)2 } with x = 0.25 gives GE = 0.1021RT . Therefore, since
mix G(actual) mix G
=
mix G(ideal) + nG
E
= nRT (xA ln xA + xB ln xB ) + nGE = nRT (0.25 ln 0.25 + 0.75 ln 0.75) + nGE = 0.562nRT + 0.1021nRT = 0.460nRT
Since n = 4 mol and RT = (8.314 J K 1 mol1 ) (303.15 K) = 2.52 kJ mol1 ,
mix G
= (0.460) (4 mol) (2.52 kJ mol1 ) = 4.6 kJ
Solutions to theoretical problems
P7.18 xA dA + xB dB = 0 [7.11, GibbsDuhem equation] Therefore, after dividing through by dxA xA A + xB xA p,T A  xB xA p,T B =0 xA p,T B =0 xB p,T
or, since dxB = dxA , as xA + xB = 1 xA or,
dx B d ln x = ln xB p,T x f ln fA  Then, since =  + RT ln  , = p ln xA p,T
A = ln xA p,T
ln fB ln xB p,T
ln pA ln pB = ln xA p,T ln xB p,T If A satisfies Raoult's law, we can write pA = xA pA , which implies that On replacing f by p,
ln pA ln pA ln xA = + =1+0 ln xA p,T ln xA ln xA
ln pB =1 ln xB p,T which is satisfied if pB = xB pB (by integration, or inspection). Hence, if A satisfies Raoult's law, so does B. Therefore,
108
INSTRUCTOR'S MANUAL
P7.20
ln xA =
 fus G (Section 7.5 analogous to equation for ln xB used in derivation of eqn 7.39) RT
fus G
d ln xA 1 d = dT R dT
xA 1
T
fus H dT RT 2
=
fus H RT 2 fus H T T
[GibbsHelmholtz equation] dT T2
d ln xA =
T T
R
ln xA =
 fus H R
1 1  T T
The approximations ln xA xB and T T then lead to eqns 33 and 36, as in the text. P7.22 Retrace the argument leading to eqn 7.40 of the text. Exactly the same process applies with aA in place of xA . At equilibrium (p) = A (xA , p + A )
which implies that, with = + RT ln a for a real solution, (p) = (p + A A
p+
) + RT ln aA = (p) + A Vm dp = RT ln aA
p+ p
Vm dp + RT ln aA
and hence that
p
For an incompressible solution, the integral evaluates to In terms of the osmotic coefficient (Problem 7.21) Vm = rRT r= xB nB = xA nA =
Vm , so
Vm = RT ln aA
xA 1 ln aA =  ln aA xB r
For a dilute solution, nA Vm V Hence, V = nB RT and therefore, with [B] = nB V = [B]RT
Solutions to applications
P7.24 By the van't Hoff equation [7.40] cRT = [B]RT = M Division by the standard acceleration of free fall, g, gives c(R/g)T = 8 M (a) This expression may be written in the form cR T = M which has the same form as the van't Hoff equation, but the unit of osmotic pressure ( ) is now force/area (mass length)/(area time2 ) mass = = 2 2 area length/time length/time
SIMPLE MIXTURES
109
This ratio can be specified in g cm2 . Likewise, the constant of proportionality (R ) would have the units of R/g (mass length2 /time2 ) K1 mol1 energy K 1 mol1 = = mass length K 1 mol1 2 length/time length/time2 This result may be specified in g cm K1 mol1 R = 8.314 51 J K1 mol1 R = g 9.806 65m s2 103 g kg 102 cm m
= 0.847 844 kg m K1 mol1 R = 84 784.4 g cm K1 mol1
In the following we will drop the primes giving cRT = M and use the units of g cm2 and the R units g cm K1 mol1 . (b) By extrapolating the low concentration plot of /c versus c (Fig. 7.4 (a)) to c = 0 we find the intercept 230 g cm2 /g cm3 . In this limit van't Hoff equation is valid so RT RT = intercept or M n = intercept Mn (84 784.4 g cm K1 mol1 ) (298.15 K) Mn = (230 g cm2 )/(g cm3 ) M n = 1.1 105 g mol
1
500
450
400
350
300
250
200 0.000
0.010
0.020
0.030
0.040
Figure 7.4(a)
110
INSTRUCTOR'S MANUAL
(c) The plot of /c versus c for the full concentration range (Fig. 7.4(b)) is very nonlinear. We may conclude that the solvent is good . This may be due to the nonpolar nature of both solvent and solute.
7000
6000
5000
4000
3000
2000
1000
0 0.00 0.050 0.100 0.150 0.200 0.250 0.300
Figure 7.4(b)
(d) /c = (RT /M n )(1 + B c + C c2 ) Since RT /M n has been determined in part (b) by extrapolation to c = 0, it is best to determine the second and third virial coefficients with the linear regression fit (/c)/(RT /M n )  1 =B +C c c R = 0.9791 B = 21.4 cm3 g1 , C = 211 cm6 g2 , standard deviation = 2.4 cm3 g1 standard deviation = 15 cm6 g2
(e) Using 1/4 for g and neglecting terms beyond the second power, we may write 1/2 = c RT 1/2 Mn
1 (1 + 2 B c)
SIMPLE MIXTURES
111
We can solve for B , then g(B )2 = C . 1/2
c RT Mn 1/2
1  1 = 2B c
RT /M n has been determined above as 230 g cm2 /g cm3 . We may analytically solve for B from one of the data points, say, /c = 430 g cm2 /g cm3 at c = 0.033 g cm3 . 430 g cm2 /g cm3 1/2 1  1 = 2 B (0.033 g cm3 ) 230 g cm2 /g cm3 2 (1.367  1) B = = 22.2 cm3 g1 0.033 g cm3 C = g(B )2 = 0.25 (22.2 cm3 g1 )2 = 123 cm6 g2 Better values of B and C can be obtained by plotting
1/2
RT 1/2 Mn
c
against c. This plot
is shown in Fig. 7.4(c). The slope is 14.03 cm3 g1 . B = 2 slope = 28.0 cm3 g1 C is then 196 cm6 g2 The intercept of this plot should thereotically be 1.00, but it is in fact 0.916 with a standard deviation of 0.066. The overall consistency of the values of the parameters confirms that g is roughly 1/4 as assumed.
6.0
5.0
4.0
n
3.0
2.0
1.0
0.0 0.00
0.05
0.10
0.15
0.20
0.25
0.30
Figure 7.4(c)
8
Phase diagrams
Solutions to exercises
Discussion questions
E8.1(b) What factors determine the number of theoretical plates required to achieve a desired degree of separation in fractional distillation? The principal factor is the shape of the twophase liquidvapor region in the phase diagram (usually a temperaturecomposition diagram). The closer the liquid and vapour lines are to each other, the more theoretical plates needed. See Fig. 8.15 of the text. But the presence of an azeotrope could prevent the desired degree of separation from being achieved. Incomplete miscibility of the components at specific concentrations could also affect the number of plates required. See Figs 8.1(a) and 8.1(b).
E8.2(b)
Liquid A&B Solid B
Liquid A and B
Liquid A&B Liquid A & B Solid AB2 Eutectic Solid AB2 and Solid A 0.33 Solid A
Solid B and Solid AB2
Figure 8.1(a)
B
Vapor
dl iqu id
d
Liquid
0.67
Va p
d li
qui
or
Vap o
r an
an
Figure 8.1(b)
PHASE DIAGRAMS
113
E8.3(b)
See Fig. 8.2.
Liquid A&B Liquid (A & B)
Solid B
Liquid (A & B) Solid A Solid A Solid A2B
Liquid Liquid (A & B) (A & B) Solid A2B Solid A2B
Liquid (A & B) Solid B2A
Solid B Solid B2A Two solid phases
Solid A2B Solid B2A Two solid phases 0.666
Two solid phases 0.333
Figure 8.2
Numerical exercises
E8.4(b)
p = pA + pB = xA pA + (1  xA )pB p  pB  p pA B
xA = xA = yA =
19 kPa  18 kPa = 0.5 20 kPa  18 kPa
A is 1,2dimethylbenzene
x A pA (0.5) (20 kPa) + (p  p )x = 18 kPa + (20 kPa  18 kPa)0.5 = 0.526 0.5 pB B A A
yB = 1  0.526 = 0.474 0.5 E8.5(b)
pA = yA p = 0.612p = xA pA = xA (68.8 kPa) pB = yB p = (1  yA )p = 0.388p = xB pB = (1  xA ) 82.1 kPa xA pA yA p = yB p xB p B
and
68.8xA 0.612 = 0.388 82.1(1  xA )
(0.388) (68.8)xA = (0.612) (82.1)  (0.612) (82.1)xA 26.694xA = 50.245  50.245xA xA = 50.245 26.694 + 50.245 = 0.653 xB = 1  0.653 = 0.347
p = xA pA + xB pB = (0.653) (68.8 kPa) + (0.347) (82.1 kPa) = 73.4 kPa
114
INSTRUCTOR'S MANUAL
E8.6(b)
(a) If Raoult's law holds, the solution is ideal.
pA = xA pA = (0.4217) (110.1 kPa) = 46.43 kPa pB = xB pB = (1  0.4217) (94.93 kPa) = 54.90 kPa
p = pA + pB = (46.43 + 54.90) kPa = 101.33 kPa = 1.000 atm Therefore, Raoult's law correctly predicts the pressure of the boiling liquid and the solution is ideal . 46.43 kPa pA (b) yA = = = 0.4582 p 101.33 kPa yB = 1  yA = 1.000  0.4582 = 0.5418 E8.7(b) Let B = benzene and T = toluene. Since the solution is equimolar zB = zT = 0.500 (a) Initially xB = zB and xT = zT ; thus p = xB pB + xT pT [8.3] = (0.500) (74 Torr) + (0.500) (22 Torr) = 37 Torr + 11 Torr = 48 Torr pB 37 Torr = 0.77 (b) yB = [4] = 48 Torr p (c) Near the end of the distillation yB = zB = 0.500 y T = 1  0.77 = 0.23
and y T = zT = 0.500
Equation 5 may be solved for xA [A = benzene = B here] xB =
y B pT (0.500) (22 Torr) + (p  p )y = (75 Torr) + (22  74) Torr (0.500) = 0.23 pB T B B
xT = 1  0.23 = 0.77 This result for the special case of zB = zT = 0.500 could have been obtained directly by realizing that yB (initial) = xT (final) yT (initial) = xB (final)
p(final) = xB pB + xT pT = (0.23) (74 Torr) + (0.77) (22 Torr) = 34 Torr
Thus in the course of the distillation the vapour pressure fell from 48 Torr to 34 Torr. E8.8(b) See the phase diagram in Fig. 8.3. (a) (b) E8.9(b) yA = 0.81 xA = 0.67 AlCl3 + 3H2 O AlCl3 H2 O Al
3+
yA = 0.925 Al(OH)3 + 3HCl + 3Cl
Al3+ , H+ , AlCl3 , Al(OH)3 , OH , Cl , H2 O giving seven species. There are also three equilibria
H+ + OH
and one condition of electrical neutrality [H+ ] + 3[Al3+ ] = [OH ] + [Cl ] Hence, the number of independent components is C = 7  (3 + 1) = 3
PHASE DIAGRAMS
115
155 150
145
140 135 130
125 120 0 0.2 0.4 0.6 0.8 1.0
Figure 8.3 E8.10(b) NH3 (g) + HCl(g)
NH4 Cl(s)
(a) For this system C = 1 [Example 8.1] and P = 2 (s and g). (b) If ammonia is added before heating, C = 2 (because NH4 Cl, NH3 are now independent) and P = 2 (s and g). E8.11(b) (a) Still C = 2 (Na2 SO4 , H2 O), but now there is no solid phase present, so P = 2 (liquid solution, vapour). (b) The variance is F = 2  2 + 2 = 2 . We are free to change any two of the three variables, amount of dissolved salt, pressure, or temperature, but not the third. If we change the amount of dissolved salt and the pressure, the temperature is fixed by the equilibrium condition between the two phases. E8.12(b) E8.13(b) See Fig. 8.4. See Fig. 8.5. The phase diagram should be labelled as in Fig. 8.5. (a) Solid Ag with dissolved Sn begins to precipitate at a1 , and the sample solidifies completely at a2 . (b) Solid Ag with dissolved Sn begins to precipitate at b1 , and the liquid becomes richer in Sn. The peritectic reaction occurs at b2 , and as cooling continues Ag3 Sn is precipitated and the liquid becomes richer in Sn. At b3 the system has its eutectic composition (e) and freezes without further change. See Fig. 8.6. The feature denoting incongruent melting is circled. Arrows on the tie line indicate the decomposition products. There are two eutectics: one at xB = 0.53 , T = T2 ; another at xB = 0.82 , T = T3 . E8.15(b) The cooling curves corresponding to the phase diagram in Fig. 8.7(a) are shown in Fig. 8.7(b). Note the breaks (abrupt change in slope) at temperatures corresponding to points a1 , b1 , and b2 . Also note the eutectic halts at a2 and b3 .
E8.14(b)
116
INSTRUCTOR'S MANUAL
Figure 8.4
(a)
(b)
800
460
e
200
Figure 8.5
0
0.33
0.67
Figure 8.6
PHASE DIAGRAMS
117
(a)
(b)
0
0.33
0.67
1
Figure 8.7 E8.16(b) E8.17(b) Rough estimates based on Fig. 8.37 of the text are (a) xB 0.75 (b) xAB2 0.8 (c) xAB2 0.6 The phase diagram is shown in Fig. 8.8. The given data points are circled. The lines are schematic at best.
1000
900
800
700
0
0.2
0.4
0.6
0.8
Figure 8.8 A solid solution with x(ZrF4 ) = 0.24 appears at 855 C. The solid solution continues to form, and its ZrF4 content increases until it reaches x(ZrF4 ) = 0.40 and 820 C. At that temperature, the entire sample is solid.
118
INSTRUCTOR'S MANUAL
E8.18(b)
The phase diagram for this system (Fig. 8.9) is very similar to that for the system methyl ethyl ether and diborane of Exercise 8.12(a). (See the Student's Solutions Manual.) The regions of the diagram contain analogous substances. The solid compound begins to crystallize at 120 K. The liquid becomes progressively richer in diborane until the liquid composition reaches 0.90 at 104 K. At that point the liquid disappears as heat is removed. Below 104 K the system is a mixture of solid compound and solid diborane.
140
130
120
110
100
90 0 1
Figure 8.9
E8.19
Refer to the phase diagram in the solution to Exercise 8.17(a). (See the Student's Solutions Manual.) The cooling curves are sketched in Fig. 8.10.
95 93 91 89 87 85 83
Figure 8.10
E8.20
(a) When xA falls to 0.47, a second liquid phase appears. The amount of new phase increases as xA falls and the amount of original phase decreases until, at xA = 0.314, only one liquid remains. (b) The mixture has a single liquid phase at all compositions. The phase diagram is sketched in Fig. 8.11.
Solutions to problems
Solutions to numerical problems
P8.2 (a) The phase diagram is shown in Fig. 8.12. (b) We need not interpolate data, for 296.0 K is a temperature for which we have experimental data. The mole fraction of N, Ndimethylacetamide in the heptanerich phase (, at the left of the phase diagram) is 0.168 and in the acetamiderich phase (, at right) 0.804. The proportions of the two
PHASE DIAGRAMS
119
54 52 50 48 46 44 42 40 38 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Figure 8.11
310
305
300
295 290 0.0 0.2 0.4 0.6 0.8 1.0
Figure 8.12 phases are in an inverse ratio of the distance their mole fractions are from the composition point in question, according to the lever rule. That is n /n = l / l = (0.804  0.750)/(0.750  0.168) = 0.093 The smooth curve through the data crosses x = 0.750 at 302.5 K , the temperature point at which the heptanerich phase will vanish. P8.6 See Fig. 8.13(a). The number of distinct chemical species (as opposed to components) and phases present at the indicated points are, respectively b(3, 2), d(2, 2), e(4, 3), f (4, 3), g(4, 3), k(2, 2) [Liquid A and solid A are here considered distinct species.] The cooling curves are shown in Fig. 8.13(b). P8.8 The information has been used to construct the phase diagram in Fig. 8.14(a). In MgCu2 the mass 24.3 48.6 = 16 , and in Mg2 Cu it is (100) = 43 . percentage of Mg is (100) 24.3 + 127 48.6 + 63.5 The initial point is a1 , corresponding to a liquid singlephase system. At a2 (at 720 C) MgCu2 begins to come out of solution and the liquid becomes richer in Mg, moving toward e2 . At a3 there is solid
120
INSTRUCTOR'S MANUAL
Liquid A & B Liquid A & B Solid B b Liquid A & B Solid A c e d f Liquid A & B Solid AB2 Solid A and Solid AB2 g
k Solid AB2 and Solid B B
A 16% 23% xB 57% 67% 84%
Figure 8.13(a)
0.16
0.23
0.57
0.67
0.84
Figure 8.13(b)
MgCu2 + liquid of composition e2 (33 per cent by mass of Mg). This solution freezes without further change. The cooling curve will resemble that shown in Fig. 8.14(b). P8.10 (a) eutectic: 40.2 at % Si at 1268 C eutectic: 69.4 at % Si at 1030 C [8.7] melts into CaSi(s) and liquid [8.6]
congruent melting compounds: Ca2 Si mp = 1314 C CaSi mp = 1324 C incongruent melting compound: CaSi2 (68 at % Si) mp = 1040 C
PHASE DIAGRAMS
121
(a) 1200 a
(b)
800 e1
a1 a2 a3 e2 e3
400
Figure 8.14 (b) At 1000 C the phases at equilibrium will be Ca(s) and liquid (13 at % Si). The lever rule gives the relative amounts: lliq 0.2  0 nCa = 2.86 = = nliq lCa 0.2  0.13 (c) When an 80 at % Si melt it cooled in a manner that maintains equilibrium, Si(s) begins to appear at about 1250 C. Further cooling causes more Si(s) to freeze out of the melt so that the melt becomes more concentrated in Ca. There is a 69.4 at % Si eutectic at 1030 C. Just before the eutectic is reached, the lever rule says that the relative amounts of the Si(s) and liquid (69.4% Si) phases are: lliq 0.80  0.694 nSi = = 0.53 = relative amounts at T slightly higher than 1030 C = nliq lSi 1.0  0.80 Just before 1030 C, the Si(s) is 34.6 mol% of the total heterogeneous mixture; the eutectic liquid is 65.4 mol%. At the eutectic temperature a third phase appears  CaSi2 (s). As the melt cools at this temperature both Si(s) and CaSi2 (s) freeze out of the melt while the concentration of the melt remains constant. At a temperature slightly below 1030 C all the melt will have frozen to Si(s) and CaSi2 (s) with the relative amounts: lCaSi2 0.80  0.667 nsi = = nCaSi2 1.0  0.80 lSi = 0.665 = relative amounts at T slightly higher than 1030 C Just under 1030 C, the Si(s) is 39.9 mol% of the total heterogeneous mixture; the CaSi2 (s) is 60.1 mol%. A graph of mol% Si(s) and mol% CaSi2 (s) vs. mol% eutectic liquid is a convenient way to show relative amounts of the three phases as the eutectic liquid freezes. Equations for the graph are derived with the law of conservation of mass. For the silicon mass, n zSi = nliq wSi + nSi xSi + nCaSi2 ySi where n = total number of moles.
122
INSTRUCTOR'S MANUAL
wSi = Si fraction in eutectic liquid = 0.694 xSi = Si fraction in Si(s) = 1.000 ySi = Si fraction in CaSi2 (s) = 0.667 zSi = Si fraction in melt = 0.800 This equation may be rewritten in mole fractions of each phase by dividing by n: zSi = (mol fraction liq) wSi + (mol fraction Si) xSi + (mol fraction CaSi2 ) ySi Since, (mol fraction liq) + (mol fraction Si) + (mol fraction CaSi2 ) = 1 or (mol fraction CaSi2 ) = 1  (mol fraction liq + mol fraction Si), we may write : zSi = (mol fraction liq) wSi + (mol fraction Si) xSi +[1  (mol fraction liq + mol fraction Si)] ySi Solving for mol fraction Si: mol fraction Si := (zSi  ySi )  (wSi  ySi )(mol fraction liq) xSi  ySi mol fraction CaSi2 := 1  (mol fraction liq + mol fraction Si)
These two eqns are used to prepare plots of the mol fraction of Si and the mol fraction of CaSi2 against the mol fraction of the melt in the range 00.65.
Freezing of Eutectic Melt at 1030C 0.7 0.6 0.5 0.4 mol fraction CaSi2 mol fraction Si 0.3 0.2 0.1 0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
mol fraction liq Freezing Proceeds toward Left
Figure 8.15
Solutions to theoretical problems
P8.12 The general condition of equilibrium in an isolated system is dS = 0. Hence, if and constitute an isolated system, which are in thermal contact with each other dS = dS + dS = 0 (a)
PHASE DIAGRAMS
123
Entropy is an additive property and may be expressed in terms of U and V . S = S(U, V ) The implication of this problem is that energy in the form of heat may be transferred from one phase to another, but that the phases are mechanically rigid, and hence their volumes are constant. Thus, dV = 0, and dS = S dU + U V S 1 1 dU = dU + dU [5.4] U V T T 1 1 = or T = T T T
But, dU = dU ; therefore
Solutions to applications
P8.14 C = 1; hence, F =CP +2=3P Since the tube is sealed there will always be some gaseous compound in equilibrium with the condensed phases. Thus when liquid begins to form upon melting, P = 3 (s, l, and g) and F = 0, corresponding to a definite melting temperature. At the transition to a normal liquid, P = 3 (l, l , and g) as well, so again F = 0. P8.16 The temperaturecomposition lines can be calculated from the formula for the depression of freezing point [7.33]. T RT 2 xB fus H
For bismuth (8.314 J K1 mol1 ) (544.5 K)2 RT 2 = 227 K = 10.88 103 J mol1 fus H For cadmium (8.314 J K1 mol1 ) (594 K)2 RT 2 = = 483 K 6.07 103 J mol1 fus H We can use these constants to construct the following tables
x(Cd) T /K Tf /K x(Bi) T /K Tf /K 0.1 22.7 522 0.1 48.3 546 0.2 45.4 499 0.2 96.6 497 0.3 68.1 476 0.3 145 449 0.4 90.8 454 0.4 193 401 ( T = x(Cd) 227 K) (Tf = Tf  T ) ( T = x(Bi) 483 K) (Tf = Tf  T )
These points are plotted in Fig. 8.16(a). The eutectic temperature and concentration are located by extrapolation of the plotted freezing point lines until they intersect at e, which corresponds to TE 400 K and xE (Cd) 0.60. Liquid at a cools without separation of a solid until a is reached (at 476 K). Solid Bi then seperates, and the liquid becomes richer in Cd. At a (400 K) the composition is pure solid Bi + liquid of composition x(Bi) = 0.4. The whole mass then solidfies to solid Bi + solid Cd.
124
INSTRUCTOR'S MANUAL
(a) 600
(b)
500 s
l
Solid precipitates Eutectic halt
400 0 1 Time
Figure 8.16
(a) At 460 K (point a ),
l(s) n(l) = 5 by the lever rule. n(s) l(l)
(b) At 375 K (point a ) there is no liquid . The cooling curve is shown in Fig. 8.16(b). Comment. The experimental values of TE and xE (Cd) are 417 K and 0.55. The extrapolated values can be considered to be remarkably close to the experimental ones when one considers that the formulas employed apply only to dilute (ideal) solutions. P8.17 (a) The data are plotted in Fig. 8.17.
10 8 6 4 2 0 0.2 0.4 0.6 0.8 1.0
Figure 8.17 (b) We need not interpolate data, for 6.02 MPa is a pressure for which we have experimental data. The mole fraction of CO2 in the liquid phase is 0.4541 and in the vapour phase 0.9980. The proportions of the two phases are in an inverse ratio of the distance their mole fractions are from the composition point in question, according to the lever rule. That is nliq v 0.9980  0.5000 = = = 10.85 nvap l 0.5000  0.4541 P8.19 (a) As the solutions become either pure methanol (xmethanol = 1) or pure TAME (xmethanol = 0), the activity coefficients should become equal to 1 (Table 7.3). This means that the extremes in the range of ln (x) curves should approach zero as they do in the above plot (Fig. 8.18(a)).
PHASE DIAGRAMS
125
2
1.5 methanol ln 1 TAME
0.5
0
0
0.2
0.4 xmethanol
0.6
0.8
1
Figure 8.18(a)
1000
800
GE / J mol1
600
400
200
0
0
0.2
0.4 xmethanol
0.6
0.8
1
Figure 8.18(b)
(b) The large positive deviation of GE from the ideal mixture (GE ideal = 0, Section 7.4) indicates that the mixing process is unfavorable. This may originate from the breakage of relatively strong methanol hydrogen bonding upon solution formation. GE for a regular solution is expected to be symmetrical about the point xmethanol = 0.5. Visual inspection of the GE (xmethanol ) plot reveals that methanol/TAME solutions are approximately E "regular". The symmetry expectation can be demonstrated by remembering that Hm = W xA xB E E E and S = 0 for a regular solution (Section 7.4b). Then, for a regular solution Gm = Hm TS E = m E E H = W xA xB , which is symmetrical about x = 0.5 in the sense that Gm at x = 0.5  equals GE at x = 0.5 + . m (c) Azeotrope composition and vapor pressure: xmethanol = ymethanol = 0.682 P = 11.59 kPa when xmethanol = 0.2, P = 10.00 kPa. (d) The vapor pressure plot shows positive deviations from ideality. The escaping tendency is stronger than that of an ideal solution. To get the Henry's law constants, estimate values for the targets of Pmethanol at xmethanol = 0 and PTAME at xmethanol = 1.
126
INSTRUCTOR'S MANUAL
12 Total Vapor Pressure / kPa
Soln composition (xmethanol) 10
8
Vapor composition (ymethanol)
6
0
0.2
0.4
0.6
0.8
1
Methanol Mole Fraction (xmethanol or ymethanol) 15
Figure 8.18(c)
P = Pmethanol + PTAME 10 Vapor Pressure / kPa Pmethanol
5 PTAME
0
0
0.2
0.4 xmethanol
0.6
0.8
1
Figure 8.18(d)
For methanol in TAME (eqn 7.26): Kmethanol = dPmethanol = 45.1 kPa dxmethanol xmethanol =0 dPTAME dPTAME = = 25.3 kPa dxTAME xTAME =0 dxmethanol xmethanol =1
For TAME in methanol: KTAME =
(e) According to eqn 6.3, the vapor pressure should increase when the applied pressure is increased. For TAME: P = P eVm P /RT = 6.16 kPa The applied pressure increases the vapor pressure by about 1%, molecules have been "squeezed" out of the liquid phase and into the gas phase but only to a slight extent.
1 1 1 cm3 mol bar)/[(83.1451 3 )(288.15 cm bar K mol K)] = (6.09 kPa) e(131.78 )(2.0
9
Chemical equilibrium
Solutions to exercises
Discussion questions
E9.1(b) The thermodynamic equilibrium constant involves activities rather than pressures. See eqn 9.18 and Example 9.1. For systems involving gases, the activities are the dimensionless fugacities. At low pressures, the fugacity may be replaced with pressures with little error, but at high pressures that is not a good approximation. The difference between the equilibrium constant expressed in activities and the constant expressed in pressures is dependent upon two factors: the stoichiometry of the reaction and the magnitude of the partial pressures. Thus there is no one answer to this question. For the example of the ammonia synthesis reaction, in a range of pressures where the fugacity coefficients are greater than one, an increase in pressure results in a greater shift to the product side than would be predicted by the constant expressed in partial pressures. For an exothermic reaction, such as the ammonia synthesis, an increase in temperature will shift the reaction to the reactant side, but the relative shift is independent of the fugacity coefficients. The ratio ln(K2 /K1 ) depends only on r H . See eqn 6.26. The physical basis of the dependence of the equilibrium constant on temperature as predicted by the    van't Hoff equation can be seen when the expression r G = r H   T r S  is written in the     form R ln K =  r H /T + r S . When the reaction is exothermic and the temperature is raised, ln K and hence K decrease, since T occurs in the denominator, and the reaction shifts to favor the reactants. When the reaction is endothermic, increasing T makes ln K less negative, or K more positive, and products are favored. Another factor of importance when the reaction is endothermic is the increasing entropy of the reacting system resulting in a more positive ln K, favoring products. A typical pH curve for the titration of a weak base with a strong acid is shown in Figure 9.1. The stoichiometric point S occurs on the acidic side of pH = 7 because the salt formed by the neutralization reaction has an acid cation. Buffers work best when S A , that is when the concentrations of the salt and acid are not widely different. An abundant supply of A ions can remove by reaction any H3 O+ supplied by the addition of an acid; likewise an abundant supply of HA can remove by reaction any OH supplied by addition of base. Indicators are weak acids which in their undissociated acid form have one colour, and in their dissociated anion form, another. In acidic solution, the indicator exists in the predominantly acid form (one colour), in basic solution in the predominantly anion form (the other colour). The ratio of the two forms is very pH sensitive because of the small value of pKa of the indicator, so the colour change can occur very rapidly with change in pH.
E9.2(b)
E9.3(b)
E9.4(b)
Numerical exercises
E9.5(b)
rG  
= RT ln K = (8.314 J K1 mol1 ) (1600 K) ln(0.255) = +18.177 kJ mol1 = +18.18 kJ mol1
E9.6(b)
rG
 
= RT ln K (0.178 103 J mol1 ) (8.314 J K1 mol1 ) (1173 K)
 K = e( r G /RT ) = exp 
= 0.982 = 0.98
128
INSTRUCTOR'S MANUAL
14
12
Strong acid
10
8 pH 7 6 S 4 Weak base
2
0
0
10
20
30
Volume of acid added (mL)
Figure 9.1
2NO2 (g) 2n 2 1+ 2p 1+
E9.7(b)
Amount at equilibrium Mole fraction Partial pressure
N2 O4 (g) (1  )n 1 1+ (1  )p 1+
Assuming that the gases are perfect, aJ = K=
 (pNO2 /p  )2 4 2 p =   (pN2 O4 /p  ) (1  2 )p 
pJ  p
 For p = p  , K =
4 2 1  2
(a) (b) (c) E9.8(b)
rG
= 0 at equilibrium K=
= 0.201
4(0.201)2 = 0.16841 1  0.2012   = RT ln K = (8.314 J K1 mol1 ) (298 K) ln(0.16841) rG = 4.41 kJ mol1
Br2 (g) 2Br(g) = 0.24 2n 2 1+ 2p 1+
(a)
Amount at equilibrium Mole fraction Partial pressure
(1  )n 1 1+ (1  )p 1+
CHEMICAL EQUILIBRIUM
129
Assuming both gases are perfect aJ = K = = (b)
rG  
pJ  p
 [p = p  ]
 (pBr /p  )2 4 2 p 4 2 = =    pBr2 /p (1  2 )p  1  2
4(0.24)2 = 0.2445 = 0.24 1  (0.24)2 = RT ln K = (8.314 J K1 mol1 ) (1600 K) ln(0.2445) = 19 kJ mol1
rH  
(c)
ln K(2273 K) = ln K(1600 K)  = ln(0.2445)  = 1.084 K(2273 K) = e1.084 = 2.96
R
1 1  2273 K 1600 K (1.851 104 )
112 103 J mol1 8.314 J K1 mol1
E9.9(b)
(CHCl3 ) = 1, (a)
rG  
(HCl) = 3,
(CH4 ) = 1,
(Cl2 ) = 3
   = f G (CHCl3 , l) + 3 f G (HCl, g)  f G (CH4 , g) = (73.66 kJ mol1 ) + (3) (95.30 kJ mol1 )  (50.72 kJ mol1 )
= 308.84 kJ mol1 ln K = 
rG  
RT
[8] =
(308.84 103 J mol1 ) = 124.584 (8.3145 J K1 mol1 ) (298.15 K)
K = 1.3 1054 (b)
   = f H  (CHCl3 , l) + 3 f H  (HCl, g)  f H  (CH4 , g) 1 1 = (134.47 kJ mol ) + (3) (92.31 kJ mol )  (74.81 kJ mol1 ) = 336.59 kJ mol1   1 1 rH  [9.28] ln K(50 C) = ln K(25 C)  R 323.2 K 298.2 K rH  
= 124.584  K(50 C) = 3.5 1049
rG  
336.59 103 J mol1 8.3145 J K1 mol1
(2.594 104 K 1 ) = 114.083
(50 C) = RT ln K(50 C)[18] = (8.3145 J K 1 mol1 )(323.15 K)(114.083) = 306.52 kJ mol1
E9.10(b)
Draw up the following table
Initial amounts/mol Stated change/mol Implied change/mol Equilibrium amounts/mol Mole fractions A 2.00 0.79 1.21 0.1782 + B 1.00 0.79 0.21 0.0309 C + 0 +0.79 +0.79 0.79 0.1163 2D 3.00 +1.58 4.58 0.6745 Total 6.00
6.79 0.9999
130
INSTRUCTOR'S MANUAL
(a) Mole fractions are given in the table. (b) Kx =
J xJ J
Kx =
(0.1163) (0.6745)2 (0.1782) (0.0309)
= 9.6 pJ , so  p = Kx when p = 1.00 bar
(c) pJ = xJ p. Assuming the gases are perfect, aJ = K=
  (pC /p  ) (pD /p  )2 = Kx   (pA /p  ) (pB /p  )
p  p
K = Kx = 9.6 (d)
rG  
= RT ln K = (8.314 J K1 mol1 ) (298 K) ln(9.609) = 5.6 kJ mol1
E9.11(b)
At 1120 K,
rG
 
= +22 103 J mol1
 (22 103 J mol1 )  r G = 2.363 = RT (8.314 J K1 mol1 ) (1120 K)  
ln K(1120 K) =
K = e2.363 = 9.41 102 ln K2 = ln K1 
rH
R
1 1  T2 T1
Solve for T2 at ln K2 = 0 (K2 = 1) 1 R ln K1 1 (8.314 J K1 mol1 ) (2.363) 1 = 7.36 104 + = + =  T2 H T1 1120 K (125 103 J mol1 ) r T2 = 1.4 103 K E9.12(b)
  rH  d(ln K) = d(1/T ) R
Use
We have ln K = 2.04  1176 K
rH  
1 T
+ 2.1 107 K 3 1 2 T
1 3 T

R
= 1176 K + 3 (2.1 107 K 3 )
T = 450 K so 
rH  
R
rH  
= 1176 K + 3 (2.1 107 K 3 )
2 1 = 865 K 450 K
= +(865 K) (8.314 J mol1 K 1 ) = 7.191 kJ mol1
CHEMICAL EQUILIBRIUM
131
Find
rS rG
 
from
rG
 
 
= RT ln K = (8.314 J K1 mol1 ) (450 K) 2.04  = 16.55 kJ mol1 1176 K 2.1 107 K 3 + 450 K (450 K)3
rG rS
 
= =
rH rH
   
  T rS
 
 T
rG
 
=
7.191 kJ mol1  16.55 kJ mol1 = 20.79 J K1 mol1 450 K = 21 J K1 mol1
E9.13(b)
3 U(s) + 2 H2 (g)
UH3 (s),
rG
 
= RT ln K p . The activities of the  p
At this low pressure, hydrogen is nearly a perfect gas, a(H2 ) = solids are 1. Hence, ln K = ln p p 3/2 3 =  2 ln   p p p  3 G = 2 RT ln  p =
3 2
(8.314 J K 1 mol1 ) (500 K) ln
1.04 Torr 750 Torr
 [p  = 1 bar 750 Torr]
= 41.0 kJ mol1 E9.14(b) Kx =
J xJ J [analogous to 17]
The relation of Kx to K is established in Illustration 9.4 K =
J
pJ J  p
xJ J
9.18 with aJ =
J J
pJ  p p  p
J
=
J
p  p
[pJ = xJ p] = Kx
J
Therefore, Kx = K
p   , Kx p [K and p  are constants]  p thus Kx (2 bar) = Kx (1 bar)
= 1 + 1  1  1 = 0, E9.15(b) N2 (g) + O2 (g)
2NO(g) K = 1.69 103 at 2300 K 5.0 g = 0.2380 mol N2 Initial moles N2 = 28.01 g mol1 2.0 g Initial moles O2 = = 6.250 102 mol O2 32.00 g mol1
132
INSTRUCTOR'S MANUAL
N2 Initial amount/mol Change/mol Equilibrium amount/mol Mole fractions 0.2380 z 0.2380  z 0.2380  z 0.300
O2 0.0625 z 0.0625  z 0.0625  z 0.300
NO 0 +2z 2z 2z 0.300
Total 0.300 0 0.300 (1)
K = Kx
p  p
=
J
J = 0 , then
K = Kx =
(2z/0.300)2
0.2380z 0.300
0.0625z 0.300
=
4z2 = 1.69 103 (0.2380  z) (0.0625  z)
4z2 = 1.69 103 0.01488  0.3005z + z2 = 2.514 105  (5.078 104 )z + (1.69 103 )z2 4.00  1.69 103 = 4.00 4z + (5.078 10
2 4
so
1/2
)z  2.514 105 = 0
5.078 104 (5.078 104 )2  4 (4) (2.514 105 ) z= 8 = 1 (5.078 104 2.006 102 ) 8 z>0 [z < 0 is physically impossible] so z = 2.444 103 xNO = E9.16(b)
rG  
2z 2(2.444 103 ) = 1.6 102 = 0.300 0.300 = RT ln K [9.8]
rG  
Hence, a value of (a) (b) E9.17(b)
rG rG    
< 0 at 298 K corresponds to K > 1. K<1 K>1 ) = (690.00)  (33.56)  (2) (120.35) = 415.74,
/(kJ mol1 ) = (2) (33.56)  (166.9) = +99.8, /(kJ mol
1
Le Chatelier's principle in the form of the rules in the first paragraph of Section 9.4 is employed.   Thus we determine whether r H  is positive or negative using the f H  values of Table 2.6. (a) (b)
rH rH    
/(kJ mol1 ) = (2) (20.63)  (178.2) = +136.9 /(kJ mol1 ) = (813.99)  (20.63)  (2) (187.78) = 417.80
Since (a) is endothermic, an increase in temperature favours the products, which implies that a reduction in temperature favours the reactants; since (b) is exothermic, an increase in temperature favours the reactants, which implies that a reduction in temperature favours the products (in the sense of K increasing).
CHEMICAL EQUILIBRIUM
133
E9.18(b)
K = ln K
rH
 
R
1 1  T T T = 325 K;
so
rH
 
=
R ln K K
1 T 1 T
T = 310 K, Now
rH  
let
=
(8.314 J K1 mol1 )
1 310 K rH rH
K = K

1 325 K
ln = 55.84 kJ mol1 ln
(a) (b) E9.19(b)
= 2,
1 = 2,
   
= (55.84 kJ mol1 ) (ln 2) = 39 kJ mol1
1 = (55.84 kJ mol1 ) ln 2 = 39 kJ mol1
NH4 Cl(s) K=
J
NH3 (g) + HCl(g)
aJ J [17];
p = p(NH3 ) + p(HCl) = 2p(NH3 ) (a)
[p(NH3 ) = p(HCl)] pJ a(gases) =  ; a(NH4 Cl, s) = 1 p p(NH3 )2 1 =  4 p 2 608 kPa 2 = 9.24 100 kPa 1115 kPa 2 = 31.08 100 kPa = p 2  p
K=
p(NH3 )  p
At 427 C (700 K), At 459 C (732 K), (b)
rG  
p(HCl)  p 1 K= 4 1 K= 4
= RT ln K[8] = (8.314 J K1 mol1 ) (700 K) (ln 9.24) = 12.9 kJ mol1 (at 427 C) R ln K K
1 T 1 T
(c)
rH
 
[26]
rS  
(8.314 J K1 mol1 ) ln 31.08 9.24
1 700 K rH  

1 732 K
= +161 kJ mol1
(d) E9.20(b)
=
 T
rG
 
=
(161 kJ mol1 )  (12.9 kJ mol1 ) = +248 J K1 mol1 700 K
The reaction is CuSO4 5H2 O(s) CuSO4 (s) + 5H2 O(g)
For the purposes of this exercise we may assume that the required temperature is that temperature at which the K = 1 which corresponds to a pressure of 1 bar for the gaseous products. For K = 1,  ln K = 0, and r G = 0.
rG  
=
rH
 
  T rS = 0
when
rH
 
 = T rS
Therefore, the decomposition temperature (when K = 1) is T =
  rH   rS
134
INSTRUCTOR'S MANUAL
CuSO4 5H2 O(s)
rH   rS  
CuSO4 (s) + 5H2 O(g)
= [(771.36) + (5) (241.82)  (2279.7)] kJ mol1 = +299.2 kJ mol1 = [(109) + (5) (188.83)  (300.4)] J K 1 mol1 = 752.8 J K1 mol1
299.2 103 J mol1 = 397 K 752.8 J K1 mol1 Question. What would the decomposition temperature be for decomposition defined as the state at 1 which K = 2 ? Therefore, T = E9.21(b) (a) The halfway point corresponds to the condition [acid] = [salt], for which pH = pKa
Thus pKa = 4.82 and Ka = 104.82 = 1.5 105 (b) When [acid] = 0.025 M
1 1 1 1 pH = 2 pKa  2 log[acid] = 2 (4.82)  2 (1.60) = 3.21
E9.22(b)
(a) The HCO ion acts as a weak base. 2 HCO (aq) + H2 O(l) 2 HCOOH(aq) + OH (aq)
Then, since [HCOOH] [OH ] and [HCO ] S, the nominal concentration of the salt, 2 Kb [OH ]2 S and [OH ] = (SKb )1/2
1 1 Therefore pOH = 2 pKb  2 log S However, pH + pOH = pKw , so pH = pKw  pOH and pKa + pKb = pKw , so pKb = pKw  pKa 1 1 1 1 1 Thus pH = pKw  2 (pKw  pKa ) + 2 log S = 2 pKw + 2 pKa + 2 log S 1 1 1 = 2 (14.00) + 2 (3.75) + 2 log(0.10) = 8.37
(b) The same expression is obtained
1 1 1 pH = 2 pKw + 2 pKa + 2 log S 1 1 1 = 2 (14.00) + 2 (4.19) + 2 log(0.20) = 8.74
(c)
0.150 M HCN(aq) HCN(aq) + H2 O(l) H3 O+ (aq) + CN (aq) Ka = [H3 O+ ][CN ] [HCN]
Since we can ignore water autoprotolysis, [H3 O+ ] = [CN ], so Ka = [H3 O+ ]2 A
where A = [HCN], the nominal acid concentration.
CHEMICAL EQUILIBRIUM
135
1 1 Thus [H3 O+ ] (AKa )1/2 and pH 2 pKa  2 log A 1 1 pH = 2 (9.31)  2 log(0.150) = 5.07
E9.23(b)
The pH of a solution in which the nominal salt concentration is S is
1 1 1 pH = 2 pKw + 2 pKa + 2 log S
The volume of solution at the stoichiometric point is V = (25.00 mL) + (25.00 mL) S = (0.100 M) 25.00 mL 39.286 mL 0.100 M 0.175 M = 39.286 mL
= 6.364 102 M
pKa = 1.96 for chlorous acid.
1 1 1 pH = 2 (14.00) + 2 (1.96) + 2 log(6.364 102 )
= 7.38 E9.24(b)
1 1 1 When only the salt is present, use pH = 2 pKa + 2 pKw + 2 log S 1 1 1 pH = 2 (4.19) + 2 (14.00) + 2 log(0.15) = 8.68
(a)
When A S, use the HendersonHasselbalch equation pH = pKa  log A A = 3.366  log A = 4.19  log 0.15 S S, use (c) (b)
When so much acid has been added that A
1 1 pH = 2 pKa  2 log A
We can make up a table of values
A/(mol L1 ) pH Formula 0 8.68 (a) 0.06 4.59 0.08 4.46 0.10 4.36 (b) 0.12 4.29 0.14 4.21 0.6 0.8 1.0 2.21 2.14 2.09 (c)
These values are plotted in Fig. 9.2. E9.25(b) According to the HendersonHasselbalch equation the pH of a buffer varies about a central value given [acid] by pKa . For the ratio to be neither very large nor very small we require pKa pH (buffer) [salt] (a) For pH = 4.6, use aniline and anilinium ion , pKa = 4.63. (b) For pH = 10.8, use ethylammonium ion and ethylamine , pKa = 10.81
136
INSTRUCTOR'S MANUAL
8.00
6.00
4.00
2.00 0 0.2 0.4 0.6 0.8 1.0
Figure 9.2
Solutions to problems
Solutions to numerical problems
P9.2 CH4 (g) C(s) + 2H2 (g) This reaction is the reverse of the formation reaction. (a)
 =  f G     = f H   T f S fG = 74 850 J mol1  298 K (80.67 J K 1 mol1 ) rG  
= 5.08 104 J mol1 ln K =
rG  
RT
[9.8] =
5.08 104 J mol1 = 20.508 8.314 J K1 mol1 298 K
K = 1.24 109 (b)
 =  f H  = 74.85 kJ mol1   1 1 rH  ln K(50 C) = ln K(298 K)  R 323 K 298 K rH  
[9.28]
= 20.508  K(50 C) = 1.29 108 (c) Draw up the equilibrium table
CH4 (g) (1  )n 1 1+ 1 p 1+
7.4850 104 J mol1 8.3145 J K1 mol1
(2.597 104 ) = 18.170
Amounts Mole fractions Partial pressures
H2 (g) 2n 2 1+ 2p 1+
CHEMICAL EQUILIBRIUM
137
K=
J
aJ J [9.18]
=
pH2 2 p pCH4 p
1.24 109 = =
(2)2 1  2
p  p
4 2 p
[
1]
1.24 109 = 1.8 104 4 0.010
(d) Le Chatelier's principle provides the answers: As pressure increases, decreases, since the more compact state (less moles of gas) is favoured at high pressures. As temperature increases the side of the reaction which can absorb heat is  favoured. Since r H  is positive, that is the righthand side, hence increases. This can also be seen from the results of parts (a) and (b), K increased from 25 C to 50 C, implying that increased. P9.3
3 U(s) + 2 H2 (g) fH  
UH3 (s)
 K = (p/p  )3/2 [Exercise 9.13(b)]  3  2 ln p/p 
= RT 2
d ln K d [9.26] = RT 2 dT dT d ln p 3 =  2 RT 2 dT
3 =  2 RT 2
14.64 103 K 5.65  T T2
3 =  2 R(14.64 103 K  5.65T )
= (2.196 104 K  8.48T )R
 d( f H  ) =   r Cp dT
[from 2.44]
or P9.5
  r Cp
=
 fH  = 8.48R T p
CaCl2 NH3 (s)
rG  
CaCl2 (s) + NH3 (g)
K=
p  p
p = RT ln K = RT ln  p = (8.314 J K1 mol1 ) (400 K) ln = +13.5 kJ mol1
 
12.8 Torr 750 Torr
 [p  = 1 bar = 750.3 Torr]
at 400 K
rG  
Since r G and ln K are related as above, the dependence of determined from the dependence of ln K on temperature.
rG  (T ) 
on temperature can be
T

rG
 (T 
)
T
=
rH
 
1 1  T T
[26]
138
INSTRUCTOR'S MANUAL
Therefore, taking T = 400 K,
rG  
(T ) =
T 400 K
(13.5 kJ mol1 ) + (78 kJ mol1 ) 1  (13.5  78) kJ mol1 400 T K
T 400 K
= (78 kJ mol1 ) + That is, P9.7
rG  
(T )/(kJ mol1 ) = 78  0.161(T /K) 2A(g). A = acetic acid
The equilibrium we need to consider is A2 (g)
It is convenient to express the equilibrium constant in terms of , the degree of dissociation of the dimer, which is the predominant species at low temperatures.
A 2n 2 1+ 2p 1+ A2 (1  )n 1 1+ 1 p 1+ Total (1 + )n 1 p
At equilibrium Mole fraction Partial pressure
The equilibrium constant for the dissociation is Kp =
pA 2 p pA2 p 2 4 2 pp  pA = =   pA2 p 1  2
We also know that pV = ntotal RT = (1 + )nRT , In the first experiment, = (764.3 Torr) (21.45 103 L) (120.1 g mol1 ) pV M  1 = 0.392 1= mRT (0.0519 g) (62.364 L Torr K 1 mol1 ) (437 K)
764.3 (4) (0.392)2 750.1
implying that
=
pV 1 nRT
and
n=
m M
1  (0.392)2 In the second experiment, =
Hence, K =
= 0.740
(764.3 Torr) (21.45 103 L) (120.1 g mol1 ) pV M 1=  1 = 0.764 mRT (0.038 g) (62.364 L Torr K 1 mol1 ) (471 K)
764.3 (4) (0.764)2 750.1
1  (0.764)2 The enthalpy of dissociation is
rH  
Hence, K =
= 5.71
=
R ln K K
1 T

1 T
[9.28, Exercise 9.18(a)] =
5.71 R ln 0.740 1 437 K

1 471 K
= +103 kJ mol1
The enthalpy of dimerization is the negative of this value, or 103 kJ mol1 (i.e. per mole of dimer).
CHEMICAL EQUILIBRIUM
139
P9.9
Draw up the following equilibrium table
Initial amounts/mol Stated change/mol Implied change/mol Equilibrium amounts/mol Mole fractions A 1.00 0.60 0.40 0.087 B 2.00 0.30 1.70 0.370 C 0 +0.90 +0.90 0.90 0.196 D 1.00 +0.60 1.60 0.348 Total 4.00
4.60 1.001
The mole fractions are given in the table. Kx =
J v xJ J [analogous to eqn 9.18 and Illustration 9.4]
Kx =
(0.196)3 (0.348)2 = 0.326 = 0.33 (0.087)2 (0.370) p = 1 bar,
 p  = 1 bar
pJ = xJ p,
Assuming that the gases are perfect, aJ = K=
  (pC /p  )3 (pD /p  )2   (pA /p  )2 (pB /p  )
pJ , hence  p
x3 x2 = C D 2 xA x B P9.10
p 2 = Kx  p
when p = 1.00 bar = 0.33
The equilibrium I2 (g)
2I(g) is described by the equilibrium constant
p 4 2 p   x(I)2 p K= [Problem 9.7]  = x(I2 ) p  1  2
If p0 = =
nRT , then p = (1 + )p 0 , implying that V p  p0 p0
We therefore draw up the following table
973 K 0.06244 2.4709 0.05757 0.08459 1.800 10
 H  = RT 2
3
p/atm 104 nI p0 /atm K
1073 K 0.07500 2.4555 0.06309 0.1888 1.109 10
2
1173 K 0.09181 2.4366 0.06844 0.3415 4.848 102 p0 =
nRT V
d ln K dT
= (8.314 J K1 mol1 ) (1073 K)2 = +158 kJ mol1
3.027  (6.320) 200 K
140
INSTRUCTOR'S MANUAL
P9.13
The reaction is Si(s) + H2 (g) SiH2 (g)
The equilibrium constant is K = exp
  r G RT
= exp
fH  
  rH  RT
exp
  rS R
Let h be the uncertainty in low value is
, so that the high value is h+ the low value. The K based on the
rS      r Hhigh
   r Hlow Klow H = exp RT
exp
R
= exp
RT
exp
h RT
exp
rS
 
R
= exp
h RT
Khigh H
So
Klow H h = exp RT Khigh H At 298 K, (289  243) kJ mol1 KlowH = exp KhighH (8.3145 103 kJ K1 mol1 ) (298 K) KlowH (289  243) kJ mol1 = exp KhighH (8.3145 103 kJ K1 mol1 ) (700 K) = 1.2 108 = 2.7 103
(a)
(b)
At 700 K,
Solutions to theoretical problems
P9.16 K= p(NO2 )2  p(N2 O4 )p  with p(NO2 ) + p(N2 O4 ) = p
 Since p(NO2 )2 + p(NO2 )K  pK = 0 [p p/p  ]
p(NO2 ) =
1 + 4p K
2 K
1/2
1
We choose the root with the positive sign because p must be positive. For equal absorptions l1 p1 (NO2 ) = l2 p2 (NO2 ), Therefore 1+ 1+ 4p1 1/2  = (1 + 4p2 /K)1/2  1 K 4p1 1/2 4p2 1/2 =1+ 1+ K K 4p1 K = (  1)2 + 1 + 4p2 K + 2(  1) 1 + 4p2 1/2 K or p1 = p2 [ = l1 / l2 ]
2 1 +
CHEMICAL EQUILIBRIUM
141
1+
4p2 1/2 2(p1 2  p2 ) = (  1) 1 + K K
2
2(p1 2  p2 ) 1+ K
= (  1)2 1 +
4p2 K
(p1 2  p2 )2 (  1) (p1 2  p2 )  (  1)2 p2 + =0 K K2 Hence, K = (p1 2  p2 )2  [reinstating p  ]  (  1) (p2  p1 )p  395 mm = 5.27 Since = 75 mm
 p K =
(27.8p1  p2 )2 22.5(p2  5.27p1 )
We can therefore draw up the following table
Absorbance 0.05 0.10 0.15 p1 /Torr 1.00 2.10 3.15 p2 /Torr 5.47 12.00 18.65 pK/Torr 110.8 102.5 103.0 Mean: 105
 Hence, since p  = 750 Torr (1 bar), K = 0.140
P9.18
The five conditions are: (a) Electrical neutrality: [BH+ ] + [H3 O+ ] = [A ] + [OH ] Bo VB (b) Conservation of B groups: [B] + [BH+ ] = VA + V B where VB is the (fixed) initial volume of base and VA is the volume of titrant (acid) added. Ao V A (c) Concentration of A groups : [A ] = VA + V B (d) Protonation equilibrium of B: [B]Kb = [BH+ ][OH ] (e) Autoprotolysis equilibrium: Kw = [H3 O+ ][OH ] First we express condition (b) in terms of [BH+ ] and [OH ] by using condition (d) to eliminate [B] Bo Kb VB [BH+ ] = (VA + VB )([OH ] + Kb ) Next we use this relation and condition (c), and at the same time we use condition (e) to eliminate [H3 O+ ] B o K b VB Kw Ao VA   ] + K ) + [OH ] = V + V + [OH ] (VA + VB )([OH B A b Now we multiply through by terms in = VA + VB VB [OH ], expand the fraction VA + VB , and collect VB
VA Bo Kb [OH ] + (Kw  [OH ]2 )([OH ] + Kb ) and obtain = VB ([OH ] + Kb )([OH ]2 + Ao [OH ]  Kw )
142
INSTRUCTOR'S MANUAL
If desired, this formula for can be rewritten in terms of [H3 O+ ] and pH by using relation (e) and the definition pH =  log[H3 O+ ], or [H3 O+ ] = 10pH
Solutions to applications
P9.20 Refer to Box 9.2 for information necessary to the solution of this problem. The biological standard value of the Gibbs energy for ATP hydrolysis is 30 kJ mol1 . The standard Gibbs energy of combustion of glucose is 2880 kJ mol1 . (a) If we assume that each mole of ATP formed during the aerobic breakdown of glucose produces 30 kJ mol1 , then efficiency = 38 (30 kJ mol1 ) 2880 kJ mol1 100% 40%
(b) For the oxidation of glucose under the biological conditions of pCO2 = 5.3 102 atm, pO2 = 0.132 atm, and [glucose] = 5.6 102 mol L1 we have
rG
=
rG
 
+ RT ln Q
where Q =
 (pCO2 /p  )6 (5.3 102 )6 =  [glucose] (pO2 /p  )9 5.6 102 (0.132)9
= 32.5 Then
rG
= 2880 kJ mol1 + 8.314 J K1 mol1 310 K ln(32.5) = 2871 kJ mol1
which is not much different from the standard value. For the ATP ADP conversion under the given conditions
rG
=
rG
+ RT ln
Q Q
[ADP][Pi][H3 O+ ] 1 1 107 = = 107 [ATP] 1 1.0 104 1.0 104 107.4 = 1011.4 and Q = 1.0 104 then where Q =
rG
= 30 kJ mol1 + RT ln(104.4 ) = 30 kJ mol1 + 8.314 J K1 mol1 310 K (10.1) = 56 kJ mol1
With this value for efficiency =
rG
, the efficiency becomes = 74%
38 (56 kJ mol1 ) 2871 kJ mol1
CHEMICAL EQUILIBRIUM
143
(c) The theoretical limit of the diesel engine is =1 Tc 873 K = 55% =1 Th 1923 K
75% of the theoretical limit is 41%. We see that the biological efficiency under the conditions given is greater than that of the diesel engine. What limits the efficiency of the diesel engine, or any heat engine, is that heat engines must convert heat (q c H ) into useful work (Wadd,max = r G). Because of the second law, a substantial fraction of that heat is wasted. The biological process involves r G directly and does not go through a heat step. P9.22 (a) The equilibrium constant is given by K = exp so ln K = 
  r G RT r  H   rH  RT rS  
= exp
r  S
exp
R
+ RT R  A plot of ln K against 1/T should be a straight line with a slope of  r H  /R and a yintercept   of r S /R (Fig. 9.3).
20 18 16 14 12 10 3.2 3.4 3.6 3.8 4.0 4.2 4.4
Figure 9.3 So
rH  
= R slope = (8.3145 103 kJ mol1 K 1 ) (8.71 103 K) = 72.4 kJ mol1 = R intercept = (8.3145 J K 1 mol1 ) (17.3) = 144 J K1 mol1
and (b)
rH
  rS    
     = f H  ((ClO)2 )  2 f H  (ClO) so f H  ((ClO)2 ) = r H  + 2 f H  (ClO),
fH
((ClO)2 ) = [72.4 + 2(101.8)] kJ mol1 = 131.2 kJ mol1
 S  ((ClO)2 ) = [144 + 2(226.6)] J K 1 mol1 = 309.2 J K1 mol1
P9.24
A reaction proceeds spontaneously if its reaction Gibbs function is negative.
rG
=
rG
 
+ RT ln Q )=
rG  
Note that under the given conditions, RT = 1.58 kJ mol1 . (1) (2)
r G/(kJ mol r G/(kJ mol 1 1  ) = r G (2)  RT ln pH2 O pHNO3 = 57.2  1.58 ln[(1.3 107 ) (4.1 1010 )] = +2.0
(1)  RT ln pH2 O = 23.6  1.58 ln 1.3 107 = +1.5
144
INSTRUCTOR'S MANUAL
(3) (4)
r G/(kJ mol
1
)= )=
2 (3)  RT ln pH2 O pHNO3 = 85.6  1.58 ln[(1.3 107 )2 (4.1 1010 )] = 1.3 rG 3 (4)  RT ln pH2 O pHNO3 = 85.6  1.58 ln[(1.3 107 )3 (4.1 1010 )] = 3.5 rG  
 
r G/(kJ mol
1
So both the dihydrate and trihydrate form spontaneously from the vapour. Does one convert spontaneously into the other? Consider the reaction HNO3 2H2 O(s) + H2 O(g) HNO3 3H2 O(s)
rG
which may be considered as reaction (4)  reaction (3). Therefore
rG
for this reaction is
=
r G(4) 
r G(3)
= 2.2 kJ mol1
We conclude that the dihydrate converts spontaneously to the trihydrate , the most stable solid (at least of the four we considered). P9.26 (a) The following four equilibria are needed for the construction of the Ellingham diagram for the smelting reduction of silica with graphite (Box 9.1). (1)
1 2 Si(s 1 1 or l) + 2 O2 (g) 2 SiO2 (s or l)
1 G(T )
= 0.5 GH SiO2 (l) (T )  GH Si(l) (T )  GH O2 (T ) if T > mpSiO2 = 0.5 GH SiO2 (s) (T )  GH Si(l) (T )  GH O2 (T ) if mpSi T mpSiO2 = 0.5 GH SiO2 (s) (T )  GH Si(s) (T )  GH O2 (T ) if T < mpSi
(2)
1 1 2 C(s) + 2 O2 (g) 2 G(T )
1 2 CO2 (g)
= 0.5 GH CO2 (g) (T )  GH C(s) (T )  GH O2 (T )
(3)
1 C(s) + 2 O2 (g) CO(g) 3 G(T ) 1 = GH CO(g) (T )  GH C(s) (T )  2 GH O2 (T )
(4)
1 CO(g) + 2 O2 (g) CO2 (g) 4 G(T ) 1 = GH CO2 (g) (T )  GH CO(g) (T )  2 GH O2 (T ) 1 G(T ) and then only above 1900 K. Thus, the smelting reaction.
3 G(T ) alone lies above
(5)
1 2 SiO2
1 1 + 2 C(s) 2 Si + CO(g) 3 G(T )  1 G(T ))
( 5 G(T ) =
will have an equilibrium that lies to the right at temperatures higher than the temperature for which 5 G(T ) = 0. Algebra or the root function can be used to show that this temperature equals 1892 K. The minimum smelting temperature of silica is about 1892 K. Furthermore, 2 G never lies above 1 G so we do not expect appreciable amounts of CO2 is formed during smelting of silica. (b) This problem is related to P8.18. Begin by making the definition GH (T ) = G(T )  HSER = a + b T . Write the important equilibria and calculate equilibrium contents at 2000 K. Silica and
CHEMICAL EQUILIBRIUM
145
Ellingham Diagram: Reduction of Silica 350 1G 3G 250 2G rG / kJ 4G
150
50 1600 1800 2000 Temperature / K 2200 2400
Figure 9.4
silicon are molten at this temperature. We assume that carbon forms an ideal solution with molten silicon and make the initial estimate: {initial estimate of carbon mole function in molten Si} = xest = 0.02 according to eqn 7.27,
mix G(C)
= RT xest ln xest
and
mix G(Si)
= RT (1  xest ) ln(1  xest )
There are three unknowns (xC , PCO , PSiO ) so we select three independent equilibria that involve the silicon melt and solve them selfconsistently with the ideal solution estimate. The estimate is used to calculate the small mixing Gibbs energy only. GH C in melt = GH graphite + GH Si in melt = GH Si(l) +
mix G(C)
GH C
mix G(Si)
GH Si
The independent equilibria are used to calculate a new estimate for the mole fraction of carbon in silicon, xC . The new value is used in a repeat calculation in order to have a better estimate for xest . This iteration procedure is repeated until the estimate and the calculated value of xC agree to within 1%. With the initial estimate: (1) SiO2 (l) + 2C(Si melt) Si(melt) + 2CO(g)
1G
= GH Si + 2GH CO(g)  GH SiO2 (l)  2GH C = 37.69 kJ mol1 and
2 2 xSi PCO = K1 xC
K1 = e 1 G/RT = 9.646 (2)
SiO2 (l) + 3C(Si melt) SiC(s) + 2CO(g)
2G
= GH SiC(s) + 2GH CO(g)  GH SiO2 (e)  3GH C = 85.72 kJ mol1 and
2 3 PCO = K2 xC
K2 = e 2 G/RT = 173.26
146
INSTRUCTOR'S MANUAL
Dividing the equilibrium constant expression of Reaction (1) by the one for Reaction (2), and using xC = 1  xSi , gives (1  xSi )(xSi ) = K1 /K2 Solving for xSi gives:
1 xSi = 2 1 +
1  4K1 /K2 = 0.9408
xC = 1  xSi = 0.0592 The initial estimate of xC (0.02) and the calculated value do not agree to within 1%, so the calculation is repeated (iterated) with the new estimate: xest = 0.0592. After several additional iterations, it is found that with xest = 0.0695 the calculated value is xC = 0.0698 . Since these do agree to within 1%, the calculation is selfconsistent and further iteration is unnecessary. The equilibrium expression for reaction (2) gives: PCO =
3 K2 xC bar =
(125.66)(0.0698)3 bar
PCO = 0.207 bar The third equilibrium is used to acquire PSi , it is: SiO2 (l) + C(Si melt) SiO(g) + CO(g)
3G
(3)
= GH SiO(g) + GH CO(g)  GH SiO2 (l)  GH C = 8.415 KJ mol1 K3 x C PCO 1.659(0.0698) bar 0.207
K3 = e G3 /RT = 1.659 PSiO = bar 2 =
PSiO = 0.559 bar
10
Equilibrium electrochemistry
Solutions to exercises
Discussion questions
E10.1(b) The DebyeH ckel theory is a theory of the activity coefficients of ions in solution. It is the coulombic u (electrostatic) interaction of the ions in solution with each other and also the interaction of the ions with the solvent that is responsible for the deviation of their activity coefficients from the ideal value of 1. The electrostatic ionion interaction is the stronger of the two and is fundamentally responsible for the deviation. Because of this interaction there is a build up of charge of opposite sign around any given ion in the overall electrically neutral solution. The energy, and hence, the chemical potential of any given ion is lowered as a result of the existence of this ionic atmosphere. The lowering of the chemical potential below its ideal value is identified with a nonzero value of RT ln . This nonzero value implies that will have a value different from unity which is its ideal value. The role of the solvent is more indirect. The solvent determines the dielectric constant, , of the solution. Looking at the details of the theory as outlined in Justification 10.2 we see that enters into a number of the basic equations, in particular, Coulomb's law, Poisson's equation, and the equation for the Debye length. The larger the dielectric constant, the smaller (in magnitude) is ln . The potential difference between the electrodes in a working electrochemical cell is called the cell potential. The cell potential is not a constant and changes with time as the cell reaction proceeds. Thus the cell potential is a potential difference measured under nonequilibrium conditions as electric current is drawn from the cell. Electromotive force is the zerocurrent cell potential and corresponds to the potential difference of the cell when the cell (not the cell reaction) is at equilibrium. The pH of an aqueous solution can in principle be measured with any electrode having an emf that is sensitive to H+ (aq) concentration (activity). In principle, the hydrogen gas electrode is the simplest and most fundamental. A cell is constructed with the hydrogen electrode being the righthand electrode and any reference electrode with known potential as the lefthand electrode. A common choice is the saturated calomel electrode. The pH can then be obtained from eqn 10.43 by measuring the emf (zerocurrent potential difference), E, of the cell. The hydrogen gas electrode is not convenient to use, so in practice glass electrodes are used because of ease of handling.
E10.2(b)
E10.3(b)
Numerical exercises
E10.4(b) NaCl(aq) + AgNO3 (aq) AgCl(s) + NaNO3 (aq) NaCl, AgNO3 and NaNO3 are strong electrolytes; therefore the net ionic equation is Ag+ (aq) + Cl (aq) AgCl(s)
rH  
=
fH
 
(AgCl, s) 
fH
 
(Ag+ , aq) 
fH
 
(Cl , aq)
= (127.07 kJ mol1 )  (105.58 kJ mol1 )  (167.16 kJ mol1 ) = 65.49 kJ mol1 E10.5(b) PbS(s) KS =
J
aJ J
Pb2+ (aq) + S2 (aq)
Since the solubility is expected to be low, we may (initially) ignore activity coefficients. Hence KS = b(Pb2+ ) b(S2 )   b b b(Pb2+ ) = b(S2 ) = S
148
INSTRUCTOR'S MANUAL
KS =
S2  (b  )2
  r G to obtain KS RT fG  
 S = (KS )1/2 b 
Use ln KS =
rG  
=
(S2 , aq) +
fG
 
(Pb2+ , aq) 
rG
 
(PbS, s)
= (+85.8 kJ mol1 ) + (24.43 kJ mol1 )  (98.7 kJ mol1 ) = 160.07 kJ mol1 ln KS = 160.07 103 J mol1 (8.314 J K1 mol1 ) (298 K) = 64.61
KS = e64.61 = 8.7 1029 KS = E10.6(b) S2
 b
2
 S = (KS )1/2 b  = (8.735 1029 )1/2 = 9.3 1015 mol kg1
The ratio of hydration Gibbs energies is
hyd G hyd  (NO )  3  G (Cl ) hyd G    
=
r(NO ) 3
r(Cl )
=
181 pm = 0.958 189 pm
We have So E10.7(b)
hyd G
(Cl ) = 379 kJ mol1 [Exercise 10.6a]
(NO ) = (0.958) (379 kJ mol1 ) = 363 kJ mol1 3
 2 (bi /b  )zi [10.18]
1 I=2
i
    and for an Mp Xq salt, b+ /b  = pb/b  , b /b  = qb/b  , so 2 2  1 I = 2 (pz+ + qz )b/b 
(a) (b) (c) E10.8(b)
  1 I (MgCl2 ) = 2 (1 22 + 2 1)b/b  = 3b/b    1 I (Al2 (SO4 )3 ) = 2 (2 32 + 3 22 )b/b  = 15b/b    1 I (Fe2 (SO4 )3 ) = 2 (2 32 + 3 22 )b/b  = 15b/b 
b(K3 [Fe(CN)6 ]) b(KCl) b(NaBr) 1 + + I = I (K3 [Fe(CN)6 ]) + I (KCl) + I (NaBr) = 2 (3 + 32 )    b b b = (6) (0.040) + (0.030) + (0.050) = 0.320 Question. Can you establish that the statement in the comment following the solution to Exercise 10.8a (in the Student's Solutions Manual) holds for the solution of this exercise? b I = I (KNO3 ) =  (KNO3 ) = 0.110 b Therefore, the ionic strengths of the added salts must be 0.890. (a) b I (KNO3 ) =  , so b(KNO3 ) = 0.890 mol kg1 b and (0.890 mol kg1 ) (0.500 kg) = 0.445 mol KNO3 So (0.445 mol) (101.11 g mol1 ) = 45.0 g KNO3 must be added.
E10.9(b)
EQUILIBRIUM ELECTROCHEMISTRY
149
(b)
b b 1 I (Ba(NO3 )2 ) = 2 (22 + 2 12 )  = 3  = 0.890  b b b= 0.890  b  = 0.2967 mol kg1 3
and (0.2967 mol kg1 ) (0.500 kg) = 0.1484 mol Ba(NO3 )2 So (0.1484 mol) (261.32 g mol1 ) = 38.8 g Ba(NO3 )2 E10.10(b)
  1 I (Al2 (SO4 )3 ) = 2 ((2 33 ) + (3 22 ))b/b  = 15b/b    1 I (Ca(NO3 )2 ) = 2 (22 + 2)b/b  = 3b/b 
3(0.500 mol kg1 ) = 15(b(Al2 (SO4 )3 ))
3 b(Al2 (SO4 )3 ) = 15 (0.500 mol kg1 ) = 0.100 mol kg1
E10.11(b)
= (+  )1/s
p q
s =p+q
For Al2 (SO4 )3 p = 2, q = 3, s = 5
2 3 = (+  )1/5
E10.12(b) Since the solutions are dilute, use the DebyeH ckel limiting law u log = z+ z AI 1/2
1 I=2 i  2 1 zi (bi /b  ) = 2 {1 (0.020) + 1 (0.020) + 4 (0.035) + 2 (0.035)}
= 0.125 log = 1 1 0.509 (0.125)1/2 = 0.17996 (For NaCl) = 100.17996 = 0.661 E10.13(b)
1   I (CaCl2 ) = 2 (4 + 2)b/b  = 3b/b 
log = 2 1 0.509 (0.300)1/2 = 0.5576 = 100.5576 = 0.2770 = 0.277 0.524  0.277 100 per cent = 47.1 per cent 0.524 Az+ z I 1/2 E10.14(b) The extended DebyeH ckel law is log =  u 1 + BI 1/2 Solving for B Error = B= 1 Az+ z  + log I 1/2 = 1 0.509 +  log (b/b  )1/2
150
INSTRUCTOR'S MANUAL
Draw up the following table
b/(mol kg1 ) B 5.0 103 0.927 1.32 10.0 103 0.902 1.36 50.0 103 0.816 1.29
B = 1.3 E10.15(b) PbI2 (s)
rG rG  
PbI2 (aq)
KS = 1.4 108 = 44.83 kJ mol1
= RT ln KS = (8.314 J K1 mol1 ) (298.15 K) ln(1.4 108 ) =
fG  
   
(PbI2 , aq) 
rG  
fG fG 1
   
(PbI2 , s) (PbI2 , s)
fG
(PbI2 , aq) =
+
= 44.83 kJ mol
 173.64 kJ mol1
= 128.8 kJ mol1 E10.16(b) The Nernst equation may be applied to halfcell potentials as well as to overall cell potentials. E(H+ /H2 ) = RT a(H+ ) ln  F (fH2 /p  )1/2 RT RT a2 (H+ ) b2 [f is constant] = ln ln + ) H2 F a1 (H F b 1 (0.830) (5.0 102 ) (0.929) (5.0 103 ) = +56.3 mV
E = E2  E1 =
= (25.7 mV) ln
E10.17(b) Identify electrodes using species with the desired oxidation states. L: R: Cd(s) + 2OH (aq) Cd(OH)2 (s) + 2e Ni(OH)3 (s) + e Ni(OH)2 (s) + OH (aq)
Cd(s)Cd(OH)2 (s)OH (aq)Ni(OH)2 (s)Ni(OH)3 (s)Pt E10.18(b) The cell notation specifies the right and left electrodes. Note that for proper cancellation we must equalize the number of electrons in halfreactions being combined. (a) R: Ag2 CrO4 (s) + 2e 2Ag(s) + CrO2 (aq) 4 L: Cl2 (g) + 2e 2Cl (aq) Overall (R  L): Ag2 CrO4 (s) + 2Cl (aq) 2Ag(s) + CrO2 (aq) + Cl2 (g) 4 R: Sn4+ (aq) + 2e Sn2+ (aq) L: 2Fe3+ (aq) + 2e 2Fe2+ (aq) Overall (R  L): Sn4+ (aq) + 2Fe2+ (aq) Sn2+ (aq) + 2Fe3+ (aq) R: MnO2 (s) + 4H+ (aq) + 2e Mn2+ (aq) + 2H2 O(l) L: Cu2+ (aq) + 2e Cu(s) Overall (R  L): Cu(s) + MnO2 (s) + 4H+ (aq) Cu2+ (aq) + Mn2+ (aq) + 2H2 O(l) +0.45 V +1.36 V 0.91 V +0.15 V +0.77 V 0.62 V +1.23 V +0.34 V +0.89 V
(b)
(c)
EQUILIBRIUM ELECTROCHEMISTRY
151
 Comment. Those cells for which E  > 0 may operate as spontaneous galvanic cells under standard   conditions. Those for which E < 0 may operate as nonspontaneous electrolytic cells. Recall that  E  informs us of the spontaneity of a cell under standard conditions only. For other conditions we require E.
E10.19(b) The conditions (concentrations, etc.) under which these reactions occur are not given. For the purposes of this exercise we assume standard conditions. The specification of the right and left electrodes is determined by the direction of the reaction as written. As always, in combining halfreactions to form an overall cell reaction we must write halfreactions with equal number of electrons to ensure proper cancellation. We first identify the halfreactions, and then set up the corresponding cell. (a) R: 2H2 O(l) + 2e 2OH (aq) + H2 (g) 0.83 V L: 2Na+ (aq) + 2e 2Na(s) 2.71 V and the cell is Na(s)Na+ (aq), OH (aq)H2 (g)Pt +1.88 V or more simply Na(s)NaOH(aq)H2 (g)Pt (b) R: I2 (s) + 2e 2I (aq) +0.54 V L: 2H+ (aq) + 2e H2 (g) 0 and the cell is PtH2 (g)H+ (aq), I (aq)I2 (s)Pt +0.54 V or more simply PtH2 (g)HI(aq)I2 (s)Pt (c) R: 2H+ (aq) + 2e H2 (g) 0.00 V L: 2H2 O(l) + 2e H2 (g) + 2OH (aq) 0.083 V and the cell is PtH2 (g)H+ (aq), OH (aq)H2 (g)Pt 0.083 V or more simply PtH2 (g)H2 O(l)H2 (g)Pt
 Comment. All of these cells have E  > 0, corresponding to a spontaneous cell reaction under   standard conditions. If E had turned out to be negative, the spontaneous reaction would have been the reverse of the one given, with the right and left electrodes of the cell also reversed.     E10.20(b) See the solutions for Exercise 10.18(b), where we have used E  = ER  EL , with standard electrode potentials from Table 10.7.     E10.21(b) See the solutions for Exercise 10.19(b), where we have used E  = ER  EL , with standard electrode potentials from Table 10.7.     E10.22(b) In each case find E  = ER  EL from the data in Table 10.7, then use rG    = F E  [10.32]
(a)
R: L:
S2 O2 (aq) + 2e 2SO2 (aq) +2.05 V 8 4 + 1.51 V I2 (s) + 2e 2I (aq) +0.54 V
rG  
= (2) (96.485 kC mol1 ) (1.51 V) = 291 kJ mol1
152
INSTRUCTOR'S MANUAL
(b)
Pb2+ (aq) + 2e Pb(s)
rG  
Zn2+ (aq) + 2e Zn(s) 0.76 V 0.13 V
 E  = 0.63 V
= (2) (96.485 kC mol1 ) (0.63 V) = +122 kJ mol1
E10.23(b) (a) A new halfcell may be obtained by the process (3) = (1)  (2), that is (3) 2H2 O(l) + Ag(s) + e H2 (g) + 2OH (aq) + Ag+ (aq)
   But, E3 = E1  E2 , for the reason that the reduction potentials are intensive, as opposed to  extensive, quantities. Only extensive quantities are additive. However, the r G values of the halfreactions are extensive properties, and thus   r G3
=
  r G1

  r G2
   3 F E3 = 1 F E1  (2 F E2 )  Solving for E3 we obtain  E3 =   1 E1  2 E2 (2) (0.828 V)  (1) (0.799 V) = 2.455 V = 3 1
(b) The complete cell reactions is obtained in the usual manner. We take (2) (2)  (1) to obtain 2Ag+ (aq) + H2 (g) + 2OH (aq) 2Ag(s) + 2H2 O(l)
      E  (cell) = ER  EL = E2  E1 = (0.799 V)  (0.828 V) = +1.627 V  Comment. The general relation for E  of a new halfcell obtained from two others is  E3 =   1 E1 2 E2 3
E10.24(b) (a)
 E = E 
RT ln Q = 2 F [all other activities = 1] b b  here and below b [16, b+ = b, b = b]
Q=
J
2 2 aJ J = aH+ aCl
2 2 = a+ a = (+ b+ )2 ( b )2 4 = (+  )2 (b+ b )2 = b4  Hence, E = E  
2RT RT 4  ln( b) ln( b4 ) = E   F 2F
(b) (c)
rG
= F E[10.32] = (2)(9.6485104 C mol1 )(0.4658 V) = 89.89 kJ mol1
log = z+ z AI 1/2 [19] = (0.509) (0.010)1/2 [I = b for HCl(aq)] = 0.0509 = 0.889
 E = E +
2RT ln( b) = (0.4658 V) + (2) (25.693 103 V) ln(0.889 0.010) F = +0.223 V
The value compares favourably to that given in Table 10.7.
EQUILIBRIUM ELECTROCHEMISTRY
153
E10.25(b)
R: L:
Fe2+ (aq) + 2e Fe(s) 2Ag+ (aq) + 2e 2Ag(s) 2Ag(s) + Fe2+ (aq) 2Ag+ (aq) + Fe(s)
 = F E  = 2 (9.65 104 C mol1 ) (1.24 V)
R  L:
rG  
    E  = ER  EL = (0.44 V)  (0.80 V) = 1.24 V
= +239 kJ mol1
rH    = 2 f H  (Ag+ , aq)  fH  
(Fe2+ , aq) = [(2) (105.58)  (89.1)] kJ mol1
rH  
= +300.3 kJ mol1
 r G  =  rS = T p rG  
 T
 [ r G =
rH
 T r S]
= Therefore,
rG  
(239  300.3) kJ mol1 = 0.206 kJ mol1 K 1 298.15 K
(308 K) (239) + (10 K) (0.206 K 1 ) kJ mol1 +237 kJ mol1
 F E  [10.36] RT
E10.26(b) In each case ln K = (a)
Sn(s) + CuSO4 (aq) Cu(s) + SnSO4 (aq) 2+  R: Cu (aq) + 2e Cu(s) +0.34 V + 0.48 V L: Sn2+ (aq) + 2e Sn(s) 0.14 V ln K = (2) (0.48 V) = +37.4, 25.693 mV K = 1.7 1016
(b)
Cu2+ (aq) + Cu(s) 2Cu+ (aq) 2+  R: Cu (aq) + e Cu+ (aq) +0.16 V  0.36 V +0.52 V L: Cu+ (aq) + e Cu(s) ln K = 0.36 V = 14.0, 25.693 mV K = 8.2 107
 E10.27(b) We need to obtain E  for the couple
(3) Co3+ (aq) + 3e Co(s)
 from the values of E  for the couples
(1) Co3+ (aq) + e Co2+ (aq) (2) Co2+ (aq) + 2e Co(s)
 E1 = 1.81 V  E2 = 0.28 V
We see that (3) = (1) + (2); therefore (see the solution to Exercise 10.23(b)) E3 =
  1 E1 + 2 E2 (1) (1.81 V) + (2) (0.28 V) = = 0.42 V 3 3
154
INSTRUCTOR'S MANUAL
Then, R: Co3+ (aq) + 3e Co(s) L: 3AgCl(s) + 3e 3Ag(s) + 3Cl (aq) R  L: Co3+ (aq) + 3Cl (aq) + 3Ag(s) 3AgCl(s) + Co(s)     E  = ER  EL = (0.42 V)  (0.22 V) = +0.20 V
  ER = 0.42 V  EL = 0.22 V
E10.28(b) First assume all activity coefficients are 1 and calculate KS , the ideal solubility product constant.
AgI(s) Ag+ (aq) + I (aq) S(AgI) = b(Ag+ ) = b(I ) because all stoichiometric coefficients are 1. b(Ag+ )b(I ) S2 =  2 = (1.2 108 )2 = 1.44 1016 Thus KS =  b 2 b 3+ 2Bi (aq) + 3S2 (aq) (2) Bi2 S3 (s) (1) b(Bi3+ ) = 2S(Bi2 S3 ) b(S2 ) = 3S(Bi2 S3 )
KS =
(b(Bi3+ ))2 (b(S2 ))3 (2S)2 (3S)3 S 5 = = 108    b b 5 b 5
= 1.13 1097
2 For AgI, KS = KS
log = z+ z AI 1/2 I = Sb , = 0.9999
 
A = 0.509 so
z+ z  = 1
log = (0.509) (1.2 108 )1/2 = 5.58 105
KS = (0.9999)2 KS = 0.9997KS   For Bi2 S3 , I = 15b/b  = 15Sb  , z+ z  = 6
so log = (0.509) (6) [15(1.6 1020 )]1/2 = 1.496 109 = 1.0
5 KS = KS = KS
Neglect of activity coefficients is not significant for AgI and Bi2 S3 . E10.29(b) The Nernst equation applies to halfreactions as well as whole reactions; thus for 8H+ + MnO (aq) + 5e Mn2+ (aq) + 4H2 O 4
 E = E 
a(Mn2+ ) RT ln 5F a(MnO )a(H+ )8 4
E10.30(b)
R: L:
2AgI(s) + 2e 2Ag(s) + 2I (aq) 0.15 V 2H+ (aq) + 2e H2 (g) 0V 2AgI(s) + H2 (g) 2Ag(s) + 2H+ (aq) + 2I (aq) =2
Overall(R  L):
Q = a(H+ )2 a(I )2
Assume a(H+ ) = a(I ), Q = a(H+ )4
EQUILIBRIUM ELECTROCHEMISTRY
155
 E = E 
RT 2RT   ln a(H+ )4 = E   ln a(H+ ) = E  + 2 (2.303) 2F F
 (E  E  ) =
RT F
pH
pH =
F 2 (2.303RT )
1.15 V E + 0.15 V = = 9.72 0.1183 V 0.1183 V
E10.31(b) The electrode reactions are L: R: Ag+ (aq) + e Ag(s) AgI(s) + e Ag(s) + I (aq) AgI(s) Ag+ (aq) + I (aq)
Overall(R  L):
Since the cell reaction is a solubility equilibrium, for a saturated solution there is no further tendency to dissolve and so E = 0 E10.32(b) R: L: 2Bi3+ (aq) + 6e 2Bi(s) Bi2 S3 (s) + 6e 2Bi(s) + 3S2 (aq) 2Bi3+ (aq) + 3S2 (aq) Bi2 S3 (s) =6
 F E  RT 6(0.96 V) = (25.693 103 V)
Overall(R  L): ln K =
= 224 K = e224 It is convenient to give the solution for (b) first. (b) KS = K 1 = e224 1098 , since the cell reaction is the reverse of the solubility equilibrium. (a) KS 1098 = S= 1098 108
2 3 b b (Bi3+ )  (S2 ) = (2S)2 (3S)3 = 108S 5  b b
1/5
1020 mol L1
Solutions to problems
Solutions to numerical problems
P10.1 We require two halfcell reactions, which upon subtracting one (left) from the other (right), yields the given overall reaction (Section 10.4). The halfreaction at the right electrode corresponds to reduction, that at the left electrode to oxidation, though all halfreactions are listed in Table 10.7 as reduction reactions.
 E
R: Hg2 SO4 (s) + 2e 2Hg(l) + SO2 (aq) +0.62 V 4 L: PbSO4 (s) + 2e Pb(s) + SO2 (aq) 0.36 V 4 R  L: Pb(s) + Hg2 SO4 (s) PbSO4 (s) + 2Hg(l) +0.98 V
156
INSTRUCTOR'S MANUAL
Hence, a suitable cell would be Pb(s)PbSO4 (s)H2 SO4 (aq)Hg2 SO4 (s)Hg(l) or, alternatively, Pb(s)PbSO4 (s)H2 SO4 (aq) H2 SO4 (aq) Hg2 SO4 (s)Hg(l) For the cell in which the only sources of electrolyte are the slightly soluble salts, PbSO4 and Hg2 SO4 , the cell would be Pb(s)PbSO4 (s)PbSO4 (aq) Hg2 SO4 (aq)Hg2 SO4 (s)Hg(l) The potential of this cell is given by the Nernst equation [10.34].
 E = E 
RT ln Q [10.34]; = 2 F aPb2+ aSO2 KS (PbSO4 ) 4 Q= = aHg2+ aSO2 KS (Hg2 SO4 )
2 4
E = (0.98 V)  = (0.98 V) 
RT KS (PbSO4 ) ln KS (Hg2 SO4 ) 2F 25.693 103 V 2 ln 1.6 108 6.6 107
[Table 10.6, 4th Edition, or CRC Handbook] = (0.98 V) + (0.05 V) = +1.03 V P10.6 PtH2 (g)NaOH(aq), NaCl(aq)AgCl(s)Ag(s) H2 (s) + 2AgCl(s) 2Ag(s) + 2Cl (aq) + 2H+ (aq)
 E = E 
=2
RT  ln Q, Q = a(H+ )2 a(Cl )2 [f/p  = 1] 2F RT RT RT Kw a(Cl ) Kw b(Cl )    = E  = E  ln a(H+ )a(Cl ) = E   ln ln F F F a(OH ) b(OH ) Kw b(Cl ) b(Cl ) RT RT RT  = E  ln ln Kw  ln F F F b(OH ) b(OH ) RT RT b(Cl ) pKw  ln F F b(OH )

 = E 
 = E  + (2.303)
pKw =  log Kw =
 ln Kw 2.303
b(Cl  ln b(OH)) E  E Hence, pKw = + 2.303RT /F 2.303
=
 E  E + 0.05114 2.303RT /F
      E  = ER  EL = E  (AgCl, Ag)  E  (H+ /H2 ) = +0.22 V  0 [Table 10.7]  We then draw up the following table with the more precise value for E  = +0.2223 V [Problem 10.8]
/ C E/V
2.303RT F
20.0 1.04774 0.05819 14.23
25.0 1.04864 0.05918 14.01
30.0 1.04942 0.06018 13.79
V pKw
EQUILIBRIUM ELECTROCHEMISTRY
157
  d ln Kw rH [9.26] = dT RT 2
Hence, then with
rH
rH
 
= (2.303)RT 2 pKw T
d (pKw ) dT
d pKw dT
 
(2.303) (8.314 J K 1 mol1 ) (298.15 K)2 = +74.9 kJ mol1
13.79  14.23 10 K
rG rS
 
= RT ln Kw = 2.303RT pKw = +80.0 kJ mol1 =
rH  
 
 T
rG
 
= 17.1 J K1 mol1
See the original reference for a careful analysis of the precise data. P10.7 The cells described in the problem are backtoback pairs of cells each of the type Ag(s)AgX(s)MX(b1 )Mx Hg(s) R: L: M+ (b1 ) + e  Mx Hg(s)
Hg
(Reduction of M+ and formation of amalgam)
Hg
AgX(s) + e Ag(s) + X (b1 ) Ag(s) + M+ (b1 ) + X (b1 )  Mx Hg(s) + AgX(s) =1
R  L: Q=
a(Mx Hg) a(M+ )a(X ) RT  E = E  ln Q F
For a pair of such cells back to back, Ag(s)AgX(s)MX(b1 )Mx Hg(s)MX(b2 )AgX(s)Ag(s) RT RT   ER = E   EL = E   ln QR ln QL F F QL (a(M+ )a(X ))L RT RT ln ln E= = F QR F (a(M+ )a(X ))R (Note that the unknown quantity a(Mx Hg) drops out of the expression for E.) a(M+ )a(X ) = + b+  b  b  b
2 =
b 2  b
(b+ = b )
With L = (1) and R = (2) we have E= (1) 2RT b1 2RT ln + ln b2 F (2) F
b1 Take b2 = 0.09141 mol kg1 (the reference value), and write b =  b E= 2RT F ln b + ln 0.09141 (ref)
158
INSTRUCTOR'S MANUAL
For b = 0.09141, the extended DebyeH ckel law gives u log (ref) = (1.461) (0.09141)1/2 + (0.20) (0.09141) = 0.2735 (1) + (1.70) (0.09141)1/2 b + ln 0.09141 0.5328
(ref) = 0.5328 then E = (0.05139 V) ln ln =
E b  ln (0.09141) (0.05328) 0.05139 V
We then draw up the following table
b/(mol/kg1 ) E/V 0.0555 0.0220 0.572 0.09141 0.0000 0.533 0.1652 0.0263 0.492 0.2171 0.0379 0.469 1.040 0.1156 0.444 1.350 0.1336 0.486
A more precise procedure is described in the original references for the temperature dependence of  E  (Ag, AgCl, Cl ), see Problem 10.10. P10.10 The method of the solution is first to determine
1 2 H2 (g) + AgCl(s) rG  
,
rH
 
, and
rS
 
for the cell reaction
Ag(s) + HCl(aq)
   and then, from the values of these quantities and the known values of f G , f H  , and S  for         all the species other than Cl (aq), to calculate f G , f H , and S for Cl (aq). rG    = F E 
At 298.15 K(25.00 C)
 E  /V = (0.23659)  (4.8564 104 ) (25.00)  (3.4205 106 ) (25.00)2
+ (5.869 109 ) (25.00)3 = +0.22240 V Therefore,
rS    G = (96.485 kC mol1 ) (0.22240 V) = 21.46 kJ mol1  r G = T p  E  F = F T p C  E  p K
=
[d/ C = dT /K]
(a)
E  p
V
E  p V/ C
= (4.8564 104 / C)  (2) (3.4205 106 /( C)2 ) + (3) (5.869 109 2 /( C)3 ) = (4.8564 104 )  (6.8410 106 (/ C)) + (1.7607 108 (/ C)2 )
Therefore, at 25.00 C,
 E  = 6.4566 104 V/ C p
and
 E  = (6.4566 104 V/ C) ( C/K) = 6.4566 104 V K1 T p
EQUILIBRIUM ELECTROCHEMISTRY
159
Hence, from equation (a)
rS  
= (96.485 kC mol1 ) (6.4566 104 V K1 ) = 62.30 J K1 mol1
 
and
rH
=
rG
 
 + T rS
= (21.46 kJ mol1 ) + (298.15 K) (62.30 J K 1 mol1 ) = 40.03 kJ mol1 For the cell reaction
1 2 H2 (g) + AgCl(s) Ag(s) + HCl(aq)     = f G (H+ ) + f G (Cl )  rG   = f G (Cl )  f G (AgCl)  
fG
(AgCl)
 
[ f G (H+ ) = 0]
Hence,
fG
 
(Cl ) =
rG
 
+
fG
 
(AgCl) = [(21.46)  (109.79)] kJ mol1 = 131.25 kJ mol1
Similarly,
fH
 
(Cl ) =
rH
 
+
fH
 
(AgCl) = (40.03)  (127.07 kJ mol1 ) = 167.10 kJ mol1
For the entropy of Cl in solution we use with S
  rS  
(H+ ) = 0. Then,
rS  
    1  = S  (Ag) + S  (H+ ) + S  (Cl )  2 S  (H2 )  S  (AgCl)  1    S  (Ag) + 2 S  (H2 ) + S  (AgCl)
 S  (Cl ) =
1 = (62.30)  (42.55) + 2 (130.68) + (96.2) = +56.7 J K1 mol1
P10.12
(a) From
G = V [5.10] p T rG we obtain = rV p T Substituting r G = F E [10.32] yields E rV = p T ,n F
(b) The plot (Fig. 10.1) of E against p appears to fit a straight line very closely. A linear regression analysis yields Slope = 2.840 103 mV atm1 , Intercept = 8.5583 mV, R = 0.999 997 01 (an extremely good fit) From r V E (2.666 106 m3 mol1 ) = p T ,n 1 9.6485 104 C mol1 Since J = V C = Pa m3 , C = m3 V Pa m3 or = V C Pa standard deviation = 3 106 mV atm1 standard deviation = 2.8 103 mV
160
INSTRUCTOR'S MANUAL
13
12
11
10
9
8 0 500 1000 1500
Figure 10.1 Therefore E = p T ,n 2.666 106 9.6485 104 V 1.01325 105 Pa = 2.80 106 V atm1 Pa atm = 2.80 103 mV atm1 This compares closely to the result from the potential measurements. (c) A fit to a secondorder polynomial of the form E = a + bp + cp 2 yields
a = 8.5592 mV, b = 2.835 103 mV atm1 , c = 3.02 109 mV atm2 , R = 0.999 997 11 standard deviation = 0.0039 mV standard deviation = 0.012 103 mV atm1 standard deviation = 7.89 109 mV atm1
This regression coefficient is only marginally better than that for the linear fit, but the uncertainty in the quadratic term is > 200 per cent. E = b + 2cp p T The slope changes from to E = b = 2.835 103 mV atm1 p min
E = b + 2c(1500 atm) = 2.836 103 mV atm1 p max E are very good. We conclude that the linear fit and constancy of p
EQUILIBRIUM ELECTROCHEMISTRY
161
(d) We can obtain an order of magnitude value for the isothermal compressibility from the value of c. rV = 2c p T 1 rV 2cF (T )cell =  = V p T V 1 2E = 2 F p (T )cell =
3 2(1) (3.02 1012 V atm2 ) (9.6485 104 C mol1 ) 82.058 cm Jatm 8.3145
1 cm3 0.996 g
18.016 g 1 mol
= 3.2 107 atm1
standard deviation 200 per cent
where we have assumed the density of the cell to be approximately that of water at 30 C. Comment. It is evident from these calculations that the effect of pressure on the potentials of cells involving only liquids and solids is not important; for this reaction the change is only 3 106 V atm1 . The effective isothermal compressibility of the cell is of the order of magnitude typical of solids rather than liquids; other than that, little significance can be attached to the calculated numerical value. P10.15 The equilibrium is K= a(H2 O)4 a(V4 O12 4 ) (V4 O12 4 )b(V4 O12 4 ) a(H2 VO4  )4 (H2 VO4  )4 b(H2 VO4  )4
Let x be b(H2 VO4  ); then b(V4 O12 4 ) = (0.010  x)/4. Then the equilibrium equation can be expressed as x4 K (H2 VO4  )4 (V4 O12 4 ) = (0.010  x)/4
which can be solved numerically once the constants are determined. The activity coefficients are log (H2 VO4  ) =  0.5373 = 0.269 2 so (H2 VO4  ) = 0.538 so (V4 O12 4 ) = 0.0842
and log (V4 O12 4 ) =  The equation is
0.5373(42 ) = 1.075 2
x 4 (2.5 106 ) = (0.010  x)/4 Its solution is x = 0.0048 mol kg1 = b(H2 VO4  ) and b(V4 O12 4 ) = 0.010  (0.010  0.0048)/4 = 0.0013 mol kg1 P10.18 The reduction reaction is Sb2 O3 (s) + 3H2 O(l) + 6e 2Sb(s) + 6OH (aq) Therefore (a) RT RT 2.303RT   ln a(OH )6 = E   ln a(OH ) = E  + pOH 6F F F [ln a(OH ) = 2.303 log a(OH ) = 2.303pOH]
 E = E 
Q = a(OH )6
=6
162
INSTRUCTOR'S MANUAL
(b) Since pOH + pH = pKw
 E = E +
2.303RT (pKw  pH) F
(c) The change in potential is E= 2.303RT (pOHf  pOHi ) = (59.17 mV) (pOHf  pOHi ) F
pOHf =  log(0.050 ) =  log 0.050  log =  log 0.050 + A (0.050) = 1.415 pOHi =  log(0.010 ) =  log 0.010  log =  log 0.010 + A (0.010) = 2.051 Hence, P10.19 E = (59.17 mV) (1.415  2.051) = 37.6 mV
rH  
We need to obtain
for the reaction
1 + 2 H2 (g) + Uup (aq)
Uup(s) + H+ (aq)
We draw up the thermodynamic cycle shown in Fig. 10.2. Data are obtained from Table 13.4, 14.3, 2.6, and 2.6b. The conversion factor between eV and kJ mol1 is 1 eV = 96.485 kJ mol1 The distance from A to B in the cycle is given by
rH   1 = x = (3.22 eV) + 2 (4.5 eV) + (13.6 eV)  (11.3 eV)  (5.52 eV)  (1.5 eV)
= 0.75 eV
rS  
=S
 
  1  (Uup, s) + S  (H+ , aq)  2 S  (H2 , g)  S  (Uup+ , aq)
1 = (0.69) + (0)  2 (1.354)  (1.34) meV K 1 = 1.33 meV K1
i
Figure 10.2
rG
 
=
rH
 
  T r S  = (0.75 eV) + (298.15 K) (1.33 meV K 1 ) = +1.15 eV
which corresponds to +111 kJ mol1 The electrode potential is therefore
  r G , with = 1, or 1.15 V F
EQUILIBRIUM ELECTROCHEMISTRY
163
Solutions to theoretical problems
P10.21 MX(s) M+ (aq) + X (aq), b(X ) = S + C or S 2 + CS  Ks = 0 Ks b(M+ )b(X ) b b  b
b(M+ ) = S, Ks = S(S + C),
4Ks 1/2 1 1 1 1 which solves to S = 2 (C 2 + 4Ks )1/2  2 C or S = 2 C 1 + 2  2C C If 4Ks C2, Ks 1 1  2 C (1 + x)1/2 1 + 2 x + C b(M+ ) = S ,
1/2
2Ks 1 S 2C 1 + 2 C P10.22
2 Ks = a(M+ )a(X ) = b(M+ )b(X ) ;
b(X ) = S + C
log = AI 1/2 = AC 1/2 = e2.303AC
1/2
ln = 2.303AC 1/2
1/2
2 = e4.606AC
Ks = S (S + C) e4.606AC We solve S 2 + S C  Ks
2
=0
1/2
4Ks 1 to get S = C2 + 2 2
Ks 1 [as in Problem 10.21]  C 2 2 C
1/2
2 Therefore, since = e4.606AC
S
Ks e4.606AC C
1/2
P10.25
The halfreactions involved are:
  R: cyt ox + e cytred Ecyt   L: Dox + e Dred ED The overall cell reaction is:
R  L = cyt ox + Dred
cytred + Dox
     E   = Ecyt  ED
(a) The Nernst equation for the cell reaction is E=E RT [cyt red ][Dox ] ln F [cyt ox ][Dred ]
    Ecyt  ED
at equilibrium, E = 0; therefore ln ln [cytred ]eq [Dox ]eq F = [cyt ox ]eq [Dred ]eq RT [Dox ]eq [Dred ]eq = ln
[cyt]ox [cyt]red [Dox ]eq [Dred ]eq
+
F RT
    Ecyt  ED
Therefore a plot of ln intercept of F RT
against ln
[cyt]ox [cyt]red
is linear with a slope of one and an
    Ecyt  ED
164
INSTRUCTOR'S MANUAL
(b) Draw up the following table:
[Dox ]eq [Dred ]eq [cytox ]eq ln [cyt red ]eq ln 5.882 4.547 4.776 3.772 3.661 2.415 3.002 1.625 2.593 1.094 1.436 0.2120 0.6274 0.3293
The plot of ln 1.2124. Hence
  Ecyt =
[Dox ]eq [Dred ]eq
against ln
[cytox ]eq [cyt red ]eq
is shown in Fig. 10.3. The intercept is
RT (1.2124) + 0.237 V F = 0.0257 V (1.2124) + 0.237 V
= +0.206 V
0 1 ln([Dox]eq / [Dred]eq) 2 3 4 5 6 5
y = 1.2124 + 1.0116x
R = 0.99427
4
3
2
1
0
1
ln([cytox]eq / [cytred]eq)
Figure 10.3
Solutions to application
P10.27 (a) where d is density in g cm3 at 25 C, a = 14.523 mol kg1 (g cm3 )1 , c = 25.031 mol kg1 (g cm3 )2 , and d25 = 0.99707 g cm3 . For 1 kg solvent (mH2 O = 1 kg): mass %H2 SO4 = mass %H2 SO4 (d) = mH2 SO4 mH2 SO4 + mH2 O 100 b(d) 100 = b 100 mH O b + mH 2mH 2 SO4 2 O molalityH2 SO4 = b(d) = a(d  d25 ) + c(d  d25 )2
where mH2 SO4 = 0.09807 kg mol1 b(d) + mH 1SO 2 4 an equation for the solution molarity is deduced with a unit analysis. molarityH2 SO4 (d) = b(d) 1  mass %H2 SO4 (d) 103 3 cm d 100 L kg 103 g
EQUILIBRIUM ELECTROCHEMISTRY
165
Sulfuric Acid Solutions 10 Molality / (mol/kg) 8 6 4 2 0
1
1.1
1.2 Density/(g / mL)
1.3
1.4
Figure 10.4(a)
Sulfuric Acid Solutions 50 Mass Percentage Sulfuric Acid 40 30 20 10 0 1 1.1 1.2 Density/(g / mL) 1.3 1.4
Figure 10.4(b)
Sulfuric Acid Solutions 7
Molarity/(mol / L)
6 5 4 3 2 1 0 1 1.1 1.2 Density/(g / mL) 1.3 1.4
Figure 10.4(c)
(b)
cell: Pb(s)  PbSO4 (s)  H2 SO4 (aq)  PbO2 (s)  PbSO4 (s)  Pb(s) cathode: PbO2 (s) + 3H+ (aq) + HSO (aq) + 2e PbSO4 (s) + 2H2 O(l) 4   Ecathode = 1.6913 V anode: PbSO4 (s) + H+ (aq) + 2e Pb(s) + HSO 4   Eanode = 0.3588 V net: PbO2 (s) + Pb(s) + 2H+ (aq) + 2HSO (aq) 2PbSO4 (s) + 2H2 O(l) 4    E  = Ecathode  Eanode = 2.0501V (eqn 10.38)
rG    = F E  = (2)(9.64853 104 C mol1 )(2.0501 V)
= 3.956 105 C V mol1 = 3.956 105 J mol1 = 395.6 kJ mol1
166
INSTRUCTOR'S MANUAL
  fH   rH
values of Table 2.6 and the CRC Handbook of Chemistry and Physics are used in the calculation.
    = 2 f H  (PbSO4 ) + 2 f H  (H2 O(l))  fH  
rH
(PbO2 ) 
fH
 
(Pb)
2 f H
 
(H )  2 f H
+
 
(HSO ) 4
= 2(919.94 kJ mol1 ) + 2(285.83 kJ mol1 )  (277.4 kJ mol1 ) 2(887.34 kJ mol1 )
rH rS  
= 359.5 kJ mol1 =
rH  
 
 T
rG
 
=
359.5 kJ mol1  (395.6 kJ mol1 ) 298.15 K (eqn 4.39)
 
= 121 J K1 mol1
  E  (15 C) = E  (25 C) +
  E  = E  (25 C) +
rS
K mol1 (121 J 1 ) = 2.0501 V + K) (10 mol1 2(96485 C ) = 2.0501V + 0.006V = 2.0507V
F
T
(eqn 10.45)
The temperature difference makes a negligibly small difference in the cell potential. When Q = 6.0 105 ,
 E = E 
RT ln Q (eqn 10.34) F K mol1 (8.31451 J 1 )(298.15 K) = 2.0501 V  ln(6.0 105 ) mol1 2(96485 C ) = 2.1750 V
(c) The general form of the reduction halfreaction is: ox + e + H H+ + aA red + xX using eqn 10.34,
 E = E  x ared aX RT RT  ln Q = E   ln H a F F aox aH+ aA
 = E 
1 RT ln H F aH+ (all species other than acids are at unit activity in a Pourboix diagram) H RT H RT ln(10)  ln aH+ = E  + log aH+ F F H RT ln(10) F pH (eqn 9.29)
 = E +  = E 
 E = E   (0.05916V)
H pH
EQUILIBRIUM ELECTROCHEMISTRY
167
For the PbO2  PbSO4 couple, PbO2 (s) + 4H+ + SO2 (aq) + 2e PbSO4 (s) + 2H2 O(l) 4
 E  = 1.6913 V, H = 4, = 2
E = 1.6913 V  (0.11832 V)pH For pH = 5, E = 1.0997 V For pH = 8, E = 0.7447 V For the PbSO4 /Pb couple, PbSO4 (s) + 2e Pb(s) + SO2 (aq) 4
 Since H = O, E = E  = 0.3588 V at all pH values in the Pourboix diagram.
Part 2: Structure
11
Quantum theory: introduction and principles
Solutions to exercises
Discussion questions
E11.1(b) A successful theory of blackbody radiation must be able to explain the energy density distribution of the radiation as a function of wavelength, in particular, the observed drop to zero as 0. Classical theory predicts the opposite. However, if we assume, as did Planck, that the energy of the oscillators that constitute electromagnetic radiation are quantized according to the relation E = nh = nhc/, we see that at short wavelengths the energy of the oscillators is very large. This energy is too large for the walls to supply it, so the shortwavelength oscillators remain unexcited. The effect of quantization is to reduce the contribution to the total energy emitted by the blackbody from the highenergy shortwavelength oscillators, for they cannot be sufficiently excited with the energy available. In quantum mechanics all dynamical properties of a physical system have associated with them a corresponding operator. The system itself is described by a wavefunction. The observable properties of the system can be obtained in one of two ways from the wavefunction depending upon whether or not the wavefunction is an eigenfunction of the operator. When the function representing the state of the system is an eigenfunction of the operator , we solve the eigenvalue equation (eqn 11.30) = in order to obtain the observable values, , of the dynamical properties. When the function is not an eigenfunction of , we can only find the average or expectation value of dynamical properties by performing the integration shown in eqn 11.39 = E11.3(b) No answer.
E11.2(b)
d.
Numerical exercises
E11.4(b) The power is equal to the excitance M times the emitting area P = MA = T 4 (2rl) = (5.67 108 W m2 K 4 ) (3300 K)4 (2 ) (0.12 103 m) (5.0 102 m) = 2.5 102 W Comment. This could be a 250 W incandescent light bulb. E11.5(b) Wien's displacement law is T max = c2 /5 E11.6(b) so max = 1.44 102 m K c2 = = 1.15 106 m = 1.15 m 5T 5(2500 K)
The de Broglie relation is = h h = p mv so v= h 6.626 1034 J s = m (1.675 1027 kg) (3.0 102 m)
v = 1.3 105 m s1
172
INSTRUCTOR'S MANUAL
E11.7(b)
The de Broglie relation is = h h = p mv so v= h 6.626 1034 J s = m (9.11 1031 kg) (0.45 109 m)
v = 1.6 106 m s1 E11.8(b) The momentum of a photon is p= 6.626 1034 J s h = = 1.89 1027 kg m s1 350 109 m 1.89 1027 kg m s1 p = m 2(1.0078 103 kg mol1 /6.022 1023 mol1 )
The momentum of a particle is p = mv so v=
v = 0.565 m s1 E11.9(b) The energy of the photon is equal to the ionization energy plus the kinetic energy of the ejected electron Ephoton = Eionize + Eelectron hc Eionize +
1 2 2 mv
so
hc 1 = Eionize + 2 mv 2 (6.626 1034 J s) (2.998 108 m s1 )
and =
=
1 5.12 1018 J + 2 (9.11 1031 kg) (345 103 m s1 )2
= 3.48 108 m = 38.4 nm E11.10(b) The uncertainty principle is
1 p x 2h
so the minimum uncertainty in position is x = h 2 p = h 1.0546 1034 J s = 31 kg) (0.000 010) (995 103 m s1 ) 2m v 2(9.11 10
= 5.8 106 m E11.11(b) E = h = hc ; E(per mole) = NA E = NA hc
hc = (6.62608 1034 J s) (2.99792 108 m s1 ) = 1.986 1025 J m NA hc = (6.02214 1023 mol1 ) (1.986 1025 J m) = 0.1196 J m mol1 Thus, E = 1.986 1025 J m 0.1196 J m mol1 ; E(per mole) = We can therefore draw up the following table
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
173
(a) 200 nm (b) 150 pm (c) 1.00 cm
E/J 9.93 1019 1.32 1015 1.99 1023
E/(kJ mol1 ) 598 7.98 105 0.012
E11.12(b) Assuming that the 4 He atom is free and stationary, if a photon is absorbed, the atom acquires its momentum p, achieving a speed v such that p = mv. p v= m = 4.00 1.6605 1027 kg = 6.642 1027 kg m h p= (a) 6.626 1034 J s = 3.313 1027 kg m s1 200 109 m 3.313 1027 kg m s1 p = = 0.499 m s1 v= m 6.642 1027 kg p= p= 6.626 1034 J s = 4.417 1024 kg m s1 150 1012 m 4.417 1024 kg m s1 p = 665 m s1 = v= m 6.642 1027 kg 6.626 1034 J s = 6.626 1032 kg m s1 1.00 102 m p 6.626 1032 kg m s1 = 9.98 106 m s1 v= = m 6.642 1027 kg
(b)
(c)
p=
E11.13(b) Each emitted photon increases the momentum of the rocket by h/. The final momentum of the Nh rocket will be N h/, where N is the number of photons emitted, so the final speed will be . mrocket The rate of photon emission is the power (rate of energy emission) divided by the energy per photon (hc/), so N = v = tP hc and v= tP hc h mrocket = tP cmrocket
(10.0 yr) (365 day yr 1 ) (24 h day1 ) (3600 s h1 ) (1.50 103 W) (2.998 108 m s1 ) (10.0 kg)
= 158 m s1 E11.14(b) Rate of photon emission is rate of energy emission (power) divided by energy per photon (hc/) (a) (b) rate = (0.10 W) (700 109 m) P = 3.52 1017 s1 = hc (6.626 1034 J s) (2.998 108 m s1 ) (1.0 W) (700 109 m) = 3.52 1018 s1 rate = (6.626 1034 J s) (2.998 108 m s1 ) c2 1.44 102 m K = 1800 K = 5max 5(1600 109 m)
E11.15(b) Wien's displacement law is T max = c2 /5 so T =
174
INSTRUCTOR'S MANUAL
E11.16(b) Conservation of energy requires Ephoton = + EK = h = hc/ so 2EK 1/2 me EK = hc/ 
1 and EK = 2 me v 2 so v =
(a)
(6.626 1034 J s) (2.998 108 m s1 )  (2.09 eV) (1.60 1019 J eV1 ) 650 109 m But this expression is negative, which is unphysical. There is no kinetic energy or velocity because the photon does not have enough energy to dislodge the electron. EK = EK = (6.626 1034 J s) (2.998 108 m s1 )  (2.09 eV) (1.60 1019 J eV1 ) 195 109 m
(b)
= 6.84 1019 J and v = 2(3.20 1019 J) 9.11 1031 kg so
1/2
= 1.23 106 m s1
E11.17(b) E = h = h/ , (a) (b) (c)
E = 6.626 1034 J s/2.50 1015 s = 2.65 1019 J = 160 kJ mol1 E = 6.626 1034 J s/2.21 1015 s = 3.00 1019 J = 181 kJ mol1 E = 6.626 1034 J s/1.0 103 s = 6.62 1031 J = 4.0 1010 kJ mol1
E11.18(b) The de Broglie wavelength is = h p
The momentum is related to the kinetic energy by EK = p2 2m so p = (2mEK )1/2
The kinetic energy of an electron accelerated through 1 V is 1 eV = 1.60 1019 J, so = h (2mEK )1/2
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
175
(a)
=
6.626 1034 J s (2(9.11 1031 kg) (100 eV) (1.60 1019 J eV1 ))1/2 6.626 1034 J s (2(9.11 1031 kg) (1.0 103 eV) (1.60 1019 J eV1 ))1/2 6.626 1034 J s (2(9.11 1031 kg) (100 103 eV) (1.60 1019 J eV1 ))1/2
= 1.23 1010 m (b) =
= 3.9 1011 m (c) =
= 3.88 1012 m E11.19(b) The minimum uncertainty in position is 100 pm . Therefore, since p v= h 2 x = 1.0546 1034 J s = 5.3 1025 kg m s1 2(100 1012 m)
1 x p 2h
5.3 1025 kg m s1 p = = 5.8 105 m s1 m 9.11 1031 kg
1 Ebinding = hc/  2 me v 2
E11.20(b) Conservation of energy requires
1 Ephoton = Ebinding + 2 me v 2 = h = hc/
so
and Ebinding =
(6.626 1034 J s) (2.998 108 m s1 ) 121 1012 m 1  2 (9.11 1031 kg) (5.69 107 m s1 )2
= 1.67 1016 J Comment. This calculation uses the nonrelativistic kinetic energy, which is only about 3 per cent less than the accurate (relativistic) value of 1.52 1015 J. In this exercise, however, Ebinding is a small difference of two larger numbers, so a small error in the kinetic energy results in a larger error in Ebinding : the accurate value is Ebinding = 1.26 1016 J.
Solutions to problems
Solutions to numerical problems
P11.3 h J s s1 , [E ] = =K k J K1 In terms of E the Einstein equation [11.9] for the heat capacity of solids is E = CV = 3R E 2 T eE /2T eE /T  1
2
,
classical value = 3R E or when
It reverts to the classical value when T
h 1 as demonstrated in the text kT E . (Section 11.1). The criterion for classical behaviour is therefore that T E = h (6.626 1034 J Hz1 ) = 4.798 1011 (/Hz)K = k 1.381 1023 J K1
176
INSTRUCTOR'S MANUAL
(a) For = 4.65 1013 Hz, E = (4.798 1011 ) (4.65 1013 K) = 2231 K (b) For = 7.15 1012 Hz, E = (4.798 1011 ) (7.15 1012 K) = 343 K Hence (a) CV = 3R CV = 3R 2231 K 2 298 K 343 K 2 298 K e2231/(2298) e2231/298  1 e343/(2298) e343/298  1
2 2
= 0.031 = 0.897
(b)
Comment. For many metals the classical value is approached at room temperature; consequently, the failure of classical theory became apparent only after methods for achieving temperatures well below 25 C were developed in the latter part of the nineteenth century. P11.5 The hydrogen atom wavefunctions are obtained from the solution of the Schr dinger equation in o Chapter 13. Here we need only the wavefunction which is provided. It is the square of the wavefunction that is related to the probability (Section 11.4). 2 = 1
3 a0
e2r/a0 ,
=
4 3 r , 3 0
r0 = 1.0 pm
If we assume that the volume is so small that does not vary within it, the probability is given by 2 =
3 4r0 2r/a0 4 e = 3 3 3a0
1.0 3 2r/a0 e 53 1.0 3 = 9.0 106 53 1.0 3 2 e = 1.2 106 53
(a) (b)
r=0: r = a0 :
2 =
4 3 4 3
2 =
Question. If there is a nonzero probability that the electron can be found at r = 0 how does it avoid destruction at the nucleus? (Hint. See Chapter 13 for part of the solution to this difficult question.) P11.7 According to the uncertainty principle,
1 p q 2 h,
where
q and p are rootmeansquare deviations: and p = ( p 2  p 2 )1/2 .
q = ( x 2  x 2 )1/2
To verify whether the relationship holds for the particle in a state whose wavefunction is = (2a/ )1/4 eax ,
2
We need the quantummechanical averages x , x 2 , p , and p 2 .
x =
2
x
d =

2a 1/4 ax 2 e x
2a 1/4 ax 2 e dx,
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
177
x =
2 2a 1/2 xe2ax dx = 0; 
x
2
=

2a 1/4 ax 2 2 e x
1/2
2a 1/4 ax 2 e dx = 1 ; 4a
2 2a 1/2 x 2 e2ax dx, 
x2 =
2a q = p =
1/2 2(2a)3/2
=
so
1 . 2a 1/2

hd i dx
dx
and
p
2
=

 2 h
d2 dx 2
dx.
We need to evaluate the derivatives: d = dx and d2 = dx 2
2 2a 1/4 (2ax)eax 2 2 2a 1/4 [(2ax)2 eax + (2a)eax ] = 2 2a 1/4 (4a 2 x 2  2a)eax .
So
p =

2a 1/4 ax 2 e
h i
2 2a 1/4 (2ax)eax dx
2 h = i
2 2a 1/2 xe2ax dx = 0; 
p
2
=

2 2a 1/4 ax 2 2a 1/4 e ( 2 ) h (4a 2 x 2  2a)eax dx,
p
2
2 2a 1/2 = (2a ) h (2ax 2  1)e2ax dx, 2 
p2 and Finally,
= (2a 2 ) h
2a
1/2
2a
1/2 1/2  2(2a)3/2 (2a)1/2
= a 2 ; h
p = a 1/2 h. q p=
1 a 1/2 h = 1/2 , h 2a 1/2 which is the minimum product consistent with the uncertainty principle.
178
INSTRUCTOR'S MANUAL
Solutions to theoretical problems
P11.9 We look for the value of at which is a maximum, using (as appropriate) the shortwavelength (highfrequency) approximation = 8hc 5 1 ehc/kT  1 [11.5] at = max
d 5 hc = + 2 d kT Then, 5 +
ehc/kT =0 ehc/kT  1
hc ehc/kT =0 hc/kT kT e 1 hc hc/kT e =0 kT
Hence, 5  5ehc/kT + If
hc 1 [short wavelengths, high frequencies], this expression simplifies. We neglect the initial 5, kT cancel the two exponents, and obtain hc = 5kT for = max and hc kT 1
or max T =
hc = 2.88 mm K , in accord with observation. 5k
Comment. Most experimental studies of blackbody radiation have been done over a wavelength range of a factor of 10 to 100 of the wavelength of visible light and over a temperature range of 300 K to 10 000 K. Question. Does the shortwavelength approximation apply over all of these ranges? Would it apply to the cosmic background radiation of the universe at 2.7 K where max 0.2 cm? 8hc 1 = [11.5] hc/kT  1 5 e hc decreases, and at very long wavelength hc/kT kT the exponential in a power series. Let x = hc/kT , then As increases, ex = 1 + x + = 8hc 5 1 2 1 x + x3 + 2! 3! 1. Hence we can expand
P11.10
1 1 1 1 + x + 2! x 2 + 3! x 3 +  1 1 hc/kT
lim =
1 8 hc 8hc = 1+x1 5 5 8kT = 4
This is the RayleighJeans law [11.3].
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
179
P11.12
=
8hc 1 [11.5] hc/kT 5 e 1
hc ehc/kT  2 kT
40hc = 6 = 8hc 5 5
1 ehc/kT  1 1

8 hc 5
ehc/kT  1
2
ehc/kT  1

ehc/kT 5 hc + 2 kT ehc/kT  1
=  =0 hc 5max kT 5max kT hc Let x =
1
hc 1 hc/kT 5kT 1  e and
when = max
1 1  ehc/max kT 1  ehc/max kT then
=1
=1 or 5 1 = x 1  ex
hc ; max kT
5 1  ex = 1 x
The solution of this equation is x = 4.965. Then h = However M = T 4 = 4.965max kT c 2 5 k 4 15c2 h3 T4 (1)
(2)
Substituting (1) into (2) yields M k 2 5 k 4 15c2 2 5 ckT 1835.93 max
3 c T4 4.965max kT
1835.93 M max 2 5 cT 1835.9(1.451 106 m)3 (904.48 103 W) 2 5 (2.998 108 m s1 ) (2000 K) (1.000 m2 ) (3)
k 1.382 1023 J K1
180
INSTRUCTOR'S MANUAL
Substituting (3) into (1) h 5(1.451 106 m) (1.382 1023 J K1 ) (2000 K) 2.998 108 m s1
h 6.69 1034 J s Comment. These calculated values are very close to the currently accepted values for these constants. P11.14 In each case form N ; integrate (N ) (N ) d set the integral equal to 1 and solve for N . (a) =N 2 r a0 er/2a0 r 2 r/a0 e a0
0
2 = N2 2  2 d = N 2
4r 2 
4r 3 r4 + 2 a0 a0
er/a0 dr
0
sin d
2
d
0
3 = N 2 4 2a0  4 1/2
4 6a0 24a 5 + 20 a0 a0
3 (2) (2 ) = 32 a0 N 2 ;
hence N =
1
3 32a0
where we have used
0
n! x n eax dx = n+1 [Problem 11.13] a
(b)
= N r sin cos er/(2a0 ) 2 d = N 2
0
r 4 er/a0 dr
1 1
0
sin2 sin d
2 0
cos2 d
5 = N 2 4!a0
(1  cos2 ) d cos N= 1
5 32 a0 1 1 1/2
=N
2
5 4!a0
2 5 2 2 = 32 a0 N0 ; 3 cosn sin d = 
1 1
hence
where we have used
0
cosn d cos =
x n dx
and the relations at the end of the solution to Problem 11.8. [See Student's solutions manual.]
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
181
P11.16
Operate on each function with i; if the function is regenerated multiplied by a constant, it is an eigenfunction of i and the constant is the eigenvalue. (a) f = x 3  kx i(x 3  kx) = x 3 + kx = f Therefore, f is an eigenfunction with eigenvalue, 1 f = cos kx i cos kx = cos(kx) = cos kx = f Therefore, f is an eigenfunction with eigenvalue, +1 f = x 2 + 3x  1 i(x 2 + 3x  1) = x 2  3x  1 = constant f Therefore, f is not an eigenfunction of i.
(b)
(c)
P11.19
^ The kinetic energy operator, T , is obtained from the operator analogue of the classical equation EK = that is, (p)2 ^ ^ T = 2m px = ^ Then T = N2
 2 h 2m  2 h 2m
p2 2m
h d [11.32]; i dx
hence
h px =  2 ^2
d2 dx 2
and
h2 d2 ^ T = 2m dx 2
px ^2 d = 2m
p2 ^ 2m d
d
N2 =
1 d
= = px =
d2 dx 2 (eikx cos + eikx sin ) d
d
(k 2 ) (eikx cos + eikx sin ) d d
=
h2 k 2 d h2 k 2 = d 2m 2m
P11.20
h d [11.32] i dx px dx; ^
h i
px = N 2 = (a)
N2 =
d dx
1 dx d
px dx ^ = dx
dx
= eikx , Hence,
d = ik dx
h ik dx px = i = k h dx
182
INSTRUCTOR'S MANUAL
(b)
= cos kx,

d = k sin kx dx
d dx = k cos kx sin kx dx = 0 dx 
Therefore, px = 0 2 2 d = 2xex (c) = ex , dx

2 d xe2x dx = 0 dx = 2 dx 
[by symmetry, since x is an odd function]
Therefore, px = 0 P11.23 No solution.
Solution to applications
P11.27 (a) Consider any infinitesimal volume element dx dy dz within the hemisphere (Figure 11.1) that has a radius equal to the distance traveled by light in the time dt (c dt). The objective is to find the total radiation flux perpendicular to the hemisphere face at its center. Imagine an infinitesimal area A at that point. Let r be the distance from dx dy dz to A and imagine the infinitesimal area A perpendicular to r. E is the total isotropic energy density in dx dy dz. E dx dy dz is the energy emitted in dt. A /4 r 2 is the fraction of this radiation that passes through A . The radiation flux that originates from dx dy dz and passes through A in dt is given by:
A 4 r 2
A dt The contribution of JA to the radiation flux through A, JA , is given by the expression JA ( cos )/ = JA cos . The integration of this expression over the whole hemisphere gives an A A
JA =
E dx dy dz
=
E dx dy dz 4 r 2 dt
A c dt A c dt
JA
JA
dx dy dz
Figure 11.1
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
183
expression for JA . Spherical coordinates facilitate to integration: dx dy dz = r 2 sin d d dr = r 2 d(cos ) d dr where 0 2 and 0 /2. JA =
hemisphere
cos( )
E dx dy dz 4 r 2 dt E r2 4 dt {r 2 d(cos ) d dr}
2 c dt
=
hemisphere
cos( )
E = 4 dt E 4 dt cE 4 cE 4
cos(/2)
cos( ) d(cos ) 0 w dw (2) (c dt)
1 cos(0) 0
d
0
dr
= JA = 
1 (2) {Subscript "A" has been a bookkeeping device. It may be dropped.} 2 or dJ = c dE 4
J =
8hc d dE = 5 hc/RT (e  1)
[eqn 11.5]
~ ~ By eqn 16.1 = 1/. Taking differentials to be positive, d = d/2 or d = 2 d = d /~ 2 . ~ ~ The substitution of for gives: ~ 8hc 3 ~ d ~ dE = hc /kT e ~ 1 2 hc2 3 ~ Thus, dJ = f(~ ) d where f(~ ) = hc /kT ~ e ~ 1
The value of the StefanBoltzmann constant is defined by the low n = 0 dJ (~ ) = T 4 . n is called the total exitance. Let x = hc /kT (or = kT x/ hc), substitute the above equation for ~ ~ dJ (~ ) into the StefanBoltzmann low, and integrate.
n =
o
2 k 4 T 4 2hc2 3 d ~ ~ = ~ h3 c 2 ehc /kT  1 4 15 =
0
x 3 dx ex  1 T4
=
2k 4 T 4 h3 c 2
2 5 k 4 15 h3 c2
2 5 k 4 = 5.6704 108 W m2 K 4 15h3 c2 The function f (~ ) gives radiation density in units that are compatible with those often used in discussions of infrared radiation which lies between about 33 cm1 and 12 800 cm1 (Fig. 11.2). Thus, =
184
INSTRUCTOR'S MANUAL
Blackbody radiation density at 288.16 K. 6
4.5 F( ) /1011 J m2
3
1.5
0 0 500 1000 1500 2000
1
2500
3000
Wavenumber, /cm
Figure 11.2
By graphing f (~ ) at the observed average temperature of the Earth's surface (288.16 K) we easily see that the Earth's blackbody emissions are in the infrared with a maximum at about 600 cm1 . (b) Let R represent the radius of the Earth. Assuming an average balance between the Earth's absorption of solar radiation and Earth's emission of blackbody radiation into space gives: Solar energy absorbed = blackbody energy lost R 2 (1  albedo)(solar energy flux) = (4 R 2 )( T 4 ) Solving for T gives: T = = (1  albedo)(solar energy flux) 1/4 4 (1  0.29)(0.1353 W cm2 ) 4(5.67 1012 W cm2 K 4
1/4
= 255 K
This is an estimate of what the Earth's temperature would be in the absence of the greenhouse effect.
12
Quantum theory: techniques and applications
Solutions to exercises
Discussion questions
E12.1(b) The correspondence principle states that in the limit of very large quantum numbers quantum mechanics merges with classical mechanics. An example is a molecule of a gas in a box. At room temperature, the particleinabox quantum numbers corresponding to the average energy of the gas 1 molecules ( 2 kT per degree of freedom) are extremely large; consequently the separation between the levels is relatively so small (n is always small compared to n2 , compare eqn 12.10 to eqn 12.4) that the energy of the particle is effectively continuous, just as in classical mechanics. We may also look at these equations from the point of view of the mass of the particle. As the mass of the particle increases to macroscopic values, the separation between the energy levels approaches zero. The quantization disappears as we know it must. Tennis balls do not show quantum mechanical effects. (Except those served by Pete Sampras.) We can also see the correspondence principle operating when we examine the wavefunctions for large values of the quantum numbers. The probability density becomes uniform over the path of motion, which is again the classical result. This aspect is discussed in more detail in Section 12.1(c). The harmonic oscillator provides another example of the correspondence principle. The same effects mentioned above are observed. We see from Fig. 12.22 of the text that probability distribution for large values on n approaches the classical picture of the motion. (Look at the graph for n = 20.) E12.2(b) The physical origin of tunnelling is related to the probability density of the particle which according to the Born interpretation is the square of the wavefunction that represents the particle. This interpretation requires that the wavefunction of the system be everywhere continuous, event at barriers. Therefore, if the wavefunction is nonzero on one side of a barrier it must be nonzero on the other side of the barrier and this implies that the particle has tunnelled through the barrier. The transmission probability depends upon the mass of the particle (specifically m1/2 , through eqns 12.24 and 12.28): the greater the mass the smaller the probability of tunnelling. Electrons and protons have small masses, molecular groups large masses; therefore, tunnelling effects are more observable in process involving electrons and protons. The essential features of the derivation are: (1) The separation of the hamiltonian into large (unperturbed) and small (perturbed) parts which are independent of each other. (2) The expansion of the wavefunctions and energies as a power series in an unspecified parameters, , which in the end effectively cancels or is set equal to 1. (3) The calculation of the firstorder correction to the energies by an integration of the perturbation over the zeroorder wavefunctions. (4) The expansion of the firstorder correction to the wavefunction in terms of the complete set of functions which are a solution of the unperturbed Schrodinger equation. (5) The calculation of the secondorder correction to the energies with use of the corrected first order wavefunctions. See Justification 12.7 and Further reading for a more complete discussion of the method.
E12.3(b)
186
INSTRUCTOR'S MANUAL
Numerical exercises
E12.4(b) E= n2 h 2 8me L2
h2 (6.626 1034 J s)2 = = 2.678 1020 J 2 8me L 8(9.109 1031 kg) (1.50 109 m)2 Conversion factors E kJ mol1 = NA E/J 103
1 eV = 1.602 1019 J 1 cm1 = 1.986 1023 J (a) E3  E1 = (9  1) h2 = 8(2.678 1020 J) 8me L2
= 2.14 1019 J = 129 kJ mol1 = 1.34 eV = 1.08 104 cm1 (b) E7  E6 = (49  36) h2 8me L2 = 13(2.678 1020 J) = 3.48 1019 J = 210 kJ mol1 = 2.17 eV = 1.75 104 cm1 E12.5(b) The probability is P = where dx = 2 L sin2 2 x n x n x dx = sin2 L L L
x = 0.02L and the function is evaluated at x = 0.66L.
(a) For n = 1 P = 2(0.02L) 2 sin (0.66) = 0.031 L
(b) For n = 2 P = E12.6(b) 2(0.02L) 2 sin [2(0.66 )] = 0.029 L
The expectation value is p = ^ ^ p dx
but first we need p ^ p = i ^ d 2 1/2 n x sin dx L L = i 2 1/2 n n x cos L L L
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
187
so p = ^
n x 2i n L n x dx = 0 for all n cos sin L L L2 0
h2 n2 ^ p2 = 2m H = 2mEn = ^ 4L2 for all n. So for n = 2 p2 = ^ E12.7(b) n=5 5 = h2 L2
2 1/2 5x sin L L 2 5x 2 P (x) 5 sin L dP (x) =0 dx sin 10 x L (2 sin cos = sin 2)
Maxima and minima in P (x) correspond to d d 2 5 x P (x) sin dx dx L cos
5 x L
sin = 0 when = 0, , 2, . . . = n (n = 0, 1, 2, . . .) 10x = n n 10 L nL x= 10 Minima at x = 0, x = L Maxima and minima alternate: maxima correspond to n = 1, 3, 5, 7, 9 E12.8(b) The energy levels are En1 ,n2 ,n3 = (n2 + n2 + n2 )h2 1 2 3 8mL2 = E1 (n2 + n2 + n2 ) 1 2 3 x= L 3L L 7L 9L , , , , 10 2 10 10 10
where E1 combines all constants besides quantum numbers. The minimum value for all the quantum numbers is 1, so the lowest energy is E1,1,1 = 3E1 The question asks about an energy 14/3 times this amount, namely 14E1 . This energy level can be obtained by any combination of allowed quantum numbers such that n2 + n2 + n2 = 14 = 32 + 22 + 12 1 2 3 The degeneracy, then, is 6 , corresponding to (n1 , n2 , n3 ) = (1, 2, 3), (2, 1, 3), (1, 3, 2), (2, 3, 1), (3, 1, 2), or (3, 2, 1).
188
INSTRUCTOR'S MANUAL
E12.9(b)
3 E = 2 kT is the average translational energy of a gaseous molecule (see Chapter 20). 3 E = 2 kT =
(n2 + n2 + n2 )h2 1 2 3 8mL2
=
n2 h2 8mL2
3 E = 2 (1.381 1023 J K1 ) (300 K) = 6.214 1021 J
n2 =
8mL2 E h2
If L3 = 1.00 m3 , L2 = 1.00 m2 (6.626 1034 J s)2 h2 = = 1.180 1042 J 0.02802 kg mol1 8mL2 2) (8) 6.0221023 mol1 (1.00 m n2 = 6.214 1021 J n = 7.26 1010 = 5.265 1021 ; 1.180 1042 J E = En+1  En = E7.261010 +1  E7.261010 E = (2n + 1) h2 8mL2 = [(2) (7.26 1010 + 1)] h2 8mL2 = 14.52 1010 h2 8mL2
= (14.52 1010 ) (1.180 1042 J) = 1.71 1031 J The de Broglie wavelength is obtained from = h h = [Section 11.2] p mv
The velocity is obtained from
1 3 EK = 2 mv 2 = 2 kT = 6.214 1021 J
v2 =
6.214 1021 J
1 2 0.02802 kg 6.0221023mol 1 mol
1
= 2.671 105 ;
v = 517 m s1
=
6.626 1034 J s = 2.75 1011 m = 27.5 pm (4.65 1026 kg) (517 m s1 )
The conclusion to be drawn from all of these calculations is that the translational motion of the nitrogen molecule can be described classically. The energy of the molecule is essentially continuous, E 1. E E12.10(b) The zeropoint energy is E0 =
1 2
=
1 2
k 1/2 1 = 2 (1.0546 1034 J s) m = 3.92 1021 J
285 N m1 5.16 1026 kg
1/2
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
189
E12.11(b) The difference in adjacent energy levels is E= = k = 260 N m1 E12.12(b) The difference in adjacent energy levels, which is equal to the energy of the photon, is E = = h and = hc so k 1/2 hc = m k 1/2 m so k= m( E)2
2
=
(2.88 1025 kg) (3.17 1021 J)2 (1.0546 1034 J s)2
m 1/2 k 1/2 = 2c m k
8 1
= 2(2.998 10 m s
)
(15.9949 u) (1.66 1027 kg u1 ) 544 N m1
1/2
= 1.32 105 m = 13.2 m E12.13(b) The difference in adjacent energy levels, which is equal to the energy of the photon, is E = = h and = hc so hc k 1/2 = m
m 1/2 k 1/2 = 2c m k
Doubling the mass, then, increases the wavelength by 21/2 . So taking the result from Ex. 12.12(b), the new wavelength is = 21/2 (13.2 m) = 18.7 m E12.14(b) = g 1/2 [elementary physics] l
E = = h (a) (b) E = h = (6.626 1034 J Hz1 ) (33 103 Hz) = 2.2 1029 J 1 k 1/2 1 1 = + with m1 = m2 meff meff m1 m2 For a twoparticle oscillator meff , replaces m in the expression for . (See Chapter 16 for a more complete discussion of the vibration of a diatomic molecule.) E= = E= 2k 1/2 = (1.055 1034 J s) m = 3.14 1020 J (2) (1177 N m1 ) (16.00) (1.6605 1027 kg)
1/2
190
INSTRUCTOR'S MANUAL
E12.15(b) The first excitedstate wavefunction has the form
1 = 2N1 y exp  2 y 2
m 1/2 . To see if it satisfies Schr dinger's equation, o where N1 is a collection of constants and y x we see what happens when we apply the energy operator to this function ^ H =  d2 1 + 2 m2 x 2 2m dx 2
2
We need derivatives of d dy m 1/2 d 1 (2N1 ) (1  y 2 ) exp  2 y 2 = = dx dy dx and d2 d2 = 2 dx dy 2
2
dy 2 m m 1 (2N1 ) (3y + y 3 ) exp  2 y 2 = (y 2  3) = dx m
1 (y 2  3) + 2 m2 x 2
^ So H = 
2m
3 1 1 =  2 (y 2  3) + 2 y 2 = 2 ^ Thus, is an eigenfunction of H (i.e. it obeys the Schr dinger equation) with eigenvalue o 3 E = 2 h
E12.16(b) The zeropoint energy is
1 1 E0 = 2 = 2
k 1/2 meff
For a homonuclear diatomic molecule, the effective mass is half the mass of an atom, so E0 =
34 1 J s) 2 (1.0546 10
2293.8 N m1
1 27 kg u1 ) 2 (14.0031 u) (1.66054 10
1/2
E0 = 2.3421 1020 J E12.17(b) Orthogonality requires that
m n d = 0
if m = n. Performing the integration
m n d = 2 0
N eim Nein d = N 2
2 0
ei(nm) d
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
191
If m = n, then
m n d
N2 ei(nm) = i(n  m)
2
=
0
N2 (1  1) = 0 i(n  m)
Therefore, they are orthogonal. E12.18(b) The magnitude of angular momentum is ^ L2 1/2 = (l(l + 1))1/2 = (2(3))1/2 (1.0546 1034 J s) = 2.58 1034 J s Possible projections on to an arbitrary axis are ^ Lz = ml where ml = 0 or 1 or 2. So possible projections include 0, 1.0546 1034 J s and 2.1109 1034 J s E12.19(b) The cones are constructed as described in Section 12.7(c) and Fig. 12.36 of the text; their edges are of length {6(6 + 1)}1/2 = 6.48 and their projections are mj = +6, +5, . . . , 6. See Fig. 12.1(a). The vectors follow, in units of . From the highestpointing to the lowestpointing vectors (Fig. 12.1(b)), the values of ml are 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, and 6.
Figure 12.1(a)
Figure 12.1(b)
192
INSTRUCTOR'S MANUAL
Solutions to problems
Solutions to numerical problems
P12.4 E= = l(l + 1) 2 l(l + 1) 2 [12.65] = 2I 2meff R 2 [I = meff R 2 , meff in place of m] 1 1 + 1.008 126.90
l(l + 1) (1.055 1034 J s)2 (2) (1.6605 1027 kg) (160 1012 m)2 1 1 1 = + m2 meff m1
The energies may be expressed in terms of equivalent frequencies with = Therefore, E = l(l + 1) (1.31 1022 J) = l(l + 1) (198 GHz) Hence, the energies and equivalent frequencies are
l 10 E/J
22
E = 1.509 1033 E. h
0 0 0
1 2.62 396
2 7.86 1188
3 15.72 2376
/GHz
P12.6
Treat the gravitational potential energy as a perturbation in the energy operator: H (1) = mgx. The firstorder correction to the groundstate energy, E1 , is:
L L (0) (1) 1 H 0 L (0) 1 dx
E1
(1)
=
=
0
2 1/2 x mgx sin L L
2 1/2 x dx, sin L L
E1 (1) =
2mg L
x sin2
0
x dx, L
L
E1
(1)
2mg = L
x x2 xL x x L2 cos2  cos sin  4 2 L L L 4 2
,
0
1 E1 (1) = 2 mgL
Not surprisingly, this amounts to the energy perturbation evaluated at the midpoint of the box. For m = me , E1 (1) /L = 4.47 1030 J m1 .
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
193
Solutions to theoretical problems
P12.8 
2
2m
2 2 2 + 2+ 2 2 x y z
= E
[V = 0]
We try the solution = X(x)Y (y)Z(z)  
2
2m
2
(X Y Z + XY Z + XY Z ) = EXY Z X Y Z + + Z X Y =E
2m
X depends only on x; therefore, when x changes only this term changes, but the sum of the three X X terms is constant. Therefore, must also be constant. We write X  X = EX , 2m X
2 2
with analogous terms for y, z
Hence we solve  X = E X 2m 2 X Y Z Y  Y = E Y E = E + E + E , 2m 2 Z  Z = E Z 2m
X
= XY Z
The threedimensional equation has therefore separated into three onedimensional equations, and we can write E= h2 8m n2 n2 + 2 + 3 L2 L2 L2 1 2 3 n2 1 n1 , n2 , n3 = 1, 2, 3, . . . n2 y L2 n3 z L3
=
1/2 8 n1 x sin L1 L2 L3 L1
sin
sin
For a cubic box E = (n2 + n2 + n2 ) 1 2 3 P12.10 h2 8mL2
The wavefunctions in each region (see Fig. 12.2(a)) are (eqns 12.2212.25): 1 (x) = eik1 x + B1 eik2 x 2 (x) = A2 ek2 x + B2 ek2 x 3 (x) = A3 eik3 x with the above choice of A1 = 1 the transmission probability is simply T = A3 2 . The wavefunction coefficients are determined by the criteria that both the wavefunctions and their first derivatives w/r/t
194
INSTRUCTOR'S MANUAL
x be continuous at potential boundaries 1 (0) = 2 (0); 2 (L) = 3 (L) d1 (0) d2 (0) d2 (L) d3 (L) = ; = dx dx dx dx These criteria establish the algebraic relationships: 1 + B 1  A2  B2 = 0 (ik1  k2 )A2 + (ik1 + k2 )B2 + 2ik1 = 0 A2 ek2 L + B2 ek2 L  A3 eik3 L = 0 A2 k2 ek2 L  B2 k2 ek2 L  iA3 k3 eik3 L = 0
V
V2
V3
0
V1
x 0 L
Figure 12.2(a)
Solving the simultaneous equations for A3 gives A3 = 4k1 k2 eik3 L (ia + b) ek2 L  (ia  b) ek2 L
2 where a = k2  k1 k3 and b = k1 k2 + k2 k3 . since sinh(z) = (ez  ez )/2 or ez = 2 sinh(z) + ez , substitute ek2 L = 2 sinh(k2 L) + ek2 L giving:
A3 =
2k1 k2 eik3 L (ia + b) sinh(k2 L) + b ek2 L
2 2 4k1 k2
T = A3 2 = A3 A3 =
(a 2 + b2 ) sinh2 (k2 L) + b2
2 2 2 2 2 where a 2 + b2 = (k1 + k2 )(k2 + k3 ) and b2 = k2 (k1 + k3 )2
(b) In the special case for which V1 = V3 = 0, eqns 12.22 and 12.25 require that k1 = k3 . Additionally, k1 2 E = where = E/V2 . = k2 V2  E 1 a +b =
2 2 2 (k1 2 + k 2 )2
=
4 k2
1+
2 k1 2 k2
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
195
2 2 b2 = 4k1 k2
a2
+ b2 b2
=
2 2 k1 2 k2 1 + k2 2 4k1
=
1 4(1  ) = 1 1+
a 2 +b2 b2
T =
b2 b2 + (a 2 + b2 ) sinh2 (k2 L) sinh2 (k2 L) 1+ 4(1  )
1
sinh2 (k2 L)
1
T =
(ek2 L  ek2 L )2 = 1+ 16(1  )
This proves eqn 12.28a where V1 = V3 = 0 In the high wide barrier limit k2 L 1. This implies both that ek2 L is negligibly small k2 L and that 1 is negligibly small compared to e2 k2 L /{16(1  )}. The previous compared to e equation simplifies to T = 16 (1  )e2 k2 L [eqn12.28b]
(c)
0.25
E = 10 kJ/mol, V1 = V3 = 0, L = 50 pm
0.2
0.15 T 0.1
0.05
0 1 1.2 1.4 1.6 1/ (i.e., V2/E) 1.8 2
Figure 12.2(b)
P12.12
d2 1 + 2 kx 2 = E 2m dx 2 2 2 d and we write = egx , so = 2gxegx dx The Schr dinger equation is  o
2
2 2 d2 = 2gegx + 4g 2 x 2 egx = 2g + 4g 2 x 2 2 dx
196
INSTRUCTOR'S MANUAL
2g
m
2g

2 2 g2 m
1 x 2 + 2 kx 2 = E
m
E +
1 2k

2 2 g2 m
x2 = 0
This equation is satisfied if E=
2g
m
and
1 2 2 g 2 = 2 mk,
or
1 g=2
mk 1/2
2
Therefore,
1 E=2
k 1/2 1 =2 m
+ 
if =
k 1/2 m
+ 
P12.14
x n = n y n = n x3
+ 
y n dx = n+1
2 y n dy
[x = y]
2 y 3 dy = 0 by symmetry [y 3 is an odd function of y] y 4 dy
2
x 4 = 5
+
y 4 = y 4 N Hv ey /2 1 1 1 1 y 4 Hv = y 3 2 Hv+1 + vHv1 = y 2 2 2 Hv+2 + (v + 1)Hv + v 2 Hv + (v  1)Hv2
1 1 = y 2 4 Hv+2 + v + 2 Hv + v(v  1)Hv2 1 1 1 1 = y 4 2 Hv+3 + (v + 2)Hv+1 + v + 2 2 Hv+1 + vHv1 1 +v(v  1) 2 Hv1 + (v  2)Hv3 3 3 = y 1 Hv+3 + 4 (v + 1)Hv+1 + 2 v 2 Hv1 + v(v  1) (v  2)Hv3 8

Only yHv+1 and yHv1 lead to Hv and contribute to the expectation value (since Hv is orthogonal to all except Hv ) [Table 12.1]; hence
3 y 4 Hv = 4 y{(v + 1)Hv+1 + 2v 2 Hv1 } + 3 1 1 = 4 (v + 1) 2 Hv+2 + (v + 1)Hv + 2v 2 2 Hv + (v  1)Hv2 3 = 4 {(v + 1)2 Hv + v 2 Hv } + 3 = 4 (2v 2 + 2v + 1)Hv +
+
Therefore
+  3 y 4 dy = 4 (2v 2 + 2v + 1)N 2 +  2 Hv ey dy =
2
3 (2v 2 + 2v + 1) 4
and so x 4 = ( 5 ) 3 4
3 (2v 2 + 2v + 1) = 4 (2v 2 + 2v + 1) 4
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
197
P12.17
e2 1 [13.5 with Z = 1] = x b 40 r Since 2 T = b V [12.45, T EK ] V = 2 T = V
1 Therefore, T =  2 V
with b = 1
[x r]
P12.18
In each case, if the function is an eigenfunction of the operator, the eigenvalue is also the expectation value; if it is not an eigenfunction we form = (a) (b) (c) (d) ^ d [11.39]
d i e = ei ; hence Jz = + i d d 2i ^ e lz e2i = = 2 e2i ; hence Jz = 2 i d ^ lz ei = lz
2 d cos d  cos sin d = 0 i d i 0 0 2 d (cos ei + sin ei ) d lz = N 2 (cos ei + sin ei ) i d 0 2
cos
=
i
N2
2
2 0 2 0
(cos ei + sin ei ) (i cos ei  i sin ei ) d
= N
2 0
(cos2  sin2 + cos sin [e2i  e2i ]) d
= N 2 (cos2  sin2 ) (2 ) = 2 N 2 cos 2 N2 (cos ei + sin ei ) (cos ei + sin ei ) d
2 0
= N2
(cos2 + sin2 + cos sin [e2i + e2i ]) d if N 2 = 1 2
= 2N 2 (cos2 + sin2 ) = 2 N 2 = 1 Therefore lz = cos 2 [ is a parameter]
2 d2 ^ J2 ^ ^ For the kinetic energy we use T EK = z [12.47] =  [12.52] 2I d 2 2I
(a)
^ T ei = 
2
2I
(i2 ei ) =
2
2
2I
ei ;
hence
T = hence
2
2I T = T = 2 2 I
2
(b)
^ T e2i =  ^ T cos = 
2I
2
(2i)2 e2i = (cos ) =
4 2 2i e ; 2I
2
(c)
2I
2I
cos ;
hence
2I
198
INSTRUCTOR'S MANUAL
(d)
^ T (cos ei + sin ei ) =  and hence T =
2
2
2I
(cos ei  sin ei ) =
2
2I
(cos ei + sin ei )
2I
Comment. All of these functions are eigenfunctions of the kinetic energy operator, which is also the total energy or Hamiltonian operator, since the potential energy is zero for this system.
2 0
P12.20
0
Y3,3 Y3,3 sin d d =
0
1 64
35
2 35 d [Table 12.3] sin6 sin d 0
1 = 64
(2 )
1 1
(1  cos2 )3 d cos
[sin d = d cos , sin2 = 1  cos2 ]
35 = 32 1 1
(1  3x 2 + 3x 4  x 6 ) dx
[x = cos ]
1 35 32 35 = 32 x  x 3 + 3 x 5  1 x 7 = 1 = 7 5 1 32 35
P12.22
2 =
2 2 2 + 2+ 2 2 x y z 2 f = b2 f y 2 2 f = c2 f y 2
2 f = a 2 f x 2
and f is an eigenfunction with eigenvalue (a 2 + b2 + c2 ) P12.25 (a) Suppose that a particle moves classically at the constant speed v. It starts at x = 0 at t = 0 and L at t = is at position x = L. v = and x = vt. x = 1 x dt = t=0 v = t dt = t=0 1 vt dt t=0 v 2 t 2 t=0
=
L v 2 v = = x = 2 2 2 1 2 v2 2 x dt = t dt t=0 t=0 v2 3 t 3
x2 = =
=
t=0
(v )2 L2 = 3 3
L x 2 1/2 = 1/2 3
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
199
(b)
n = x n=
n x 2 L dx x sin2 L 0 L x=0 x=L x sin 2n x cos 2n x L L 2 x2 =   L 4 4(n/L) 8(n/L)2
n xn dx = x=0
n x 2 1/2 sin L L
L
for 0 x L [12.7]
L 2 L2 = = = x n L 4 2 This agrees with the classical result. x2 n =
L x=0 n x 2 n dx =
2 x3 =  L 6 = =
1/2
2 L 2 2 n x x sin L x=0 L
dx x cos 2n x L 8(n/L)2 x=L
x=0
x2 1  4(n/L) 8(n/L)3 L 8(n/L)2
2n x sin L

2 L
L3 6

L2 1  3 4(n/L)2 =
1/2
x2 n
1 L2  3 4(n/L)2
1/2
L = 1/2 3 This agrees with the classical result. lim x 2 n n P12.27 (a) The energy levels are given by: h2 n2 , 8mL2 and we are looking for the energy difference between n = 6 and n = 7: En = h2 (72  62 ) . 8mL2 Since there are 12 atoms on the conjugated backbone, the length of the box is 11 times the bond length: E= L = 11(140 1012 m) = 1.54 109 m, (6.626 1034 J s)2 (49  36) = 3.30 1019 J . 8(9.11 1031 kg)(1.54 109 m)2 (b) The relationship between energy and frequency is: so E= E = h so = 3.30 1019 J E = 4.95 1014 s1 . = h 6.626 1034 J s
200
INSTRUCTOR'S MANUAL
(c) The frequency computed in this problem is about twice that computed in problem 12.26b, suggesting that the absorption spectrum of a linear polyene shifts to lower frequency as the number of conjugated atoms increases . The reason for this is apparent if we look at the terms in the energy expression (which is proportional to the frequency) that change with the number of conjugated atoms, N . The energy and frequency are inversely proportional to L2 and directly proportional to (n + 1)2  n2 = 2n + 1, where n is the quantum number of the highest occupied state. Since n is proportional to N (equal to N/2) and L is approximately proportional to N (strictly to N  1), the energy and frequency are approximately proportional to N 1 . P12.29 In effect, we are looking for the vibrational frequency of an O atom bound, with a force constant equal to that of free CO, to an infinitely massive and immobile protein complex. The angular frequency is = k 1/2 , m
where m is the mass of the O atom. m = (16.0 u)(1.66 1027 kg u1 ) = 2.66 1026 kg, and k is the same force constant as in problem 12.2, namely 1902 N m1 : = 1902 N m1 2.66 1026 kg
1/2
= 2.68 1014 s1 .
13
Atomic structure and atomic spectra
Solutions to exercises
Discussion questions
E13.1(b) (1) The principal quantum number, n, determines the energy of a hydrogenic atomic orbital through eqn 13.13. (2) The azimuthal quantum number, l, determines the magnitude of the angular momentum of a hydrogenic atomic orbital through the relation {l(l + 1)}1/2 h. (3) The magnetic quantum number, ml , determines the zcomponent of the angular momentum of a hydrogenic orbital through the relation ml h. (4) The spin quantum number, s, determines the magnitude of the spin angular momentum through the relation {s(s + 1)}1/2 h. For a hydrogenic atomic orbitals, s can only be 1/2. (5) The spin quantum number, ms , determines the zcomponent of the spin angular momentum through the relation ms h. For hydrogenic atomic orbitals, ms can only be 1/2. (a) A boundary surface for a hydrogenic orbital is drawn so as to contain most (say 90%) of the probability density of an electron in that orbital. Its shape varies from orbital to orbital because the electron density distribution is different for different orbitals. (b) The radial distribution function gives the probability that the electron will be found anywhere within a shell of radius r around the nucleus. It gives a better picture of where the electron is likely to be found with respect to the nucleus than the probability density which is the square of the wavefunction. E13.3(b) The first ionization energies increase markedly from Li to Be, decrease slightly from Be to B, again increase markedly from B to N, again decrease slightly from N to O, and finally increase markedly from N to Ne. The general trend is an overall increase of I1 with atomic number across the period. That is to be expected since the principal quantum number (electron shell) of the outer electron remains the same, while its attraction to the nucleus increases. The slight decrease from Be to B is a reflection of the outer electron being in a higher energy subshell (larger l value) in B than in Be. The slight decrease from N to O is due to the halffilled subshell effect; halffilled subshells have increased stability. O has one electron outside of the halffilled p subshell and that electron must pair with another resulting in strong electronelectron repulsions between them. An electron has a magnetic moment and magnetic field due to its orbital angular momentum. It also has a magnetic moment and magnetic field due to its spin angular momentum. There is an interaction energy between magnetic moments and magnetic fields. That between the spin magnetic moment and the magnetic field generated by the orbital motion is called spinorbit coupling. The energy of interaction is proportional to the scalar product of the two vectors representing the spin and orbital angular momenta and hence depends upon the orientation of the two vectors. See Fig. 13.29. The total angular momentum of an electron in an atom is the vector sum of the orbital and spin angular momenta as illustrated in Fig. 13.30 and expressed in eqn 13.46. The spinorbit coupling results in a splitting of the energy levels associated with atomic terms as shown in Figs 13.31 and 13.32. This splitting shows up in atomic spectra as a fine structure as illustrated in Fig. 13.32.
E13.2(b)
E13.4(b)
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INSTRUCTOR'S MANUAL
Numerical exercises
E13.5(b) The energy of the photon that struck the Xe atom goes into liberating the bound electron and giving it any kinetic energy it now possesses Ephoton = I + Ekinetic I = ionization energy
The energy of a photon is related to its frequency and wavelength Ephoton = h = hc
and the kinetic energy of an electron is related to its mass and speed
1 Ekinetic = 2 me s 2
So
hc hc 1 1  2 me s 2 = I + 2 me s 2 I = I = (6.626 1034 J s) (2.998 108 m s1 ) 58.4 109 m
1  2 (9.11 1031 kg) (1.79 106 m s1 )2
= 1.94 1018 J = 12.1 eV E13.6(b) The radial wavefunction is [Table 13.1] 2Zr R3,0 = A 6  2 + 1 2 e/6 where , and A is a collection of constants. Differentiating 9 a0 with respect to yields dR3,0 = 0 = A(6  2 + 1 2 )  1 e/6 + 2 + 2 Ae/6 9 9 6 d = Ae/6  + 5  3 9 54
2
This is a quadratic equation 0 = a 2 + b + c The solution is = so r = b (b2  4ac)1/2 = 15 3 7 2a 15 3(71/2 ) 2 2 a0 . Z where a =  5 1 , b = , and c = 3. 54 9
Numerically, this works out to = 7.65 and 2.35, so r = 11.5a0 /Z and 3.53a0 /Z . Substituting Z = 1 and a0 = 5.292 1011 m, r = 607 pm and 187 pm. The other maximum in the wavefunction is at r = 0 . It is a physical maximum, but not a calculus maximum: the first derivative of the wavefunction does not vanish there, so it cannot be found by differentiation.
ATOMIC STRUCTURE AND ATOMIC SPECTRA
203
E13.7(b)
The radial wavefunction is [Table 13.1] R3,1 = A 4  1 e/6 3 where = 2Zr a0
The radial nodes occur where the radial wavefunction vanishes. This occurs at = 0, and when 4  1 = 0, 3 then r = or = 4, 3 or = 12 r=0
E13.8(b)
a0 12a0 a0 = = = 6a0 = 3.18 1010 m 2Z 2 2 Normalization requires 2 d = 1 = 1 = N2
0 0 0 0 2
[N (2  r/a0 )er/2a0 ]2 d sin d r 2 dr
0
er/a0 (2  r/a0 )2 r 2 dr
sin d
2
d
0
Integrating over angles yields 1 = 4N 2 = 4N 2
0 0
er/a0 (2  r/a0 )2 r 2 dr
2 3 er/a0 (4  4r/a0 + r 2 /a0 )r 2 dr = 4 N 2 (8a0 )
In the last step, we used
0
er/k r 2 dr = 2k 3 ,
0
er/k r 3 dr = 6k 4 , and
0
er/k r 4 dr = 24k 5
So N = E13.9(b)
1
3 4 2a0
The average kinetic energy is ^ EK = ^ EK d 1 4 Z3
3 2 a0 1/2
where = N (2  )e/2 with N = h 2 ^ EK =  2m
2
and
Zr a0
here
a 3 2 sin d d d d = r 2 sin dr d d = 0 Z3
In spherical polar coordinates, three of the derivatives in 2 are derivatives with respect to angles, so those parts of 2 vanish. Thus 2 = 2 2 2 + = r r r 2 2 2 2Z + a0 r 2 = r Z 2 a0 2 2 + 2
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INSTRUCTOR'S MANUAL
1 1 = N (2  )  2 e/2  N e/2 = N 2  2 e/2 r 2 1 1 1 3 1 = N 2  2  2 e/2 + 2 N e/2 = N 2  4 e/2 2 2 = and ^ EK =
0 0 0 2
Z 2 N e/2 (4/ + 5/2  /4) a0 Z 2 a0  2 h 2m
N (2  )e/2
a 3 d sin d 2 d N e/2 (4/ + 5/2  /4) 0 Z3 The integrals over angles give a factor of 4 , so a0 h2 ^  EK = 4N 2 Z 2m
0 5 1 (2  ) 4 + 2  4 2 e d 0
The integral in this last expression works out to 2, using ^ EK = 4 Z3
3 32a0
e n d = n! for n = 1, 2, and 3. So
a0 Z
h2 m
=
h2 Z 2
2 8ma0
The average potential energy is V = and V = V d
0 0 0
where V = 
2
Z 2 e2 Ze2 = 4 0 r 4 0 a0 Z 2 e2 4 0 a0 a 3 2 sin d d d N (2  )e/2 0 Z3
N (2  )e/2 
The integrals over angles give a factor of 4 , so V = 4N 2  Z 2 e2 40 a0
3 a0 0
Z3
(2  )2 e d
The integral in this last expression works out to 2, using
0
e n d = n! for n = 1, 2, 3, and 4. So Z 2 e2 16 0 a0
V = 4
Z3
3 32a0

Z 2 e2 40 a0
3 a0
Z3
(2) = 
E13.10(b) The radial distribution function is defined as P = 4r 2 2 P3s = 4r 2 where so 1 4 P3s = 4r 2 (Y0,0 R3,0 )2 , 1 243 here. Z 3 (6  6 + 2 )2 e a0
2Zr 2Zr = na0 3a0
ATOMIC STRUCTURE AND ATOMIC SPECTRA
205
But we want to find the most likely radius, so it would help to simplify the function by expressing it in terms either of r or , but not both. To find the most likely radius, we could set the derivative of P3s equal to zero; therefore, we can collect all multiplicative constants together (including the factors of a0 /Z needed to turn the initial r 2 into 2 ) since they will eventually be divided into zero P3s = C 2 2 (6  6 + 2 )2 e Note that not all the extrema of P are maxima; some are minima. But all the extrema of (P3s )1/2 correspond to maxima of P3s . So let us find the extrema of (P3s )1/2 d(P3s )1/2 d =0= C(6  6 + 2 )e/2 d d
1 = C[(6  6 + 2 ) ( 2 ) + (6  12 + 3 2 )]e/2 1 0 = C(6  15 + 6 2  2 3 )e/2
so
12  30 + 12 2  3 = 0
Numerical solution of this cubic equation yields = 0.49, 2.79, and 8.72 corresponding to r = 0.74a0 /Z, 4.19a0 /Z, and 13.08a0 /Z Comment. If numerical methods are to be used to locate the roots of the equation which locates the extrema, then graphical/numerical methods might as well be used to locate the maxima directly. That is, the student may simply have a spreadsheet compute P3s and examine or manipulate the spreadsheet to locate the maxima. E13.11(b) Orbital angular momentum is ^ L2 1/2 = h(l(l + 1))1/2 There are l angular nodes and n  l  1 radial nodes ^ (a) n = 4, l = 2, so L2 1/2 = 61/2 h = 2.45 1034 J s ^ (b) n = 2, l = 1, so L2 1/2 = 21/2 h = 1.49 1034 J s ^ (c) n = 3, l = 1, so L2 1/2 = 21/2 h = 1.49 1034 J s E13.12(b) For l > 0, j = l 1/2, so (a) (b) l = 1, l = 5, so so j = 1/2 or 3/2 j = 9/2 or 11/2 2 angular nodes 1 angular node 1 angular node 1 radial node 0 radial nodes 1 radial node
E13.13(b) Use the ClebschGordan series in the form J = j1 + j2 , j1 + j2  1, . . . , j1  j2  Then, with j1 = 5 and j2 = 3 J = 8, 7, 6, 5, 4, 3, 2
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INSTRUCTOR'S MANUAL
E13.14(b) The degeneracy g of a hydrogenic atom with principal quantum number n is g = n2 . The energy E of hydrogenic atoms is E= hcZ 2 RH hcZ 2 RH = g n2
so the degeneracy is g= (a) (b) (c) hcZ 2 RH E hc(2)2 RH = 1 4hcRH hc(4)2 RH
1  4 hcRH
g= g= g=
= 64
hc(5)2 RH = 25 hcRH
E13.15(b) The letter F indicates that the total orbital angular momentum quantum number L is 3; the superscript 3 is the multiplicity of the term, 2S + 1, related to the spin quantum number S = 1; and the subscript 4 indicates the total angular momentum quantum number J . E13.16(b) The radial distribution function varies as 4 P = 4r 2 2 = 3 r 2 e2r/a0 a0 The maximum value of P occurs at r = a0 since dP 2r 2 2r  a0 dr e2r/a0 = 0 at r = a0 and Pmax = 4 2 e a0
P falls to a fraction f of its maximum given by f =
4r 2 2r/a0 3 e a0 4 2 a0 e
r2 = 2 e2 e2r/a0 a0
and hence we must solve for r in f 1/2 r = er/a0 e a0 f = 0.50 r 0.260 = er/a0 solves to r = 2.08a0 = 110 pm and to r = 0.380a0 = 20.1 pm a0 (b) f = 0.75 r 0.319 = er/a0 solves to r = 1.63a0 = 86 pm and to r = 0.555a0 = 29.4 pm a0 (a) In each case the equation is solved numerically (or graphically) with readily available personal computer software. The solutions above are easily checked by substitution into the equation for f . The radial distribution function is readily plotted and is shown in Fig. 13.1.
ATOMIC STRUCTURE AND ATOMIC SPECTRA
207
0.15
0.10
0.05
0.00 0.0 0.5 1.0 1.5 2.0 2.5
Figure 13.1 E13.17(b) (a) 5d 2s is not an allowed transition, for (b) 5p 3s is allowed , since (c) 5p 3f is not allowed, for l = 1. l = +2 ( l must equal 1). l = 2 ( l must equal 1).
E13.18(b) For each l, there are 2l + 1 values of ml and hence 2l + 1 orbitalseach of which can be occupied by two electrons, so maximum occupancy is 2(2l + 1) (a) 2s: l = 0; maximum occupancy = 2 (b) 4d: l = 2; maximum occupancy = 10 (c) 6f : l = 3; maximum occupancy = 14 (d) 6h: l = 5; maximum occupancy = 22 E13.19(b) V2+ : 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 = [Ar]3d 3
3 3 1 The only unpaired electrons are those in the 3d subshell. There are three. S = 2 and 2  1 = 2 . 1 3 3 For S = 2 , MS = 2 and 2 1 1 for S = 2 , MS = 2
E13.20(b) (a) Possible values of S for four electrons in different orbitals are 2, 1, and 0 ; the multiplicity is 2S + 1, so multiplicities are 5, 3, and 1 respectively. (b) Possible values of S for five electrons in different orbitals are 5/2, 3/2, and 1/2 ; the multiplicity is 2S + 1, so multiplicities are 6, 4, and 2 respectively. E13.21(b) The coupling of a p electron (l = 1) and a d electron (l = 2) gives rise to L = 3 (F), 2 (D), and 1 (P) terms. Possible values of S include 0 and 1. Possible values of J (using RussellSaunders coupling) are 3, 2, and 1 (S = 0) and 4, 3, 2, 1, and 0 (S = 1). The term symbols are
1
F3 ; 3 F4 , 3 F3 , 3 F2 ; 1 D2 ; 3 D3 , 3 D2 , 3 D1 ; 1 P1 ; 3 P2 , 3 P1 , 3 P0 .
Hund's rules state that the lowest energy level has maximum multiplicity. Consideration of spinorbit coupling says the lowest energy level has the lowest value of J (J + 1)  L(L + 1)  S(S + 1). So the lowest energy level is 3 F2
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INSTRUCTOR'S MANUAL
E13.22(b) (a) 3 D has S = 1 and L = 2, so J = 3, 2, and 1 are present. J = 3 has 7 states, with MJ = 0, 1, 2, or 3; J = 2 has 5 states, with MJ = 0, 1, or 2; J = 1 has 3 states, with MJ = 0, or 1. (b) 4 D has S = 3/2 and L = 2, so J = 7/2, 5/2, 3/2, and 1/2 are present. J = 7/2 has 8 possible states, with MJ = 7/2, 5/2, 3/2 or 1/2; J = 5/2 has 6 possible states, with MJ = 5/2 3/2 or 1/2; J = 3/2 has 4 possible states, with MJ = 3/2 or 1/2; J = 1/2 has 2 possible states, with MJ = 1/2. (c) 2 G has S = 1/2 and L = 4, so J = 9/2 and 7/2 are present. J = 9/2 had 10 possible states, with MJ = 9/2, 7/2, 5/2, 3/2 or 1/2; J = 7/2 has 8 possible states, with MJ = 7/2, 5/2, 3/2 or 1/2. E13.23(b) Closed shells and subshells do not contribute to either L or S and thus are ignored in what follows.
1 5 3 (a) Sc[Ar]3d 1 4s 2 : S = 2 , L = 2; J = 2 , 2 , so the terms are 2 D5/2 and 2 D3/2
(b) Br[Ar]3d 10 4s 2 4p 5 . We treat the missing electron in the 4p subshell as equivalent to a single 1 1 3 1 "electron" with l = 1, s = 2 . Hence L = 1, S = 2 , and J = 2 , 2 , so the terms are 2 P3/2 and 2 P1/2
Solutions to problems
Solutions to numerical problems
P13.2 All lines in the hydrogen spectrum fit the Rydberg formula 1 = RH 1 n2 1 1  2 n2 13.1, with = ~ 1 RH = 109 677 cm1
Find n1 from the value of max , which arises from the transition n1 + 1 n1 1 1 1 2n1 + 1 = 2 = 2 max RH (n1 + 1)2 n1 n1 (n1 + 1)2 n2 (n1 + 1)2 = (656.46 109 m) (109 677 102 m1 ) = 7.20 max RH = 1 2n1 + 1 and hence n1 = 2, as determined by trial and error substitution. Therefore, the transitions are given by = ~ 1 = (109 677 cm1 ) 1 1  2 4 n2 1 1  4 49 , n2 = 3, 4, 5, 6
The next line has n2 = 7, and occurs at = ~ 1 = (109 677 cm1 ) = 397.13 nm
The energy required to ionize the atom is obtained by letting n2 . Then = ~ 1 = (109 677 cm1 ) 1  0 = 27 419 cm1 , 4 or 3.40 eV
ATOMIC STRUCTURE AND ATOMIC SPECTRA
209
(The answer, 3.40 eV, is the ionization energy of an H atom that is already in an excited state, with n = 2.) Comment. The series with n1 = 2 is the Balmer series. P13.4 The lowest possible value of n in 1s 2 nd 1 is 3; thus the series of 2 D terms correspond to 1s 2 3d, 1s 2 4d, etc. Figure 13.2 is a description consistent with the data in the problem statement.
413 nm
I
670 nm
460 nm
610 nm
Figure 13.2 If we assume that the energies of the d orbitals are hydrogenic we may write E(1s 2 nd 1 , 2 D) =  hcR n2 [n = 3, 4, 5, . . .]
Then for the 2 D 2 P transitions = ~ 1 E(1s 2 2p 1 , 2 P) R  2 = hc n E = h = hc E = hc , = ~ ~ hc
R 1 610.36 107 cm + 9 E(1s 2 2p 1 , 2 P) 1 R 1 R = + 2 = + 460.29 107 cm hc n 16 1 R + 25 413.23 107 cm (b)  (a) solves to R = 109 886 cm1 Then (a)  (c) solves to R = 109 910 cm1 Mean = 109 920 cm1 (b)  (c) solves to R = 109 963 cm1 The binding energies are therefore E(1s 2 3d 1 , 2 D) = R = 12 213 cm1 9 1  12 213 cm1 = 28 597 cm1 E(1s 2 2p, 2 P) =  610.36 107 cm 1  28 597 cm1 = 43 505 cm1 E(1s 2 2s 1 , 2 S) =  670.78 107 cm
from which we can write
(a) (b) (c)
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INSTRUCTOR'S MANUAL
Therefore, the ionization energy is I (1s 2 2s 1 ,2 S) = 43 505 cm1 , P13.5 or 5.39 eV
The 7p configuration has just one electron outside a closed subshell. That electron has l = 1, s = 1/2, and j = 1/2 or 3/2, so the atom has L = 1, S = 1/2, and J = 1/2 or 3/2. The term symbols are P1/2 and 2 P3/2 , of which the former has the lower energy. The 6d configuration also has just one electron outside a closed subshell; that electron has l = 2, s = 1/2, and j = 3/2 or 5/2, so the atom
2
has L = 2, S = 1/2, and J = 3/2 or 5/2. The term symbols are 2 D3/2 and 2 D5/2 , of which the former has the lower energy. According to the simple treatment of spinorbit coupling, the energy is given by
1 El,s,j = 2 hcA[j (j + 1)  l(l + 1)  s(s + 1)]
where A is the spinorbit coupling constant. So
1 1 1 E(2 P1/2 ) = 2 hcA[ 2 (1/2 + 1)  1(1 + 1)  2 (1/2 + 1)] = hcA 1 3 1 3 and E(2 D3/2 ) = 2 hcA[ 2 (3/2 + 1)  2(2 + 1)  2 (1/2 + 1)] =  2 hcA
This approach would predict the ground state to be 2 D3/2 Comment. The computational study cited above finds the 2 P1/2 level to be lowest, but the authors caution that the error of similar calculations on Y and Lu is comparable to the computed difference between levels. P13.7 RH = kH , RD = kD , R = k [18] where R corresponds to an infinitely heavy nucleus, with = me . me mN [N = p or d] Since = me + m N RH = kH = kme R me = me 1 + mp 1 + mp
Likewise, RD =
R me where mp is the mass of the proton and md the mass of the deuteron. The 1 + md two lines in question lie at 1 1 3 = RH 1  4 = 4 RH H 1 1 3 = RD 1  4 = 4 RD D
and hence D H ~ RH = = RD H D ~ Then, since 1+ RH = RD 1+
me md me mp
,
md =
me
me 1 + mp RH RD
1
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211
and we can calculate md from md = me 1+
me mp D H
1
=
me 1+
me mp H ~ D ~
1 = 3.3429 1027 kg
=
9.10939 1031 kg
1 9.109391031 1 + 1.672621027 kg 82 259.098 cm1  1 kg 82 281.476 cm
Since I = Rhc, ID RD D ~ 82 281.476 cm1 = = = = 1.000 272 IH RH H ~ 82 259.098 cm1 P13.10 If we assume that the innermost electron is a hydrogenlike 1s orbital we may write r = 52.92 pm a0 [Example 13.3] = = 0.420 pm Z 126
Solutions to theoretical problems
P13.12 Consider 2pz = 2,1,0 which extends along the zaxis. The most probable point along the zaxis is where the radial function has its maximum value (for 2 is also a maximum at that point). From Table 13.1 we know that R21 e/4 dR 1 = 1  4 e/4 = 0 when = 4. d 2a0 2a0 Therefore, r = , and the point of maximum probability lies at z = = 106 pm Z Z Comment. Since the radial portion of a 2p function is the same, the same result would have been obtained for all of them. The direction of the most probable point would, however, be different. and so P13.13 In each case we need to show that
1 2 d = 0 0 0 2
all space
(a)
0
1s 2s r 2 dr sin d d = 0
?
1 1/2 1s = R1,0 Y0,0 Y0,0 = [Table 12.3] 2s = R2,0 Y0,0 4 Since Y0,0 is a constant, the integral over the radial functions determines the orthogonality of the functions.
0
R1,0 R2,0 r 2 dr = 2Zr a0 Zr a0 eZr/2a0 = 2Zr a0
R1,0 e/2 = eZr/a0
R2,0 (2  /2)e/4 = 2 
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INSTRUCTOR'S MANUAL
0
R1,0 R2,0 r 2 dr = =
0 0
eZr/a0 2 
Zr a0
eZr/2a0 r 2 dr
0 4
2e(3/2)Zr/a0 r 2 dr   Z a0 3!
3 Z 2 a0
Z (3/2)Zr/a0 3 e r dr a0
2 2!
3 Z 2 a0 3
= 0
Hence, the functions are orthogonal. (b) We use the px and py orbitals in the form given in Section 13.2(f ), eqn 25 px x, Thus px py dx dy dz
+  +  + 
py y
all space
xy dx dy dz
This is an integral of an odd function of x and y over the entire range of variable from  to +, therefore, the integral is zero . More explicitly we may perform the integration using the orbitals in the form (Section 13.2(f ), eqn 13.25) px = f (r) sin cos
all space
py = f (r) sin sin
0
px py r 2 dr sin d d =
f (r)2 r 2 dr
0
sin2 d
2 0
cos sin d
. 2 The third factor is zero. Therefore, the product of the integrals is zero and the functions are orthogonal. The first factor is nonzero since the radial functions are normalized. The second factor is P13.14  h2 2 2 d d2 + r dr dr 2 + Veff R = ER [13.11] (1)
where Veff = 
l(l + 1) 2 h l(l + 1) 2 h Ze2 Z 2 h + + = 2 2 40 r a0 r 2r 2r Using = Zr/a0 , the derivative term of the Hamiltonian can be written in the form d2 2 d = + 2 r dr dr d2 2 d Z 2 + 2 a0 d d Dop (2)
To determine E2s and E2p , we will evaluate the left side of (1) and compare the result to the right side. 2s orbital. R2s = N2s (2  )e/2 where Zr/a0 here 4 2 e/2 = 4 R2s 2(2  )
dR2s 1 = N2s 1  2 (2  ) e/2 = N2s d
d2 R2s /2 6 1 1 3 e R2s = N2s 2  4 (  4) e/2 = N2s 2  = 4 4(2  ) d 2
ATOMIC STRUCTURE AND ATOMIC SPECTRA
213

h2 h2 Dop R2s =  2 2 =  h2 2
4 6 + 4(2  ) (2  ) 8 4 h2 Z 2 R2s a0 1 R2s
Z 2 R2s a0
Veff R2s =  
Z 2 h Z 2 R2s =  a0 r a0
h2 Dop + Veff R2s = 2 =
Z 2 h2  a0 2 Z 2 h2
2 2a0
8 2 + R2s 4
1 R2s 4
1 Therefore E2s =  4
Z 2 h2
2 2a0
(3)
2p orbital.
R2p = N2p e/2
where Zr/a0 here
dR2p /2 2 = N2p 1  e R2p = d 2 2 d2 R2p 1 1 = N2p  2  2 1  2 2 d  h2 h2 Dop R2p =  2 2 = Veff R2p =  =  h2 2 e/2 = 4 R2p 4 Z 2 R2p a0
 4 4  2 + 4 2 2 2  8 + 8 4 2
Z 2 R2p a0 h2 1 + Z 2 a0 h2 1 R2p 2
Z 2 h Z 2 h2 + 2 R2p =  a0 r a0 r 2(  1) R2p 2
Z 2 h2  a0 2
h2 Dop + Veff R2p = 2 =
Z 2 h2  a0 2 Z 2 h2
2 2a0
2  8 + 8 2(  1) R2p + 4 2 2 Z 2 h2
2 2a0
2 4 2
R2p = 
1 R2p 4
1 Therefore E2p =  4
Z 2 h2
2 2a0
(4)
Comparison of eqns (3) and (4) reveals that E2s = E2p .
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INSTRUCTOR'S MANUAL
P13.15
(a) We must show that coordinate (Fig. 11).
3px 2 d = 1. The integrations are most easily performed in spherical
2
3px
2
d =
0 0 0 2
3px
2 2
r sin( ) dr d d
2
=
0 0 0
R31 ()
Y11  Y11 2
r 2 sin( ) dr d d (Table 13.1, eqn 13.25)
where = 2r/a0 , r = a0 /2, dr = (a0 /2) d. 1 = 2
2
a0 3 2
1 27(6)1/2
0 0 0
1 3/2 1 4  e/6 a0 3
2 3 1/2 2 sin( ) cos() 2 sin( ) d d d 8 2 2 1 4  e/6 sin( ) cos() 2 sin( ) d d d 3
1 = 46 656 1 = 46 656
0 0 0 2
cos () d
0 0
2
sin ( ) d
0 4/3
3
1 2 4  4 e/3 d 3
34992
=1
Thus, 3px is normalized to 1. 3px 3dxy d = 0
We must also show that
Using Tables 12.3 and 13.1, we find that 3px = 1 54(2)1/2 1 3/2 1 4  e/6 sin( ) cos() a0 3
3dxy = R32 =
Y22  Y22 2i 1 3/2 2 /6 2 e sin ( ) sin(2) a0
1 32(2)1/2
where = 2r/a0 , r = a0 /2, dr = (a0 /2)d.
2
3px 3dxy d = constant
0
e
5 /3
d
0
cos() sin(2)d
0 0
sin4 ( )d
Since the integral equals zero, 3px and 3dxy are orthogonal.
ATOMIC STRUCTURE AND ATOMIC SPECTRA
215
(b) Radial nodes are determined by finding the values ( = 2r/a0 ) for which the radial wavefunction equals zero. These values are the roots of the polynomial portion of the wavefunction. For the 3s orbital 6  6 + 2 = 0 when node = 3 + 3 and node = 3  3 . The 3s orbital has these two spherically symmetrical modes. There is no node at = 0 so we conclude that there is a finite probability of finding a 3s electron at the nucleus. For the 3px orbital (4  )() = 0 when node = 0 and node = 4 . There is a zero probability of finding a 3px electron at the nucleus. For the 3dxy orbital node = 0 is the only radial node. (c) r 3s =
R10 Y00 2 r d =
2 2 R10 r 3 dr 0 0 0
R10 Y00 2 r 3 sin( ) dr d d
=
Y00 2 sin( ) d d
1
a0 = 3 888
6  2 + 2 /9
0 52488
2
3 e/3 d
r 3s = (d)
0.12
27a0 2
Radial distribution functions of atomic hydrogen
3dxy 0.1
3px
3s 0.08
r 2R2a0
0.06
0.04
0.02
0 0 5 10 15 r /a0 20 25 30
Figure 13.3(a)
The plot shows that the 3s orbital has larger values of the radial distribution function for r < a0 . This penetration of inner core electrons of multielectron atoms means that a 3s electron
216
INSTRUCTOR'S MANUAL
experiences a larger effective nuclear charge and, consequently, has a lower energy than either a 3px or 3dxy electron. This reasoning also lead us to conclude that a 3px electron has less energy than a 3dxy electron. E3s < E3px < E3dxy . (e) Polar plots with = 90
The s Orbital 90 120 150 180 00.20.40.60.8 210 240 270 300 330 60 30 0
The p Orbital 90 120 60
150
30
180 0 0.2 0.4 0.6 0.8
0
210
330
240 270
300
The d Orbital 90 120 60
150
30
180 0 0.2 0.4
0
210
330
240 270
300
Figure 13.3(b)
ATOMIC STRUCTURE AND ATOMIC SPECTRA
217
Boundary surface plots
s  Orbital boundary surface p  Orbital boundary surface
d  Orbital boundary surface
f  Orbital boundary surface
Figure 13.3(c) P13.20 The attractive Coulomb force = Ze2 1 4 0 r 2 (angular momentum)2 (n )2 h The repulsive centrifugal force = = [postulated] me r 3 me r 3 The two forces balance when Ze2 1 n2 h2 2 = , 40 r me r 3 The total energy is E = EK + V = = n2 h2 2me implying that r= 4 n2 h2 0 Ze2 me
Ze2 n2 h2 (angular momentum)2 1 Ze2  [postulated] =  2I 4 0 r 4 0 r 2me r 2 Ze2 me 4n2 h2
0 2

Ze2 4 0
Ze2 me 4 n2 h2
0
= 
Z 2 e 4 me
2 32 2 0 h2
1 2 n
218
INSTRUCTOR'S MANUAL
P13.21
(a) The trajectory is defined, which is not allowed according to quantum mechanics. h (b) The angular momentum of a threedimensional system is given by {l(l + 1)}1/2 h, not by n . In the Bohr model, the ground state possesses orbital angular momentum (n , with n = 1), but h the actual ground state has no angular momentum (l = 0). Moreover, the distribution of the electron is quite different in the two cases. The two models can be distinguished experimentally by (a) showing that there is zero orbital angular momentum in the ground state (by examining its magnetic properties) and (b) examining the electron distribution (such as by showing that the electron and the nucleus do come into contact, Chapter 18). Justification 13.5 noted that the transition dipole moment, fi had to be nonzero for a transition to be allowed. The Justification examined conditions that allowed the z component of this quantity to be nonzero; now examine the x and y components. x,fi = e f x i d and y,fi = e f y i d
P13.25
As in the Justification, express the relevant Cartesian variables in terms of the spherical harmonics, Yl,m . Start by expressing them in spherical polar coordinates: x = r sin cos and y = r sin sin . Note that Y1,1 and Y1,1 have factors of sin . They also contain complex exponentials that can be related to the sine and cosine of through the identities (eqns FI1.20 and FI1.21) cos = 1/2(ei + ei ) and sin =1/2i(ei ei ).
These relations motivate us to try linear combinations Y1,1 +Y1,1 and Y1,1 +Y1,1 (form Table 12.3; note c here corresponds to the normalization constant in the table): Y1,1 + Y1,1 = c sin (ei + ei ) = 2c sin cos = 2cx/r, so x = (Y1,1 + Y1,1 )r/2c; Y1,1  Y1,1 = c sin (ei  ei ) = 2ic sin sin = 2icy/r, so y = (Y1,1  Y1,1 )r/2ic. Now we can express the integrals in terms of radial wavefunctions Rn,l and spherical harmonics Yl,ml e x,fi = 2c
2
Rnf ,lf rRni ,li r dr
0 0 0
2
Y lf ,mlf (Y1,1 + Y1,1 )Yli ,mli sin d d.
The angular integral can be broken into two, one of which contains Y1,1 and the other Y1,1 . According to the "triple integral" relation below Table 12.3, the integral
2
Y lf ,mlf Y1,1 Yli ,mli sin d d
0 0
vanishes unless lf = li 1 and mf = mi 1. The integral that contains Y1,1 introduces no further constraints; it vanishes unless lf = li 1 and mlf = mli 1. Similarly, the y component introduces no further constraints, for it involves the same spherical harmonics as the x component. The whole set of selection rules, then, is that transitions are allowed only if l = 1 and ml = 0 or 1 .
ATOMIC STRUCTURE AND ATOMIC SPECTRA
219
P13.26
(a) The speed distribution in the molecular beam is related to the speed distribution within the chamber by a factor of v cos as shown in Fig. 13.4. Since an integration over all possible must be performed, the cos factor may be absorbed into the constant of proportionality. fbeam (v) = Cvfchamber (v) where C is to be determined
Molecular beam Chamber
Figure 13.4 By normalization over the possible beam speeds (0 < vbeam < )
2 fbeam = Cv v 2 e(mv /2kT ) 2 = Cv 3 e(mv /2kT )
v=0
fbeam dv = 1 = C
2
v=0
2 v 3 e(mv /2kT ) dv = C
1 2(m/2kT )2
C = 2(m/2kT ) v2 =
v=0
v 2 fbeam (v) dv = C
2 v 5 e(mv /2kT ) dv
1 (m/2kT )2 =2 (m/2kT )3 (m/2kT )3 4kT = m m 2 m 4kT v = = 2kT EK = 2 2 m =C (b) or x= 2B L2 4EK dB dz
dB 4EK x 4(2kT ) x = = dz 2B L2 2B L2 = = 4kT x B L2 4(1.3807 1023 J K1 ) (1000 K) (1.00 103 m) (9.27402 1024 J T1 ) (50 102 m)2
dB = 23.8 T m1 dz
14
Molecular structure
Solutions to exercises
Discussion questions
E14.1(b) Consider the case of the carbon atom. Mentally we break the process of hybridization into two major steps. The first is promotion, in which we imagine that one of the electrons in the 2s orbital of carbon (2s 2 2p 2 ) is promoted to the empty 2p orbital giving the configuration 2s2p 3 . In the second step we mathematically mix the four orbitals by way of the specific linear combinations in eqn 14.3 corresponding to the sp3 hybrid orbitals. There is a principle of conservation of orbitals that enters here. If we mix four unhybridized atomic orbitals we must end up four hybrid orbitals. In the construction of the sp2 hybrids we start with the 2s orbital and two of the 2p orbitals, and after mixing we end up with three sp 2 hybrid orbitals. In the sp case we start with the 2s orbital and one of the 2p orbitals. The justification for all of this is in a sense the first law of thermodynamics. Energy is a property and therefore its value is determined only by the final state of the system, not by the path taken to achieve that state, and the path can even be imaginary. It can be proven that if an arbitrary wavefunction is used to calculate the energy of a system, the value calculated is never less than the true energy. This is the variation principle. This principle allows us an enormous amount of latitude in constructing wavefunctions. We can continue modifying the wavefunctions in any arbitrary manner until we find a set that we feel provide an energy close to the true minimum in energy. Thus we can construct wavefunctions containing many parameters and then minimize the energy with respect to those parameters. These parameters may or may not have some chemical or physical significance. Of course, we might strive to construct trial wavefunctions that provide some chemical and physical insight and interpretation that we can perhaps visualize, but that is not essential. Examples of the mathematical steps involved are illustrated in Sections 14.6(c) and (d), Justification 14.3, and Section 14.7. These are all terms originally associated with the Huckel approximation used in the treatment of conjugated electron molecules, in which the electrons are considered independent of the electrons. electron binding energy is the sum the energies of each electron in the molecule. The delocalization energy is the difference in energy between the conjugated molecule with n double bonds and the energy of n ethene molecules, each of which has one double bond. The bond formation energy is the energy released when a bond is formed. It is obtained from the total electron binding energy by subtracting the contribution from the Coulomb integrals, . In ab initio methods an attempt is made to evaluate all integrals that appear in the secular determinant. Approximations are still employed, but these are mainly associated with the construction of the wavefunctions involved in the integrals. In semiempirical methods, many of the integrals are expressed in terms of spectroscopic data or physical properties. Semiempirical methods exist at several levels. At some levels, in order to simplify the calculations, many of the integrals are set equal to zero. The HartreeFock and DFT methods are similar in that they are both regarded as ab initio methods. In HF the central focus is the wavefunction whereas in DFT it is the electron density. They are both iterative self consistent methods in that the process are repeated until the energy and wavefunctions (HF) or energy and electron density (DFT) are unchanged to within some acceptable tolerance.
E14.2(b)
E14.3(b)
E14.4(b)
Numerical exercises
E14.5(b)
 Use Fig. 14.23 for H2 , 14.30 for N2 , and 14.28 for O2 .  (a) H2 (3 electrons):
1 2 2 1
b = 0.5
MOLECULAR STRUCTURE
221
(b) N2 (10 electrons): (c) O2 (12 electrons): E14.6(b)
1 2 2 2 1 4 3 2
b=3 b=2
1 2 2 2 3 2 1 4 2 2
ClF is isoelectronic with F2 , CS with N2 . (a) ClF (14 electrons): (b) CS (10 electrons): (c) O (13 electrons): 2 1 2 2 2 3 2 1 4 2 4 1 2 2 2 1 4 3 2 b=3 b = 1.5 b=1
1 2 2 2 3 2 1 4 2 3
E14.7(b)
Decide whether the electron added or removed increases or decreases the bond order. The simplest procedure is to decide whether the electron occupies or is removed from a bonding or antibonding orbital. We can draw up the following table, which denotes the orbital involved (a) AB Change in bond order (b) AB+ Change in bond order N2 2 1/2 3 1/2 NO 2 1/2 2 +1/2 O2 2 1/2 2 +1/2 C2 3 +1/2 1 1/2 F2 4 1/2 2 +1/2 CN 3 +1/2 3 1/2
(a) Therefore, C2 and CN are stabilized (have lower energy) by anion formation. (b) NO, O2 and F2 are stabilized by cation formation; in each of these cases the bond order increases. E14.8(b) E14.9(b) Figure 14.1 is based on Fig. 14.28 of the text but with Cl orbitals lower than Br orbitals. BrCl is likely 1 to have a shorter bond length than BrCl ; it has a bond order of 1, while BrCl has a bond order of 2 . O+ (11 electrons) : 2 O2 (12 electrons) : O (13 electrons) : 2 O2 (14 electrons) : 2 1 2 2 2 3 2 1 4 2 1 1 2
2 2 2
b = 5/2 b=2 b = 3/2 b=1
3 1 2
2 4
2
4
2 3
1 2
2
3 1 2
1 2 2 2 3 2 1 4 2 4
Figure 14.1
222
INSTRUCTOR'S MANUAL
Each electron added to O+ is added to an antibonding orbital, thus increasing the length. So the 2 sequence O+ , O2 , O , O2 has progressively longer bonds. 2 2 2 E14.10(b) 2 d = N2 (A + B )2 d = 1 = N2
2 2 (A + 2 B + 2A B ) d = 1
= N 2 (1 + 2 + 2S) Hence N =
1/2 1 1 + 2S + 2
A B d = S
E14.11(b) We seek an orbital of the form aA + bB, where a and b are constants, which is orthogonal to the orbital N (0.145A + 0.844B). Orthogonality implies (aA + bB)N (0.145A + 0.844B) d = 0 0=N [0.145aA2 + (0.145b + 0.844a)AB + 0.844bB 2 ] d AB d is the overlap integral S, so a=  0.145S + 0.844 b 0.145 + 0.844S
The integrals of squares of orbitals are 1 and the integral 0 = (0.145 + 0.844S)a + (0.145S + 0.844)b so
This would make the orbitals orthogonal, but not necessarily normalized. If S = 0, the expression simplifies to a= 0.844 b 0.145
and the new orbital would be normalized if a = 0.844N and b = 0.145N. That is N (0.844A  0.145B) E14.12(b) The trial function = x 2 (L  2x) does not obey the boundary conditions of a particle in a box, so it is not appropriate . In particular, the function does not vanish at x = L. E14.13(b) The variational principle says that the minimum energy is obtained by taking the derivate of the trial energy with respect to adjustable parameters, setting it equal to zero, and solving for the parameters: Etrial = e2 3a 2 h  2 0 a 1/2 2 3 so 3 2 h e2 dEtrial =  da 2 20
1/2 1 = 0. 2 3 a
Solving for a yields: 3 2 h e2 = 2 20
1/2 1 2 3 a
so a =
e2 3 2 0 h
2
1 2 3
=
2 e4 18 3h4 e02
.
Substituting this back into the trial energy yields the minium energy: 3 2 h Etrial = 2 2 e 4 2 18 3h4 e0 e2  0 2 e 4 2 18 3h4 e0 2 3
1/2
=
e4
2 12 3 0 h2
.
MOLECULAR STRUCTURE
223
E14.14(b) The molecular orbitals of the fragments and the molecular oribitals that they form are shown in Fig. 14.2.
Figure 14.2
E14.15(b) We use the molecular orbital energy level diagram in Fig. 14.38. As usual, we fill the orbitals starting with the lowest energy orbital, obeying the Pauli principle and Hund's rule. We then write
 (a) C6 H6 (7 electrons): 2 4 1 a2u e1g e2u
E = 2( + 2) + 4( + ) + (  ) = 7 + 7
+ (b) C6 H6 (5 electrons): 2 3 a2u e1g
E = 2( + 2) + 3( + ) = 5 + 7 E14.16(b) The secular determinants from E14.16(a) can be diagonalized with the assistance of generalpurpose mathematical software. Alternatively, programs specifically designed of H ckel calculau tions (such as the one at Austrialia's Northern Territory University, http://www.smps.ntu.edu.au/ modules/mod3/interface.html) can be used. In both molecules, 14 electrons fill seven orbitals. (a) In anthracene, the energies of the filled orbitals are + 2.41421, + 2.00000, + 1.41421 (doubly degenerate), + 1.00000 (doubly degenerate), and + 0.41421, so the total energy is 14 + 19.31368 and the energy is 19.31368 . (b) For phenanthrene, the energies of the filled orbitals are + 2.43476, + 1.95063, + 1.51627, + 1.30580, + 1.14238, + 0.76905, + 0.60523, so the total energy is 14 + 19.44824 and the energy is 19.44824 .
224
INSTRUCTOR'S MANUAL
Solutions to problems
Solutions to numerical problems
P14.1 A = cos kx measured from A, B = cos k (x  R) measuring x from A. Then, with = A + B = cos kx + cos k (x  R) = cos kx + cos k R cos k x + sin k R sin k x [cos(a  b) = cos a cos b + sin a sin b] ; cos k R = cos = 0; k=k = 2R 2 x x + sin = cos 2R 2R sin k R = sin =1 2
(a)
1 1 1 1 For the midpoint, x = 2 R, so 2 R = cos 4 + sin 4 = 21/2 and there is constructive interference ( > A , B ). 3 3 , k = ; cos k R = cos = 0, sin k R = 1. (b) k= 2R 2R 2 x 3x = cos  sin 2R 2R 1 1 For the midpoint, x = 2 R, so 2 R interference ( < A , B ). 1 3 = cos 4  sin 4 = 0 and there is destructive
P14.5
2 2 We obtain the electron densities from + = + and  =  with + and  as given in Problem 14.4 2 = N
1
3 a0
{ez/a0 ezR/a0 }2
We evaluate the factors preceding the exponentials in + and  N+ 1
3 a0 1/2
= 0.561 1
3 a0 1/2
1/2 1 1 = 3 (52.9 pm) 1216 pm3/2
Likewise, N Then + =
=
1 621 pm3/2
1 {ez/a0 + ezR/a0 }2 (1216)2 pm3 1 and  = {ez/a0 + ezR/a0 }2 (622)2 pm3 The "atomic" density is
1 1 = 2 {1s (A)2 + 1s (B)2 } = 2
1
3 a0
{e2rA /a0 + e2rB /a0 }
=
e(2rA /a0 ) + e(2rB /a0 ) e(2z/a0 ) + e(2zR/a0 ) = 9.30 105 pm3 9.30 105 pm3
The difference density is = 
MOLECULAR STRUCTURE
225
Draw up the following table using the information in Problem 14.4 z/pm + 10 /pm  107 /pm3 107 /pm3 + 107 /pm3  107 /pm3 z/pm + 10 /pm  107 /pm3 107 /pm3 + 107 /pm3  107 /pm3
7 3 7 3
100 0.20 0.44 0.25 0.05 0.19 60 3.73 0.25 3.01 0.71 2.76
80 0.42 0.94 0.53 0.11 0.41 80 4.71 4.02 4.58 0.13 0.56
60 0.90 2.01 1.13 0.23 0.87 100 7.42 14.41 8.88 1.46 5.54
40 1.92 4.27 2.41 0.49 1.86 120 5.10 11.34 6.40 1.29 4.95
20 4.09 9.11 5.15 1.05 3.96 140 2.39 5.32 3.00 0.61 2.33
0 8.72 19.40 10.93 2.20 8.47 160 1.12 2.50 1.41 0.29 1.09
20 5.27 6.17 5.47 0.20 0.70 180 0.53 1.17 0.66 0.14 0.51
40 3.88 0.85 3.26 0.62 2.40 200 0.25 0.55 0.31 0.06 0.24
The densities are plotted in Fig. 14.3(a) and the difference densities are plotted in Fig. 14.3(b).
20
15
10
5 0 100
0 z/pm
100
200
Figure 14.3(a)
10 8 6 4 2 0 2 4 6 100
0 z/pm
100
200
Figure 14.3(b)
226
INSTRUCTOR'S MANUAL
R/2 A Z=0
R/2 Z
P14.6
(a) With spatial dimensions in units (multiples) of a0 , the atomic arbitals of atom A and atom B may be written in the form (eqn 13.24):
2 2 2 1 pz,A = (z + R/2) e x +y +(z+R/2) 4 2 1/2
2
2 2 2 1 pz,B = (z  R/2) e x +y +(zR/2) 4 2
1/2
2
according to eqn 14.98, the LCAOMO's have the form: pz,A + pz,B u = 2(1 + s) pz,A + pz,B and g = 2(1  s)
where s =
  
pz,A pz,B dx dy dz
(eqn14.17)
computations and plots are readily prepared with mathematical software such as mathcad.
Probability densities along internuclear axis (x = y = 0) with R = 3. (all distances in units of a0) 0.02
0.015
g
2
0.01
0.005
u
0 10
5
0 z
5
10
Figure 14.4(a)
(b) With spatial dimensions in units of a0 , the atomic orbitals for the construction of molecular orbitals are: px,A = 1 4 2
2 2 2 xe x +y +(z+R/2) 1/2
2
MOLECULAR STRUCTURE
227
R=3 Amplitude of Sigma Antibonding MO in xz Probability Density of Sigma Antibonding MO
Amplitude of Sigma bonding MO in xz
Probability Density of Sigma Bonding MO
Amplitude of Sigma Antibonding MO in xz
Amplitude of Sigma bonding MO in xz
Figure 14.4(b)
228
INSTRUCTOR'S MANUAL
R=3 2p Pi Bonding Amplitude Surface 2p Pi Bonding Probability Density Surface
2p Pi Antibonding Amplitude Surface
2p Pi Antibonding Probability Density Surface
2p Pi Bonding
2p Pi Antibonding
Figure 14.4(c)
MOLECULAR STRUCTURE
229
px , B =
2 2 2 1 x e x +y +(zR/2) 4 2
1/2
2
The MO's are: px,A + px,B u = 2(1 + s)
and
px,A  px,B g = 2(1  s)
where s =
  
px,A px,B dx dy dz
The various plot clearly show the constructive interference that makes a bonding molecular orbital. Nodal planes created by destructive interference are clearly seen in the antibonding molecular orbitals. When calculations and plots are produced for the R = 10 case, constructive and destructive interference is seen to be much weaker because of the weak atomic orbital overlap. P14.7 P = 2 d 2 , (a) From Problem 14.5
2 + (z = 0) = + (z = 0) = 8.7 107 pm3
= 1.00 pm3
Therefore, the probability of finding the electron in the volume at nucleus A is P = 8.6 107 pm3 1.00 pm3 = 8.6 107 (b) By symmetry (or by taking z = 106 pm) P = 8.6 107
2 1 (c) From Fig. 14.4(a), + 2 R = 3.7 107 pm3 , so P = 3.7 107
(d) From Fig. 14.5, the point referred to lies at 22.4 pm from A and 86.6 pm from B.
pm
86.
10.0 pm
22
.4
6p
m
B
A
20.0 pm
86.0 pm
Figure 14.5
Therefore, =
e22.4/52.9 + e86.6/52.9 0.65 + 0.19 = = 6.98 104 pm3/2 3/2 1216 pm 1216 pm3/2 so P = 4.9 107
2 = 4.9 107 pm3 ,
For the antibonding orbital, we proceed similarly. (a)
2  (z = 0) = 19.6 107 pm3 [Problem 14.5],
so
P = 2.0 106
(b) By symmetry, P = 2.0 106 (c)
2 1  2 R = 0,
so
P = 0
230
INSTRUCTOR'S MANUAL
(d) We evaluate  at the point specified in Fig. 14.5  = 0.65  0.19 = 7.41 104 pm3/2 621 pm3/2 so P = 5.5 107
2  = 5.49 107 pm3 ,
P14.10
(a) To simplify the mathematical expressions, atomic units (a.u.) are used for which all distances are in units of a0 and e2 /(40 a0 ) is the energy unit.
(x,y,z) rA A R/2 z=0 rB R /2 B z
Figure 14.6(a)
1 1 2 2 2 A = erA = e x +y +(z+R/2) 1 1 2 2 2 B = erB = e x +y +(zR/2) H = 2 1 1 1  +  2 rA rB R (eqn 14.6)
(eqn 14.8)
= = =
AH A d A  A 
(coulomb integral, eqn 14.24) A d A2 1 d + rB R
j
2 1 1 1  +  2 rA rB R 2 1  2 rA
E1s =1/2
A d 
A2 d
1/R
(eqns 13.13,13.15)
(BornOppenheimer approx.)
1 1 = j + 2 R
= =
AH B d A 
(Resonance integral, eqn 14.24) B d
2 1 1 1  +  2 rA rB R
MOLECULAR STRUCTURE
231
=
A 
1 2  2 rB
B d 
AB 1 d + rA R
k
AB d
S
1 E1s B= 2 B
=
1 2
AB d k +
S
S R
=
1 1  Sk R 2
+ according to eqn 14.28, E1g = . In order to numerically calculate E as a function of 1+S R we must devise a method by which S, j , and k are evaluated with numerical integrations at specified R values. In the cartesian coordinate system drawn above, d = dx dy dz and triple integrals are required. Numerical integration may proceed slowly with this coordinate system. However, the symmetry of the wavefunction may be utilized to reduce the problem to double integrals by using the spherical coordinate system of Fig. 14.15 and eqn 14.9. The numerical integration will proceed more rapidly. A 1 = er and
1 1 2 2 B = erB = e r +R 2rR cos
(eqn 14.9)
2
S(R) =
AB d =
0  0
A(r)B(r, , R)r 2 sin( ) d dr d
= 2
 0
A(r)B(r, , R)r 2 sin( ) d dr
(x,y,z) rA= r rB = r2+R22rR cos( ) B R z
A z=0
Figure 14.6(b)
The numerical integration, Snumerical (R), may be performed with mathematical software (mathcad, TOL = 0.001) and compared with the exact analytic solution (eqn 14.12), Sexact (R). As shown in the following plot, the percentage deviation of the numerical integration is never more than 0.01% below R = L/ao . This is satisfactory. The numerical integrals of j and k are setup in the same way.
j (R) = 2
 0
A(r)2 r 2 sin( ) d dr rB (r, , R)
232
INSTRUCTOR'S MANUAL
0.005
Snumerical (R)Sexact (R) 100 Sexact (R)
0
0.005
0.01 0 1 2 R 3 4
Figure 14.7(a)
k(R) = 2
 0
A(r)B(r, , R)r sin( ) d dr
The coulomb and resonance integrals are: 1 1 (R) =   j (R) + 2 R and (R) = 1 1  S(R)  k(R) R 2
(R) + (R) 1 + S(R) This numerical calculation of the energy, Enumerical (R), may be performed and compared with the exact analytic solution (eqns 14.11 and 14.12), Eexact (R). The following plot shows that the numerical integration method correctly gives energy values within about 0.06% of the exact value in the range a0 R 4a0 . This orbital energy is: E1g (R) = (b) The minimum energy, as determined by a numerical computation, may be evaluated with several techniques. When the computations do not consume excessive lengths of time, E(R)
0.02
Enumerical (R)Eexact (R) 100 Eexact (R)
0
0.02
0.04
0.06 1 1.5 2 2.5 R 3 3.5 4
Figure 14.7(b)
MOLECULAR STRUCTURE
233
0.2
0.3
Enumerical
0.4
0.5
1
1.5
2
2.5 R
3
3.5
4
Figure 14.7(c)
may be calculated at many R values as is done in the above figure. The minimum energy and corresponding R may be read from a table of calculated values. Values of the figure give: Emin = 0.5647(a.u.) = 15.367e and Re = 2.4801(a.u.) = 131.24 pm . Alternatively, lengthy computations necessitate a small number of numerical calculations near the minimum after which an interpolation equation is devised for calculating E at any value of d R. The minimum is determined by the criteria that Einterpolation (R) = 0. dR + The spectroscopic dissociation constant, De , for H2 is referenced to a zero electronic energy when a hydrogen atom and a proton are at infinite separation. 1 De = Emin  EH + Eproton = 0.5647   + 0 2 De = 0.0647 (a.u.) = 1.76 eV P14.12 The internuclear distance r n n2 a0 , would be about twice the average distance ( 1.06 106 pm) of a hydrogenic electron from the nucleus when in the state n = 100. This distance is so large that each of the following estimates are applicable. Resonance integral,  (where 0) Overlap integral, S (where 0) Coulomb integral, En=100 for atomic hydrogen Binding energy = 2{E+  En=100 } = 2 +  En=100 1S (a.u.)
= 2{  En=100 } 0 Vibrational force constant, k 0 because of the weak binding energy. Rotational constant, B = h2 h2 2 = 0 because rAB is so large. 2 2hcl 2hcrAB
234
INSTRUCTOR'S MANUAL
The binding energy is so small that thermal energies would easily cause the Rydberg molecule to break apart. It is not likely to exist for much longer than a vibrational period. P14.13 In the simple H ckel approximation u
:O :
:
1O N 4 O 3
O N O O
N O:
: :O
O 2
O  E 0 0
:
(E  O )2 (E  O ) (E  N )  3 2 = 0 Therefore, the roots are E  O = 0 (twice) and (E  O ) (E  N )  3 2 = 0
Each equation is easily solved (Fig. 14.8(a)) for the permitted values of E in terms of O , N , and . The quadratic equation is applicable in the second case.
In contrast, the energies in the absence of resonance are derived for N==O, that is, just one of the three O  E N  E =0
Expanding the determinant and solving for E gives the result in Fig.14.8(b). Delocalization energy = 2 {E (with resonance)  E (without resonance)} = If 2 (O  N )2 , then 4 2 . (O  N ) (O  N )2 + 12 2
1/2
Delocalization energy
:
0 O  E 0
0 0 O  E
N  E
=0
Figure 14.8(a)
 (O  N )2 + 4 2
1/2
MOLECULAR STRUCTURE
235
Figure 14.8(b) P14.17 In all of the molecules considered in P14.16, the HOMO was bonding with respect to the carbon atoms connected by double bonds, but antibonding with respect to the carbon atoms connected by single bonds. (The bond lengths returned by the modeling software suggest that it makes sense to talk about double bonds and single bonds. Despite the electron delocalization, the nominal double bonds are consistently shorter than the nominal single bonds.) The LUMO had just the opposite character, tending to weaken the C==C bonds but strengthen the C C bonds. To arrive at this conclusion, examine the nodal surfaces of the orbitals. An orbital has an antibonding effect on atoms between which nodes occur, and it has a binding effect on atoms that lie within regions in which the orbital does not change sign. The transition, then, would lengthen and weaken the double bonds and shorten and strengthen the single bonds, bringing the different kinds of polyene bonds closer to each other in length and strength. Since each molecule has more double bonds than single bonds, there is an overall weakening of bonds.
HOMO HOMO
LUMO
LUMO
Figure 14.9(a)
Solutions to theoretical problems
P14.19 Since 1 2s = R20 Y00 = 2 2
1 =4
/4 Z 3/2 e 2 a0 2
1 1/2 [Tables 13.1, 12.3] 4
1 1/2 2
Z 3/2 /4 e 2 a0 2
236
INSTRUCTOR'S MANUAL
HOMO
LUMO
HOMO
LUMO
Figure 14.9(b) 1 2px = R21 (Y1,1  Y11 ) [Section 13.2] 2 1 = 12 1 = 12 = 2py = = Therefore, 1 1 = 4 3 = 1 4 1 1/2 Z 3/2 a0 1 4 Z 3/2 /4 e a0 2 Z 3/2 /4 e a0 2 3 1/2 sin (ei + ei ) [Tables 13.1, 12.3] 8 3 1/2 sin cos 8
1 1/2 2
Z 3/2 /4 e sin cos 2 a0
1 R21 (Y1,1 + Y11 ) [Section 13.2] 2i 1 4 1 1/2 2 Z 3/2 /4 e sin sin [Tables 13.1, 12.3] a0 2
3 1 1  sin cos + sin sin e/4 2 2 22 2 2 2 Z 3/2 1 2  sin cos + a0 2 22 3 sin sin e/4 22
1 1/2 6
MOLECULAR STRUCTURE
237
= =
1 4 1 4
1 1/2 6 1 1/2 6
Z 3/2 2 a0 2 Z 3/2 2 a0 2
3 1 1 + sin cos  sin sin e/4 2 2 [cos  3 sin ] 1+ sin e/4 2
The maximum value of occurs when sin has its maximum value (+1), and the term multiplying /2 has its maximum negative value, which is 1, when = 120 . P14.21 The normalization constants are obtained from 2 d = 1, N2 = N (A B )
2 2 (A + B 2A B ) d = N 2 (1 + 1 2S) = 1
(A B )2 d = N 2 1 2(1 S)
Therefore, N 2 = H =
h2 2 e2 1 e2 1 e2 1   + 2m 40 rA 4 0 rB 4 0 R
H = E implies that  h2 2 e2 1 e2 1 e2 1  +  = E 2m 40 rA 4 0 rB 4 0 R
Multiply through by (= ) and integrate using   h2 2 e2 1 A = EH A A  2m 40 rA h2 2 e2 1 B  B = EH B 2m 40 rB
Then for = N (A + B ) N EH A + EH B  e2 1 40 R e2 1 e2 1 e2 1 B  A + (A + B ) d = E 4 0 rA 4 0 rB 4 0 R 2 d  e2 N 4 0 B A + rA rB d = E d = E
hence EH and so EH + Then use A
2 d +
e2 1 1 1 1 e2 1 A B + B B + A A + B A  N2 40 R 4 0 rA rA rB rA 1 1 A B d = B A d [by symmetry] = V2 /(e2 /4 0 ) rA rB B 1 B d [by symmetry] = V1 /(e2 /4 0 ) rA 1 1+S (V1 + V2 ) = E
1 A d = rB
which gives EH +
e2 1  40 R
238
INSTRUCTOR'S MANUAL
or E = EH 
e2 V1 + V2 1 + = E+ 1+S 40 R
as in Problem 14.8. The analogous expression for E is obtained by starting from = N (A  B ) 1 2(1  S) and the following through the stepwise procedure above. The result is with N 2 = E = EH  V1  V2 e2 = E + 1S 40 R e2 h2 2  2 4 0 r
as in Problem 14.9. P14.22 (a) = ekr 2 d = H =
0
1 d = r 2 d = = Therefore H d = and E=
h2 2k
d = 3 k 0 0 2 re2kr dr sin d d = 2 k 0 0 0 2k 1 d2 d (rekr ) d = k 2  r dr 2 r 2  = k k k r 2 e2kr dr sin d
2
h2 e2 2  2 k 4 0 k
e  4 k 2 0 /k 3
2
=
h2 k 2 e2 k  2 4 0 when k= e2 4 0 h2
h2 e2 dE =2 k =0 dk 2 40
The optimum energy is therefore e4 E= = hcRH the exact value 2 32 2 0 h2 (b) = ekr , H as before.
2
2 d =
0
2 e2kr r 2 dr
0
sin d
2 0
d =
2
1/2 2k 3
2 2 1 d = re2kr dr sin d d = r k 0 0 0
MOLECULAR STRUCTURE
239
2 d = 2 = 2 = 8 Therefore E=
(3k  2k 2 r 2 ) d
0
(3kr 2  2k 2 r 4 )e2kr dr
2
0
sin d
2
d
0
3k 8
1/2 3k 2  16 2k 3
1/2 2k 5
3 2 k h e2 k 1/2  2 0 (2)1/2 when k = e 4 2
2 18 3 0 h4
dE =0 dk
and the optimum energy is therefore E= e4
2 12 3 0 h2
= 
8 hcRH 3
Since 8/3 < 1, the energy in (a) is lower than in (b), and so the exponential wavefunction is better than the Gaussian. P14.23 (a) The variation principle selects parameters so that energy is minimized. We begin by finding the cirteria for selecting best at constant R( = R) dEel (best ) = 0 d = 2F1 + 2 d dF1 d dF2 + F2 + d d d d dF1 dF2 = 2F1 + 2 R + F2 + R d d dF2 () F2 ()  d = dF1 () 2F1 ()  d
best ()
We must now select R so as to minimize the total energy, E. Using Hartree atomic units for which length is in units of a0 and energy is in units of e2 /4 0 a0 , the total energy equation is: E() = Eel () + 1 best () 2 = best F1 () + best ()F2 () + R()
where R() = /best (). Mathematical software provides numerical methods for easy evaluation of derivatives within best (). We need only setup the software to calculate E() and R() over a range of values. The value of R for which E is a minimum is the solution. The following plot is generated with 1.5 8.0 The plots indicates an energy minimum at about 0.58 au and an Re value of about 2.0 au. More precise values can be determined by generating a plot over a more restricted range, say, 2.478 2.481. A table of , R(), and E() may be examined for the minimum energy and corresponding and R values.
240
INSTRUCTOR'S MANUAL
Total energy vs internuclear distance
0.45
0a0
E
2/4
0.5
0.55
0.6 0 1 2 3 4 R / a0 5 6 7 8
Figure 14.10
best = 2.4802a0 Re = 2.0033a0 = 106.011 pm E(Re ) = 0.5865 au = 15.960 eV best = 1.2380 De for H+ is referenced to a zero electronic energy when a hydrogen atom and a proton are at rest at infinite separation.
+ De =  E(Re )  EH  Eproton
=  [0.5865 au + 0.5 au  0 au] De = 0.0865 au = 2.35 eV The experimental value of De is 2.78 eV and that of Re is 2.00a0 . The equilibrium internuclear distance is in excellent agreement with the experimental value but the spectroscopic dissociation energy is off by 15.3%. (b) The virial theorem (Atkins Eq. 12.46) states that the potential energy is twice the negative of the kinetic energy. In the electronic energy equation, Eel = 2 F1 () + F2 () the term 2 F1 () is the electron kinetic energy and, consequently, the total kinetic energy because the nuclei do not move the BornOppenheimer approximation. The term F2 () is the electron potential energy only so the nuclear potential (1/Re in au) must be added to get the total potential energy. The wavefunction approximation satisfies the virial theorem when
2 f = best F2 (best ) + 1/Re + 2best F1 (best ) = 0
MOLECULAR STRUCTURE
241
Since numerology has been used, we will calculate the fraction f/E(Re ). If the fraction is very small the virial theorem is satisfied.
1 2 2best F1 (best ) + best F2 (best ) + Re
E(Re )
= 4.996 106
The fraction is so small that we conclude that the virial theorem is satisfied. (c) A = 3 rA /a0 e ; 3 a0 A B d = 3
3 a0 2 1
B = 3
2 rB /a0 e 3 a0
S =
3 a0
e(rA +rB )/a0 d R3 2 (  2 ) d d d 8
=
eR/a0
0 1 1
=
1
2 1 2 R/a0 d d e d 3 R 0 1 1 1 2a0 2  d 2 d eR/a0 d 4 a 0 3 R 3 R 3 2a0
0 1 2 2a0
=
1
+ 2R a0 + 2 R 2 eR/a0
1
=
1
R 3 2a0
3 2 a0
3 R 3
2 a0 2 eR/a0 3 R 2 2 4 + 4R + R eR/a0 2 a0 a0 2 2 R 2 R/a0 e  2 3 a0 where = R/a0
1 = 4 4 + 4 + 4 2 e 3
S = 1 + + 1 2 e 3 P14.25
m The secular determinant for a cyclic species HN has the form
1 x 1 0 0 . . . . . . 1
2 1 x 1 0 . . . . . . 0
3 0 1 x 1 . . . . . . 0
... ... ... N  1 ... ... ... ... ... ... 1 ... ... x 1 ... . . . . . . . . . . . . . . . . . . 0 0 ... 0 0 0 0 . . . . . . 1
N 1 0 0 0 . . . 1 x
242
INSTRUCTOR'S MANUAL
E or E =  x Expanding the determinant, finding the roots of the polynomial, and solving for the total binding energy yields the following table. Note that < 0 and < 0. where x =
Species Number of e H4 4
+ H5
4 5 6 6 6
H5
 H5
H6 + H7
Permitted x 2,0,0,2 1 1 1 1 1+ 5 1 5 , 1 5 , 1+ 5 , 2, 2 2 2 2 1 1 1 1 1 5 , 1+ 5 , 1+ 5 1 5 , 2, 2 2 2 2 1 1 1 1 1 5 , 1+ 5 , 1+ 5 1 5 , 2, 2 2 2 2 2,1,1,1,1,2 2,1.248,1.248,1.248,1.248,0.445,0.445,0.445
Total binding energy 4 + 4 4 + 3 + 5 5 + 1 5+3 5 2 6 + 2 + 2 5 6 + 8 6 + 8.992
H4 2H2
+ H5
rU rU
= 4( + )  (4 + 4) = 0 = 2( + ) + (2 + 4)  (4 + 5.236) = 0.764 < 0
+ H2 H3
The above
rU
+ values indicate that H4 and H5 are unstable. rU
  H5 H2 + H3
= 2( + )  (4 + 2)  (6 + 6.472) = 2.472 > 0 = 6( + )  (6 + 8) = 2 > 0 = 4( + ) + (2 + 4)  (6 + 8.992) = 0.992 > 0
H6 3H2
+ H7
rU
+ 2H2 + H3
rU
The
rU
 + values for H5 , H6 , and H7 indicate that they are stable.
Species H4 , 4e + H5 , 4e  H5 , 6e H6 , 6e + H7 , 6e
Statisfies H ckel's 4n + 2 low energy rule u Correct number of e Stable No No No No Yes Yes Yes Yes Yes Yes
H ckel's 4n + 2 rule successfully predicts the stability of hydrogen rings. u
15
Molecular symmetry
Solutions to exercises
Discussion questions
E15.1(b)
Symmetry operations 1. Identity, E 2. nfold rotation 3. Reflection 4. Inversion 5. nfold improper rotation Symmetry elements 1. The entire object 2. nfold axis of symmetry, Cn 3. Mirror plane, 4. Centre of symmetry, i 5. nfold improper rotation axis, Sn
E15.2(b)
A molecule may be chiral, and therefore optically active, only if it does not posses an axis of improper rotation, Sn . An improper rotation is a rotation followed by a reflection and this combination of operations always converts a righthanded object into a lefthanded object and viceversa; hence an Sn axis guarantees that a molecule cannot exist in chiral forms. See Sections 15.4(a) and (b). The direct sum is the decomposition of the direct product. The procedure for the decomposition is the set of steps outlined in Section 15.5(a) on p. 471 and demonstrated in Illustration 15.1.
E15.3(b) E15.4(b)
Numerical exercises
E15.5(b) CCl4 has 4 C3 axes (each CCl axis), 3 C2 axes (bisecting ClCCl angles), 3 S4 axes (the same as the C2 axes), and 6 dihedral mirror planes (each ClCCl plane). E15.6(b) Only molecules belonging to Cs , Cn , and Cnv groups may be polar, so . . . (a) CH3 Cl (C3v ) may be polar along the CCl bond; (b) HW2 (CO)10 (D4h ) may not be polar (c) SnCl4 (Td ) may not be polar E15.7(b) The factors of the integrand have the following characters under the operations of D6h
px z pz Integrand E 2 1 1 2 2C6 1 1 1 1 2C3 1 1 1 1 C2 2 1 1 2 3C2 0 1 1 0 3C2 0 1 1 0 i 2 1 1 2 2S3 1 1 1 1 2S6 1 1 1 1 h 2 1 1 2 3d 0 1 1 0 3v 0 1 1 0
The integrand has the same set of characters as species E1u , so it does not include A1g ; therefore the integral vanishes E15.8(b) We need to evaluate the character sets for the product A1g E2u q, where q = x, y, or z
A1g E2u (x, y) Integrand E 1 2 2 4 2C6 1 1 1 1 2C3 1 1 1 1 C2 1 2 2 4 3C2 1 0 0 0 3C2 1 0 0 0 i 1 2 2 4 2S3 1 1 1 1 2S6 1 1 1 1 h 1 2 2 4 3d 1 0 0 0 3v 1 0 0 0
244
INSTRUCTOR'S MANUAL
To see whether the totally symmetric species A1g is present, we form the sum over classes of the number of operations times the character of the integrand c(A1g ) = (4) + 2(1) + 2(1) + (4) + 3(0) + 3(0) + (4) +2(1) + 2(1) + (4) + 3(0) + 3(0) = 0 Since the species A1g is absent, the transition is forbidden for x or ypolarized light. A similar analysis leads to the conclusion that A1g is absent from the product A1g E2u z; therefore the transition is forbidden. E15.9(b) The classes of operations for D2 are: E, C2 (x), C2 (y), and C2 (z). How does the function xyz behave under each kind of operation? E leaves it unchanged. C2 (x) leaves x unchanged and takes y to y and z to z, leaving the product xyz unchanged. C2 (y) and C2 (z) have similar effects, leaving one axis unchanged and taking the other two into their negatives. These observations are summarized as follows
xyz E 1 C2 (x) 1 C2 (y) 1 C2 (z) 1
A look at the character table shows that this set of characters belong to symmetry species A1 E15.10(b) A molecule cannot be chiral if it has an axis of improper rotation. The point group Td has S4 axes and mirror planes (= S1 ) , which preclude chirality. The Th group has, in addition, a centre of inversion (= S2 ) . E15.11(b) The group multiplication table of group C4v is
+  C4 C2 v (x) v (y) d (xy) d (xy) E C4 +  E E C4 C4 C2 v (x) v (y) d (xy) d (xy) + +  C4 C4 C2 E C4 d (xy) (xy) v (y) v (x)  +  C4 C4 d (xy) (xy) v (x) v (y) C4 E C2  + C2 C2 C4 C4 E v (y) v (x) d (xy) d (xy)  + v (x) v (x) d (xy) d (xy) v (y) E C2 C4 C4 +  v (y) v (y) d (xy) d (xy) v (x) C2 E C4 C4 +  d (xy) d (xy) v (x) v (y) d (xy) C4 C4 E C2  + d (xy) d (xy) v (y) v (x) d (xy) C4 C4 C2 E
E15.12(b) See Fig. 15.1. (a) Sharpened pencil: E, C , v ; therefore Cv (b) Propellor: E, C3 , 3C2 ; therefore D3 (c) Square table: E, C4 , 4v ; therefore C4v ; Rectangular table: E, C2 , 2v ; therefore C2v (d) Person: E, v (approximately); therefore Cs E15.13(b) We follow the flow chart in the text (Fig. 15.14). The symmetry elements found in order as we proceed down the chart and the point groups are (a) Naphthalene: E, C2 , C2 , C2 , 3h , i; D2h (b) Anthracene: E, C2 , C2 , C2 , 3h , i; D2h
MOLECULAR SYMMETRY
245
(a)
(b)
(c)
(d)
Figure 15.1 (c) Dichlorobenzenes: (i) 1,2dichlorobenzene: E, C2 , v , v ; C2v (ii) 1,3dichlorobenzene: E, C2 , v , v ; C2v (iii) 1,4dichlorobenzene: E, C2 , C2 , C2 , 3h , i; D2h E15.14(b) (a) H F
(b) F F I F F F F OC (e) (f) Td F F OC Fe CO F F (c) F O Xe O OC OC Fe CO CO (d) OC CO
F
F
The following responses refer to the text flow chart (Fig. 15.14) for assigning point groups. (a) HF: linear, no i, so Cv (b) IF7 : nonlinear, fewer than 2Cn with n > 2, C5 , 5C2 perpendicular to C5 , h , so D5h
246
INSTRUCTOR'S MANUAL
(c) XeO2 F2 : nonlinear, fewer than 2Cn with n > 2, C2 , no C2 perpendicular to C2 , no h , 2v , so C2v (d) Fe2 (CO)9 : nonlinear, fewer than 2Cn with n > 2, C3 , 3C2 perpendicular to C3 , h , so D3h (e) cubane (C8 H8 ): nonlinear, more than 2Cn with n > 2, i, no C5 , so Oh (f) tetrafluorocubane (23): nonlinear, more than 2Cn with n > 2, no i, so Td E15.15(b) (a) Only molecules belonging to Cs , Cn , and Cnv groups may be polar. In Exercise 15.13b orthodichlorobenzene and metadichlorobenzene belong to C2v and so may be polar; in Exercise 15.10b, HF and XeO2 F2 belong to Cnv groups, so they may be polar. (b) A molecule cannot be chiral if it has an axis of improper rotationincluding disguised or degenerate axes such as an inversion centre (S2 ) or a mirror plane (S1 ). In Exercises 15.9b and 15.10b, all the molecules have mirror planes, so none can be chiral. E15.16(b) In order to have nonzero overlap with a combination of orbitals that spans E, an orbital on the central atom must itself have some E character, for only E can multiply E to give an overlap integral with a totally symmetric part. A glance at the character table shows that px and py orbitals available to a bonding N atom have the proper symmetry. If d orbitals are available (as in SO3 ),
2 all d orbitals except dz could have nonzero overlap.
E15.17(b) The product f () i must contain A1 (Example 15.7). Then, since i = B1 , () = (y) = B2 (C2v character table), we can draw up the following table of characters
B2 B1 B 1 B2 E 1 1 1 C2 1 1 1 v 1 1 1 v 1 1 1
= A2
Hence, the upper state is A2 , because A2 A2 = A1 . E15.18(b) (a) Anthracene
H H H H H H H H H D 2h H
The components of span B3u (x), B2u (y), and B1u (z). The totally symmetric ground state is Ag . Since Ag = in this group, the accessible upper terms are B3u (xpolarized), B2u (ypolarized), and B1u (zpolarized). (b) Coronene, like benzene, belongs to the D6h group. The integrand of the transition dipole moment must be or contain the A1g symmetry species. That integrand for transitions from the ground state is A1g qf , where q is x, y, or z and f is the symmetry species of the upper state. Since the ground state is already totally symmetric, the product qf must also have A1g symmetry for the entire integrand to have A1g symmetry. Since the different symmetry species are orthogonal, the only way qf can have A1g symmetry is if q and f have the same symmetry. Such combinations include zA2u , xE1u , and yE1u . Therefore, we conclude that transitions are allowed to states with A2u or E1u symmetry.
MOLECULAR SYMMETRY
247
E15.19(b)
A1 A2 E sin cos Product
E 1 1 2 1 1 1
2C3 1 1 1 Linear combinations of sin and cos 1
3v 1 1 0 1 1 1
The product does not contain A1 , so yes the integral vanishes.
Solutions to problems
P15.3 Consider Fig. 15.2. The effect of h on a point P is to generate h P , and the effect of C2 on h P is to generate the point C2 h P . The same point is generated from P by the inversion i, so C2 h P = iP for all points P . Hence, C2 h = i , and i must be a member of the group.
Figure 15.2 P15.6 Representation 1 D(C3 )D(C2 ) = 1 1 = 1 = D(C6 ) and from the character table is either A1 or A2 . Hence, either D(v ) = D(d ) = +1 or 1 respectively. Representation 2 D(C3 )D(C2 ) = 1 (1) = 1 = D(C6 ) and from the character table is either B1 or B2 . Hence, either D(v ) = D(d ) = 1 or D(v ) = D(d ) = 1 respectively. P15.8 A quick rule for determining the character without first having to set up the matrix representation is to count 1 each time a basis function is left unchanged by the operation, because only these functions give a nonzero entry on the diagonal of the matrix representative. In some cases there is a sign change, (. . . f . . .) (. . . f . . .); then 1 occurs on the diagonal, and so count 1. The character of the identity is always equal to the dimension of the basis since each function contributes 1 to the trace.
248
INSTRUCTOR'S MANUAL
E: all four orbitals are left unchanged; hence = 4 C3 : One orbital is left unchanged; hence = 1 C2 : No orbitals are left unchanged; hence = 0 S4 : No orbitals are left unchanged; hence = 0 d : Two orbitals are left unchanged; hence = 2 The character set 4, 1, 0, 0, 2 spans A1 + T2 . Inspection of the character table of the group Td shows that s spans A1 and that the three p orbitals on the C atom span T2 . 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 T2 set (dxy , dyz , dzx ) may contribute to molecular orbital formation with the H orbitals. P15.9 (a) In C3v symmetry the H1s orbitals span the same irreducible representations as in NH3 , which is A1 + A1 + E. There is an additional A1 orbital because a fourth H atom lies on the C3 axis. In C3v , the d orbitals span A1 + E + E [see the final column of the C3v character table]. Therefore, all five d orbitals may contribute to the bonding. (b) In C2v symmetry the H1s orbitals span the same irreducible representations as in H2 O, but one "H2 O" fragment is rotated by 90 with respect to the other. Therefore, whereas in H2 O the H1s orbitals span A1 + B2 [H1 + H2 , H1  H2 ], in the distorted CH4 molecule they span A1 + B2 + A1 + B1 [H1 + H2 , H1  H2 , H3 + H4 , H3  H4 ]. In C2v the d orbitals span 2A1 + B1 + B2 + A2 [C2v character table]; therefore, all except A2 (dxy ) may participate in bonding. P15.10 P15.12 The most distinctive symmetry operation is the S4 axis through the central atom and aromatic nitrogens on both ligands. That axis is also a C2 axis. The group is S4 . (a) Working through the flow diagram (Fig. 15.14) in the text, we note that there are no Cn axes with n > 2 (for the C3 axes present in a tetrahedron are not symmetry axes any longer), but it does have C2 axes; in fact it has 2C2 axes perpendicular to whichever C2 we call principal; it has no h , but it has 2d . So the point group is D2d . (b) Within this point group, the distortion belongs to the fully symmetric species A1 , for its motion is unchanged by the S4 operation, either class of C2 , or d . (c) The resulting structure is a square bipyramid, but with one pyramid's apex farther from the base than the other's. Working through the flow diagram in Fig. 15.14, we note that there is only one Cn axis with n > 2, namely a C4 axis; it has no C2 axes perpendicular to the C4 , and it has no h , but it has 4v . So the point group is C4v . (d) Within this point group, the distortion belongs to the fully symmetric species A1 . The translation of atoms along the given axis is unchanged by any symmetry operation for the motion is contained within each of the group's symmetry elements. P15.14 (a) xyz changes sign under the inversion operation (one of the symmetry elements of a cube); hence it does not span A1g and its integral must be zero (b) xyz spans A1 in Td [Problem 15.13] and so its integral need not be zero (c) xyz xyz under z z (the h operation in D6h ), and so its integral must be zero
MOLECULAR SYMMETRY
249
P15.16
We shall adapt the simpler subgroup C6v of the full D6h point group. The six orbitals span A1 + B1 + E1 + E2 , and are 1 a1 = (1 + 2 + 3 + 4 + 5 + 6 ) 6 1 b1 = (1  2 + 3  4 + 5  6 ) 6 1 (2   + 2   ) 4 1 2 3 5 6 12 e2 = 1 2 (2  3 + 5  6 ) 1 (2 +   2  + ) 2 3 4 1 5 6 12 e1 = 1 2 (2 + 3  5  6 ) The hamiltonian transforms as A1 ; therefore all integrals of the form H d vanish unless
and belong to the same symmetry species. It follows that the secular determinant factorizes into four determinants A1 : B1 : E1 : Hence E2 : Hence P15.17 Ha1 a1 = 1 6
1 Hb1 b1 = 6
(1 + + 6 )H (1 + + 6 ) d = + 2 (1  2 + )H (1  2 + ) d =  2
He1 (a)e1 (a) =  , He1 (b)e1 (b) =  , He1 (a)e1 (b) = 0   0 0 = 0 solves to =  (twice)  
He2 (a)e2 (a) = + , He2 (b)e2 (b) = + , He2 (a)e2 (b) = 0 +  0 0 = 0 solves to = + (twice) + 
a c d g e f g f e b a b c d
Consider phenanthrene with carbon atoms as labeled in the figure below
(a) The 2p orbitals involved in the system are the basis we are interested in. To find the irreproducible representations spanned by this basis, consider how each basis is transformed under the symmetry operations of the C2v group. To find the character of an operation in this basis, sum the coefficients of the basis terms that are unchanged by the operation.
14 0 0 14
E C2 v v
a a a a a
a a a a a
b b b b b
b b b b b
c c c c c
c c c c c
d d d d d
d d d d d
e e e e e
e e e e e
f f f f f
f f f f f
g g g g g
g g g g g
250
INSTRUCTOR'S MANUAL
To find the irreproducible representations that these orbitals span, multiply the characters in the representation of the orbitals by the characters of the irreproducible representations, sum those products, and divide the sum by the order h of the group (as in Section 15.5(a)). The table below illustrates the procedure, beginning at left with the C2v character table.
A1 A2 B1 B2 E 1 1 1 1 C2 1 1 1 1 v 1 1 1 1 v 1 1 1 1 product E 14 14 14 14 C2 0 0 0 0 v 0 0 0 0 v 14 14 14 14 sum/h 0 7 7 0
The orbitals span 7A2 + B2 . To find symmetryadapted linear combinations (SALCs), follow the procedure described in Section 15.5(c). Refer to the table above that displays the transformations of the original basis orbitals. To find SALCs of a given symmetry species, take a column of the table, multiply each entry by the character of the species' irreproducible representation, sum the terms in the column, and divide by the order of the group. For example, the characters of species A1 are 1, 1, 1, 1, so the columns to be summed are identical to the columns in the table above. Each column sums to zero, so we conclude that there are no SALCs of A1 symmetry. (No surprise here: the orbitals span only A2 and B1 .) An A2 SALC is obtained by multiplying the characters 1, 1, 1, 1 by the first column:
1 4 (a 1  a  a + a) = 2 (a  a ).
The A2 combination from the second column is the same. There are seven distinct A2 combinations in all: 1/2(a  a ), 1/2(b  b ), . . . , 1/2(g  g ) . The B1 combination from the first column is:
1 4 (a 1 + a + a + a) = 2 (a + a ).
The B1 combination from the second column is the same. There are seven distinct B1 com1 1 1 binations in all: 2 (a + a ), 2 (b + b ), . . . , 2 (g + g ) . There are no B2 combinations, as the columns sum to zero. (b) The structure is labeled to match the row and column numbers shown in the determinant. The H ckel secular determinant of phenanthrene is: u
a b c d e f g g f e d c b a
a E 0 0 0 0 0 0 0 0 0 0 0
b E 0 0 0 0 0 0 0 0 0 0
c 0 E 0 0 0 0 0 0 0 0 0 0
d 0 0 E 0 0 0 0 0 0 0 0 0
e 0 0 0 E 0 0 0 0 0 0 0 0
f 0 0 0 0 E 0 0 0 0 0 0 0
g 0 0 0 0 E 0 0 0 0 0 0
g 0 0 0 0 0 0 E 0 0 0 0
f 0 0 0 0 0 0 0 E 0 0 0 0
e 0 0 0 0 0 0 0 0 E 0 0 0
d 0 0 0 0 0 0 0 0 0 E 0 0
c 0 0 0 0 0 0 0 0 0 0 E 0
b 0 0 0 0 0 0 0 0 0 0 E
a 0 0 0 0 0 0 0 0 0 0 0 E
This determinant has the same eigenvalues as as in exercise 14.16(b)b.
MOLECULAR SYMMETRY
251
(c) The ground state of the molecule has A1 symmetry by virtue of the fact that its wavefunction is the product of doubly occupied orbitals, and the product of any two orbitals of the same symmetry has A1 character. If a transition is to be allowed, the transition dipole must be nonzero, which in turn can only happen if the representation of the product f i includes the totally symmetric species A1 . Consider first transitions to another A1 wavefunction, in which case we need the product A1 A1 . Now A1 A1 = A1 , and the only character that returns A1 when multiplied by A1 is A1 itself. The z component of the dipole operator belongs to species A1 , so zpolarized A1 A1 transitions are allowed. (Note: transitions from the A1 ground state to an A1 excited state are transitions from an orbital occupied in the ground state to an excitedstate orbital of the same symmetry.) The other possibility is a transition from an orbital of one symmetry (A2 or B1 ) to the other; in that case, the excitedstate wavefunction will have symmetry of A1 B1 = B2 from the two singly occupied orbitals in the excited state. The symmetry of the transition dipole, then, is A1 B2 = B2 , and the only species that yields A1 when multiplied by B2 is B2 itself. Now the y component of the dipole operator belongs to species B2 , so these transitions are also allowed (ypolarized). P15.21 (a) Following the flow chart in Fig. 15.14, not that the molecule is not linear (at least not in the mathematical sense); there is only one Cn axis (a C2 ), and there is a h . The point group, then, is C2h .
b a c d e f g h i j k k j i h g f e d c b a
(b) The 2pz orbitals are transformed under the symmetry operations of the C2h group as follows.
22 0 0 22
E C2 i h
a a a a a
a a a a a
b b b b b
b b b b b
c c c c c
c c c c c
... ... ... ... ...
j j j j j
j j j j j
k k k k k
k k k k k
To find the irreproducible representations that these orbitals span, we multiply the characters of orbitals by the characters of the irreproducible representations, sum those products, and divide the sum by the order h of the group (as in Section 15.5(a)). The table below illustrates the procedure, beginning at left with the C2h character table.
E 1 1 1 1 C2 1 1 1 1 i 1 1 1 1 h 1 1 1 1 E 22 22 22 22 C2 0 0 0 0 i 0 0 0 0 h 22 22 22 22
product
Ag Au Bg Bu
sum/h 0 11 11 0
The orbitals span 11Au + 11Bg . To find symmetryadapted linear combinations (SALCs), follow the procedure described in Section 15.5(c). Refer to the above that displays the transformations of the original basis orbitals. To find SALCs of a given symmetry species, take a column of the table, multiply each entry by the character of the species' irreproducible representation, sum the terms in the column, and divide by the order of the group. For example, the characters of species Au are 1, 1, 1, 1, so the
252
INSTRUCTOR'S MANUAL
columns to be summed are identical to the columns in the table above. Each column sums to zero, so we conclude that there are no SALCs of Ag symmetry. (No surprise: the orbitals span only Au and Bg ). An Au SALC is obtained by multiplying the characters 1, 1, 1, 1 by the first column:
1 4 (a 1 + a + a + a) = 2 (a + a ).
The Au combination from the second column is the same. There are 11 distinct Au combinations in all: 1/2(a + a ), 1/2(b + b ), . . . 1/2(k + k ) . The Bg combination from the first column is:
1 4 (a 1  a  a + a) = 2 (a  a ).
The Bg combination from the second column is the same. There are 11 distinct Bg combinations in all: 1/2(a  a ), 1/2(b  b ), . . . 1/2(k  k ) . There are no Bu combinations, as the columns sum to zero. (c) The structure is labeled to match the row and column numbers shown in the determinant. The H ckel secular determinant is: u
a b c ... i j k k j i ... c b a a E 0 ... 0 0 0 0 0 0 ... 0 0 0 b E ... 0 0 0 0 0 0 ... 0 0 0 c 0 E ... 0 0 0 0 0 0 ... 0 0 0 ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... i 0 0 0 ... E 0 0 0 0 ... 0 0 0 j 0 0 0 ... E 0 0 0 ... 0 0 0 k 0 0 0 ... 0 E 0 0 ... 0 0 0 k 0 0 0 ... 0 0 E 0 ... 0 0 0 j 0 0 0 ... 0 0 0 E ... 0 0 0 i 0 0 0 ... 0 0 0 0 E ... 0 0 0 ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... c 0 0 0 ... 0 0 0 0 0 0 ... E 0 b 0 0 0 ... 0 0 0 0 0 0 ... E a 0 0 0 ... 0 0 0 0 0 0 ... 0 E
The energies of the filled orbitals are +1.98137, +1.92583, +1.83442, +1.70884, + 1.55142, + 1.36511, + 1.15336, + 0.92013, + 0.66976, + 0.40691, and + 0.13648. The energy is 27.30729. (d) The ground state of the molecule has Ag symmetry by virtue of the fact that its wavefunction is the product of doubly occupied orbitals, and the product of any two orbitals of the same symmetry has Ag character. If a transition is to be allowed, the transition dipole must be nonzero, which in turn can only happen if the representation of the product f i includes the totally symmetric species Ag . Consider first transitions to another Ag wavefunction, in which case we need the product Ag Ag . Now Ag Ag = Ag , and the only character that returns Ag when multiplied by Ag is Ag itself. No component of the dipole operator belongs to species Ag , so no Ag Ag transitions are allowed. (Note: such transitions are transitions from an orbital occupied in the ground state to an excitedstate orbital of the same symmetry.) The other possibility is a transition from an orbital of one symmetry (Au or Bg ) to the other; in that case, the excitedstate wavefunction will have symmetry of Au Bg = Bu from the two singly occupied orbitals in the excited state. The symmetry of the transition dipole, then, is Ag Bu = Bu , and the only species that yields Ag when multiplied by Bu is Bu itself. The x and y components of the dipole operator belongs to species Bu , so these transitions are allowed.
16
Spectroscopy 1: rotational and vibrational spectroscopy
Solutions to exercises
Discussion questions
E16.1(b) (1) Doppler broadening. This contribution to the linewidth is due to the Doppler effect which shifts the frequency of the radiation emitted or absorbed when the atoms or molecules involved are moving towards or away from the detecting device. Molecules have a wide range of speeds in all directions in a gas and the detected spectral line is the absorption or emission profile arising from all the resulting Doppler shifts. As shown in Justification 16.3, the profile reflects the distribution of molecular velocities parallel to the line of sight which is a bellshaped Gaussian curve. (2) Lifetime broadening. The Doppler broadening is significant in gas phase samples, but lifetime broadening occurs in all states of matter. This kind of broadening is a quantum mechanical effect related to the uncertainty principle in the form of eqn 16.25 and is due to the finite lifetimes of the states involved in the transition. When is finite, the energy of the states is smeared out and hence the transition frequency is broadened as shown in eqn 16.26. (3) Pressure broadening or collisional broadening. The actual mechanism affecting the lifetime of energy states depends on various processes one of which is collisional deactivation and another is spontaneous emission. The first of these contributions can be reduced by lowering the pressure, the second cannot be changed and results in a natural linewidth. E16.2(b) (1) Rotational Raman spectroscopy. The gross selection rule is that the molecule must be anisotropically polarizable, which is to say that its polarizability, , depends upon the direction of the electric field relative to the molecule. Nonspherical rotors satisfy this condition. Therefore, linear and symmetric rotors are rotationally Raman active. (2) Vibrational Raman spectroscopy. The gross selection rule is that the polarizability of the molecule must change as the molecule vibrates. All diatomic molecules satisfy this condition as the molecules swell and contract during a vibration, the control of the nuclei over the electrons varies, and the molecular polarizability changes. Hence both homonuclear and heteronuclear diatomics are vibrationally Raman active. In polyatomic molecules it is usually quite difficult to judge by inspection whether or not the molecule is anisotropically polarizable; hence group theoretical methods are relied on for judging the Raman activity of the various normal modes of vibration. The procedure is discussed in Section 16.17(b) and demonstrated in Illustration 16.7. E16.3(b) The exclusion rule applies to the benzene molecule because it has a center of symmetry. Consequently, none of the normal modes of vibration of benzene can be both infrared and Raman active. If we wish to characterize all the normal modes we must obtain both kinds of spectra. See the solutions to Exercises 16.29(a) and 16.29(b) for specified illustrations of which modes are IR active and which are Raman active.
Numerical exercises
E16.4(b) The ratio of coefficients A/B is (a) 8h 3 8(6.626 1034 J s) (500 106 s1 )3 A = = = 7.73 1032 J m3 s B c3 (2.998 108 m s1 )3
254
INSTRUCTOR'S MANUAL
(b) The frequency is = E16.5(b) c so A 8h 8(6.626 1034 J s) = 3 = = 6.2 1028 J m3 s B (3.0 102 m)3
A source approaching an observer appears to be emitting light of frequency approaching = Since s [16.22, Section 16.3] 1 c
1 s , obs = 1  c For the light to appear green the speed would have to be s = 1 obs c = (2.998 108 m s1 ) 1  520 nm 660 nm = 6.36 107 m s1
or about 1.4 108 m.p.h. (Since s c, the relativistic expression obs = 1+ 1
s c s c 1/2
should really be used. It gives s = 7.02 107 m s1 .) E16.6(b) The linewidth is related to the lifetime by = ~ 5.31 cm1 [16.26] /ps so = 5.31 cm1 ps ~
(a) We are given a frequency rather than a wavenumber = /c ~ or 1.59 ns (b) E16.7(b) = 5.31 cm1 ps = 2.48 ps 2.14 cm1 (5.31 cm1 )c /ps so = (5.31 cm1 ) (2.998 1010 cm s1 ) ps = 1.59 103 ps 100 106 s1
The linewidth is related to the lifetime by = ~ 5.31 cm1 /ps so =
(a) If every collision is effective, then the lifetime is 1/(1.0109 s1 ) = 1.0109 s = 1.0103 ps = ~ (5.31 cm1 ) (2.998 1010 cm s1 ) = 1.6 108 s1 = 160 MHz 1.0 103
(b) If only one collision in 10 is effective, then the lifetime is a factor of 10 greater, 1.0 104 ps = ~ (5.31 cm1 ) (2.998 1010 cm s1 ) = 1.6 107 s1 = 16 MHz 1.0 104
SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
255
E16.8(b)
The frequency of the transition is related to the rotational constant by h = E = hc F = hcB[J (J + 1)  (J  1)J ] = 2hcBJ
where J refers to the upper state (J = 3). The rotational constant is related to molecular structure by B= h h = 4cI 4cmeff R 2
where I is moment of inertia, meff is effective mass, and R is the bond length. Putting these expressions together yields = 2cBJ = hJ 2meff R 2
The reciprocal of the effective mass is m1 = m1 + m1 = C O eff So = E16.9(b) (12 u)1 + (15.9949 u)1 1.66054 1027 kg u1 = 8.78348 1025 kg1
(8.78348 1025 kg1 ) (1.0546 1034 J s) (3) = 3.4754 1011 s1 2(112.81 1012 m)2 (a) The wavenumber of the transition is related to the rotational constant by hc = ~ E = hc F = hcB[J (J + 1)  (J  1)J ] = 2hcBJ
where J refers to the upper state (J = 1). The rotational constant is related to molecular structure by B= h 4cI
where I is moment of inertia. Putting these expressions together yields = 2BJ = ~ hJ 2cI so I= hJ (1.0546 1034 J s) (1) = c ~ 2(2.998 1010 cm s1 ) (16.93 cm1 )
I = 3.307 1047 kg m2 (b) The moment of inertia is related to the bond length by I = meff R 2 so R=
1/2 I meff
m1 = m1 + m1 = H Br eff
(1.0078 u)1 + (80.9163 u)1 1.66054 1027 kg u1
= 6.0494 1026 kg1
1/2
and R = (6.0494 1026 kg1 ) (3.307 1047 kg m2 ) = 1.414 1010 m = 141.4 pm
256
INSTRUCTOR'S MANUAL
E16.10(b) The wavenumber of the transition is related to the rotational constant by hc = ~ E = hc F = hcB[J (J + 1)  (J  1)J ] = 2hcBJ
where J refers to the upper state. So wavenumbers of adjacent transitions (transitions whose upper states differ by 1) differ by = 2B = ~ h 2cI so I= h 2 c ~
where I is moment of inertia, meff is effective mass, and R is the bond length. So I = (1.0546 1034 J s) = 5.420 1046 kg m2 2(2.9979 1010 cm s1 ) (1.033 cm1 ) The moment of inertia is related to the bond length by I = meff R 2 so R =
1/2 I meff
m1 = m1 + m1 = F eff Cl
(18.9984 u)1 + (34.9688 u)1 1.66054 1027 kg u1
= 4.89196 1025 kg1
1/2
and R = (4.89196 1025 kg1 ) (5.420 1046 kg m2 ) = 1.628 1010 m = 162.8 pm E16.11(b) The rotational constant is B= h h = 4cI 4c(2mO R 2 ) so R=
1/2 h 8 cmO B
where I is moment of inertia, meff is effective mass, and R is the bond length. R= (1.0546 1034 J s) 8(2.9979 1010 cm s1 ) (15.9949 u) (1.66054 1027 kg u1 ) (0.39021)
1/2
= 1.1621 1010 m = 116.21 pm E16.12(b) This exercise is analogous to Exercise 16.12(a), but here our solution will employ a slightly different algebraic technique. Let R = ROC , R = RCS , O = 16 O, C = 12 C. I= h [Footnote 6, p. 466] 4B 1.05457 1034 J s = 1.3799 1045 kg m2 = 8.3101 1019 u m2 (4) (6.0815 109 s1 ) 1.05457 1034 J s = 1.4145 1045 kg m2 = 8.5184 1019 u m2 (4) (5.9328 109 s1 )
I (OC32 S) = I (OC34 S) =
The expression for the moment of inertia given in Table 16.1 may be rearranged as follows. I m = mA mR 2 + mC mR 2  (mA R  mC R )2 = mA mR 2 + mC mR 2  m2 R 2 + 2mA mC RR  m2 R 2 A C = mA (mB + mC )R 2 + mC (mA + mB )R 2 + 2mA mC RR
SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
257
Let mC = m32 S and mC = m34 S Im mA = (mB + mC )R 2 + (mA + mB )R 2 + 2mA RR mC mC Im mA = (mB + mC )R 2 + (mA + mB )R 2 + 2mA RR mC mC Subtracting Im I m  = mC mC Solving for R 2 R =
2 Im mC mA mC
(a) (b)
mA mC
(mB + mC ) 
mA mC
(mB + mC ) R 2
 Imm
C
(mB + mC ) 
mA mC
(mB + mC )
=
mC I m  mC I m mB mA (mC  mC )
Substituting the masses, with mA = mO , mB = mC , mC = m32 S , and mC = m34 S m = (15.9949 + 12.0000 + 31.9721) u = 59.9670 u m = (15.9949 + 12.0000 + 33.9679) u = 61.9628 u R2 = (33.9679 u) (8.3101 1019 u m2 ) (59.9670 u) (12.000 u) (15.9949 u) (33.9679 u  31.9721 u)  = (31.9721 u) (8.5184 1019 u m2 ) (61.9628 u) (12.000 u) (15.9949 u) (33.9679 u  31.9721 u)
51.6446 1019 m2 = 1.3482 1020 m2 383.071
R = 1.1611 1010 m = 116.1 pm = ROC Because the numerator of the expression for R 2 involves the difference between two rather large numbers of nearly the same magnitude, the number of significant figures in the answer for R is certainly no greater than 4. Having solved for R, either equation (a) or (b) above can be solved for R . The result is R = 1.559 1010 m = 155.9 pm = RCS E16.13(b) The wavenumber of a Stokes line in rotational Raman is Stokes = i  2B(2J + 3) [16.49a] ~ ~ where J is the initial (lower) rotational state. So Stokes = 20 623 cm1  2(1.4457 cm1 ) [2(2) + 3] = 20 603 cm1 ~
1 E16.14(b) The separation of lines is 4B, so B = 4 (3.5312 cm1 ) = 0.88280 cm1 1/2 h Then we use R = [Exercise16.11(a)] 4meff cB
258
INSTRUCTOR'S MANUAL
1 1 with meff = 2 m(19 F) = 2 (18.9984 u) (1.6605 1027 kg u1 ) = 1.577 342 1026 kg
R =
1.0546 1034 J s 26 kg) (2.998 1010 cm s1 ) (0.88280 cm 1 ) 4(1.577 342 10
1/2
= 1.41785 1010 m = 141.78 pm E16.15(b) Polar molecules show a pure rotational absorption spectrum. Therefore, select the polar molecules based on their wellknown structures. Alternatively, determine the point groups of the molecules and use the rule that only molecules belonging to Cn , Cnv , and Cs may be polar, and in the case of Cn and Cnv , that dipole must lie along the rotation axis. Hence all are polar molecules. Their point group symmetries are (a) H2 O, C2v , (b) H2 O2 , C2 , (c) NH3 , C3v , (d) N2 O, Cv
All show a pure rotational spectrum. E16.16(b) A molecule must be anisotropically polarizable to show a rotational Raman spectrum; all molecules except spherical rotors have this property. So CH2 Cl2 , CH3 CH3 , and N2 O can display rotational Raman spectra; SF6 cannot. E16.17(b) The angular frequency is = k 1/2 = 2 m so k = (2 )2 m = (2 )2 (3.0 s1 )2 (2.0 103 kg)
k = 0.71 N m1 k 1/2 meff k meff
1/2
E16.18(b)
=
=
[prime = 2 H37 Cl]
The force constant, k, is assumed to be the same for both molecules. The fractional difference is  =  =
k meff 1/2
 mk eff
1/2
1/2
k meff
=
1 meff
1/2
 m1 eff
1/2
1/2
1 meff
=
meff meff
1/2
1
meff meff
1/2
1=
mH mCl (m2 H + m37 Cl ) 1/2 1 mH + mCl (m2 H m37 Cl ) (1.0078 u) (34.9688 u) (2.0140 u) + (36.9651 u) 1/2 1 (1.0078 u) + (34.9688 u) (2.0140 u) (36.9651 u)
=
= 0.284
Thus the difference is 28.4 per cent
SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
259
E16.19(b) The fundamental vibrational frequency is = k 1/2 = 2 = 2 c ~ meff so k = (2 c )2 meff ~
We need the effective mass m1 = m1 + m1 = (78.9183 u)1 + (80.9163 u)1 = 0.025 029 8 u1 1 2 eff k= [2(2.998 1010 cm s1 ) (323.2 cm1 )]2 (1.66054 1027 kg u1 ) 0.025 029 8 u1
= 245.9 N m1 E16.20(b) The ratio of the population of the ground state (N0 ) to the first excited state (N1 ) is h N0 = exp N1 kT (a) = exp hc ~ kT = 0.212 = 0.561
(6.626 1034 J s) (2.998 1010 cm s1 ) (321 cm1 ) N0 = exp N1 (1.381 1023 J K1 ) (298 K) N0 (6.626 1034 J s) (2.998 1010 cm s1 ) (321 cm1 ) = exp N1 (1.381 1023 J K1 ) (800 K)
(b)
E16.21(b) The relation between vibrational frequency and wavenumber is k 1/2 = 2 = 2 c ~ meff 1 = ~ 2 c km1 k 1/2 eff = meff 2 c
1/2
=
so
The reduced masses of the hydrogen halides are very similar, but not identical m1 = m1 + m1 D X eff We assume that the force constants as calculated in Exercise 16.21(a) are identical for the deuterium halide and the hydrogen halide. For DF m1 = eff (2.0140 u)1 + (18.9984 u)1 1.66054 1027 kg u1 = 3.3071 1026 kg1
1/2
= ~ For DCl
(3.3071 1026 kg1 ) (967.04 kg s2 ) 2(2.9979 1010 cm s1 ) (2.0140 u)1 + (34.9688 u)1 1.66054 1027 kg u1
= 3002.3 cm1
m1 = eff
= 3.1624 1026 kg1
1/2
= ~
(3.1624 1026 kg1 ) (515.59 kg s2 ) 2(2.9979 1010 cm s1 )
= 2143.7 cm1
260
INSTRUCTOR'S MANUAL
For DBr m1 = eff (2.0140 u)1 + (80.9163 u)1 1.66054 1027 kg u1 = 3.0646 1026 kg1
1/2
= ~ For DI
(3.0646 1026 kg1 ) (411.75 kg s2 ) 2(2.9979 1010 cm s1 ) (2.0140 u)1 + (126.9045 u)1 1.66054 1027 kg u1
= 1885.8 cm1
m1 = eff
= 3.0376 1026 kg1
1/2
= ~
(3.0376 1026 kg1 ) (314.21 kg s2 ) 2(2.9979 1010 cm s1 )
= 1640.1 cm1
E16.22(b) Data on three transitions are provided. Only two are necessary to obtain the value of and xe . The ~ third datum can then be used to check the accuracy of the calculated values. G(v = 1 0) =  2 xe = 2345.15 cm1 [16.64] ~ ~ G(v = 2 0) = 2  6~ xe = 4661.40 cm1 [16.65] ~ Multiply the first equation by 3, then subtract the second. = (3) (2345.15 cm1 )  (4661.40 cm1 ) = 2374.05 cm1 ~ Then from the first equation xe =  2345.15 cm1 ~ (2374.05  2345.15) cm1 = = 6.087 103 2 ~ (2) (2374.05 cm1 )
~ xe data are usually reported as xe which is xe = 14.45 cm1 ~ G(v = 3 0) = 3~  12xe = (3) (2374.05 cm1 )  (12) (14.45 cm1 ) = 6948.74 cm1 which is close to the experimental value. E16.23(b) Gv+1/2 =  2(v + 1)xe [16.64] ~ ~ Therefore, since Gv+1/2 = (1  2xe )~  2vxe ~ a plot of Gv+1/2 against v should give a straight line which gives (1  2xe )~ from the intercept at v = 0 and 2xe from the slope. We draw up the following table ~
v G(v)/cm Gv+1/2 /cm1
1
where
Gv+1/2 = G(v + 1)  G(v)
0 1144.83 2230.07
1 3374.90 2150.61
2 5525.51 2071.15
3 7596.66 1991.69
4 9588.35
SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
261
2200
2100
2000
0
1
2
3
Figure 16.1 The points are plotted in Fig. 16.1. The intercept lies at 2230.51 and the slope = 79.65 cm1 ; hence xe = 39.83 cm1 . ~ Since  2xe = 2230.51 cm1 , it follows that = 2310.16 cm1 . ~ ~ ~ The dissociation energy may be obtained by assuming that the molecule is described by a Morse potential and that the constant De in the expression for the potential is an adequate first approximation for it. Then De = 2 ~ (2310.16 cm1 )2 ~ = 33.50 103 cm1 = 4.15 eV [16.62] = = 4xe 4xe ~ (4) (39.83 cm1 )
However, the depth of the potential well De differs from D0 , the dissociation energy of the bond, by the zeropoint energy; hence
1 1 D0 = De  2 = (33.50 103 cm1 )  2 (2310.16 cm1 ) ~
= 3.235 104 cm1 = 4.01 eV E16.24(b) The wavenumber of an Rbranch IR transition is R = + 2B(J + 1) [16.69c] ~ ~ where J is the initial (lower) rotational state. So R = 2308.09 cm1 + 2(6.511 cm1 ) (2 + 1) = 2347.16 cm1 ~ E16.25(b) See Section 16.10. Select those molecules in which a vibration gives rise to a change in dipole moment. It is helpful to write down the structural formulas of the compounds. The infrared active compounds are (a) CH3 CH3 (b) CH4 (g) (c) CH3 Cl Comment. A more powerful method for determining infrared activity based on symmetry considerations is described in Section 16.15.
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INSTRUCTOR'S MANUAL
E16.26(b) A nonlinear molecule has 3N  6 normal modes of vibration, where N is the number of atoms in the molecule; a linear molecule has 3N  5. (a) C6 H6 has 3(12)  6 = 30 normal modes. (b) C6 H6 CH3 has 3(16)  6 = 42 normal modes. (c) HCC CCH is linear; it has 3(6)  5 = 13 normal modes. E16.27(b) (a) A planar AB3 molecule belongs to the D3h group. Its four atoms have a total of 12 displacements, of which 6 are vibrations. We determine the symmetry species of the vibrations by first determining the characters of the reducible representation of the molecule formed from all 12 displacements and then subtracting from these characters the characters corresponding to translation and rotation. This latter information is directly available in the character table for the group D3h . The resulting set of characters are the characters of the reducible representation of the vibrations. This representation can be reduced to the symmetry species of the vibrations by inspection or by use of the little orthogonality theorem.
D3h (translation) Unmoved atoms (total, product) (rotation) (vibration) E 3 4 12 3 6 h 1 4 4 1 4 2C3 0 1 0 0 0 2S3 2 1 2 2 2 3C2 1 2 2 1 0 3v 1 2 2 1 2
(vibration) corresponds to A1 + A2 + 2E . Again referring to the character table of D3h , we see that E corresponds to x and y, A2 to z; hence A2 and E are IR active. We also see from the character table that E and A1 correspond to the quadratic terms; hence A1 and E are Raman active . (b) A trigonal pyramidal AB3 molecule belongs to the group C3v . In a manner similar to the analysis in part (a) we obtain
C3v (total) (vibration) E 12 6 2C3 0 2 3v 2 2
(vibration) corresponds to 2A1 + 2E. We see from the character table that A1 and E are IR active and that A1 + E are also Raman active. Thus all modes are observable in both the IR and the Raman spectra. E16.28(b) (b) The boatlike bending of a benzene ring clearly changes the dipole moment of the ring, for the moving of the C H bonds out of the plane will give rise to a noncancelling component of their dipole moments. So the vibration is IR active . (a) Since benzene has a centre of inversion, the exclusion rule applies: a mode which is IR active (such as this one) must be Raman inactive . E16.29(b) The displacements span A1g + A1u + A2g + 2E1u + E1g . The rotations Rx and Ry span E1g , and the translations span E1u + A1u . So the vibrations span A1g + A2g + E1u
SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
263
Solutions to problems
Solutions to numerical problems
P16.1 Use the energy density expression in terms of wavelengths (eqn 11.5) E = d Evaluate
700109 m
8 hc where = 5 hc/kT . (e  1)
E=
400109 m
8hc d 5 (ehc/kT  1)
at three different temperatures. Compare those results to the classical, RayleighJeans expression (eqn 11.3): Eclass = class d where class = 8 kT , 4
700109 m
so
Eclass =
400109 m
8kT 8 kT 700109 m d =  . 4 33 400109 m
T /K (a) 1500 (b) 2500 (c) 5800
E/J m3 2.136 106 9.884 104 3.151 101
Eclass /J m3 2.206 3.676 8.528
The classical values are very different from the accurate Planck values! Try integrating the expressions over 400700 m or mm to see that the expressions agree reasonably well at longer wavelengths. P16.3 On the assumption that every collision deactivates the molecule we may write = 1 kT m 1/2 = 4p kT z
For HCl, with m 36 u, (1.381 1023 J K1 ) (298 K) (4) (0.30 1018 m2 ) (1.013 105 Pa) (36) (1.661 1027 kg) (1.381 1023 J K1 ) (298 K)
1/2
2.3 1010 s h E h = [24]
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INSTRUCTOR'S MANUAL
The width of the collisionbroadened line is therefore approximately 1 1 700 MHz = 2 (2) (2.3 1010 s)
The Doppler width is approximately 1.3 MHz (Problem 16.2). Since the collision width is proportional 1.3 = 0.002 before to p [ 1/ and 1/p], the pressure must be reduced by a factor of about 700 Doppler broadening begins to dominate collision broadening. Hence, the pressure must be reduced to below (0.002) (760 Torr) = 1 Torr P16.5 B= h [16.31]; 4cI mC m O = mC + m O I = meff R 2 ; R2 = h 4 cmeff B (1.66054 1027 kg u1 )
meff =
(12.0000 u) (15.9949 u) (12.0000 u) + (15.9949 u)
= 1.13852 1026 kg h = 2.79932 1044 kg m 4c
2 R0 =
2.79932 1044 kg m = 1.27303 1020 m2 (1.13852 1026 kg) (1.9314 102 m1 ) 2.79932 1044 kg m = 1.52565 1020 m2 (1.13852 1026 kg) (1.6116 102 m1 )
R0 = 1.1283 1010 m = 112.83 pm
2 R1 =
R1 = 1.2352 1010 m = 123.52 pm Comment. The change in internuclear distance is roughly 10 per cent, indicating that the rotations and vibrations of molecules are strongly coupled and that it is an oversimplification to consider them independently of each other. P16.8 = 2B(J + 1)[16.44] = 2B ~ Hence, B(1 HCl) = 10.4392 cm1 , B(2 HCl) = 5.3920 cm1 B= h [30] I = meff R 2 [Table 16.1] 4cI h h R2 = = 2.79927 1044 kg m 4cmeff B 4c (1.007825 u) (34.96885 u) meff (HCl) = (1.66054 1027 kg u1 ) (1.007825 u) + (34.96885 u) = 1.62665 1027 kg (2.0140 u) (34.96885 u) meff (DCl) = (2.0140 u) + (34.96885 u) = 3.1622 1027 kg R 2 (HCl) = 2.79927 1044 kg m = 1.64848 1020 m2 (1.62665 1027 kg) (1.04392 103 m1 ) (1.66054 1027 kg u1 )
R(HCl) = 1.28393 1010 m = 128.393 pm
SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
265
R 2 (2 HCl) =
2.79927 1044 kg m = 1.6417 1020 m2 (3.1622 1027 kg) (5.3920 102 m1 )
R(2 HCl) = 1.2813 1010 m = 128.13 pm The difference between these values of R is small but measurable. Comment. Since the effects of centrifugal distortion have not been taken into account, the number of significant figures in the calculated values of R above should be no greater than 4, despite the fact that the data is precise to 6 figures. P16.10 From the equation for a linear rotor in Table 16.1 it is possible to show that Im = ma mc (R + R )2 + ma mb R 2 + mb mc R 2 . Thus, I (16 O12 C32 S) = I (16 O12 C34 S) = m(16 O)m(32 S) m(16 O12 C32 S (R + R )2 + m(12 C){m(16 O)R 2 + m(32 S)R 2 } m(16 O12 C32 S)
m(16 O)m(34 S) m(16 O12 C34 S
(R + R )2 +
m(12 C){m(16 O)R 2 + m(34 S)R 2 } m(16 O12 C34 S)
m(16 O) = 15.9949 u, m(12 C) = 12.0000 u, m(32 S) = 31.9721 u, and m(34 S) = 33.9679 u. Hence, I (16 O12 C32 S)/u = (8.5279) (R + R )2 + (0.20011) (15.9949R 2 + 31.9721R 2 ) I (16 O12 C34 S)/u = (8.7684) (R + R )2 + (0.19366) (15.9949R 2 + 33.9679R 2 ) The spectral data provides the experimental values of the moments of inertia based on the relation h [16.31]. These values are set equal to the above equations = 2c B(J + 1) [16.44] with B = 4 cI which are then solved for R and R . The mean values of I obtained from the data are I (16 O12 C32 S) = 1.37998 1045 kg m2 I (16 O12 C34 S) = 1.41460 1045 kg m2 Therefore, after conversion of the atomic mass units to kg, the equations we must solve are 1.37998 1045 m2 = (1.4161 1026 ) (R + R )2 + (5.3150 1027 R 2 ) +(1.0624 1026 R 2 ) 1.41460 1045 m2 = (1.4560 1026 ) (R + R )2 + (5.1437 1027 R 2 ) +(1.0923 1026 R 2 ) These two equations may be solved for R and R . They are tedious to solve, but straightforward. Exercise 16.6(b) illustrates the details of the solution. The outcome is R = 116.28 pm and R = 155.97 pm . These values may be checked by direct substitution into the equations. Comment. The starting point of this problem is the actual experimental data on spectral line positions. Exercise 16.12(b) is similar to this problem; its starting points is, however, given values of the rotational constants B, which were themselves obtained from the spectral line positions. So the results for R and R are expected to be essentially identical and they are. Question. What are the rotational constants calculated from the data on the positions of the absorption lines?
266
INSTRUCTOR'S MANUAL
P16.12
The wavenumbers of the transitions with ~ ~ Gv+1/2 =  2(v + 1)xe [16.64] A plot of 2xe . ~
v = +1 are and De = 2 ~ [16.62] 4xe ~
~ Gv+1/2 against v + 1 should give a straight line with intercept at v + 1 = 0 and slope
Draw up the following table
v+1 Gv+1/2 /cm
1
1 284.50
2 283.00
3 281.50
The points are plotted in Fig. 16.2.
286
285
284
283
282
281 0 1 2 3 4
Figure 16.2 ~ The intercept is at 286.0, so = 286 cm1 . The slope is 1.50, so xe = 0.750 cm1 . It follows ~ that De = (286 cm1 )2 = 27300 cm1 , (4) (0.750 cm1 ) or 3.38 eV
The zeropoint level lies at 142.81 cm1 and so D0 = 3.36 eV . Since meff = (22.99) (126.90) u = 19.464 u (22.99) + (126.90)
the force constant of the molecule is ~ k = 4 2 meff c2 2 [Exercise 16.19(a)] = (4 2 ) (19.464) (1.6605 1027 kg) [(2.998 1010 cm s1 ) (286 cm1 )]2 = 93.8 N m1
SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
267
P16.14
The set of peaks to the left of center are the P branch, those to the right are the R branch. Within the rigid rotor approximation the two sets are separated by 4B. The effects of the interactions between vibration and rotation and of centrifugal distortion are least important for transitions with small J values hence the separation between the peaks immediately to the left and right of center will give good approximate values of B and bond length. (a) ~ Q (J ) = [46 b] = 2143.26 cm1 ~ NA hc (1071.63 cm1 ) = NA hc (1.07163 105 m1 ) = 1.28195 104 J mol1 = 12.8195 kJ mol1 (c) k = 4 2 c2 2 ~ (12 C16 O) = mC mO = mC + m O (12.0000 u) (15.9949 u) (12.0000 u) + (15.9949 u) (1.66054 1027 kg u1 )
1 ~ (b) The zeropoint energy is 2 = 1071.63 cm1 . The molar zeropoint energy in J mol1 is
= 1.13852 1026 kg k = 4 2 c2 (1.13852 1026 kg) (2.14326 105 m1 )2 = 1.85563 103 N m1 (d) 4B 7.655 cm1 B 1.91 cm1 [4 significant figures not justified] h h [Table 16.1] [16.31] = B= 4cI 4 cR 2 R2 h h = = 1.287 1020 m2 4cB (4c) (1.13852 1026 kg) (191 m1 )
(e)
R = 1.13 1010 m = 113 pm P16.15 D0 = De  ~ (a)
1 1 with = 2  4 xe [Section 16.11] ~ ~ 1 ~
1 HCl: = (1494.9)  4 (52.05) , cm1 = 1481.8 cm1 , ~
or
0.184 eV
Hence, D0 = 5.33  0.18 = 5.15 eV 2meff xe 1 2 ~ = a 2 [16.62], so xe (b) 2 HCl: ~ as a is a constant. We also have De = h meff 4xe ~ 1 1 2 [Exercise 16.23(a)]; so ~ , implying 1/2 . Reduced masses were calculated in ~ meff meff Exercises 16.21(a) and 16.21(b), and we can write ( HCl) = ~ xe (2 HCl) = ~
2
meff (1 HCl) meff (2 HCl)
1/2
(1 HCl) = (0.7172) (2989.7 cm1 ) = 2144.2 cm1 ~ ~ xe (1 HCl) = (0.5144) (52.05 cm1 ) = 26.77 cm1 0.132 eV
meff (1 HCl) meff (2 HCl)
1 1 (2 HCl) = 2 (2144.2)  4 (26.77 cm1 ) = 1065.4 cm1 , ~
Hence, D0 (2 HCl) = (5.33  0.132) eV = 5.20 eV
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INSTRUCTOR'S MANUAL
P16.19
(a) Vibrational wavenumbers (~ /cm1 ) computed by PC Spartan ProTM at several levels of theory are tabulated below, along with experimental values:
A1 Semiempirical PM3 SCF 6316G Density functional Experimental 412 592 502 525 A1 801 1359 1152 1151 B2 896 1569 1359 1336
The vibrational modes are shown graphically below.
A1
B2
Figure 16.3 (b) The wavenumbers computed by density functional theory agree quite well with experiment. Agreement of the semiempirical and SCF values with experiment is not so good. In this molecule, experimental wavenumbers can be correlated rather easily to computed vibrational modes even where the experimental and computed wavenumbers disagree substantially. Often, as in this case, computational methods that do a poor job of computing absolute transition wavenumbers still put transitions in proper order by wavenumber. That is, the modeling software systematically overestimates (as in this SCF computation) or underestimates (as in this semiempirical computation) the wavenumbers, thus keeping them in the correct order. Group theory is another aid in the assignment of tansitions: it can classify modes as forbidden, allowed only in particular polarizations, etc. Also, visual examination of the modes of motion can help to classify many modes as predominantly bondstretching, bondbending, or internal rotation; these different modes of vibration can be correlated to quite different ranges of wavenumbers (stretches highest, especially stretches involving hydrogen atoms, and internal rotations lowest.). P16.21 Summarize the six observed vibrations according to their wavenumbers (~ /cm1 ):
IR 870 Raman 877 1370 1408 2869 1435 3417 3407.
(a) If H2 O2 were linear, it would have 3N  5 = 7 vibrational modes. (b) Follow the flow chart in Fig. 15.14. Structure 2 is not linear, there is only one Cn axis (a C2 ), and there is a h ; the point group is C2h . Structure 3 is not linear, there is only one Cn axis (a C2 ),
SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
269
no h , but two v ; the point group is C2v . Structure 4 is not linear, there is only one Cn axis (a C2 ), no h , no v ; the point group is C2 . (c) The exclusion rule applies to structure 2 because it has a center of inversion: no vibrational modes can be both IR and Raman active. So structure 2 is inconsistent with observation. The vibrational modes of structure 3 span 3A1 +A2 +2B2 . (The full basis of 12 cartesian coordinates spans 4A1 + 2A2 + 2B1 + 4B2 ; remove translations and rotations.) The C2v character table says that five of these modes are IR active (3A1 + 2B2 ) and all are Raman active. All of the modes of structure 4 are both IR and Raman active. (A look at the character table shows that both symmetry species are IR and Raman active, so determining the symmetry species of the normal modes does not help here.) Both structures 3 and 4 have more active modes than were observed. This is consistent with the observations. After all, group theory can only tell us whether the transition moment must be zero by symmetry; it does not tell us whether the transition moment is sufficiently strong to be observed under experimental conditions.
Solutions to theoretical problems
P16.22 The centre of mass of a diatomic molecule lies at a distance x from atom A and is such that the masses on either side of it balance mA x = mB (R  x) and hence it is at mB x= R m = mA + mB m The moment of inertia of the molecule is I = mA x 2 + mB (R  x)2 [26] = m B m2 R 2 mA m2 R 2 mA mB 2 A B + = R m m2 m2 m A mB = meff R 2 since meff = mA + m B
P16.23
Because the centrifugal force and the restoring force balance, k(rc  re ) = 2 rc , we can solve for the distorted bond length as a function of the equilibrium bond length: rc = re 1  2 /k
Classically, then, the energy would be the rotational energy plus the energy of the stretched bond: E= J2 k(rc  re )2 J2 k 2 (rc  re )2 J2 (2 rc )2 + = + = + . 2I 2 2I 2k 2I 2k
How is the energy different form the rigidrotor energy? Besides the energy of stretching of the bond, 2 the larger moment of inertia alters the strictly rotational piece of the energy. Substitute rc for I and substitute for rc in terms of re throughtout: So E= 2 4 re 2 J 2 (1  2 /k)2 . + 2 2k(1  2 /k)2 2re
270
INSTRUCTOR'S MANUAL
Assuming that 2 /k is small (a reasonable assumption for most molecules), we can expand the expression and discard squares or higher powers of 2 /k: E
2 J 2 (1  22 /k) 2 4 re . + 2k 2re 2
(Note that the entire second term has a factor of 2 /k even before squaring and expanding the denominator, so we discard all terms of that expansion after the first.) Begin to clean up the expression by using classical definitions of angular momentum: J = I = r 2 so = J /re 2 ,
which allows us to substitute expressions involving J for all s: E J4 J4 J2 .  2 6 + 2re 2 re k 22 re 6 k
(At the same time, we have expanded the first term, part of which we can now combine with the last term.) Continue to clean up the expression by substituting I / for r 2 , and then carry the expression over to its quantum mechanical equivalent by substituting J (J + 1) 2 for J 2 : h E J 2 (J + 1)2 h4 J (J + 1) 2 h J 4 J2 .   3 E 3k 2I 2I 2I 2I k
Dividing by hc gives the rotational term, F (J ): F (J ) J 2 (J + 1)2 h4 J (J + 1) h J 2 (J + 1)2 h3 J (J + 1) 2 h   = , 2hcI 4 cI 2hcI 3 k 4 cI 3 k
where we have used h = h/2 to eliminate a common divisor of h. Now use the definition of the rotational constant, B= h 4cI F (J ) J (J + 1)B  J 2 (J + 1)2 B 3 16 2 c2 . k
Finally, use the relationship between the force constant and vibrational wavenumber: k 1/2 = vib = 2 = 2c ~ leaving F (J ) BJ (J + 1)  P16.26 so 1 = 2 c2 2 k 4 ~ where D = 4B 3 . 2 ~
4B 3 2 J (J + 1)2 = BJ (J + 1)  DJ 2 (J + 1)2 2 ~
1 ~ S(v, J ) = v + 2 + BJ (J + 1) [16.68] O ~ SJ =  2B(2J  1) S SJ
[ v = 1, J = 2]
= + 2B(2J + 3) [ v = 1, J = +2] ~
The transition of maximum intensity corresponds, approximately, to the transition with the most probable value of J, which was calculated in Problem 16.25 Jmax =
1/2 kT 1  2hcB 2
SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
271
The peaktopeak separation is then S =
S SJmax  O 1 SJmax = 2B(2Jmax + 3)  {2B(2Jmax  1)} = 8B Jmax + 2
= 8B
1/2 kT = 2hcB
32BkT 1/2 hc
To analyse the data we rearrange the relation to B= hc( S)2 32kT h , with I = 2mx R 2 (Table 16.1) for a linear rotor. This 4 cI 2kT 1/2 mx
and convert to a bond length using B = gives R=
1/2 h = 8cmx B
1 c S
We can now draw up the following table
HgCl2 T /K mx /u S/cm1 R/pm 555 35.45 23.8 227.6 HgBr 2 565 79.1 15.2 240.7 HgI2 565 126.90 11.4 253.4
Hence, the three bond lengths are approximately 230, 240, and 250 pm P16.28 The energy levels of a Morse oscillator, expressed as wavenumbers, are given by:
2 2 1 1 1 1 ~ ~ G() = + 2  + 2 xe = + 2  + 2 2/4De . ~ ~
States are bound only if the energy is less than the well depth, De , also expressed as a wavenumber: G() < De or
2 1 1 ~ + 2  + 2 2 /4De < De . ~
Solve for the maximum value of by making the inequality into an equality:
2 1 1 + 2 2 /4De  + 2 + De = 0. ~ ~
Multiplying through by 4De results in an expression that can be factored by inspection into:
2 1 ~ + 2  2De = 0
so
1 + 2 = 2De /~
and
= 2De /~  2 . 1
Of course, is an integer, so its maximum value is really the greatest integer less than this quantity.
272
INSTRUCTOR'S MANUAL
Solutions to applications
P16.29 (a) Resonance Raman spectroscopy is preferable to vibrational spectroscopy for studying the O O stretching mode because such a mode would be infrared inactive , or at best only weakly active. (The mode is sure to be inactive in free O2 , because it would not change the molecule's dipole moment. In a complex in which O2 is bound, the O O stretch may change the dipole moment, but it is not certain to do so at all, let alone strongly enough to provide a good signal.) (b) The vibrational wavenumber is proportional to the frequency, and it depends on the effective mass as follows, ~ k 1/2 , meff so (18 O2 ) ~ = (16 O2 ) ~ meff (16 O2 ) meff (18 O2 )
1/2
=
16.0 u 1/2 = 0.943, 18.0 u
and (18 O2 ) = (0.943)(844 cm1 ) = 796 cm1 . ~ Note the assumption that the effective masses are proportional to the isotopic masses. This assumption is valid in the free molecule, where the effective mass of O2 is equal to half the mass of the O atom; it is also valid if the O2 is strongly bound at one end, such that one atom is free and the other is essentially fixed to a very massive unit. (c) The vibrational wavenumber is proportional to the square root of the force constant. The force constant is itself a measure of the strength of the bond (technically of its stiffness, which correlates with strength), which in turn is characterized by bond order. Simple molecule orbital analysis of  O2 , O2 , and O2 2 results in bond orders of 2, 1.5, and 1 respectively . Given decreasing bond order, one would expect decreasing vibrational wavenumbers (and vice versa). (d) The wavenumber of the O O stretch is very similar to that of the peroxide anion, suggesting Fe3+ 2 O2 2 . (e) The detection of two bands due to 16 O18 O implies that the two O atoms occupy nonequivalent positions in the complex. Structures 7 and 8 are consistent with this observation, but structures 5 and 6 are not. P16.31 (a) The molar absorption coefficient (~ ) is given by (~ ) = A(~ ) RT A(~ ) = l[CO2 ] lxCO2 p (eqns 16.11, 1.15, and 1.18)

where T = 298 K, l = 10 cm, p = 1 bar, and xCO2 = 0.021. The absorption band originates with the 001 000 transition of the antisymmetric stretch vibrational mode at 2349 cm1 (Fig. 16.48). The band is very broad because of accompanying rotational transitions and lifetime broadening of each individual absorption (also called collisional broadening or pressure broadening, Section 16.3). The spectra reveals that the Q branch is missing so we conclude that the transition J = 0 is forbidden (Section 16.12) for the Dh point group of CO2 . The Pbranch ( J = 1) is evident at lower energies and the Rbranch ( J = +1) is evident at higher energies. 16 12 16 C O has two identical nuclei of zero spin so the CO2 wavefunction must be sym(b) O metric w/r/t nuclear interchange and it must obey BoseEinstein nuclear statistics (Section 16.8). Consequently, J takes on even values only for the = 0 vibrational state and odd values only for the = 1 state. The (, J ) states for this absorption band are (1, J + 1) (0, J ) for J = 0, 2, 4, . . . . According to eqn 16.68, the energy of the (0, J ) state is
1 S(0, J ) = 2 + BJ (J + 1),
SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
273
Carbon dioxide IR band 2
1.5 Absorption
1
0.5
0 2280
2300
2320
2340 Wavenumber / cm1
2360
2380
2400
Figure 16.4(a)
Molar absorption coefficient 20
15
m2 mol1
10
5
0 2280
2300
2320
2340 Wavenumber / cm1
2360
2380
2400
Figure 16.4(b)
where = 2349 cm I =
1
2(0.01600 kg mol1 )(116.2 1012 m)2 2mO R 2 = NA 6.022 1023 mol1
B
= 7.175 1046 kg m2 (Table 16.1) h = (eqn 16.31) 8 2 cI = 6.626 1034 J s 8 2 (2.998 108 m s1 )(7.175 1046 kg m2 )
= 39.02 m1 = 0.3902 cm1 The transitions of the P and R branches occur at P =  2BJ ~ ~ [16.69b]
274
INSTRUCTOR'S MANUAL
and ~ R = + 2B(J + 1) ~ [16.69c] where J = 0, 2, 4, 6 . . . The highest energy transition of the P branch is at  4B; the lowest energy transition of the ~ R branch is at + 2B. Transitions are separated by 4B(1.5608 cm1 ) within each branch. The ~ probability of each transition is proportional to the lower state population, which we assume to be given by the Boltzman distribution with a degeneracy of 2J + 1. The transition probability is also proportional to both a nuclear degeneracy factor (eqn 16.50) and a transition dipole moment, which is approximately independent of J . The former factors are absorbed into the constant of proportionality. transition probability (2J + 1)eS(0,J )hc/kT A plot of the righthandside of this equation, Fig. 16.4(c), against J at 298 K indicates a maximum transition probability at Jmax = 16. We "normalize" the maximum in the predicted structure, and eliminate the constant of proportionality by examining the transition probability ratio: transition probability for J th state (2J + 1)eS(0,J )hc/kT = transition probability forJmax state 33eS(0,16)hc/RT = 2J + 1 (J 2 +J 272)Bhc/kT e 33
A plot Fig. 16.4(c) of the above ratio against predicted wavenumbers can be compared to the ratio A(~ )/Amax where Amax is the observed spectrum maximum (1.677). It shows a fair degree of agreement between the experimental and simple theoretical band shapes.
Simple theoretical and exp. spectra 1
0.8
0.6 A Amax 0.4
0.2
0 2300
2320
2340 /cm1
2360
2380
2400
Figure 16.4(c)
SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
275
0.8 Transmittance 0.6 0.4 0.2 20 2300 2350 /cm1 60 2400 40 h/m 0
Figure 16.4(d)
(c) Using the equations of justificant 16.1, we may write the relationship A = (~ )
h 0
[CO2 ] dh
The strong absorption of the band suggests that h should not be a very great length and that [CO2 ] should be constant between the Earth's surface and h. Consequently, the integration gives A = (~ )[CO2 ]h xCO2 p = (~ )h RT
Dalton's law of partial pressures
p and T are not expected to change much for modest values of h so we estimate that p = 1 bar and T = 288 K. (3.3 104 ) 1 105 Pa A = (~ )h 8.31451 J 1 mol1 (288 ) K K = (0.0138 m3 mol) (~ )h
3 Transmittance = 10A = 10 0.0138 m mol (~ )h
[16.10]
The transmittance surface plot clearly shows that before a height of about 30 m has been reached all of the Earth's IR radiation in the 2320 cm1  2380 cm1 range has been absorbed by atmospheric carbon dioxide. See C.A. Meserole, F.M. Mulcalry, J. Lutz, and H.A. Yousif, J. Chem. Ed., 74, 316 (1997).
276
INSTRUCTOR'S MANUAL
P16.34
+ (a) The H3 molecule is held together by a twoelectron, threecenter bond, and hence its structure is expected to be an equilateral triangle. Looking at Fig. 16.5 and using the Law of cosines 2 2 R 2 = 2RC  2RC cos(180  2) 2 2 = 2RC (1  cos(120 )) = 3RC
Therefore
2 IC = 3mRC = 3m(R/ 3)2 = mR 2 IB = 2mRB = 2m(R/2)2 = mR 2 /2 Therefore IC = 2IB
RC = R/ 3
{
(b) B = R = = 8.764 1011 m = 87.64 pm C = R = h h = [36] 4cIC 4cmR 2 = 8.986 1011 m = 89.86 pm
{
h 2 h h = = [16.37] 4cIB 4cmR 2 2 cmR 2
1/2 1/2 h hNA = 2cmB 2 cMH B 1/2 2 (1.0546 1034 J s) (6.0221 1023 mol1 ) 10cm m = 2(2.998 108 m s1 ) (0.001 008 kg mol1 ) (43.55 cm1 )
Alternatively the rotational constant C can be used to calculate R.
1/2 1/2 h hNA = 4cmC 4 cMH C 1/2 2 (1.0546 1034 J s) (6.0221 1023 mol1 ) 10cm m = 4(2.998 108 m s1 ) (0.001 008 kg mol1 ) (20.71 cm1 )
{
Figure 16.5
SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
277
The values of R calculated with either the rotational constant C or the rotational constant B differ slightly. We approximate the bond length as the average of these two. R (c) (87.64 + 89.86) pm = 88.7 pm 2
2
(1.0546 1034 J s) (6.0221 1023 mol1 ) 10cm m h = B = 2cmR 2 2(2.998 108 m s1 ) (0.001 008 kg mol1 ) (87.32 1012 m)2 = 43.87 cm1
1 C = 2 B = 21.93 cm1
(d)
1 3 = m meff
or
meff = 1 m 3
Since mD = 2mH , meff,D = 2mH /3 2 (D+ ) = ~ 3 = = meff (H3 ) 1/2 2 (H3 ) [57] ~ meff (D3 ) mH /3 1/2 2 (H2 ) ~ 2 (H3 ) = 1/2 ~ 2mH /3 2 2521.6 cm1 = 1783.0 cm1 21/2 1 , where m = mass of H or D m MH = 43.55 cm1 MD MH = 20.71 cm1 MD 1.008 2.014 1.008 2.014 = 21.80 cm1 = 10.37 cm1
Since B and C
+ B(D+ ) = B(H3 ) 3 + C(D+ ) = C(H3 ) 3
17
Spectroscopy 2: electronic transitions
Solutions to exercises
Discussion questions
E17.1(b) The FranckCondon principle states that because electrons are so much lighter than nuclei an electronic transition occurs so rapidly compared to vibrational motions that the internuclear distance is relatively unchanged as a result of the transition. This implies that the most probable transitions f i are vertical. This vertical line will, however, intersect any number of vibrational levels f in the upper electronic state. Hence transitions to many vibrational states of the excited state will occur with transition probabilities proportional to the FrankCondon factors which are in turn proportional to the overlap integral of the wavefunctions of the initial and final vibrational states. A vibrational progression is observed, the shape of which is determined by the relative horizontal positions of the two electronic potential energy curves. The most probable transitions are those to excited vibrational states with wavefunctions having a large amplitude at the internuclear position Re . Question. You might check the validity of the assumption that electronic transitions are so much faster than vibrational transitions by calculating the time scale of the two kinds of transitions. How much faster is the electronic transition, and is the assumption behind the FranckCondon principle justified? E17.2(b) Color can arise by emission, absorption, or scattering of electromagnetic radiation by an object. Many molecules have electronic transitions that have wavelengths in the visible portion of the electromagnetic spectrum. When a substance emits radiation the perceived color of the object will be that of the emitted radiation and it may be an additive color resulting from the emission of more than one wavelength of radiation. When a substance absorbs radiation its color is determined by the subtraction of those wavelengths from white light. For example, absorption of red light results in the object being perceived as green. Color may also be formed by scattering, including the diffraction that occurs when light falls on a material with a grid of variation in texture of refractive index having dimensions comparable to the wavelength of light, for example, a bird's plumage. The characteristics of fluorescence which are consistent with the accepted mechanism are: (1) it ceases as soon as the source of illumination is removed; (2) the time scale of fluorescence, 109 s, is typical of a process in which the rate determining step is a spontaneous radiative transition between states of the same multiplicity; slower than a stimulated transition, but faster than phosphorescence; (3) it occurs at longer wavelength (higher frequency) than the inducing radiation; (4) its vibrational structure is characteristic of that of a transition from the ground vibrational level of the excited electronic state to the vibrational levels of the ground electronic state; and (5), the observed shifting and in some instances quenching of the fluorescence spectrum by interactions with the solvent. See Table 17.4 for a summary of the characteristics of laser radiation that result in its many advantages for chemical and biochemical investigations. Two important applications of lasers in chemistry have been to Raman spectroscopy and to the development of time resolved spectroscopy. Prior to the invention of lasers the source of intense monochromatic radiation required for Raman spectroscopy was a large spiral discharge tube with liquid mercury electrodes. The intense heat generated by the large current required to produce the radiation had to be dissipated by clumsy water cooled jackets and exposures of several weeks were sometimes necessary to observe the weaker Raman lines. These problems have been eliminated with the introduction of lasers as the source of the required monochromatic radiation. As a consequence, Raman spectroscopy has been revitalized and is now almost as routine as infrared spectroscopy. See Section 17.7(b). Time resolved laser spectroscopy can
E17.3(b)
E17.4(b)
SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
279
be used to study the dynamics of chemical reactions. Laser pulses are used to obtain the absorption, emission, and Raman spectrum of reactants, intermediates, products, and even transition states of reactions. When we want to study the rates at which energy is transferred from one mode to another in a molecule, we need femotosecond and picosecond pulses. These time scales are available from modelocked lasers and their development has opened up the possibility of examining the details of chemical reactions at a level which would have been unimaginable before.
Numerical exercises
E17.5(b) To obtain the parities of Fig. 14.38 of the text we recognize that what is shown in the figure are the signs (light = positive, dark = negative) of the upper (positive zdirection) lobe of the pz orbitals. The lower lobes (not shown) have opposite signs. Inversion through the centre changes + to  for the pz lobes of a2 and e2 , but the e1 and b2 lobes do not change sign. Therefore a2 and e2 are u, e1 and b2 are g. According to Hund's rule, we expect one 1u electron and one 2g electron to be unpaired. Hence S = 1 and the multiplicity of the spectroscopic term is 3 . The overall parity is u g = u since (apart from the complete core), one electron occupies a u orbital another occupies a g orbital. Use the BeerLambert law log I = [J]l = (327 L mol1 cm1 ) (2.22 103 mol L1 ) (0.15 cm) I0 = 0.10889 I = 100.10889 = 0.778 I The reduction in intensity is 22.2 per cent E17.8(b) = I 1 log [16.9, 16.10] [J]l I0 1 = log 0.655 = 787 L mol1 cm1 (6.67 104 mol L1 ) (0.35 cm) = 787 dm3 mol1 cm1 = 787 103 cm3 mol1 cm1 = 7.9 105 cm2 mol1 E17.9(b) The BeerLambert law is log I = [J]l I0 so [J] = 1 (323 L mol1 cm1 (0.750 cm) I 1 log l I0 log(1  0.523) = 1.33 103 mol L1 [1 dm = 10 cm]
E17.6(b)
E17.7(b)
[J] =
E17.10(b) Note. A parabolic lineshape is symmetrical, extending an equal distance on either side of its peak. The given data are not consistent with a parabolic lineshape when plotted as a function of either wavelength or wavenumber, for the peak does not fall at the centre of either the wavelength or the wavenumber range. The exercise will be solved with the given data assuming a triangular lineshape as a function of wavenumber.
280
INSTRUCTOR'S MANUAL
The integrated absorption coefficient is the area under an absorption peak A= d~
If the peak is triangular, this area is
1 A = 2 (base) (height) 1 = 2 [(199 109 m)1  (275 109 m)1 ] (2.25 104 L mol1 cm1 )
= 1.56 1010 L m1 mol1 cm1 =
(1.56 109 L m1 mol1 cm1 ) (100 cm m1 ) 103 L m3
= 1.56 109 m mol1 = 1.56 108 L mol1 cm2 E17.11(b) Modelling the electrons of 1,3,5hexatriene as free electrons in a linear box yields nondegenerate energy levels of En = n2 h2 8me L2
The molecule has six electrons, so the lowestenergy transition is from n = 3 to n = 4. The length of the box is 5 times the C C bond distance R. So Elinear = (42  33 )h2 8me (5R)2
Modelling the electrons of benzene as free electrons on a ring of radius R yields energy levels of Eml = m2 h2 l 2I
where I is the moment of inertia: I = me R 2 . These energy levels are doubly degenerate, except for the nondegenerate ml = 0. The six electrons fill the ml = 0 and 1 levels, so the lowestenergy transition is from ml = 1 to ml = 2 Ering = h (22  12 ) 2 (22  12 )h2 = 2me R 2 8 2 me R 2
Comparing the two shows Elinear = 7 25 h2 8me R 2 < Ering = 3 2 h2 8me R 2
Therefore, the lowestenergy absorption will rise in energy. E17.12(b) The BeerLambert law is log I = [J]l = log T I0
so a plot (Fig. 17.1) of log T versus [J] should give a straight line through the origin with a slope m of l. So = m/ l.
SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
281
The data follow
[dye]/(mol L1 ) 0.0010 0.0050 0.0100 0.0500 T 0.73 0.21 0.042 1.33 107 log T 0.1367 0.6778 1.3768 6.8761
0
2
4
6
8 0.00
0.01
0.02
0.03
0.04
0.05
0.06
Figure 17.1 The molar absorptivity is = 138 L mol1 = 552 L mol1 cm1 0.250 cm
E17.13(b) The BeerLambert law is log T = [J]l = so = 1 log T [J]l
1 log 0.32 = 128 L mol1 cm1 (0.0155 mol L1 ) (0.250 cm)
Now that we have , we can compute T of this solution with any size of cell
1 1 1 T = 10[J]l = 10{(128 L mol cm )(0.0155 mol L )(0.450 cm)} = 0.13
E17.14(b) The BeerLambert law is log I = [J]l I0 so l =  1 (30 L mol1 cm1 ) (1.0 mol L1 ) 1 (30 L mol1 cm1 ) (1.0 mol L1 ) I 1 log [J] I0 log 1 = 0.020 cm 2
(a) l =  (b) l = 
log 0.10 = 0.033 cm
282
INSTRUCTOR'S MANUAL
E17.15(b) The integrated absorption coefficient is the area under an absorption peak A= d~
We are told that is a Gaussian function, i.e. a function of the form = max exp x 2 a2
where x =  max and a is a parameter related to the width of the peak. The integrated absorption ~ ~ coefficient, then, is A=

max exp
x 2 a2
dx = max a
We must relate a to the halfwidth at halfheight, x1/2
1 2 max
= max exp
2 x1/2
a2
so
ln
1 2
=
2 x1/2
a2
and
x1/2 a= ln 2
So A = max x1/2
1/2 1/2 = (1.54 104 L mol1 cm1 ) (4233 cm1 ) ln 2 ln 2
= 1.39 108 L mol1 cm2 In SI base units A= (1.39 108 L mol1 cm2 ) (1000 cm3 L1 ) 100 cm m1
= 1.39 109 m mol1
+ + E17.16(b) F2 is formed when F2 loses an antibonding electron, so we would expect F2 to have a shorter bond than F2 . The difference in equilibrium bond length between the ground state (F2 ) and excited state + (F2 + e ) of the photoionization experiment leads us to expect some vibrational excitation in the upper state. The vertical transition of the photoionization will leave the molecular ion with a stretched bond relative to its equilibrium bond length. A stretched bond means a vibrationally excited molecular ion, hence a stronger transition to a vibrationally excited state than to the vibrational ground state of the cation.
Solutions to problems
Solutions to numerical problems
P17.3 Initially we cannot decide whether the dissociation products are produced in their ground atomic states or excited states. But we note that the two convergence limits are separated by an amount of energy exactly equal to the excitation energy of the bromine atom: 18 345 cm1  14 660 cm1 = 3685 cm1 . Consequently, dissociation at 14 660 cm1 must yield bromine atoms in their ground state. Therefore, the possibilities for the dissociation energy are 14 660 cm1 or
SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
283
14 660 cm1  7598 cm1 = 7062 cm1 depending upon whether the iodine atoms produced are in their ground or excited electronic state. In order to decide which of these two possibilities is correct we can set up the following BornHaber cycle
(1) (2) (3) (4) (5) IBr(g) 1 I (s) 2 2
1 Br 2 (l) 2 1 I (g) 2 2 1 Br 2 (g) 2
1 I (g) 2 2 1 I (g) 2 2
1 + 2 Br 2 (l)
H =  1 1 H = 2 2 H = 3 H = 4  H5 = H
1 2 1 2 1 2
 f H (IBr, g)  sub H (I2 , s)  vap H (Br 2,
1 Br 2 (g) 2
l)
I(g) Br(g) I(g) + Br(g)
H (I I) H (Br Br)
IBr(g)
    1 1 H  =  f H  (IBr, g) + 2 sub H  (I2 , s) + 2 vap H  (Br 2 , l) 1 1 + 2 H (I I) + 2 H (Br Br) 1 1 1 1 = 40.79 + 2 62.44 + 2 30.907 + 2 151.24 + 2 192.85 kJ mol1
[Table 2.6 and data provided] = 177.93 kJ mol1 = 14 874 cm1 Comparison to the possibilities 14 660 cm1 and 7062 cm1 shows that it is the former that is the correct dissociation energy. P17.5
~ ~ We write = max ex = max e~ /2 the variable being and being a constant. is measured 1 from the band centre, at which = 0. = 2 max when 2 = 2 ln 2. Therefore, the width at ~ ~ halfheight is
2 2
1/2 = 2 (2 ln 2)1/2 , ~
implying that
=
1/2 ~2 8 ln 2
Now we carry out the intregration A= d~ = max 2 1/2 ~2 8 ln 2
 2 e~ /2 d~ = max (2 )1/2 
ex dx = 1/2
2
1/2
= max
=
1/2 max 1/2 = 1.0645max 1/2 ~ ~ 4 ln 2
~ ~ ~ A = 1.0645max 1/2 , with centred on 0 Since = ~ 1 , 1/2 ~ 1/2 2 0 1/2 2 0 1/2 = 38 nm with 0 = 290 nm and max [ 0 ]
A = 1.0645max
From Fig. 17.52 of the text, we find 235 L mol1 cm1 ; hence
284
INSTRUCTOR'S MANUAL
A=
1.0645 (235 L mol1 cm1 ) (38 107 cm) = 1.1 106 L mol1 cm2 (290 107 cm)2
Since the dipole moment components transform as A1 (z), B1 (x), and B2 (y), excitations from A1 to A1 , B1 , and B2 terms are allowed. P17.8 Draw up a table like the following: Hydrocarbon hmax /eV Benzene 4.184 Biphenyl 3.654 Naphthalene 3.452 Phenanthrene 3.288 Pyrene 2.989 Anthracene 2.890
EHOMO /eV 9.7506 8.9169 8.8352 8.7397 8.2489 8.2477
Semiempirical, PM3 level, PC Spartan ProTM
Figure 17.2 shows a good correlation: r 2 = 0.972.
8.0 8.5 9.0 9.5 10.0 2.5 3.0 3.5 4.0 4.5
Figure 17.2
P17.11
Refer to Fig. 14.30 of the text. The lowest binding energy corresponds to the highest occupied orbital, the next lowest to next highest orbital, and so on. We draw up the following table N2 LineEK /eV 5.6 4.5 2.4 7.2 4.9 1.7 Binding energy/eV Assignment 15.6 3 16.7 1 18.8 2 14.0 3 16.3 1 19.5 2
CO
The spacing of the 4.5 eV lines in N2 is 0.24 eV, or about 1940 cm1 . The spacing of the 4.9 eV lines in CO is 0.23 eV, or about 1860 cm1 . These are estimates from the illustrations of the separation + of the vibrational levels of the N2 and CO+ ions in their excited states. P17.13 0.125 eV corresponds to 1010 cm1 , markedly less than the 1596 cm1 of the bending mode. This suggests that the ejected electron tended to bond between the two hydrogens of the water molecule.
SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
285
Solutions to theoretical problems
P17.14 We need to establish whether the transition dipole moments fi =
f i d
[16.20]
connecting the states 1 and 2 and the states 1 and 3 are zero or nonzero. The particle in a box 2 1/2 n x wavefunctions are n = [12.8] sin L L Thus 2,1 and 3,1 sin sin 2x L 3x L x sin x sin x dx L x dx L x cos x cos 3 x x  cos L L 2 x L  cos 4 x L dx dx
1 1 having used sin sin = 2 cos(  )  2 cos( + ). Both of these integrals can be evaluated using the standard form
1 x x(cos ax) dx = 2 cos ax + sin ax a a
L 0 0 L
x cos x cos
x x 1 cos dx = 2 L L
L
x x + sin 0 L L
L
L 0
= 2
L 0
L 2 =0 = 2 L 2 =0 3
3x L
dx =
1
3 L 2
cos
3 x L
L 0
+
x
3 L
sin
3 x L
Thus 2,1 = 0. In a similar manner 3,1 = 0. Comment. A general formula for fi applicable to all possible particle in a box transitions may be derived. The result is (n = f, m = i) eL cos(n  m)  1 cos(n + m)  1 nm =  2  (n  m)2 (n + m)2 For m and n both even or both odd numbers, nm = 0; if one is even and the other odd, nm = 0. See also Problem 17.18. Question. Can you establish the general relation for nm above? P17.16 We need to determine how the oscillator strength (Problem 17.17) depends on the length of the chain. We assume that wavefunctions of the conjugated electrons in the linear polyene can be approximated by the wavefunctions of a particle in a onedimensional box. Then f = 8 2 me 2 fi  [Problem 17.17] 3he2
L 0 n
x = e =
(x)x n (x) dx, sin
n
=
2 1/2 n x sin L L
2e L n x x sin L 0 L
n x dx L
286
INSTRUCTOR'S MANUAL
=
0 8eL + 2
n(n + 1) (2n + 1)2
if n = n + 2 if n = n + 1
The integral is standard, but may also be evaluated using 2 sin A sin B = cos(A  B)  cos(A + B) as in Problem 17.14 h = En+1  En = (2n + 1) h2 8me L2
Therefore, for the transition n + 1 n, f = 8 2 3 me he2 h 8me L2 (2n + 1) 8eL 2 n2 (n + 1)2 = 2 (2n + 1)4 64 3 2 n2 (n + 1)2 (2n + 1)3
n2 (n + 1)2 (2n + 1)3 The value of n depends on the number of bonds: each bond supplies two electrons and so n increases by 1. For large n, Therefore, f f n4 n 8 8n3 and f n
Therefore, for the longest wavelength transitions f increases as the chain length is increased. The 1 (2n + 1) ; but as n L, this energy is proportional to . energy of the transition is proportional to 2 L L n2 h2 (2n + 1)h2 [ n = +1] , E= Since En = 8me L2 8me L2 but L = 2nd is the length of the chain (Exercise 17.11(a)), with d the carboncarbon interatomic distance. Hence E=
L 2d
+ 1 h2
8me L2
h2 1 16me dL L
Therefore, the transition moves toward the red as L is increased and the apparent color of the dye shifts towards blue . P17.17 = e
v
x v dx
1x 0 dx
From Problem 12.15, 10 = e Hence, f =
= e
1/2 h 2(me k)1/2
1 8 2 me e2 h = 2 1/2 3 3he 2(me k)
2 =
k 1/2 me
P17.19
(a) Vibrational energy spacings of the lower state are determined by the spacing of the peaks of A. From the spectrum, 1800 cm1 . ~ (b) Nothing can be said about the spacing of the upper state levels (without a detailed analysis of the intensities of the lines). For the second part of the question, we note that after some vibrational
SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
287
decay the benzophenone (which does absorb near 360 nm) can transfer its energy to naphthalene. The latter then emits the energy radiatively. P17.21 (a) The BeerLambert Law is: A = log I0 = [J]l. I
The absorbed intensity is: Iabs = I0  I so I = I0  Iabs .
Substitute this expression into the BeerLambert law and solve for Iabs : log and I0 = [J]l I0  Iabs so I0  Iabs = I0 10[J]l ,
Iabs = I0 (1  10[J]l ) .
(b) The problem states that If (~ f ) is proportional to f and to Iabs (~ ), so: If (~ f ) f I0 (~ ) (1  10[J]l ). If the exponent is small, we can expand 1  10[J]l in a power series: 10[J]l = (eln 10 )[J]l 1  [J]l ln 10 + , and P17.22 If (~ f ) f I0 (~ )[J]l ln 10 .
Use the ClebschGordan series [Chapter 13] to compound the two resultant angular momenta, and impose the conservation of angular momentum on the composite system. (a) O2 has S = 1 [it is a spin triplet]. The configuration of an O atom is [He]2s 2 2p 4 , which is equivalent to a Ne atom with two electronlike "holes". The atom may therefore exist as a spin singlet or as a spin triplet. Since S1 = 1 and S2 = 0 or S1 = 1 and S2 = 1 may each combine to give a resultant with S = 1, both may be the products of the reaction. Hence multiplicities 3 + 1 and 3 + 3 may be expected. 1 3 (b) N2 , S = 0. The configuration of an N atom is [He] 2s 2 2p 3 . The atoms may have S = or . 2 2 3 3 1 1 Then we note that S1 = and S1 = can combine to give S = 0; S1 = and S2 = can 2 2 2 2 3 1 also combine to give S = 0 (but S1 = and S2 = cannot). Hence, the multiplicities 4 + 4 2 2 and 2 + 2 may be expected.
Solutions to applications
P17.24 The integrated absorption coefficient is A= (~ ) d [16.12] ~
If we can express as an analytical function of , we can carry out the integration analytically. ~ Following the hint in the problem, we seek to fit to an exponential function, which means that a
288
INSTRUCTOR'S MANUAL
plot of ln versus ought to be a straight line (Fig. 17.3). So if ~ ln = m~ + b and A = then = exp(m~ ) exp(b)
eb exp(m~ ) (evaluated at the limits integration). We draw up the following table and find m the bestfit line /nm 292.0 296.3 300.8 305.4 310.1 315.0 320.0
5 4 3 2 1 0 31 000
/(L mol1 cm1 ) /cm1 ~ 1512 34248 865 33748 477 33248 257 32748 135.9 32248 69.5 31746 34.5 31250
ln /(L mol1 cm1 ) 4.69 4.13 3.54 2.92 2.28 1.61 0.912
32 000
33 000
34 000
35 000
Figure 17.3 e38.383 1.26 103 cm exp 290 107 cm 1.26 103 cm 1.26 103 cm 320 107 cm
So A =
 exp
L mol1 cm1
= 1.24 105 L mol1 cm2 P17.25 The concentration of the hypothetical pure layer is [O3 ] = n p 1 atm = 4.46 102 mol L1 = = V RT (0.08206 L atm mol1 K 1 ) (273 K)
So for 300 DU A = cl = (476 L mol1 cm1 ) (0.300 cm) (4.46 102 mol L1 ) = 6.37 and for 100 DU A = cl = (476 L mol1 cm1 ) (0.100 cm) (4.46 102 mol L1 ) = 2.12
SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
289
P17.27
The reaction enthalpy for process (2) is
rH  
=
fH
 
(Cl) +
fH  
fH
 
(OClO+ ) +
fH  
fH
 
(e ) 
fH  
fH
 
(Cl2 O2 )
rH  
so
fH
 
(Cl2 O2 ) =
(Cl) +
(OClO+ ) +
(e ) 
fH
 
(Cl2 O2 ) = (121.68 + 1096 + 0) kJ mol1  (10.95 eV) (96.485 kJ eV1 ) = 161 kJ mol1
We see that the Cl2 O2 in process (2) is different from that in process (1), for its heat of formation is 28 kJ mol1 greater. This is consistent with the computations, which say that ClOOCl is likely to be the lowestenergy isomer. Experimentally we see that the Cl2 O2 of process (2), which is not ClOOCl, is not very much greater in energy than the lowestenergy isomer.
18
Spectroscopy 3: magnetic resonance
Solutions to exercises
Discussion questions
E18.1(b) Before the application of a pulse the magnetization vector, M, points along the direction of the static external magnetic field B0 . There are more spins than spins. When we apply a rotating magnetic field B1 at right angles to the static field, the magnetization vector as seen in the rotating frame begins to precess about the B1 field with angular frequency 1 = B1 . The angle through which M rotates is = B1 t, where t is the time for which the B1 pulse is applied. When t = /2 B1 , = /2 = 90 , and M has rotated into the xy plane. Now there are equal numbers of and spins. A 180 pulse applied for a time / B1 , rotates M antiparallel to the static field. Now there are more spins than spins. A population inversion has occurred. The basic COSY experiment uses the simplest of all twodimensional pulse sequences: a single 90 pulse to excite the spins at the end of the preparation period, and a mixing period containing just a second 90 pulse (see Fig. 18.44 of the text). The key to the COSY technique is the effect of the second 90 pulse, which can be illustrated by consideration of the four energy levels of an AX system (as shown in Fig. 18.12). At thermal equilibrium, the population of the AX level is the greatest, and that of AX level is the smallest; the other two levels have the same energy and an intermediate population. After the first 90 pulse, the spins are no longer at thermal equilibrium. If a second 90 pulse is applied at a time t1 that is short compared to the spinlattice relaxation time T1 the extra input of energy causes further changes in the populations of the four states. The changes in populations will depend on how far the individual magnetizations have precessed during the evolution period. For simplicity, let us consider a COSY experiment in which the second 90 pulse is split into two selective pulses, one applied to X and one to A. Depending on the evolution time t1 , the 90 pulse that excites X may leave the population differences across each of the two X transitions unchanged, inverted, or somewhere in between. Consider the extreme case in which one population difference is inverted and the other unchanged (Fig. 18.45). The 90 pulse that excites A will now generate an FID in which one of the two A transitions has increased in intensity, and the other has decreased. The overall effect is that precession of the X spins during the evolution period determines the amplitudes of the signals from the A spins obtained during the detection period. As the evolution time t1 is increased, the intensities of the signals from A spins oscillate at rates determined by the frequencies of the two X transitions. This transfer of information between spins is at the heart of twodimensional NMR spectroscopy and leads to the correlation of different signals in a spectrum. In this case, information transfer tells us that there is a scalar coupling between A and X. If we conduct a series of experiments in which t1 is incremented, Fourier transformation of the FIDs on t2 yields a set of spectra I (1 , 2 ) in which the A signal amplitudes oscillate as a function of t1 . A second Fourier transformation, this time on t1 , converts these oscillations into a twodimensional spectrum I (1 , 2 ). The signals are spread out in 1 according to their precession frequencies during the detection period. Thus, if we apply the COSY pulse sequence to our AX spin system (Fig. 18.44), the result is a twodimensional spectrum that contains four groups of signals centred on the two chemical shifts in 1 and 2 . Each group will show fine structure, consisting of a block of four signals separated by JAX . The diagonal peaks are signals centerd on (A A ) and (X X ) and lie along the diagonal 1 = 2 . They arise from signals that did not change chemical shift between t1 and t2 . The cross peaks (or offdiagonal peaks) are signals centred on (A X ) and (X A ) and owe their existence to the coupling between A and X.
E18.2(b)
SPECTROSCOPY 3: MAGNETIC RESONANCE
291
Consequently, cross peaks in COSY spectra allow us to map the couplings between spins and to trace out the bonding network in complex molecules. Figure 18.46 shows a simple example of a proton COSY spectrum of 1nitropropane. E18.3(b) The molecular orbital occupied by the unpaired electron in an organic radical can be identified through the observation of hyperfine splitting in the EPR spectrum of the radical. The magnitude of this splitting is proportional to the spin density of the unpaired electron at those positions in the radical having atoms with nuclear moments. In addition, the spin density on carbon atoms adjacent to the magnetic nuclei can be determined indirectly through the McConnell relation. Thus, for example, in the benzene negative ion, unpaired spin densities on both the carbon atoms and hydrogen atoms can be determined from the EPR hyperfine splittings. The next step then is to construct a molecular orbital which will theoretically reproduce these experimentally determined spin densities. A good match indicates that we have found a good molecular orbital for the radical.
Numerical exercises
E18.4(b) For 19 F = 2.62835, g = 5.2567 N gI N B = L = with = 2 h gI N B (5.2567) (5.0508 1027 J T1 ) (16.2 T) = h (6.626 1034 J s) = 6.49 108 s1 = 649 MHz E18.5(b) EmI =  hBmI = gI N BmI mI = 1, 0, 1 EmI = (0.404) (5.0508 1027 J T1 ) (11.50 T)mI = (2.3466 1026 J)mI 2.35 1026 J, 0, +2.35 1026 J E18.6(b) The energy separation between the two levels is E = h where = B (1.93 107 T1 s1 ) (15.4 T) = 2 2 = 4.73 107 s1 = 47.3 MHz E18.7(b) A 600 MHz NMR spectrometer means 600 MHz is the resonance field for protons for which the magnetic field is 14.1 T as shown in Exercise 18.4(a). In highfield NMR it is the field not the frequency that is fixed. (a) A 14 N nucleus has three energy states in a magnetic field corresponding to mI = +1, 0, 1. But E(+1 0) = E(0 1) E = Em  EmI =  hBmI  ( hBmI )
I
Hence, =
=  hB(mI  mI ) =  hB mI The allowed transitions correspond to mI = 1; hence E = h = hB = gI N B = (0.4036) (5.051 1027 J T1 ) (14.1 T) = 2.88 1026 J
292
INSTRUCTOR'S MANUAL
(b) We assume that the electron gvalue in the radical is equal to the free electron gvalue, ge = 2.0023. Then E = h = ge B B[37] = (2.0023) (9.274 1024 J T1 ) (0.300 T) = 5.57 1024 J Comment. The energy level separation for the electron in a free radical in an ESR spectrometer is far greater than that of nuclei in an NMR spectrometer, despite the fact that NMR spectrometers normally operate at much higher magnetic fields. E18.8(b) E = h = hB = gI N B Hence, B = E18.9(b) [Exercise 18.4(a)] h (6.626 1034 J Hz1 ) (150.0 106 Hz) = 3.523 T = gI N (5.586) (5.051 1027 J T1 ) In all cases the selection rule mI = 1 is applied; hence (Exercise 18.7(b)(a)) B = h 6.626 1034 J Hz1 = gI N gI 5.0508 1027 J T1
= (1.3119 107 ) Hz T = (0.13119) MHz T gI gI We can draw up the following table
B/T (a) (b) gI 300 MHz 750 MHz
14
N
19
F
31
P
0.40356 97.5 244
5.2567 7.49 18.7
2.2634 17.4 43.5
Comment. Magnetic fields above 20 T have not yet been obtained for use in NMR spectrometers. As discussed in the solution to Exercise 18.7(b), it is the field, not the frequency, that is fixed in highfield NMR spectrometers. Thus an NMR spectrometer that is called a 300 MHz spectrometer refers to the resonance frequency for protons and has a magnetic field fixed at 7.05 T.
1 E18.10(b) The relative population difference for spin  2 nuclei is given by
N  N N hB gI N B = = N + N N 2kT 2kT =
[Justification 18.1]
1.405(5.05 1027 J T1 )B = 8.62 107 (B/T) 2(1.381 1023 J K1 ) (298 K)
N = (8.62 107 ) (0.50) = 4.3 107 N N (b) For 2.5 T = (8.62 107 ) (2.5) = 2.2 106 N N = (8.62 107 ) (15.5) = 1.34 105 (c) For 15.5 T N (a) For 0.50 T
SPECTROSCOPY 3: MAGNETIC RESONANCE
293
E18.11(b) The ground state has
1 mI = + 2 = spin, 1 mI =  2 = spin
Hence, with N = N  N N  N N N  N e E/kT = = N N + N N + N e E/kT = [Justification 18.1] [for E kT ]
1  e E/kT E gI N B 1  (1  E/kT ) = 1+1 2kT 2kT 1 + e E/kT NgI N B N h N = = 2kT 2kT Thus, N N(800 MHz) 800 MHz = = 13 N(60 MHz) 60 MHz
This ratio is not dependent on the nuclide as long as the approximation (a) = 
E
kT holds.
106 [18.25] Since both and depend upon the magnetic field in the same manner, namely = gI N B h and = gI N B0 [Exercise 18.4(a)] h
is independent of both B and . (b) Rearranging [10]  = 106 and we see that the relative chemical shift is  (800 MHz) 800 MHz = 13 =  (60 MHz) 60 MHz Comment. This direct proportionality between  and is one of the major reasons for operating an NMR spectrometer at the highest frequencies possible. E18.12(b) Bloc = (1  )B  Bloc  = ( )B [(CH3 )  (CH2 )]B = 1.16  3.36 106 B = 2.20 106 B (a) (b) E18.13(b) B = 1.9 T,  Bloc  = (2.20 106 ) (1.9 T) = 4.2 106 T B = 16.5 T,  Bloc  = (2.20 106 ) (16.5 T) = 3.63 105 T
 = 106   (  )(CH2 )  (  )(CH3 ) = (CH2 )  (CH3 ) = [(CH2 )  (CH3 )] 106 = (3.36  1.16) 106 = 2.20 106
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INSTRUCTOR'S MANUAL
6.97 Hz
6.97 Hz
770 Hz
at 350 MHz
Figure 18.1
(a) (b)
= 350 MHz = 650 MHz
  = (2.20 106 ) (350 MHz) = 770 Hz [Fig. 18.1]   = (2.20 106 ) (650 MHz) = 1.43 kHz
At 650 MHz, the spinspin splitting remains the same at 6.97 Hz, but as has increased to 1.43 kHz, the splitting appears narrower on the scale. E18.14(b) The difference in resonance frequencies is = ( 106 ) = (350 s1 ) (6.8  5.5) = 4.6 102 s1 The signals will be resolvable as long as the conformations have lifetimes greater than = (2 )1 The interconversion rate is the reciprocal of the lifetime, so a resolvable signal requires an interconversion rate less than rate = (2 ) = 2(4.6 102 s1 ) = 2.9 103 s1 E18.15(b) gI N B [Exercise 18.4(a)] h g(31 P) (31 P) = 1 Hence, 1 ( H) g( H) 2.2634 500 MHz = 203 MHz or (31 P) = 5.5857 =
1 1 The proton resonance consists of 2 lines 2 2 +1 and the 31 P resonance of 5 lines 2 4 2 +1 . 1 The intensities are in the ratio 1 : 4 : 6 : 4 : 1 (Pascal's triangle for four equivalent spin 2 nuclei, 5.5857 Section 18.6). The lines are spaced = 2.47 times greater in the phosphorus region than the 2.2634 proton region. The spectrum is sketched in Fig. 18.2.
Proton resonance
Phosphorus resonance
Figure 18.2
SPECTROSCOPY 3: MAGNETIC RESONANCE
295
E18.16(b) Look first at A and M, since they have the largest splitting. The A resonance will be split into a widely spaced triplet (by the two M protons); each peak of that triplet will be split into a less widely spaced sextet (by the five X protons). The M resonance will be split into a widely spaced triplet (by the two A protons); each peak of that triplet will be split into a narrowly spaced sextet (by the five X protons). The X resonance will be split into a less widely spaced triplet (by the two A protons); each peak of that triplet will be split into a narrowly spaced triplet (by the two M protons). (See Fig. 18.3.) Only the splitting of the central peak of Fig. 18.3(a) is shown in Fig. 18.3(b).
(a)
(b)
Figure 18.3
E18.17(b) (a) Since all JHF are equal in this molecule (the CH2 group is perpendicular to the CF2 group), the H and F nuclei are both chemically and magnetically equivalent. (b) Rapid rotation of the PH3 groups about the MoP axes makes the P and H nuclei chemically and magnetically equivalent in both the cis and transforms. E18.18(b) Precession in the rotating frame follows L = B1 2 or 1 = B1
Since is an angular frequency, the angle through which the magnetization vector rotates is gI N = B1 t = B1 t h h ( ) (1.0546 1034 J s) So B1 = = = 9.40 104 T g I N t (5.586) (5.0508 1027 J T1 ) (12.5 106 s) a 90 pulse requires E18.19(b) B= =
1 2
12.5 s = 6.25 s
hc h = ge B ge B
(6.626 1034 J s) (2.998 108 m s1 ) = 1.3 T (2) (9.274 1024 J T1 ) (8 103 m) E18.20(b) The g factor is given by g= g= h ; B B h 6.62608 1034 J s = = 7.1448 1011 T Hz1 = 71.448 mT GHz1 B 9.2740 1024 J T1
71.448 mT GHz1 9.2482 GHz = 2.0022 330.02 mT
296
INSTRUCTOR'S MANUAL
E18.21(b) The hyperfine coupling constant for each proton is 2.2 mT , the difference between adjacent lines in the spectrum. The g value is given by g= (71.448 mT GHz1 ) (9.332 GHz) h = = 1.992 B B 334.7 mT
E18.22(b) If the spectrometer has sufficient resolution, it will see a signal split into eight equal parts at 1.445 1.435 1.055 mT from the centre, namely 328.865, 330.975, 331.735, 331.755, 333.845, 333.865, 334.625, and 336.735 mT If the spectrometer can only resolve to the nearest 0.1 mT, then the spectrum will appear as a sextet with intensity ratios of 1 : 1 : 2 : 2 : 1 : 1. The four central peaks of the more highly resolved spectrum would be the two central peaks of the less resolved spectrum. E18.23(b) (a) If the CH2 protons have the larger splitting there will be a triplet (1 : 2 : 1) of quartets (1 : 3 : 3 : 1). Altogether there will be 12 lines with relative intensities 1(4 lines), 2(2 lines), 3(4 lines), and 6(2 lines). Their positions in the spectrum will be determined by the magnitudes of the two proton splittings which are not given. (b) If the CD2 deuterons have the larger splitting there will be a quintet (1 : 2 : 3 : 2 : 1) of septets (1 : 3 : 6 : 7 : 6 : 3 : 1). Altogether there will be 35 lines with relative intensities 1(4 lines), 2(4 lines), 3(6 lines), 6(8 lines), 7(2 lines), 9(2 lines), 12(4 lines), 14(2 lines), 18(2 lines), and 21(1 line). Their positions in the spectrum will be determined by the magnitude of the two deuteron splittings which are not given. E18.24(b) The hyperfine coupling constant for each proton is 2.2 mT , the difference between adjacent lines in the spectrum. The g value is given by g= h B B so B= h h = 71.448 mT GHz1 , B g B
(a) (b)
B= B=
(71.448 mT GHz1 ) (9.312 GHz) = 332.3 mT 2.0024 (71.448 mT GHz1 ) (33.88 GHz) = 1209 mT 2.0024
E18.25(b) Two nuclei of spin I = 1 give five lines in the intensity ratio 1 : 2 : 3 : 2 : 1 (Fig. 18.4).
First nucleus with I = 1 second nucleus with I = 1 1 2 3 2 1
Figure 18.4
E18.26(b) The X nucleus produces four lines of equal intensity. The three H nuclei split each into a 1 : 3 : 3 : 1 quartet. The three D nuclei split each line into a septet with relative intensities 1 : 3 : 6 : 7 : 6 : 3 : 1 (see Exercise 18.23(a)). (See Fig. 18.5.)
SPECTROSCOPY 3: MAGNETIC RESONANCE
297
Figure 18.5
Solutions to problems
Solutions to numerical problems
P18.2 J 1 1 = 2 (2) ((5.2  4.0) 106 ) (60 106 Hz)
2.2 ms, corresponding to a rate of jumping of 450 s1 . When = 300 MHz J 1 = 0.44 ms (2) {(5.2  4.0) 106 } (300 106 Hz)
corresponding to a jump rate of 2.3 103 s1 . Assume an Arrheniuslike jumping process (Chapter 25) rate eEa /RT Then, ln Ea rate(T ) = rate(T ) R R ln(r /r)
1 T
1 1  T T =
3 8.314 J K1 mol1 ln 2.310 450
and therefore Ea = P18.5

1 T
1 280 K

1 300 K
= 57 kJ mol1
It seems reasonable to assume that only staggered conformations can occur. Therefore the equilibria are
When R3 = R4 = H, all three of the above conformations occur with equal probability; hence
3
JHH (methyl) = 1 3 t + 2 3 g J 3 J
[t = trans, g = gauche; CHR3 R4 = methyl]
Additional methyl groups will avoid being staggered between both R1 and R2 . Therefore
3 3 1 JHH (ethyl) = 2 (Jt + Jg )
[R3 = H, R4 = CH3 ] [R3 = R4 = CH3 ]
JHH (isopropyl) = Jt
298
INSTRUCTOR'S MANUAL
We then have three simultaneous equations in two unknowns Jt and Jg .
1 3 3 ( Jt 1 3 2 ( Jt 3
+ 23 g ) = 7.3 Hz J + 3 g ) = 8.0 Hz J
(1) (2)
Jt = 11.2 Hz
J J The two unknowns are overdetermined. The first two equations yield 3 t = 10.1, 3 g = 5.9. 3 3 However, if we assume that Jt = 11.2 as measured directly in the ethyl case then Jg = 5.4 (eqn 1) or 4.8 (eqn 2), with an average value of 5.1. Using the original form of the Karplus equation
3 3
Jt = A cos2 (180 ) + B = 11.2 Jg = A cos2 (60 ) + B = 5.1
or 11.2 = A + B 5.1 = 0.25A + B These simultaneous equations yield A = 6.8 Hz and B = 4.8 Hz. With these values of A and B, the original form of the Karplus equation fits the data exactly (at least to within the error in the values of 3 Jt and 3 g and in the measured values reported). J From the form of the Karplus equation in the text [21] we see that those values of A, B, and C cannot be determined from the data given, as there are three constants to be determined from only two values of J . However, if we use the values of A, B, and C given in the text, then Jt = 7 Hz  1 Hz(cos 180 ) + 5 Hz(cos 360 ) = 11 Hz Jg = 7 Hz  1 Hz(cos 60 ) + 5 Hz(cos 120 ) = 5 Hz The agreement with the modern form of the Karplus equation is excellent, but not better than the original version. Both fit the data equally well. But the modern version is preferred as it is more generally applicable. P18.8 Refer to the figure in the solution to Exercise 18.23(a). The width of the CH3 spectrum is 3aH = 6.9 mT . The width of the CD3 spectrum is 6aD . It seems reasonable to assume, since the hyperfine interaction is an interaction of the magnetic moments of the nuclei with the magnetic moment of the electron, that the strength of the interactions is proportional to the nuclear moments. = gI N I or z = gI N mI [18.14, 18.15]
and thus nuclear magnetic moments are proportional to the nuclear gvalues; hence aD 0.85745 aH = 0.1535aH = 0.35 mT 5.5857
Therefore, the overall width is 6aD = 2.1 mT P18.10 We write P (N2s) = 5.7 mT = 0.10 (10 percent of its time) 55.2 mT 1.3 mT = 0.38 (38 percent of its time) P (N2pz ) = 3.4 mT
SPECTROSCOPY 3: MAGNETIC RESONANCE
299
The total probability is (a) P (N) = 0.10 + 0.38 = 0.48 (48 percent of its time). (b) P (O) = 1  P (N) = 0.52 (52 percent of its time). The hybridization ratio is P (N2p) 0.38 = 3.8 = 0.10 P (B2s) The unpaired electron therefore occupies an orbital that resembles as sp 3 hybrid on N, in accord with the radical's nonlinear shape. From the discussion in Section 14.3 we can write a2 = 1 + cos 1  cos 2 cos 1  cos , implying that cos = 0.66, so = 2+
b2 = 1  a 2 = =
1 cos b2 = 1 + cos a2
Then, since = 3.8, cos P18.11
= 131
 For C6 H6 , a = Q with Q = 2.25 mT [18.52]. If we assume that the value of Q does not change from this value (a good assumption in view of the similarity of the anions), we may write
=
a a = Q 2.25 mT
Hence, we can construct the following maps
0.005 0.076 0.076 0.005
0.200 0.048 0.200
0.121
0.050 0.050
0.050 0.050
Solutions to theoretical problems
P18.14 h0 mI g I N 0 (1  3 cos2 ) [18.36] = 4R 3 4 R 3 which rearranges to Bnuc =  R = gI N 0 1/3 = 4 Bnuc 1 [mI = + , = 0, h = gI N ] 2
(5.5857) (5.0508 1027 JT1 ) (4 107 T2 J1 m3 ) (4 ) (0.715 103 T)
1/3
= (3.946 1030 m3 )1/3 = 158 pm
300
INSTRUCTOR'S MANUAL
P18.17
We have seen (Problem 18.16) that, if G cos 0 t, then I () at 0 . Therefore, if G(t) a cos 1 t + b cos 2 t we can anticipate that I () a b + 2 2 1 + 1  ) 1 + (2  )2 2
1 which peaks [1 + (0  )2 2 ]
and explicit calculation shows this to be so. Therefore, I () consists of two absorption lines, one peaking at 1 and the other at 2 . P18.21 The desired result is the linear equation: [I]0 = [E]0  K,
so the first task is to express quantities in terms of [I]0 , [E]0 , , , and K, eliminating terms such as [I], [EI], [E], I , EI , and . (Note: symbolic mathematical software is helpful here.) Begin with : = [I] [EI] [EI] [I]0  [EI] I + EI , EI = I + [I]0 [I] + [EI] [I]0 [I] + [EI]
where we have used the fact that total I (i.e., free I plus bound I) is the same as intitial I. Solve this expression for [EI]: [EI] = [I]0 (  I ) [I]0 = , EI  I
where in the second equality we notice that the frequency differences that appear are the ones defined in the problem. Now take the equilibrium constant: K= [E][I] ([E]0  [EI])([I]0  [EI]) ([E]0  [EI])[I]0 = . [EI] [EI] [EI] [E]0 ),
We have used the fact that total I is much greater than total E (from the condition that [I]0 so it must also be much greater than [EI], even if all E binds I. Now solve this for [E]0 : [E]0 = K + [I]0 [EI] = [I]0 K + [I]0 [I]0 [I]0 = (K + [I]0 ) .
The expression contains the desired terms and only those terms. Solving for [I]0 yields: [I]0 = [E]0 K ,
which would result in a straight line with slope [E]0 and yintercept K if one plots [I]0 against 1/.
19
Statistical thermodynamics: the concepts
Solutions to exercises
Discussion questions
E19.1(b) Consider the value of the partition function at the extremes of temperature. The limit of q as T approaches zero, is simply g0 , the degeneracy of the ground state. As T approaches infinity, each term in the sum is simply the degeneracy of the energy level. If the number of levels is infinite, the partition function is infinite as well. In some special cases where we can effectively limit the number of states, the upper limit of the partition function is just the number of states. In general, we see that the molecular partition function gives an indication of the average number of states thermally accessible to a molecule at the temperature of the system. The statistical entropy may be defined in terms of the Boltzmann formula, S = k ln W , where W is the statistical weight of the most probable configuration of the system. The relation between the entropy and the partition function is developed in two stages. In the first stage, we justify Boltzmann's formula, in the second, we express W in terms of the partition function. The justification for Boltzmann's formula is presented in Justification 19.6. Without repeating the details of this justification, we can see that the entropy defined through the formula has the properties we expect of the entropy. W can be thought of as a measure of disorder, hence the greater W , the greater the entropy; and the logarithmic form is consistent with the additive properties of the entropy. We expect the total disorder of a combined system to be the product of the individual disorders and S = k ln W = k ln W1 W2 = k ln W1 + k ln W2 = S1 + S2 . In the second stage the formula relating entropy and the partition function is derived. This derivation is presented in Justification 19.7. The expression for W , eqn 19.1, is recast in terms of probabilities, which in turn are expressed in terms of the partition function through eqn 10. The final expression which is eqn 19.34 then follows immediately. Since and temperature are inversely related, strictly speaking one can never replace the other. The concept of temperature is useful in indicating the direction of the spontaneous transfer of energy in the form of heat. It seems natural to us to think of the spontaneous direction for this transfer to be from a body at high T to one at low T . In terms of , the spontaneous direction would be from low to high and this has an unnatural feel. On the other hand, has a direct connection to the energy level pattern of systems of atoms and molecules. It arises in a natural, purely mathematical, manner from our knowledge of how energy is distributed amongst the particles of our atomic/molecular system. We would not have to invoke the abstract laws of thermodynamics, namely the zeroth and second laws in order to define our concept of temperature if we used as the property to indicate the natural direction of heat flow. We can easily demonstrate that is directly related to the statistical weight W through the relation = ( ln W/U )N . W, U , and N are all concrete properties of an atomic/molecular system. Identical particles can be regarded as distinguishable when they are localized as in a crystal lattice where we can assign a set of coordinates to each particle. Strictly speaking it is the lattice site that carries the set of coordinates, but as long as the particle is fixed to the site, it too can be considered distinguishable.
E19.2(b)
E19.3(b)
E19.4(b)
302
INSTRUCTOR'S MANUAL
Numerical exercises
E19.5(b) ni = Thus n2 = n1 N ei q where q =
j
ej
e2 = e(2 1 ) = e = e /kT e1 n2 1 Given = , = 300 cm1 2 n1 k = (1.38066 1023 J K1 ) n2 = e /kT n1 ln n2 n1 =  /kT 1 cm1 1.9864 1023 J = 0.69506 cm1 K 1
T = =
 = k ln(n2 /n1 ) k ln(n1 /n2 ) 300 cm1 = 622.7 K 623 K (0.69506 cm1 K 1 ) ln(2) =h
1/2 1/2 1 [19.22] = h 2m 2 mkT
E19.6(b)
(a)
= (6.626 1034 J s) = 1 (2) (39.95) (1.6605 1027 kg) (1.381 1023 J K1 ) T
1/2
276 pm (T /K)1/2 V [22] = 3 (1.00 106 m3 ) (T /K)3/2 = 4.76 1022 (T /K)3/2 (2.76 1010 m)3 = 1.59 1011 m = 15.9 pm , = 5.04 pm , q = 2.47 1026
(b)
q= (i)
T = 300 K,
(ii) T = 3000 K,
q = 7.82 1027
Question. At what temperature does the thermal wavelength of an argon atom become comparable to its diameter? E19.7(b) The translational partition function is V qtr = 3 (2kT m)3/2 h so qXe = qHe mXe 3/2 = mHe 131.3 u 3/2 = 187.9 4.003 u
STATISTICAL THERMODYNAMICS: THE CONCEPTS
303
E19.8(b)
q=
levels
gj ej = 2 + 3e1 + 2e2 1.4388(~ /cm1 ) hc ~ = T /K kT
=
Thus q = 2 + 3e(1.43881250/2000) + 2e(1.43881300/2000) = 2 + 1.2207 + 0.7850 = 4.006 E19.9(b) E = U  U (0) =  N d N dq = (2 + 3e1 + 2e2 ) q d q d N hc N ~ 31 e1  22 e2 = 3~ 1 ehc1 + 22 ehc2 ~ = q q NA hc 3(1250 cm1 ) e(1.43881250/2000) = 4.006 + 2(1300 cm1 ) e(1.43881300/2000) = NA hc 4.006 (2546 cm1 )
= (6.022 1023 mol1 ) (6.626 1034 J s) (2.9979 1010 cm s1 ) (2546 cm1 ) = 7.605 kJ mol1 E19.10(b) In fact there are two upper states, but one upper level. And of course the answer is different if the question asks when 15 per cent of the molecules are in the upper level, or if it asks when 15 per cent of the molecules are in each upper state. The solution below assumes the former. The relative population of states is given by the Boltzmann distribution  E n2 = exp n1 kT Thus T = = exp hc ~ kT so ln hc ~ n2 = n1 kT
hc ~ k ln(n2 /n1 ) Having 15 per cent of the molecules in the upper level means 0.15 2n2 = n1 1  0.15 and T = so n2 = 0.088 n1
(6.626 1034 J s) (2.998 1010 cm s1 ) (360 cm1 ) (1.381 1023 J K1 ) (ln 0.088)
= 213 K E19.11(b) The energies of the states relative to the energy of the state with mI = 0 are N hB, 0, + N hB, where N h = 2.04 1027 J T1 . With respect to the lowest level they are 0, N h, 2N h. The partition function is q=
states
eEstate /kT
where the energies are measured with respect to the lowest energy. So in this case q = 1 + exp N hB kT + exp 2N hB kT
As B is increased at any given T , q decays from q = 3 toward q = 1 as shown in Fig. 19.1(a).
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INSTRUCTOR'S MANUAL
2
Figure 19.1(a) The average energy (measured with respect to the lowest state) is E =
states Estate e Estate /kT
q
=
1 + N hB exp N hB + 2N hB exp 2N hB kT kT 1 + exp N hB + exp 2N hB kT kT
The expression for the mean energy measured based on zero spin having zero energy becomes E =
N hB  N hB exp 2N hB kT 1 + exp N hB + exp 2N hB kT kT
=
N hB 1  exp 2N hB kT 1 + exp N hB + exp 2N hB kT kT
As B is increased at constant T , the mean energy varies as shown in Fig. 19.1(b).
Figure 19.1(b) The relative populations (with respect to that of the lowest state) are given by the Boltzmann factor exp  E kT = exp N hB kT or exp 2N hB kT
Note that
N hB (2.04 1027 J T1 ) (20.0 T) = 2.95 103 K = k 1.381 1023 J K1 so the populations are (a) (b) exp exp and 2.95 103 K 1.0 K 2.95 103 K 298 exp = 0.997 and exp 2(2.95 103 K) 1.0 K = 0.994
= 1  1 105 = 1  2 105
2(2.95 103 K) 298
STATISTICAL THERMODYNAMICS: THE CONCEPTS
305
E19.12(b) (a) The ratio of populations is given by the Boltzmann factor n3  E n2 = e50.0 K/T = e25.0 K/T and = exp kT n1 n1 (1) At 1.00 K n2 25.0 K = 1.39 1011 = exp n1 1.00 K n3 50.0 K = exp n1 1.00 K (2) At 25.0 K and 25.0 K n2 = exp n1 25.0 K (3) At 100 K 25.0 K n2 = exp n1 100 K = 0.779 and n3 50.0 K = exp n1 100 K = 0.607 = 1.93 1022 n3 50.0 K = exp 25.0 K n1
= 0.368
and
= 0.135
(b) The molecular partition function is q=
states
eEstate /kT = 1 + e25.0 K/T + e50.0 K/T
At 25.0 K, we note that e25.0 K/T = e1 and e50.0 K/T = e2 q = 1 + e1 + e2 = 1.503 (c) The molar internal energy is Um = Um (0)  So Um = Um (0)  At 25.0 K Um  Um (0) =  (6.022 1023 mol1 ) (25.0 K) (1.381 1023 J K1 ) 1.503 NA q q where = (kT )1
NA (25.0 K)k e25.0 K/T + 2e50.0 K/T q
(e1 + 2e2 ) = 88.3 J mol1 (d) The molar heat capacity is CV ,m = Um 1 25.0 K/T = NA (25.0 K)k + 2e50.0 K/T e T V T q 25.0 K 25.0 K/T = NA (25.0 K)k + 4e50.0 K/T e qT 2 1  2 e25.0 K/T + 2e50.0 K/T q where q T
25.0 K 25.0 K/T q = + 2e50.0 K/T e T T2
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INSTRUCTOR'S MANUAL
so CV ,m = At 25.0 K CV ,m =
NA (25.0 K)2 k T 2q
e25.0 K/T + 4e50.0 K/T 
(e25.0 K/T + 2e50.0 K/T )2 q
(6.022 1023 mol1 ) (25.0 K)2 (1.381 1023 J K1 ) (25.0 K)2 (1.503) e1 + 4e2  (e1 + 2e2 )2 1.503
= 3.53 J K1 mol1 (e) The molar entropy is Sm = At 25.0 K Sm = 88.3 J mol1 + (6.022 1023 mol1 ) (1.381 1023 J K1 ) ln 1.503 25.0 K Um  Um (0) + NA k ln q T
= 6.92 J K1 mol1 E19.13(b) n1 g1 e1 /kT = g1 e /kT = 3ehcB/kT = n0 g0 e0 /kT n1 1 Set = and solve for T . n0 e ln 1 e = ln 3 + hcB k(1 + ln 3) 6.626 1034 J s 2.998 1010 cm s1 10.593 cm1 +1.381 1023 J K1 (1 + 1.0986) hcB kT
T = =
= 7.26 K E19.14(b) The SackurTetrode equation gives the entropy of a monoatomic gas as S = nR ln (a) At 100 K = 6.626 1034 J s 2(1.381 1023 J K1 ) (100 K) (131.3 u) (1.66054 1027 kg u1 )
1/2
e5/2 kT p 3
where
=
h 2kT m
= 1.52 1011 m
STATISTICAL THERMODYNAMICS: THE CONCEPTS
307
and Sm = (8.3145 J K1 mol1 ) ln = 147 J K1 mol1 (b) At 298.15 K =
e5/2 (1.381 1023 J K1 ) (100 K) (1.013 105 Pa) (1.52 1011 m)3
6.626 1034 J s 2(1.381 1023 J K1 ) (298.15 K) (131.3 u) (1.66054 1027 kg u1 )
1/2
= 8.822 1012 m and Sm = (8.3145 J K1 mol1 ) ln = 169.6 J K1 mol1 E19.15(b) 1 1 =  ~ 1e 1  ehc (1.4388 cm K) (321 cm1 ) hc = ~ = 0.76976 600 K 1 = 1.863 Thus q = 0.76976 1e The internal energy due to vibrational excitation is q= U  U (0) = = and hence N e 1  e
~ N hc ehc ~ N hc ~ = hc = (0.863) (N hc) (321 cm1 ) hc ~ ~ 1 1e e
e5/2 (1.381 1023 J K1 ) (298.15 K) (1.013 105 Pa) (8.822 1012 m)3
Sm U  U (0) = + ln q = (0.863) NA k NA kT =
hc kT
(321 cm1 ) + ln(1.863)
(0.863) (1.4388 K cm) (321 cm1 ) + ln(1.863) 600 K
= 0.664 + 0.62199 = 1.286 and Sm = 1.286R = 10.7 J K1 mol1 E19.16(b) Inclusion of a factor of (N !)1 is necessary when considering indistinguishable particles. Because of their translational freedom, gases are collections of indistinguishable particles. The factor, then, must be included in calculations on (a) CO2 gas .
Solutions to problems
Solutions to numerical problems
P19.4 S = k ln W or W = eS/k [19.30] S V T ,N W W = V T ,N k
308
INSTRUCTOR'S MANUAL
S = nR ln
e5/2 e5/2 V = nR ln V + ln N 3 N 3
S NR ln V nR = = nR = V T ,N V T ,N V NA V W NW N RW = = V T ,N NA kV V W V pV V N = W V kT V (1 105 Pa) (20 m3 ) (1 105 ) (1.381 1023 J K1 ) (300 K)
4.8 1021 Notice that the value of W is much larger than that of W/W . For example, at the conventional temperature the molar entropy of helium is 126 J K 1 mol1 . Therefore, S = nSm = pV RT Sm = (1 105 Pa) (20 m3 ) (126 J K 1 mol1 ) (8.315 J K1 mol1 ) (298 K)
= 1.02 105 J K1 1.02 105 J K1 S = 7.36 1027 = k 1.381 1023 J K1 W = eS/k = e7.3610 = 103.2010
27 27
P19.6
P19.8
4 n1 g1 e1 /kT 4 ~ = = e /kT = ehc /kT = 2e{(1.4388450)/300} = 0.23 0 /kT 2 2 n0 g0 e 0.30 The observed ratio is = 0.43. Hence the populations are not at equilibrium . 0.70 First we evaluate the partition function q=
j
gj ej [19.12] =
j
~ gj ehc j
1.43877 cm K = 4.041 104 cm At 3287 C = 3560 K, hc = 3560 K
4 1 4 1 q = 5 + 7e{(4.04110 cm)(170 cm )} + 9e{(4.04110 cm)(387 cm )} 4 1 + 3e{(4.04110 cm)(6557 cm )}
= (5) + (7) (0.934) + (9) (0.855) + (3) (0.0707) = 19.445 The fractions of molecules in the various states are pj =
~ gj ehc j gj ej [19.10] = q q
5 = 0.257 19.445 (9) (0.855) p(3 F4 ) = = 0.396 19.445 p(3 F2 ) =
(7) (0.934) = 0.336 19.445 (3) (0.0707) p(4 F1 ) = = 0.011 19.445 p(3 F3 ) =
STATISTICAL THERMODYNAMICS: THE CONCEPTS
309
Comment. P19.10
j
pj = 1. Note that the most highly populated level is not the ground state. E kT hcG kT hcG kT
The partition function is the sum over states of the Boltzmann factor q=
states
exp 
=
states
exp 
=
levels
g exp 
where g is the degeneracy. So, at 298 K q = 1 + 3 exp  = 1.209 At 1000 K q = 1 + 3 exp  = 3.004 P19.11 q=
i
(6.626 1034 J s) (2.998 1010 cm s1 ) (557.1 cm1 ) (1.381 1023 J K1 ) (298 K)
+
(6.626 1034 J s) (2.998 1010 cm s1 ) (557.1 cm1 ) (1.381 1023 J K1 ) (1000 K)
+
ei =
i
~ ehc i [19.11]
At 100 K, hc = (a)
1 1 and at 298 K, hc = . Therefore, at 100 K 1 207.22 cm1 69.50 cm
q = 1 + e213.30/69.50 + e435.39/69.50 + e636.27/69.50 + e845.93/69.50 = 1.049 and at 298 K (b) q = 1 + e213.30/207.22 + e425.39/207.22 + e636.27/207.22 + e845.93/207.22 = 1.55 In each case, pi = p0 = p1 = p2 =
~ ehc i [19.10] q
1 = (a) 0.953 , q
(b) 0.645 (b) 0.230 (b) 0.083
~ ehc 1 = (a) 0.044 , q ~ ehc 2 = (a) 0.002 , q
For the molar entropy we need to form Um  Um (0) by explicit summation Um  Um (0) = NA NA ~ i ei = hci ehc i [19.25, 19.26] ~ q i q i
= 123 J mol1 (at 100 K) , 1348 J mol1 (at 298 K) Sm = (a) (b) Um  Um (0) + R ln q [19.34] T
Sm =
123 J mol1 + R ln 1.049 = 1.63 J K1 mol1 100 K 1348 J mol1 Sm = + R ln 1.55 = 8.17 J K1 mol1 298 K
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INSTRUCTOR'S MANUAL
Solutions to theoretical problems
P19.13 p = kT ln Q [20.4] V T ,N ln(q N /N !) = kT [19.46] V
T ,N
= kT = N kT = N kT N kT = V P19.15
[N ln q  ln N !] = N kT V T ,N ln(V / 3 ) V
ln q V T ,N
T ,N
[ln V  ln 3 ] V or
= N kT
T ,N
ln V V T ,N
pV = N kT = nRT
We draw up the following table
0 8 7 7 7 7 6 6 6 6 6 6 6 5 5 5 5 5 5 4 4 4 4 3 3 2 2 1 0 0 1 0 0 0 2 0 0 1 1 1 0 3 0 2 2 1 1 4 3 3 2 5 4 6 5 7 9 2 0 0 1 0 0 0 2 0 0 1 0 1 0 3 1 0 2 1 0 1 0 2 0 1 0 2 1 0 3 0 0 0 1 0 0 0 3 0 0 1 1 0 1 0 1 0 2 0 0 2 1 0 1 1 0 0 0 4 0 0 0 0 1 0 0 0 2 0 0 1 0 0 0 1 1 0 0 1 0 0 1 0 0 0 0 0 5 0 0 0 0 1 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 6 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 7 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 9 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 W 9 72 72 72 72 252 252 84 252 504 504 504 504 504 1512 1512 1512 1512 630 2520 1260 3780 504 2520 252 756 72 1
The most probable configuration is the "almost exponential" {4, 2, 2, 1, 0, 0, 0, 0, 0, 0}
STATISTICAL THERMODYNAMICS: THE CONCEPTS
311
P19.16
nj = e(j 0 ) = ej , n0
which implies that j kT
j = ln nj  ln n0
and therefore that ln nj = ln n0 
Therefore, a plot of ln nj against j should be a straight line with slope  ln pj against j , since ln pj = const  j kT
. Alternatively, plot kT
We draw up the following table using the information in Problem 19.8
j nj ln nj 0 4 1.39 1 2 0.69 2 2 0.69 3 1 0
[most probable configuration]
These are points plotted in Fig. 19.2 (full line). The slope is 0.46, and since slope corresponds to a temperature T = (50 cm1 ) (2.998 1010 cm s1 ) (6.626 1034 J s) = 160 K (0.46) (1.381 1023 J K1 )
= 50 cm1 , the hc
(A better estimate, 104 K represented by the dashed line in Fig. 19.2, is found in Problem 19.18.)
1.6
1.2
0.8
0.4
0
0.4 0 1 2 j 3 4
Figure 19.2 (b) Choose one of the weight 2520 configurations and one of the weight 504 configurations, and draw up the following table
j nj ln nj nj ln nj 0 4 1.39 6 1.79 1 3 1.10 0  2 1 0 1 0 3 0  1 0 4 1 0 1 0
W = 2520 W = 504
Inspection confirms that these data give very crooked lines.
312
INSTRUCTOR'S MANUAL
P19.19
(a) The form of Stirling's approximation used in the text in the derivation of the Boltzmann distribution is ln x! = x ln x  x [19.2] ln W = N ln N 
i
or
ln N ! = N ln N  N
i
and ln ni ! = ni ln ni  ni which then leads to N is cancelled by  ni ln ni [19.3]
ni
If N ! = N N , ln N ! = N ln N , likewise ln ni ! = ni ln ni and eqn 3 is again obtained. 1 1 (b) For ln x! = x + 2 ln x  x + 2 ln 2 [Marginal note, p. 631], Since the method of undetermined multipliers requires only (Justification 19.3) d ln W , only the 1 terms d ln ni ! survive. The constant term, 2 ln 2 , drops out, as do all terms in N . The difference, 1 then, is in terms arising from ln ni ! We need to compare ni ln ni to 2 ln ni , as both these terms survive the differentiation. The derivatives are (ni ln ni ) = 1 + ln ni ln ni [large ni ] ni ni 1 ln ni 2 = 1 2ni 1 decreases and in the limit becomes negligible. For 2ni
Whereas ln ni increases as ni increases, ni = 1106 , ln ni = 13.8,
1 = 5107 ; the ratio is about 2108 which could probably not 2ni be seen in experiments. However, for experiments on, say, 1000 molecules, such as molecular dynamics simulations, there could be a measurable difference.
Solutions to applications
P19.21 N (h)/V p(h) = = e{((h)(h0 ))/kT } [19.6] p(h0 ) N (h0 )/V = emg(hh0 )/kT For p(0) p0 , p(h) = emgh/kT p0
M(O2 )gh N (8.0 km) N (8.0 km)/V = = e RT N (0) N (0)/V
 N (8.0 km) [O2 ] = e N (0)
(0.032 kg mol1 )(9.81 m s2 )(8.0103 m) (8.315 J K1 mol1 )(298 K)
= 0.36 for O2
 N (8.0 km) [H2 O] = e N (0)
(0.018 kg mol1 )(9.81 m s2 )(8.0103 m) (8.315 J K1 mol1 )(298 K)
= 0.57
for H2 O
STATISTICAL THERMODYNAMICS: THE CONCEPTS
313
P19.23
(a) The electronic partition function, qE , of a perfect, atomic hydrogen gas consists of the electronic energies En that can be written in the form: 1 En = 1  2 n hcRH , n = 1, 2, 3, . . . , ,
where we have used the state n = 1 as the zero of energy (in contrast to the usual zero being at infinite separation of the proton and electron, eqn 13.13). The degeneracy of each level is gn = 2n2 where the n2 factor is the orbital degeneracy of each shell and the factor of 2 accounts for spin degeneracy. qE =
n=1
gn eEn /kT = 2
n=1
n2 e
 1
1 n2
C,
where C = hcRH /kTphotosphere = 27.301. qE , when written as an infinite sum, is infinitely large because lim of partition function terms corresponding to large n values is clearly an error. (b) States corresponding to large n values have very large average radii and most certainly interact with other atoms, thereby, blurring the distinct energy level of the state. Blurring interaction most likely occurs during the collision between an atom in state n and an atom in the ground state n = 1. Collisional lifetime broadening (eqn 16.25) is given by: En = zn h h = , 2 2
n
n2 e
(1
1 )C n2
= lim
n
n2 eC = eC lim (n2 ) = . The inclusion
n
where zn = collisional frequency of nth state of atomic perfect gas 2n c 2n cNA = (eqn 24.12) = kT MH c = mean speed = 8RT M
1 2
= 1.106 104 m s1
(eqn 24.7)
n = collisional crosssection of nth state (Fig. 24.9) = ( r n + a0 )2 =
2 a0
32 + 2 n 2
2
(Example 13.2)
Any quantum state within E of the continuum of an isolated atom will have its energy blurred by collisions so as to be indistinguishable from the continuum. Only states having energies in the range 0 E < E  E will be a distinct atomic quantum state. The maximum term, nmax , that should be retained in the partition function of a hydrogen atom is given by Enmax = E  Enmax 1 1 2 nmax hcRH = hcRH 
2 2 a0 3n2 +2 2 max c NA h 2
2 MH
with = 1.99 104 kg m3 and MH = 0.001 kg mol1 .
314
INSTRUCTOR'S MANUAL
The root function of a calculator or mathematical software may be used to solve this equation for nmax . nmax = 28 for atomic hydrogen of the photosphere Furthermore, examination of the partition function terms n = 2, 3, . . . , nmax indicates that they are negligibly small and may be discarded. The point is that very large n values should not be included in qE because they do not reflect reality. (c) n =
0
2 n2 eEn /kT qE
where
T = 5780 K
(eqn 19.6)
log ( n)
5
10 0 5 10 15 n 20 25 30
Figure 19.3
Even at the high temperature of the Sun's photosphere only the ground electronic state is significantly populated. This leads us to expect that at more ordinary temperatures only the ground state of atom and molecules are populated at equilibrium. It would be a mistake to thoughtlessly apply equilibrium populations to a study of the Sun's photosphere, however, it is bombarded with extremely high energy radiation from the direction of the Sun's core while radiating at a much low energy. The photosphere may show significant deviations from equilibrium. See S. J. Strickler, J. Chem. Ed., 43, 364 (1966).
20
Statistical thermodynamics: the machinery
Solutions to exercises
Discussion questions
E20.1(b) The symmetry number, , is a correction factor to prevent the overcounting of rotational states when computing the high temperature form of the rotational partition function. An elementary interpretation of is that it recognizes that in a homonuclear diatomic molecule AA the orientations AA and A A are indistinguishable, and should not be counted twice, so the quantity q = kT / hcB is replaced by q = kT / hcB with = 2. A more sophisticated interpretation is that the Pauli principle allows only certain rotational states to be occupied, and the symmetry factor adjusts the high temperature form of the partition function (which is derived by taking a sum over all states), to account for this restriction. In either case the symmetry number is equal to the number of indistinguishable orientations of the molecule. More formally, it is equal to the order of the rotational subgroup of the molecule. The temperature is always high enough (provided the gas is above its condensation temperature) for 3 the mean translational energy to be 2 kT . The equipartition value. Therefore, the molar constantT 3 volume heat capacity for translation is CV ,m = 2 R. 3 Translation is the only mode of motion for a monatomic gas, so for such a gas CV ,m = 2 R = 12.47 J K1 mol1 : This result is very reliable: helium, for example has this value over a range of 2000 K. When the temperature is high enough for the rotations of the molecules to be highly excited (when T R ) we can use the equipartition value kT for the mean rotational energy (for a linear rotor) to 3 obtain CV ,m = R. For nonlinear molecules, the mean rotational energy rises to 2 kT , so the molar 3 R . Only the lowest rotational state is occupied rotational heat capacity rises to 2 R when T when the temperature is very low, and then rotation does not contribute to the heat capacity. We can calculate the rotational heat capacity at intermediate temperatures by differentiating the equation for the mean rotational energy (eqn 20.29). The resulting expression, which is plotted in Fig. 20.9 of the text shows that the contribution rises from zero (when T = 0) to the equipartition value (when T R ). Because the translational contribution is always present, we can expect the molar heat T R 3 5 capacity of a gas of diatomic molecules (CV ,m + CV ,m ) to rise from 2 R to 2 R as the temperature is increased above R . Molecular vibrations contribute to the heat capacity, but only when the temperature is high enough for them to be significantly excited. The equipartition mean energy is kT for each mode, so the maximum contribution to the molar heat capacity is R. However, it is very unusual for the vibrations to be so highly excited that equipartition is valid and it is more appropriate to use the full expression for the vibrational heat capacity which is obtained by differentiating eqn 20.32. The curve in Fig. 20.10 of the text shows how the vibrational heat capacity depends on temperature. Note that even when the temperature is only slightly above the vibrational temperature, the heat capacity is close to its equipartition value. The total heat capacity of a molecular substance is the sum of each contribution (Fig. 20.11 of the text). When equipartition is valid (when the temperature is well above the characteristic temperature of the mode T M ) we can estimate the heat capacity by counting the numbers of modes that are active. In gases, all three translational modes are always active and contribute 3/2 R to the molar heat capacity. If we denote the number of active rotational modes by R (so for most molecules at normal temperatures R = 2 for linear molecules, and 3 for nonlinear molecules), then the rotational contribution is 1/2 R R. If the temperature is high enough for V vibrational modes to be active the
E20.2(b)
316
INSTRUCTOR'S MANUAL
vibrational contribution to the molar heat capacity is R R. In most cases V 0. It follows that the total molar heat capacity is 1 CV ,m = 2 (3 + R + 2V )R
E20.3(b)
See Justification 20.4 for a derivation of the general expression (eqn 20.54) for the equilibrium constant in terms of the partition functions and difference in molar energy, r E0 , of the products and reactants in a chemical reaction. The partition functions are functions of temperature and the ratio of partition functions in eqn 20.54 will therefore vary with temperature. However, the most direct effect of temperature on the equilibrium constant is through the exponential term e r E0 /RT . The manner in which both factors affect the magnitudes of the equilibrium constant and its variation with temperature is described in detail for a simple R P gas phase equilibrium in Section 20.7(c) and Justification 20.5.
Numerical exercises
E20.4(b)
1 CV ,m = 2 (3 + R + 2V )R [20.40]
with a mode active if T > M . At low temperatures, the vibrational modes are not active, that is, V = 0; at high temperatures they are active and approach the equipartition value. Therefore (a) (b) (c) O3 : C 2 H6 : CO2 : CV ,m = 3R
5 CV ,m = 2 R
or or
6R or
(3 3  6) vibrational modes (3 8  6) vibrational modes (3 3  5) vibrational modes
CV ,m = 3R
21R 6.5R
where the first value applies to low temperatures and the second to high. E20.5(b)
1 The equipartition theorem would predict a contribution to molar heat capacity of 2 R for every translational and rotational degree of freedom and R for each vibrational mode. For an ideal gas, Cp,m = R + CV ,m . So for CO2 7.5 1 1 = 1.15 With vibrations CV ,m /R = 3 2 + 2 2 + (3 4  6) = 6.5 and = 6.5 3.5 1 1 Without vibrations CV ,m /R = 3 2 + 2 2 = 2.5 and = = 1.40 2.5 37.11 J mol1 K 1 Experimental = = 1.29 37.11  8.3145 J mol1 K 1 The experimental result is closer to that obtained by neglecting vibrations, but not so close that vibrations can be neglected entirely.
E20.6(b)
The rotational partition function of a linear molecule is qR = 0.6952(T /K) (1.381 1023 J K1 )T kT = = 34 J s) (2.998 1010 cm s1 )B hcB (6.626 10 B/cm1 qR = 0.6952(25 + 273) = 143 1.4457 0.6952(250 + 273) = 251 qR = 1.4457
(a) At 25 C (b) At 250 C E20.7(b)
The symmetry number is the order of the rotational subgroup of the group to which a molecule belongs (except for linear molecules, for which = 2 if the molecule has inversion symmetry and 1 otherwise).
STATISTICAL THERMODYNAMICS: THE MACHINERY
317
(a) CO2 : Full group Dh ; subgroup C2 = 2 (b) O3 : Full group C2v ; subgroup C2 = 2
(d) SF6 : Oh = 24 (e) Al2 Cl6 : D2d = 4
2 (c) SO3 : Full group D3h ; subgroup {E, C3 , C3 , 3C2 } = 6
E20.8(b)
The rotational partition function of nonlinear molecule is given by qR = = 1 1 2
1/2 kT 3/2 hc ABC
(1.381 1023 J K1 ) (298 K) (6.626 1034 J s) (2.998 1010 cm s1 )
3/2
1/2 = 5.84 103 (2.02736) (0.34417) (0.293535) cm3
This hightemperature approximation is valid if T R = = hc(ABC)1/3 k
R , where R , the rotational temperature, is
(6.626 1034 J s) (2.998 1010 cm s1 ) 1.381 1023 J K1 [(2.02736) (0.34417) (0.293535) cm3 ]1/3
= 0.8479 K E20.9(b) q R = 5837 [Exercise 20.8(b)]
All rotational modes of SO2 are active at 25 C; therefore
R R 3 Um  Um (0) = E R = 2 RT ER R Sm = + R ln q R T 3 = 2 R + R ln(5836.9) = 84.57 J K1 mol1
E20.10(b) (a) The partition function is q=
states
eEstate /kT =
levels
geElevel /kT
where g is the degeneracy of the level. For rotations of a symmetric rotor such as CH3 CN, the energy levels are EJ = hc[BJ (J +1)+(AB)K 2 ] and the degeneracies are gJ,K = 2(2J +1) if K = 0 and 2J + 1 if K = 0. The partition function, then, is q =1+
J =1
(2J + 1)e{hcBJ (J +1)/kT } 1 + 2
J K=1
2 e{hc(AB)K /kT }
To evaluate this sum explicitly, we set up the following columns in a spreadsheet (values for B = 5.2412 cm1 and T = 298.15 K)
318
INSTRUCTOR'S MANUAL
J 0 1 2 3 . . . 82 83
J (J + 1) 0 2 6 12 . . . 6806 6972
2J + 1 1 3 5 7 . . . 165 167
e{hcBJ (J +1)/kT } 1 0.997 0.991 0.982 . . . 4.18 105 3.27 105
J term 1 8.832 23.64 43.88 . . . 0.079 0.062
e{hc(AB)K
2 /kT }
K sum 1 2.953 4.770 6.381 . . . 11.442 11.442
J sum 1 9.832 33.47 77.35 . . . 7498.95 7499.01
1 0.976 0.908 0.808 . . . 8 1071 2 1072
The column labelled K sum is the term in large parentheses, which includes the inner summation. The J sum converges (to 4 significant figures) only at about J = 80; the K sum converges much more quickly. But the sum fails to take into account nuclear statistics, so it must be divided by the symmetry number. At 298 K, q R = 2.50 103 . A similar computation at T = 500 K yields q R = 5.43 103 . (b) The rotational partition function of a nonlinear molecule is given by qR = 1
R 1/2 kT 3/2 hc ABC
At 298 K
1 q = 3
(1.381 1023 J K1 ) (298 K) (6.626 1034 J s) (2.998 1010 cm s1 )
1/2 (5.28) (0.307)2 cm3
3/2
= 2.50 103 At 500 K qR = 1 3 (1.381 1023 J K1 ) (500 K) (6.626 1034 J s) (2.998 1010 cm s1 )
1/2 (5.28) (0.307)2 cm3 3/2
= 5.43 103 E20.11(b) The rotational partition function of a nonlinear molecule is given by qR = 1
1/2 kT 3/2 hc ABC
(a) At 25 C q
R
1 = 1
(1.381 1023 J K1 ) (298 K) (6.626 1034 J s) (2.998 1010 cm s1 )
1/2 (3.1252) (0.3951) (0.3505) cm3
3/2
= 8.03 103
STATISTICAL THERMODYNAMICS: THE MACHINERY
319
(b) At 100 C q
R
1 = 1
(1.381 1023 J K1 ) (373 K) (6.626 1034 J s) (2.998 1010 cm s1 )
1/2 (3.1252) (0.3951) (0.3505) cm3
3/2
= 1.13 104 E20.12(b) The molar entropy of a collection of oscillators is given by Sm = NA + R ln q T
1 1 hc ~ = k /T and q = = where = hc hc ~ ~ 1 e 1 1  e/T e 1e where is the vibrational temperature hc /k. Thus ~ Sm = R(/T )  R ln(1  e/T ) e/T
A plot of Sm /R versus T / is shown in Fig. 20.1.
2.5 2.0 1.5 1.0 0.5 0.0 0 2 4 6 8 10
Figure 20.1 The vibrational entropy of ethyne is the sum of contributions of this form from each of its seven normal modes. The table below shows results from a spreadsheet programmed to compute Sm /R at a given temperature for the normalmode wavenumbers of ethyne.
/cm1 ~ 612 729 1974 3287 3374 /K 880 1049 2839 4728 4853 T = 298 K T / Sm /R 0.336 0.208 0.284 0.134 0.105 0.000766 0.0630 0.00000217 0.0614 0.00000146 T = 500 K T / Sm /R 0.568 0.491 0.479 0.389 0.176 0.0228 0.106 0.000818 0.103 0.000652
The total vibrational heat capacity is obtained by summing the last column (twice for the first two entries, since they represent doubly degenerate modes).
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INSTRUCTOR'S MANUAL
(a) At 298 K (b) At 500 K
Sm = 0.685R = 5.70 J mol1 K 1 Sm = 1.784R = 14.83 J mol1 K 1
E20.13(b) The contributions of rotational and vibrational modes of motion to the molar Gibbs energy depend on the molecular partition functions Gm  Gm (0) = RT ln q The rotational partition function of a nonlinear molecule is given by 1 q =
R 1/2 1.0270 kT 3/2 = hc ABC
(T /K)3 ABC/cm3
1/2
and the vibrational partition function for each vibrational mode is given by qV = 1 1  e/T
R
where = hc /k = 1.4388(~ /cm1 )/(T /K) ~ 2983 (3.553) (0.4452) (0.3948)
1/2
At 298 K
1.0270 q = 2
= 3.35 103
and GR  GR (0) = (8.3145 J mol1 K 1 ) (298 K) ln 3.35 103 m m = 20.1 103 J mol1 = 20.1 kJ mol1 The vibrational partition functions are so small that we are better off taking ln q V =  ln(1  e/T ) e/T
V ln q1 e{1.4388(1110)/298} = 4.70 103 V ln q2 e{1.4388(705)/298} = 3.32 102 V ln q3 e{1.4388(1042)/298} = 6.53 103
so GV GV (0) = (8.3145 J mol1 K 1 )(298 K)(4.70 103 +3.32 102 +6.53103 ) m m = 110 J mol1 = 0.110 kJ mol1 E20.14(b) q=
j
gj ej ,
g = (2S + 1)
1 for 2 for
states , , . . . states
[Section 17.1]
Hence q = 3 + 2e At 400 K =
[the 3 term is triply degenerate, and the 1 degenerate]
term is doubly (orbitally)
(1.4388 cm K) (7918.1 cm1 ) = 28.48 400 K
Therefore, the contribution to Gm is Gm  Gm (0) = RT ln q [Table 20.1, n = 1] RT ln q = (8.314 J K1 mol1 ) (400 K) ln(3 + 2 e28.48 ) = (8.314 J K1 mol1 ) (400 K) (ln 3) = 3.65 kJ mol1
STATISTICAL THERMODYNAMICS: THE MACHINERY
321
5 E20.15(b) The degeneracy of a species with S = 2 is 6. The electronic contribution to molar entropy is
Sm =
Um  Um (0) + R ln q = R ln q T
(The term involving the internal energy is proportional to a temperaturederivative of the partition function, which in turn depends on excited state contributions to the partition function; those contributions are negligible.) Sm = (8.3145 J mol1 K 1 ) ln 6 = 14.9 J mol1 K 1 E20.16(b) Use Sm = R ln s [20.52] Draw up the following table
n: s Sm /R 0 1 0 1 6 1.8 o 6 1.8 2 m 6 1.8 p 3 1.1 a 6 1.8 3 b 6 1.8 c 2 0.7 o 6 1.8 4 m 6 1.8 5 p 3 1.1 6 1.8 6 1 0
where a is the 1, 2, 3 isomer, b the 1, 2, 4 isomer, and c the 1, 3, 5 isomer. E20.17(b) We need to calculate K =
J
q J,m NA
J
e E0 /RT [Justification 20.4]
=
q(79 Br 2 )q(81 Br 2 )  E0 /RT m m e q(79 Br 81 Br)2 m
Each of these partition functions is a product
T qm q R q V q E
with all q E = 1. 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 = 2, so q R (79 Br 2 )q R (81 Br 2 ) q R (79 Br 81 Br)2 The value of K 0.25 = 0.25
E0 is also very small compared with RT , so
Solutions to problems
Solutions to numerical problems
P20.2 = = gB B [18.48, Section 18.14] q = 1 + e CV ,m /R = x 2 ex [Problem 20.1], (1 + ex )2 x = 2B B [g = 2]
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INSTRUCTOR'S MANUAL
Therefore, if B = 5.0 T, x= (2) (9.274 1024 J T1 ) (5.0 T) (1.381 1023 J K1 ) T = 6.72 T /K
(a) T = 50 K, x = 0.134, CV = 4.47 103 R, implying that CV = 3.7 102 J K1 mol1 . Since the equipartition value is about 3R [R = 3, V 0], the field brings about a change of about 0.1 per cent (b) T = 298 K, x = 2.26 102 , CV = 1.3 104 R, implying that CV = 1.1 mJ K1 mol1 , a change of about 4 103 per cent
Question. What percentage change would a magnetic field of 1 kT cause? P20.4 q = 1 + 5e [gj = 2J + 1] [E = hcBJ (J + 1)] 5e
= E(J = 2)  E(J = 0) = 6hcB 1 q U  U (0) = = q N 1 + 5e CV ,m = k 2 CV ,m /R = Um [20.35] V
5 2 2 e 180(hcB)2 e6hcB = (1 + 5e )2 (1 + 5e6hcB )2
hcB = 1.4388 cm K 60.864 cm1 = 87.571 K k Hence, CV ,m /R = 1.380 106 e525.4 K/T (1 + 5e525.4 K/T ) (T /K)2
We draw up the following table
T /K CV ,m /R 50 0.02 100 0.68 150 1.40 200 1.35 250 1.04 300 0.76 350 0.56 400 0.42 450 0.32 500 0.26
These points are plotted in Fig. 20.2.
1.5
1.0
V
0.5
0 0 100 200 300 400 500
Figure 20.2
STATISTICAL THERMODYNAMICS: THE MACHINERY
323
P20.6
T qm = 2.561 102 (T /K)5/2 (M/g mol1 )3/2 [Table 20.3] NA
= (2.561 102 ) (298)5/2 (28.02)3/2 = 5.823 106 qR = 1 298 0.6950 [Table 20.3] = 51.81 2 1.9987 1 qV = [Table 20.3] = 1.00 2358/207.2 1e
Therefore q m = (5.823 106 ) (51.81) (1.00) = 3.02 108 NA
3 5 Um  Um (0) = 2 RT + RT = 2 RT
[T
T , R ]
Hence S = m q Um  Um (0) + R ln m + 1 T NA
5 = 2 R + R{ln 3.02 108 + 1} = 23.03R = 191.4 J K1 mol1
The difference between the experimental and calculated values is negligible, indicating that the residual entropy is negligible. P20.9 (a) Rotational state probability distribution,
R PJ (T ) =
J =0
(2J + 1)ehcBJ (J +1)/kT , (2J + 1)ehcBJ (J +1)/kT
[20.14]
is conveniently plotted against J at several temperatures using mathematical software. This distribution at 100 K is shown below as both a bar plot and a line plot.
Rotational distributions 0.15 100 K 0.1 PR(T) J 0.05 1000 K 300 K 600 K
0
0
5
10
15
20 J
25
30
35
40
Figure 20.3(a)
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INSTRUCTOR'S MANUAL
The plots show that higher rotational states become more heavily populated at higher temperature. Even at 100 K the most populated state has 4 quanta of rotational energy; it is elevated to 13 quanta at 1000 K. Values of the vibrational state probability distribution,
V ~ ~ P (T ) = ehc /kT (1  ehc /kT )1 ,
[20.21]
are conveniently tabulated against at several temperatures. Computations may be discontinued when values drop below some small number like 107 .
PV (T ) 0 1 2 3 4 5 100 K 1 2.77 1014 300 K 1 3.02 105 9.15 1010 600 K 0.995 5.47 103 3.01 105 1.65 107 1000 K 0.956 0.042 1.86 103 8.19 105 3.61 106 1.59 107
Only the state = 0 is appreciably populated below 1000 K and even at 1000 K only 4% of the molecules have 1 quanta of vibrational energy. (b) R = 6.626 1034 J s 3.000 108 m s1 hcB = k 1.381 1023 J K1 193.1 m1 (Section 20.2b)
R = 2.779 K T where T is the lowest temperature of current interest (100 K), we expect that the Since R classical rotational partition function,
R qclassical (T ) =
kT , hcB
[20.15a]
Classical partition function error 0
0.2 Percentage deviation
0.4
0.6
0.8
1 0 200 400 600 800 1000 Temperature / K
Figure 20.3(a)
STATISTICAL THERMODYNAMICS: THE MACHINERY
325
should agree well with the rotational partition function calculated with the discrete energy distribution,
qR =
J =0
(2J + 1)ehcBJ (J +1)/kT .
[20.14]
R A plot of the percentage deviation (qclassical q R )100/q R confirms that they agree. The maximum deviation is about 0.9% at 100 K and the magnitude decreases with increasing temperature.
(c) The translational, rotational, and vibrational contributions to the total energy are specified by eqns 20.28, 20.30, and 20.32. As molar quantities, they are:
3 U T = 2 RT ,
U R = RT ,
~ NA hc U V = hc /kT e ~ 1
The contributions to the energy change from 100 K are U T (T ) = U T (T )  U T (100 K), etc. The following graph shows the individual contributions to the total molar internal energy change from 100 K. Translational motion contributes 50% more than the rotational motion because it has 3 quadratic degrees of freedom compared to 2 quadratic degrees of freedom for rotation. Very little change occurs in the vibration energy because very high temperatures are required to populate = 1, 2, . . . . states (see Part a). CV ,m (T ) = = U (T ) = T V (U T + U R + U V ) T V [2.19]
3 dU V 5 dU V R+R+ = R+ 2 dT 2 dT
Energy change contributions 15
Total 10 Translational U kJ mol1 5 Rotational
Vibrational 0 100 200 300 400 500 T/K 600 700 800 900 1000
Figure 20.3(c)
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INSTRUCTOR'S MANUAL
The derivative dU V /dT may be evaluated numerically with numerical software (we advise exploration of the technique) or it may be calculated with the analytical function of eqn 20.39: eV /2T 1  eV /T
2
V CV ,m
dU V = =R dT
V T
where V = hc /k = 3122 K. The following graph shows the ratio of the vibrational contribu~ tion to the sum of translational and rotational contributions. Below 300 K, vibrational motions makes a small, perhaps negligible, contribution to the heat capacity. The contribution is about 10% at 600 K and grows with increasing temperature.
Relative contributions to the heat capacity 0.2
V Cv ,m 0.1 T R Cv ,m + Cv ,m
0 0 200 400 T/K 600 800 1000
Figure 20.3(d)
The molar entropy change with temperature may be evaluated by numerical integration with mathematical software. Cp,m (T ) dT T 100 K [3.20]
T
S(T ) = S(T )  S(100 K) = = =
T
[4.19]
CV ,m (T ) + R dT T 100 K
T 100 K 7 2R V + CV ,m (T )
T
dT
V T CV ,m (T ) T 7 dT + S(T ) = R ln 2 100 K T 100 K S T +R (T ) S V (T )
Even at the highest temperature the vibrational contribution to the entropy change is less than 2.5% of the contributions from translational and rotational motion. The vibrational contribution is negligible at low temperature.
STATISTICAL THERMODYNAMICS: THE MACHINERY
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Relative contributions to the entropy change 0.03
0.02 S V S T + R 0.01
0 0 200 400 T/K 600 800 1000
Figure 20.3(e)
P20.10
q(CHD3 )q(DCl)  E0 m K = m e [20.54, NA factors cancel] q (CD4 )q(HCl) m m The ratio of translational partition functions is
T T qm (CHD3 )qm (DCl) = T (CD )q T (HCl) qm 4 m
M(CHD3 )M(DCl) 3/2 = M(CD4 )M(HCl)
19.06 37.46 3/2 = 0.964 20.07 36.46
The ratio of rotational partition functions is (CD4 ) (B(CD4 )/cm1 )3/2 B(HCl)/cm1 q R (CHD3 )q R (DCl) = R (CD )q R (HCl) (CHD3 ) (A(CHD3 )B(CHD3 )2 /cm3 )1/2 B(DCl)/cm1 q 4 = 12 2.633/2 10.59 = 6.24 3 (2.63 3.282 )1/2 5.445
The ratio of vibrational partition functions is q V (CHD3 )q V (DCl) q(2993)q(2142)q(1003)3 q(1291)2 q(1036)2 q(2145) = q V (CD4 )q V (HCl) q(2109)q(1092)2 q(2259)3 q(996)3 q(2991) 1 1  e1.4388x/(T /K) We also require E0 , which is equal to the difference in zero point energies where q(x) = 1 E0 = {(2993 + 2142 + 3 1003 + 2 1291 + 2 1036 + 2145) hc 2  (2109 + 2 1092 + 3 2259 + 3 996 + 2991)} cm1 = 1053 cm1 Hence, K = 0.964 6.24 Qe+1.4388990/(T /K) = 6.02Qe+1424/(T /K)
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INSTRUCTOR'S MANUAL
where Q is the ratio of vibrational partition functions. We can now evaluate K (on a computer), and obtain the following values
T /K K 300 698 400 217 500 110 600 72 700 54 800 44 900 38 1000 34
The values of K are plotted in Fig. 20.4.
800
600
400
200
0 300 400 500 600 700 800 900 1000
Figure 20.4
Solutions to theoretical problems
P20.13 (a) V and R are the constant factors in the numerators of the negative exponents in the sums that are the partition functions for vibration and rotation. They have the dimensions of temperature which occurs in the denominator of the exponents. So high temperature means T V or R and only then does the exponential become substantial. Thus V is a measure of the temperature at which higher vibrational and rotational states become populated. R = (2.998 108 m s1 ) (6.626 1034 J s) (60.864 cm1 ) hc = k (1.381 1023 J K1 ) (1 m/100 cm) hc ~ (6.626 1034 J s) (4400.39 cm1 ) (2.998 108 m s1 ) = k (1.381 1023 J K1 ) (1 m/100 cm)
= 87.55 K V =
= 6330 K (b) and (c) These parts of the solution were performed with Mathcad 7.0 and are reproduced on the following pages. Objective: To calculate the equilibrium constant K(T ) and Cp (T ) for dihydrogen at high temperature for a system made with n mol H2 at 1 bar. H2 (g) 2H(g) At equilibrium the degree of dissociation, , and the equilibrium amounts of H2 and atomic hydrogen are related by the expressions nH2 = (1  )n and nH = 2n
STATISTICAL THERMODYNAMICS: THE MACHINERY
329
The equilibrium mole fractions are xH2 = (1  )n/{(1  )n + 2n} = (1  )/(1 + ) xH = 2n/{(1  )n + 2n} = 2/(1 + ) The partial pressures are pH2 = (1  )p/(1 + ) The equilibrium constant is
   K(T ) = (pH /p  )2 /(pH2 /p  ) = 4 2 (p/p  )/(1  2 )
and
pH = 2p/(1 + )
= 4 2 /(1  2 ) The above equation is easily solved for = (K/(K + 4))1/2
 where p = p  = 1 bar
The heat capacity at constant volume for the equilibrium mixture is CV (mixture) = nH CV ,m (H) + nH2 CV ,m (H2 ) The heat capacity at constant volume per mole of dihydrogen used to prepare the equilibrium mixture is CV = CV (mixture)/n = {nH CV ,m (H) + nH2 CV ,m (H2 )}/n = 2CV ,m (H) + (1  )CV ,m (H2 ) The formula for the heat capacity at constant pressure per mole of dihydrogen used to prepare the equilibrium mixture (Cp ) can be deduced from the molar relationship Cp,m = CV ,m + R Cp = nH Cp,m (H) + nH2 Cp,m (H2 ) /n nH nH = CV ,m (H) + R + 2 CV ,m (H2 ) + R n n nH CV ,m (H) + nH2 CV ,m (H2 ) nH + nH2 = +R n n = CV + R(1 + ) Calculations J = joule mol = mole h = 6.62608 1034 J s R = 8.31451 J K1 mol1 Molecular properties of H2 = 4400.39 cm1 1 g mol1 mH = NA hc V = k B = 60.864 cm1 mH2 = 2mH R = hcB k D = 432.1 kJ mol1 s = second g = gram c = 2.9979 108 m s1 NA = 6.02214 1023 mol1 kJ = 1000 J bar = 1 105 Pa k = 1.38066 1023 J K1 p = 1 bar
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INSTRUCTOR'S MANUAL
Computation of K(T ) and (T ) N = 200
Hi
i = 0, . . . , N
Ti = 500 K +
H2i
=
h (2mH kTi )1/2 1 1  e(V /Ti ) kTi
3 H2i
= Ti 2R
h (2 mH2 kTi )1/2
i 5500 K N
qVi =
qRi = e(D/RTi )
Hi 6
Keqi =
 p  qVi qRi
i =
1/2 Keqi Keqi + 4
See Fig. 20.5(a) and (b).
(a) 1
0.5
0 0 1000 2000 3000 4000 5000 6000
(b) 100 80 60 40 20 0 0 1000 2000 3000 4000 5000 6000
Figure 20.5
STATISTICAL THERMODYNAMICS: THE MACHINERY
331
Heat capacity at constant volume per mole of dihydrogen used to prepare equilibrium mixture (see Fig. 20.6(a))
27 26 25 24 23 22 21 20 0 100 0 2000 3000 4000 5000 6000
Figure 20.6(a)
CV (H) = 1.5R CV (H2i ) = 2.5R + V e(V /2Ti ) Ti 1  e(V /Ti )
2
R
CVi = 2i CV (H) + (1  i )CV (H2i )
The heat capacity at constant pressure per mole of dihydrogen used to prepare the equilibrium mixture is (see Fig. 20.6(b)) Cpi = CVi + R(1 + i )
42
40
38 ( 36 ( 34 32 30 28 0 1000 2000 3000 4000 5000 6000
Figure 20.6(b)
332
INSTRUCTOR'S MANUAL
P20.14
q=
1 , 1  ex
x = h = hc = ~ N q
V [Table 20.3] T
U  U (0) =  = CV =
q d = N (1  ex ) (1  ex )1 V d
N h N hex = x x 1e e 1
U U U = k 2 h = k 2 x T V = k( )2 N h ex (ex  1)2 = kN x 2 ex (ex  1)2 N h N hex = x x 1e e 1
H  H (0) = U  U (0)[q is independent of V ] = S=
N kxex U  U (0) + nR ln q =  N k ln(1  ex ) T 1  ex x  ln(1  ex ) ex  1
= Nk
A  A(0) = G  G(0) = nRT ln q = N kT ln(1  ex ) The functions are plotted in Fig. 20.7. P20.15 (a) gJ eJ /kT NJ gJ eJ /kT = = J /kT N q J gJ e For a linear molecule gJ = 2J + 1 and J = hcBJ (J + 1). Therefore, NJ (2J + 1)ehcBJ (J +1)/kT (b) Jmax occurs when dnJ /dJ = 0. dNJ N d = q dJ dJ 2  (2Jmax + 1) 2Jmax + 1 = Jmax = (2J + 1)e hcB kT

hcBJ (J +1) kT
=0
(2Jmax + 1) = 0
2kT 1/2 hcB
1/2 kT 1  2hcB 2
(c) Jmax 3 because the R branch J = 3 4 transition has the least transmittance. Solving the previous equation for T provides the desired temperature estimate.
STATISTICAL THERMODYNAMICS: THE MACHINERY
333
6 100 4 S/Nk 2 50 0 0.01 0 2 0 0.01 4 0.1 1.0 x 0.1 x 1.0 1.0 0.8 0.6 0.4 0.2 0.0 0.01 0.1 1 x 10 100
Figure 20.7
T
hcB (2Jmax + 1)2 2k
2 (6.626 1034 J s) (3.000 108 m s1 ) (10.593 cm1 ) 10mcm (7)2
2(1.38066 1023 J K1 )
T 374 K P20.17 All partition functions other than the electronic partition function are unaffected by a magnetic field; hence the relative change in K is the relative change in q E . qE =
MJ
egB BMJ ,
3 1 1 3 MJ =  2 ,  2 , + 2 , + 2 ; g = 4 3
Since gB B
1 for normally attainable fields, 1 1  gB BMJ + (gB BMJ )2 + 2
qE =
MJ
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INSTRUCTOR'S MANUAL
1 2 MJ = 4 + (gB B)2 2 MJ
MJ
MJ = 0 = 4 1 +
10 (B B)2 9
g=
4 3
Therefore, if K is the actual equilibrium constant and K 0 is its value when B = 0, we write
2 K 20 10 = 1 + (B B)2 1 + 2 2 B 2 0 9 9 B K
For a shift of 1 per cent, we require
20 2 2 2 9 B B
0.01,
or
B B 0.067
Hence B 0.067kT (0.067) (1.381 1023 J K1 ) (1000 K) = 100 T B 9.274 1024 J T1
Solutions to applications
P20.20 The standard molar Gibbs energy is given by
 q     Gm  Gm (0) = RT ln m NA    qm,tr R V E q where m = q q q NA NA
Translation:
  qm,tr
NA
=
kT
 p 3
= 2.561 102 (T /K)5/2 (M/g mol1 )3/2 = (2.561 102 ) (2000)5/2 (38.90)3/2 = 1.111 109
Rotation of a linear molecule: qR = 0.6950 T /K kT = hcB B/cm1
The rotational constant is B= h h = 4cI 4cmeff R 2 where meff = = meff = 1.296 1026 kg B= 1.0546 1034 J s 4(2.998 1010 cm s1 ) (1.296 1026 kg) (190.5 1012 m)2 0.6950 2000 = 2335 1 0.5952 = 0.5952 cm1 mB mSi mB + mSi 103 kg mol1 (10.81) (28.09) 10.81 + 28.09 6.022 1023 mol1
so q R =
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335
Vibration: q V =
1
~ 1  ehc /kT
=
1 1  exp
1.4388(~ /cm1 ) T /K
=
1 1  exp
1.4388(772) 2000
= 2.467 The Boltzmann factor for the lowestlying electronic excited state is exp (1.4388) (8000) 2000 = 3.2 103
The degeneracy of the ground level is 4 (spin degeneracy = 4, orbital degeneracy = 1), and that of the excited level is also 4 (spin degeneracy = 2, orbital degeneracy = 2), so q E = 4(1 + 3.2 103 ) = 4.013 Putting it all together yields
    Gm  Gm (0) = (8.3145 J mol1 K 1 ) (2000 K) ln(1.111 109 ) (2335)
(2.467) (4.013) = 5.135 105 J mol1 = 513.5 kJ mol1 P20.22 The standard molar Gibbs energy is given by
 q     Gm  Gm (0) = RT ln m NA    qm,tr R V E q where m = q q q NA NA
First, at 10.00 K Translation:
  qm,tr
NA
= 2.561 102 (T /K)5/2 (M/g mol1 )3/2 = (2.561 102 ) (10.00)5/2 (36.033)3/2 = 1752
Rotation of a nonlinear molecule: qR = 1
1/2 1.0270 (T /K)3/2 kT 3/2 = hc ABC (ABC/cm3 )1/2
The rotational constants are B= h 4cI so ABC = h 3 1 , 4 c I A I B IC
3
ABC =
1.0546 1034 J s 4(2.998 1010 cm s1 )
(1010 m
1 6 )
(39.340) (39.032) (0.3082) (u 2 )3 (1.66054 1027 kg u1 )3
= 101.2 cm3 so q R = 1.0270 (10.00)3/2 = 1.614 2 (101.2)1/2
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INSTRUCTOR'S MANUAL
Vibration: q V =
1
~ 1  ehc /kT
=
1 1  exp
1.4388(~ /cm1 ) T /K
=
1 1  exp
1.4388(63.4) 10.00
= 1.0001 Even the lowestfrequency mode has a vibrational partition function of 1; so the stiffer vibrations have q V even closer to 1. The degeneracy of the electronic ground state is 1, so q E = 1. Putting it all together yields
    Gm  Gm (0) = (8.3145 J mol1 K 1 ) (10.00 K) ln(1752) (1.614) (1) (1)
= 660.8 J mol1 Now at 1000 K Translation: Rotation: Vibration:
  qm,tr
qR =
1.0270 (1000)3/2 = 1614 2 (101.2)1/2 1 q V(1) = = 11.47 (1.4388)(63.4) 1  exp  1000 1 q V(2) = = 1.207 (1.4388)(1224.5) 1  exp  1000 1 V(3) = 1.056 q = 1  exp  (1.4388)(2040) 1000 q V = (11.47) (1.207) (1.056) = 14.62
NA
= (2.561 102 ) (1000)5/2 (36.033)3/2 = 1.752 108
Putting it all together yields
    Gm  Gm (0) = (8.3145 J mol1 K 1 ) (1000 K) ln(1.752 108 ) (1614)
(14.62) (1) = 2.415 105 J mol1 = 241.5 kJ mol1
21
Molecular interactions
Solutions to exercises
Discussion questions
E21.1(b) When the applied field changes direction slowly, the permanent dipole moment has time to reorientatethe whole molecule rotates into a new directionand follow the field. However, when the frequency of the field is high, a molecule cannot change direction fast enough to follow the change in direction of the applied field and the dipole moment then makes no contribution to the polarization of the sample. Because a molecule takes about 1 ps to turn through about 1 radian in a fluid, the loss of this contribution to the polarization occurs when measurements are made at frequencies greater than about 1011 Hz (in the microwave region). We say that the orientation polarization, the polarization arising from the permanent dipole moments, is lost at such high frequencies. The next contribution to the polarization to be lost as the frequency is raised is the distortion polarization, the polarization that arises from the distortion of the positions of the nuclei by the applied field. The molecule is bent and stretched by the applied field, and the molecular dipole moment changes accordingly. The time taken for a molecule to bend is approximately the inverse of the molecular vibrational frequency, so the distortion polarization disappears when the frequency of the radiation is increased through the infrared. The disappearance of polarization occurs in stages: as shown in Justification 21.3, each successive stage occurs as the incident frequency rises above the frequency of a particular mode of vibration. At even higher frequencies, in the visible region, only the electrons are mobile enough to respond to the rapidly changing direction of the applied field. The polarization that remains is now due entirely to the distortion of the electron distribution, and the surviving contribution to the molecular polarizability is called the electronic polarizability. There are three van der Waals type interactions that depend upon distance as 1/r 6 ; they are the Keesom interaction between rotating permanent dipoles, the permanentdipoleinduced dipoleinteraction, and the induceddipoleinduceddipole, or London dispersion, interaction. In each case, we can visualize the distance dependence of the potential energy as arising from the 1/r 3 dependence of the field (and hence the magnitude of the induced dipole) and the 1/r 3 dependence of the potential energy of interaction of the dipoles (either permanent or induced). The goal is to construct the radial distribution function, g(r), which gives the relative locations of the particles in the liquid (eqn 21.35). Once g(r) is known it can be used to calculate the thermodynamic properties of the liquid. This expression is nothing more than the Boltzmann distribution of statistical thermodynamics for two molecules in a field generated by all the other molecules in the system. There are several ways of building the intermolecular potential into the calculation of g(r). Numerical methods take a box of about 103 particles (the number increases as computers grow more powerful), and the rest of the liquid is simulated by surrounding the box with replications of the original box (Fig. 21.29 of the text). Then, whenever a particle leaves the box through one of its faces, its image arrives through the opposite face. When calculating the interactions of a molecule in a box, it interacts with all the molecules in the box and all the periodic replications of those molecules and itself in the other boxes. Once g(r) is known it can be used to calculate the thermodynamic properties of liquids. (a) Monte Carlo methods In the Monte Carlo method, the particles in the box are moved through small but otherwise random distances, and the change in total potential energy of the N particles in the box, VN , is calculated
E21.2(b)
E21.3(b)
338
INSTRUCTOR'S MANUAL
using one of the intermolecular potentials discussed in Sections 21.5 and 21.6. Whether or not this new configuration is accepted is then judged from the following rules: 1 If the potential energy is not greater than before the change, then the configuration is accepted. 2 If the potential energy is greater than before change, the Boltzmann factor e VN /kT is compared with a random number between 0 and 1; if the factor is larger than the random number, the configuration is accepted; if the factor is not larger, the configuration is rejected. This procedure ensures that at equilibrium the probability of occurrence of any configuration is proportional to the Boltzmann factor. The configurations generated in this way can then be used to construct g(r) simply by counting the number of pairs of particles with a separation r and averaging the result over the whole collection of configurations. (b) Molecular dynamics In the molecular dynamics approach, the history of an initial arrangement is followed by calculating the trajectories of all the particles under the influence of the intermolecular potentials. Newton's laws are used to predict where each particle will be after a short time interval (about 1 fs. which is shorter than the average time between collisions), and then the calculation is repeated for tens of thousands of such steps. The timeconsuming part of the calculation is the evaluation of the net force on the molecule arising from all the other molecules present in the system. A molecular dynamics calculation gives a series of snapshots of the liquid, and g(r) can be calculated as before. The temperature of the system is inferred by computing the mean kinetic energy 2 of the particles and using the equipartition result that 1/2 mvq = 1/2 kT for each coordinate q. E21.4(b) Describe how molecular beams are used to investigate intermolecular potentials. A molecular beam is a narrow stream of molecules with a narrow spread of velocities and, in some cases, in specific internal states or orientations. Molecular beam studies of nonreactive collisions are used to explore the details of intermolecular interactions with a view to determining the shape of the intermolecular potential. The primary experimental information from a molecular beam experiment is the fraction of the molecules in the incident beam that are scattered into a particular direction. The fraction is normally expressed in terms of dI , the rate at which molecules are scattered into a cone that represents the area covered by the "eye" of the detector (Fig. 21.21 of the text). This rate is reported as the differential scattering crosssection, , the constant of proportionality between the value of dI and the intensity, I , of the incident beam, the number density of target molecules, N , and the infinitesimal path length dx through the sample: dI = I Ndx. The value of (which has the dimensions of area) depends on the impact parameter, b, the initial perpendicular separation of the paths of the colliding molecules (Fig. 21.22), and the details of the intermolecular potential. The scattering pattern of real molecules, which are not hard spheres, depends on the details of the intermolecular potential, including the anisotropy that is present when the molecules are nonspherical. The scattering also depends on the relative speed of approach of the two particles: a very fast particle might pass through the interaction region without much deflection, whereas a slower one on the same path might be temporarily captured and undergo considerable deflection (Fig. 21.24). The variation of the scattering crosssection with the relative speed of approach therefore gives information about the strength and range of the intermolecular potential. Another phenomenon that can occur in certain beams is the capturing of one species by another. The vibrational temperature in supersonic beams is so low that van der Waals molecules may be formed, which are complexes of the form AB in which A and B are held together by van der Waals forces or
MOLECULAR INTERACTIONS
339
hydrogen bonds. Large numbers of such molecules have been studied spectroscopically, including ArHCl, (HCl)2 ArCO2 , and (H2 O)2 . More recently, van der Waals clusters of water molecules have been pursued as far as (H2 O)6 . The study of their spectroscopic properties gives detailed information about the intermolecular potentials involved.
Numerical exercises
E21.5(b) E21.6(b) A molecule that has a centre of symmetry cannot be polar. SO3 (D3h ) and XeF4 (D4h ) cannot be polar. SF4 (seesaw, C2v ) may be polar. The molar polarization depends on the polarizability through Pm = NA 30 + 2 3kT
This is a linear equation in T 1 with slope m= NA 2 90 k so = 90 km 1/2 = (4.275 1029 C m) (m/(m3 mol1 K))1/2 NA
and with yintercept b= NA 30 so = 30 b = (4.411 1035 C2 m2 J1 )b/(m3 mol1 ) NA
Since the molar polarization is linearly dependent on T 1 , we can obtain the slope m and the intercept b m= Pm,2  Pm,1 (75.74  71.43) cm3 mol1 = 5.72 103 cm3 mol1 K = 1 1 (320.0 K)1  (421.7 K)1 T1  T2
and b = Pm  mT 1 = 75.74 cm3 mol1  (5.72 103 cm3 mol1 K) (320.0 K)1 = 57.9 cm3 mol1 It follows that = (4.275 1029 C m) (5.72 103 )1/2 = 3.23 1030 C m and = (4.411 1035 C2 m2 J1 ) (57.9 106 ) = 2.55 1039 C2 m2 J1 E21.7(b) The relative permittivity is related to the molar polarization through Pm r  1 = C r + 2 M so r = 2C + 1 , 1C
C=
(1.92 g cm3 ) (32.16 cm3 mol1 ) = 0.726 85.0 g mol1 2 (0.726) + 1 = 8.97 1  0.726
r =
340
INSTRUCTOR'S MANUAL
E21.8(b)
If the permanent dipole moment is negligible, the polarizability can be computed from the molar polarization Pm = NA 30 so = 30 Pm NA
and the molar polarization from the refractive index Pm r  1 n2  1 r = = 2 M r + 2 nr + 2 = so = 30 M NA n2  1 r n2 + 2 r 1.6222  1 1.6222 + 2
3 (8.854 1012 J1 C2 m1 ) (65.5 g mol1 ) (6.022 1023 mol1 ) (2.99 106 g m3 )
= 3.40 1040 C2 m2 J1 E21.9(b) = qR [q = be, b = bond order]
For example, ionic (C F) = (1.602 1019 C) (1.41 1010 m) = 22.6 1030 C m = 6.77 D obs Then, per cent ionic character = 100 ionic values are based on Pauling electronegativities as found in any general chemistry text. We draw up the following table
Bond C F C O obs /D 1.4 1.2 ionic /D 6.77 6.87 Per cent 21 17 1.5 1.0
The correlation is at best qualitative . Comment. There are other contributions to the observed dipole moment besides the term qR. These are a result of the delocalization of the charge distribution in the bond orbitals. Question. Is the correlation mentioned in the text [21.2] any better? E21.10(b) = (2 + 2 + 21 2 cos )1/2 1 2
2 2
[21.3a]
= [(1.5) + (0.80) + (2) (1.5) (0.80) (cos 109.5 )]1/2 D = 1.4 D E21.11(b) The components of the dipole moment vector are x = and y =
i
qi xi = (4e) (0) + (2e) (162 pm) + (2e) (143 pm) (cos 30 ) = (572 pm)e
i
qi yi = (4e) (0) + (2e) (0) + (2e) (143 pm) (sin 30 ) = (143 pm)e
The magnitude is = (2 + 2 )1/2 = ((570)2 + (143)2 )1/2 pm e = (590 pm)e y x = (590 1012 m) (1.602 1019 C) = 9.45 1029 C m
MOLECULAR INTERACTIONS
341
and the direction is = tan1 the negative xaxis).
y 143 pm e = 194.0 from the xaxis (i.e., 14.0 below = tan1 x 572 pm e
E21.12(b) The induced dipole moment is = E = 40 E = 4(8.854 1012 J1 C2 m1 ) (2.22 1030 m3 ) (15.0 103 V m1 ) = 3.71 1036 C m
E21.13(b) The solution to Exercise 21.8(a) showed that = 30 M NA n2  1 r n2 + 2 r or = 3M 4NA n2  1 r n2 + 2 r
which may be solved for nr to yield nr = = nr = + 2 
1/2
with =
3M 4NA
(3) (72.3 g mol1 ) = 3.314 1029 m3 (4) (0.865 106 g m3 ) (6.022 1023 mol1 ) 33.14 + 2 2.2 33.14  2.2
1/2
= 1.10
E21.14(b) The relative permittivity is related to the molar polarization through r  1 Pm = C r + 2 M so r = 2C + 1 1C
The molar polarization depends on the polarizability through Pm = C= NA 30 + 2 3kT so C= NA 30 M 4 0 + 2 3kT
(1491 kg m3 ) (6.022 1023 mol1 ) 3(8.854 1012 J1 C2 m1 ) (157.01 103 kg mol1 ) 4(8.854 1012 J1 C2 m1 ) (1.5 1029 m3 ) + (5.17 1030 C m)2 3(1.381 1023 J K1 ) (298 K) and r = 2(0.83) + 1 = 16 1  0.83
C = 0.83
342
INSTRUCTOR'S MANUAL
E21.15(b) The rotation of planepolarized light is described by = (nR  nL ) = 2l so nR  n L = 2 l 2 360
(450 109 m) (2 192 ) 2(15 102 m)
nR  nL = 3.2 106
Solutions to problems
Solutions to numerical problems
P21.2 The energy of the dipole 1 E. To flip it over requires a change in energy of 21 E. This will occur when the energy of interaction of the dipole with the induced dipole of the Ar atom equals 21 E. The magnitude of the dipoleinduced dipole interaction is V = r6 = 2 2 1 0 r 6 [21.26] = 21 E [after flipping over]
1 2 (6.17 1030 C m) (1.66 1030 m3 ) = 20 E (2) (8.854 1012 J1 C2 m1 ) (1.0 103 V m1 ) = 1.84 1052 m6
r = 2.4 109 m = 2.4 nm Comment. This distance is about 24 times the radius of the Ar atom. r  1 M 4 NA 2 Pm = NA + and Pm = [21.15 and 21.16 with = 4 0 ] r + 2 3 90 kT The data have been corrected for the variation in methanol density, so use = 0.791 g cm3 for all entries. Obtain and from the liquid range ( > 95 C) results, but note that some molecular rotation occurs even below the freezing point (thus the 110 C value is close to the 80 C value). Draw up the following table using M = 32.0 g mol1 .
/ C T /K 1000 T /K r r  1 r + 2 Pm /(cm3 mol1 ) 80 193 5.18 57 0.949 38.4 50 223 4.48 49 0.941 38.1 20 253 3.95 42 0.932 37.7 0 273 3.66 38 0.925 37.4 20 293 3.41 34 0.917 37.1
P21.4
MOLECULAR INTERACTIONS
343
Pm is plotted against
39
1 in Fig. 21.1. T
38
m
37 3.2 3.6 4.0 4.4 4.8 5.2
Figure 21.1 1 = 0 is 34.8 (not shown in the figure) and the slope is 721 (from a The extrapolated intercept at T leastsquares analysis). It follows that = 3Pm (at intercept) (3) (35.0 cm3 mol1 ) = 1.38 1023 cm3 = 4NA (4 ) (6.022 1023 mol1 )
= (1.282 102 D) (721)1/2 [from Problem 21.3] = 0.34 D The jump in r which occurs below the melting temperature suggests that the molecules can rotate while the sample is still solid. P21.6 4 N A 2 NA + [21.16, with = 4 0 ] 3 90 kT Draw up the following table Pm = T /K 384.3 420.1 444.7 484.1 522.0 1000 2.602 2.380 2.249 2.066 1.916 T /K 53.5 50.1 46.8 43.1 Pm /(cm3 mol1 ) 57.4 The points are plotted in Fig. 21.2. The extrapolated (leastsquares) intercept is 3.44 cm3 mol1 ; the slope is 2.084 104 . = (1.282 102 D) (slope)1/2 [Problem 21.3] = 1.85 D = 3Pm (at intercept) (3) (3.44 cm3 mol1 ) = 1.36 1024 cm3 = 4NA (4 ) (6.022 1023 mol1 )
Comment. The agreement of the value of with Table 22.1 is exact, but the polarizability volumes differ by about 8 per cent.
344
INSTRUCTOR'S MANUAL
65
60
55
50
45
40 1.9 2.0 2.1 2.2 2.3 1000 K T 2.4 2.5 2.6 2.7
Figure 21.2
P21.7
If there is a simple groupadditivity relationship, then elec ought to be a linear function of the number of Si2 H4 groups. That is, a plot of elec versus N ought to be a straight line. The plot is shown in Fig. 21.3 and a table shows values of elec computed from the best fit of the data and their deviations from the reported values. The equation of the bestfit line is elec /(1040 J1 C m2 ) = 4.8008N  1.7816 so the average contribution per Si2 H4 unit is 4.80 1040 J1 C m2 N Reported Best fit elec Deviation
elec
1 3.495 3.019 0.476
2 7.766 7.820 0.054
3 12.40 12.62 0.22
4 17.18 17.42 0.24
5 22.04 22.22 0.18
6 26.92 27.02 0.098
7 31.82 31.82 0.002
8
9
36.74 41.63 36.62 41.43 0.110 0.21
50
2
40 30 20 10 0 0 2 4 6 8 10
Figure 21.3 The rootmeansquare deviation is 0.26 1040 J1 C m2
MOLECULAR INTERACTIONS
345
P21.9
D0 can be obtained by adding together all the vibrational transitions; then De = D0 +
1 2 1 1  2 xe = D0 + G(0) ~
The potential obviously has some anharmonicity, for no two transitions have the same or nearly the same energy. But we cannot compute xe without knowing De for xe = ~ 4De
For that matter, we do not know exactly either. Our best estimate at the moment is G(1)  G(0), ~ which would equal if the vibration were harmonic, but in general it is ~ G(1)  G(0) = 1 +
1 2
 1+ ~
1 2
2
xe  ~
1 2
 2 2 xe = (1  2xe ) ~ 1 ~ ~
Our solution is first to compute De as if the potential were harmonic, then to compute xe based on the harmonic De and to recompute from G(1)  G(0) and xe . De can then be recomputed ~ based on the improved and xe and the process repeated until the values stop changing in successive ~ approximations. In the harmonic approximation
1 De = 1909.3 + 1060.3 + 386.3 + 2 (1909.3) m1 = 4310.6 m1
and the parameter a is given by a = = meff 1/2 = 2hcDe 2meff c 1/2 ~ hDe
1/2
2(2.2128 1026 kg) (2.998 108 m s1 ) (6.626 1034 J s) (4310.6 m1 )
(1909.3 m1 )
= 1.293 1010 m1 The anharmonicity constant is substantial xe = 1909.3 m1 = 0.1107 4(4310.6 m1 )
A spreadsheet may be used to recompute the parameters, which converge to xe = 0.1466, = 2701 m1 , ~ De = 4607 m1 , and a = 1.769 1010 m1
or De = 46.07 cm1 and a = 1.769 108 cm1 P21.11 An electric dipole moment may be considered as charge +q and q separated by a distance l such that = ql so q = / l = (1.77 D) (3.336 1030 C m/D) = 1.97 020 C 299 1012 m
In units of the electron charge q/e = (1.97 1020 C)/(1.602 1019 C) = 0.123
346
INSTRUCTOR'S MANUAL
P21.12
Neglecting the permanent dipole moment contribution NA [21.16] 30 (6.022 1023 mol1 ) (3.59 1040 J1 C2 m2 ) 3(8.854 1012 J1 C2 m1 )
Pm
= =
= 8.14 106 m3 mol1 = 8.14 cm3 mol1 r  1 Pm [21.17] = M r + 2 = (0.7914 g cm3 ) (8.14 cm3 mol1 ) 32.04 g mol1 r = 1.76 = 0.201
r  1 = 0.201r + 0.402; nr = r
1/2
= (1.76)1/2 = 1.33 [21.19]
The neglect of the permanent dipole moment contribution means that the results are applicable only to the case for which the applied field has a much larger frequency than the rotational frequency. Since red light has a frequency of 4.3 1014 Hz and a typical rotational frequency is about 1 1012 Hz, the results apply in the visible.
Answers to theoretical problems
P21.15 The timescale of the oscillations is about 1 = 2 109 s for benzene and toluene, and 0.55 GHz 2.5 109 s for the additional oscillations in toluene. Toluene has a permanent dipole moment; benzene does not. Both have dipole moments induced by fluctuations in the solvent. Both have anisotropic polarizabilities (so that the refractive index is modulated by molecular reorientation). Both benzene and toluene have rotational constants of 0.2 cm1 , which correspond to the energies of microwaves in this frequency range. Pure rotational absorption can occur for toluene, but not for benzene. An `exponential6' LennardJones potential has the form V = 4 Aer/  6 r
P21.18
and is sketched in Fig. 21.4. The minimum occurs where dV = 4 dr A r/ 6 6 + 7 e r =0
which occurs at the solution of A 7 = er/ 7 6 r Solve this equation numerically. As an example, when A = = 1, a minimum occurs at r = 1.63 .
MOLECULAR INTERACTIONS
347
Figure 21.4 N d = N d . The energy of interaction of V these molecules with one at a distance r is V N d . The total interaction energy, taking into account the entire sample volume, is therefore The number of molecules in a volume element d is u= V N d = N V d [V is the interaction, not the volume]
1 2
P21.19
1 The total interaction energy of a sample of N molecules is 2 N u (the counting), and so the cohesive energy density is
is included to avoid double
U =
1Nu U 1 1 = 2 = 2Nu = 2N2 V V
V d
C6 For V =  6 and d = 4r 2 dr r dr U N 2 C6 2  = 2N 2 C6 = V 3 r4 d3 a NA , where M is the molar mass; therefore However, N = M U= P21.21 2 3 NA 2 M C6 d3
Once again (as in Problem 21.20) we can write b  2 arcsin b R1 + R2 (v) (v) = R1 + R2 (v) 0 b > R1 + R2 (v) but R2 depends on v R2 (v) = R2 ev/v
1 1 Therefore, with R1 = 2 R2 and b = 2 R2
348
INSTRUCTOR'S MANUAL
(a)
(v) =  2 arcsin
1 1 (The restriction b R1 + R2 (v) transforms into 2 R2 2 R2 + R2 ev/v , which is valid for all v.) This function is plotted as curve a in Fig. 21.5. 1 The kinetic energy of approach is E = 2 mv 2 , and so
1 1 + 2ev/v
160
120
80
40
0 0 2 4 6 8 10
Figure 21.5 1
1/2
(b)
1 + 2 e(E/E ) 2 1 with E = 2 mv . This function is plotted as curve b in Fig. 21.5.
(E) =  2 arcsin
22
Macromolecules and aggregates
Solutions to exercises
Discussion questions
E22.1(b) (a) S is the change in conformational entropy of a random coil of a polymer chain. It is the statistical entropy arising from the arrangement of bonds, when a coil containing N bonds of length l is stretched or compressed by nl, where n is a numerical factor giving the amount of stretching in units of l. The amount of stretching relative to the number of monomer units in the chain is = n/N.
(b) Rrms is one of several measures of the size of a random coil. For a polymer of N monomer units each of length l, the root mean square separation, Rrms , is a measure of the average separation of the ends of a random coil. It is the square root of the average value of R 2 , calculated by weighting each possible value of R 2 with the probability that R occurs. (c) Rg , the radius of gyration, is another measure of the size of a random coil. It is the radius of a thin hollow spherical shell of the same mass and moment of inertia as the polymer molecule. All of these expressions are derived for the freely jointed random coil model of polymer chains which is the simplest possibility for the conformation of identical units not capable of forming hydrogen bonds or any other type of specific bond. In this model, any bond is free to make any angle with respect to the preceding one (Fig. 22.3 of the text). We assume that the residues occupy zero volume, so different parts of the chain can occupy the same region of space. We also assume in the derivation of the expression for the probability of the ends of the chain being a distance nl apart, that the chain is compact in the sense that n N . This model is obviously an oversimplification because a bond is actually constrained to a cone of angles around a direction defined by its neighbour (Fig. 22.4). In a hypothetical onedimensional freely jointed chain all the residues lie in a straight line, and the angle between neighbours is either 0 or 180 . The residues in a threedimensional freely jointed chain are not restricted to lie in a line or a plane. The random coil model ignores the role of the solvent: a poor solvent will tend to cause the coil to tighten; a good solvent does the opposite. Therefore, calculations based on this model are best regarded as lower bounds to the dimensions of a polymer in a good solvent and as an upper bound for a polymer in a poor solvent. The model is most reliable for a polymer in a bulk solid sample, where the coil is likely to have its natural dimensions. E22.2(b) E22.3(b) No solution. The formation of micelles is favored by the interaction between hydrocarbon tails and is opposed by charge repulsion of the polar groups which are placed close together at the micelle surface. As salt concentration is increased, the repulsion of head groups is reduced because their charges are partly shielded by the ions of the salt. This favors micelle formation causing the micelles to be larger and the critical micelle concentration to be smaller. A surfactant is a species that is active at the interface of two phases or substances, such as the interface between hydrophilic and hydrophobic phases. A surfactant accumulates at the interface and modifies the properties of the surface, in particular, decreasing its surface tension. A typical surfactant consists of a long hydrocarbon tail and other nonpolar materials, and a hydrophilic head group, such as the carboxylate group, CO , that dissolves in a polar solvent, typically water. In other words, a 2 surfactant is an amphipathic substance, meaning that it has both hydrophobic and hydrophilic regions. How does the surfactant decrease the surface tension? Surface tension is a result of cohesive forces and the solute molecules must weaken the attractive forces between solvent molecules. Thus molecules
E22.4(b)
350
INSTRUCTOR'S MANUAL
with bulky hydrophobic regions such as fatty acids can decrease the surface tension because they attract solvent molecules less strongly than solvent molecules attract each other. See Section 22.15(b) for an analysis of the thermodynamics involved in this process.
Numerical Exercises
E22.5(b) For a random coil, the r.m.s. separation is Rrms = l(N )1/2 = (1.125 nm) (1200)1/2 = 38.97 nm E22.6(b) Polypropylene is (CH(CH3 )CH2 ) n , where n is given by n= Mpolymer 174 kg mol1 = = 4.13 103 Mmonomer 42.1 103 kg mol1
The repeat length is the length of two C C bonds. The contour length is Rc = nl = (4.13 103 ) (2 1.53 1010 m) = 1.26 106 m The r.m.s. separation is Rrms = ln1/2 = (2 1.53 1010 m) (4.13 103 )1/2 = 1.97 108 m = 19.7 nm E22.7(b) The numberaverage molar mass is Mn = 1 [3 (62) + 2 (78)] kg mol1 = 68 kg mol1 Ni M i = N i 5
The massaverage molar mass is Mw = E22.8(b)
i
Ni Mi2 3 (62)2 + 2 (78)2 = kg mol1 = 69 kg mol1 3 (62) + 2 (78) i Ni Mi
For a random coil, the radius of gyration is Rg = l(N/6)1/2 so N = 6(Rg / l)2 = 6 (18.9 nm/0.450 nm)2 = 1.06 104
E22.9(b)
(a) Osmometry gives the numberaverage molar mass, so N 1 M1 + N 2 M2 = Mn = N1 + N 2 = 100 g
25 g 22 kg mol1 m1 M1 m M1 + M2 M2 2 m1 M1 m + M2 2
=
m1 + m2
m1 M1 m + M2 2
+
75 g 7.33 kg mol1
[assume 100 g of solution] = 8.8 kg mol1
(b) Lightscattering gives the massaverage molar mass, so Mw = m1 M1 + m2 M2 = [(0.25) (22) + (0.75) (7.33)] kg mol1 = 11 kg mol1 m1 + m 2
MACROMOLECULES AND AGGREGATES
351
E22.10(b)
=
4a 3 [see E22.10(a)] 3kT
(H2 O, 20 C) = 1.00 103 kg m1 s1 [Handbook of Chemistry and Physics, 81st Edition] = 4 (4.5 109 m)3 1.00 103 kg m1 s1 3 1.381 1023 J K1 293 K
= 9.4 108 s E22.11(b) The rate of sedimentation is proportional to the sedimentation constant S S= bM n f NA
The frictional coefficient f is proportional to the radius a of the sedimenting substance. The buoyancy b is the same for both of our substances because the density of the polymers, and therefore their specific volumes, are the same. The mass of a particle varies with its volume, and therefore with the cube of its radius. Thus S a 3 /a = a 2 so rate1 = rate2 a1 2 = (8.4)2 = 71 a2
with the larger particle sedimenting faster. E22.12(b) The molar mass is related to the sedimentation constant Mn = SRT SRT = bD (1  water vsolute )D (7.46 1013 s) (8.3145 J K 1 mol1 ) (298 K) [1  (1000 kg m3 ) (8.01 104 m3 kg1 )] (7.72 1011 m2 s1 )
where we have assumed the data refer to aqueous solution at 298 K. Mn =
= 120 kg mol1 E22.13(b) The drift speed is the speed s at which the frictional force f s precisely balances the gravitational force meff g s= (1  solution /solute )mg meff g = f 6asolv = solute V = 4solute a 3 /3 = 4 (1250 kg m3 ) (15.5 (106 m)3 /3 = 1.95 1011 kg So s = [1  (1000 kg m3 )/(1250 kg m3 )] (1.95 1011 kg) (9.81 m s2 ) 6 (15.5 106 m) (8.9 104 kg m1 s1 )
The mass of the particle is m
= 1.47 104 m s1
352
INSTRUCTOR'S MANUAL
E22.14(b) The molar mass is related to the sedimentation constant SRT SRT = Mn = bD (1  solution vsolute )D where we have assumed the data refer to aqueous solution at 298 K. Mn = (5.1 1013 s) (8.3145 J K 1 mol1 ) (293 K) = 56 kg mol1 [1  (0.997 g cm3 ) (0.721 cm3 g1 )] (7.9 1011 m2 s1 ) c2 c1
2 2 M w (r2  r1 )b2 c2 = c1 2RT
E22.15(b) In a sedimentation experiment, the massaverage molar mass is given by Mw = 2RT
2 (r2 2  r1 )b2
ln
so
ln
This implies that ln c = M w r 2 b2 + constant 2RT 2RT m b2
so the plot of ln c versus r 2 has a slope m equal to m= M w b2 2RT and Mw =
Mw =
2 (8.3145 J K1 mol1 ) (293 K) (821 cm2 ) (100 cm m1 )2 [1  (1000 kg m3 ) (7.2 104 m3 kg1 )] [(1080 s1 ) (2 )]2
= 3.1 103 kg mol1 E22.16(b) The centrifugal acceleration is a = r2 a/g = so a/g = r2 /g
(5.5 cm) [2 (1.32 103 s1 )]2 = 3.9 105 (100 cm m1 ) (9.8 m s2 )
Solutions to problems
Solutions to numerical problems
P22.1 For a rigid rod, Rg l [Problem 22.15] M, but for a random coil Rg N 1/2 [22.7] M 1/2 . Therefore, poly ( benzylLglutamate) is rodlike whereas polystyrene is a random coil (in butanol).
2 2 2 2 mb2 (r1  r2 ) 2 2 M w b 2 (r1  r2 ) c1 = [22.42] = c2 2kT RT and hence 1/2 c1 RT ln c2 = 2 2 2 2 M w b(r1  r2 )
P22.3
ln
[ = 2 ]
=
(8.314 J K 1 mol1 ) (298 K) (ln 5) 2 2 (1 102 kg mol1 ) (1  0.75) (7.02  5.02 ) 104 m2 or 3500 r.p.m.
1/2
= 58 Hz,
MACROMOLECULES AND AGGREGATES
353
Question. What would the concentration gradient be in this system with a speed of operation of 70 000 r.p.m. in an ultracentrifuge? P22.4 b = 1  s = 1  (0.765 g cm3 ) (0.93 cm3 g1 ) = 0.289; T = 308.15 K D/(cm2 s1 ) = (1.3 104 ) (Mw /g mol1 )0.497 bD 1 (1 + 2B c + 3gB 2 c2 + ) = Mw SRT or b 3gB 2 2 1 2B c+ c + = + Mw D Mw D SRT Mw D
c/(mg cm3 ) S/(10 s) (b/SRT )/(g1 cm2 s mol)
13
2.0 14.8 7.62
3.0 13.9 8.11
4.0 13.1 8.61
5.0 12.4 9.10
6.0 11.8 9.56
7.0 11.2 10.07
The regression fit of the form (b/SRT ) A = 6.639, (g1 cm2 s mol) = A + Bc + Cc2 cm3 , yields
standard deviation = 0.040 standard deviation = 0.019 mg1 cm3 standard deviation = 0.002 112 mg2 cm6
B = 0.494 mg
1
C = 0.000 697 mg2 cm6 ,
R = 0.999 940 (Note that the standard deviation of C is greater than its value.) 1 = A(g1 cm2 s mol) M wD 1 = 6.639 cm2 s 0.497 Mw Mw 4 cm 2 s1 ) g/mol (1.3 10 g/mol 1
Mw g/mol 0.503
= 8.631 104
M w = 1.23 106 g mol1 2B M wD B = = B(g1 cm2 s mol) = (0.494 mg1 cm3 ) (g1 cm2 s mol) 2 (6.639 g1 cm2 s mol)
B(g1 cm2 s mol) 2(1/M w D)
B = 3.72 102 mg1 cm3 We might test the significance of the form of the third term in the expression, namely C= 3gB 2 M wD
354
INSTRUCTOR'S MANUAL
by using the value of C obtained by the fitting process to calculate the value of g. But we must note again that the standard deviation of C is greater than C itself; hence the value of g obtained by this calculation could not be considered significant. g is about 1/4 for a good solvent, but cyclohexane is a theta solvent for polystyrene. There is no reason for them to agree; they are different samples; there is no fixed value of M for polystyrene. P22.11 = RT 1+ B Mn c [Example 7.5]
c Mn = gh; so
h BRT RT + c = 2 c gM n gM n h and we should plot against c. Draw up the following table c
c/(g/100 cm3 ) h/cm h (100 cm4 g1 ) c 0.200 0.48 2.4 0.400 1.12 2.80 0.600 1.86 3.10 0.800 2.76 3.45 1.00 3.88 3.88
The points are plotted in Fig. 22.1, and give a leastsquares intercept at 2.043 and a slope 1.805
4.0
3.0
2.0 0 0.2 0.4 0.6 0.8 1.0
Figure 22.1 Therefore, and hence Mn = (8.314 J K 1 mol1 ) (298 K) (0.798 103 kg m3 ) (9.81 m s2 ) (2.043 103 m4 kg1 ) 100 cm4 g1 g/(100 cm3 ) = 155 kg mol1 RT gM n = (2.043) (100 cm4 g1 ) = 2.043 103 m4 kg1
From the slope BRT gM n = (1.805) 2 = 1.805 104 cm7 g2 = 1.805 104 m7 kg2
MACROMOLECULES AND AGGREGATES
355
and hence B = = gM n RT M n (1.805 104 m7 kg2 )
(155 kg mol1 ) (1.805 104 m7 kg2 ) 2.043 103 m4 kg1
= 13.7 m3 mol1
Solutions to theoretical problems
P22.12
2 dN e(MM) /2 dM
We write the constant of proportionality as K, and evaluate it by requiring that M  M = (2 )1/2 x, and N = K(2 )1/2 K(2 )1/2 so
a 0
dN = N . Put
dM = (2 )1/2 dx
2
ex dx
2
a=
M (2 )1/2
1 ex dx[a 0] = K(2 )1/2 2 1/2
Hence, K =
2 1/2 N. It then follows that
2 2 1/2 Me(MM )/2 dM 0
Mn = =
2 2 2 1/2 M xex + (2 ) ex dx (2 )1/2 0
=
8 1/2 1 + 2
1/2 M = M+ 8
2 1/2
P22.14
A simple procedure is to generate numbers in the range 1 to 8, and to step north 1 or 2, east 3 or 4, south for 5 or 6, and west for 7 or 8 on a uniform grid. One such walk is shown in Fig. 22.2. Roughly, they would appear to vary as N 1/2 . We use the definition of the radius of gyration given in Footnote 4 and Problem 22.17, namely
2 Rg =
P22.15
1 R2 N j j
(a) For a sphere of uniform density, the centre of mass is at the centre of the sphere. We may visualize the sphere as a collection of a very large number, N . of small particles distributed with equal
356
INSTRUCTOR'S MANUAL
Figure 22.2 number density throughout the sphere. Then the summation above may be replaced with an integration.
1 a N r 2 P (r) dr 2 Rg = N 0 a 0 P (r) dr
P (r) is the probability per unit distance that a small particle will be found at distance r from the centre, that is, within a spherical shell of volume 4 r 2 dr. Hence, P (r) = 4 r 2 dr. The denominator ensures normalization. Hence
a 2 a 1 5 a r P (r) dr 4 r 4 dr 3 2 Rg = 0 a = 0 = 5 = a2, a 1 3 5 P (r) dr 4 r 2 dr 0 0 3a
Rg =
3 1/2 a 5
(b) For a long straight rod of uniform density the centre of mass is at the centre of the rod and P (r) is constant for a rod of uniform radius; hence,
1 1 (1/2)l 2 2 0 r dr 3 2l 2 Rg = = 1 (1/2)l 2 0 dr 2l 3
=
1 2 l , 12
l Rg = 2 3
For a spherical macromolecule a= and so Rg = = 3 1/2 5 3 1/2 5 3v s M 1/3 4NA (3vs /cm3 g1 ) cm3 g1 (M/g mol1 ) g mol1 (4 ) (6.022 1023 mol1 )
1/3
3Vm 1/3 = 4NA
3vs M 1/3 4NA
= (5.690 109 ) (vs /cm3 g1 )1/3 (M/g mol1 )1/3 cm = (5.690 1011 m) {(vs /cm3 g1 ) (M/g mol1 )}1/3 cm That is, Rg /nm = 0.05690 {(vs /cm3 g1 (M/g mol1 )}1/3
MACROMOLECULES AND AGGREGATES
357
When M = 100 kg mol1 and s = 0.750 cm3 g1 , Rg /nm = (0.05690) {0.750 1.00 105 }1/3 = 2.40 For a rod, vmol = a 2 l, so Rg = = vmol 1 v sM = 2 3 NA 2a 2 a 2 3 (0.750 cm3 g1 ) (1.00 105 g mol1 ) (6.022 1023 mol1 ) (2 ) (0.5 107 cm)2
6
3
= 4.6 10
cm = 46 nm
Comment. Rg may also be defined through the relation
2 Rg = i
mi ri2 i mi
Question. Does this definition lead to the same formulas for the radii of gyration of the sphere and the rod as those derived above? P22.17 Refer to Fig. 22.3.
2
1
Figure 22.3 Ri = 0,
Since Ri = R1 + hi and N R1 +
i
i
hi = 0 1 hi N i R1
i
and hence R1 = 
2 R1 =
1 hi hj , N 2 ij
hi = 
1 hi hj N ij
2 Rg =
1 1 2 Ri [new defintion] = {(R1 + hi ) (R1 + hi )} N i N i 1 1 1 2 = N R1 + h2 + 2R1 hi = h2  hi h j i i N N N ij i i i 1 2 (h + h2  h2 ) [cosine rule] j ij 2 i
Since hi hj =
358
INSTRUCTOR'S MANUAL
1 1 1 1 2 h2 + h2  h2  h2 Rg = i N 2N ij ij 2 i i 2 j j i = 1 h2 [the original definition] 2N 2 ij ij
(In the last two terms, the summation over the second index contributes a factor N .) P22.21 (a) We seek an expression for a ratio of scattering intensities of a macromolecule in two different conformations, a rigid rod or a closed circle. The dependence on scattering angle is contained in the Rayleigh ratio R . The definition of this quantity, in eqn 22.25, may be inverted to give an expression for the scattering intensity at scattering angle : I = R I0 sin2 , r2
where is an angle related to the polarization of the incident light and r the distance between sample and detector. Thus, for any given scattering angle, the ratio of scattered intensity of two conformations is the same as the ratio of their Rayleigh ratios: Prod Prod Irod = = . Icc Rcc Pcc The last equality, stems from eqn 22.28, which related the Rayleigh ratios to a number of angle independent factors that would be the same for both conformations and the structure factor (P ) that depends on both conformation and scattering angle. Finally, eqn 22.30 gives an approximate value of the structure factor as a function of the macromolecule's radius of gyration Rg , the wavelength of light, and the scattering angle: P 1 
1 2 16 2 Rg sin2 ( 2 )
32
=
1 2 32  16 2 Rg sin2 ( 2 )
32
.
The radius of gyration of a rod of length l is Rrod = l/[2(3)1/2 ]. For a closed circle, the radius of gyration, which is the rms distance from the center of mass, is simply the radius of a circle whose circumference is l: l = 2Rcc so Rcc = 1 . 2
The intensity ratio is:
1 32  4 2 l 2 sin2 ( 2 ) Irod 3 . = 1 Icc 32  4l 2 sin2 ( 2 )
Putting the numbers in yields:
/ Irod Icc 20 0.976 45 0.876 90 0.514
MACROMOLECULES AND AGGREGATES
359
(b) I would work at a detection angle at which the ratio is smallest, i.e., most different from unity, provided I had sufficient intensity to make accurate measurements. Of the angles considered in part a, 90 is the best choice. With the help of a spreadsheet or symbolic mathematical program, the ratio can be computed for a large range of scattering angles and plotted:
1.0
0.5 Irod /Icc 0.0 0.5 0 45 90 / 135 180
Figure 22.4
A look at the results of such a calculation shows that both the intensity ratio and the intensities themselves decrease with increasing scattering angle from 0 through 180 , that of the closed circle conformation changing much more slowly than that of the rod. Note: the approximation used above yields negative numbers for Prod at large scattering angles; this is because the approximation, which depends on the molecule being much smaller than the wavelength, is shaky at best, particularly at large angles. P22.22 ln c = const. + of slope r/cm c/(mg cm ) r 2 /(cm2 ) ln(c/mg cm3 )
3
M wb . We draw up the following table RT 5.0 0.536 25.0 0.624 5.1 0.284 26.0 1.259 5.2 0.148 27.0 1.911
M w b2 r 2 [22.42, rearranged] and a plot of ln c against r 2 should be a straight line 2RT
5.3 0.077 28.1 2.564
5.4 0.039 29.2 3.244
The points are plotted in Fig. 22.5. The leastsquares slope is 0.623. Therefore M w (1  vs )2 = 0.623 cm2 = 0.623 104 m2 2RT It follows that Mw = P22.24 (0.623 104 m2 ) (2) (8.314 J K 1 mol1 ) (293 K) = 65.6 kg mol1 {(1)  (1.001 g cm3 ) (1.112 cm3 g1 )} [(2 ) (322 s1 )]2
The sedimentation constant S must first be calculated from the experimental data. S= s 1 d ln r [Problem 22.2] [11] = 2 2 r dt
Therefore, a plot of ln r against t will give S. We draw up the following table
360
INSTRUCTOR'S MANUAL
0
1
2
3
4 25 26 27 28 29 30
Figure 22.5 t/s r/cm ln(r/cm) 0 6.127 1.813 300 6.153 1.817 600 6.179 1.821 900 6.206 1.826 1200 6.232 1.830 1500 6.258 1.834 1800 6.284 1.838
The leastsquares slope is 1.408 105 s1 , so S= 1.408 105 s1 = 5.14 1013 s = 5.14 Sv [(2) (50 103 /60 s)]2 SRT (5.14 1013 s) (8.314 J K 1 mol1 ) (293 K) = [22.41] = bD (1  0.9981 0.728) (7.62 1011 m2 s)2
Then M n =
60.1 kg mol1 We need to determine the ratio of the actual frictional coefficient, f , of the macromolecule to that of the frictional coefficient, f0 , of a sphere of the same volume, so that by interpolating in Table 23.1 we can obtain the dimensions of the molecular ellipsoid. f = kT (1.381 1023 J K1 ) (293 K) = 5.31 1011 kg s1 = D 7.62 1011 m2 s1
Vm = (0.728 cm3 g1 ) (60.1 103 g mol1 ) = 43.8 103 cm3 mol1 = 4.38 102 m3 mol1 Then, a = 3 Vm 1/3 = 4NA (3) (4.38 102 m3 mol1 ) (4) (6.022 1023 mol1 )
1/3
= 2.59 nm
f0 = 6a = (6) (2.59 109 m) (1.00 103 kg m1 s1 ) = 4.89 1011 kg s1 which gives f 5.31 = 1.09 = 4.89 f0 Therefore, the molecule is either prolate or oblate, with an axial ratio of about 2.8 (Table 22.3).
P22.25
The peaks are separated by 104 g mol1 , so this is the molar mass of the repeating unit of the polymer. This peak separation is consistent with the identification of the polymer as polystyrene, for the repeating group of CH2 CH(C6 H5 ) (8 C atoms and 8 H atoms) has a molar mass of 8 (12 + 1) g mol1 = 104 g mol1 . A consistent difference between peaks suggests a pure system
MACROMOLECULES AND AGGREGATES
361
and against different numbers of subunits (of different molecular weight) being incorporated into the polymer molecules. The most intense peak has a molar mass equal to that of n repeating groups plus that of a silver cation plus that of terminal groups: M(peak) = nM(repeat) + M(Ag+ ) + M(terminal). If the both ends of the polymer have terminal tbutyl groups, then M(terminal) = 2M(tbutyl) = 2(4 12 + 9) g mol1 = 114 g mol1 , M(peak)  M(Ag+ )  M(terminal) 25578  108  114 = = 243.8. M(repeat) 104 Obviously this is not an integer. Revisit the assumption of two tbutyl groups on the ends: and n = M(terminal) = M(peak)  nM(repeat)  M(Ag+ ). If n = 243, then M(terminal) = (25578  243 104  108) g mol1 = 198 g mol1 , which does not correspond to a whole number of tbutyl groups! Try again, supposing this time that there is a single tbutyl group at one end and a hydrogen atom at the other. Then: n= M(peak)  M(Ag+ )  M(terminal) 25578  108  58 = 244.3. = 104 M(repeat)
This is not an integer either. If n = 244, then M(terminal) = (25578  244 104  108) g mol1 = 94 g mol1 , not a whole number of butyl or butane groups. t (a) = 1 + c + k 2 c2 = t t/t  1 = + k 2 c or F = c A linear regression of F against c yields an intercept equal to and a slope equal to k 2 . (1) In toluene: Linear regression (R = 0.999 54) yields = 0.08566 L g1 = 0.086 L g1 ; k 2 = 0.002 688 L2 g2 ; Then k = 0.002 688 L2 g2 (0.085 66 L g1 )2 = 0.37 standard deviation = 0.000 20 L g1
P22.26
standard deviation = 0.000 057 L2 g2
(2) In cyclohexane: Linear regression (R = 0.981 98) yields = 0.04150 L g1 = 0.042 L g1 ; k 2 = 0.000 600 1 L2 g2 ; standard deviation = 0.000 18 L g1
standard deviation = 0.000 116 L2 g2
362
INSTRUCTOR'S MANUAL
Then k = 0.000 6001 L2 g2 (0.041 50 L g1 )2
a
= 0.35 1/a K
(1/0.72)
(b)
= KM v (1) In toluene Mv =
or
Mv =
0.085 66 L g1 1.15 105 L g1
g mol1 = 2.4 105 g mol1
(2) In cyclohexane Mv = (c) /(L g1 ) = rrms = 0.041 50 L g1 8.2 105 L g1 (rrms /m)3 /M, m,
1 1/2
g mol1 = 2.6 105 g mol1
= 2.84 1026
M 1/3
where rrms = r 2 1/2
1/3
(1) In toluene: rrms =
0.08566 2.39 105 2.84 1026
m = 42 nm
1/3
(2) In cyclohexane: rrms = (d)
0.04150 2.56 105 2.84 1026
m = 33 nm
M(styrene) = 104 g mol1 average number of monomeric units, n is n = Mv M(styrene) 2.39 105 g mol1 104 g mol1 2.56 105 g mol1 104 g mol1
(1) In toluene n = = 2.3 103
(2) In cyclohexane n = = 2.5 103
(e) Consider the geometry in Fig. 22.6. For a polymer molecule consisting of n monomers, the maximum molecular length, Lmax , is Lmax = 2l n cos = 2(0.154 nm) n cos 35 = (0.2507 nm) n
MACROMOLECULES AND AGGREGATES
363
Figure 22.6 (1) In toluene: Lmax = (0.2507 nm) (2.30 103 ) = 5.8 102 nm (2) In cyclohexane: Lmax = (0.2507 nm) (2.46 103 ) = 6.2 102 nm (f) Rg = n 3
1/2
l = (0.0889 nm) n 1/2 M 1/3 n 1/2 l =
1/3 M 2.84 1026
KR KirkwoodRiseman : rrms =
rrms = (2 n )1/2 l[39]
or
Solvent Toluene Cyclohexane
n 2.30 103 2.46 103
Rg / nm 4.3 4.4
KR rrms /nm
rrms /nm 10.4 or 7.4 10.8 or 7.6
42 33
(g) There is no reason for them to agree; they are different samples; there is no fixed value of M for polystyrene. The manufacturer's claim appears to be valid.
23
The solid state
Solutions to exercises
Discussion questions
E23.1(b) We can use the DebyeScherrer powder diffraction method, follow the procedure of Example 23.3, and in particular look for systematic absences in the diffraction patterns. We can proceed through the following sequence 1. Measure distances of the lines in the diffraction pattern from the centre. 2. From the known radius of the camera, convert the distances to angles. 3. Calculate sin2 . 4. Find the common factor A = 2 /4a2 in sin2 = (2 /4a 2 )(h2 + k 2 + l 2 ). 5. Index the lines using sin2 /A = h2 + k 2 + l 2 . 6. Look for the systematic absences in (hkl). See Fig. 23.22 of the text. For bodycentred cubic, diffraction lines corresponding to h + k + l that are odd will be absent. For facecentred cubic, only lines for which h, k, and l are either all even or all odd will be present, other will be absent. 7. Solve A = 2 /4a 2 for a. E23.2(b) The phase problem arises with the analysis of data in Xray diffraction when seeking to perform a Fourier synthesis of the electron density. In order to carry out the sum it is necessary to know the signs of the structure factors; however, because diffraction intensities are proportional to the square of the structure factors, the intensities do not provide information on the sign. For noncentrosymmetric crystals, the structure factors may be complex, and the phase in the expression Fhkl = Fhkl ei is indeterminate. The phase problem may be evaded by the use of a Patterson synthesis or tackled directly by using the socalled direct methods of phase allocation. The Patterson synthesis is a technique of data analysis in Xray diffraction which helps to circumvent the phase problem. In it, a function P is formed by calculating the Fourier transform of the squares of the structure factors (which are proportional to the intensities): P (r) = 1 Fhkl 2 e2 i(hx+ky+lz) V hkl
The outcome is a map of the separations of the atoms in the unit cell of the crystal. If some atoms are heavy (perhaps because they have been introduced by isomorphous replacement), they dominate the Patterson function, and their locations can be deduced quite simply. Their locations can then be used in the determination of the locations of lighter atoms. E23.3(b) In a facecentred cubic closepacked lattice, there is an octahedral hole in the centre. The rocksalt structure can be thought of as being derived from an fcc structure of Cl ions in which Na+ ions have filled the octahedral holes. The caesiumchloride structure can be considered to be derived from the ccp structure by having Cl ions occupy all the primitive lattice points and octahedral sites, with all tetrahedral sites occupied by Cs+ ions. This is exceedingly difficult to visualize and describe without carefully constructed figures or models. Refer to S.M. Ho and B. E. Douglas, J. Chem. Educ. 46, 208, 1969, for the appropriate diagrams. E23.4(b) A metallic conductor is a substance with a conductivity that decreases as the temperature is raised. A semiconductor is a substance with a conductivity that increases as the temperature is raised. A
THE SOLID STATE
365
semiconductor generally has a lower conductivity than that typical of metals, but the magnitude of the conductivity is not the criterion of the distinction. It is conventional to classify semiconductors with very low electrical conductivities, such as most synthetic polymers, as insulators. We shall use this term. But it should be appreciated that it is one of convenience rather than one of fundamental significance. The conductivity of these three kinds of materials is explained by band theory. When each of N atoms of a metallic element contributes one atomic orbital to the formation of molecular orbitals, the resulting N molecular orbitals form an almost continuous band of levels. The orbital at the bottom of the band is fully bonding between all neighbours, and the orbital at the top of the band is fully antibonding between all immediate neighbours. If the atomic orbitals are sorbitals, then the resulting band is called an sband; if the original orbitals are porbitals, then they form a pband. In a typical case, there is so large an energy difference between the s and p atomic orbitals that the resulting s and pbands are separated by a region of energy in which there are no orbitals. This region is called the band gap, and its whidth is denoted Eg . When electrons occupy the orbitals in the bands, they do so in accord with the Pauli principle. If insufficient electrons are present to fill the band, the electrons close to the top of the band are mobile and the solid is a metallic conductor. An unfilled band is called a conduction band and the energy of the highest occupied orbital at T = 0 K is called the Fermi level. Only the electrons close to the Fermi level can contribute to conduction and to the heat capacity of a metal. If the band is full, then the electrons cannot transport a current readily, and the solid is an insulator; more formally, it is a species of semiconductor with a large band gap. A full band is called a valence band. The detailed population of the levels in a band taking into account the role of temperature is expressed by the FermiDirac distribution. The distinction between metallic conductors and semiconductors can be traced to their band structure: a metallic conductor has an incomplete band, its conductance band, and a semiconductor has full bands, and hence lacks a conductance band. The decreasing conductance of a metallic conductor with temperature stems from the scattering of electrons by the vibrating atoms of the metal lattice. The increasing conductance of a semiconductor arises from the increasing population of an upper empty band as the temperature is increased. Many substances, however, have such large band gaps that their ability to conduct an electric current remains very low at all temperatures: it is conventional to refer to such solids as insulators. The ability of a semiconductor to transport charge is enhanced by doping it, or adding substances in controlled quantities. If the dopant provides additional electrons, then the semiconductor is classified as ntype. If it removes electrons from the valence band and thereby increases the number of positive holes, it is classified as ptype. E23.5(b) The FermiDirac distribution is a version of the Boltzmann distribution that takes into account the effect of the Pauli exclusion principle. It can therefore be used to calculate the population, P , of a state of given energy in a manyelectron system at a temperature T : 1 P = (E)/kT +1 e In this expression, is the Fermi energy, or chemical potential, the energy of the level for which P = 1/2. The Fermi energy should be distinguished from the Fermi level, which is the energy of the highest occupied state at T = 0. See Fig. 23.53 of the text. From thermodynamics (Chapter 5) we know that dU = p dV +T dS + dn for a onecomponent system. This may also be written dU = p dV + T dS + dN , and this is the chemical potential per particle that appears in the FD distribution law. The term in dU containing is the chemical work and gives the change in internal energy with change in the number of particles. Thus, has a wider significance than its interpretation as a partial molar Gibbs energy and it is not surprising that
366
INSTRUCTOR'S MANUAL
it occurs in the FD expression in comparision to the energy of the particle. The Helmholtz energy, A, and are related through dA = p dV  S dT + dN , and so also gives the change in the Helmholtz energy with change in number of particles. To fully understand how the chemical potential enters into the FD expression for P , we must examine its derivation (see Further reading) which makes use of the relation between and A and of that between A and the partition function for FD particles.
Numerical exercises
E23.6(b)
1 1 0 and 0, 2 , 2 . There are six faces to each cube, but each face is shared by two cubes. So other face midpoints can be described by one of these three sets of coordinates on an adjacent unit cell. 1 1 Taking reciprocals of the coordinates yields 1, 1 , 1 and 2 , 1 , 4 respectively. Clearing the 3 3 1 2, 1 0, 2 is the midpoint of a face. All face midpoints are alike, including 1 1 2, 2,
E23.7(b)
fractions yields the Miller indices (313) and (643) E23.8(b) The distance between planes in a cubic lattice is dhkl = (h2 + k2 a + l 2 )1/2
This is the distance between the origin and the plane which intersects coordinate axes at (h/a, k/a, l/a). d121 = 523 pm = 214 pm (1 + 22 + 1)1/2 523 pm = 174 pm d221 = 2 (2 + 22 + 1)1/2 523 pm d244 = 2 = 87.2 pm (2 + 42 + 42 )1/2
E23.9(b)
The Bragg law is n = 2d sin Assuming the angle given is for a firstorder reflection, the wavelength must be = 2(128.2 pm) sin 19.76 = 86.7 pm
E23.10(b) Combining the Bragg law with Miller indices yields, for a cubic cell sin hkl = 2 (h + k 2 + l 2 )1/2 2a
In a facecentred cubic lattice, h, k, and l must be all odd or all even. So the first three reflections would be from the (1 1 1), (2 0 0), and (2 2 0) planes. In an fcc cell, the face diagonal of the cube is
THE SOLID STATE
367
4R, where R is the atomic radius. The relationship of the side of the unit cell to R is therefore (4R)2 = a 2 + a 2 = 2a 2 so 4R a= 2
Now we evaluate 154 pm = = = 0.189 2a 4 2R 4 2(144 pm) We set up the following table hkl sin 111 0.327 200 0.378 220 0.535 / 19.1 22.2 32.3 2/ 38.2 44.4 64.6
E23.11(b) In a circular camera, the distance between adjacent lines is D = R (2), where R is the radius of the camera (distance from sample to film) and is the diffraction angle. Combining these quantities with the Bragg law ( = 2d sin , relating the glancing angle to the wavelength and separation of planes), we get D = 2R = 2R sin1 2d = 0.054 cm
= 2(5.74 cm) sin1
96.035 95.401 pm  sin1 2(82.3 pm) 2(82.3 pm)
E23.12(b) The volume of a hexagonal unit cell is the area of the base times the height c. The base is equivalent to two equilateral triangles of side a. The altitude of such a triangle is a sin 60 . So the volume is
1 V = 2 2 a a sin 60 c = a 2 c sin 60 = (1692.9 pm)2 (506.96 pm) sin 60
= 1.2582 109 pm3 = 1.2582 nm3 E23.13(b) The volume of an orthorhombic unit cell is V = abc = (589 pm) (822 pm) (798 pm) = The mass per formula unit is m= 135.01 g mol1 6.022 1023 mol1 = 2.24 1022 g 3.86 108 pm3 = 3.86 1022 cm3 (1010 pm cm1 )3
The density is related to the mass m per formula unit, the volume V of the unit cell, and the number N of formula units per unit cell as follows d= Nm V so N = dV (2.9 g cm3 ) (3.86 1022 cm3 ) = 5 = m 2.24 1022 g
A more accurate density, then, is d= 5(2.24 1022 g) = 2.90 g cm3 3.86 1022 cm3
368
INSTRUCTOR'S MANUAL
E23.14(b) The distance between the origin and the plane which intersects coordinate axes at (h/a, k/b, l/c) is given by dhkl = h2 k2 l2 + 2+ 2 a2 b c
1/2
=
32 22 22 + + (679 pm)2 (879 pm)2 (860 pm)2
1/2
d322 = 182 pm E23.15(b) The fact that the 111 reflection is the third one implies that the cubic lattice is simple, where all indices give reflections. The 111 reflection would be the first reflection in a facecentred cubic cell and would be absent from a bodycentred cubic The Bragg law sin hkl = 2 (h + k 2 + l 2 )1/2 2a
can be used to compute the cell length a= 137 pm (12 + 12 + 12 )1/2 = 390 pm (h2 + k 2 + l 2 )1/2 = 2 sin 17.7 2 sin hkl
With the cell length, we can predict the glancing angles for the other reflections expected from a simple cubic hkl = sin1 2 (h + k 2 + l 2 )1/2 2a = sin1 (0.176(h2 + k 2 + l 2 )1/2 )
100 = sin1 (0.176(12 + 0 + 0)1/2 ) = 10.1 (checks) 110 = sin1 (0.176(12 + 12 + 0)1/2 ) = 14.4 (checks) 200 = sin1 (0.176(22 + 0 + 0)1/2 ) = 20.6 (checks) These angles predicted for a simple cubic fit those observed, confirming the hypothesis of a simple lattice; the reflections are due to the (1 0 0), (1 1 0), (1 1 1), and (2 0 0) planes. E23.16(b) The Bragg law relates the glancing angle to the separation of planes and the wavelength of radiation = 2d sin so = sin1 2d
The distance between the orgin and plane which intersects coordinate axes at (h/a, k/b, l/c) is given by dhkl = h2 k2 l2 + 2+ 2 a2 b c
1/2
So we can draw up the following table hkl 100 010 111 dhkl /pm 574.1 796.8 339.5 hkl / 4.166 3.000 7.057
THE SOLID STATE
369
E23.17(b) All of the reflections present have h + k + l even, and all of the even h + k + l are present. The unit cell, then, is bodycentred cubic . E23.18(b) The structure factor is given by Fhkl = fi eii where i = 2(hxi + kyi + lzi )
i
All eight of the vertices of the cube are shared by eight cubes, so each vertex has a scattering factor of f/8. The coordinates of all vertices are integers, so the phase is a multiple of 2 and ei = 1. The bodycentre point belongs exclusively to one unit cell, so its scattering factor is f . The phase is
1 1 1 = 2 2 h + 2 k + 2 l = (h + k + l)
When h + k + l is even, is a multiple of 2 and ei = 1; when h + k + l is odd, is + a multiple of 2 and ei = 1. So ei = (1)h+k+l and Fhkl = 8(f/8)(1) + f (1)h+k+l = 2f for h + k + l even and 0 for h + k + l odd
E23.19(b) There are two smaller (white) triangles to each larger (grey) triangle. Let the area of the larger triangle 1 1 1 be A and the area of the smaller triangle be a. Since b = 2 B(base) and h = 2 H (height), a = 4 A. The white space is then 2N A/4, for N of the larger triangles. The total space is then N A + NA = 2 3N A/2. Therefore the fraction filled is N A/(3N A/2) = 2 3 E23.20(b) See Fig. 23.1.
Figure 23.1 The body diagonal of a cube is a 3. Hence a 3 = 2R + 2r r = 0.732 R or 3R = R + r [a = 2R]
370
INSTRUCTOR'S MANUAL
E23.21(b) The ionic radius of K + is 138 pm when it is 6fold coordinated, 151 pm when it is 8fold coordinated. (a) The smallest ion that can have 6fold coordination with it has a radius of 57 pm . (b) The smallest ion that can have 8fold coordination with it has a radius of 111 pm . E23.22(b) The diagonal of the face that has a lattice point in its centre is equal to 4r, where r is the radius of the atom. The relationship between this diagonal and the edge length a is 4r = a 2 so a = 2 2r The volume of the unit cell is a 3 , and each cell contains 2 atoms. (Each of the 8 vertices is shared among 8 cells; each of the 2 face points is shared by 2 cells.) So the packing fraction is 2(4/3) r 3 2Vatom = = = 0.370 3 Vcell 3(2)3/2 (2 2r) E23.23(b) The volume of an atomic crystal is proportional to the cube of the atomic radius divided by the packing fraction. The packing fractions for hcp, a closepacked structure, is 0.740; for bcc, it is 0.680. So for titanium Vbcc 0.740 = Vhcp 0.680 122 pm 3 = 0.99 126 pm 3  1 (151 pm) = 2  1 (138 pm) =
The bcc structure has a smaller volume, so the transition involves a contraction . (Actually, the data are not precise enough to be sure of this. 122 could mean 122.49 and 126 could mean 125.51, in which case an expansion would occur.) E23.24(b) Draw points corresponding to the vectors joining each pair of atoms. Heavier atoms give more intense contributions than light atoms. Remember that there are two vectors joining any pair of atoms (AB
 
and AB); don't forget the AA zero vectors for the centre point of the diagram. See Fig. 23.2 for C6 H6 .
Figure 23.2
THE SOLID STATE
371
h 1 1 E23.25(b) Combine E = 2 kT and E = 2 mv 2 = 2m2 , to obtain
2
= E23.26(b) =
6.626 1034 J s h = = 252 pm (mkT )1/2 [(1.675 1027 kg) (1.381 1023 JK1 ) (300 K)]1/2 h h = p me v =e so
1/2 v = 2e e m 1/2
1 2 2 me v
and =
h2 2me e
= =
6.626 1034 J s [(2) (9.109 1031 kg) (1.602 1019 C) ( )]1/2 1.227 nm ( /V )1/2
(a) (b) (c)
= 1.0 kV, = 10 KV, = 40 KV,
1.227 nm = 39 pm (1.0 103 )1/2 1.227 nm = = 12 pm (1.0 104 )1/2 1.227 nm = = 6.1 pm (4.0 104 )1/2 =
E23.27(b) The lattice enthalpy is the difference in enthalpy between an ionic solid and the corresponding isolated ions. In this exercise, it is the enthalpy corresponding to the process MgBr 2 (s) Mg2+ (g) + 2Br  (g) The standard lattice enthalpy can be computed from the standard enthalpies given in the exercise by considering the formation of MgBr 2 (s) from its elements as occuring through the following steps: sublimation of Mg(s), removing two electrons from Mg(g), vaporization of Br 2 (1), atomization of Br 2 (g), electron attachment to Br(g), and formation of the solid MgBr 2 lattice from gaseous ions
fH  
(MgBr 2 , s) =
sub H
 
(Mg, s) +
ion H
 
(Mg, g) + (Br, g) 
vap H LH
 
(Br 2 , 1)
+ at H So the lattice enthalpy is
LH  
 
(Br 2 , g) + 2 eg H
 
 
(MgBr 2 , s)
(MgBr 2 , s) =
sub H
 
(Mg, s) +
ion H
 
(Mg, g) +
vap H fH
 
(Br 2 , 1)
  + at H  (Br 2 , g) + 2 eg H  (Br, g)  LH  
 
(MgBr 2 , s)
(MgBr 2 , s) = [148 + 2187 + 31 + 193  2(331) + 524] kJ mol1 = 2421 kJ mol1
E23.28(b) Tension reduces the disorder in the rubber chains; hence, if the rubber is sufficiently stretched, crystallization may occur at temperatures above the normal crystallization temperature. In unstretched rubber the random thermal motion of the chain segments prevents crystallization. In stretched rubber these random thermal motions are drastically reduced. At higher temperatures the random motions may still have been sufficient to prevent crystallization even in the stretched rubber, but lowering the
372
INSTRUCTOR'S MANUAL
temperature to 0 C may have resulted in a transition to the crystalline form. Since it is random motion of the chains that resists the stretching force and allows the rubber to respond to forced dimensional changes, this ability ceases when the motion ceases. Hence, the seals failed. Comment. The solution to the problem of the cause of the Challenger disaster was the final achievement, just before his death, of Richard Feynman, a Nobel prize winner in physics and a person who loved to solve problems. He was an outspoken person who abhorred sham, especially in science and technology. Feynman concluded his personal report on the disaster by saying, `For a successful technology, reality must take precedence over public relations, for nature cannot be fooled' (James Gleick, Genius: the life and science of Richard Feynman. Pantheon Books, New York (1992).) E23.29(b) Young's modulus is defined as: E= normal stress normal strain
where stress is deforming force per unit area and strain is a fractional deformation. Here the deforming force is gravitational, mg, acting across the crosssectional area of the wire, r 2 . So the strain induced in the exercise is strain = stress 4mg 4(10.0 kg)(9.8 m s2 ) mg = = = 5.8 102 = E (d/2)2 E d 2E (0.10 103 m)2 (215 109 Pa)
The wire would stretch by 5.8%. E23.30(b) Poisson's ratio is defined as: P = transverse strain normal strain
where normal strain is the fractional deformation along the direction of the deforming force and transverse strain is the fractional deformation in the directions transverse to the deforming force. Here the length of a cube of lead is stretched by 2.0 per cent, resulting in a contraction by 0.41 2.0 per cent, or 0.82 per cent, in the width and height of the cube. The relative change in volume is: V + V = (1.020)(0.9918)(0.9918) = 1.003 V and the absolute change is: V = (1.003  1)(1.0 dm3 ) = 0.003 dm3 E23.31(b) m = ge {S(S + 1)}1/2 B [23.34, with S in place of s]
Therefore, since m = 4.00B
1 S(S + 1) = 4 (4.00)2 = 4.00,
implying that
S = 1.56
3 Since S 2 , implying three unpaired spins.
In actuality most Mn2+ compounds have 5 unpaired spins.
THE SOLID STATE
373
E23.32(b)
m = Vm =
(7.9 106 ) (84.15 g mol1 ) M = 0.811 g cm3
= 8.2 104 cm3 mol1 E23.33(b) The molar susceptibility is given by
2 NA ge 0 2 S(S + 1) B 3kT NO2 is an oddelectron species, so it must contain at least one unpaired spin; in its ground state it 1 has one unpaired spin, so S = 2 . Therefore,
m =
m = (6.022 1023 mol1 ) (2.0023)2 (4 107 T2 J1 m3 )
1 1 (9.274 1024 J T1 )2 2 2 + 1 3(1.381 1023 J K1 ) (298 K)
= 1.58 108 m3 mol1 The expression above does not indicate any pressuredependence in the molar susceptibility. However, the observed decrease in susceptibility with increased pressure is consistent with the fact that NO2 has a tendency to dimerize, and that dimerization is favoured by higher pressure. The dimer has no unpaired electrons, so the dimerization reaction effectively reduced the number of paramagnetic species. E23.34(b) The molar susceptibility is given by m =
2 NA ge 0 2 S(S + 1) B 3kT
so
S(S + 1) =
3kT m
2 N A g e 0 2 B
S(S + 1) =
3(1.381 1023 J K1 ) (298 K) (6.022 1023 mol1 ) (2.0023)2
(6.00 108 m3 mol1 ) (4 107 T2 J1 m3 ) (9.274 1024 J T1 )2 1 + 1 + 4(2.84) = 1.26 = 2.84 so S = 2 corresponding to 2.52 effective unpaired spins. The theoretical number is 2 . The magnetic moments in a crystal are close together, and the interact rather strongly. The discrepancy is most likely due to an interaction among the magnetic moments. E23.35(b) The molar susceptibility is given by
2 NA ge 0 2 S(S + 1) B 3kT
m =
374
INSTRUCTOR'S MANUAL
Mn2+ has five unpaired spins, so S = 2.5 and m = (6.022 1023 mol1 ) (2.0023)2 (4 107 T2 J1 m3 ) 3(1.381 1023 J K1 ) (9.274 1024 J T1 )2 (2.5) (2.5 + 1) (298 K)
= 1.85 107 m3 mol1
E23.36(b) The orientational energy of an electron spin system in a magnetic field is E = ge B M S B The Boltzmann distribution says that the population ratio r of the various states is proportional to r = exp  E kT
where E is the difference between them. For a system with S = 1, the MS states are 0 and 1. So between adjacent states r = exp ge B MS B kT = exp (2.0023) (9.274 1024 J T1 ) (1) (15.0 T) (1.381 1023 J K1 ) (298 K)
= 0.935
The population of the highestenergy state is r 2 times that of the lowest; r 2 = 0.873
Solutions to problems
Solutions to numerical problems
P23.1 = 2dhkl sin hkl = (h2 2a sin hkl [eqn 23.5, inserting eqn 23.2] = 2a sin 6.0 = 0.209a + k 2 + l 2 )1/2
In an NaCl unit cell (Fig. 23.3) the number of formula units is 4 (each corner ion is shared by 8 cells, each edge ion by 4, and each face ion by 2).
Figure 23.3
THE SOLID STATE
375
Therefore, = a = 4M NM = 3 , V NA a NA implying that a= 4M 1/3 NA
1/3
[Exercise 23.13(a)] = 563.5 pm
(4) (58.44 g mol1 ) (2.17 1016 g m3 ) (6.022 1023 mol1 )
and hence = (0.209) (563.5 pm) = 118 pm P23.4 Note that since R = 28.7 mm, /deg = following sequence: 1. Measure the distances from the figure. 2. Convert from distances to angle using /deg = D/mm. 3. Calculate sin2 . 4. Find the common factor A = 2 in sin2 = 4a 2 2 4a 2 (h2 + k 2 + l 2 ). D 2R 180 = D/mm. Then proceed through the
5. Index the lines using 6. Solve A = (a)
sin2 = h2 + k 2 + l 2 . A
2 for a. 4a 2
22 22 140 30 30 250 36 36 345 44 44 482 50 50 587 58 58 719 67 67 847 77 77 949
D/mm /deg 103 sin2
(h k l) 104 A
(1 1 1) 467
Analysis of facecentred cubic possibility (2 0 0) (2 1 1) (3 1 1) (2 2 2) (4 0 0) 625 431 438 489 449
(3 3 1) 446
(4 2 0) 475
(h k l) 104 A
(1 1 0) 700
Analysis of bodycentred cubic possibility (2 0 0) (2 1 1) (2 2 0) (3 1 0) (2 2 2) 625 575 603 587 599
(3 2 1) 605
(4 0 0) 593
Begin by performing steps 13 in order to determine D, , and sin2 and place them in tabular form as above. It is now possible to reject the primitive (simple) cubic cell possibility immediately because the separation between the sixth and seventh lines is not significantly larger than the separation between the fifth and sixth lines (see Problem 23.2 and Fig. 23.22). The relatively large uncertainties of the separation measurements force the modification of steps 4 and 5 for the identification of the unit cell as being either facecentred cubic or bodycentred cubic. We analyse both possibilities by calculating the common factor A = sin2 / h2 + k 2 + l 2 ) for each datum in each case. Comparison of the standard deviations of the average of A determines the unit cell type.
376
INSTRUCTOR'S MANUAL
The analysis of both the facecentred cubic and bodycentred cubic possibilities is found in the above table. Successive reflective planes are determined with the rules found in Fig. 23.22. fcc possibility: bcc possibility: Aav. = 0.0478, Aav. = 0.0611, A = 0.0063 (13 per cent) A = 0.0016 (6 per cent)
These standard deviations (A ) indicate that the cell type is bodycentred cubic The Q test of the (1 1 0) reflection datum for A yields Q = 0.6. Consequently this datum may be rejected with better than 95 per cent confidence. This yields a better average value for A. Aav. = 0.0598, A = 0.0016 (3 per cent)
Then a =
154 pm = = 315 pm 1/2 (2) (0.0598)1/2 2A 4R = 3a, so R = 136 pm [Fig. 23.1 above with r = R]
D/mm /deg 103 sin2 21 21 128 25 25 179 37 37 362 45 45 500 47 47 535 59 59 735 67 67 847 72 72 905
(b)
(h k l) 104 A
(1 1 1) 427
Analysis of facecentred cubic possibility (2 0 0) (2 2 0) (3 1 1) (2 2 2) (4 0 0) 448 453 455 446 459
(3 3 1) 446
(4 2 0) 453
(h k l) 104 A
(1 1 0) 640
Analysis of bodycentred cubic possibility (2 0 0) (2 1 1) (2 2 0) (3 1 0) (2 2 2) 448 603 625 535 613
(3 2 1) 605
(4 0 0) 566
Following the procedure established in part (a), the above table is constructed. fcc possibility: bcc possibility: Aav. = 0.0448, Aav. = 0.0579, A = 0.0010 (2 per cent) A = 0.0063 (11 per cent)
The standard deviations indicate that the cell type is facecentred cubic Then 4R = P23.6 a= 2a, 154 pm = = 364 pm 1/2 2A (2) (0.0448)1/2 so R = 129 pm
When a very narrow Xray beam (with a spread of wavelenths) is directed on the centre of a genuine pearl, all the crystallites are irradiated parallel to a trigonal axis and the result is a Laue photograph with sixfold symmetry. In a cultured pearl the narrow beam will have an arbitrary orientation with respect to the crystallite axes (of the central core) and an unsymmetrical Laue photograph will result. (See J. Bijvoet et al., Xray analysis of crystals. Butterworth (1951).)
THE SOLID STATE
377
P23.8
V = abc sin and the information given tells us that a = 1.377b, c = 1.436b, and = 122 49 ; hence V = (1.377) (1.436b3 ) sin 122 49 = 1.662b3 Since = b= = 2M NM = we find that V NA 1.662b3 NA
1/3 2M 1.662NA
(2) (128.18 g mol1 ) (1.662) (1.152 106 g m3 ) (6.022 1023 mol1 )
1/3
= 605.8 pm
Therefore, a = 834 pm , b = 606 pm , c = 870 pm P23.10 In a monoclinic cell, the area of parallelogram faces whose sides are a and c is A = ca cos(  90 ) so the volume of the unit cell is V = abc cos(  90 ) = (1.0427 nm) (0.8876 nm) (1.3777 nm) cos(93.254  90 ) = 1.2730 nm3 The mass per unit cell is m = V = (2.024 g cm3 ) (1.2730 nm3 ) (107 cm nm1 )3 = 2.577 1021 g The monomer is CuC7 H13 N5 O8 S, so its molar mass is M = 63.546 + 7(12.011) + 13(1.008) + 5(14.007) + 8(15.999) + 32.066 g mol1 = 390.82 g mol1 The number of monomer units, then, is the mass of the unit cell divided by the mass of the monomer N= P23.12 (2.577 1021 g) (6.022 1023 mol1 ) mNA = 3.97 = M 390.82 g mol1 or 4
 The problem asks for an estimate of f H  (CaCl). A BornHaber cycle would envision formation of CaCl(s) from its elements as sublimation of Ca(s), ionization of Ca(g), atomization of Cl2 (g) electrom gain of Cl(g), and formation of CaCl(s) from gaseous ions. Therefore fH  
(CaCl, s) =
sub H
 
(Ca, s) +
ion H
 
 (Ca, g) + 2 f H  (Cl, g)
 +2 eg H  (Cl, g) 
LH
 
(CaCl, s)
Before we can estimate the lattice enthalpy of CaCl, we select a lattice with the aid of the radiusratio rule. The ionic radius for Cl is 181 pm; use the ionic radius of K + (138 pm) for Ca+ = 138 pm = 0.762 181 pm
suggesting the CsCl structure. We can interpret the BornMayer equation (eqn 23.15) as giving the negative of the lattice enthalpy
LH  
Az1 z2 NA e2 40 d
1
d d
378
INSTRUCTOR'S MANUAL
The distance d is d = 138 + 181 pm = 319 pm so
LH LH    
(1.763)(1)(1)(6.022 1023 mol1 )(1.602 1019 C)2 4(8.854 1012 J1 C2 m1 )(319 1012 m)
1
34.5 pm 319 pm
6.85 105 J mol1 = 685 kJ mol1
The enthalpy of formation, then, is
fH  
(CaCl, s) [176 + 589.7 + 2(121.7  348.7)  685] kJ mol1 = 373 kJ mol1 .
Although formation of CaCl(s) from its elements is exothermic, formation of CaCl2 (s) is still more favoured energetically. Consider the reaction 2 CaCl(s) Ca(s) + CaCl2 (s) for which H
 
=
fH
 
(Ca) +
fH
 
 (CaCl2 )  2 f H  (CaCl)
[0  795.8  2(373)] kJ mol1
 H  50 kJ mol1
Note: Using the tabulated ionic radius of Ca (i.e., that of Ca2+ ) would be less valid than using the atomic radius of a neighbouring monovalent ion, for the problem asks about a hypothetical compound of monovalent calcium. Predictions with the smaller Ca2+ radius (100 pm) differ substantially from those listed above: the expected structure changes to rocksalt, the lattice enthalpy to 758 kJ mol1 ,   1 and the final reaction enthalpy to +96 kJ mol1 . f H (CaCl) to 446 kJ mol
Solutions to theoretical problems
P23.15 If the sides of the unit cell define the vectors a, b, and c, then its volume is V = a b c [given]. Introduce the orthogonal set of unit vectors ^ ^ k so that i, j, ^ ^ i j a = ax ^ + ay ^ + az k ^ b = bx ^ + by ^ + bz k i j ^ c = cx ^ + cy ^ + cz k i j ax Then V = a b c = bx cx Therefore ay by cy az bz cz
ax ay az ax ay az V 2 = bx by bz bx by bz c x cy c z c x c y c z ax ay az ax ay az = bx by bz bx by bz c x cy c z c x c y c z [interchange rows and columns, no change in value]
THE SOLID STATE
379
ax ax + ay ay + az az = bx ax + b y ay + b z az cx a x + c y a y + c z a z a2 a b a c = b a b2 b c = c a c b c2
ax bx + a y by + a z bz ax cx + a y cy + a z cz by bx + b y by + b z bz bx cx + b y cy + b z cz cx b x + c y b y + c z b z cx c x + c y cy + c z cz a2 ab cos ac cos bc cos ab cos b2 ac cos bc cos c2
= a 2 b2 c2 (1  cos2  cos2  cos2 + 2 cos cos cos )1/2 Hence V = abc(1  cos2  cos2  cos2 + 2 cos cos cos )1/2 For a monoclinic cell, = = 90 V = abc(1  cos2 )1/2 = abc sin For an orthorhombic cell, = = = 90 , and V = abc P23.18 Fhkl = fi e2 i(hxi +kyi lzi ) [23.7]
i
For each A atom use 1 fA (each A atom shared by eight cells) but use fB for the central atom (since 8 it contributes solely to the cell). Fhkl = 1 fA 1 + e2 ih + e2 ik + e2 il + e2 i(h+k) + e2 i(h+l) + e2 i(k+l) + e2 i(h+k+l) 8 +fB e2 i(h+k+l) = fA + (1)(h+k+l) fB (a) fA = f, fB = 0; [h, k, l are all integers, ei = 1]
Fhkl = f no systematic absences
1 (b) fB = 2 fA ;
1 Fhkl = fA 1 + 2 (1)(h+k+l) 1 = 2 fA , and when h + k + l is even,
1 Therefore, when h + k + l is odd, Fhkl = fA 1  2
3 Fhkl = 2 fA . That is, there is an alternation of intensity (I F 2 ) according to whether h + k + l is odd or even .
(c) fA = fB = f ;
Fh+k+l = f 1 + (1)h+k+l = 0
if h + k + l is odd.
Thus, all h + k + l odd lines are missing . P23.20 Write t = aT , then t = a, T l U = t  aT [Problem 23.19] = 0 l T
and the internal energy is independent of the extension. Therefore t = T S [Problem 23.19] l T
and the tension is proportional to the variation of entropy with extension. The extension reduces the disorder of the chains, and they tend to revert to their disorderly (nonextended) state.
380
INSTRUCTOR'S MANUAL
P23.22
(a) The density of energy levels is: (E) = dk = dE dE 1 dk
d k 2 k dE = + 2 cos = sin dk dk N +1 N +1 N +1 1 N +1 k so (E) =  sin 2 N +1 Unlike the expression just derived, the relationship the problem asks us to derive has no trigonometric functions and it contains E and within a square root. This comparison suggests that the trigonometric identity sin2 + cos2 = 1 will be of use here. Let = k/(N + 1); then where sin = 1(1  cos2 )1/2 however, cos is related to the energy E = + 2 cos so cos = E 2
and sin = 1  Finally, (E) =
1/2 E 2 2
 N +1 2
2 1/2 1  E 2
(b) The denominator of this expression vanishes as the energy approaches 2. Near those limits, E  becomes 2, making the quantity under the square root zero, and (E) approach infinity. P23.23 = e2 2 r 6me
0
r2 =
r 2 2 d
0 0
with =
1
3 a0
1/2
er/a0
= 4 = 4
3 a0
r 4 2 dr
[d = 4 r 2 dr]
0
2 r 4 e2r/a0 dr = 3a0
n! x n eax dx = n+1 a
Therefore, =
2 e2 a0 2me
Then, since m = NA 0 m =
2 NA 0 e2 a0 2me
[23.32, m = 0]
THE SOLID STATE
381
P23.25
If the proportion of molecules in the upper level is P , where they have a magnetic moment of 2B (which replaces {S(S + 1)}1/2 B in eqn 23.35), the molar susceptibility m = (6.3001) [S(S + 1)] cm3 mol1 [Illustration 23.1] T /K
is changed to m = 25.2P (6.3001) (4) P cm3 mol1 cm3 mol1 [22 replaces S(S + 1)] = T /K T /K
The proportion of molecules in the upper state is P =
~ ehc /kT 1 [Boltzmann distribution] = hc /kT hc /kT ~ 1+e ~ 1+e
and
hc ~ (1.4388 cm K) (121 cm1 ) 174 = = T /K kT T
25.2 cm3 mol1 (T /K) (1 + e174/(T /K) ) This function is plotted in Fig. 23.4 Therefore, m =
4
3
2
0 0 100 200 300 400 500
Figure 23.4 Comment. The explanation of the magnetic properties of NO is more complicated and subtle than indicated by the solution here. In fact the full solution for this case was one of the important triumphs of the quantum theory of magnetism which was developed about 1930. See J. H. van Vleck, The theory of electric and magnetic susceptibilities. Oxford University Press (1932).
Solutions to applications
P23.29 (100 K) = 22 2 25 , (300 K) = 21 57 59
sin (100 K) = 0.37526, sin (300 K) = 0.37406 sin (300 K) a(100 K) [see Problem 21.7] = 0.99681 = a(300 K) sin (100 K)
382
INSTRUCTOR'S MANUAL
(154.062 pm) 3 3 = = 356.67 pm a(300 K) = 2 sin (2) (0.37406) a(100 K) = (0.99681) (356.67 pm) = 355.53 pm a 356.67  355.53 = = 3.206 103 a 355.53 356.673  355.533 V = = 9.650 103 V 355.533 volume = volume = 9.560 103 1 V = = 4.8 105 K 1 V T 200 K 3.206 103 1 a = = 1.6 105 K 1 a T 200 K
Part 3: Change
24
Molecules in motion
Solutions to exercises
Discussion questions
E24.1(b) Diffusion is the migration of particles (molecules) down a concentration gradient. Diffusion can be interpreted at the molecular level as being the result of the random jostling of the molecules in a fluid. The motion of the molecules is the result of a series of short jumps in random directions, a socalled random walk. In the random walk model of diffusion, although a molecule may take many steps in a given time, it has only a small probability of being found far from its starting point because some of the steps lead it away from the starting point but others lead it back. As a result, the net distance traveled increases only as the square root of the time. There is no net flow of molecules unless there is a concentration gradient in the fluid, alse there are just as many molecules moving in one direction as another. The rate at which the molecules spread out is proportional to the concentration gradient. The constant of proportionality is called the diffusion coefficient. On the molecular level in a gas, thermal conduction occurs because of random molecular motions in the presence of a temperature gradient. Across any plane in the gas, there is a net flux of energy from the high temperature side, because molecules coming from that side carry a higher average energy per molecule across the plane than those coming from the low temperature side. In solids, the situation is more complex as energy transport occurs through quantized elastic waves (phonons) and, in metals, also by electrons. Conduction in liquids can occur by all the mechanisms mentioned. At the molecular (ionic) level, electrical conduction in an electrolytic solution is the net migration of ions in any given direction. When a gradient in electrical potential exists in a conductivity cell there will be a greater flow of positive ions in the direction of the negative electrode than in the direction of the positive electrode, hence there is a net flow of positive charge toward the region of low electrical potential. Likewise a net flow of negative ions in the direction of the positive electrode will occur. In metals, only negatively charged electrons contribute to the current. To see the connection between the flux of momentum and the viscosity, consider a fluid in a state of Newtonian flow, which can be imagined as occurring by a series of layers moving past one another (Fig. 24.11 of the text). The layer next to the wall of the vessel is stationary, and the velocity of successive layers varies linearly with distance, z, from the wall. Molecules ceaselessly move between the layers and bring with them the xcomponent of linear momentum they possessed in their original layer. A layer is retarded by molecules arriving from a more slowly moving layer because they have a low momentum in the xdirection. A layer is accelerated by molecules arriving from a more rapidly moving layer. We interpret the net retarding effect as the fluid's viscosity. E24.2(b) According to the Grotthuss mechanism, there is an effective motion of a proton that involves the rearrangement of bonds in a group of water molecules. However, the actual mechanism is still highly contentious. Attention now focuses on the H9 O4 + unit in which the nearly trigonal planar H3 O+ ion is linked to three strongly solvating H2 O molecules. This cluster of atoms is itself hydrated, but the hydrogen bonds in the secondary sphere are weaker than in the primary sphere. It is envisaged that the ratedetermining step is the cleavage of one of the weaker hydrogen bonds of this secondary sphere (Fig. 24.19a of the text). After this bond cleavage has taken place, and the released molecule has rotated through a few degrees (a process that takes about 1 ps), there is a rapid adjustment of bond lengths and angles in the remaining cluster, to form an H5 O2 + cation of structure H2 O H+ OH2 (Fig. 24.19b). Shortly after this reorganization has occurred, a new H9 O4 + cluster forms as other molecules rotate into a position where they can become members of a secondary hydration sphere,
386
INSTRUCTOR'S MANUAL
but now the positive charge is located one molecule to the right of its initial location (Fig. 24.19c). According to this model, there is no coordinated motion of a proton along a chain of molecules, simply a very rapid hopping between neighbouring sites, with a low activation energy. The model is consistent with the observation that the molar conductivity of protons increases as the pressure is raised, for increasing pressure ruptures the hydrogen bonds in water. E24.3(b) Because the drift speed governs the rate at which charge is transported, we might expect the conductivity to decrease with increasing solution viscosity and ion size. Experiments confirm these predictions for bulky ions, but not for small ions. For example, the molar conductivities of the alkali metal ions increase from Li+ to Cs+ (Table 24.6) even though the ionic radii increase. The paradox is resolved when we realize that the radius a in the Stokes formula is the hydrodynamic radius (or "Stokes radius") of the ion, its effective radius in the solution taking into account all the H2 O molecules it carries in its hydration sphere. Small ions give rise to stronger electric fields than large ones, so small ions are more extensively solvated than big ions. Thus, an ion of small ionic radius may have a large hydrodynamic radius because it drags many solvent molecules through the solution as it migrates. The hydrating H2 O molecules are often very labile, however, and NMR and isotope studies have shown that the exchange between the coordination sphere of the ion and the bulk solvent is very rapid. The proton, although it is very small, has a very high molar conductivity (Table 24.6)! Proton and ONMR show that the times characteristic of protons hopping from one molecule to the next are about 1.5 ps, which is comparable to the time that inelastic neutron scattering shows it takes a water molecule to reorientate through about 1 rad (12 ps).
17
Numerical exercises
E24.4(b) (a) The mean speed of a gas molecule is c= so 8RT 1/2 M M(Hg) 1/2 = M(He) 200.59 1/2 = 7.079 4.003
c(He) = c(Hg)
1 (b) The mean kinetic energy of a gas molecule is 2 mc2 , where c is the root mean square speed
c=
3RT 1/2 M
1 So 2 mc2 is independent of mass, and the ratio of mean kinetic energies of He and Hg is 1
E24.5(b)
(a) The mean speed can be calculated from the formula derived in Example 24.1. c= 8 RT 1/2 = M 8 (8.314 J K 1 mol1 ) (298 K) (28.02 103 kg mol1 )
1/2
= 4.75 102 m s1
kT [24.14] (b) The mean free path is calculated from = 1/2 2 p with = d 2 = (3.95 1010 m)2 = 4.90 1019 m2 (1.381 1023 J K1 ) (298 K) Then, = 5 1 21/2 (4.90 1019 m2 ) (1 109 Torr) 760atm 1.01310 Pa 1 atm Torr = 4 104 m
MOLECULES IN MOTION
387
E24.6(b)
(c) The collision frequency could be calculated from eqn 31, but is most easily obtained from eqn 32, 4.75 102 m s1 c = 1 102 s1 since and c have already been calculated z = = 4.46 104 m Thus there are 100 s between collisions, which is a very long time compared to the usual timescale of molecular events. The mean free path is much larger than the dimensions of the pumping apparatus used to generate the very low pressure. kT p = 1/2 [24.14] 2 = d 2 , p = d= 1/2 = 0.36 nm2
1/2
= 0.34 nm
(1.381 1023 J K1 ) (298 K) = 2.4 107 Pa (21/2 ) (0.36 1018 m2 ) (0.34 109 m)
This pressure corresponds to about 240 atm, which is comparable to the pressure in a compressed gas cylinder in which argon gas is normally stored. E24.7(b) The mean free path is kT (1.381 1023 J K1 ) (217 K) = 1/2 = = 4.1 107 m 2 p 21/2 [0.43 (109 m)2 ] (12.1 103 Pa atm1 ) E24.8(b) Obtain data from Exercise 24.7(b) The expression for z obtained in Exercise 24.8(a) is z =
1/2 16 p mkT
Substituting = 0.43 nm2 , p = 12.1 103 Pa, m = (28.02 u), and T = 217 K we obtain z = 4 (0.43 1018 m2 ) (12.1 103 Pa) [ (28.02) (1.6605 1027 kg) (1.381 1023 J K1 ) (217 K)]1/2
= 9.9 108 s1 E24.9(b) The mean free path is kT (1.381 1023 J K1 ) (25 + 273) K 5.50 103 m Pa = 1/2 = = 1/2 [0.52 (109 m)2 ]p p 2 p 2 (a) = (b) = (c) = 5.50 103 m Pa (15 atm) (1.013 105 Pa atm1 ) 5.50 103 m Pa (1.0 bar) (105 Pa bar 1 ) 5.50 103 m Pa (1.0 Torr)
1.013105 Pa atm1 760 Torr atm1
= 3.7 109 m
= 5.5 108 m = 4.1 105 m
388
INSTRUCTOR'S MANUAL
E24.10(b) The fraction F of molecules in the speed range from 200 to 250 m s1 is F =
250 m s1 200 m s1
f (v)dv
where f (v) is the Maxwell distribution. This can be approximated by F f (v) v = 4
3/2 Mv 2 M v 2 exp 2RT 2RT
v,
with f (v) evaluated in the middle of the range F 4 44.0 103 kg mol1 2(8.3145 J K 1 mol1 ) (300 K) 2(8.3145 J K1 mol1 ) (300 K)
3/2
(225 m s1 )2 (50 m s1 ),
exp
(44.0 103 kg mol1 ) (225 m s1 )2
F 9.6 102 Comment. The approximation we have employed, taking f (v) to be nearly constant over a narrow range of speeds, may not be accurate enough, for that range of speeds includes about 10 per cent of the molecules.
Numerical exercises
E24.11(b) The number of collisions is N = ZW At = = pAt (2mkT )1/2 (111 Pa) (3.5 103 m) (4.0 103 m) (10 s) {2 (4.00 u) (1.66 1027 kg u1 ) (1.381 1023 J K1 ) (1500 K)}1/2
= 1.1 1021 E24.12(b) The mass of the sample in the effusion cell decreases by the mass of the gas which effuses out of it. That mass is the molecular mass times the number of molecules that effuse out m = mN = mZW At = mpAt m 1/2 = pAt = pAt 1/2 2 kT (2mkT )
2 1/2 1/2 M 2 RT
1 = (0.224 Pa) 2 3.00 103 m
(24.00 h) (3600 s h1 )
300 103 kg mol1 2 (8.3145 J K 1 mol1 ) (450 K)
= 4.89 104 kg
MOLECULES IN MOTION
389
E24.13(b) The flux is J =  dT dT 1 =  CV ,m v [X] dz 3 dz
where the minus sign indicates flow toward lower temperature and = 1 2N , v = 8kT 1/2 = m 8RT 1/2 , M and [M] = n/V = N/NA
So J = 
2CV ,m 3 NA
RT 1/2 dT M dz 2 (28.832  8.3145) J K 1 mol1
=
3 [0.27 (109 m)2 ] (6.022 1023 mol1 ) (8.3145 J K1 mol1 ) (260 K) (2.016 103 kg mol1 )
1/2
(3.5 K m1 )
= 0.17 J m2 s1 E24.14(b) The thermal conductivity is = 1 2CV ,m CV ,m v [X] = 3 3 NA RT 1/2 M so = 2CV ,m 3NA RT 1/2 M
= (0.240 mJ cm2 s1 ) (K cm1 )1 = 0.240 101 J m1 s1 K 1 so = 2 (29.125  8.3145) J K 1 mol1 3 (0.240 101 J m1 s1 K 1 ) (6.022 1023 mol1 ) (8.3145 J K1 mol1 ) (298 K) (28.013 103 kg mol1 )
1/2
= 1.61 1019 m2 E24.15(b) Assuming the space between sheets is filled with air, the flux is J =  dT = [(0.241 103 J cm2 s1 ) (K cm1 )1 ] dz = 1.45 103 J cm2 s1 . So the rate of energy transfer and energy loss is J A = (1.45 103 J cm2 s1 ) (1.50 m2 ) (100 cm m1 )2 = 22 J s1 [50  (10)] K 10.0 cm
390
INSTRUCTOR'S MANUAL
E24.16(b) The time dependence of the pressure of a gas effusing without replenishment is p = p0 et/ where m The time t it takes for the pressure to go from any initial pressure p0 to a prescribed fraction of that pressurefp0 is t = ln fp0 = ln f p0
so the time is proportional to and therefore also to m. Therefore, the ratio of times it takes two different gases to go from the same initial pressure to the same final pressure is related to their molar masses as follows t1 = t2 M1 1/2 M2 and M 2 = M1 t2 2 t1
82.3 s 2 = 554 g mol1 18.5 s E24.17(b) The time dependence of the pressure of a gas effusion without replenishment is So Mfluorocarbon = (28.01 g mol1 ) p = p0 et/ where = = V A0 so t = ln p0 /p 2 M 1/2 RT 2 (28.0 103 kg mol1 ) (8.3145 J K1 mol1 ) (293 K)
1/2
2m 1/2 V = kT A0
22.0 m3 (0.50 103 m)2
= 2.4 105 s
122 kPa = 1.5 104 s 105 kPa E24.18(b) The coefficient of viscosity is so t = (8.6 105 s) ln 2 = 1 mN v = 3 3 mkT 1/2 so = 2 3 mkT 1/2
= 1.66 P = 166 107 kg m1 s1 so = 2 3 (166 107 kg m1 s1 ) (28.01 103 kg mol1 ) (1.381 1023 J K1 ) (273 K) (6.022 1023 mol1 )
1/2
= 3.00 1019 m2 E24.19(b) The rate of fluid flow through a tube is described by
2 (p2  pout ) r 4 dV = in dt 16lp0
so
pin =
1/2 16lp0 dV 2 + pout r 4 dt
Several of the parameters need to be converted to MKS units
1 r = 2 (15 103 m) = 7.5 103 m dV and = 8.70 cm3 (102 m cm1 )3 s1 = 8.70 106 m3 s1 . dt
MOLECULES IN MOTION
391
Also, we have the viscosity at 293 K from the table. According to the T 1/2 temperature dependence, the viscosity at 300 K ought to be (300 K) = (293 K) 300 K 1/2 = (176 107 kg m1 s1 ) 293 K 300 1/2 293
= 1.78 107 kg m1 s1 pin = 16(10.5 m) (178 107 kg m1 s1 ) (1.00 105 Pa) (7.5 103 m)4
1/2
(8.70 106 m3 s1 ) + (1.00 105 Pa)2 = 1.00 105 Pa Comment. For the exercise as stated the answer is not sensitive to the viscosity. The flow rate is so low that the inlet pressure would equal the outlet pressure (to the precision of the data) whether the viscosity were that of N2 at 300 K or 293 Kor even liquid water at 293 K! E24.20(b) The coefficient of viscosity is 2 = 1 mN v = 3 3 = mkT 1/2 (78.12 103 kg mol1 ) (1.381 1023 J K1 )T (6.022 1023 mol1 )
1/2
2 3[0.88 (109 m)2 ]
= 5.72 107 (T /K)1/2 kg m1 s1 (a) At 273 K (b) At 298 K (c) At 1000 K = (5.72 107 ) (273)1/2 kg m1 s1 = 0.95 105 kg m1 s1 = (5.72 107 ) (298)1/2 kg m1 s1 = 0.99 105 kg m1 s1 = (5.72 107 ) (1000)1/2 kg m1 s1 = 1.81 105 kg m1 s1
E24.21(b) The thermal conductivity is 2CV ,m = 1 CV ,m v [X] = 3 3 NA (a) = RT 1/2 M
2 [(20.786  8.3145) J K 1 mol1 ] 3[0.24 (109 m)2 ] (6.022 1023 mol1 ) (8.3145 J K 1 mol1 ) (300 K) (20.18 103 kg mol1 )
1/2
= 0.0114 J m1 s1 K 1
392
INSTRUCTOR'S MANUAL
The flux is J =  dT = (0.0114 J m1 s1 K 1 ) dz (305  295) K 0.15 m = 0.76 J m2 s1
so the rate of energy loss is J A = (0.76 J m2 s1 ) (0.15 m)2 = 0.017 J s1 (b) = 2 [(29.125  8.3145) J K 1 mol1 ] 3[0.43 (109 m)2 ] (6.022 1023 mol1 ) 8.3145 J K 1 mol1 ) (300 K) (28.013 103 kg mol1 )
1/2
= 9.0 103 J m1 s1 K 1 The flux is J =  dT = (9.0 103 J m1 s1 K 1 ) dz (305  295) K 0.15 m = 0.60 J m2 s1
so the rate of energy loss is J A = (0.60 J m2 s1 ) (0.15 m)2 = 0.014 J s1 E24.22(b) The rate of fluid flow through a tube is described by
2 (p2  pout ) r 4 dV = in dt 16lp0
so the rate is inversely proportional to the viscosity, and the time required for a given volume of gas to flow through the same tube under identical pressure conditions is directly proportional to the viscosity 1 t2 t1 (208 P) (18.0 s) = 52.0 P = 52.0 107 kg m1 s1 CFC = 72.0 s The coefficient of viscosity is so 2 = = 1 mN v = 3 2 3 mkT 1/2 = 2 3 d 2 mkT 1/2 t1 1 = t2 2
so the molecular diameter is d = = 2 1/2 3 mkT 1/4 2 3(52.0 107 kg m1 s1 ) (200 103 kg mol1 ) (1.381 1023 J K1 ) (298 K) (6.022 1023 mol1 )
1/4 1/2
= 9.23 1010 m = 923 pm
MOLECULES IN MOTION
393
E24.23(b)
2CV ,m = 1 CV ,m v [X] = 3 3 NA =
RT 1/2 M (8.3145 J K1 mol1 ) (300 K) (28.013 103 kg mol1 )
1/2
2 (29.125  8.3145) J K 1 mol1 3[0.43 (109 m)2 ] (6.022 1023 mol1 )
= 9.0 103 J m1 s1 K 1 E24.24(b) The diffusion constant is 2(RT )3/2 D = 1 v = 3 3pNA ( M)1/2 = 2[(8.3145 J K 1 mol1 ) (298 K)]3/2 3[0.43 (109 m)2 ]p(6.022 1023 mol1 ) (28.013 103 kg mol1 ) 1.07 m2 s1 p/Pa d[X] d = D dx dx n V D RT dp dx
1/2
=
The flux due to diffusion is J = D =
where the minus sign indicates flow from high pressure to low. So for a pressure gradient of 0.10 atm cm1 J = D/(m2 s1 ) (8.3145 J K1 mol1 ) (298 K) 1.07 m2 s1 = 0.107 m2 s1 10.0 1.07 m2 s1 = 1.07 105 m2 s1 100 103 1.07 m2 s1 = 7.13 108 m2 s1 15.0 106 (0.20 105 Pa m1 )
= (8.1 mol m2 s1 ) (D/(m2 s1 )) (a) D=
and J = (8.1 mol m2 s1 ) (0.107) = 0.87 mol m2 s1 (b) D=
and J = (8.1 mol m2 s1 ) (1.07 105 ) = 8.7 105 mol m2 s1 (c) D=
and J = (8.1 mol m2 s1 ) (7.13 108 ) = 5.8 107 mol m2 s1 E24.25(b) Molar ionic conductivity is related to mobility by = zuF = (1) (4.24 108 m2 s1 V1 ) (96 485 C mol1 ) = 4.09 103 S m2 mol1 E24.26(b) The drift speed is given by s = uE = u (4.01 108 m2 s1 V1 ) (12.0 V) = = 4.81 105 m s1 l 1.00 102 m
394
INSTRUCTOR'S MANUAL
E24.27(b) The limiting transport number for Cl in aqueous NaCl at 25 C is
t =
u 7.91 = 0.604 = u+ + u  5.19 + 7.91
(The mobilities are in 108 m2 s1 V1 .) E24.28(b) The limiting molar conductivity of a dissolved salt is the sum of that of its ions, so
m (MgI2 )
= (Mg2+ ) + 2(I ) =
m (Mg(C2 H3 O2 )2 ) + 2
m (NaI)  2
m (NaC2 H3 O2 )
= (18.78 + 2(12.69)  2(9.10)) mS m2 mol1 = 25.96 mS m2 mol1 E24.29(b) Molar ionic conductivity is related to mobility by = zuF F : Cl : Br  : u= u= u= so u = zF = 5.74 108 m2 V1 s1 = 7.913 108 m2 V1 s1
5.54 103 S m2 mol1 (1) (96 485 C mol1 ) (1) (96 485 C mol1 )
7.635 103 S m2 mol1 7.81 103 S m2 mol1
= 8.09 108 m2 V1 s1 (1) (96 485 C mol1 ) E24.30(b) The diffusion constant is related to the mobility by D = (4.24 108 m2 s1 V1 ) (8.3145 J K 1 mol1 ) (298 K) uRT = zF (1) (96 485 C mol1 )
= 1.09 109 m2 s1 E24.31(b) The mean square displacement for diffusion in one dimension is x 2 = 2Dt In fact, this is also the mean square displacement in any direction in two or threedimensional diffusion from a concentrated source. In three dimensions r 2 = x 2 + y 2 + z2 so r 2 = x 2 + y 2 + z2 = 3 x 2 = 6Dt r 2 is
So the time it takes to travel a distance t=
r2 (1.0 102 m)2 = 4.1 103 s = 6D 6(4.05 109 m2 s1 )
E24.32(b) The diffusion constant is related to the viscosity of the medium and the size of the diffusing molecule as follows D= kT 6a so a= kT (1.381 1023 J K1 ) (298 K) = 6D 6(1.00 103 kg m1 s1 ) (1.055 109 m2 s1 )
a = 2.07 1010 m = 207 pm
MOLECULES IN MOTION
395
E24.33(b) The EinsteinSmoluchowski equation related the diffusion constant to the unit jump distance and time 2 2 so = 2 2D If the jump distance is about one molecular diameter, or two effective molecular radii, then the jump distance can be obtained by use of the StokesEinstein equation D= D= and = kT kT = 6a 3 so = kT 3 D
[(1.381 1023 J K1 ) (298 K)]2 (kT )2 = 18( )2 D 3 18[(0.387 103 kg m1 s1 )]2 (3.17 109 m2 s1 )3
= 2.00 1011 s = 20 ps E24.34(b) The mean square displacement is (from Exercise 24.31(b)) r 2 = 6Dt so t = (1.0 106 m)2 r2 = 1.7 102 s = 6D 6(1.0 1011 m2 s1 )
Solutions to problems
Solutions to numerical problems
P24.3 X = (a) h = 1 Ni Xi [See Problem 24.2] N i
1 {1.80 m + 2 (1.82 m) + + 1.98 m} 1.89 m 53 1 (1.80 m)2 + 2 (1.82 m)2 + + (1.98 m)2 = 3.57 m2 (b) h2 = 53 h2 = 1.89 m P24.4 = 1 cCV ,m [A] [24.28] 3 8kT 1/2 c= [24.7] T 1/2 m Hence, T 1/2 CV ,m , so = T T
1/2
CV ,m CV ,m
P24.7
3 5 3 At 300 K, CV ,m 2 R + R = 2 R At 10 K, CV ,m 2 R [rotation not excited] 5 300 1/2 = 9.1 = Therefore, 10 3 The atomic current is the number of atoms emerging from the slit per second, which is ZW A with A = 1 107 m2 . We use p [24.15] ZW = (2mkT )1/2 p/Pa = 1 ) (1.6605 1027 kg) (1.381 1023 J K 1 ) (380 K)]1/2 [(2) (M/g mol
= (1.35 1023 m2 s1 )
p/Pa (M/g mol1 )1/2
396
INSTRUCTOR'S MANUAL
(a) Cadmium: ZW A = (1.35 1023 m2 s1 ) (1 107 m2 ) (b) Mercury: ZW A = (1.35 1023 m2 s1 ) (1 107 m2 ) P24.10 c= c
m
0.13 (112.4)1/2 152 (200.6)1/2
= 2 1014 s1
= 1 1017 s1
[24.98]
m
[c small, conductivity of water allowed for in the data] [Exercise 24.28(a)]
1.887 106 S cm1 138.3 S cm2 mol1
1.36 108 mol cm3 = solubilit y = 1.36 105 M P24.12 u(H+ ) 3.623 = 0.82 [24.61] = + ) + u(Cl ) 3.623 + 0.791 u(H When a third ion is present we use t (H+ ) = t (H+ ) = I (H+ ) + I (Na+ ) + I (Cl ) I (H+ ) [24.58]
For each I , I = zucF AE = constant cu. Hence, when NaCl is added t (H+ ) = = c(H+ )u(H+ ) c(H+ )u(H+ ) + c(Na+ )u(Na+ ) + c(Cl )u(Cl ) (1.0 103 ) (3.623) = 0.0028 (1.0 103 ) (3.623) + (1.0) (0.519) + (1.001) (0.791) x t [Problem 24.13]
P24.14
t+ =
zcAF I
The density of the solution is 0.682 g cm3 ; the concentration c is related to the molality m by c/(mol L1 ) = /(kg L1 ) m/(mol kg1 ) which holds for dilute solutions such as these. A = r 2 = (2.073 103 m)2 = 1.350 105 m2 czAF (1.350 105 m2 ) (9.6485 104 C mol1 ) c = (0.1042 m2 mol1 ) c = I t (5.000 103 A) (2500 s) = (0.1042 m2 mol1 ) m = (0.1042 m2 mol1 ) (682 kg m3 ) m = (71.06 kg m1 mol1 ) m = (0.07106 kg mm1 mol1 ) m and so t+ = (0.07106 kg mm1 mol1 ) x m
In the first solution t+ = (0.07106 kg mm1 mol1 ) (286.9 mm) (0.01365 mol kg1 ) = 0.278
In the second solution t+ = (0.07106 kg mm1 mol1 ) (92.03 mm) (0.04255 mol kg1 ) = 0.278
MOLECULES IN MOTION
397
Therefore, t (H+ ) = 0.28, a value much less than in pure water where t (H+ ) = 0.63. Hence, the mobility is much less relative to its counterion, NH . 2 P24.17 If diffusion is analogous to viscosity [Section 24.5, eqn 24.36] in that it is also an activation energy controlled process, then we expect D eEa /RT Therefore, if the diffusion constant is D at T and D at T , Ea =  R ln D D
1 T

1 T
=
(8.314 J K1 mol1 ) ln 2.89 2.05
1 298 K 1  273 K
= 9.3 kJ mol1
That is, the activation energy for diffusion is 9.3 kJ mol1 P24.19 x 2 = 2Dt [24.91], Hence, = D= kT [24.83] 6 a
kT kT t 1.381 1023 J K1 ) (298.15 K) t = = 6Da 3a x 2 (3 ) (2.12 107 m) x 2 t x2
= (2.06 1015 J m1 ) and therefore /(kg m1 s1 ) = We draw up the following table
t/s 108 x 2 /cm2 103 /(kg m1 s1 ) 30 88.2 0.701
2.06 1011 (t/s) ( x 2 /cm2 )
60 113.4 1.09
90 128 1.45
120 144 1.72
Hence, the mean value is 1.2 103 kg m1 s1 . P24.21 The viscosity of a perfect gas is = 1 N mc = 3 The mass is m= (a) 17.03 103 kg mol1 6.022 1023 mol1 2 3(9.08 106 kg m1 s1 ) (2.828 1026 kg) (1.381 1023 J K1 ) (270 K)
19 1/2
mc 2 = 3 3 2
mkT 1/2
so
=
2 3
mkT 1/2
= 2.828 1026 kg
=
= 4.25 10
m = d
2
2
so
d=
4.25 1019 m2
1/2
= 3.68 1010 m
398
INSTRUCTOR'S MANUAL
(b)
=
2 3(17.49 106 kg m1 s1 ) (2.828 1026 kg) (1.381 1023 J K1 ) (490 K) so d= 2.97 1019 m2
1/2 1/2
= 2.97 1019 m2 = d 2
= 3.07 1010 m
Comment. The change in diameter with temperature can be interpreted in two ways. First, it shows the approximate nature of the concept of molecular diameter, with different values resulting from measurements of different quantities. Second, it is consistent with the idea that, at higher temperatures, more forceful collisions contract a molecule's perimeter. P24.22 The diffusion constant of an ion in solution is related to the mobility of the ion and to its radius in separate relations D= a= kT uRT = zF 6a so a= zF k ze = 6 uR 6 u = 8.3 1010 m = 830 pm
(1) (1.602 1019 C) 6(0.93 103 kg m1 s1 ) (1.1 108 m2 V1 s1 )
Solutions to theoretical problems
P24.25 Write the mean velocity initially as a; then in the emerging beam vx = K
a 0
vx f (vx ) dvx where
K is a constant which ensures that the distribution in the emergent beam is also normalized. That is, a m 1/2 a mv 2 /2kT x 1=K f (vx ) dvx = K e dvx 2kT 0 0 This integral cannot be evaluated analytically but it can be related to the error function by defining x2 =
2 mvx 2kT
which gives dvx = 1= K =
2kT 1/2 dx. Then m 2kT 1/2 b x 2 e dx m 0 [b = (m/2kT )1/2 a]
m 1/2 2kT
b 2 K 1 ex dx = 2 Kerf(b) 1/2 0
where erf (z) is the error function [Table 12.2]: erf(z) = 2 Therefore, K = erf(b) The mean velocity of the emerging beam is vx = K
2 1/2
0
z
ex dx
2
2 m 1/2 a m 1/2 vx emvx /2kT dvx = K 2kT 2 kT 0
kT m
d mv 2 /2kT x (e dvx dvx 0
a
MOLECULES IN MOTION
399
= K
kT 1/2 ma 2 /2kT (e  1) 2m
2kT 1/2 m This expression for the average magnitude of the onedimensional velocity in the x direction may be obtained from Now use a = vx initial = vx = 2
0
vx f (vx )dvx = 2 =
0
vx
m 1/2 mv 2 /2kT x e dvx 2 kT 2kT m = 2kT 1/2 m
m 1/2 2 kT
It may also be obtained very quickly by setting a = in the expression for vx in the emergent beam with erf(b) = erf() = 1. Substituting a = erf 1 1/2
2 2kT 1/2 into vx in the emergent beam ema /2kT = e1/ and erf(b) = m
Therefore, vx =
2kT 1/2 1  e1/ m erf 1 1/2 From tables of the error function (expanded version of Table 12.2), or from readily available software, or by interpolating Table 12.2. erf 1 1/2 = erf(0.56) = 0.57 and e1/ = 0.73
Therefore, vx = 0.47 vx initial P24.27 The most probable speed, c , was evaluated in Problem 24.23 and is c = v(most probable) = Consider a range of speeds 2kT 1/2 m v around c and nc , then with v = c
2 2
2 2 2 f (nc ) (nc )2 emn c /2kT [24.4] = n2 e(n 1)mc /2kT = n2 e(1n ) = ) 2 emc2 /2kT f (c c
P24.28
f (4c ) f (3c ) = 9 e8 = 3.02 103 = 16 e15 = 4.9 106 ) f (c f (c ) The current Ij carried by an ion j is proportional to its concentration cj , mobility uj , and charge number zj . [Justification 24.9] Therefore Therefore, Ij = Acj uj zj where A is a constant. The total current passing through a solution is I=
j
Ij = A
j
cj uj z j
400
INSTRUCTOR'S MANUAL
The transport number of the ion j is therefore tj = Ij Acj uj zj = = A j cj u j z j I c j uj z j j c j u j zj
If there are two cations in the mixture t cuz = t c u z = cu c u if z = z
P24.29
2c c = D 2 [24.84] t x a bx 2 /t or c = 1/2 e t then 1 c = t 2 a t 1/2
with c =
n0 ex /4Dt [24.88] A( Dt)1/2
2
a t 3/2 2bx t
2 ebx /t +
a t 1/2
bx 2 t2
2 bx 2 c ebx /t =  + 2 c 2t t
c = x
2 ebx /t
2c 2b = 2 t x = = 1 2Dt
a t 1/2
2 ebx /t +
a t 1/2
2bx 2 bx 2 /t 2b e = c+ t t
2bx 2 c t
c+
bx 2 Dt 2
c
1 c as required D t
Initially the material is concentrated at x = 0. Note that c = 0 for x > 0 when t = 0 on 2 1 account of the very strong exponential factor (ebx /t 0 more strongly than 1/2 ). When t 2 x = 0, ex /4Dt = 1. We confirm the correct behaviour by noting that x = 0 and x 2 = 0 at t = 0 [24.90], and so all the material must be at x = 0 at t = 0. P24.31 Draw up the following table based on the third and last equations of Justification 24.12
N P (6)Exact P (6)Approx. N P (6)Exact P (6)Approx. 4 0 0.004 30 0.0806 0.0799 6 0.016 0.162 40 0.0807 0.0804 8 0.0313 0.0297 60 0.0763 0.0763 10 0.0439 0.0417 100 0.0666 0.0666 20 0.0739 0.0725
The points are plotted in Fig. 24.1. The discrepancy is less than 0.1 per cent when N > 60
MOLECULES IN MOTION
401
0.10
0.05
0 0 20 40 60 80 100
Figure 24.1
Solutions to applications
P24.33 The work required for a mass, m, to go from a distance r from the centre of a planet of mass m to infinity is w=
r
F dr
where F is the force of gravity and is given by Newton's law of universal gravitation, which is F = Gmm r2
r
G is the gravitational constant (not to be confused with g). Then w = Gmm Gmm dr = r r2
Since according to Newton's second law of motion, F = mg, we may make the identification g= Gm r2
Thus, w = grm. This is the kinetic energy that the particle must have in order to escape the planet's 1 gravitational attraction at a distance r from the planet's centre; hence w = 2 mv 2 = mgr ve = (2g Rp )1/2 [Rp = radius of planet]
which is the escape velocity. (a) ve = [(2) (9.81 m s2 ) (6.37 106 m)]1/2 = 11.2 km s1 (b) g(Mars) = R(Earth)2 m(Mars) g(Earth) = (0.108) m(Earth) R(Mars)2 = 3.76 m s2 Hence, ve = [(2) (3.76 m s2 ) (3.38 106 m)]1/2 = 5.0 km s1 6.37 2 (9.81 m s2 ) 3.38
402
INSTRUCTOR'S MANUAL
Since c =
Mc2 8RT 1/2 ,T = M 8R and we can draw up the following table
103 T /K Earth Mars H2 11.9 2.4 He 23.7 4.8 O2 190 38
[c = 11.2 km s1 ] [c = 5.0 km s1 ]
In order to calculate the proportion of molecules that have speeds exceeding the escape velocity, ve , we must integrate the Maxwell distribution [24.4] from ve to infinity. P =
ve
f (v)dv =
ve
4
m M m 3/2 2 mv 2 /2kT = v e dv R k 2 kT
This integral cannot be evaluated analytically and must be expressed in terms of the error function. We proceed as follows. m Defining = and y 2 = v 2 gives v = 1/2 y, v 2 = 1 y 2 , ve = 1/2 ye , 2kT ye = 1/2 ve , P = 4 = 4 1/2
0
and
dv = 1/2 dy
2 2 3/2 1 1/2 4 y 2 ey dy = 1/2 y 2 ey dy 1/2 v 1/2 v e e
y 2 ey dy 
2
1/2 ve 0
y 2 ey dy
2
The first integral can be evaluated analytically; the second cannot.
0
y 2 ey dy =
2
1/2 , hence 4
1/2 ve
P =1
2 1/2 0
ye
y 2
(2y dy) = 1 
2 1/2 0
1/2 ve
y d(ey )
2
This integral may be evaluated by parts P =1 2 1/2 y(e
y 2
)
1/2 ve 0

1/2 ve 0
(ey ) dy
2
1/2 ve 2 2 2 1/2 2 1/2 ve P =1+2 ve e  1/2 ey dy = 1 + 2 ve eve  erf( 1/2 ve ) 0
= erfc( 1/2 ve ) + 2 From =
2 1/2 ve eve
[erfc(z) = 1  erf(z)]
m M = and ve = (2gRp )1/2 2kT 2RT MgRp 1/2 RT
1/2 ve =
MOLECULES IN MOTION
403
For H2 on Earth at 240 K 1/2 ve = (0.002016 kg mol1 ) (9.807 m s2 ) (6.37 106 m) (8.314 J K1 mol1 ) (240 K) 7.94 1/2
2
1/2
= 7.94
P = erfc(7.94) + 2 at 1500 K
1/2
e(7.94) = (2.9 1029 ) + (3.7 1027 ) = 3.7 1027
ve =
(0.002016 kg mol1 ) (9.807 m s2 ) (6.37 106 m) (8.314 J K1 mol1 ) (1500 K) 3.18 1/2
2
1/2
= 3.18
P = erfc(3.18) + 2 For H2 on Mars at 240 K 1/2 ve =
e(3.18) = (6.9 106 ) + (1.46 104 ) = 1.5 104
(0.002016 kg mol1 ) (3.76 m s2 ) (3.38 106 m) (8.314 J K1 mol1 ) (240 K) 3.58 1/2
2
1/2
= 3.58
P = erfc(3.58) + 2 at 1500 K, 1/2 ve = 1.43
e(3.58) = (4.13 107 ) + (1.10 105 ) = 1.1 105
P = erfc(1.43) + (1.128) (1.43) e(1.43) = 0.0431 + 0.209 = 0.25
2
For He on Earth at 240 K 1/2 ve = (0.004003 kg mol1 ) (9.807 m s2 ) (6.37 106 m) (8.314 J K1 mol1 ) (240 K)
2
1/2
= 11.19
P = erfc(11.2) + (1.128) (11.2) e(11.2) = 0 + (4 1054 ) = 4 1054 at 1500 K, 1/2 ve = 4.48 P = erfc(4.48) + (1.128) (4.48) e(4.48) = (2.36 1010 ) + (9.71 109 )
2
= 1.0 108 For He on Mars at 240 K
1/2 1/2
ve =
(0.004003 kg mol1 ) (3.76 m s2 ) (3.38 106 m) (8.314 J K1 mol1 ) (240 K)
2
= 5.05
P = erfc(5.05) + (1.128) (5.05) e(5.05) = (9.21 1013 ) + (4.79 1011 ) = 4.9 1011 at 1500 K, 1/2 ve = 2.02 P = erfc(2.02) + (1.128) (2.02) e(2.02) = (4.28 103 ) + (0.0401) = 0.044
2
For O2 on Earth it is clear that P 0 at both temperatures.
404
INSTRUCTOR'S MANUAL
For O2 on Mars at 240 K, 1/2 ve = 14.3 P = erfc(14.3) + (1.128) (14.3) e(14.3) = 0 + (2.5 1088 ) = 2.5 1088 0
2
at 1500 K, 1/2 ve = 5.71 P = erfc(5.71) + (1.128) (5.71) e(5.71) = (6.7 106 ) + (4.46 1014 )
2
= 4.5 1014 Based on these numbers alone, it would appear that H2 and He would be depleted from the atmosphere of both Earth and Mars only after many (millions?) years; that the rate on Mars, though still slow, would be many orders of magnitude larger than on Earth; that O2 would be retained on Earth indefinitely; and that the rate of O2 depletion on Mars would be very slow (billions of years?), though not totally negligible. The temperatures of both planets may have been higher in past times than they are now. In the analysis of the data, we must remember that the proportions, P , are not rates of depletion, though the rates should be roughly proportional to P . The results of the calculations are summarized in the following table
240 K H2 P (Earth) P (Mars) 3.7 10 1.1 105
27
He 4 10 4.9 1011
54
O2 0 0
H2 1.5 10 0.25
4
1500 K He 1.0 10 0.044
8
O2 0 4.5 1014
25
The rates of chemical reactions
Solutions to exercises
Discussion questions
E25.1(b) The determination of a rate law is simplified by the isolation method in which the concentrations of all the reactants except one are in large excess. If B is in large excess, for example, then to a good approximation its concentration is constant throughout the reaction. Although the true rate law might be = k[A][B], we can approximate [B] by [B]0 and write = k [A] k = k[B]0 [25.8] which has the form of a firstorder rate law. Because the true rate law has been forced into firstorder form by assuming that the concentration of B is constant, it is called a pseudofirstorder rate law. The dependence of the rate on the concentration of each of the reactants may be found by isolating them in turn (by having all the other substances present in large excess), and so constructing a picture of the overall rate law. In the method of initial rates, which is often used in conjunction with the isolation method, the rate is measured at the beginning of the reaction for several different initial concentrations of reactants. We shall suppose that the rate law for a reaction with A isolated is = k[A]a ; then its initial rate, 0 is given by the initial values of the concentration of A, and we write 0 = k[A]a . Taking logarithms 0 gives: log 0 = log k + a log[A]0 [25.9] For a series of initial concentrations, a plot of the logarithms of the initial rates against the logarithms of the initial concentrations of A should be a straight lime with slope a. The method of initial rates might not reveal the full rate law, for the products may participate in the reaction and affect the rate. For example, products participate in the synthesis of HBr, where the full rate law depends on the concentration of HBr. To avoid this difficulty, the rate law should be fitted to the data throughout the reaction. The fitting may be done, in simple cases at least, by using a proposed rate law to predict the concentration of any component at any time, and comparing it with the data. Because rate laws are differential equations, we must integrate them if we want to find the concentrations as a function of time. Even the most complex rate laws may be integrated numerically. However, in a number of simple cases analytical solutions are easily obtained, and prove to be very useful. These are summarized in Table 25.3. In order to determine the rate law, one plots the right hand side of the integrated rate laws shown in the table against t in order to see which of them results in a straight line through the origin. The one that does is the correct rate law. E25.2(b) The ratedetermining step is not just the slowest step: it must be slow and be a crucial gateway for the formation of products. If a faster reaction can also lead to products, then the slowest step is irrelevant because the slow reaction can then be sidestepped. The ratedetermining step is like a slow ferry crossing between two fast highways: the overall rate at which traffic can reach its destination is determined by the rate at which it can make the ferry crossing. If the first step in a mechanism is the slowest step with the highest activation energy, then it is ratedetermining, and the overall reaction rate is equal to the rate of the first step because all subsequent steps are so fast that once the first intermediate is formed it results immediately in the formation
406
INSTRUCTOR'S MANUAL
of products. Once over the initial barrier, the intermediates cascade into products. However, a ratedetermining step may also stem from the low concentration of a crucial reactant or catalyst and need not correspond to the step with highest activation barrier. A ratedetermining step arising from the low activity of a crucial enzyme can sometimes be identified by determining whether or not the reactants and products for that step are in equilibrium: if the reaction is not at equilibrium it suggests that the step may be slow enough to be ratedetermining. E25.3(b) The parameter A, which corresponds to the intercept of the line at 1/T = 0 (at infinite temperature), is called the preexponential factor or the frequency factor. The parameter Ea , which is obtained from the slope of the line (Ea /R), is called the activation energy. Collectively, the two quantities are called the Arrhenius parameters. The temperature dependence of some reactions is not Arrheniuslike, in the sense that a straight line is not obtained when ln k is plotted against 1/T . However, it is still possible to define an activation energy as Ea = RT 2 d ln k dT
This definition reduces to the earlier one (as the slope of a straight line) for a temperatureindependent activation energy. However, this latter definition is more general, because it allows Ea to be obtained from the slope (at the temperature of interest) of a plot of ln k against 1/T even if the Arrhenius plot is not a straight line. NonArrhenius behaviour is sometimes a sign that quantum mechanical tunnelling is playing a significant role in the reaction. E25.4(b) The expression k = ka kb [A]/(kb + ka [A]) for the effective rate constant of a unimolecular reaction A P is based on the validity of the assumption of the existence of the preequilibrium A + A A + A(ka , ka ). This can be a good assumption if both ka and ka are much larger than kb . The expression for the effective rateconstant, k, can be rearranged to 1 1 k = a + k ka kb ka [A] Hence, a test of the theory is to plot 1/k against 1/[A], and to expect a straight line. Another test is based on the prediction from the LindemannHinshelwood mechanism that as the concentration (and therefore the partial pressure) of A is reduced, the reaction should switch to overall second order kinetics. Whereas the mechanism agrees in general with the switch in order of unimolecular reactions, it does not agree in detail. A typical graph of 1/k against 1/[A] has a pronounced curvature, corresponding to a larger value of k (a smaller value of 1/k) at high pressures (low 1/[A]) than would be expected by extrapolation of the reasonably linear low pressure (high 1/[A]) data.
Numerical exercises
E25.5(b) Rate of reaction =  d[A] 1 d[B] d[C] 1 d[D] = = = = 1.00 mol L1 s1 so dt 3 dt dt 2 dt
Rate of consumption of A = 1.0 mol L1 s1 Rate of consumption of B = 3.0 mol L1 s1 Rate of formation of C = 1.0 mol L1 s1 Rate of formation of D = 2.0 mol L1 s1
THE RATES OF CHEMICAL REACTIONS
407
E25.6(b)
Rate of consumption of B = 
d[B] = 1.00 mol L1 s1 . dt 1 d[D] d[A] d[C] 1 d[B] = = = 0.33 mol L1 s1 = Rate of reaction =  dt 2 dt dt 3 dt
Rate of formation of C = 0.33 mol L1 s1 Rate of formation of D = 0.66 mol L1 s1 Rate of consumption of A = 0.33 mol L1 s1 E25.7(b) The dimensions of k are amount length3 time1 dim of v = (dim of [A]) (dim of [B])2 (amount length3 )3 = length6 amount 2 time1 In mol, L, s units, the units of k are L2 mol2 s1 (a) v =  (b) v = E25.8(b) d[A] = k[A][B]2 dt so so d[A] = k[A][B]2 dt
d[C] dt
d[C] = k[A][B]2 dt
The dimensions of k are amount length3 time1 dim of v = time1 = 3 ) dim of [A] dim of [B] (dim of [C] )1 (amount length The units of k are s1 v= d[C] = k[A][B][C]1 dt
E25.9(b)
The rate law is v = kp a = kp0 (1  f )a where a is the reaction order, and f the fraction reacted (so that 1  f is the fraction remaining). Thus kp0 (1  f1 )a v1 = = v2 kp0 (1  f2 )a 1  f1 a 1  f2 and a= ln(v1 /v2 )
1f1 ln 1f2
=
ln(9.71/7.67) ln 10.100 10.200
= 2.00
E25.10(b) The halflife changes with concentration, so we know the reaction order is not 1. That the halflife increases with decreasing concentration indicates a reaction order <1. Inspection of the data shows the halflife roughly proportional to concentration, which would indicate a reaction order of 0 according to Table 25.3.
408
INSTRUCTOR'S MANUAL
More quantitatively, if the reaction order is 0, then t1/2 p and t1/2
(2) t1/2 (1)
=
p1 p2
We check to see if this relationship holds t1/2
(2) t1/2 (1)
=
340 s = 1.91 178 s
and
55.5 kPa p1 = = 1.92 p2 28.9 kPa
so the reaction order is 0 E25.11(b) The rate law is v= 1 d[A] = k[A] 2 dt
The halflife formula in the text, however, is based on a rate constant for the rate of change of the reactant. That is, it would be accurate to say t1/2 = ln 2 k
provided the k here referred to a rate law  d[A] = k [A] = 2k[A] dt so t1/2 = ln 2 = 1.80 106 s 2(2.78 107 s1 )
The concentration of our reactant (pressure in this case) is [A] = [A]0 e2kt (a) Therefore, after 10 h, we have [A] = (32.1 kPa) exp[2 (2.78 107 s1 ) (3.6 104 s)] = 31.5 kPa (b) and after 50 h, we have [A] = (32.1 kPa) exp[2 (2.78 107 s1 ) (1.8 105 s)] = 29.0 kPa E25.12(b) From Table 25.3, we see that for A + 2B P the integrated rate law is kt = 1 [A]0 ([B]0  2x) ln ([A]0  x)[B]0 [B]0  2[A]0
(a) Substituting the data after solving for k k= 1 0.075 (0.080  0.060) ln (0.075  0.030) 0.080 (3.6 103 s) (0.080  2 0.075) (mol L1 )
= 3.47 103 L mol1 s1
THE RATES OF CHEMICAL REACTIONS
409
(b) The halflife in terms of A is [A]0 [B]0  2[A]0 1 2 t1/2 (A) = ln [A]0 [B]0 k([B]0  2[A]0 )
2
which reduces to t1/2 (A) = = 2[A]0 1 ln 2  [B]0 k([B]0  2[A]0 ) 1 (3.47 103 L mol1 s1 ) (0.070 mol L1 ) ln 2  0.150 0.080
= 8561 s = 2.4 h The halflife in terms of B is [A]0 [B]0  [B]0 1 2 t1/2 (B) = ln [B]0 k([B]0  2[A]0 ) [A]0  4 [B]0 which reduces to t1/2 (B) = = [A]0 /2 1 ln k([B]0  2[A]0 ) [A]0  [B]0 /4 1 (3.47 103 L mol1 s1 ) (0.070 mol L1 )
ln
0.075/2 0.075  (0.080/4)
= 1576 s = 0.44 h E25.13(b) The dimensions of a secondorder rate constant are dim of v amount length3 time = (dim of [A] )2 (amount length3 )2
1
= length3 amount 1 time1
In molecule, m, s units, the units of k are m3 molecule1 s1 The dimensions of a secondorder rate constant in pressure units are dim of v pressure time1 = = pressure1 time1 (dim of p)2 (pressure)2 In SI units, the pressure unit is N m2 = Pa, so the units of k are Pa1 s1 The dimensions of a thirdorder rate constant are dim of v amount length3 time1 = = length6 amount 2 time1 (dim of [A])3 (amount length3 )3 In molecule, m, s units, the units of k are m6 molecule2 s1
410
INSTRUCTOR'S MANUAL
The dimensions of a thirdorder rate constant in pressure units are pressure time1 dim of v = = pressure2 time1 (dim of p)3 (pressure)3 In SI units, the pressure unit is N m2 = Pa, so the units of k are Pa2 s1 E25.14(b) The integrated rate law is kt = [A]0 ([B]0  2[C]) 1 ln ([A]0  [C])[B]0 [B]0  2[A]0
Solving this for [C] yields [C] = (a) [A]0 [B]0 {exp[kt ([B]0  2[A]0 )]  1} [B]0 exp[kt ([B]0  2[A]0 )]  2[A]0 (0.025) (0.150) {exp[(0.21 s1 ) (10 s) (0.150  2 0.025)]  1} [C] = 6.5 103 mol L1 (b) [C]/(mol L1 ) = (0.025) (0.150) {exp[(0.21 s1 ) (600 s) (0.150  2 0.025)]  1} [C] = 0.025 mol L1 E25.15(b) The rate law is v= 1 d[A] = k[A]3 2 dt (0.150) exp[(0.21 s1 ) (600 s) (0.150  2 0.025)]  2(0.025) (0.150) exp[(0.21 s1 ) (10 s) (0.150  2 0.025)]  2(0.025)
[C]/(mol L1 ) =
which integrates to 2kt = t= 1 1 1 so t =  2 2 4k [A] [A]0 1 4 L2 mol2 s1 ) 4(3.50 10 1 2 1 1  2 [A] [A]2 0 1 1  1 )2 (0.021 mol L (0.077 mol L1 )2
= 1.5 106 s E25.16(b) The rate constant is given by k = A exp so at 30 C it is 1.70 102 L mol1 s1 = A exp Ea (8.3145 J K1 mol1 ) [(24 + 273) K] Ea RT
THE RATES OF CHEMICAL REACTIONS
411
and at 50 C it is 2.01 102 L mol1 s1 = A exp Dividing the two rate constants yields 1.70 102 = exp 2.01 102 so ln 1.70 102 2.01 102 = Ea 8.3145 J K1 mol1 Ea 8.3145 J K1 mol1 1 1  297 K 310 K 1 1  297 K 310 K (8.3145 J K 1 mol1 ) Ea 1 mol1 ) [(37 + 273) K] (8.3145 J K
and Ea = 
1 1 1 1.70 102  ln 297 K 310 K 2.01 102
= 9.9 103 J mol1 = 9.9 kJ mol1 With the activation energy in hand, the prefactor can be computed from either rate constant value A = k exp Ea RT = (1.70 102 L mol1 s1 ) exp = 0.94 L mol1 s1 E25.17(b) (a) Assuming that the ratedetermining step is the scission of a C H bond, the ratio of rate constants for the tritiated versus protonated reactant should be kT = e kH where = kf 2kT
1/2
9.9 103 J mol1 (8.3145 J K1 mol1 ) (297 K)
1
1/2 CH

1
1/2 CT
The reduced masses will be roughly 1 u and 3 u respectively, for the protons and 3 H nuclei are far lighter than the rest of the molecule to which they are attached. So = (1.0546 1034 J s) (450 N m1 )1/2 2 (1.381 1023 J K1 ) (298 K) (1.66 1027 kg u1 )1/2 = 2.8 so kT = 0.06 1/16 kH 1 1  1/2 (1 u) (3 u)1/2
(b) The analogous expression for 16 O and 18 O requires reduced masses for C16 O and C18 O bonds. These reduced masses could vary widely depending on the size of the whole molecule. I will use 12 C16 O, for example 16 = = (16.0 u) (12.0 u) = 6.86 u (16.0 + 12.0) u and 18 = (18.0 u) (12.0 u) = 7.20 u (18.0 + 12.0) u
(1.0546 1034 J s) (1750 N m1 )1/2 2 (1.381 1023 J K1 ) (298 K) 1 1  (6.86 u)1/2 (7.20 u)1/2 so k18 = 0.89 k16 (1.66 1027 kg u1 )1/2
= 0.12
412
INSTRUCTOR'S MANUAL
E25.18(b) A reaction nthorder in A has the following rate law  d[A] = k[A]n dt so d[A] = k dt = [A]n d[A] [A]n
Integration yields [A]1n  [A]1n 0 = kt 1n Let t1/3 be the time at which [A] = 1 [A]0 , so 3 [A]0 kt1/3 = 3
1 1n
 [A]1n [A]1n [ 1 0 0 3 = 1n 1n
1n
 1]
and t1/3 =
3n1  1 [A]1n 0 k(n  1)
E25.19(b) The effective rate constant is related to the individual steps by 1 k 1 = a + k ka kb ka p ka = = 1 1  p1 p2 so 1 1 1  = k1 k2 ka 1 1  p1 p2
1 1 1  k1 k2
1 1 1  1.7 103 s1 2.2 104 s1
1 1  1.09 103 Pa 25 Pa
= 9.9 106 s1 Pa1 E25.20(b) The equilibrium constant of the reaction is K= kf kr so kf = Kkr
The relaxation time for the temperature jump is = {kf + kr ([B] + [C])}1 so kf = 1  kr ([B] + [C])
Setting these two expressions for kf equal yields 1 (K + [B] + [C]) 1 kr = (3.0 106 s) (2.0 1016 + 2.0 104 + 2.0 104 ) mol L1 Kkr = 1  kr ([B] + [C]) so kr = = 8.3 108 L mol1 s1 and kf = (2.0 1016 mol L1 ) (8.3 108 L mol1 s1 ) = 1.7 107 s1
THE RATES OF CHEMICAL REACTIONS
413
Solutions to problems
Solutions to numerical problems
P25.2 The procedure is that described in solution to Problem 25.1. Visual inspection of the data seems to indicate that the halflife is roughly independent of the concentration. Therefore, we first try to fit [A] the data to eqn 10b. As in Example 25.3 we plot ln against time to see if a straight line is [A]0 obtained. We draw up the following table (A = (CH3 )3 CBr)
t/h [A]/(10 mol L ) [A] [A]0 [A] ln [A]0 1 (L mol1 ) [A]
2 1
0 10.39 1 0 9.62
3.15 8.96 0.862 0.148 11.16
6.20 7.76 0.747 0.292 12.89
10.00 6.39 0.615 0.486 15.65
18.30 3.53 0.340 1.080 28.3
30.80 2.07 0.199 1.613 48.3
The data are plotted in Fig. 25.1. The fit to a straight line is only fair. The least squares value of k is 0.0542 h1 = 1.51 105 s1 with a correlation coefficient of 0.996. If we try to fit the data to eqn 12b, which corresponds to a secondorder reaction, the fit is not as good. The correlation coefficient is 0.985. Thus we conclude that the reaction is most likely firstorder . A more complex order, which is neither first nor second, is possible, but not likely. At 43.8 h ln [A] [A]0 = 2.359
[A] = 9.82 103 mol L1
0
1.0
2.0 0 10 20 30
Figure 25.1 P25.4 Examination of the data shows that the halflife remains constant at about 2 minutes. Therefore, the reaction is firstorder . This can be confirmed by fitting any two pairs of data to the integrated firstorder rate law, solving for k from each pair, and checking to see that they are the same to within experimental error.
414
INSTRUCTOR'S MANUAL
ln
[A] [A]0
= kt
[10b, A = N2 O5 ]
Solving for k, k= ln [A]0 [A] t ln 1.000 0.705 1.00 min
at t = 1.00 min, [A] = 0.705 mol L1 k= = 0.350 min1 = 5.83 103 s1
at t = 3.00 min, [A] = 0.399 mol L1 = 0.351 min1 = 5.85 103 s1 3.00 min Values of k may be determined in a similar manner at all other times. The average value of k obtained k= is 5.84 103 s1 . The constancy of k, which varies only between 5.83 and 5.85 103 s1 confirms that the reaction is firstorder . A linear regression of ln[A] against t yields the same result. t1/2 = P25.7 0.693 ln 2 = 118.7 s = 1.98 min [11] = k 5.84 103 s1 ln 1.000 0.349
1 [B]0 = 2 [A]0 ; hence [A]0 = 0.624 mol L1 . For the reaction 2A B, [A] = [A]0  2[B]. We can therefore draw up the following table
t/s [B]/(mol L1 ) [A]/(mol L1 )
0 0 0.624
600 0.089 0.446
1200 0.153 0.318
1800 0.200 0.224
2400 0.230 0.164
The data are plotted in Fig. 25.2(a).
0.6
0.4
0.2
0
0
1200
2400
Figure 25.2(a) We see that the halflife of A from its initial concentration is approximately 1200 s, and that its halflife from the concentration at 1200 s is also 1200 s. This indicates a firstorder reaction. We confirm this conclusion by plotting the data accordingly, using ln [A]0 = kA t [A] if d[A] = kA [A] dt
THE RATES OF CHEMICAL REACTIONS
415
First, draw up the table
t/s [A]0 ln [A]
1.4
0 0
600 0.34
1200 0.67
1800 1.02
2400 1.34
and plot the points (Fig. 25.2(b)).
1.0
0.6
0.2 0 0 1200 2400
Figure 25.2(b) The points lie as a straight line, which confirms firstorder kinetics. Since the slope of the line is 5.6 104 , we conclude that kA = 5.6 104 s1 . To express the rate law in the form v = k[A]
1 we note that v =  2
d[A] 1 1 =  2 (kA [A]) = 2 kA [A] dt
P25.8
1 and hence k = 2 kA = 2.8 104 s1 The data do not extend much beyond one halflife; therefore, we cannot see whether the halflife is constant over the course of the reaction as a preliminary step in guessing a reaction order. In a firstorder reaction, however, not only the halflife but any other similarlydefined fractional lifetime remains constant. (That is a property of the exponential function.) In this problem, we can see that the 2/3life is not constant. (It takes less than 1.6 ms for [ClO] to drop from the first recorded value (8.49 mol L1 ) by more than 1/3 of that value (to 5.79 mol L1 ); it takes more than 4.0 more ms for the concentration to drop by not even 1/3 of that value (to 3.95 mol L1 ). So our working assumption is that the reaction is not firstorder but secondorder. Draw up the following table
t/ms 0.12 0.62 0.96 1.60 3.20 4.00 5.75
[CIO]/(mol L1 ) 8.49 8.09 7.10 5.79 5.20 4.77 3.95
(1/[CLO])/(L mol1 ) 0.118 0.124 0.141 0.173 0.192 0.210 0.253
The plot of (1/[ClO])/(L mol1 ) vs. t/ms yields a good straight line; the linear least squares fit is: (1/[ClO]/(L mol1 ) = 0.118 + 0.0237(t/ms) r 2 = 0.974
416
INSTRUCTOR'S MANUAL
0.30 1 (L mol1) [C1O] 0.25 0.20 0.15 0.10 0 1 2 3 t/ms 4 5 6
Figure 25.3
The rate constant is equal to the slope k = 0.0237 L mol1 ms1 = 2.37 107 L mol1 s1 The lifetime or time constant is the time required for the concentration to drop to 1/e of its initial value. Use the integrated secondorder rate law 1 1 = kt  [ClO] [ClO]0 to solve for the time when [ClO] = [ClO]0 /e 1 e  = kt [ClO]0 [ClO]0 t= e1 e1 = 8.56 103 s . = 7 L mol1 s1 )(8.47 106 mol L1 ) k[ClO]0 (2.37 10
So
Note: [ClO]0 was taken from the intercept of the bestfit equation (1/[ClO]0 )/(L mol1 ) = 0.118 so [ClO]0 = 8.47 mol L1
P25.11
Using spreadsheet software to evaluate eqn 25.36, one can draw up a plot like the following. The curves in this plot represent the concentration of the intermediate [I] as a function of time. They are labeled with the ratio k1 /k2 , where k2 = 1 s1 for all curves and k1 varies. The thickest curve, labeled 10, corresponds to k1 = 10 s1 , as specified in part a of the problem. As the ratio k1 /k2 gets smaller (or, as the problem puts it, the ratio k2 /k1 gets larger), the concentration profile for I becomes lower, broader, and flatter; that is, [I] becomes more nearly constant over a longer period of time. This is the nature of the steadystate approximation , which becomes more and more valid as consumption of the intermediate becomes fast compared with its formation.
THE RATES OF CHEMICAL REACTIONS
417
0.8
10
0.6 [I] / (mol / L)
3
0.4
0.99
0.2
0.3 0.1
0.0 0 1
2 t/s
3
4
5
Figure 25.4
P25.13
Ea =
eff R ln keff
k
1 T

1 T
[Exercise 25.16(a) from eqn 25.25] =
R ln 3
1 343 K 1  292 K
= 18 kJ mol1
But keff = kK1 K2 [Problem 25.12] ln keff = ln k + ln K1 + ln K2 d ln keff Ea = R [25.26] = Ea + d(1/T ) since
r H1
+
r H2
 rH d ln K = [van't Hoff equation, Chapter 9] Therefore, d(1/T ) R
r H1
Ea = Ea  P25.15

r H2
= [(18) + (14) + (14)] kJ mol1 = +10 kJ mol1
1 1 k = a + [25.63] k ka k b ka [A] or, in terms of pressure of A 1 1 k = a + k ka kb ka p and we expect a straight line when
p/Torr 1/(p/Torr) 104 /(k/s1 )
1 1 is plotted against . We draw up the following table k p
0.569 1.76 1.17 0.120 8.33 2.55 0.067 14.9 3.30
84.1 0.012 0.336
11.0 0.091 0.448
2.89 0.346 0.629
These points are plotted in Fig. 25.5. There are marked deviations at low pressures, indicating that the Lindemann theory is deficient in that region.
418
INSTRUCTOR'S MANUAL
4
3 s 2 1 0 0
4
8
12
16
Figure 25.5
Solutions to theoretical problems
P25.18 We assume a preequilibrium (as the initial step is fast), and write K= [A]2 , [A2 ] implying that [A] = K 1/2 [A2 ]1/2
The ratedetermining step then gives v= d[P] = k2 [A][B] = k2 K 1/2 [A2 ]1/2 [B] = keff [A2 ]1/2 [B] dt
P25.20
where keff = k2 K 1/2 . d[P] = k[A][B] dt Let the initial concentrations be A0 , B0 , and [P]0 = 0. Then, when an amount x of P is formed, the amount of A changes to A0  2x and that of B changes to B0  3x. Therefore d[P] dx = = k(A0  2x)(B0  3x) dt dt
t 0
with x = 0 at t = 0.
k dt = = =
dx (A0  2x) (B0  3x) 0
x 0
x
6 2B0  3A0
1 1  dx 3(A0  2x) 2(B0  3x)
x dx dx  x  (1/2)A0 x  (1/3)B0 0 0 x
1 (2B0  3A0 )
kt = = =
1 (2B0  3A0 ) 1 2B0  3A0 ln
ln
1 x  2 A0 1  2 A0
 ln
x  1 B0 3  1 B0 3
(2x  A0 )B0 A0 (3x  B0 )
(2x  A0 )B0 1 ln (3A0  2B0 ) A0 (3x  B0 )
THE RATES OF CHEMICAL REACTIONS
419
P25.23
kt =
1 n1
1 1  n1 [A] [A]n1 0 1 n1
[Exercise 25.18(a), n = 1] 1 n1 A0
At t = t1/2 , kt1/2 =
2 n1  A0
3 At t = t3/4 , [A] = 4 [A]0
kt3/4 =
1 n1
4 n1  3A0
1 n1 A0
Hence,
t1/2 = t3/4
2n1  1
4 3 n1
1
P25.24
Let the forward rates be written as r1 = k1 [A], r2 = k2 [B], r3 = k3 [C]
and the reverse rates as r1 = k1 [B], r2 = k2 [C], r3 = k3 [D]
The net rates are then R1 = k1 [A]  k1 [B], R2 = k2 [B]  k2 [C], R3 = k3 [C]  k3 [D]
But [A] = [A]0 and [D] = 0, so that the steadystate equations for the rates of the intermediates are k1 [A]0  k1 [B] = k2 [B]  k2 [C] = k3 [C] From the second of these equations we find [C] = k2 [B] k2 + k 3
After inserting this expression for [C] into the first of the steadystate equations we obtain [B] = [A]0 k1 k2 + k 1  k 2 2 2 +k3
k k
Thus, at the steady state R1 = R2 = R3 = [A]0 k1 1  k1 k2 + k 1 
k2 k2 k2 +k3
420
INSTRUCTOR'S MANUAL
P25.25
v = k([A]0  x)([B]0 + x) dv = k([A]0  x)  k([B]0 + x) dx The extrema correspond to [A]0  x = [B]0 + x Substitute into v to obtain vmax = k [B]0 [A]0 + 2 2 [B]0 [A]0 + 2 2 = k [A]0 + [B]0 2 2 dv = 0, or dx or 2x = [A]0  [B]0 or x= [A]0  [B]0 2
Since v and x cannot be negative in the reaction, [B]0 [A]0 To see the variation of v with x, let [B]0 = [A]0 . The rate equation becomes v = k([A]0  x)([A]0 + x) = k([A]2  x 2 ) = k[A]2  kx 2 0 0 or x2 = 1 k[A]2 [A]2 0 0 v v = 1+ x [A]0 1 from x [A]0 x =0 [A]0
Thus we plot
x2 against 1  k[A]2 [A]2 0 0
The plot is shown in Fig. 25.6 in which X =
x x . 1 corresponds to reality [A]0 [A]0
1.0
0.8
0.6 X 0.4 0.2 0.0 0.0 0.2 0.4 X 0.6 0.8 1.0
Figure 25.6
THE RATES OF CHEMICAL REACTIONS
421
P25.26
For A B C [I] = ka (eka t  ekb t )[A]0 [25.36] kb  k a
ka d[I] (kb ekb t  ka eka t ) = dt kb  k a [I] reaches a maximum when d[I]/dt=0. This occurs when t satisfies the equation kb ekb tmax  ka eka tmax = 0 kb ekb tmax 1  ka e(ka kb )tmax kb
=0
1
ka (ka kb )tmax e =0 kb
e(ka kb )tmax = kb /ka (ka  kb )tmax = ln(kb /ka ) tmax = ln(kb /ka ) (ka /kb )ln(ka /kb ) = ka (kb  ka ) ka kb  1
For ka = 1.0 min1 , the times at which [I] is a maximum are
ka /kb tmax /min 5 2.01 1 1 0.5 0.693
The evaluation for tmax when ka /kb = 1 requires special care. Imagine ka /kb > 0 and take (tmax ). In this limit the value of
ka  1 in the denominator becomes very small (call this value x) kb and can be viewed as being part of the Taylor series expansion of ln (1 + x) ln(1 + x) = x  x2 x3 + + x 2 3 (ka /kb ) ln(ka /kb ) 1 lim ka ka /kb 1 ln(ka /kb ) 1 ka
ka /kb 1
lim
ka /kb 1
lim (tmax ) = =
Plots of
[I] for ka /kb = 5, 1, and 0.5 are shown in Fig. 25.6. [A]0
422
INSTRUCTOR'S MANUAL
0.7 5 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0 1 2 3 4 5 0.5 1
Figure 25.6(a) For A + B P dx v= = k[A][B] = k([A]0  x)([B]0  x) dt
t dx k dt = kt = t=0 x=0 ([A]0  x)([B]0  x) x dx = kt [A]0 [B]0  ([A]0 + [B]0 )x + x 2 x=0 x
The integral on the left may be found in standard mathematics handbooks. dz = + bz + c 1 b2  4ac ln 2ax + b  2ax + b + b2  4ac b2  4ac
az2
The transformations to our working equation are
a1 c [A]0 [B]0 b b+ b2  4ac 2[A]0 b2  4ac 2[B]0
x x=0
b ([A]0 + [B]0 ) b2  4ac [A]0  [B]0
1 x  [A]0 ln [A]0  [B]0 x  [B]0
= kt = kt
1 [A]0  x [A]0 ln  ln [A]0  [B]0 [B]0  x [B]0 1 [B]0 ([A]0  x) ln = kt [A]0  [B]0 [A]0 ([B]0  x) This can also be written in the form 1 [A]0 ([B]0  x) ln [B]0  [A]0 [B]0 ([A]0  x) = kt
THE RATES OF CHEMICAL REACTIONS
423
We will now solve this for x [B]0 {([A]0 [B]0 )kt} [B]0  x f (t) e = [A]0 [A]0  x [B]0  x = [A]0 f (t)  f (t)x (1  f (t))x = [B]0  [A]0 f (t) x= [B]0  [A]0 f (t) [B]0  [B]0 e{([A]0 [B]0 )kt} = 1  f (t) 1  ([B]0 /[A]0 )e{([A]0 [B]0 )kt}
Solutions to applications
P25.28 The firstorder halflife is related to the rate constant by t1/2 = ln 2 k so k = ln 2 ln 2 = = 2.47 102 y1 t1/2 28.1 y
The integrated rate law tells us [90 Sr] = [90 Sr]0 ekt so m = m0 ekt
where m is the mass of 90 Sr. (a) After 18 y : (b) After 70 y : P25.30 m = (1.00 g) exp[(2.47 102 y1 ) (18 y)] = 0.642 g m = (1.00 g) exp[(2.47 102 y1 ) (70 y)] = 0.177 g
We assume a preequilibrium (as the initial step is fast), and write K= [unstable helix] , [A][B] implying that [unstable helix] = K[A][B]
The ratedetermining step then gives v= d[double helix] = k2 [unstable helix] = k2 K[A][B] = k[A][B] dt [k = k2 K]
The equilibrium constant is the outcome of the two processes A+B
k1 k1
unstable helix,
K=
k1 k1
Therefore, with v = k[A][B], k = P25.33 (a) The rate of reaction is v = k[CH4 ][OH]
k1 k2 k1
= (1.13 109 L mol1 s1 ) exp
14.1 103 J mol1 (8.3145 J K1 mol1 ) (263 K)
(4.0 108 mol L1 ) (1.5 1015 mol L1 ) = 1.1 1016 mol L1 s1
424
INSTRUCTOR'S MANUAL
(b) The mass is the amount consumed (in moles) times the molar mass; the amount consumed is the rate of consumption times the volume of the "reaction vessel" times the time m = MvV t = (0.01604 kg mol1 ) (1.1 1016 mol L1 s1 ) (4 1021 L) (365 24 3600 s) = 2.2 1011 kg or 220 Tg P25.35 The initial rate is v0 = (3.6 106 L3 mol3 s1 ) (5 104 mol L1 )2 (104.5 mol L1 )2 = 9 1010 mol L1 s1 The halflife for a secondorder reaction is t1/2 = 1 k [HSO ]0 3
where k is the rate constant in the expression  d[HSO ] 3 = k [HSO ]2 3 dt
Comparison to the given rate law and rate constant shows k = 2k[H+ ]2 = 2(3.6 106 L3 mol3 s1 ) (104.5 mol L1 )2 = 7.2 103 L mol1 s1 and t1/2 = 1 (7.2 103 L mol1 s1 ) (5 104 mol L1 ) = 2.8 105 s = 3 days
26
The kinetics of complex reactions
Solutions to exercises
Discussion questions
E26.1(b) In the analysis of stepwise polymerization, the rate constant for the secondorder condensation is assumed to be independent of the chain length and to remain constant throughout the reaction. It follows, then, that the degree of polymerization is given by n = 1 + kt[A]0 Therefore, the average molar mas can be controlled by adjusting the initial concentration of monomer and the length of time that the polymerization is allowed to proceed. Chain polymerization is a complicated radical chain mechanism involving initiation, propagation, and termination steps (see Section 26.4 for the details of this mechanism). The derivation of the overall rate equation utilizes the steady state approximation and leads to the following expression for the average number of monomer units in the polymer chain: n = 2k[M][I]1/2 , where k =1/2 kP (f ki kt )1/2 , with kP , ki , and kt , being the rate constants for the propagation, initiation, and termination steps, and f is the fraction of radicals that successfully initiate a chain. We see that the average molar mass of the polymer is directly proportional to the monomer concentration, and inversely proportional to the square root of the initiator concentration and to the rate constant for initiation. Therefore, the slower the initiation of the chain, the higher the average molar mass of the polymer. E26.2(b) Refer to eqns 26.26 and 26.27, which are the analogues of the MichaelisMenten and Lineweaver Burk equations (26.21 and 26.22), as well as to Fig. 26.12. There are three major modes of inhibition that give rise to distinctly different kinetic behaviour (Fig. 26.12). In competitive inhibition the inhibitor binds only to the active site of the enzyme and thereby inhibits the attachment of the substrate. This condition corresponds to > 1 and = 1 (because ESI does not form). The slope of the LineweaverBurk plot increases by a factor of relative to the slope for data on the uninhibited enzyme ( = = 1). The yintercept does not change as a result of competitive inhibition. In uncompetitive inhibition, the inhibitor binds to a site of the enzyme that is removed from the active site, but only if the substrate is already present. The inhibition occurs because ESI reduces the concentration of ES, the active type of the complex. In this case = 1 (because EI does not form) and > 1. The y intercept of the LineweaverBurk plot increases by a factor of relative to the yintercept for data on the uninhibited enzyme, but the slope does not change. In noncompetitive inhibition, the inhibitor binds to a site other than the active site, and its presence reduces the ability of the substrate to bind to the active site. Inhibition occurs at both the E and ES sites. This condition corresponds to > 1 and > 1. Both the slope and yintercept of the LineweaverBurk plot increase upon addition of the inhibitor. Figure 26.12c shows the special case of KI = KI and = , which results in intersection of the lines at the xaxis. In all cases, the efficiency of the inhibitor may be obtained by determining KM and max from a control experiment with uninhibited enzyme and then repeating the experiment with a known concentration of inhibitor. From the slope and yintercept of the LineweaverBurk plot for the inhibited enzyme (eqn 26.27), the mode of inhibition, the values of or , and the values of KI , or KI may be obtained.
426
INSTRUCTOR'S MANUAL
E26.3(b)
The steadystate approximation is applied to reactive intermediates in consecutive reactions and is the assumption that their concentrations do not change much with time. It is a good approximation if the rate constant for the reaction of the intermediate, in either the forward or backward direction, is large compared to the rate constant in the other direction. This approximation is applicable when chemical production and chemical consumption are closely balanced. In the steadystate condition of a chemical reaction, a reactant or product is maintained at a constant concentration throughout the course of the reaction by supplying it to or withdrawing it from the reaction vessel. Steadystates are not equilibrium states in the thermodynamic sense and in fact are the other extreme from equilibrium. Bistability is a condition in which two distinct, far from equilibrium, steadystates are chemically available to the reacting system. In some systems, bistability is a necessary condition for chemical oscillation to occur. Under the right conditions, the system may jump periodically between the two steady states as the reaction progresses. Refer to Figs 26.19 and 26.20 of the text for an illustration of the process. However, bistability alone is not a sufficient condition to achieve oscillation in an autocatalytic reaction. In order for the oscillation to occur, it is necessary to have a feedback mechanism involving a third species Z that reacts with the intermediates X and Y according to: Y + Z X and X + Z Y. Thus Z reacts with X to produce Y and with Y to produce X. As a result the system can switch periodically between the upper and lower steady states.
E26.4(b)
The shortening of the lifetime of an excited state is called quenching. Quenching effects may be studied by monitoring the emission from the excited state that is involved in the photochemical process. The addition of a quencher opens up an additional channel for the deactivation of the excited singlet state. Three common mechanisms for bimolecular quenching of an excited singlet (or triplet) state are: Collisional deactivation: S + Q S + Q Energy transfer: S + Q S + Q Electron transfer: S + Q S+ + Q 
or
S + Q +
Collisional quenching is particularly efficient when Q is a heavy species, such as iodide ion, which receives energy from S and then decays primarily by internal conversion to the ground state. Pure collisional quenching can be detected by the appearance of vibrational and rotational excitation in the spectrum of the acceptor. In many cases, it is possible to prove that energy transfer is the predominant mechanism of quenching if the excited state of the acceptor fluoresces or phosphoresces at a characteristic wavelength. In a pulsed laser experiment, the rise in fluorescence intensity from Q with a characteristic time which is the same as that for the decay of the fluorescence of S is often taken as indication of energy transfer from S to Q. Electron transfer can be studied by timeresolved spectroscopy (Section 17.7e). The oxidized and reduced products often have electronic absorption spectra distinct from those of their neutral parent compounds. Therefore, the rapid appearance of such known features in the absorption spectrum after excitation by a laser pulse may be taken as indication of quenching by electron transfer.
Numerical exercises
In the following exercises and problems, it is recommended that rate constants are labelled with the number of the step in the proposed reaction mechanism and that any reverse steps are labelled similarly but with a prime.
THE KINETICS OF COMPLEX REACTIONS
427
E26.5(b)
The intermediates are NO and NO3 and we apply the steadystate approximation to each of their concentrations k2 [NO2 ][NO3 ]  k3 [NO][N2 O5 ] = 0 k1 [N2 O5 ]  k1 [NO2 ][NO3 ]  k2 [NO2 ][NO3 ] = 0 Rate =  1 d[N2 O5 ] 2 dt
d[N2 O5 ] = k1 [N2 O5 ] + k1 [NO2 ][NO3 ]  k3 [NO][N2 O5 ] dt From the steady state equations k3 [NO][N2 O5 ] = k2 [NO2 ][NO3 ] k1 [N2 O5 ] [NO2 ][NO3 ] = k1 + k 2 Substituting, k k1 d[N2 O5 ] k 2 k1 2k1 k2 [N2 O5 ]  [N2 O5 ] =  [N2 O5 ] = k1 [N2 O5 ] + 1 k1 + k 2 k1 + k 2 k1 + k 2 dt k1 k2 Rate = [N2 O5 ] = k[N2 O5 ] k1 + k 2 E26.6(b) d[R] = 2k1 [R2 ]  k2 [R][R2 ] + k3 [R ]  2k4 [R]2 dt d[R ] = k2 [R][R2 ]  k3 [R ] dt Apply the steadystate approximation to both equations 2k1 [R2 ]  k2 [R][R2 ] + k3 [R ]  2k4 [R]2 = 0 k2 [R][R2 ]  k3 [R ] = 0 The second solves to [R ] = k2 [R][R2 ] k3
1/2 k1 [R2 ] k4
and then the first solves to [R] = Therefore, E26.7(b)
d[R2 ] = k1 [R2 ]  k2 [R2 ][R] = k1 [R2 ]  k2 dt
k1 1/2 [R2 ]3/2 k4
(a) The figure suggests that a chainbranching explosion does not occur at temperatures as low as 700 K. There may, however, be a thermal explosion regime at pressures in excess of 106 Pa. (b) The lower limit seems to occur when log(p/Pa) = 2.1 so p = 102.1 Pa = 1.3 102 Pa
There does not seem to be a pressure above which a steady reaction occurs. Rather the chainbranching explosion range seems to run into the thermal explosion range around log(p/Pa) = 4.5 so p = 104.5 Pa = 3 104 Pa
428
INSTRUCTOR'S MANUAL
E26.8(b)
The rate of production of the product is d[BH+ ] = k2 [HAH+ ][B] dt HAH+ is an intermediate involved in a rapid preequilibrium k1 [HAH+ ] = [HA][H+ ] k1 and so [HAH+ ] = k1 [HA][H+ ] k1
k 1 k2 d[BH+ ] = [HA][H+ ][B] dt k1
This rate law can be made independent of [H+ ] if the source of H+ is the acid HA, for then H+ is given by another equilibrium [H+ ]2 [H+ ][A ] = Ka = [HA] [HA] and E26.9(b) k 1 k2 K a d[BH+ ] = dt k1 d[A2 ] = k1 [A2 ] dt Consequently, the rate of consumption of [A2 ] is first order in A2 and the rate is independent of intermediate concentrations. E26.10(b) The maximum velocity is kb [E]0 and the velocity in general is v = k[E]0 = vmax = kb [S][E]0 KM + [S] so vmax = kb [E]0 = KM + [S] v [S]
1/2
so
[H+ ] = (Ka [HA])1/2
[HA]3/2 [B]
A2 appears in the initiation step only.
(0.042 + 0.890) mol L1 0.890 mol L1
(2.45 104 mol L1 s1 ) = 2.57 104 mol L1 s1
E26.11(b) The quantum yield tells us that each mole of photons absorbed causes 1.2 102 moles of A to react; the stoichiometry tells us that 1 mole of B is formed for every mole of A which reacts. From the yield of 1.77 mmol B, we infer that 1.77 mmol A reacted, caused by the absorption of 1.77 103 mol/(1.2 102 mol Einstein1 ) = 1.5 105 moles of photons E26.12(b) The quantum efficiency is defined as the amount of reacting molecules nA divided by the amount of photons absorbed nabs . The fraction of photons absorbed fabs is one minus the fraction transmitted ftrans ; and the amount of photons emitted nphoton can be inferred from the energy of the light source (power P times time t) and the energy of the photons (hc/). = = nA hcNA (1  ftrans )P t (0.324 mol) (6.626 1034 J s) (2.998 108 m s1 ) (6.022 1023 mol1 ) (1  0.257) (320 109 m) (87.5 W) (28.0 min) (60 s min1 )
= 1.11
THE KINETICS OF COMPLEX REACTIONS
429
Solutions to problems
Solutions to numerical problems
P26.2 O + Cl2 ClO + Cl [O] [O]0 ek t That being so, ln [O]0 d = k t = k[Cl2 ]t = k[Cl2 ] [O] v where k = [Cl2 ]k, v is the flow rate, and d is the distance along the tube. We draw up the following table
d/cm [O]0 ln [O] 0 0.27 2 0.31 4 0.34 6 0.38 8 0.45 10 0.46 12 0.50 14 0.55 16 0.56 18 0.60
p(Cl2 ) constant [Cl2 at high pressure]
Therefore, the reaction is probably pseudofirst order, and
The points are plotted in Fig. 26.1.
0.6
0.5
0.4
0.3
0.2 0 10 20
Figure 26.1 k[Cl2 ] = 0.0189 cm1 . v (0.0189 cm1 ) v Therefore, k = [Cl2 ] The slope is 0.0189, and so (0.0189 cm1 ) (6.66 102 cm s1 ) = 5.0 107 L mol1 s1 2.54 107 mol L1 (There is a very fast O + ClO Cl + O2 reaction, and so the answer given here is actually twice the true value.) = P26.5 H2 2H H + O2 OH + O O + H2 OH + H H + O2 HO2 HO2 + H2 H2 O + OH HO2 + wall destruction H + M destruction initiation, branching, branching, propagation, propagation, termination, termination, v v v v v v v = vinit = k1 [H][O2 ] = k2 [O][H2 ] = k3 [H][O2 ] = k4 [HO2 ][H2 ] = k5 [HO2 ] = k6 [H][M]
430
INSTRUCTOR'S MANUAL
We identify the onset of explosion with the rapid increase in the concentration of radicals which we initially identify with [H]. Then vrad = vinit  k1 [H][O2 ] + k2 [O][H2 ]  k3 [H][O2 ]  k6 [H][M] Intermediates are examined with the steadystate approximation. d[O] = k1 [H][O2 ]  k2 [O][H2 ] 0 dt k1 [H][O2 ] [O]SS k2 [H2 ] Therefore, vrad = vinit  k1 [H][O2 ] + k2 k1 [H][O2 ] [H2 ]  k3 [H][O2 ]  k6 [H][M] k2 [H2 ]
= vinit  (k3 [O2 ] + k6 [M])[H] The factor (k3 [O2 ] + k6 [M]) is always positive and, therefore, vrad always decreases for all values of [H]. No explosion is possible according to this mechanism, or at least no exponential growth of [H] is observed. Let us try a second approach for which the concentration of radicals is identified with [O]. vrad = k1 [H][O2 ]  k2 [O][H2 ] Using the steadystate approximation to describe [H], we find that [H]SS = vrad = vinit + k2 [H2 ][O] (k1 + k3 )[O2 ] + k6 [M]
k1 k2 [H2 ][O2 ] vinit k1 [O2 ] +  k2 [H2 ] [O] (k1 + k3 )[O2 ] + k6 [M] (k1 + k3 )[O2 ] + k6 [M]
This has the form vrad = d[O] = C1 + {C2  C3 }[O] dt
where C1 , C2 , and C3 are always positive. This means that the mechanism predicts exponential growth k1 [O2 ] > 1. But of radicals, and explosion, when C2 > C3 . This will occur when (k1 + k3 )[O2 ] + k6 [M] this is not possible. So no exponential growth of [O] can occur. The proposed mechanism is inconsistent with the existence of an explosion on the assumption that the steadystate approximation can be applied to the intermediates H and O. It is, however, unlikely that the steadystate approximation can be applied to explosive reactions, and this is where the analysis breaks down. P26.8 M + hi M , Ia [M = benzophenone] M + Q M + Q, kq M M + hf , kf ] d[M = Ia  kf [M ]  kq [Q][M ] 0 [steady state] dt Ia and hence [M ] = kf + kq [Q]
THE KINETICS OF COMPLEX REACTIONS
431
Then If = kf [M ] = and so
kf Ia kf + kq [Q]
kq [Q] 1 1 = + If Ia kf Ia
If the exciting light is extinguished, [M ], and hence If , decays as ekf t in the absence of a quencher. Therefore we can measure kq /kf Ia from the slope of 1/If plotted against [Q], and then use kf to determine kq . We draw up the following table
103 [Q]/M 1 If 1 2.4 5 4.0 10 6.3
The points are plotted in Fig. 26.2.
8
6
4
2
0 0 0.005 0.010
Figure 26.2 The intercept lies at 2.0, and so Ia = kq = 430 L mol1 kf Ia Then, since Ia = 0.50 and kf = ln 2 , t1/2 ln 2 29 106 s = 5.1 106 L mol1 s1 1 = 0.50. The slope is 430, and so 2.0
kq = (0.50) (430 L mol1 )
Solutions to theoretical problems
P26.11 d[CH3 CH3 ] = ka [CH3 CH3 ]  kb [CH3 ][CH3 CH3 ]  kd [CH3 CH3 ][H] + ke [CH3 CH2 ][H] dt We apply the steadystate approximation to the three intermediates CH3 , CH3 CH2 , and H. d[CH3 ] = 2ka [CH3 CH3 ]  kb [CH3 CH3 ][CH3 ] = 0 dt which implies that [CH3 ] = 2ka . kb
432
INSTRUCTOR'S MANUAL
d[CH3 CH2 ] = kb [CH3 ][CH3 CH3 ]  kc [CH3 CH2 ] dt + kd [CH3 CH3 ][H]  ke [CH3 CH2 ][H] = 0 d[H] = kc [CH3 CH2 ]  kd [CH3 CH3 ][H]  ke [CH3 CH2 ][H] = 0 dt These three equations give [H] = kc
[CH3 CH3 ke + kd [CH3 CH2 ] ]
[CH3 CH2 ]2 
ka kc ka 2kc
[CH3 CH3 ][CH3 CH2 ]  ka 2 + 2kc k a kd kc ke
k a kd kc ke
1/2
[CH3 CH3 ]2 = 0
or [CH3 CH2 ] = which implies that [H] = , ke + kd kc
+
[CH3 CH3 ]
=
ka 2kc
+
ka 2 + 2kc
k a kd kc k e
1/2
If ka is small in the sense that only the lowest order need be retained, [CH3 CH2 ] [H] ka kd 1/2 [CH3 CH3 ] kc ke kc
kc ke ke + kd ka kd
1/2
ka kc 1/2 k d ke
The rate of production of ethene is therefore d[CH2 CH2 ] = kc [CH3 CH2 ] = dt ka kc kd 1/2 [CH3 CH3 ] ke
The rate of production of ethene is equal to the rate of consumption of ethane (the intermediates all have low concentrations), so d[CH3 CH3 ] = k[CH3 CH3 ], dt k= ka kc kd 1/2 ke
Different orders may arise if the reaction is sensitized so that ka is increased. P26.12 CH3 CHO CH3 + CHO, ka CH3 + CH3 CHO CH4 + CH2 CHO, kb CH2 CHO CO + CH3 , kc CH3 + CH3 CH3 CH3 , kd d[CH4 ] = kb [CH3 ][CH3 CHO] dt d[CH3 CHO] = ka [CH3 CHO]  kb [CH3 CHO][CH3 ] dt
THE KINETICS OF COMPLEX REACTIONS
433
d[CH3 ] = ka [CH3 CHO]  kb [CH3 CHO][CH3 ] + kc [CH2 CHO]  2kd [CH3 ]2 = 0 dt d[CH2 CHO] = kb [CH3 ][CH3 CHO]  kc [CH2 CHO] = 0 dt Adding the last two equations gives ka [CH3 CHO]  2kd [CH3 ]2 = 0, Therefore d[CH4 ] = kb dt ka 1/2 [CH3 CHO]3/2 2kd ka 1/2 [CH3 CHO]3/2 2kd or [CH3 ] = ka 1/2 [CH3 CHO]1/2 2kd
d[CH3 CHO] = ka [CH3 CHO]  kb dt Note that, to lowestorder in ka , d[CH3 CHO] kb dt Mn = M3
3 3
ka 1/2 [CH3 CHO]3/2 2kd n3 p n1 [Pn = pn1 (1  p), Problem 26.13]
and the reaction is threehalves order in CH3 CHO. P26.14 (a)
n
n3 Pn = M 3 (1  p)
n
d d d d n2 p n = M 3 (1  p) p p pn = M (1  p) dp n dp dp dp n d M 3 (1 + 4p + p 2 ) d d = M 3 (1  p) p p (1  p)1 = dp dp dp (1  p)3 2 (1 + p) M 2 Mn = [Problem 26.13] (1  p)2 Therefore, (b) Mn
3
Mn 1 [26.8], n = 1p
3
= 2
M(1 + 4p + p 2 ) 1  p2 so p =1 1 n
P26.16
= (6 n 2  6 n + 1) n 2 Mn d[A] = k[A]2 [OH] = k[A]3 because dt d[A]3 = k dt [A] since
[A]
Mn
[A] = [OH].
and
t d[A] = k dt = kt 3 [A]0 [A] 0
1 dx = 2 , the equation becomes 3 x 2x or [A] = [A]0 (1 + 2kt[A]0 )1/2
1 1  = 2kt 2 [A] [A]2 0
434
INSTRUCTOR'S MANUAL
By eqn 26.8a the degree of polymerization, n , is given by n = [A]0 = (1 + 2kt[A]0 )1/2 [A]
P26.18
d[P] = k[A][P]2 dt dx = k(A0  x)(P0 + x)2 [x = P  P0 ] dt x dx kt = 2 0 (A0  x)(P0 + x) Integrate by partial fractions (as in Problem 26.17) kt = = = 1 A0 + P 0 1 A0 + P 0 1 A0 + P 0
x 0 2 1 + P0 + x
1 A0 + P 0
1 1 + P0 + x A0  x
dx
1 P0 + x A0 1 1 ln + ln  + P0 P0 + x A0 + P 0 P0 A0  x x (P0 + x)A0 1 ln + P0 (P0 + x) A0 + P 0 P0 (A0  x) P0 x , and p = [A]0 A0 y p(p + y) + 1 p+y ln 1+p p(1  y)
Therefore, with y =
A0 (A0 + P0 )kt =
As in Problem 26.6, the rate is maximum when d[P] dvP = 2k[A][P] dt dt +k d[A] [P]2 dt
= 2k[A][P]vP  k[P]2 vP = k[P](2[A]  [P])vP = 0
1 That is, at [A] = 2 [P]
On substitution of this condition into the integrated rate law, we find A0 (A0 + P0 )ktmax = 2p 2p(1 + p) + 1 2 ln 1+p p
or (A0 + P0 )2 ktmax = P26.20 (i)
2p 2 + ln 2p p
d[X] = ka [A][Y]  kb [X][Y] + kc [A][X]  2kd [X]2 dt d[Y] = ka [A][Y]  kb [X][Y]  ke [Z] (ii) dt
THE KINETICS OF COMPLEX REACTIONS
435
Express these differential equations as finitedifference equations (i) X(ti+1 ) = X(ti ) + {ka [A]Y(ti )  kb X(ti )Y(ti ) + kc [A]X(ti )  2kd X2 (ti )} t (ii) Y(ti+1 ) = Y(ti ) + {kc [Z]  ka [A]Y(ti )  kb X(ti )Y(ti )} t Solve these equations by iteration. d[B] = Ia AB dt d[B] = k[B]2 BA dt In the photostationary state Ia  k[B]2 = 0. Hence, [B] = Ia 1/2 [A]1/2 k [because I [A]]
P26.21
The illumination may increase the rate of the forward reaction without affecting the reverse reaction. Hence the position of equilibrium may be shifted toward products. P26.23 Cl2 + h 2Cl Cl + CHCl3 CCl3 + HCl CCl3 + Cl2 CCl4 + Cl 2CCl3 + Cl2 2CCl4 (i) Ia k2 k3 k4
d[CCl4 ] = 2k4 [CCl3 ]2 [Cl2 ] + k3 [CCl3 ][Cl2 ] dt d[CCl3 ] (ii) = k2 [Cl][CHCl3 ]  k3 [CCl3 ][Cl2 ]  2k4 [CCl3 ]2 [Cl2 ] = 0 dt d[Cl] (iii) = 2Ia  k2 [Cl][CHCl3 ] + k3 [CCl3 ][Cl2 ] = 0 dt d[Cl2 ] = Ia  k3 [CCl3 ][Cl2 ]  k4 [CCl3 ]2 [Cl2 ] (iv) dt Therefore, Ia = k4 [CCl3 ]2 [Cl2 ] [(ii) + (iii)] which implies that [CCl3 ] = Then, with (i), d[CCl4 ] k3 Ia [Cl]1/2 = 2Ia + 1/2 dt k
4 1/2
1 1/2 k4
Ia 1/2 [Cl2 ]
When the pressure of chlorine is high, and the initiation rate is slow (in the sense that the lowest powers of Ia dominate), the second term dominates the first, giving d[CCl4 ] k3 Ia 1/2 = 1/2 [Cl2 ]1/2 = kIa [Cl2 ]1/2 dt k4 k3 with k = 1/2 . It seems necessary to suppose that Cl + Cl recombination (which needs a third body) k4 is unimportant.
1/2
436
INSTRUCTOR'S MANUAL
Solutions to applications
P26.26 The mechanism considered is E+S
ka ka
(ES)
ka ka
P+E
We apply the steadystate approximation to [(ES)]. d[ES] = ka [E][S]  ka [(ES)]  kb [(ES)] + kb [E][P] = 0 dt Substituting [E] = [E]0  [(ES)] we obtain ka ([E]0  [(ES)])[S]  ka [(ES)]  kb [(ES)] + kb ([E]0  [(ES)])[P] = 0 (ka [S]  ka  kb  kb [P])[(ES)] + ka [E]0 [S]  kb [E]0 [P] = 0 [E]0 [S] + kb [E]0 [P] ka [E]0 [S] + kb [E]0 [P] a [(ES)] = = k ka [S] + ka + kb + kb [P] KM + [S] + kb [P] a [E]0 [S] + kb [E]0 [P] d[P] a Then, = kb [(ES)]  kb [P][E] = kb k dt KM + [S] + kb [P] a k [E]0 [S] + kb a [E]0  KM + [S] +
k k k
k + kb KM = a ka
 kb [P] [E]0 [P]
kb ka
[P]
=
kb [E]0 [S] + kb [E]0 [P]  kb [E]0 [P ]KM a KM + [S] + kb [P] a
k
Substituting for KM in the numerator and rearranging
a kb [E]0 [S] + ka b [E]0 [P] d[P] = k dt KM + [S] + kb [P] a
k k
For large concentrations of substrate, such that [S] d[P] = kb [E]0 dt
KM and [S]
[P],
which is the same as the unmodified mechanism. For [S] [S]  (k/kb )[P] d[P] = kb [E]0 dt [S] + (k/ka )[P] k= ka kb ka
KM , but [S] [P]
For [S] 0,
ka kb [E]0 [P] ka [E]0 [P] d[P] = = ka + kb + kb [P] dt kP + [P]
k + kb where kP = a kb
THE KINETICS OF COMPLEX REACTIONS
437
Comment. The negative sign in the expression for
d[P] for the case [S] 0 is to be interpreted dt to mean that the mechanism in this case is the reverse of the mechanism for the case [P] 0. The roles of P and S are interchanged. Question. Can you demonstrate the last statement in the comment above?
P26.28
max [26.21] 1 + KM [S]0 Taking the inverse and multiplying by max , we find that = max = + KM Thus, = max  KM [S]0 or max =  [S]0 KM KM [S]0
The regression slope and intercept of the EadieHofstee data plot of against /[S]0 gives KM and max , respectively. Alternatively, the regression slope and intercept of an EadieHofstee data plot of /[S]0 against gives 1/KM , and max /KM , respectively. The slope and intercept of the latter plot can be used to in the calculation of KM and max . P26.32 The rate of reaction is the rate at which ozone absorbs photons times the quantum yield. The rate at which ozone absorbs photons is the rate at which photons impinge on the ozone times the fraction of photons absorbed. That fraction is 1  T , where T is the transmittance. T is related to the absorbance A by A =  log T = cl so 1  T = 1  10cl
1 1 9 1 5 1  T = 1  10{(260 L mol cm )(810 mol L )(10 cm)} = 0.38
If we let F stand for the flux of photons (the rate at which photons impinge on our sample of ozone), then the rate of reaction is v= (1  T )F = (0.94) (0.38) (1 1014 cm2 s1 ) (1000 cm3 L1 ) (6.022 1023 mol1 ) (105 cm)
= 5.9 1013 mol L1 s1
27
Molecular reaction dynamics
Solutions to exercises
Discussion questions
E27.1(b) A reaction in solution can be regarded as the outcome of two stages: one is the encounter of two reactant species, which is followed by their reaction, the second stage, if they acquire their activation energy. If the ratedetermining step is the former, then the reaction is said to be diffusioncontrolled. If the ratedetermining step is the latter, then the reaction is activation controlled. For a reaction of the form A + B P that obeys the secondorder rate law = k2 [A][B], in the diffusioncontrolled regime, k2 = 4R DNA where D is the sum of the diffusion coefficients of the two reactant species and R is the distance at which reaction occurs. A further approximation is that each molecule obeys the StokesEinstein relation and Stokes' law, and then k2 8RT 3
where is the viscosity of the medium. The result suggests that k2 is independent of the radii of the reactants. E27.2(b) In the kinetic salt effect, the rate of a reaction in solution is changed by modification of the ionic strength of the medium. If the reactant ions have the same sign of charge (as in cation/cation or anion/anion reactions), then an increase in ionic strength increases the rate constant. If the reactant ions have opposite signs (as in cation/anion reactions), then an increase in ionic strength decreases the rate constant. In the former case, the effect can be traced to the denser ionic atmosphere (see the DebyeHuckel theory) that forms round the newly formed and highly charged ion that constitutes the activated complex and the stronger interaction of that ion with the atmosphere. In the latter case, the ion corresponding to the activated complex has a lower charge than the reactants and hence it has a more diffuse ionic atmosphere and interacts with it more weakly. In the limit of low ionic strength the rate constant can be expected to follow the relation log k = log k + 2AzA zB I 1/2 E27.3(b) Refer to Figs 27.21 and 27.22 of the text. The first of these figures shows an attractive potential energy surface, the second, a repulsive surface. (a) Consider Fig. 27.21. If the original molecule is vibrationally excited, then a collision with an incoming molecule takes the system along the floor of the potential energy valley (trajectory C). This path is bottled up in the region of the reactants, and does not take the system to the saddle point. If, however, the same amount of energy is present solely as translational kinetic energy, then the system moves along a successful encounter trajectory C and travels smoothly over the saddle point into products. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion. Moreover, the potential surface shows that once past the saddle point the trajectory runs up the steep wall of the product valley, and then rolls from side to side as it falls to the foot of the valley as the products separate. In other words, the products emerge in a vibrationally excited state.
MOLECULAR REACTION DYNAMICS
439
(b) Now consider the repulsive surface (Fig. 27.22). On trajectory C the collisional energy is largely in translation. As the reactants approach, the potential energy rises. Their path takes them up the opposing face of the valley, and they are reflected back into the reactant region. This path corresponds to an unsuccessful encounter, even though the energy is sufficient for reaction. On a successful trajectory C , some of the energy is in the vibration of the reactant molecule and the motion causes the trajectory to weave from side to side up the valley as it approaches the saddle point. This motion may be sufficient to tip the system round the corner to the saddle point and then on to products. In this case, the product molecule is expected to be in an unexcited vibrational state. Reactions with repulsive potential surfaces can therefore be expected to proceed more efficiently if the excess is present as vibrations.
Numerical exercises
E27.4(b) The collision frequency is z = 21/2 v p kT where = d 2 = 4 r 2 and v = 8RT 1/2 M
so z = =
21/2 p 8RT 1/2 16pNA r 2 1/2 (4r 2 ) = kT M (RT M)1/2 16 (100 103 Pa) (6.022 1023 mol1 ) (180 1012 m)2 ( )1/2 [(8.3145 J K1 mol1 ) (298 K) (28.01 103 kg mol1 )]1/2
= 6.64 109 s1 The collision density is 1 zp (6.64 109 s1 ) (100 103 Pa) = 8.07 1034 m3 s1 ZAA = zN/V = = 2 2kT 2(1.381 1023 J K1 ) (298 K) Raising the temperature at constant volume means raising the pressure in proportion to the temperature ZAA T so the per cent increase in z and ZAA due to a 10 K increase in temperature is 1.6 per cent , same as Exercise 27.4(a). E27.5(b) The appropriate fraction is given by f = exp Ea RT
The values in question are (a) (i) f = exp (ii) f = exp (b) (i) f = exp (ii) f = exp 15 103 J mol1 (8.3145 J K1 mol1 ) (300 K) 15 103 J mol1 (8.3145 J K1 mol1 ) (800 K) 150 103 J mol1 (8.3145 J K1 mol1 ) (300 K) 150 103 J mol1 (8.3145 J K1 mol1 ) (800 K) = 2.4 103 = 0.10 = 7.7 1027 = 1.6 1010
440
INSTRUCTOR'S MANUAL
E27.6(b)
A straightforward approach would be to compute f = exp
Ea at the new temperature and RT compare it to that at the old temperature. An approximate approach would be to note that f changes Ea Ea from f0 = exp to f = exp , where x is the fractional increase in the RT RT (1 + x) Ea Ea temperature. If x is small, the exponent changes from to approximately (1  x) and f RT RT x Ea Ea Ea (1  x) Ea x = f 0 f0 . exp changes from exp to exp = exp RT RT RT RT x Thus the new Boltzmann factor is the old one times a factor of f0 . The factor of increase is
x (ii) f0 = (0.10)10/800 = 1.03 x (b) (i) f0 = (7.7 1027 )10/300 = 7.4 x (a) (i) f0 = (2.4 103 )10/300 = 1.2
x (ii) f0 = (1.6 1010 )10/800 = 1.3
E27.7(b)
The reaction rate is given by v = P 8kB T 1/2 NA exp(Ea /RT )[D2 ][Br2 ] 8kB T 1/2 NA exp(Ea /RT ) 8(1.381 1023 J K1 ) (450 K) (3.930 u) (1.66 1027 kg u1 ) 200 103 J mol1 (8.3145 J K1 mol1 ) (450 K)
1/2
so, in the absence of any estimate of the reaction probability P , the rate constant is k =
= [0.30 (109 m)2 ]
(6.022 1023 mol1 ) exp
= 1.71 1015 m3 mol1 s1 = 1.7 1012 L mol1 s1 E27.8(b) The rate constant is kd = 4R DNA where D is the sum of two diffusion constants. So kd = 4(0.50 109 m) (2 4.2 109 m2 s1 ) (6.022 1023 mol1 ) = 3.2 107 m3 mol1 s1 In more common units, this is kd = 3.2 1010 L mol1 s1 E27.9(b) (a) A diffusioncontrolled rate constant in decylbenzene is kd = 8 (8.3145 J K1 mol1 ) (298 K) 8RT = 1.97 106 m3 mol1 s1 = 3 3 (3.36 103 kg m1 s1 )
MOLECULAR REACTION DYNAMICS
441
(b) In concentrated sulfuric acid kd = 8 (8.3145 J K 1 mol1 ) (298 K) 8RT = 2.4 105 m3 mol1 s1 = 3 3 (27 103 kg m1 s1 )
E27.10(b) The diffusioncontrolled rate constant is kd = 8RT 8 (8.3145 J K1 mol1 ) (298 K) = 1.10 107 m3 mol1 s1 = 3 3 (0.601 103 kg m1 s1 )
In more common units, kd = 1.10 1010 L mol1 s1 The recombination reaction has a rate of v = kd [A][B] with [A] = [B]
so the halflife is given by t1/2 = 1 1 = = 5.05 108 s 10 L mol1 s1 ) (1.8 103 mol L1 ) k[A]0 (1.10 10
E27.11(b) The reactive crosssection is related to the collision crosssection by = P so P = /.
The collision crosssection is related to effective molecular diameters by = d 2 so d = (/ )1/2
2 1/2 1/2 2 2 1 1 Now AB = dAB = 2 (dA + dB ) = 4 AA + BB
so P =
1 4
AA + BB
1/2
1/2 2
8.7 1022 m = 1 = 2.22 103 [((0.88)1/2 + (0.40)1/2 ) 109 m]2 4 E27.12(b) The diffusioncontrolled rate constant is kd = 8 (8.3145 J K1 mol1 ) (293 K) 8RT = 5.12 106 m3 mol1 s1 = 3 3 (1.27 103 kg m1 s1 )
In more common units, kd = 5.12 109 L mol1 s1 The recombination reaction has a rate of v = kd [A][B] = (5.12 109 L mol1 s1 ) (0.200 mol L1 ) (0.150 mol L1 ) = 1.54 108 mol L1 s1
442
INSTRUCTOR'S MANUAL
E27.13(b) The enthalpy of activation for a reaction in solution is
H = Ea  RT = (8.3145 J K1 mol1 ) (6134 K)  (8.3145 J K 1 mol1 ) (298 K) = 4.852 104 J mol1 = 48.52 kJ mol1
The entropy of activation is
S = R ln
A 1 B
where B =
kRT 2  hp 
B=
(1.381 1023 J K1 ) (8.3145 J K 1 mol1 ) (298 K)2 (6.626 1034 J s) (1.00 105 Pa) 8.72 1012 L mol1 s1 (1000 L m3 ) (1.54 1011 m3 mol1 s1 )
= 1.54 1011 m3 mol1 s1 so
S = (8.3145 J K1 mol1 ) ln = 32.2 J K1 mol1
1
Comment. In this connection, the enthalpy of activation is often referred to as `energy' of activation. E27.14(b) The Gibbs energy of activation is related to the rate constant by k2 = B exp  G RT where B = kRT 2  hp  so
G = RT ln
k2 B
k2 = (6.45 1013 L mol1 s1 )e{(5375 K)/(298 K)} = 9.47 105 L mol1 s1 = 947 m3 mol1 s1 Using the value of B computed in Exercise 27.13(b), we obtain
G = (8.3145 103 kJ K1 mol1 ) (298 K) ln = 46.8 kJ mol1
947 m3 mol1 s1 1.54 1011 m3 mol1 s1
E27.15(b) The entropy of activation for a bimolecular reaction in the gas phase is
S = R ln
A 2 B
where B =
kRT 2  hp 
B=
(1.381 1023 J K1 ) (8.3145 J K 1 mol1 ) [(55 + 273) K]2 (6.626 1034 J s) (1.00 105 Pa)
= 1.86 1011 m3 mol1 s1 The rate constant is k2 = A exp Ea RT so A = k2 exp Ea RT 49.6 103 J mol1 (8.3145 J K1 mol1 ) (328 K)
A = (0.23 m3 mol1 s1 ) exp = 1.8 107 m3 mol1 s1
MOLECULAR REACTION DYNAMICS
443
and
S = (8.3145 J K1 mol1 ) ln = 93 J K1 mol1
1.8 107 m3 mol1 s1 1.86 1011 m3 mol1 s1
2
E27.16(b) The entropy of activation for a bimolecular reaction in the gas phase is
S = R ln
A 2 B
where B =
kRT 2  hp 
For the collision of structureless particles, the rate constant is k2 = NA 8kT 1/2  E0 exp RT 8kT 1/2 = 4NA RT 1/2 M
so the prefactor is A = NA
1 where we have used the fact that = 2 m for identical particles and k/m = R/M. So
A = 4 (6.022 10
23
mol
1
)
(8.3145 J K 1 mol1 ) (500 K) (78 103 kg mol1 )
1/2
(0.68 1018 m2 )
= 2.13 108 m3 mol1 s1 B = (1.381 1023 J K1 ) (8.3145 J K 1 mol1 ) (500 K)2 (6.626 1034 J s) (1.00 105 Pa) 2.13 108 m3 mol1 s1 4.33 1011 m3 mol1 s1
= 4.33 1011 m3 mol1 s1 and
S = (8.3145 J K1 mol1 ) ln = 80.0 J K1 mol1
2
E27.17(b) (a) The entropy of activation for a unimolecular gasphase reaction is
S = R ln
A 1 B
where B = 1.54 1011 m3 mol1 s1 [See Exercise 27.17(a)] 2.3 1013 L mol1 s1 (1000 L m3 ) (1.54 1011 m3 mol1 s1 ) 1
so S = (8.3145 J K1 mol1 ) ln = 24.1 J K1 mol1 (b) The enthalpy of activation is
H = Ea  RT = 30.0 103 J mol1  (8.3145 J K 1 mol1 ) (298 K) = 27.5 103 J mol1 = 27.5 kJ mol1
(c) The Gibbs energy of activation is
G =
H  T S = 27.5 kJ mol1  (298 K) (24.1 103 kJ K1 mol1 )
= 34.7 kJ mol1
444
INSTRUCTOR'S MANUAL
E27.18(b) The dependence of a rate constant on ionic strength is given by
log k2 = log k2 + 2AzA zB I 1/2 At infinite dilution, I = 0 and k2 = k2 , so we must find log k2 = log k2  2AzA zB I 1/2 = log(1.55)  2 (0.509) (+1) (+1) (0.0241)1/2
= 0.323
and
k2 = 1.08 L2 mol2 min1
Solutions to problems
Solutions to numerical problems
P27.1 A = NA
8kT
1 [Section 27.1 and Exercise 27.16(a); = 2 m(CH3 )] 23 1
= ( ) (6.022 10
mol
)
(8) (1.381 1023 J K1 ) (298 K) ( ) (1/2) (15.03 u) (1.6605 1027 kg/u) 2.4 107 mol1 m3 s1 5.52 1026 mol1 m s1
1/2
= (5.52 1026 ) ( mol1 m s1 ) (a) = 2.4 1010 mol1 dm3 s1 5.52 1026 mol1 m s1 = = 4.4 1020 m2
(b) Take d 2 and estimate d as 2 bond length; therefore = ( ) (154 2 1012 m)2 = 3.0 1019 m2 Hence P = P27.3 4.35 1020 = 0.15 = 3.0 1019
For radical recombination it has been found experimentally that Ea 0. The maximum rate of recombination is obtained when P = 1 (or more), and then k2 = A = NA 8kT 1/2 = 4 NA kT 1/2 m
1 = 2m
d 2 = (308 1012 m)2 = 3.0 1019 m2 Hence k2 = (4) (3.0 1019 m2 ) (6.022 1023 mol1 ) (1.381 1023 J K1 ) (298 K) ( ) (15.03 u) (1.6605 1027 kg/u)
1/2
= 1.7 108 m3 mol1 s1 = 1.7 1011 M1 s1 The rate constant is for the rate law v = k2 [CH3 ]2 Therefore d[CH3 ] = 2k2 [CH3 ]2 dt
MOLECULAR REACTION DYNAMICS
445
and its solution is
1 1  = 2k2 t [CH3 ] [CH3 ]0 For 90 per cent recombination, [CH3 ] = 0.10 [CH3 ]0 , which occurs when 2k2 t = 9 [CH3 ]0 or t= 9 2k2 [CH3 ]0
The mole fractions of CH3 radicals in which 10 mol% of ethane is dissociated is (2) (0.10) = 0.18 1 + 0.10 The initial partial pressure of CH3 radicals is thus p0 = 0.18 p = 1.8 104 Pa 1.8 104 Pa RT 9RT (9) (8.314 J K 1 mol1 ) (298 K) Therefore t = = (2k2 ) (1.8 104 Pa) (1.7 108 m3 mol1 s1 ) (3.6 104 Pa) and [CH3 ]0 = = 3.6 ns P27.6 Figure 27.1 shows that log k is proportional to the ionic strength for neutral molecules.
0.19 ) 0.17 k/( 0.15 0 0.05 0.10 0.15
Figure 27.1 From the graph, the intercept at I = 0 is 0.182, so k = 0.658 L mol1 min1 Comment. In comparison to the effect of ionic strength on reactions in which two or more reactants are ions, the effect when only one is an ion is slight, in rough qualitative agreement with eqn 27.69. P27.7 e2 40 d(I  Eea )
2
[Example 27.2]
Taking = d 2 gives
e2 40 [I (M)  Eea (X2 )]
2
=
6.5 nm2 (I  Eea )/eV
446
INSTRUCTOR'S MANUAL
Thus, is predicted to increase as I  Eea decreases. The data let us construct the following table
/nm2 Na K Rb Cs Cl2 0.45 0.72 0.77 0.97 Br 2 0.42 0.68 0.72 0.90 I2 0.56 0.97 1.05 1.34
All values of in the table are smaller than the experimental ones, but they do show the correct trends down the columns. The variation with Eea across the table is not so good, possibly because the electron affinities used here are poor estimates. Question. Can you find better values of electron affinities and do they improve the horizontal trends in the table? P27.10 A + A A2
v = 1 A
kT h RT p
[27.63]
m 4.07 105 M1 s1 103 L
3
S = R ln
+ 2
= (8.3145 J K 1 mol1 ) ln
(1.3811023 J K 1 )(300 K)2 (8.3145 J K1 mol1 ) (6.6261034 J s)(1.013105 Pa)
+ 2
= (8.3145 J K1 mol1 ) [ln(2.631 109 ) + 2]
S = 148 J K1 mol1
H = Ea  2RT = 65.43 kJ mol1  2 (8.3145 J K 1 mol1 ) (300 K) 103 kJ [27.60, 27.61] J H = 60.44 kJ mol1 H = U =
U+ H
(pV ) (pV ) =
H
vRT 103 kJ J
= (60.44 kJ mol1 )  (1) (8.3145 J K 1 mol1 ) (300 K)
U = 62.9 kJ mol1
G=
H  T S = 60.44 kJ mol1  (300 K) (148 J K 1 mol1 )
103 kJ J
G = 104.8 kJ mol1 [27.59]
MOLECULAR REACTION DYNAMICS
447
P27.12
(a) The multilinear hypothesis is Ea = c1 Gb + c2 I + c3 where the constants c1 , c2 , and c3 are independent of temperature. The substitutions Ea = RT ln(k/A) and Gb = RT ln(Kb ) (eqns 25.25 and 9.19) give RT ln k A = c1 RT ln(Kb ) + c2 I + c3
c3 may be eliminated by subtracting the analogous equation for the methylbenzene reference. Assuming that the preexponential A values for the reference and members of the series are comparable, the working equation becomes RT ln k ktoluene
pk
= c1 RT ln k ktoluene ,
Kb Kb,toluene
p Kb
+ c2 (I  Itoluene ) Kb Kb,toluene , I = I  Itoluene gives
substituting
pk
=  log
=  log
= c1 p kb +
c2 I RT ln(10)
pk
(1) depends upon
p Kb .
The temperature dependence of RT ln(10) p Kb = RT ln(10) p Kb = Gb =
Hb  T S b Sb Hb (assuming that Sb = C Hb ) (2)
Hb  T
= (cT  1)
Evaluating the above equation at T = T0 = 273.15 K gives
o Hb = o RT0 ln(10) p Kb cT0  1
(3)
o o where Hb (T0 ) = Hb and p Kb (T0 ) = p Kb . Assuming that Hb is approxio Hb . Substitute equation (3) into (2) mately independent of temperature gives Hb = and substitute the result into equation (1) to get pk
=
o a T0 (cT  1) p Kb b I + T (cT0  1) RT ln(10)
where the symbols c1 and c2 have been replaced with the symbols a and b. (b) The activation parameters for the ring destruction of pxylene are determined with a linear regression analysis of the experimental data plotted as ln(k) versus 1/T (eqn 25.25 and Example 25.5). The regression first gives: slope = 8.875 103 K intercept = 25.53
448
INSTRUCTOR'S MANUAL
Ea = R slope = 73.8 kJ mol1 A = eintercept L mol1 s1 = 1.223 1011 L mol1 s1 H and S values for solution reactions may be calculated with the following equation (k is the Boltzmann constant in these equations).
H = Ea  RT
and
S = R ln
hA KekT
where the transmission coefficient K is assumed to equal 1. These equations may be deduced by modification of Sections 27.4 and 27.5 concepts to the solution phase. Eqn 27.42 becomes [C ] = K [A][B]; eqn 27.44 becomes k2 = k K = kK (eqn 27.45). Eqns 27.5 and 27.53 become k2 = K(kT K / h) = KkT K / h.
Eqn 27.58 becomes k2 = (KkT / h) e G/RT and eqn 27.60 becomes
k2 = (K kT / h) e
S/R
e H /RT . According to the last equation,
ln k2 1 1 = + T T R p
p p S H H 1 1 1 (eqn 4.16) T = +  + R T RT T T RT 2 p p
S T
+
1  2 RT RT
H
H T
=
H 1 + T RT 2
Substitution into the formal definition of activation energy (eqn 25.26) Ea = RT 2 ( ln k2 /T )p , gives Ea = H + RT or H = Ea  RT . Subtitution of this condusion into the k2 equation gives k2 = (K RT / h)e = (Ke kT / h)e
S/R e H /RT = (K kT / h)e S/R
e(Ea RT )/RT
S/R
Substituting of k2 = AeEa /RT (eqn 25.25) and solving for
S gives the final result.
S = R ln
hA Ke kT
T /K 293.15 303.15 313.15 323.15 333.15 343.15
k/102 L mol1 s1 0.86 2.5 5.4 13 47 59
H /k J mol1 71.4 71.3 71.2 71.1 71.0 70.9
S/J K1 mol1 40.8 41.1 41.4 41.6 41.9 42.1
Entropy decreases upon formation of the transition state. (c) The 6 temperatures at which rate constants are measured may be indexed as i = 0, 1, 2, . . . , 5. o The 7 arenes studied may be indexed as j = 0, 1, 2, . . . , 6. p K, p Kb , and I values may be calculated for each arene at each temperature. Methybenzene (toluene is the reference arene). The values for p kexp (T ) are calcluated with the arrhenius parameters. The constants a, b, and c that appear in the equation deduced in part (a) are determined by systematically altering their values so that the sum of the squares of errors (SSE) between p kexp and the fitted equation p kfit is minimized.
MOLECULAR REACTION DYNAMICS
449
6
5 i=0 p kexp (Ti )  p kfit (a, b, c, Ti )
SSE(a, b, c) =
j =0
2 j
Mathematical software like mathcad's given/minerr solve block easily perform the minimization. We find a best fit when a = 0.413 b = 0.192 c = 1.39 103 K 1
The goodness of the fit may be graphically evaluated by plotting the ratio p kfit / p kexp against p Kexp for the 7 arenes at a temperature of choice. A good fit gives a ratio of 1 to within experimental error. The following plot gives the ratio at 293.15 K.
1.05
p kfit p kexp
1
0.95
0.9 5 4 3 2 1 p kexp 0 1 2 3
Figure 27.2
p kfit volumes are within about 3% of the experimental values. This is a good fit, which confirms that the activation energy for arene distinction is multilinear in the basicity constant and the ionization energy. This is also evidence for the proposed arene ring oxidation mechanism.
Solutions to theoretical problems
P27.14 Programs for numerical integration using, for example, Simpson's rule are readily available for personal computers and handheld calculators. Simplify the form of eqn 27.40 by writing z2 = kx 2 , 4D = kt, j= A n0 D 1/2 [J] k
Then evaluate j=
0
1 1/2 z2 /  e e d +
1 1/2 z2 /  e e
for various values of k.
450
INSTRUCTOR'S MANUAL
P27.16
Ka =
[H+ ][A ] 2 [H+ ][A ] 2 [HA]HA [HA] [HA]Ka
2 [A ]
Therefore, [H+ ] =
and log[H+ ] = log Ka + log Write v = k2 [H+ ][B]
[HA] [HA]  2 log = log Ka + log  + 2AI 1/2 [A ] [A ]
then log v = log(k2 [B] + log[H+ ] [HA] = log(k2 [B]) + log  + 2AI 1/2 + log Ka [A ] [B][HA]Ka = log v + 2AI 1/2 , v = k2 [A ] That is, the logarithm of the rate should depend linearly on the square root of the ionic strength, log v I 1/2 P27.18 k1 = kT q e E [Problem 27.17] h q kT 2 R q h
V V R q = qz qy qx
qR
(T /K)3/2 1.027 [Table 20.4, A = B = C] 80 (B/cm1 )3/2 kT 3 h
V V V q = qz qy qx
3 Therefore, k1 80 2 e E0 80 5.4 104 s1 [Problem 27.15] = 4 106 s1 Consequently, D (80) (2.7 1015 m2 s1 ) = P27.20 2 1013 m2 s1 if = and
1 9 1013 m2 s1 if = 2 . It follows that, since N s and l are the same for the two experiments,
(CH2 F2 ) ln 0.6 = [Problem 27.17] = 5 (Ar) ln 0.9 CH2 F2 is a polar molecule; Ar is not. CsCl is a polar ion pair and is scattered more strongly by the polar CH2 F2 .
Solutions to applications
P27.22 Collision theory gives for a rate constant with no energy barrier k = P 8kT 1/2 NA so P = k NA 1/2 8kT
MOLECULAR REACTION DYNAMICS
451
P =
k/(L mol1 s1 ) (103 m3 L1 ) (/nm2 ) (109 m)2 (6.022 1023 mol1 ) (/u) (1.66 1027 kg) 8 (1.381 1023 J K1 ) (298 K)
1/2
=
(6.61 1013 )k/(L mol1 s1 ) (/nm2 ) (/u)1/2
The collision crosssection is AB =
2 dAB
+ 1 where dAB = (dA + dB ) = A 1/2B 2 2
1/2
1/2
so
AB =
(A
1/2
+ B )2 4
1/2
The collision crosssection for O2 is listed in the Data Section. We would not be far wrong if we took that of the ethyl radical to equal that of ethene; similarly, we will take that of cyclohexyl to equal that of benzene. For O2 with ethyl (0.401/2 + 0.641/2 )2 nm2 = 0.51 nm2 4 (32.0 u) (29.1 u) m O me = = 15.2 u = mO + m e (32.0 + 29.1) u = so P = (6.61 1013 ) (4.7 109 ) = 1.6 103 (0.51) (15.2)1/2 For O2 with cyclohexyl (0.401/2 + 0.881/2 )2 nm2 = 0.62 nm2 4 mO m C (32.0 u) (77.1 u) = = = 22.6 u mO + m C (32.0 + 77.1) u = so P = (6.61 1013 ) (8.4 109 ) = 1.8 103 (0.62) (22.6)1/2
28
Processes at solid surfaces
Solutions to exercises
Discussion questions
E28.1(b) The motion of one section of a crystal past another (a dislocation) results in steps and terraces. See Figures 28.2 and 28.3 of the text. A special kind of dislocation is the screw dislocation shown in Fig. 28.3. Imagine a cut in the crystal, with the atoms to the left of the cut pushed up through a distance of one unit cell. The surface defect formed by a screw dislocation is a step, possibly with kinks, where growth can occur. The incoming particles lie in ranks on the ramp, and successive ranks reform the step at an angle to its initial position. As deposition continues the step rotates around the screw axis, and is not eliminated. Growth may therefore continue indefinitely. Several layers of deposition may occur, and the edges of the spirals might be cliffs several atoms high (Fig. 28.4). Propagating spiral edges can also give rise to flat terraces (Fig. 28.5). Terraces are formed if growth occurs simultaneously at neighbouring left and righthanded screw dislocations (Fig. 28.6). Successive tables of atoms may form as counterrotating defects collide on successive circuits, and the terraces formed may then fill up by further deposition at their edges to give flat crystal planes. E28.2(b) Consult the appropriate sections of the textbook (listed below) for the advantages and limitations of each technique. AFM: 28.2(h) and Box 28.1; FIM: 28.5(c); LEED: 28.2(g); MBRS: 28.6(c); MBS: 28.2(i); SAM: 28.2(e); SEM: 28.2(h); and STM: 28.2(h). E28.3(b) In the LangmuirHinshelwood mechanism of surface catalysed reactions, the reaction takes place by encounters between molecular fragments and atoms already adsorbed on the surface. We therefore expect the rate law to be secondorder in the extent of surface coverage: A+BP = kA B
Insertion of the appropriate isotherms for A and B then gives the reaction rate in terms of the partial pressures of the reactants. For example, if A and B follow Langmuir isotherms (eqn 28.5), and adsorb without dissociation, then it follows that the rate law is = kKA KB pA pB (1 + KA pA + KB pB )2
The parameters K in the isotherms and the rate constant k are all temperature dependent, so the overall temperature dependence of the rate may be strongly nonArrhenius (in the sense that the reaction rate is unlikely to be proportional to exp(Ea /RT ). In the EleyRideal mechanism (ER mechanism) of a surfacecatalysed reaction, a gasphase molecule collides with another molecule already adsorbed on the surface. The rate of formation of product is expected to be proportional to the partial pressure, pB of the nonadsorbed gas B and the extent of surface coverage, A , of the adsorbed gas A. It follows that the rate law should be A+BP = kpA B
The rate constant, k, might be much larger than for the uncatalysed gasphase reaction because the reaction on the surface has a low activation energy and the adsorption itself is often not activated. If we know the adsorption isotherm for A, we can express the rate law in terms of its partial pressure, pA . For example, if the adsorption of A follows a Langmuir isotherm in the pressure range
PROCESSES AT SOLID SURFACES
453
of interest, then the rate law would be = kKpA pB . 1 + KpA
If A were a diatomic molecule that adsorbed as atoms, we would substitute the isotherm given in eqn 28.8 instead. According to eqn 28.24, when the partial pressure of A is high (in the sense KpA l, there is almost complete surface coverage, and the rate is equal to kpB . Now the ratedetermining step is the collision of B with the adsorbed fragments. When the pressure of A is low (KpA 1), perhaps because of its reaction, the rate is equal to kKpA pB ; and now the extent of surface coverage is important in the determination of the rate. In the Mars van Krevelen mechanism of catalytic oxidation, for example in the partial oxidation of propene to propenal, the first stage is the adsorption of the propene molecule with loss of a hydrogen to form the allyl radical, CH2 =CHCH2 . An O atom in the surface can now transfer to this radical, leading to the formation of acrolein (propenal, CH2=CHCHO) and its desorption from the surface. The H atom also escapes with a surface O atom, and goes on to form H2 O, which leaves the surface. The surface is left with vacancies and metal ions in lower oxidation states. These vacancies are attacked by O2 molecules in the overlying gas, which then chemisorb as O ions, so reforming the 2 catalyst. This sequence of events involves great upheavals of the surface, and some materials break up under the stress. E28.4(b) Zeolites are microporous aluminosilictes, in which the surface effectively extends deep inside the solid. Mn+ cations and H2 O molecules can bind inside the cavities, or pores, of the AlOSi framework (see Fig. 28.31 of the text). Small neutral molecules, such as CO2 , NH3 , and hydrocarbons (including aromatic compounds), can also adsorb to the internal surfaces and this partially accounts for the utility of zeolites as catalysts. Like enzymes, a zeolite catalyst with a specific composition and structure is very selective toward certain reactants and products because only molecules of certain sizes can enter and exit the pores in which catalysis occurs. It is also possible that zeolites derive their selectivity from the ability to bind to stabilize only transition states that fit properly in the pores.
Numerical exercises
E28.5(b) The number collisions of gas molecules per unit surface area is ZW = NA p (2MRT )1/2 (6.022 1023 mol1 ) (10.0 Pa) (2 (28.013 103 kg mol1 ) (8.3145 J K 1 mol1 ) (298 K))1/2 = 2.88 1023 m2 s1 = 2.88 1019 cm2 s1 (ii) ZW = (6.022 1023 mol1 ) (0.150 106 Torr) (1.01 105 Pa/760 Torr) (2 (28.013 103 kg mol1 ) (8.3145 J K 1 mol1 ) (298 K))1/2 = 5.75 1017 m2 s1 = 5.75 1013 cm2 s1
(a) For N2 (i) ZW =
454
INSTRUCTOR'S MANUAL
(b) For methane (i) ZW = (6.022 1023 mol1 ) (10.0 Pa) (2 (16.04 103 kg mol1 ) (8.3145 J K 1 mol1 ) (298 K))1/2
= 3.81 1024 m2 s1 = 3.81 1020 cm2 s1 (ii) ZW = (6.022 1023 mol1 ) (0.150 106 Torr) (1.01 105 Pa/760 Torr) (2 (16.04 103 kg mol1 ) (8.3145 J K 1 mol1 ) (298 K))1/2
= 7.60 1017 m2 s1 = 7.60 1013 cm2 s1 E28.6(b) The number of collisions of gas molecules per unit surface area is ZW = p= NA p (2MRT )1/2 so p = ZW A(2 MRT )1/2 NA A
(5.00 1019 s1 ) (6.022 1023 mol1 ) (1/2 2.0 103 m)2 (2 (28.013 103 kg mol1 ) (8.3145 J mol1 K 1 ) (525 K))1/2
= 7.3 102 Pa E28.7(b) The number of collisions of gas molecules per unit surface area is ZW = NA p (2MRT )1/2
so the rate of collision per Fe atom will be ZW A where A is the area per Fe atom. The exposed surface consists of faces of the bcc unit cell, with one atom per face. So the area per Fe is A = c2 and rate = ZW A = NA pc2 (2 MRT )1/2
where c is the length of the unit cell. So rate = (6.022 1023 mol1 ) (24 Pa) (145 1012 m)2 (2 (4.003 103 kg mol1 ) (8.3145 J K 1 mol1 ) (100 K))1/2
= 6.6 104 s1 E28.8(b) The number of CO molecules adsorbed on the catalyst is N = nNA = (1.00 atm) (4.25 103 L) (6.022 1023 mol1 ) pV NA = RT (0.08206 L atm K1 mol1 ) (273 K) = 1.14 1020 The area of the surface must be the same as that of the molecules spread into a monolayer, namely, the number of molecules times each one's effective area A = N a = (1.14 1020 ) (0.165 1018 m2 ) = 18.8 m2
PROCESSES AT SOLID SURFACES
455
E28.9(b)
If the adsorption follows the Langmuir isotherm, then = Kp 1 + Kp so K= V /Vmon = p(1  ) p(1  V /Vmon )
Setting this expression at one pressure equal to that at another pressure allows solution for Vmon V2 /Vmon p1 (Vmon  V1 ) V1 /Vmon p2 (Vmon  V2 ) = so = p1 (1  V1 /Vmon ) p2 (1  V2 /Vmon ) V1 V2 p1  p 2 (52.4  104) kPa Vmon = = = 9.7 cm3 p1 /V1  p2 /V2 (52.4/1.60  104/2.73) kPa cm3 E28.10(b) The mean lifetime of a chemisorbed molecule is comparable to its half life: t1/2 = 0 exp Ed RT (1014 s) exp 155 103 J mol1 (8.3145 J K1 mol1 ) (500 K) = 200 s
E28.11(b) The desorption rate constant is related to the mean lifetime (halflife) by t = (ln 2)/k d so k d = (ln 2)/t Ed RT
The desorption rate constant is related to its Arrhenius parameters by k d = A exp and E d = E d RT so = ln k d = ln A 
(ln k1  ln k2 )R
1 1 T2  T 1
(ln 1.35  ln 0) (8.3145 J K 1 mol1 ) (600 K)1  (1000 K)1
E d = 3.7 103 J mol1 E28.12(b) The Langmuir isotherm is = (a) (b) Kp 1 + Kp so p= K(1  )
p=
0.20 = 0.32 kPa (0.777 kPa1 ) (1  0.20) 0.75 = 3.9 kPa p= 1 ) (1  0.75) (0.777 kPa Kp 1 + Kp m/mmon = p(1  ) p(1  m/mmon )
E28.13(b) The Langmuir isotherm is =
We are looking for , so we must first find K or mmon K=
Setting this expression at one pressure equal to that at another pressure allows solution for mmon m1 /mmon p2 (mmon  m2 ) m2 /mmon p1 (mmon  m1 ) = = so p1 (1  m1 /mmon ) p2 (1  m2 /mmon ) m1 m2 p1  p 2 (36.0  4.0) kPa mmon = = = 0.84 mg p1 /m1  p2 /m2 (36.0/0.63  4.0/0.21) kPa mg1 So 1 = 0.63/0.84 = 0.75 and 2 = 0.21/0.84 = 0.25
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INSTRUCTOR'S MANUAL
E28.14(b) The mean lifetime of a chemisorbed molecule is comparable to its halflife t1/2 = 0 exp Ed RT 20 103 J mol1 (8.3145 J K1 mol1 ) (400 K) 20 103 J mol1 (8.3145 J K1 mol1 ) (800 K)
(a) At 400 K : t1/2 = (0.12 1012 s) exp = 4.9 1011 s At 800 K : t1/2 = (0.12 1012 s) exp = 2.4 1012 s (b) At 400 K : t1/2 = (0.12 1012 s) exp = 1.6 1013 s At 800 K : t1/2 = (0.12 1012 s) exp = 1.4 s E28.15(b) The Langmuir isotherm is = Kp 1 + Kp so p= K(1  )
200 103 J mol1 (8.3145 J K1 mol1 ) (400 K) 200 103 J mol1 (8.3145 J K1 mol1 ) (800 K)
For constant fractional adsorption pK = constant But K exp so p1 K1 = p2 K2 and p2 = p1 so
  ad H  K1 = exp K2 R
K1 K2 1 1  T1 T2
  ad H  RT
p2 = p1 exp
  ad H  R
1 1  T1 T2 12.2 103 J mol1 8.3145 J K1 mol1 1 1  298 K 318 K = 6.50 kPa
= (8.86 kPa) exp
E28.16(b) The Langmuir isotherm would be (a) (b) (c) = = = Kp 1 + Kp (Kp)1/2 1 + (Kp)1/2 (Kp)1/3 1 + (Kp)1/3
PROCESSES AT SOLID SURFACES
457
A plot of versus p at low pressures (where the denominator is approximately 1) would show progressively weaker dependence on p for dissociation into two or three fragments. E28.17(b) The Langmuir isotherm is = Kp 1 + Kp so p= K(1  )
For constant fractional adsorption pK = constant But K exp and
ad H  
so p1 K1 = p2 K2
and
p2 K1 = p1 K2 1 1  T1 T2
  ad H  RT
so
  ad H  p2 = exp p1 R
=R
1 1 1 p1  ln , T1 T2 p2
1 1 1 350 kPa  ln 180 K 240 K 1.02 103 kPa
ad H
 
= (8.3145 J K1 mol1 )
= 6.40 104 J mol1 = 6.40 kJ mol1 E28.18(b) The time required for a given quantity of gas to desorb is related to the activation energy for desorption by t exp Ed RT so Ed t1 = exp t2 R 1 1  T1 T2
and E d = R
1 1 1 t1  ln T1 T2 t2
1 1856 s 1 1 ln  873 K 1012 K 8.44 s
E d = (8.3145 J K1 mol1 ) = 2.85 105 J mol1
(a) The same desorption at 298 K would take t = (1856 s) exp 2.85 105 J mol1 8.3145 J K1 mol1 1 1  298 K 873 K = 1.48 1036 s
(b) The same desorption at 1500 K would take t = (8.44 s) exp 2.85 105 J mol1 8.3145 J K1 mol1 1 1  1500 K 1012 K
= 1.38 104 s
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INSTRUCTOR'S MANUAL
Solutions to problems
Solutions to numerical problems
P28.2 Refer to Fig. 28.1.
Figure 28.1 The (100) and (110) faces each expose two atoms, and the (111) face exposes four. The areas of the faces of cell are (a) (352 pm)2 = 1.24 1015 cm2 , (b) 2 (352 pm)2 = 1.75 1015 cm2 , each and (c) 3 (352 pm)2 = 2.15 1015 cm2 . The numbers of atoms exposed per square centimetre are therefore (a) (b) (c) 2 = 1.61 1015 cm2 1.24 1015 cm2 2 = 1.14 1015 cm2 1.75 1015 cm2 4 = 1.86 1015 cm2 2.15 1015 cm2
For the collision frequencies calculated in Exercise 28.5(a), the frequency of collision per atom is calculated by dividing the values given there by the number densities just calculated. We can therefore draw up the following table
Z/(atom1 s1 ) (100) (110) (111) Hydrogen 100 Pa 107 Torr 6.8 105 9.6 105 5.9 105 8.7 102 1.2 101 7.5 102 Propane 100 Pa 107 Torr 1.4 105 2.0 105 1.2 105 1.9 102 2.7 102 1.7 102
P28.4
cz Vmon (1  z){1  (1  c)z} This rearranges to = z 1 (c  1)z = + (1  z)V cVmon cVmon
V
28.10, BET isotherm, z =
p p
Therefore a plot of the lefthand side against z should result in a straight line if the data obeys the
PROCESSES AT SOLID SURFACES
459
BET isotherm. We draw up the following tables (a) 0 C, p = 3222 Torr
p/Torr 103 z 103 z (1  z)(V /cm3 ) 105 32.6 3.04 282 87.5 7.10 492 152.7 12.1 594 184.4 14.1 620 192.4 15.4 755 234.3 17.7 798 247.7 20.0
(b)
18 C, p = 6148 Torr
p/Torr 103 z 103 z (1  z)(V /cm3 ) 39.5 6.4 0.70 62.7 10.2 1.05 108 17.6 1.74 219 35.6 3.27 466 75.8 6.36 555 90.3 7.58 601 97.8 8.09 765 124.4 10.08
The points are plotted in Fig. 28.2, but we analyse the data by a leastsquares procedure. The intercepts are at (a) 0.466 and (b) 0.303. Hence 1 = (a) 0.466 103 cm3 , (b) 0.303 103 cm3 cVmon The slopes of the lines are (a) 76.10 and (b) 79.54. Hence c1 = (a) 76.10 103 cm3 , (b) 79.54 103 cm3 cVmon Solving the equations gives c  1 = (a) 163.3, (b) 262.5 and hence c = (a) 164 , (b) 264
(a) 20 10
Vmon = (a) 13.1 cm3 , (b) 12.5 cm3
(b)
10
5
0 0 0.1 0.2
0 0 0.05
Figure 28.2
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INSTRUCTOR'S MANUAL
P28.7
We assume that the Langmuir isotherm applies. = Kp [28.5] 1 + Kp and 1 = 1 1 + Kp
For a strongly adsorbed species, Kp
1 . Since the reaction rate is proportional to Kp the pressure of ammonia and the fraction of sites left uncovered by the strongly adsorbed hydrogen product, we can write 1 and 1  = kc pNH3 dpNH3 = kc pNH3 (1  )  dt KpH2
To solve the rate law, we write
3 pH2 = 2 {p0NH3  pNH3 } 1 3 NH3 2 N2 + 2 H2
from which it follows that, with p = pNH3 kp dp = , dt p0  p k= 2kc 3K
This equation integrates as follows
p p0
1
p0 p
dp = k
t
dt
0
or
p  p0 p0 p =k+ ln t t p0
We write F =
p0 p  p0 p ln , G = t p0 t and obtain G = k + F = p0 F Hence, a plot of G against F should give a straight line with intercept k at F = 0. Alternatively, the difference G  F should be a constant, k. We draw up the following table
t/s p/Torr G/(Torr s1 ) F /(Torr s1 ) (G  F )/(Torr s1 ) 0 100 30 88 0.40 0.43 0.03 60 84 0.27 0.29 0.02 100 80 0.20 0.22 0.02 160 77 0.14 0.16 0.02 200 74 0.13 0.15 0.02 250 72 0.11 0.13 0.02
Thus, the data fit the rate law, and we find k = 0.02 Torr s1 . P28.9 Taking the log of the isotherm gives ln cads = ln K + (ln csol )/n so a plot of ln cads versus ln csol would have a slope of 1/n and a yintercept of ln K. The transformed data and plot are shown in Fig. 28.3.
PROCESSES AT SOLID SURFACES
461
5
4
3
2 1 2.0 2.5 3.0 3.5 4.0
Figure 28.3
csol /(mg g1 ) cads /(mg g1 ) ln csol ln cads
8.26 4.4 2.11 1.48
15.65 19.2 2.75 2.95
25.43 35.2 3.24 3.56
31.74 52.0 3.46 3.95
40.00 67.2 3.69 4.21
K = e1.9838 mg g1 = 0.138 mg g1
and
n = 1/1.71 = 0.58
In order to express this information in terms of fractional coverage, the amount of adsorbate corresponding to monolayer coverage must be known. This saturation point, however, has no special significance in the Freundlich isotherm (i.e. it does not correspond to any limiting case). P28.11 The Langmuir isotherm is = Kp n = 1 + Kp n so n(1 + Kp) = n Kp and p p 1 = + n n Kn
So a plot of p/n against p should be a straight line with slope 1/n and yintercept 1/Kn . The transformed data and plot (Fig. 28.4) follow
p/kPa n/(mol kg1 ) p/n kPa mol1 kg 31.00 1.00 31.00 38.22 1.17 53.03 1.54 76.38 2.04 37.44 101.97 2.49 40.95 130.47 2.90 44.99 165.06 3.22 51.26 182.41 3.30 55.28 205.75 3.35 61.42 219.91 3.36 65.45
32.67 34.44
70
60
50
40
30 0 50 100 150 200 250 300
Figure 28.4
462
INSTRUCTOR'S MANUAL
n =
1 = 5.78 mol kg1 0.17313 mol1 kg
The yintercept is b= 1 Kn so K= 1 1 = bn (24.641 kPa mol1 kg) (5.78 mol kg1 )
K = 7.02 103 kPa1 = 7.02 Pa1 P28.12 For the Langmuir adsorption isotherm we must alter eqn 5 so that it describes adsorption from solution. This can be done with the transforms p concentration, c V amount adsorbed per gram adsorbent, s Langmuir isotherm and regression analysis 1 c c + = s s Ks 1 = 0.163 g mmol1 , standard deviation = 0.017 g mmol1 s 1 = 35.6 (mmol L1 ) (g mmol1 ), standard deviation = 5.9 (mmol L1 ) (g mmol1 ) Ks R (Langmuir) = 0.973 K= 0.163 g mmol1 = 0.0046 L mmol1 35.6 (mmol L1 ) (g mmol1 )
Freundlich isotherm and regression analysis s = c1 c1/c2 c1 = 0.139, standard deviation = 0.012 1 = 0.539, standard deviation = 0.003 c2 R (Freundlich) = 0.999 94 Temkin isotherm and regression analysis s = c1 ln(c2 c) c1 = 1.08, c2 = 0.074, standard deviation = 0.14 standard deviation = 0.023
R (Temkin) = 0.9590 The correlation coefficients and standard deviations indicate that the Freundlich isotherm provides the best fit of the data.
PROCESSES AT SOLID SURFACES
463
Solutions to theoretical problems
P28.17 = Kp , 1 + Kp = V V
p=
V = K(1  ) K(V  V ) V dp 1 V = = + 2 dV K(V  V ) K(V  V ) K(V  V )2 RT RT RT V d ln p = RT V dp p V K(V  V )2 dV
d =  = =
K(V  V ) V V V dV V  V
Therefore, we can adopt any of several forms, d =  P28.18
RT
V
V  V
dV = 
RT
1
dV = 
RT V
1
d =
RT V d ln(1  )
For the Langmuir and BET isotherm tests we draw up the following table (using p = 200 kPa = 1500 Torr) [Examples 28.1 and 28.3]
p/Torr p (Torr cm3 ) V3 10 z 103 z (1  z)(V /cm3 ) 100 5.59 67 4.01 200 6.06 133 4.66 300 6.38 200 5.32 400 6.58 267 5.98 500 6.64 333 6.64 600 6.57 400 7.30
103 z p is plotted against p in Fig. 28.5(a), and is plotted against z in Fig. 28.5(b). V (1  z)V
7.0
6.0
5.0 100 200 300 400 500 600
Figure 28.5(a)
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INSTRUCTOR'S MANUAL
We see that the BET isotherm is a much better representation of the data than the Langmuir 1 isotherm. The intercept in Fig. 28.5(b) is at 3.33 103 , and so = 3.33 103 cm3 . The cVmon slope of the graph is 9.93, and so c1 = 9.93 103 cm3 cVmon Therefore, c  1 = 2.98, and hence c = 3.98 , Vmon = 75.4 cm3
8 7
6
5
4
3 0 0.1 0.2 0.3 0.4
Figure 28.5(b)
P28.22
(a) Kunit : (gR L
1 1
)
[gR = mass (grams) of rubber]
1 KF unit : (mg) 11/n gR L1/n KL unit : (mg L1 )1 M unit : (mg g1 ) R
(b) Linear sorption isotherm q = Kceq K= q so K is best determined as an average of all q/ceq data pairs. ceq standard deviation = 0.041(gR L1 )1
Kav = 0.126(gR L1 )1 ,
95 per cent confidence limit: (0.083  0.169)(gR L1 )1 If this is done as a linear regression, the result is significantly different. K (linear) = 0.0813(gR L1 )1 , R (linear) = 0.9612 Freundlich sorption isotherm: q = KF ceq , using a power regression analysis, we find that KF = 0.164 , standard deviation = 0.317
1/n
standard deviation = 0.0092(gR L1 )1
PROCESSES AT SOLID SURFACES
465
1 = 0.877, n
standard deviation = 0.113; n = 1.14
R (Freundlich) = 0.9682 Langmuir sorption isotherm q= 1 = q KL Mceq 1 + KL ceq 1 KL M 1 ceq + 1 M
1 = 8.089 gR L1 , standard deviation = 1.031; KL = 0.00053(gR L1 )1 KL M 1 = 0.0043 gR mg1 , standard deviation = 0.1985; M = 233 mg g1 R M R (Langmuir) = 0.9690 All regression fits have nearly the same correlation coefficient so that cannot be used to determine which is the best fit. However, the Langmuir isotherm give a negative value for KL . If KL is to represent an equilibrium constant, which must be positive, the Langmuir description must be rejected. The standard deviation of the slope of the Freundlich isotherm is twice as large as the slope itself. This would seem to be unfavourable. Thus, the linear description seems to be the best , but not excellent choice. However, the Freundlich isotherm is usually preferred for this kind of system, even though that choice is not supported by the data in this case. (c)
1.14 0.164 ceq qrubber 0.46 = = 0.164 ceq 1.6 qcharcoal ceq The sorption efficiency of ground rubber is much less than that of activated charcoal and drops significantly with increasing concentration. The only advantage of the ground rubber is its exceedingly low cost relative to activated charcoal, which might convert to a lower cost per gram of contaminant adsorbed.
29
Dynamics of electron transfer
Solutions to exercises
Discussion questions
E29.1(b) E29.2(b) No solution. The net current density at an electrode is j ; j0 is the exchange current density; is the transfer coefficient; f is the ratio F /RT ; and is the overpotential. (a) j = j0 f is the current density in the low overpotential limit. (b) j = j0 e(1)f applies when the overpotential is large and positive. (c) j = j0 ef applies when the overpotential is large and negative. E29.3(b) In cyclic voltammetry, the current at a working electrode is monitored as the applied potential difference is changed back and forth at a constant rate between preset limits (Figs 29.20 and 29.21). As the  potential difference approaches E  (Ox, Red) for a solution that contains the reduced component  (Red), current begins to flow as Red is oxidized. When the potential difference is swept beyond E  (Ox, Red), the current passes through a maximum and then falls as all the Red near the electrode is consumed and converted to Ox, the oxidized form. When the direction of the sweep is reversed and the  potential difference passes through E  (Ox, Red), current flows in the reverse direction. This current is caused by the reduction of the Ox formed near the electrode on the forward sweep. It passes through the maximum as Ox near the electrode is consumed. The forward and reverse current maxima bracket  E  (Ox, Red), so the species present can be identified. Furthermore, the forward and reverse peak currents are proportional to the concentration of the couple in the solution, and vary with the sweep rate. If the electron transfer at the electrode is rapid, so that the ratio of the concentrations of Ox and Red at the electrode surface have their equilibrium values for the applied potential (that is, their relative concentrations are given by the Nernst equation), the voltammetry is said to be reversible. In this case, the peak separation is independent of the sweep rate and equal to (59 mV)/n at room temperature, where n is the number of electrons transferred. If the rate of electron transfer is low, the voltammetry is said to be irreversible. Now, the peak separation is greater than (59 mV)/n and increases with increasing sweep rate. If homogeneous chemical reactions accompany the oxidation or reduction of the couple at the electrode, the shape of the voltammogram changes, and the observed changes give valuable information about the kinetics of the reactions as well as the identities of the species present. Corrosion is an electrochemical process. We will illustrate it with the example of the rusting of iron, but the same principles apply to other corrosive processes. The electrochemical basis of corrosion in the presence of water and oxygen is revealed by comparing the standard potentials of the metal reduction, such as Fe2+ (aq) + 2e Fe(s)
 E  = 0.44 V
E29.4(b)
with the values for one of the following halfreactions In acidic solution (a) 2 H+ (aq) + 2 e H2 (g)
 E = 0 V  E  = +1.23 V
(b) 4 H+ (aq) + O2 (g) + 4 e 2H2 O(l) In basic solution: (c) 2H2 O(l) + O2 (g) + 4 e 4OH (aq)
 E  = +0.40 V
DYNAMICS OF ELECTRON TRANSFER
467
 Because all three redox couples have standard potentials more positive than E  (Fe2+ /Fe), all three can drive the oxidation of iron to iron(II). The electrode potentials we have quoted are standard values, and they change with the pH of the medium. For the first two  E(a) = E  (a) + (RT /F ) ln a(H+ ) = (0.059 V)pH  E(b) = E  (b) + (RT /F ) ln a(H+ ) = 1.23 V  (0.059 V)pH
These expressions let us judge at what pH the iron will have a tendency to oxidize (see Chapter 10). A thermodynamic discussion of corrosion, however, only indicates whether a tendency to corrode exists. If there is a thermodynamic tendency, we must examine the kinetics of the processes involved to see whether the process occurs at a significant rate. The effect of the exchange current density on the corrosion rate can be seen by considering the specific case of iron in contact with acidified water. Thermodynamically, either the hydrogen or oxygen reduction reaction (a) or (b) is effective. However, the exchange current density of reaction (b) on iron is only about 1014 A cm2 , whereas for (a) it is 106 A cm2 . The latter therefore dominates kinetically, and iron corrodes by hydrogen evolution in acidic solution. For corrosion reactions with similar exchange current densities, eqn 29.62 predicts that the rate of corrosion is high when E is large. That is, rapid corrosion can be expected when the oxidizing and reducing couples have widely differing electrode potentials. Several techniques for inhibiting corrosion are available. First, from eqn 62 we see that the rate of corrosion depends on the surfaces exposed: if either A or A is zero, then the corrosion current is zero. This interpretation points to a trivial, yet often effective, method of slowing corrosion: cover the surface with some impermeable layer, such as paint, which prevents access of damp air. Paint also increases the effective solution resistance between the cathode and anode patches on the surface. Another form of surface coating is provided by galvanizing, the coating of an iron object with zinc. Because the latter's standard potential is 0.76 V, which is more negative than that of the iron couple, the corrosion of zinc is thermodynamically favoured and the iron survives (the zinc survives because it is protected by a hydrated oxide layer). Another method of protection is to change the electric potential of the object by pumping in electrons that can be used to satisfy the demands of the oxygen reduction without involving the oxidation of the metal. In cathodic protection, the object is connected to a metal with a more negative standard potential (such as magnesium, 2.36 V). The magnesium acts as a sacrificial anode, supplying its own electrons to the iron and becoming oxidized to Mg2+ in the process.
Numerical exercises
E29.5(b) Equation 29.14 holds for a donoracceptor pair separated by a constant distance, assuming that the reorganization energy is constant: ln ket = 
   ( r G  )2 rG  + constant, 4RT 2RT
or equivalently ln ket = 
   ( r G  )2 rG  + constant, 4kT 2kT
if energies are expressed as molecular rather than molar quantities. Two sets of rate constants and reaction Gibbs energies can be used to generate two equation (eqn 29.14 applied to the two sets) in
468
INSTRUCTOR'S MANUAL
two unknowns: and the constant. ln ket, l +
      ( r G1 )2 ( r G2 )2 r G1 r G2 + = constant = ln ket,2 + + , 4kT 2kT 4kT 2kT   r G2   r G1 2kT
so
  ( r G1 )2  ( r G2 )2 ket,2 + = ln ket,1 4kT   ( r G1 )2  ( r G2 )2
and =
4 kT ln ket,2 + et,1 =
k
 r G2 
2
 r G1
,
(0.665 eV)2  (0.975 eV)2
4(1.3811023 J K1 )(298 K) 1.6021019 J eV1 3.33106 ln 2.02105  2(0.975  0.665) eV
= 1.531 eV
If we knew the activation Gibbs energy, we could use eqn 29.12 to compute HDA from either rate constant, and we can compute the activation Gibbs energy from eqn 29.4:
G=
 ( r G + )2 [(0.665 + 1.531)eV]2 = = 0.122 eV. 4 4(1.531 eV)
Now
ket
2 HDA 2 = h hket 1/2 2
3 4kT
1/2
exp
 G , kT
G
so
HDA =
4kT 1/4 exp 3
2kT
,
1/2
HDA =
(6.626 1034 J s)(2.02 105 s1 ) 2
4(1.531 eV)(1.602 1019 J eV1 )(1.381 1023 J K1 )(298 K) 3 exp E29.6(b) (0.122 eV)(1.602 1019 J eV1 ) 2(1.381 1023 J K1 )(298 K) = 9.39 1024 J
1/4
Equation 29.13 applies. In E29.6(a), we found the parameter to equal 12 nm1 , so: ln ket /s1 = r + constant so constant = ln ket /s1 + r, and constant = ln 2.02 105 + (12 nm1 )(1.11 nm) = 25. Taking the exponential of eqn 29.13 yields: ket = er+constant s1 = e(12/nm)(1.48 nm)+25 s1 = 1.4 103 s1 .
E29.7(b)
Disregarding signs, the electric field is the gradient of the electrical potential E= d 0.12 C m2 = 2.8 108 V m1 = = = dx d r 0 (48) (8.854 1012 J1 C2 m1 )
DYNAMICS OF ELECTRON TRANSFER
469
E29.8(b)
In the high overpotential limit j = j0 e(1)f so j1 = e(1)f (1 2 ) j2 where f = F 1 = RT 25.69 mV 7255 mA cm2 17.0 mA cm2
The overpotential 2 is 2 = 1 + j2 1 = 105 mV + ln f (1  ) j1 25.69 mV 1  0.42 ln
= 373 mV E29.9(b) In the high overpotential limit j = j0 e(1)f so j0 = j e(1)f j0 = (17.0 mA cm2 ) e{(0.421)(105 mV)/(25.69 mV)} = 1.6 mA cm2 E29.10(b) In the high overpotential limit j = j0 e(1)f so j1 = e(1)f (1 2 ) j2 and j2 = j1 e(1)f (2 1 ) .
So the current density at 0.60 V is j2 = (1.22 mA cm2 ) e{(10.50)(0.60 V0.50 V)/(0.02569 V)} = 8.5 mA cm2 Note. The exercise says the data refer to the same material and at the same temperature as the previous exercise (29.10(a)), yet the results for the current density at the same overpotential differ by a factor of over 5! E29.11(b) (a) The ButlerVolmer equation gives j = j0 e(1)f  ef = (2.5 103 A cm2 ) e{(10.58)(0.30 V)/(0.02569 V)}  e{(0.58)(0.30 V)/(0.02569 V)} = 0.34 A cm2 (b) According to the Tafel equation j = j0 e(1)f = (2.5 103 A cm2 )e{(10.58)(0.30 V)/(0.02569 V)} = 0.34 A cm2 The validity of the Tafel equation improves as the overpotential increases. E29.12(b) The limiting current density is zF Dc but the diffusivity is related to the ionic conductivity (Chapter 24) jlim = RT D= 2 2 z F jlim = so jlim = c zf
(1.5 mol m3 ) (10.60 103 S m2 mol1 ) (0.02569 V) (0.32 103 m) (+1)
= 1.3 A m2
470
INSTRUCTOR'S MANUAL
 E29.13(b) For the iron electrode E  = 0.44 V (Table 10.7) and the Nernst equation for this electrode (Section 10.5) is  E = E 
RT 1 ln F [Fe2+ ]
=2
Since the hydrogen overpotential is 0.60 V evolution of H2 will begin when the potential of the Fe electrode reaches 0.60 V. Thus 0.60 V = 0.44 V + ln[Fe2+ ] = 0.02569 V ln[Fe2+ ] 2
0.16 V = 12.5 0.0128 V
[Fe2+ ] = 4 106 mol L1 Comment. Essentially all Fe2+ has been removed by deposition before evolution of H2 begins. E29.14(b) The zerocurrent potential of the electrode is given by the Nernst equation
 E = E 
1 a(Fe2+ ) 1 a(Fe2+ ) RT  = 0.77 V  ln ln Q = E   ln f F f a(Fe3+ ) a(Fe3+ )
The ButlerVolmer equation gives j = j0 (e(1)f  ef ) = j0 (e(0.42)f  e0.58f ) where is the overpotential, defined as the working potential E minus the zerocurrent potential E. = E  0.77 V + 1 a(Fe2+ ) 1 = E  0.77 V + ln r, ln 3+ ) f f a(Fe
where r is the ratio of activities; so j = j0 (e(0.42)E /f e{(0.42)(0.77 V)/(0.02569 V)} r 0.42  e(0.58)E /f e{(0.58)(0.77 V)/(0.02569 V)} r 0.58 ) Specializing to the condition that the ions have equal activities yields j = (2.5 mA cm2 ) [e(0.42)E /f (3.41 106 )  e(0.58)E /f (3.55 107 )] E29.15(b) Note. The exercise did not supply values for j0 or . Assuming = 0.5, only j/j0 is calculated. From Exercise 29.14(b)
  j = j0 (e(0.50)E /f e(0.50)E /f r 0.50  e(0.50)E /f e(0.50)E /f r 0.50 )
1 1 1  = 2j0 sinh 2 f E  2 f E  + 2 ln r ,
so, if the working potential is set at 0.50 V, then
1 1 j = 2j0 sinh 2 (0.91 V)/(0.02569 V) + 2 ln r 1 j/j0 = 2 sinh 8.48 + 2 ln r 1 At r = 0.1: j/j0 = 2 sinh 8.48 + 2 ln 0.10 = 1.5 103 mA cm2 = 1.5 A cm2
DYNAMICS OF ELECTRON TRANSFER
471
At r = 1: j/j0 = 2 sinh(8.48 + 0.0) = 4.8 103 mA cm2 = 4.8 A cm2
1 At r = 10: j/j0 = 2 sinh 8.48 + 2 ln 10 = 1.5 104 mA cm2 = 15 A cm2
E29.16(b) The potential needed to sustain a given current depends on the activities of the reactants, but the overpotential does not. The ButlerVolmer equation says j = j0 (e(1)f  ef ) This cannot be solved analytically for , but in the highoverpotential limit, it reduces to the Tafel equation j = j0 e(1)f = 0.61 V This is a sufficiently large overpotential to justify use of the Tafel equation. E29.17(b) The number of singly charged particles transported per unit time per unit area at equilibrium is the exchange current density divided by the charge N= j0 e so = 0.02569 V j 15 mA cm2 1 = ln ln (1  )f j0 1  0.75 4.0 102 mA cm2
The frequency f of participation per atom on an electrode is f = Na where a is the effective area of an atom on the electrode surface. For the Cu, H2 H+ electrode N= j0 1.0 106 A cm2 = 6.2 1012 s1 cm2 = e 1.602 1019 C = 4.2 103 s1 For the PtCe4+ , Ce3+ electrode N= j0 4.0 105 A cm2 = = 2.5 1014 s1 cm2 e 1.602 1019 C
f = N a = (6.2 1012 s1 cm2 ) (260 1010 cm)2
The frequency f of participation per atom on an electrode is f = N a = (2.5 1014 s1 cm2 ) (260 1010 cm)2 = 0.17 s1 E29.18(b) The resistance R of an ohmic resistor is R= potential = current jA j fj0
where A is the surface area of the electrode. The overpotential in the low overpotential limit is = so R = 1 fj0 A
472
INSTRUCTOR'S MANUAL
(a) (b)
R= R=
0.02569 V (5.0 1012 A cm2 ) (1.0 cm2 )
= 5.1 109
= 5.1 G
0.02569 V = 10 (2.5 103 A cm2 ) (1.0 cm2 )
E29.19(b) No reduction of cations to metal will occur until the cathode potential is dropped below the zerocurrent potential for the reduction of Ni2+ (0.23 V at unit activity). Deposition of Ni will occur at an appreciable rate after the potential drops significantly below this value; however, the deposition of Fe will begin (albeit slowly) after the potential is brought below 0.44 V. If the goal is to deposit pure Ni, then the Ni will be deposited rather slowly at just above 0.44 V; then the Fe can be deposited rapidly by dropping the potential well below 0.44 V. E29.20(b) As was noted in Exercise 29.10(a), an overpotential of 0.6 V or so is necessary to obtain significant deposition or evolution, so H2 is evolved from acid solution at a potential of about 0.6 V. The reduction potential of Cd2+ is more positive than this (0.40 V), so Cd will deposit (albeit slowly) from Cd2+ before H2 evolution. E29.21(b) Zn can be deposited if the H+ discharge current is less than about 1 mA cm2 . The exchange current, according to the high negative overpotential limit, is j = j0 ef At the standard potential for reduction of Zn2+ (0.76 V) j = (0.79 mA cm2 ) e{(0.5)(0.76 V)/(0.02569 V)} = 2.1 109 mA cm2 much too large to allow deposition . (That is, H2 would begin being evolved, and fast, long before Zn began to deposit.) E29.22(b) Fe can be deposited if the H+ discharge current is less than about 1 mA cm2 . The exchange current, according to the high negative overpotential limit, is j = j0 ef At the standard potential for reduction of Fe2+ (0.44 V) j = (1 106 A cm2 ) e{(0.5)(0.44 V)/(0.02569 V)} = 5.2 103 A cm2 a bit too large to allow deposition . (That is, H2 would begin being evolved at a moderate rate before Fe began to deposit.) E29.23(b) The lead acid battery halfcells are Pb4+ + 2e Pb2+ and Pb2+ + 2e Pb for a total of E
 
1.67 V 0.13 V,
= 1.80 V . Power is
P = I V = (100 103 A) (1.80 V) = 0.180 W if the cell were operating at its zerocurrent potential yet producing 100 mA.
DYNAMICS OF ELECTRON TRANSFER
473
E29.24(b) The thermodynamic limit to the zerocurrent potential under standard conditions is the standard  potential E  , which is related to the standard Gibbs energy by
rG    = F E 
so
E=
  r G F
The reaction is C3 H8 (g) + 7O2 (g) 3CO2 (g) + 4H2 O(l) with = 14
rG     = 3 f G (CO2 ) + 4 f G (H2 O)  fG    (C3 H8 )  7 f G (O2 )
= (3 (394.36) + 4 (237.13)  (23.49)  0) kJ mol1 = 1319.4 kJ mol1 1319.39 103 J mol1 = 0.97675 V 14 (96485 C mol1 ) E29.25(b) Two electrons are lost in the corrosion of each zinc atom, so the number of zinc atoms lost is half the number of electrons which flow per unit time, i.e. half the current divided by the electron charge. The volume taken up by those zinc atoms is their number divided by their number density; their number density is their mass density divided by molar mass times Avogadro's number. Dividing the volume of the corroded zinc over the surface from which they are corroded gives the linear corrosion rate; this affects the calculation by changing the current to the current density. So the rate of corrosion is
 so E  =
rate =
(1.0 A m2 ) (65.39 103 kg mol1 ) jM = 2eNA 2(1.602 1019 C) (7133 kg m3 ) (6.022 1023 mol1 )
= 4.8 1011 m s1 = (4.8 1011 m s1 ) (103 mm m1 ) (3600 24 365 s y1 ) = 1.5 mm y1
Solutions to problems
Solutions to numerical problems
P29.3 RT ln a(M+ ) zF Deposition may occur when the potential falls to below E and so simultaneous deposition will occur if the two potentials are the same; hence the relative activities are given by
 E = E +  E  (Sn, Sn2+ ) +
RT RT  ln a(Sn2+ ) = E  (Pb, Pb2+ ) + ln a(Pb2+ ) 2F 2F
  {E  (Pb, Pb2+ )  E  (Sn, Sn2+ )} =
or ln
a(Sn2+ ) = a(Pb2+ )
2F RT
(2) (0.126 + 0.136) V = 0.78 0.0257 V
That is, we require a(Sn2+ ) 2.2a(Pb2+ ) P29.8
1/2 RT [22.50]  2F 2 I b  1   where I = z2 (bi /b  ), b  = 1 mol kg1 [10.18] 2 i i
rD =
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INSTRUCTOR'S MANUAL
 For NaCl: I b  = bNaCl [NaCl] assuming 100 per cent dissociation.  1 For Na2 SO4 : I b  = 2 (1)2 (2bNa2 SO4 ) + (2)2 bNa2 SO4
= 3bNa2 SO4 3[Na2 SO4 ] assuming 100 per cent dissociation. 1/2 78.54 (8.854 1012 J1 C2 m1 ) (8.315 J K 1 mol1 ) (298.15 K) rD
3 6 3 2 (1.00 g cm3 ) 10 g kg 10mcm (96485 C mol1 )2 3
1/2 1  I b
3.043 1010 m mol1/2 kg1/2  (I b  )1/2 304.3 pm mol1/2 kg1/2  (I b  )1/2
These equations can be used to produce the graph of rD against bsalt shown in Fig. 29.1. Note the contraction of the double layer with increasing ionic strength.
5000
4000
3000
2000
1000
0 0 20 40 60 80 100
Figure 29.1 P29.9 This problem differs somewhat from the simpler oneelectron transfers considered in the text. In place of Ox + e Red we have here In3+ + 3e In namely, a threeelectron transfer. Therefore eqns 29.25, 29.26, and all subsequent equations including the ButlerVolmer equation [29.35] and the Tafel equations [29.3829.41] need to be modified by
DYNAMICS OF ELECTRON TRANSFER
475
including the factor z (in this case 3) in the equation. Thus, in place of eqn 29.26, we have
Gc =
Gc (0) + zF
and in place of eqns 29.39 and 29.41 ln j = ln j0 + z(1  )f ln(j ) = ln j0  zf We draw up the following table
j/(A m2 ) 0 0.590 1.438 3.507 E/V 0.388 0.365 0.350 0.335 /V 0 0.023 0.038 0.053 ln(j/(A m2 )) 0.5276 0.3633 1.255
anode cathode
We now do a linear regression of ln j against with the following results (see Fig. 29.2)
1.5
1.0
0.5
0.0
0.5
1.0 0.020 0.025 0.030 0.035 0.040 0.045 0.050 0.055
Figure 29.2 z(1  )f = 59.42 V1 , ln j0 = 1.894, R = 1 (almost exact) Thus, although there are only three data points, the fit to the Tafel equation is almost exact. Solving for from z(1  )f = 59.42 V1 , we obtain 59.42 V1 59.42 V1 =1 (0.025262 V) =1 3f 3 = 0.4996 = 0.50 standard deviation = 0.0154
standard deviation = 0.0006
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INSTRUCTOR'S MANUAL
which matches the usual value of exactly. j0 = e1.894 = 0.150 A m2 The cathodic current density is obtained from ln(jc ) = ln j0  zf = 3.259 jc = e3.259 = 0.0384 A m2 jc = 0.038 A m2 P29.12 At large positive values of the overpotential the current density is anodic. j = j0 e(1)f  ef j0 e(1)f = ja ln j = ln j0 + (1  )f Performing a linear regression analysis of ln j against , we find ln(j0 /(mA m2 )) = 10.826, (1  )f = 19.550 V R = 0.999 01 j0 = e10.826 mA m2 = 2.00 105 mA m2 =1 19.550 V1 19.550 V1 =1 f (0.025693 V)1
1
= 0.023 V at  E/V = 0.365
= 1.894  (3 0.4996 0.023)/(0.025 262)
[29.35]
[29.34]
standard deviation = 0.287
,
standard deviation = 0.355
= 0.498 The linear regression explains 99.90 per cent of the variation in a ln j against plot and standard deviations are low. There are no deviations from the Tafel equation/plot.
Solutions to theoretical problems
P29.14 (a) First, assume that eqn 4 applies to the bimolecular processes under consideration in this problem. (Cf. P29.1.) Thus,
G11 =
  ( r G11 + 11 )2 , 411
G22 =
  r G22
+ 22 )2 , 422
G12 =
  ( r G12 + 12 )2 412
Because the standard free energy for elctron selfexchange is zero, these simplify to:
G11 = G12 =
2 11 = 11 /4 411
and
G22 = 22 /4.
    ( r G12 )2 + 2 + 212 r G12 12 412
DYNAMICS OF ELECTRON TRANSFER
477
(b) If
  r G12
12 , then we may drop the quadratic term in the numerator, leaving:
  r G12 /2.
G12 12 /4 +
Assume that 12 = (11 + 22 )/2, so 12 /4 = (11 /4 + 22 /4)/2 = ( G11 + Thus, we have:
  r G12 )/2.
G22 )/2.
G12 ( G11 +
G22 +
(c) According to activated complex theory, we can write for the selfexchange reactions: k11 = 11 11 exp  G11 RT and k22 = 22 22 exp  G22 RT .
(d) According to activated complex theory, we can write: k12 = 12 12 exp  G12 RT 12 12 exp  G11 
G 22

  r G12
2RT
.
(e) Finally, we simplify by assuming that all terms are identical, so: k12 exp  G11 RT exp  G22 RT exp
   r G12 RT 1
/2.
The final exponential is the equilibrium constant; the first two exponentials with their factors of are electron selfexchange rate constants, so: k12 (k11 k22 K)1/2 . P29.16 Let oscillate between + and  around a mean value 0 . Then  is large and positive (and + >  ), j j0 e(1)f = j0 e(1/2)f [ = 0.5]
and varies as depicted in Fig. 29.3(a).
Figure 29.3(a) Therefore, j is a chain of increasing and decreasing exponential functions, j = j0 e( + t)f/2 et/ during the increasing phase of , where = j = j0 e(+  t)f/2 et/ 2RT , a constant, and F
478
INSTRUCTOR'S MANUAL
during the decreasing phase. This is depicted in Fig. 29.3(b).
Figure 29.3(b) P29.17 j= cF D
c c (1  ef )[29.51; z = 1] = jL (1  eF /RT )
The form of this expression is illustrated in Fig. 29.4. For the anion current, the sign of c is changed, and the current of anions approaches its limiting value as c becomes more positive (Fig. 29.4).
Cations
0
Anions
Figure 29.4
P29.19
Does eqn 29.13 ln ket = r + constant apply to these data? Draw the follwing table:
r/nm 0.48 0.95 0.96 1.23 1.35 2.24 ket /s1 1.58 1012 3.98 109 1.00 109 1.58 108 3.98 107 6.31 101 ln ket /s1 28.1 22.1 20.7 18.9 17.5 4.14
and plot ln ket vs. r
30
ln (ket / s1)
20
10
0
0
1 r/nm
2
3
Figure 29.5
DYNAMICS OF ELECTRON TRANSFER
479
The data fall on a good straight line, so the equation appears to apply . The least squares linear fit equation is: ln ket /s = 34.7  13.4r/nm so we identify = 13.4 nm1 . P29.20 The theoretical treatment of section 29.1 applies only at relatively high temperatures. At temperatures above 130 K, the reaction in question is observed to follow a temperature dependence consistent with eqn 29.12, namely increasing rate with increasing temperature. Below 130 K, the temperaturedependent terms in eqn 29.12 are replaced by FrankCondon factors; that is, temperaturedependent terms are replaced by temperatureindependent wavefunction overlap integrals. (a) The electrode potentials of halfreactions (a), (b), and (c) are (Section 29.8) (a) E(H2 , H+ ) = 0.059 V pH = (7) (0.059 V) = 0.14 V (b) E(O2 , H+ ) = (1.23 V)  (0.059 V)pH = +0.82 V (c) E(O2 , OH ) = (0.40 V) + (0.059 V)pOH = 0.81 V
 E(M, M+ ) = E  (M, M+ ) +
r 2 (correlation coefficient) = 0.991
P29.21
0.059 V 0.35 V  log 106 = E  (M, M+ )  z+ z+
Corrosion will occur if E(a), E(b), or E(c) > E(M, M+ )  (i) E  (Fe, Fe2+ ) = 0.44 V, z+ = 2 E(Fe, Fe2+ ) = (0.44  0.18) V = 0.62 V < E(a, b, and c) > E(a) (ii) E(Cu, Cu+ ) = (0.52  0.35) V = 0.17 V < E(b and c) > E(a) E(Cu, Cu2+ ) = (0.34  0.18) V = 0.16 V < E(b and c) (iii) E(Pb, Pb2+ ) = (0.13  0.18) V = 0.31 V > E(a) < E(b and c)
(iv) E(Al, Al3+ ) = (1.66  0.12) V = 1.78 V < E(a, b, and c) > E(a) (v) E(Ag, Ag+ ) = (0.80  0.35) V = 0.45 V < E(b and c) (vi) E(Cr, Cr 3+ ) = (0.74  0.12) V = 0.86 V < E(a, b, and c) (vii) E(Co, Co2+ ) = (0.28  0.15) V = 0.43 V < E(a, b, and c) Therefore, the metals with a thermodynamic tendency to corrode in moist conditions at pH = 7 are Fe, Al, Co, Cr if oxygen is absent, but, if oxygen is present, all seven elements have a tendency to corrode. (b) A metal has a thermodynamic tendency to corrosion in moist air if the zerocurrent potential for the reduction of the metal ion is more negative than the reduction potential of the halfreaction 4H+ + O2 + 4e 2H2 O
 E  = 1.23 V
The zerocurrent cell potential is given by the Nernst equation
 E = E 
[Mz+ ]/z RT RT  ln Q = E   ln + F F [H ] p(O2 )/4
We are asked if a tendency to corrode exists at pH 7 ([H+ ] = 107 ) in moist air (p(O2 ) 0.2 bar), and are to answer yes if E 0 for a metal ion concentration of 106 , so for = 4
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INSTRUCTOR'S MANUAL
and 2+ cations
  E = 1.23 V  EM 
(106 )2 0.02569 V   = 0.983 V  EM ln (1 107 )4 (0.2)
In the following, z = 2
 For Ni: E  = 0.983 V  (0.23 V) > 0
corrodes corrodes corrodes corrodes corrodes
For Cd: E For Mg: E For Ti: For Mn: E P29.22
   
= 0.983 V  (0.40 V) > 0 = 0.983 V  (2.36 V) > 0 = 0.983 V  (1.18 V) > 0 [29.62]
 E  = 0.983 V  (1.63 V) > 0   f E/4
Icorr = Aj0 e
with E = 0.62  (0.94) V = 0.32 V [as in Problem 29.21] Icorr (0.25 106 A) (e0.32/40.0257) ) 6 A
...
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