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12 Chemical Reaction Mechanisms I: Rate Laws and Mechanisms
One possibility is that the inelastic collision in step 1 excites various vibrations in the
A molecule that can eventually combine to produce a transition state (activated complex) of some kind. We write two differential equations:
d[A ∗ ]
dt
k 1 [A]
2
− k
1 [A][A
∗ ]− k 2 [A
∗ ]
(12.4-14)
d[B]
dt
k 2 [A
∗ ]
(12.4-15)
where we follow the practice of labeling rate constants by the number of the step,
with reverse rate constants labeled with a prime symbol ( ). Imposing the steady-state
approximation by setting d[A*]/dt equal to zero gives
k 1 [A]
2
− k
1 [A][A
∗ ]− k 2 [A
∗ ] 0
We solve this equation for [A*] to obtain
[A
∗ ]
k 1 [A] 2
k 2 + k
1 [A]
(12.4-16)
We substitute this equation into Eq. (12.4-15) to give
rate
d[B]
dt
k 1 k 2 [A] 2
k 2 + k
1 [A]
(12.4-17)
This equation agrees with Eq. (12.4-12), with the parameters k, k , and k identified in
terms of the rate constants for the steps of the mechanism.
Exercise 12.11
Derive the rate law that results if Eq. (12.4-13a) is replaced by Eq. (12.4-13c).
We write Eq. (12.4-17) in the form
rate
d[B]
dt
k 1 k 2 [A]
k 2 + k
1 [A]
[A] k uni [A]
(12.4-18)
where k uni depends on [A] as indicated. If the pressure or concentration of substance A is
large enough that k
1 [A] k 2 , k uni approaches a constant and the rate law is first order,
which corresponds to our general assumption about unimolecular elementary reactions.
If the pressure or concentration of substance A is small enough that k 2 k
1 [A], the
rate law approaches second order. This region of low pressure or concentration is called
the fall-off region.
E X A M P L E 12.11
a. Find the rate law predicted by the mechanism of Eq. (12.4-13) if the second step is
rate-determining. What condition turns the expression of Eq. (12.4-14) into this result?
b. Find the rate law predicted by the mechanism of Eq. (12.4-13) if the forward reaction of
the first step is rate-determining.
12 Chemical Reaction Mechanisms I: Rate Laws and Mechanisms
One possibility is that the inelastic collision in step 1 excites various vibrations in the
A molecule that can eventually combine to produce a transition state (activated complex) of some kind. We write two differential equations:
d[A ∗ ]
dt
k 1 [A]
2
− k
1 [A][A
∗ ]− k 2 [A
∗ ]
(12.4-14)
d[B]
dt
k 2 [A
∗ ]
(12.4-15)
where we follow the practice of labeling rate constants by the number of the step,
with reverse rate constants labeled with a prime symbol ( ). Imposing the steady-state
approximation by setting d[A*]/dt equal to zero gives
k 1 [A]
2
− k
1 [A][A
∗ ]− k 2 [A
∗ ] 0
We solve this equation for [A*] to obtain
[A
∗ ]
k 1 [A] 2
k 2 + k
1 [A]
(12.4-16)
We substitute this equation into Eq. (12.4-15) to give
rate
d[B]
dt
k 1 k 2 [A] 2
k 2 + k
1 [A]
(12.4-17)
This equation agrees with Eq. (12.4-12), with the parameters k, k , and k identified in
terms of the rate constants for the steps of the mechanism.
Exercise 12.11
Derive the rate law that results if Eq. (12.4-13a) is replaced by Eq. (12.4-13c).
We write Eq. (12.4-17) in the form
rate
d[B]
dt
k 1 k 2 [A]
k 2 + k
1 [A]
[A] k uni [A]
(12.4-18)
where k uni depends on [A] as indicated. If the pressure or concentration of substance A is
large enough that k
1 [A] k 2 , k uni approaches a constant and the rate law is first order,
which corresponds to our general assumption about unimolecular elementary reactions.
If the pressure or concentration of substance A is small enough that k 2 k
1 [A], the
rate law approaches second order. This region of low pressure or concentration is called
the fall-off region.
E X A M P L E 12.11
a. Find the rate law predicted by the mechanism of Eq. (12.4-13) if the second step is
rate-determining. What condition turns the expression of Eq. (12.4-14) into this result?
b. Find the rate law predicted by the mechanism of Eq. (12.4-13) if the forward reaction of
the first step is rate-determining.