Rate equations from a mechanism: when the slow step comes second
The kinetics idea IB Chemistry HL students find hardest, in five steps, with original worked examples.
A rate equation comes from the slow step of a mechanism, because the slowest step sets the pace. If the slow step contains an intermediate, the intermediate cannot stay in the equation. Replace it with the reactants that made it. For P + Q forming X quickly, then X + R forming S slowly, rate = k[P][Q][R].
Why this one is hard
The rule most students learn is "use the slow step". That works until the slow step contains something that is not a reactant. Kinetics is in all 16 HL papers since May 2025, worth 99 marks across Papers 1A and 2, and two-step mechanisms came up in 6 of the 8 sittings (our count). On one May 2025 Paper 2, about a quarter of students wrote the right rate equation when the slow step came second.
The idea: one slow checkout
A shop with one open checkout serves people at the speed of that checkout, however fast they walk in. A reaction works the same way. The slowest elementary step sets the rate of the whole reaction, which is why it is called the rate-determining step.
The five steps
- Find the slow step.
- Write the rate equation from the species in that step.
- Look for an intermediate: a species made in one step and used up in a later one. It never appears in the overall equation.
- Replace the intermediate with the reactants that made it.
- Collect any repeated concentration into a power.
Worked example: the slow step comes second
Step 1: P + Q ⇌ X fast
Step 2: X + R → S slow
Overall: P + Q + R → S
The slow step suggests rate = k[X][R]. X is an intermediate, because step 1 makes it and step 2 uses it up. It came from P and Q, so swap it for them:
rate = k[P][Q][R]
That is first order in each reactant and third order overall, so the units of k are mol−2 dm6 s−1.
When the slow step comes first
Now take P + Q → X as the slow step, followed by a fast X + R → S. The slow step contains only reactants, so rate = k[P][Q]. R reacts after the slow step, so its concentration does not change the rate, and it is zero order.
The mistakes to stop making
Using the overall equation. Its coefficients say nothing about orders. Rate equations are found by experiment, and a proposed mechanism has to fit them.
Leaving the intermediate in. A rate equation is written in terms of species you can add and measure, and the intermediate is made and used up inside the reaction.
Guessing the units of k. They follow from the overall order. Under 43% got them right on one May 2025 Paper 1A question.
Check yourself
A mechanism has a fast step M + N ⇌ Y, then a slow step Y + N → Z. Write the rate equation.
Show the answer
The slow step gives rate = k[Y][N]. Y came from M and N, so rate = k[M][N][N], which is rate = k[M][N]2.
Go through it with us, free
We run a free 45-minute live session on rate equations and the slow step: one original mechanism and one initial-rates table worked through to k. SL students get the kinetics they need in the first half.
See the free session dates Practise R2.2 questions Book a free 30-minute session
Sources
Examiner evidence: May 2025 Chemistry subject reports, in our words. Mark counts: our own, from the HL papers since May 2025. Syllabus: guide statements R2.2.6 and R2.2.9. This page was developed independently from and is not endorsed by the International Baccalaureate Organization. Examples and wording are ours.