45
Chapter 22
Reaction Mechanisms
Let’s begin with an example and introduce/review some language.
For the stoichiometric reaction:
N
2
O
5
+
NO
3 NO
2
Step 1:
N
2
O
5
→
NO
2
+
NO
3
“unimolecular”
Step 2:
NO
2
+
NO
3
→
N
2
O
5
bimolecular reverse process
Step 3:
NO
+
NO
3
→
2 NO
2
also bimolecular
Rate Laws for these elementary processes
Step 1:
rate
1
=
k
1
[N
2
O
5
]
Step 2:
rate
2
=
k
2
[NO
2
] [NO
3
]
Step 3:
rate
3
=
k
3
[NO] [NO
3
]
The idea is that we have been studying what to do in the laboratory, how to determine rate constants
and reaction orders and the like.
Now we need to figure out how to devise mechanisms and also how to connect those mechanisms back
to rate laws.
Thus, figuring out the rate laws for elementary reaction steps is an important first topic.
There are three “tricks” we use to make the connection between the mechanism and the rate law.
1. The rate determining step approximation
2. The equilibrium approximation
3. The steadystate approximation
Consider the following chemical reaction:
A
+
B
→
C
+
D
A possible mechanism
Step 1:
A
+
B
→
X
Slow (rate determining)
Step 2:
X
→
C
+
D
Fast
k
1
k
2
k
3
Reaction
Mechanism
Laboratory
Work: Rate
Laws, Rate
Constants
Is Mechanism
Consistent
k
1
k
2
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The molecularity of step 1 is bimolecular since there are two molecules on the left side of the equation.
The molecularity of step 2 is unimolecular
Step 1:
rate
1
=
k
1
[A] [B]
Step 2:
rate
1
=
k
1
[X]
Remember that the rate law for an elementary reaction step depends ONLY on the reactants, NOT the
products.
The Rate Determining Step Approximation
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 Spring '11
 LoBue
 Physical chemistry, Mole, pH, Reaction, Rate equation, Raterxn

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