Physical Chemistry Third Edition

13.1 Catalysis 573

Homogeneous Catalysis

Catalytic reactions in this class occur in a single gas or liquid phase.

Gas-Phase Catalysis

An example of a gas-phase homogeneously catalyzed reaction is the decomposition of

ozone

2O 3 −→3O 2 (13.1-28)

This reaction is catalyzed by N 2 O 5. The proposed catalyzed mechanism is^5

(1) N 2 O 5 −→NO 3 +NO 2 (13.1-29a)

(2) NO 2 +O 3 −→NO 3 +O 2 (13.1-29b)

(3) 2NO 3 −→ 2NO 2 +O 2 (13.1-29c)

(1′)NO 3 +NO 2 −→ N 2 O 5 (13.1-29d)

The fourth step is the reverse of step 1, so we label it as 1′. We have written it

separately to emphasize that the catalyst N 2 O 5 is regenerated. Step 2 must be doubled

for the equations of the mechanism to add up to the stoichiometric equation.

EXAMPLE13.4

Find the rate law for the forward reaction of the O 3 decomposition according to the above

mechanism, using the steady-state approximation.

Solution

Since there are three independent steps in the mechanism, we write three differential equa-

tions: one for the rate of change of [O 3 ], which gives the rate of the reaction, and two for the

rates of change of [NO 3 ] and [NO 2 ]. The rates of change of the reactive intermediates [NO 3 ]

and [NO 2 ] are set equal to zero in the steady-state approximation:

rate−

1

2

d[O 3 ]

dt



k 2

2

[NO 2 ][O 3 ] (13.1-30a)

d[NO 2 ]

dt

k 1 [N 2 O 5 ]−k 1 ′[NO 3 ][NO 2 ]+ 2 k 3 [NO 3 ]^2 −k 2 [NO 2 ][O 3 ]≈ 0 (13.1-30b)

d[NO 3 ]

dt

k 1 [N 2 O 5 ]−k 1 ′[NO 3 ][NO 2 ]− 2 k 3 [NO 3 ]^2 +k 2 [NO 2 ][O 3 ]≈ 0 (13.1-30c)

Subtraction of Eq. (13.1-30c) from Eq. (13.1-30b) gives an equation that is solved to obtain

[NO 2 ]

2 k 3 [NO 3 ]^2

k 2 [O 3 ]

(13.1-31)

(^5) H. S. Johnston,Gas Phase Reaction Rate Theory, Ronald Press, New York, 1966.