Consider a mechanism with first step as rate

Consider a mechanism with first step as rate-determining step.
Example: 2 NO
2
+ F
2
→ 2 NO
2
F
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
Rate Determining Step
Reaction of alkyl halides with water. Overall reaction for t -butyl bromide::
(CH
3
)
3
CBr + H
2
O → (CH
3
)
3
COH + H
+
+ Br
–
Experimental rate law: rate = k [(CH
3
)
3
CBr]
Accepted mechanism :
(1) (CH
3
)
3
C–Br → (CH
3
)
3
C
+
+ Br
–
(slow)
(2) (CH
3
)
3
C
+
+ H
2
O → (CH
3
)
3
C–OH
2
+
(fast)
(3) (CH
3
)
3
C–OH
2
+
→ H
+
+ (CH
3
)
3
C–OH (fast)
Show that reaction is consistent with experimental rate law.
Why does H
2
O not appear in the rate law?
What are the reaction intermediates?

Intro Chemistry II – 030.102 - Chapter 13 Spring, 2008
4
Reaction Mechanisms and Rate
Mechanisms in which rate-determining step is after one or more fast
steps often revealed by reaction order > 2, non-integral order, or
inverse concentration dependence of a species.
Example: 2 NO + O
2
→ 2 NO
2
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.
Reaction Mechanisms and Rate
Since 2nd step determines overall rate, we can write
However, concentration of N
2
O
2
intermediate cannot be controlled.
since 2nd 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
Substituting, 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
.

Intro Chemistry II – 030.102 - Chapter 13 Spring, 2008
5
Determining Reaction Mechanisms
Another mechanism has been proposed for oxidation of NO:
NO + O
2
k
–3
k
3
!
NO
3
(fast equilibrium)
NO
3
+ NO
k
4
!
2 NO
2
(slow)
Since 2nd step again determines overall rate, we can write
rate = k
4
[NO
3
][NO]
Assume equilibrium in the first step:
[NO
3
]
[NO][O
2
]
=
k
3
k
–3
= K
3
OR [NO
3
] = K
3
[NO][O
2
]
Substituting, we have rate = k
4
K
3
[NO]
2
[O
2
]