Physical Chemistry Third Edition

(C. Jardin) #1

12.4 Reaction Mechanisms and Rate Laws 549


The proposed mechanism is
(1) N 2 O 5 NO 2 +NO 3
(2) NO 2 +NO 3 −→NO+O 2 +NO 2
(3) NO+NO 3 −→2NO 2

(12.4-30)

Find the rate law using the steady-state approximation.
Solution
We write three differential equations and impose the steady-state approximation on the reactive interme-
diates NO 3 and NO:


d[N 2 O 5 ]
dt
k 1 [N 2 O 5 ]−k′ 1 [NO 2 ][NO 3 ] (12.4-31)

d[NO 3 ]
dt k^1 [N^2 O^5 ]−k


1 [NO^2 ][NO^3 ]−k^2 [NO^2 ][NO^3 ]−k^3 [NO][NO^3 ]≈^0 (12.4-32a)

d[NO]
dt
k 2 [NO 2 ][NO 3 ]−k 3 [NO][NO 3 ]≈ 0 (12.4-32b)

The second equalities in Eqs. (12.4-32a) and (12.4-32b) give two simultaneous algebraic equations,
which can be solved for [NO 3 ] and [NO]. The results are

[NO]k^2 [NO^2 ]
k 3
(12.4-33a)

and

[NO 3 ]( k^1 [N^2 O^5 ]
k 1 ′+ 2 k 2
)
[NO 2 ]
(12.4-33b)

When these two equations are substituted into Eq. (12.4-31), we obtain

rate
2 k 1 k 2
k 1 ′+ 2 k 2 [N^2 O^5 ]kapp[N^2 O^5 ] (12.4-34)
so that the reaction is first order. Although the steady-state approximation usually leads to a rate law with
a two-term denominator, this rate law corresponds to a definite order.

Exercise 12.13
a.Verify Eq. (12.4-34).
b.Assume that the second step of the mechanism of Example 12.13 is rate-limiting. Find the
rate law for the reaction.
c.What assumptions will cause the steady-state result to become the same as the result of
part b?

Deducing a Mechanism from a Rate Law


It is not a routine matter to decide what mechanism to propose once a rate law has been
determined. However, you can sometimes use the following procedure:^14

(^14) J. O. Edwards, E. F. Greene, and J. Ross,J. Chem. Educ., 45 , 381 (1968).

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