Physical Chemistry Third Edition

572 13 Chemical Reaction Mechanisms II: Catalysis and Miscellaneous Topics

of sites with adsorbed A molecules andθBis the fraction of sites with adsorbed B

molecules:

rate of adsorption of Ak 1 [A](1−θA−θB) (13.1-21)

rate of adsorption of Bk 2 [B](1−θA−θB) (13.1-22)

The rates of desorption are

rate of desorption of Ak′ 1 θA (13.1-23)

rate of desorption of Bk′ 2 θB (13.1-24)

When the rate of adsorption is equated to the rate of desorption for each substance

and the resulting equations are solved simultaneously forθAandθB, we obtain the

equilibrium relations:

θA

K 1 [A]

1 +K 1 [A]+K 2 [B]

(13.1-25)

θB

K 2 [B]

1 +K 1 [A]+K 2 [B]

(13.1-26)

whereK 1 k 1 /k′ 1 andK 2 k 2 /k′ 2.

Exercise 13.7

Verify Eqs. (13.1-25) and (13.1-26).

If step 3 in the Langmuir–Hinshelwood mechanism is rate-limiting both absorption

processes will be assumed to be at equilibrium. The rate law is

ratek 3 θAθB

k 3 K 1 K 2 [A][B]

( 1 +K 1 [A]+K 2 [B])^2

(Langmuir–

Hinshelwood

mechanism)

(13.1-27)

0 [A]

Rate

Figure 13.5 Schematic Plot of the

Rate of a Catalyzed Reaction A+B

→Products as a Function of [A] with

Fixed [B].

Figure 13.5 shows a schematic plot of the rate as a function of [A] for a fixed value

of [B]. For small values of [A] the rate is roughly proportional to [A], but as [A] is

increased the rate passes through a maximum and then drops, becoming proportional

to 1/[A] for large values of [A]. This decline in the rate corresponds to a value of

K 1 [A] that is larger than the other two terms in the denominator, so that the denomi-

nator becomes proportional to [A]^2. The physical reason for the decline is that as the

A molecules compete more and more successfully for the surface sites, there are fewer

B molecules adsorbed on the surface. The reaction then slows down because of the

scarcity of adsorbed B molecules. In the reaction of CO with O 2 on platinum (one of

the reactions carried out in an automobile’s catalytic converter), the rate is inversely

proportional to [CO].^4 This indicates that CO is bonded much more strongly on the

catalyst surface than is O 2 , so thatK 1 [CO] is much larger than the other terms in the

denominator of the rate expression.

(^4) K. J. Laidler,op. cit., p. 249 (note 1).