Physical Chemistry Third Edition

514 11 The Rates of Chemical Reactions

We have

d[F]

dt

k 1 [A]k 1 [A] 0 e−(k^1 +k^2 )t (11.6-5)

The right-hand side of this equation does not contain [F], so we can multiply bydtand

integrate to obtain

[F]t′−[F] 0 [F]t′

−k 1 [A] 0

k 1 +k 2

(e−(k^1 +k^2 )t

′

−1) (11.6-6)

where we assume that[F] 0 0. A similar treatment for[G]gives the same result except

thatk 1 is replaced byk 2 in the numerator:

[G]t′−[G] 0 [G]t′

−k 2 [A] 0

k 1 +k 2

(e−(k^1 +k^2 )t

′

−1) (11.6-7)

where we also assume that[G] 0  0 .The ratio[F]/[G]has the same value at any time:

[F]

[G]



k 1

k 2

(reverse reactions negligible) (11.6-8)

Exercise 11.22

For the reactions shown in Eq. (11.6-1) assume that[A] 0  0 .500 mol L−^1 , thatk 1  0 .100 s−^1

and thatk 2  0 .0100 s−^1. Construct a graph showing [A], [F], and [G] fortranging from

0to20s.

If the reverse reactions cannot be neglected, the situation can be different. If the

system comes to equilibrium,

[F]eq

[A]eq



k 1

k′ 1

K 1 (11.6-9a)

[G]eq

[A]eq



k 2

k′ 2

K 2 (11.6-9b)

so that

[F]eq

[G]eq



k 1

k 1 ′

k′ 2

k 2



k 1

k 2

k′ 2

k′ 1



K 1

K 2

(at equilibrium) (11.6-10)

Depending on the values of the four rate constants, this ratio might differ significantly

from the ratio in Eq. (11.6-8). If F is a desired product and G is an undesired product,

the ratio of[F]to[G]might be optimized by allowing the system to come to equi-

librium (usingthermodynamic control) or by stopping the reaction before it comes to

equilibrium (usingkinetic control).