IB Chemistry HLTopic 5 — How Fast? The Rate of ReactionPaper 1 & 2HL only | Core idea~12 min read
Reaction Mechanisms
A balanced equation is a summary, not a description. It tells you what went in and what came out, but almost never what actually happened. Reactions get there in small steps, and one of those steps is slower than the rest — which is why the rate equation so rarely looks like the equation you were given.
📘 What you need to know
Most reactions happen as a series of elementary steps. The whole sequence is the mechanism.
The steps must add up to the overall equation. Anything that appears on both sides cancels.
An intermediate is made in one step and used up in a later one. It never appears in the overall equation.
The slowest step is the rate-determining step (RDS). It sets the pace for everything.
The rate equation comes from the RDS: the order with respect to a species is how many particles of it are in the RDS.
Intermediates cannot appear in a rate equation. Swap them for whatever made them.
A catalyst can appear in the rate equation even though it is missing from the overall equation.
Kinetics can disprove a mechanism, but never prove one.
Why reactions go in steps
Look at an equation like 2NO + O2 → 2NO2. Taken literally, it says three molecules meet at one point, at the same instant, all facing the right way. In a gas that is close to impossible. Two-particle collisions are common; three-particle ones are not.
So the reaction cheats. Two molecules meet, do something small, and then the product of that meets the third. Each of those little events is an elementary step, and an elementary step really does happen exactly as written.
Key distinction: the overall equation is bookkeeping and its numbers tell you nothing about rate. An elementary step is a real physical event, so for that step alone the numbers do give you the powers.
The rules a mechanism has to obey
🧩 Checking any proposed mechanism
Add the steps together. Cancel anything appearing on both sides.
Compare with the overall equation. If they do not match, the mechanism is wrong, full stop.
Find the slow step. That is the rate-determining step.
Write the rate equation from the RDS, using its species and its coefficients as powers.
Remove any intermediate by replacing it with the species that made it.
Compare with the experimental rate equation. If they disagree, the mechanism is wrong.
Add the two left-hand sides and the two right-hand sides, then strike out anything that appears on both. Whatever is left has to be the overall equation, or the mechanism is not a candidate.
The rate-determining step
A reaction can only go as fast as its slowest step. Speeding up any of the fast steps changes nothing, because they are already sitting around waiting.
🤔 Why only the slow step matters
Picture a production line where one machine takes a minute per item and the rest take a second. Buy a faster version of the quick machines and the line still turns out one item a minute — the bottleneck has not moved. The same is true here. Anything you do to the fast steps is wasted; only the slow step controls the output. So the only concentrations that show up in the rate equation are the ones that affect the slow step.
This is the sentence that makes the whole topic click: the rate equation is a photograph of the rate-determining step. If you can see the slow step, you can write the rate equation without doing a single experiment.
Reading the rate equation off the mechanism
Take the slow step and treat it as a real collision. Every particle taking part contributes a power of one.
Slow step
What it means physically
Rate equation it predicts
A → products
One A particle falls apart on its own
rate = k[A]
A + B → products
One A meets one B
rate = k[A][B]
2A → products
Two A particles meet each other
rate = k[A]2
A + 2B → products
One A meets two B at once (rare)
rate = k[A][B]2
X + B → products, where X is an intermediate from 2A
The X has to be made from two A first
rate = k[A]2[B]
The last row is the tricky one. If the slow step uses an intermediate, you cannot leave it in the answer — nobody can measure its concentration. Replace it with whatever the earlier step made it from, and the powers follow.
Energy profiles with more than one hump
Each elementary step gets its own hump. The dip between the humps is the intermediate: a real substance that exists for a moment, sitting in an energy valley. The step with the bigger activation energy is the slow one, so it gets the bigger climb.
Two humps means two steps and one intermediate. Count the humps and you have counted the steps — a one-mark question that comes up regularly.
Worked examples
WORKED EXAMPLE
From mechanism to rate equation
Hydrogen peroxide decomposes in the presence of iodide ions by this mechanism:
Step 1 (slow): H2O2 + I– → H2O + IO–
Step 2 (fast): H2O2 + IO– → H2O + O2 + I–
(a) Show the steps give the overall equation. (b) Deduce the rate equation. (c) State the role of I– and of IO–.
(a) Add the two stepsLeft: 2H₂O₂ + I⁻ + IO⁻. Right: 2H₂O + O₂ + IO⁻ + I⁻. Cancel I⁻ and IO⁻ from both sides.2H2O2 → 2H2O + O2(b) Use the slow step onlyStep 1 involves one H₂O₂ particle and one I⁻ particle, so each gets a power of 1.rate = k[H2O2][I–](c) RolesI⁻ is used then remade, so it is a catalyst. IO⁻ is made then used, so it is an intermediate.Note the catalyst appears in the rate equation but not in the overall equationthat last point surprises people every year – it is only possible because I⁻ is in the slow step
WORKED EXAMPLE
From rate equation to mechanism
For 2NO(g) + O2(g) → 2NO2(g), experiment gives rate = k[NO]2[O2].
Explain why a single-step mechanism is unlikely, and propose a two-step mechanism that fits.
Step 1: Why one step is unlikelyA single step would need two NO and one O₂ to collide at the same instant, all correctly lined up. Three-particle collisions are very rare.Step 2: What the rate equation is telling usThird order overall, so the slow step must effectively involve two NO and one O₂. The two NO must join up first.Step 3: Propose the stepsStep 1 (fast): NO + NO ⇌ N2O2Step 2 (slow): N2O2 + O2 → 2NO2Step 4: Check it predicts the right rate equationSlow step gives rate = k[N₂O₂][O₂]. N₂O₂ is an intermediate, so replace it: it comes from two NO.rate = k[NO]2[O2] — matches experimentalso check the steps add up: 2NO + N₂O₂ + O₂ gives N₂O₂ + 2NO₂, which cancels to 2NO + O₂ → 2NO₂
WORKED EXAMPLE
Ruling a mechanism out
For CH3Br + OH– → CH3OH + Br–, experiment gives rate = k[CH3Br][OH–].
A student suggests:
Step 1 (slow): CH3Br → CH3+ + Br–
Step 2 (fast): CH3+ + OH– → CH3OH
Explain why this mechanism must be rejected.
Step 1: The steps do add up correctlyCH₃⁺ cancels, leaving CH₃Br + OH⁻ → CH₃OH + Br⁻. So that test is passed.Step 2: Now check the rate equation it predictsThe slow step involves only one CH₃Br particle and nothing else.predicted rate = k[CH3Br]Step 3: Compare with experimentExperiment shows the rate does depend on [OH⁻], but this mechanism says it should not.Rejected — OH– must be involved in the slow step, so the reaction happens in one bimolecular stepadding up correctly is necessary but not enough – the rate equation is the second hurdle
🧠 Two tests, both must pass
Every mechanism question is really the same question twice. Test 1: do the steps add up to the overall equation? Test 2: does the slow step predict the experimental rate equation? Fail either and the mechanism is out. Pass both and all you can say is that it is consistent — never that it is proved.
💡 Exam tip
Write the words slow and fast next to your steps. Examiners look for them.
When adding steps up, do the left-hand sides and right-hand sides in two separate lists, then cancel. It is much harder to lose something that way.
Never leave an intermediate in a rate equation. Trace it back to the species that formed it.
If the question says “suggest” or “propose”, you are allowed to invent an intermediate, as long as everything balances.
Use the phrase “consistent with the experimental rate equation” rather than “proves the mechanism”.
On an energy profile, count humps to count steps, and count dips to count intermediates.
⚠ Common mix-up
Taking the powers from the overall equation. Only an elementary step lets you do that.
Leaving an intermediate in the rate equation. You cannot measure its concentration, so it cannot be there.
Assuming the first step is always the slow one. It often is, but plenty of mechanisms have a fast first step.
Forgetting the catalyst can appear in the rate equation. If it is in the slow step, it belongs there.
Saying kinetics proves a mechanism. It can only rule mechanisms out.
Reading the tallest peak as the rate-determining step. What matters is the size of the climb from the valley before it.
Up next: Molecularity — the proper name for how many particles take part in a single step, and why it is not the same thing as order.
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