Reaction Mechanisms and Rate Slides and Notes - Reaction...

Reaction Mechanisms and Rate
In many mechanisms, one step (called the rate-determining step ) is
significantly slower than all others. Overall reaction can only proceed
as fast as its slowest step (analogy to automobile traffic flow).
Consider a mechanism with first step as rate-determining step.
Example:
Rate law found experimentally to be
rate = k
obs
[NO
2
][F
2
]
Possible mechanism:
NO
2
+ F
2
k
1
→
NO
2
F + F (slow)
NO
2
+ F
k
2
→ NO
2
F (fast)
Overall rate
determined by
first step
2 NO
2
+ F
2
→ 2 NO
2
F

Reaction Mechanisms and Rate
Mechanisms in which rate-determining step is after one or more fast
steps often revealed by reaction order more than 2, non-integral order,
or inverse concentration dependence of a species.
Example:
Rate law found experimentally to be
rate = k
obs
[NO]
2
[O
2
]
Alternative mechanism:
NO + NO
k
–1
k
1

N
2
O
2
(fast equilibrium)
N
2
O
2
+ O
2
k
2
→
2 NO
2
(slow)
Rate law consistent with single termolecular step, but these are rare.
2 NO + O
2
→ 2 NO
2

Reaction Mechanisms and Rate
Since second step determines overall rate, we can write
However, concentration of N
2
O
2
intermediate cannot be controlled.
since second step is slow, we can assume elementary reactions
before rate-determining step are in equilibrium. Hence, for 1st step,
rate = k
2
[N
2
O
2
][O
2
]
[N
2
O
2
]
[NO]
2
=
k
1
k
–1
= K
1
OR [N
2
O
2
] = K
1
[NO]
2
Substiuting, we have rate = k
2
K
1
[NO]
2
[O
2
]
This is consistent with observed reaction order, with k
obs
= k
2
K
1
.

Problem
Consider the following reaction:
The following three-step mechanism is proposed:
What rate law is predicted by this mechanism?
NH
3
+ OCl
–
k
–1
k
1

NH
2
Cl + OH
–
(fast equilibrium)
NH
2
Cl + NH
3
k
2
→ N
2
H
5
+
+ Cl
–
(slow)
N
2
H
5
+
+ OH
–
k
3
→
N
2
H
4
+ H
2
O (fast)
2 NH
3
( aq ) + OCl
–
( aq ) → N
2
H
4
( aq ) + H
2
O( l ) + Cl
–
( aq )

Reaction Mechanisms and Rate
An inverse dependence of overall rate on a concentration is often a
clue of a rapid equilibrium before rate-determining step. Similarly
for fractional orders of reaction.
Example:
Initial rate of this reaction (before HBr buildup) is
rate = k
obs
[H
2
][Br
2
]
1/2