Steady-State Approximation: Three Solved Kinetics Problems

What the approximation says, and when it is allowed

In a multi-step mechanism, a reactive intermediate is produced and consumed. If it is consumed almost as fast as it is formed, its concentration rises quickly to a small value and then stays roughly constant for most of the reaction. Over that period we may write:

This is not a claim that the intermediate is at equilibrium, and it is not a claim that nothing is happening. It is a statement that formation and consumption rates are nearly equal, so the net rate of change is negligible compared with the rates of the individual steps.

Problem 1 — a simple two-step mechanism

For the mechanism

with B a reactive intermediate, derive the rate of formation of C.

Solution

Rate of change of B is formation minus consumption:

Therefore [B] = k₁[A]/k₂, and

The result is first order in A, and k₂ cancels entirely. That cancellation is the physical point: when the second step is fast, the first step controls the overall rate, so the rate constant of the fast step cannot appear in the answer.

Problem 2 — a reversible first step

For

Solution

B is now formed by one route and consumed by two:

The two limiting cases are what examiners ask about:

The second limit is exactly the pre-equilibrium result. Pre-equilibrium is therefore a special case of the steady state, not a competing method — a distinction worth stating explicitly in a written answer.

Problem 3 — the Lindemann mechanism

Unimolecular gas-phase reactions appear first order at high pressure and second order at low pressure. The Lindemann scheme explains this:

Solution

At high pressure [A] is large, so k₋₁[A] >> k₂ and the rate becomes (k₁k₂/k₋₁)[A] — first order. At low pressure k₂ >> k₋₁[A] and the rate becomes k₁[A]² — second order. One mechanism reproduces both observed orders, which is why this remains a standard examination piece.

How to avoid the usual errors

• Identify the intermediate first. It is the species that appears in the mechanism but in neither the overall reactants nor products.
• Account for every route. The commonest error in Problem 2 is omitting k₋₁[B] from the consumption terms.
• Never leave an intermediate in the final rate law. A rate law expressed in terms of a species that cannot be measured is not an answer.
• State the validity condition when the question carries derivation marks. It is frequently worth a mark on its own.

Frequently asked questions

How is the steady-state approximation different from pre-equilibrium?

Pre-equilibrium assumes the first step stays at equilibrium, which requires the reverse step to dominate. The steady state makes no such assumption and reduces to the pre-equilibrium result in that limit, so it is the more general treatment.

When does the approximation break down?

During the induction period before the intermediate concentration stabilises, and whenever the intermediate is stable enough to accumulate. In both cases d[I]/dt is not negligible.

Can it be applied to more than one intermediate?

Yes. Write one steady-state equation per intermediate and solve the resulting simultaneous equations. Chain reactions are handled exactly this way.

Which exams ask this most?

It appears across IIT-JAM, CSIR-NET and GATE Chemistry. JAM and GATE lean toward deriving a rate law from a given mechanism; CSIR-NET more often asks which mechanism is consistent with an observed rate law, which is the same skill in reverse.