5.9 Steady-State Approximation

What is the steady-state approximation?

In multi-step mechanisms with a reactive intermediate, the steady-state approximation assumes the intermediate's concentration stays roughly constant over time — its rate of formation equals its rate of consumption. This lets you set d[intermediate]/dt ≈ 0 and solve for [intermediate] algebraically, then substitute it into the rate law for the overall reaction.

Why use it?

Many mechanisms involve unstable intermediates whose concentrations are hard to measure directly. Setting the net rate of change of the intermediate to zero lets chemists derive a rate law expressed only in terms of reactants and products (measurable species), even when the mechanism has multiple elementary steps of comparable speed rather than one clear rate-determining step.

Steady-state vs. fast pre-equilibrium

The steady-state approximation is an alternative to assuming a fast initial equilibrium step. Both methods eliminate intermediate concentrations from the rate law, but steady-state applies more broadly — it doesn't require any step to be much faster than another, just that the intermediate's concentration stays low and nearly constant throughout the reaction.

Key terms

Steady-state approximation

Assumption that a reactive intermediate's concentration stays nearly constant, so its rate of formation ≈ rate of consumption (d[intermediate]/dt ≈ 0).

Why use the steady-state approximation?

To eliminate hard-to-measure intermediate concentrations from a derived rate law, expressing it only in terms of reactants/products.

Steady-state vs. pre-equilibrium

Steady-state doesn't require one step to be much faster; pre-equilibrium assumes a fast reversible step reaches equilibrium before a slow step.

Mathematical condition for steady-state

d[intermediate]/dt ≈ 0 — formation rate equals consumption rate for the intermediate.

When is steady-state approximation most useful?

In multi-step mechanisms with comparable step rates and a low, roughly constant concentration of a reactive intermediate.

Try an AP-style question

What does the steady-state approximation assume?

1. An intermediate's concentration stays nearly constant
2. All of the reactants are used up immediately
3. The products don't form until the very end
4. The first step is always the slowest step