This document discusses analyzing reaction mechanisms with slow first steps or fast equilibrium steps. It provides an example of a mechanism with a slow first step that matches the observed rate law. It also works through an example of a mechanism with a fast equilibrium step, showing how to substitute the equilibrium expression for the intermediate and derive a rate law that matches what is observed experimentally.

1. Chemical Kinetics (Pt. 7)
Analyzing Reaction
Mechanisms
By Shawn P. Shields, Ph.D.
This work is licensed by Shawn P. Shields-Maxwell under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0
International License.

2. Reaction Mechanisms
Recall: A reaction mechanism is a
sequence of elementary reactions
(called “steps”) involved in the
conversion of reactants to products.

3. Recall: Rate-Determining Step
The slowest step in a chemical
reaction determines the kinetics for
the entire reaction.
Steps that are faster than the slow
step are “invisible.”

4. Recall: Reaction Intermediates
An “intermediate” is a transient
species involved in a mechanism step
(or more than one) that is not a
reactant or a product.

5. Analyzing Mechanisms
We are going to analyze two types of
reaction mechanisms:
Reactions where the first step in
the mechanism is the slow step.
Reactions involving one (or more)
fast equilibrium steps

6. General Procedure
Start by writing the rate law for the
slow step.
Steps after the slow step will not be
used in writing the rate law.
Substitute for intermediates using
other steps, if necessary.

7. Example: Analyzing Reaction Mechanisms with
a Slow First Step
Overall reaction:
NO2 + CO NO + CO2
Proposed mechanism:
Step 1: NO2 + NO2 NO3 + NO (slow)
Step 2: NO3 + CO NO2 + CO2
Experimental rate law Rate = k[NO2]2
Could this mechanism be correct?
k1
k2

8. Example Mechanism with a Slow First Step
Overall reaction:
NO2 + CO NO + CO2
Proposed mechanism:
Step 1: NO2 + NO2 NO3 + NO (slow)
Step 2: NO3 + CO NO2 + CO2
First, add the two steps together to make sure
that the overall reaction is produced.
k1
k2

9. Example Mechanism with a Slow First Step
Overall reaction:
NO2 + CO NO + CO2
Proposed mechanism:
Step 1: NO2 + NO2 NO3 + NO (slow)
Step 2: NO3 + CO NO2 + CO2
NO2 + NO2 + NO3 + CO  NO3 + NO + NO2 + CO2
Cancel out intermediates and species on both sides of
the overall reaction.
k1
k2

10. Example Mechanism with a Slow First Step
Overall reaction:
NO2 + CO NO + CO2
Proposed mechanism:
Step 1: NO2 + NO2 NO3 + NO (slow)
Step 2: NO3 + CO NO2 + CO2
NO2 + NO2 + NO3 + CO  NO3 + NO + NO2 + CO2
NO2 + CO NO + CO2
.
k1
k2
First hurdle is crossed…

11. Example (Continued)
Overall reaction:
NO2 + CO NO + CO2
Proposed mechanism:
Step 1: NO2 + NO2 NO3 + NO (slow)
Step 2: NO3 + CO NO2 + CO2
Given the rate law: Rate = k [NO2]2
Write the rate law for the slow step in the mechanism…
.
k1
k2

12. Example (Continued)
Overall reaction:
NO2 + CO NO + CO2
Proposed mechanism:
Step 1: NO2 + NO2 NO3 + NO (slow)
Step 2: NO3 + CO NO2 + CO2
Rate law derived from mechanism: Rate = k1[NO2] [NO2]
Or…. Rate = k1[NO2]2
Does it match the observed rate law?
Yes! We’re done. The mechanism may be correct.
k1
k2

13. Analyzing Mechanisms
Two types of reaction mechanisms (in
this course):
Reactions where the first step in
the mechanism is the slow step.
Reactions involving one (or more)
fast equilibrium steps

14. Analyzing Reaction Mechanisms with
One (or More) Fast Equilibrium Step(s)
What is a fast equilibrium step?
For the reaction
A B
The forward reaction is A B
The backward reaction is A B
(or alternatively B A)
k1
k1
k1
k1
k1

15. Analyzing Reaction Mechanisms with
One (or More) Fast Equilibrium Step(s)
For the reaction
A B
The rate law for the forward reaction is
Rate = k1[A]
The rate law for the backward reaction is
Rate = k1[B]
k1
k1

16. Analyzing Reaction Mechanisms with
One (or More) Fast Equilibrium Step(s)
For the reaction
A B
The forward reaction rate is equal to
the backward rate at equilibrium, so set
them equal to each other.
k1[A] = k1[B]
rate of fwd rxn = rate of back rxn
k1
k1

17. Example: Mechanism with a Fast Equilibrium Step
Overall reaction:
2NO + O2 2NO2
Proposed mechanism:
Step 1: NO + NO N2O2 (fast equil)
Step 2: N2O2 + O2 2 NO2 (slow)
Observed rate law: Rate = kobs [NO]2[O2]
k2
k1
k1
Question:
Could this
mechanism
be correct?

18. Example: Mechanism with a Fast Equilibrium Step
Overall reaction:
2NO + O2 2NO2
Proposed mechanism:
Step 1: NO + NO N2O2 (fast equil)
Step 2: N2O2 + O2 2 NO2 (slow)
The two steps add to the overall reaction.
k2
k1
k1

19. Example (Continued)
Proposed mechanism:
Step 1: NO + NO N2O2 (fast equil)
Step 2: N2O2 + O2 2 NO2 (slow)
Write the rate law for the slow step:
Rate = k2 [N2O2][O2]
k2
k1
k1
There is an intermediate in
the rate law…not allowed!

20. Example (Continued)
Step 1: NO + NO N2O2 (fast equil)
Step 2: N2O2 + O2 2 NO2 (slow)
Use the fast equilibrium step to substitute for the
intermediate.
Write the forward rate equal to the backward rate
k1[NO][NO] = k1[N2O2]
rate of fwd rxn = rate of back rxn
k2
k1
k1

21. Example (Continued)
Step 1: NO + NO N2O2 (fast equil)
Step 2: N2O2 + O2 2 NO2 (slow)
Solve for the intermediate [N2O2] by dividing
both sides by k1
𝐍 𝟐 𝐎 𝟐 =
𝐤 𝟏 𝐍𝐎 𝟐
𝐤−𝟏
k2
k1
k1
Plug this quantity into the rate law for [N2O2]

22. Example (Continued)
Substitute
𝐤 𝟏 𝐍𝐎 𝟐
𝐤−𝟏
into the rate law for
[N2O2]
Rate law for slow step: Rate = k2 [N2O2][O2]
Substitute… Rate =
k2k1
k−1
[NO]2[O2]
Observed rate law: Rate = kobs [NO]2[O2]
Does the rate law match? Yes!

23. Example Problems
will be posted separately.
Next up,
Collision Theory and Activation
Energy (Pt 8)