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 intermediates 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 + 2k 2
[NO 2 ]
(12.4-33b)
When these two equations are substituted into Eq. (12.4-31), we obtain
rate
2k 1 k 2
k
1 + 2k 2
[N 2 O 5 ] k app [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).
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 intermediates 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 + 2k 2
[NO 2 ]
(12.4-33b)
When these two equations are substituted into Eq. (12.4-31), we obtain
rate
2k 1 k 2
k
1 + 2k 2
[N 2 O 5 ] k app [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).