Mistake Master
Mechanism and rate law
In a multistep reaction, one step is the bottleneck. That slowest step sets the pace for everything, so the overall rate law comes from it, not from the tidy overall equation.
§1
The bottleneck sets the rate.
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When a reaction has a mechanism, the overall rate law is governed by the rate-determining step — the slowest elementary step. It is the bottleneck: the whole reaction can go no faster than its slowest step.
Because that slow step is elementary, its rate law comes directly from its molecularity (its reactant coefficients become the orders). This rate law, not the overall equation's coefficients, is the reaction's rate law.
If the slow step involves an intermediate, that intermediate cannot appear in the final rate law (it is not measurable). It must be re-expressed using an earlier fast step, so the rate law ends up in terms of the actual reactants.
§2
Deriving the rate law from a mechanism.
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Start from the slow step; remove any intermediate.
- Find the rate-determining (slowest) step. The overall rate is limited by this step.
- Write its rate law from molecularity. Since the slow step is elementary, its reactant coefficients are the orders.
- Check for intermediates. If the slow step's rate law contains an intermediate, it cannot stay there.
- Substitute the intermediate. Use an earlier fast step (often a fast equilibrium) to express the intermediate in terms of reactants.
§3
The pieces you'll meet.
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A few ideas connect mechanism to rate law.
§4
Worked example: rate law with a slow first step.
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Mechanism. Step 1 (slow): A + A → C. Step 2 (fast): C + B → products.
Rate-determining step. Step 1 is slow, so it governs the rate.
Rate law from the slow step. Step 1 is elementary and bimolecular in A, so rate = k[A]².
Check. The slow step uses only A (no intermediate needed on the reactant side), so the overall rate law is rate = k[A]² — note that B does not appear, even though it is in the overall equation. The mechanism, not the overall equation, sets this.
§5
Mistakes that cost real points.
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"The rate law comes from the overall balanced equation."
For a multistep reaction, the rate law comes from the rate-determining (slowest) step, not the overall equation. A reactant that appears only in a fast step (after the slow one) may not appear in the rate law at all.
Fix. Derive the rate law from the slow step's molecularity; only for a single elementary reaction does the overall equation give it.
"Just copy the coefficients from the overall equation as the orders."
The orders come from the rate-determining step, which may involve different species than the overall equation suggests. Copying overall coefficients ignores the mechanism and usually gives the wrong rate law.
Fix. Use the slow step's reactant coefficients as the orders, after removing any intermediate.
"An intermediate can stay in the final rate law."
A rate law must be written in terms of measurable reactant concentrations, so an intermediate appearing in the slow step's rate law must be substituted out using an earlier fast step. Leaving an intermediate in is not an acceptable final rate law.
Fix. Express any intermediate via a prior fast step so the final rate law contains only reactants.
§6
Skill Check.
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Ten scenarios. Pick the chips that match your answer, then check. A scenario marks complete the first time every part is right. Progress saves on this device.