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.

Reaction Mechanisms

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

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.”

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.

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

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.

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

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

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

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…

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

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

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

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

k1

k1

k1

k1

Analyzing Reaction Mechanisms with One (or More) Fast Equilibrium Step(s)

For the reactionA B

The rate law for the forward reaction is Rate = k1[A]

The rate law for the backward reaction is Rate = k1[B]

k1

k1

Analyzing Reaction Mechanisms with One (or More) Fast Equilibrium Step(s)

For the reactionA B

The forward reaction rate is equal to the backward rate at equilibrium, so set them equal to each other.

k1[A] = k1[B]rate of fwd rxn = rate of back rxn

k1

k1

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

k1

Question:Could this mechanism be correct?

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

k1

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

k1

There is an intermediate in the rate law…not allowed!

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] = k1[N2O2]rate of fwd rxn = rate of back rxn

k2

k1

k1

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 k1

k2

k1

k1

Plug this quantity into the rate law for [N2O2]

Example (Continued)

Substitute into the rate law for [N2O2]

Rate law for slow step: Rate = k2 [N2O2][O2]

Substitute… Rate = [NO]2[O2]Observed rate law: Rate = kobs [NO]2[O2]

Does the rate law match? Yes!

Example Problemswill be posted separately.

Next up, Collision Theory and

Activation Energy (Pt 8)