Physical Chemistry Third Edition

518 11 The Rates of Chemical Reactions

[C]

[A]

[B]

[B] 0

[A] 0

[C] 0

Concentration

t 50

Time

Figure 11.9 The Behavior of a System in a T-Jump Experiment.

The differential equation for the net rate is

rate

d∆[C]

dt

k 1 [A][B]−k′ 1 [C] (11.7-5)

Using Eq. (11.7-4),

d∆[C]

dt

k 1 ([A]eq−∆[C])([B]eq−∆[C])−k 1 ′([C]eq+∆[C])

k 1 [A]eq[B]eq−k′ 1 [C]eq−k 1 ([A]eq+[B]eq)∆[C]

−k′ 1 ∆[C]+k 1 (∆[C])^2 (11.7-6)

The first two terms on the right-hand side of the final version of Eq. (11.7-6) cancel

because the first term is the forward rate at equilibrium and the second term is the

reverse rate at equilibrium. We assume that|∆[C]|is small, since it is not possible to

change the equilibrium composition very much with a temperature jump. We neglect

the final term on the right-hand side since if|∆[C]|is small (∆[C])^2 will be even

smaller.

d∆[C]

dt

−(k 1 ([A]eq+[B]eq)+k′ 1 )∆[C] (11.7-7)

Equation (11.7-7) is exactly like Eq. (11.2-2) except for the symbols used, so we can

transcribe its solution and write

∆[C]∆[C] 0 e−t/τ (11.7-8)

where

1

τ

k 1 ([A]eq+[B]eq)+k′ 1 (11.7-9)

The quantityτis therelaxation timefor∆[C]. Although the reaction is not first order in

both directions,∆[C] decays exponentially. This is because we linearized Eq. (11.7-6)

by neglecting the (∆[C])^2 term.