Physical Chemistry Third Edition

570 13 Chemical Reaction Mechanisms II: Catalysis and Miscellaneous Topics

Exercise 13.4

Verify Eq. (13.1-15).

The Rate of a Heterogeneously Catalyzed Reaction

Consider the mechanism of Eqs. (13.1-2) and (13.1-3):

(1) A+surface site
A(adsorbed) (fast) (13.1-16a)

(2) A(adsorbed)−→ products (slow) (13.1-16b)

We assume that the second step is rate-limiting. Using Eq. (13.1-8) for the assumed

equilibrium of the first step, the rate is given by

ratek 2 θ

k 2 K[A]

1 +K[A]

(13.1-17)

For sufficiently small values of [A] the rate becomes first order in A, but for large

enough values of [A] it is zero order in A. This limit corresponds to the fully covered

catalytic surface so that the rate is determined by the number of surface sites and not

by the concentration of A.

EXAMPLE13.2

Derive an expression for the rate of the reaction of Eq. (13.1-16) assuming the steady-state

approximation instead of the rate-limiting step approximation.

Solution

Assume thatθis approximately in a steady state:

dθ

dt

k 1 [A](1−θ)−k′ 1 θ−k 2 θ≈ 0

θ≈

k 1 [A]

k 1 [A]+k′ 1 +k 2

ratek 2 θ

k 2 k 1 [A]

k 1 [A]+k′ 1 +k 2



k 2 [A]

[A]+k′ 1 /k 1 +k 2 /k 1

Exercise 13.5

Tell what condition would make the result of the previous example become the same as that of

Eq. (13.1-17).

EXAMPLE13.3

Assume that a reactant A 2 dissociates on adsorption. Find the rate law for the forward rate of

the catalyzed reaction

A 2 −→products