5.8 Mechanism → Rate Law

Reaction Mechanism → Rate Law

A slow-first mechanism throttles particles through a bottleneck up top while a rate-vs-[X] curve responds live. Sliders reveal orders 0, 1, and 2 straight from the slow step.

Slow Step = RDS3 MechanismsRate-vs-[X] CurveMolecularity → Order
Topic 5.8

Reaction Mechanism and Rate Law

Identify the rate law for a reaction from a mechanism in which the first step is rate limiting.

For reaction mechanisms in which each elementary step is irreversible, or in which the first step is rate limiting, the rate law of the reaction is set by the molecularity of the slowest elementary step — the rate-determining step.

The physical picture is a bottleneck: no matter how fast the later steps run, the overall reaction cannot outpace its slowest step. Everything after the bottleneck simply waits.

The procedure when step 1 is slow:

  1. Identify the slow step.
  2. Write its rate law directly from its coefficients (it is elementary, so 5.4 applies).
  3. Done — no substitution needed, because a first slow step involves only genuine reactants.

The diagnostic signal. If a species appears in the overall equation but not in the experimental rate law, it must enter after the rate-determining step. Conversely, if a species appears in the rate law with an order that does not match its overall coefficient, the mechanism is multistep.

Using the rate law to test a mechanism. This is the direction the AP Exam most often asks for. Given a proposed mechanism and an experimental rate law, you must check both requirements: do the steps sum to the overall equation, and does the predicted rate law match the measured one? A mechanism failing either test is rejected.

Key points

  • The rate law comes from the slow step’s molecularity when that step is first.
  • A species in the overall equation but absent from the rate law enters after the RDS.
  • A valid mechanism must satisfy two independent tests: sum and rate law.
  • A rate law may never contain an intermediate.

Common mistakes

  • Using the overall equation coefficients. Use the slow step’s.
  • Assuming the first step is always slow. Read the labels.
  • Leaving an intermediate in the answer. If the slow step contains one, you need topic 5.9.

Worked example

The reaction 2 NO₂(g) + F₂(g) → 2 NO₂F(g) has the experimental rate law rate = k[NO₂][F₂]. Evaluate the proposed mechanism: Step 1 (slow): NO₂ + F₂ → NO₂F + F Step 2 (fast): NO₂ + F → NO₂F

Test 1 — do the steps sum to the overall equation?
NO₂ + F₂ + NO₂ + F → NO₂F + F + NO₂F
F cancels (produced in step 1, consumed in step 2 — an intermediate).
Overall: 2 NO₂ + F₂ → 2 NO₂F ✓

Test 2 — does it predict the observed rate law?
Step 1 is the slow, rate-determining step and it is elementary and bimolecular, so
rate = k[NO₂][F₂] ✓

Conclusion: the mechanism is consistent with both the overall stoichiometry and the experimental rate law, so it is a plausible mechanism. (It is not proven — another mechanism could also fit.)

Note the tell: the overall equation has a coefficient of 2 on NO₂, but the rate law is first order in NO₂. That mismatch is exactly what signals a multistep mechanism with only one NO₂ in the slow step.

Full notes for topic 5.8 →