Steady-State Approximation

Sometimes the rate-determining step of a mechanism involves an intermediate - a species that does not appear in the overall reaction. Since intermediates are not something you can easily measure or control in an experiment, their concentrations cannot appear in the final rate law. You need a way to replace the intermediate with reactant concentrations. That is what the steady-state and pre-equilibrium approximations do.

The Problem: Intermediates in the Rate Law

Consider this mechanism:

Step 1: A + B ⇌ C (fast, reversible)

Step 2: C + D → E (slow)

The slow step determines the rate law: rate = k₂[C][D]

But C is an intermediate. We cannot report a rate law with [C] in it because C is not a reactant we can measure directly. We need to express [C] in terms of [A], [B], and other measurable quantities.

The Pre-Equilibrium Approximation

The pre-equilibrium approximation applies when a fast, reversible step occurs before the slow step. Since Step 1 is fast and reversible, it reaches equilibrium much faster than Step 2 consumes C. We can treat Step 1 as being at equilibrium:

K_eq = k₁/k₋₁ = [C] / ([A][B])

Solving for [C]: [C] = (k₁/k₋₁)[A][B]

Substituting into the rate law:

rate = k₂[C][D] = k₂(k₁/k₋₁)[A][B][D]

Combining constants: rate = $k_{\text{obs}}$[A][B][D]

where $k_{\text{obs}}$ = k₁k₂/k₋₁

When to Use Pre-Equilibrium

Use the pre-equilibrium approximation when:

• A fast, reversible step occurs before the slow step
• The fast step has time to reach equilibrium because the slow step is much slower

This is the most common scenario on the MCAT.

The Steady-State Approximation

The steady-state approximation is a more general approach. It assumes that after a brief initial period, the concentration of an intermediate remains approximately constant - its rate of production equals its rate of consumption.

For the intermediate C:

Rate of production of C = Rate of consumption of C

k₁[A][B] = k₋₁[C] + k₂[C][D]

Solve for [C] and substitute into the rate law for the slow step.

The steady-state approximation is more mathematically involved than pre-equilibrium. On the MCAT, pre-equilibrium is far more common, but you should recognize the steady-state concept if it appears in a passage.

Comparison of Approaches

Worked Example

Consider the mechanism for the reaction 2 NO + O₂ → 2 NO₂:

Step 1: 2 NO ⇌ N₂O₂ (fast, reversible)

Step 2: N₂O₂ + O₂ → 2 NO₂ (slow)

Rate from slow step: rate = k₂[N₂O₂][O₂]

N₂O₂ is an intermediate. Using pre-equilibrium on Step 1:

K = [N₂O₂] / [NO]²

[N₂O₂] = K[NO]²

Substituting: rate = k₂ · K · [NO]² · [O₂] = $k_{\text{obs}}$[NO]²[O₂]

The experimentally observed rate law is rate = k[NO]²[O₂], which matches. This confirms the proposed mechanism is consistent with the data.