Deduce reaction order from a mechanism
Use the rate-determining step and preceding equilibria to predict orders from a multi-step mechanism.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 26. Rate laws are empirical unless an elementary step is explicitly stated. Written practice is self-checked
Before you begin
- Use AS collision theory, concentration, graph gradients, and balanced reaction equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Why replace an intermediate in a slow-step rate expression?
What you will learn
- Write a rate expression for an elementary rate-determining step
- Use a fast equilibrium to remove intermediate concentrations
- Deduce reactant orders from the resulting expression
Start with the rate-determining elementary step
For a proposed elementary slow step, the rate expression is proportional to the concentrations of its reactants with powers from that step. If the slow step contains an intermediate, its concentration must be related to measurable reactants using earlier mechanism steps.
For fast A + B ⇌ I followed by slow I + B → products, [I] is proportional to [A][B] at equilibrium. Substitution gives rate proportional to [A][B]².
Mechanism order is a prediction to test
The predicted rate equation should match experimental orders. A mechanism can be rejected when its predicted law conflicts with observed data.
Do not infer that every elementary step's exponents equal the overall reaction coefficients. Only the rate-determining step and any established pre-equilibria are used for this derivation.
Pause and explain
If [I] ∝ [A]² and the slow step is I + B → products, what is the predicted rate law?
Worked example
Fast 2A ⇌ I; slow I + B → P. If [I] is proportional to [A]², deduce the rate equation.
Show the worked solution
- The slow step gives rate = k[I][B].
- Substitute [I] ∝ [A]².
- Collect measurable reactant concentrations.
Answer rate = k′[A]²[B], so the predicted order is 2 in A and 1 in B.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Fast A ⇌ I with [I]∝[A]; slow I + A → P. Deduce the rate law.
Give me a hint
Use rate∝[I][A] and replace [I].
Compare my reasoning
- Slow step gives rate∝[I][A].
- Fast equilibrium gives [I]∝[A].
- Multiply the A factors.
rate∝[A]².
Look for these in your work
- Slow step translated
- Intermediate replaced
- Order deduced
Fast A + B ⇌ I; slow I + A → products. Deduce the predicted orders if [I]∝[A][B].
Give me a hint
The slow step contains one I and one A.
Compare my reasoning
- Write rate∝[I][A].
- Substitute [I]∝[A][B].
- Combine powers.
rate∝[A]²[B], second order in A and first order in B.
Look for these in your work
- Slow-step reactants included
- Fast equilibrium substituted
- Orders combined
A mechanism-derived rate law does not match experiment. What conclusion is justified?
Give me a hint
Mechanisms are hypotheses tested against evidence.
Compare my reasoning
- Identify the mismatched order.
- State that the proposed mechanism is inconsistent with those data.
- Avoid claiming the experimental rate law is wrong without new evidence.
The proposed mechanism is not consistent with the measured rate law under the stated conditions; another mechanism or assumption is needed.
Look for these in your work
- Mismatch identified
- Mechanism treated as a hypothesis
- Conclusion calibrated to evidence
Common mistakes
- Leaving an intermediate in the final empirical rate law
- Using the slow step without its molecularity
- Forgetting powers from the fast equilibrium
- Calling a predicted law experimental proof
What can you explain now?
Why replace an intermediate in a slow-step rate expression?
Compare with the explanation
The experimental rate law uses measurable reactant concentrations
The intermediate concentration must be related to observable species.
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Make it stick
How does an earlier equilibrium let you convert a slow-step expression into measurable reactant orders?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.