Pre-Equilibrium Approximation

Learning Objective 5.9.A Identify the rate law for a reaction from a mechanism in which the first step is not rate limiting.

Quick Notes

If a reaction’s first step is fast and not rate-limiting, it may reach a pre-equilibrium before the slow step.
To eliminate intermediates from the rate law:
1. Write an expression for the fast equilibrium.
2. Substitute that expression into the rate law for the slow step.
The resulting rate law may include concentrations from the fast step but still reflects control by the slow step.

Full Notes

Reactions Where the First Step Is Fast

Not all reaction mechanisms start with the rate-determining step. In some cases, the first step is fast and reversible, quickly reaching a pre-equilibrium before a slower, rate-limiting step occurs.

The fast initial step establishes a dynamic equilibrium between intermediates and reactants.
The slow second step determines the overall reaction rate.
To write the rate law, we use the pre-equilibrium approximation:
- Express the intermediate concentration using the fast step’s equilibrium expression.
- Substitute it into the slow step’s rate law.

This method is closely related to the steady-state assumption, where the concentration of intermediates is assumed to remain relatively constant throughout the reaction.

Worked Example

Pre-Equilibrium Approximation (NO + O₂ → NO₂)

Overall reaction (gas phase):
2NO (g) + O₂ (g) → 2NO₂ (g)

Experimental rate law: Rate = k[NO]²[O₂]

Proposed mechanism
Step 1 (fast, reversible): NO + NO ⇌ N₂O₂
Step 2 (slow): N₂O₂ + O₂ → 2NO₂

Rate from the slow step
Rate = k₂[N₂O₂][O₂]
(contains the intermediate N₂O₂, which we must remove)

Use pre-equilibrium from Step 1
Forward rate: k_f[NO]² Backward rate: k_b[N₂O₂]
At equilibrium: k_f[NO]² = k_b[N₂O₂]
meaning [N₂O₂] = (k_f/k_b)[NO]²

Substitute into the slow-step rate law
Rate = k₂[(k_f/k_b)[NO]²][O₂]
= (k₂·k_f/k_b)[NO]²[O₂]

Combine constants
Let k′ = (k₂·k_f/k_b), so the final rate law is:
Rate = k′[NO]²[O₂]
This matches experiment and uses only reactant concentrations.

Takeaway: When a fast reversible step precedes a slow step, write the slow-step rate, express any intermediate using the fast equilibrium (e.g., [N₂O₂] = (k_f/k_b)[NO]²), and substitute to eliminate the intermediate.

Steps to Determine the Rate Law

Identify the slow (rate-limiting) step.
Write the rate law for the slow step (using its reactants).
If that rate law contains an intermediate, write an equilibrium expression for the fast step.
Solve the equilibrium expression for the intermediate.
Substitute the result into the rate law so that it contains only concentrations of stable species (not intermediates).

Important Considerations

This method only applies when the first step is fast and followed by a slower, rate-limiting step.
Intermediates should always be replaced using equilibrium expressions.
The final rate law should only contain concentrations of reactants and/or products, not intermediates.

Matt’s exam tip

If an intermediate appears in the rate law from a slow step, it must be removed using a fast pre-equilibrium expression. Look for a reversible fast step and apply equilibrium reasoning.

Summary

In reaction mechanisms where the first step is fast and not rate-limiting, the rate law must be determined using the pre-equilibrium approximation. Write the rate law for the slow step, identify intermediates, and substitute them using expressions derived from the fast equilibrium. This ensures the final rate law reflects only stable, measurable species while still accounting for the mechanism’s steps.