Page 218 - Physical Chemistry

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PROBLEMS

Section 6.1 6.11 True or false for ideal-gas reactions? (a) If G° for re-
6.1 Use m m° RT ln (P /P°) to calculate G when the action 1 is less than G° for reaction 2, then K° 300
for reac-
i
i
i
P,300
300
pressure of 3.00 mol of a pure ideal gas is isothermally de- tion 1 must be greater than K° for reaction 2. (b) If G° for
P,300
creased from 2.00 bar to 1.00 bar at 400 K. reaction 1 is greater than G° for reaction 1, then K° 300 for
400 P,300
6.2 True or false? (a) The chemical potential of ideal gas i in reaction 1 must be less than K° P,400 for reaction 1.
an ideal gas mixture at temperature T and partial pressure P i
equals the chemical potential of pure gas i at temperature T and Section 6.3
pressure P . (b) m of a pure ideal gas goes to q as P → 0 and 6.12 For the reaction N O (g) ∆ 2NO (g), measurements of
2
4
2
i
goes to q as P → q. (c) The entropy of a mixture of N 2 the composition of equilibrium mixtures gave K° 0.144 at
P
and O gases (assumed ideal) is equal to the sum of the 25.0°C and K° 0.321 at 35.0°C. Find H°, S°, and G°at
P
2
entropies of the pure gases, each at the same temperature and 25°C for this reaction. State any assumptions made. Do not use
volume as the mixture. Appendix data.
6.13 For PCl (g) ∆ PCl (g) Cl (g), observed equilibrium
2
3
5
Section 6.2 constants (from measurements on equilibrium mixtures at low
6.3 For the gas-phase reaction 2SO O ∆ 2SO , observed pressure) vs. T are
2
2
3
mole fractions for a certain equilibrium mixture at 1000 K
and 1767 torr are x 0.310, x 0.250, and x 0.440. K° P 0.245 1.99 4.96 9.35
SO 2 O 2 SO 3
(a) Find K° and G° at 1000 K, assuming ideal gases. (b) Find T/K 485 534 556 574
P
K at 1000 K. (c) Find K° at 1000 K.
c
P
(a) Using only these data, find H°, G°, and S° at 534 K for
6.4 An experimenter places 15.0 mmol of A and 18.0 mmol this reaction. (b) Repeat for 574 K.
of B in a container. The container is heated to 600 K, and the
gas-phase equilibrium A B ∆ 2C 3D is established. The 6.14 For the ideal-gas reaction PCl (g) ∆ PCl (g) Cl (g),
3
2
5
equilibrium mixture is found to have pressure 1085 torr and to use Appendix data to estimate K° at 400 K; assume that H°is
P
contain 10.0 mmol of C. Find K° and G° at 600 K, assuming independent of T.
P
ideal gases.
6.15 The ideal-gas reaction CH (g) H O(g) ∆ CO(g)
2
4
3
6.5 A 1055-cm container was evacuated, and 0.01031 mol of 3H (g) at 600 K has H° 217.9 kJ/mol, S° 242.5 J/
2
NO and 0.00440 mol of Br were placed in the container; the (mol K), and G° 72.4 kJ/mol. Estimate the temperature at
2
equilibrium 2NO(g) Br (g) ∆ 2NOBr(g) was established at which K° 26 for this reaction. State approximations made.
P
2
323.7 K, and the final pressure was measured as 231.2 torr. 6.16 For the reaction N O (g) ∆ 2NO (g) in the range 298 to
Find K° and G° at 323.7 K, assuming ideal gases. (Hint: 900 K, 2 4 2
P
Calculate n .)
tot
b c>1T>K2
K° P a1T>K2 e
6.6 The reaction N (g) ∆ 2N(g) has K° 3 10 6 at 4000
P
2
13
K. A certain gas mixture at 4000 K has partial pressures P where a 1.09 10 , b 1.304, and c 7307. (a) Find
N 2
720 torr, P 0.12 torr, and P He 320 torr. Is the mixture in expressions for G°, H°, S°, and C° as functions of T for
P
N
reaction equilibrium? If not, will the amount of N(g) increase this reaction. (b) Calculate H° at 300 K and at 600 K.
or decrease as the system proceeds to equilibrium at 4000 K in
a fixed volume? 6.17 Complete the work of part (b) of Example 6.2 in Sec. 6.3
as follows. Show that if C° is assumed independent of T, then
P
6.7 Derive Eq. (6.27) relating K and K°.
x
P
K° P 1T 2 2 ¢H°1T 1 2 1 1
6.8 Evaluate 4 j 1 j ( j 1). ln K° P 1T 1 2 R a T 1 T 2 b
6.9 Use Appendix data to find K° P, 298 for the ideal-gas reaction ¢C° P 1T 1 2
O (g) ∆ 2O(g). a ln T 2 T 1 1b
2
R T 1 T 2
6.10 True or false for ideal-gas reactions? (a) K° is always
P
dimensionless. (b) K is always dimensionless. (c) K is never Use this equation and Appendix data to estimate K° P,600 for
P
P
dimensionless. (d) K° for the reverse reaction is the negative of N O (g) ∆ 2NO (g).
2
4
2
P
K° for the forward reaction. (e) K° for the reverse reaction is 6.18 (a) Replacing T by T and considering T as a fixed tem-
P
P
2
1
the reciprocal of K° for the forward reaction. (f) Doubling the perature, we can write the approximate equation (6.39) in the
P
coefficients doubles K°. (g) Doubling the coefficients squares form ln K°(T) H°/RT C, where the constant C equals
P
P
K°. (h) K° for a particular reaction is a function of temperature ln K°(T ) H°/RT . Derive the following exact equation:
P
P
but is independent of pressure and of the initial composition of P 1 1
the reaction mixture. ln K° P 1T 2 ¢H° T >RT ¢S° T >R