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Mechanisms and Rate-determining Step

Learn and apply Mechanisms and Rate-determining Step in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Mechanisms and Rate-determining Step: Orientation

Mechanism questions are where kinetics becomes “proof”: you must show that a proposed set of steps can produce both (i) the overall equation and (ii) the observed rate law. This lesson trains the consistency check and the intermediate-elimination method.

Link this page to Rate Equations, Orders, and Rate Constant and the Reaction Kinetics hub to keep mechanism and data interpretation consistent.

Definitions (Must Know)

A. Mechanism

A mechanism is a sequence of elementary steps that add up to the overall reaction.

B. Elementary step

An elementary step is a single step in a mechanism with its own molecularity and rate law.

C. Rate-determining step (RDS)

The rate-determining step is the slow step that controls the overall rate.

D. Intermediate

An intermediate is a species formed in one step and consumed in a later step, so it does not appear in the overall equation.

E. Fast pre-equilibrium

A fast pre-equilibrium is a fast reversible step that establishes an equilibrium before the slow step. It is used to express an intermediate concentration in terms of reactant concentrations.

Detailed Explanations

A. The consistency check (use this in exam answers)

  1. Sum the steps to see if they give the overall equation.
  2. Check that intermediates cancel (appear on both sides when steps are added).
  3. Write the rate law from the slow step.
  4. If an intermediate appears, eliminate it using a fast equilibrium/earlier step.
  5. Compare with the experimental rate equation.

B. Rate law from a slow elementary step

If the slow step is: A + B → products then the simplest expected rate law is: rate = k[A][B]

C. Eliminating an intermediate (classic pattern)

Overall reaction: 2NO(g) + Br₂(g) → 2NOBr(g)

Proposed mechanism:

  1. NO(g) + Br₂(g) ⇌ NOBr₂(g) fast equilibrium
  2. NOBr₂(g) + NO(g) → 2NOBr(g) slow

Slow step gives: rate = k₂[NOBr₂][NO]

Fast equilibrium implies: [NOBr₂] = K[NO][Br₂]

Substitute: rate = k₂K[NO]²[Br₂]

So the predicted rate equation is second order in NO and first order in Br₂.

Worked Examples

Modelled example 1

Write a slow-step rate law

Core

Problem

If the slow elementary step is A + B → products, what is the simplest expected rate law?
Study the worked solution

  1. Identify slow-step reactants

    Method

    List A and B as the species colliding in the stated elementary rate-determining step.

    Reason

    The molecularity of an elementary slow step supplies its rate-law concentration powers.

    Working

    One A and one B participate.

  2. Write the law

    Method

    Include each slow-step reactant to power 1.

    Reason

    Each has coefficient 1 in this elementary step.

    Working

    rate = k[A][B].

Guided practice 2

Eliminate an intermediate from the rate law

About 8 min

Problem

A mechanism has fast equilibrium A + B ⇌ C followed by slow C + A → products. Derive the rate equation in terms of [A] and [B] only.

Try this before viewing the solution

Hints

Hint 1: start from the slow step
Write rate = k₂[C][A].
Hint 2: remove C
The fast equilibrium supplies [C] = K[A][B]; substitute this before simplifying constants.
View solution step by step

  1. Write the slow-step law

    Method

    Use C and A from the slow elementary step.

    Reason

    That step controls the overall rate.

    Working

    rate = k₂[C][A].

  2. Express the intermediate

    Method

    Use the fast pre-equilibrium relationship.

    Reason

    The final experimental rate law must not retain intermediate C.

    Working

    [C] = K[A][B].

  3. Substitute and simplify

    Method

    Replace C and combine k₂K into an observed constant.

    Reason

    The concentration dependence is [A][B][A].

    Working

    rate = k₂K[A]²[B] = k[A]²[B].

Common misconception 3

Correct an overall-equation rate law

Find and correct the mistake

Learner method

The overall reaction is 2A + B → products. A learner writes rate = k[A]²[B] before examining the proposed elementary steps. Explain why that procedure is invalid.

Choose the controlling evidence

A mechanism-derived rate law is controlled by

View solution step by step

  1. Separate the equations

    Method

    State that the overall equation is the sum of all mechanism steps.

    Reason

    It does not reveal which elementary collisions occur in the slow step.

    Working

    Overall coefficients alone do not establish reaction orders.

  2. Use the valid route

    Method

    Write the slow-step rate law, eliminate any intermediate using an earlier relationship, then compare with experiment.

    Reason

    This preserves the kinetic dependence predicted by the proposed pathway.

    Working

    k[A]²[B] is acceptable only if that mechanism analysis produces it.

Examiner practice 4

Reject an inconsistent mechanism

3 marks

Problem

The experimental law is rate = k[A][B]. A proposed mechanism has slow step A → X followed by fast X + B → products. Is it kinetically consistent? [3 marks]

Try this before viewing the solution

View solution step by step

  1. Predict from the slow step

    1 mark

    Method

    Write the law for unimolecular slow conversion of A.

    Reason

    B appears only after the rate-determining event.

    Working

    rate_predicted = k[A].

  2. Compare with experiment

    1 mark

    Method

    Contrast the missing B dependence with the observed law.

    Reason

    Experiment shows first-order dependence on both A and B.

    Working

    k[A] ≠ k[A][B] in concentration dependence.

  3. Judge

    1 mark

    Method

    Reject the proposed mechanism.

    Reason

    It cannot reproduce the experimental rate equation.

    Working

    The mechanism is not kinetically consistent.

Challenge 5

Choose between two proposed mechanisms

Minimal support

Problem

For overall 2A + B → P, the observed law is rate = k[A]²[B]. Proposal I has slow A + B → I then fast I + A → P. Proposal II has fast equilibrium 2A ⇌ J then slow J + B → P. Determine which proposal is kinetically consistent.

Try this before viewing the solution

Hints

Hint 1: test proposal I
Its slow elementary step predicts dependence on one A and one B.
Hint 2: eliminate J in proposal II
The fast equilibrium makes [J] proportional to [A]² before J appears in the slow-step law.
View solution step by step

  1. Test proposal I

    Method

    Write the law from its slow elementary step.

    Reason

    The later fast step cannot add A dependence to the rate-determining event.

    Working

    rate_I = k_I[A][B]: inconsistent with second order in A.

  2. Test proposal II

    Method

    Write rate = k₂[J][B] and eliminate J using [J] = K[A]².

    Reason

    The fast equilibrium expresses the intermediate through overall reactants.

    Working

    rate_II = k₂K[A]²[B]: matches the observed powers.

  3. Check and conclude

    Method

    Confirm that proposal II’s steps add to the overall equation with J cancelling.

    Reason

    A valid proposal must satisfy both stoichiometric and kinetic evidence.

    Working

    Proposal II is consistent; proposal I is not.

Mind Stretchers

Mind stretcher 1Extension

The overall reaction is: A + 2B → products and the experimental rate law is: rate = k[A][B]

Suggest a two-step mechanism that is consistent with both.

Show Hint

The observed rate equation can contain reactants from or before the slow step, but an intermediate must be eliminated.

Show Answer

Mark scheme (one valid example):

  • Step 1 (slow): A + B → I
  • Step 2 (fast): I + B → products
  • Rate law from slow step: rate = k[A][B].
  • Adding steps gives A + 2B → products and intermediate I cancels.

Mind stretcher 2: Rejecting a proposed mechanismExtension

Question. The observed law is rate = k[A][B]. A proposed slow elementary step is A + I → products, where I is formed rapidly from two B molecules. Explain what must be shown before accepting the mechanism.

Show Hint

Write the slow-step rate expression, then express the intermediate concentration using the preceding fast step.

Show Answer

The slow step gives rate proportional to [A][I]. The fast step must supply a relationship that makes [I] proportional to [B], not [B]², if the observed law is to result. Without that elimination the proposal is not established.