Page 221 - Physical Chemistry

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General (d) For this reaction, state whether each of these quantities in-
6.53 The synthesis of ammonia from N and H is an exother- creases or decreases as T increases: G°, K°, and G°/T.
2 2 P
mic reaction. Hence the equilibrium yield of ammonia de- (e) Is it possible for G° of a reaction to increase with T while
creases as T increases. Explain why the synthesis of ammonia at the same time K° also increases with T?
P
from its elements (Haber process) is typically run at the high
6.60 Given the data points (x , y ), where i 1, . . . , n, we
temperature of 800 K rather than at a lower temperature. (Haber i i
want to find the slope m and intercept b of the straight line y
developed the use of Cl as a poison gas in World War I. His
2 mx b that gives the best fit to the data. We assume that (1)
wife, also a chemist, tried to dissuade him from this work, but
there is no significant error in the x values; (2) the y measure-
failed. She then committed suicide.) i i
ments each have essentially the same relative precision; (3) the
6.54 For the gas-phase reaction errors in the y values are randomly distributed according to the
i
normal distribution law. With these assumptions, it can be
I 2 cyclopentene ∆ cyclopentadiene 2HI shown that the best values of m and b are found by minimizing
the sum of the squares of the deviations of the experimental y
i
measured K° values in the range 450 to 700 K are fitted by values from the calculated y values. Show that minimization of
P
3
2
log K° 7.55 (4.83 10 )(K/T). Calculate G°, H°, S°, (y mx b) (by setting
/
m and
/
b of the sum equal
P i i i
and C° for this reaction at 500 K. Assume ideal gases. to zero) leads to mD n x y x y and bD x 2
P i i i i i i i i i
2
2
y x x y , where D n x ( x ) . Condition
i
i
i
i
i
i
i
i
i
i i
6.55 A certain ideal-gas dissociation reaction A ∆ 2B has (1) is usually met in physical chemistry because the x ’s are
i
1
G° 1000 4000 J mol , which gives K° 0.6 at 1000 K. If things like reciprocals of temperature or time, and these quan-
P
pure A is put in a vessel at 1000 K and 1 bar and held at con- tities are easily measured accurately. However, condition (2) is
stant T and P, then A will partially dissociate to give some B. often not met because the y values are things like ln K°,
i
P
Someone presents the following chain of reasoning. “The sec- whereas it is the K° values that have been measured and that
P
ond law of thermodynamics tells us that a process in a closed have the same precision. Therefore, don’t put too much faith in
system at constant T and P that corresponds to G 0 is for- least-squares-calculated quantities.
bidden [Eq. (4.16)]. The standard Gibbs free-energy change for
the reaction A ∆ 2B is positive. Therefore, any amount of dis- 6.61 Consider the ideal-gas reaction N 3H ∆ 2NH run
2
2
3
sociation of A to B at constant T and P corresponds to an in- at constant T and P with T 500 K and P 4 bar, and with
crease in G and is forbidden. Hence gas A held at 1000 K and the initial composition n 1 mol, n 3 mol, n 0.
N 2 H 2 NH 3
1 bar will not give any B at all.” Point out the fallacy in this ar- (a) Express the mole fractions in terms of the extent of reaction
gument. j. (b) Use the italicized statement at the end of Sec. 6.1 and Eq.
(6.4) for m to express G and H of the reaction mixture in terms
i
6.56 An ideal-gas reaction mixture is in a constant-temperature
of the m°’s, j, P, T, and the H° ’s. (c) Conventional values of
m,i
i
bath. State whether each of the following will change the value
G° m,i m° (Sec. 5.8) at 500 K are 97.46 kJ/mol for N ,
i
2
of K°. (a) Addition of a reactant. (b) Addition of an inert gas.
P 66.99 kJ/mol for H , and 144.37 kJ/mol for NH . Conven-
(c) Change in pressure for a reaction with n 0. (d) Change 2 3
tional values of H° (Sec. 5.4) at 500 K are 5.91 kJ/mol for N ,
m,i
2
in temperature of the bath.
5.88 kJ/mol for H , and 38.09 kJ/mol for NH . Calculate G
3
2
and H of the reaction mixture for j values of 0, 0.2, 0.3, 0.4,
6.57 Suppose we have a mixture of ideal gases reacting ac-
cording to A B ∆ C 2D. The mixture is held at constant 0.6, 0.8, 1.0. Then use G H TS to calculate TS. Check your
T and at a constant (total) pressure of 1 bar. Let one mole of A results against Fig. 6.8. Part (c) is a lot more fun if done on a
react. (a) Is the observed H per mole of reaction in the mix- computer or programmable calculator.
ture equal to H° for the reaction? (b) Is the observed S per 6.62 Give a specific example of an ideal-gas reaction for
mole of reaction equal to S°? (c) Is the observed G per mole which (a) the equilibrium position is independent of pressure;
of reaction equal to G°? (b) the equilibrium position is independent of temperature.
6.58 Suppose that the standard pressure had been chosen as 6.63 (a) Give a specific example of a gas-phase reaction mix-
1000 torr instead of 1 bar. With this definition, what would be ture for which the mole fraction of one of the reactants in-
the values of K and K° at 25°C for N O (g) ∆ 2NO (g) ? Use
P P 2 4 2 creases when the reaction proceeds a small extent to the right.
Appendix data. If you can’t think of an example, see part (b) of this problem.
(b) For a reaction mixture containing only gases that participate
6.59 For cis-EtHCPCHPr(g) ∆ trans-EtHCPCHPr(g) in the reaction, use x n /n and dn n dj [Eq. (4.97)] to
i
i
i
i
tot
(where Et is C H and Pr is CH CH CH ), H° 300 0.9 show that the infinitesimal change dx in the mole fraction
3
5
2
2
2
i
kcal/mol, S° 300 0.6 cal/(mol K), and C° P,300 0. [K. W. of gas i due to a change dj in the extent of reaction is dx
Egger, J. Am. Chem. Soc., 89, 504 (1967).] Assume that C° n 1 [n x ( n/mol)] dj. i
P
0 for all temperatures above 300 K so that H° and S° remain tot i i
constant as T increases. (a) The equilibrium amount of which 6.64 Rodolfo states that the equation d ln K°/dT H°/RT 2
P
isomer increases as T increases? (b) In the limit of very high T, shows that the sign of H° determines whether K° increases or
P
which isomer is present in the greater amount? (c) Explain any decreases as T increases. Mimi states that the equations G°
apparent contradiction between the answers to (a) and (b). RT ln K° and d G°/dT S° show that the sign of S°
P