Physical Chemistry Third Edition

546 12 Chemical Reaction Mechanisms I: Rate Laws and Mechanisms

One possibility is that the inelastic collision in step 1 excites various vibrations in the

A molecule that can eventually combine to produce a transition state (activated com-

plex) of some kind. We write two differential equations:

d[A∗]

dt

k 1 [A]^2 −k′ 1 [A][A∗]−k 2 [A∗] (12.4-14)

d[B]

dt

k 2 [A∗] (12.4-15)

where we follow the practice of labeling rate constants by the number of the step,

with reverse rate constants labeled with a prime symbol (′). Imposing the steady-state

approximation by settingd[A*]/dtequal to zero gives

k 1 [A]^2 −k′ 1 [A][A∗]−k 2 [A∗] 0

We solve this equation for [A*] to obtain

[A∗]

k 1 [A]^2

k 2 +k′ 1 [A]

(12.4-16)

We substitute this equation into Eq. (12.4-15) to give

rate

d[B]

dt



k 1 k 2 [A]^2

k 2 +k 1 ′[A]

(12.4-17)

This equation agrees with Eq. (12.4-12), with the parametersk,k′, andk′′identified in

terms of the rate constants for the steps of the mechanism.

Exercise 12.11

Derive the rate law that results if Eq. (12.4-13a) is replaced by Eq. (12.4-13c).

We write Eq. (12.4-17) in the form

rate

d[B]

dt



k 1 k 2 [A]

k 2 +k′ 1 [A]

[A]kuni[A] (12.4-18)

wherekunidepends on [A] as indicated. If the pressure or concentration of substance A is

large enough thatk′ 1 [A]k 2 ,kuniapproaches a constant and the rate law is first order,

which corresponds to our general assumption about unimolecular elementary reactions.

If the pressure or concentration of substance A is small enough thatk 2 k 1 ′[A], the

rate law approaches second order. This region of low pressure or concentration is called

thefall-off region.

EXAMPLE12.11

a.Find the rate law predicted by the mechanism of Eq. (12.4-13) if the second step is

rate-determining. What condition turns the expression of Eq. (12.4-14) into this result?

b.Find the rate law predicted by the mechanism of Eq. (12.4-13) if the forward reaction of

the first step is rate-determining.