1. Chapter 14: Chemical Kinetics
Lecture 5: Reaction Mechanisms

2. Collision Model Key Idea: Molecules must collide to react.
However, only a small fraction of collisions produces a reaction. Why?

3. Collision Model
Collisions must have sufficient energy to produce the reaction (equal to or exceeding the activation energy).
Colliding particles must be correctly oriented to one another in order to produce a reaction.

4. Factors Affecting Rate
Increasing temperature always increases the rate of a reaction.
Particles collide more frequently
Particles collide more energetically
Increasing surface area increases the rate of a reaction
Increasing Concentration USUALLY increases the rate of a reaction
Presence of Catalysts, which lower the activation energy by providing alternate pathways

5. Reaction Mechanism Reactions that occur in a single step.
ELEMENTARY REACTIONS:
Reactions that occur in a single step.
The number of molecules that participate is the molecularity:
Unimolecular: One molecule involved
A  B
Bimolecular: Two molecules involved
A + B  C
Termolecular: Three molecules involved
A + B + C  D

6. Reaction Mechanisms
Unimolecular
Termolecular
Bimolecular

7. Rate Law for Elementary Reactions
The order for the rate law for an elementary reaction is the molecularity.
Unimolecular:
Rate = k[A]1
Bimolecular:
Rate = k[A]2 or Rate = k[A][B]
Termolecular:
Rate = k[A]3 or Rate = k[A][B][C]

8. Reaction Mechanism
The reaction mechanism is the series of elementary steps by which a chemical reaction occurs.
The sum of the elementary steps must give the overall balanced equation

9. Reaction Mechanisms Elementary step 1: H2O2 (aq) +I  H2O (l) +IO
Elementary step 2: H2O2 (aq) +IO  H2O (l) + O2 (g) +I
Overall rxn: H2O2 (aq)  2 H2O (l) + O2 (g)
IO is the intermediate. It forms in the first elementary reaction and then reacts to produce the final products.
I is the catalyst. It is an initial reactant and a final product.

10. Reaction Mechanisms Elementary step 1: H2O2 (aq) +I  H2O (l) +IO
Elementary step 2: H2O2 (aq) +IO  H2O (l) + O2 (g) +I
Overall rxn: H2O2 (aq)  2 H2O (l) + O2 (g)
The rate law for each elementary step is:
Elementary step 1: Rate = k [H2O2] [I]
Elementary step 2: Rate = k [H2O2][IO]
Rate = [concentration of reactants] coefficient

11. Reaction Mechanisms Elementary step 1: NO +NO  N2O2
+ Elementary step 2: N2O2 + O2  2NO2
Overall step: 2 NO + O2  2NO2
The rate law for each elementary step is:
Elementary step 1: Rate = k [NO]2
Elementary step 2: Rate = k [N2O2][O2]
Rate = [concentration of reactants] coefficient

12. Rate-Determining Step
In a multi-step reaction, the slowest step is the rate-determining step.
The slowest step:
Will have the HIGHEST activation energy.
Determines the rate law for the reaction: Rate = [reactants]Coefficients for slow step

13. Reaction Mechanisms
Elementary Rxn
Reactants→Intermediates→Products

14. Reaction Mechanisms Reactants→Intermediates→Products
Second step is rate limiting step: It has the higher activation energy.

15. Rate Law with Slow Initial Step
For the reaction:
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
Step #1 is the slowest step. What is the rate law?
Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g)
Step #2 N2O(g) + H2(g)  N2(g) + H2O(g)
N2O(g) is the intermediate.
The experimental rate law is the rate law for the slowest step:
rate = k[NO]2[H2]

16. Slow Initial Step Rate = k [NO2]2
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)
The rate law is the one for the slow elementary step:
Rate = k [NO2]2
Overall: NO2 (g) + CO (g)  NO (g) + CO2 (g)

17. Step 2: N2O(g) + O (g)  N2(g)+ O2 (g) (slow)
Fast Initial Step
Step 1: N2O (g)  N2(g)+ O(g) (fast)
Step 2: N2O(g) + O (g)  N2(g)+ O2 (g) (slow)
Write the rate law for the overall reaction:
The rate law for the overall reaction is the rate law for the slow step: Rate = k2 [N2O] [O]

18. Fast Initial Step Rate = k [NOBr2] [NO]
Step 1: NO (g) + Br2 (g)  NOBr2(g) (fast)
Step 2: NOBr2 (g) + NO (g)  2 NOBr(g) (slow)
Overall: NO (g) + Br2 (g)  2 NOBr(g)
The rate law is the one for the slow elementary step:
Rate = k [NOBr2] [NO]
If the rate law contains a catalyst or intermediate, need to use fast elementary step to put in terms of initial reactants.
Write the rate law in terms initial reactants or final products, because we can measure these. Intermediates are usually very difficult to measure.

19. Fast Initial Step Rate = k [NOBr2] [NO]
Step 1: NO (g) + Br2 (g)  NOBr2(g) (fast)
Step 2: NOBr2 (g) + NO (g)  2 NOBr(g) (slow)
Overall: NO (g) + Br2 (g)  2 NOBr(g)
The rate law is the one for the slow elementary step:
Rate = k [NOBr2] [NO]

20. Fast Initial Step NOBr2 can react two ways:
With NO to form NOBr
By decomposition to reform NO and Br2
The reactants and products of the first step are in equilibrium with each other.
Therefore,
Ratef = Rater

21. Fast Initial Step Because Ratef = Rater , k1 [NO] [Br2] = k−1 [NOBr2]
Solving for [NOBr2] gives us k1
k−1
[NO] [Br2] = [NOBr2]

22. Fast Initial Step
Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives
k2k1
k−1
Rate =
[NO] [Br2] [NO]
= k [NO]2 [Br2]