AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/5/5-8)
Unit 5 · Topic 5.8
5.8 Reaction Mechanism and Rate Law
In a multistep mechanism, the slowest step, called the rate-determining or rate-limiting step, sets the overall rate. When the first step is the slow one, the rate law for the whole reaction is just the rate law of that first step.
Key terms
- rate-determining step
- rate-limiting step
- molecularity
- overall rate law
The slowest step controls the rate
Picture a car wash with a quick rinse station, then a very slow drying station, then a quick wax station. Cars come out only as fast as the dryer can handle them, no matter how fast the other stations are. In a mechanism, the slowest elementary step plays the role of the dryer. It's the rate-determining step (also called the rate-limiting step), and the overall rate can't be faster than it.
When the first step is slow
If the first step is the slow one, the overall rate law is the rate law for that step, written from its coefficients (topic 5.4). Steps after it are fast, so they have no effect on the rate.
If a later step is the slow one, its rate law usually contains an intermediate made in an earlier step. Rate laws for overall reactions are written in terms of reactants, not intermediates, so you need a way to rewrite it. When the earlier step is a fast equilibrium, the pre-equilibrium approximation in topic 5.9 does exactly that.
Checking a mechanism against data
Ask these questions about a proposed mechanism. If the answer to every one is yes, the mechanism is consistent with the evidence. More than one mechanism can be consistent with the same rate law, so data can rule mechanisms out but can't prove one correct.
- Do the steps add up to the overall balanced equation?
- Does the slow step's rate law, written from its coefficients, match the experimentally determined rate law?
- Is every step reasonable (no more than two or, rarely, three particles colliding)?
Species that don't appear in the rate law
A reactant that enters only after the slow step doesn't show up in the rate law; it's zero order. That explains why a rate law can lack a reactant from the balanced equation. In NO₂ + CO → NO + CO₂, CO first reacts in step 2, after the slow step, so the rate law doesn't include [CO].
What the overall k means
When the first step is rate determining, the rate constant in the overall rate law is simply the rate constant of that slow step. Speeding up any fast step, for example with a catalyst that only affects step 2, won't change the overall rate at all. Only changes that affect the slow step matter, which is why chemists focus on the rate-determining step when they try to speed up a reaction.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Rate law from a slow first step
For NO₂ + CO → NO + CO₂, the proposed mechanism is Step 1 (slow): NO₂ + NO₂ → NO₃ + NO; Step 2 (fast): NO₃ + CO → NO₂ + CO₂. Write the rate law predicted by this mechanism.
Show the solutionHide the solution
- Step 1: The rate-determining step is step 1, the slow step.
- Step 2: Step 1 is elementary, with two NO₂ colliding, so its rate law is k[NO₂]².
- Step 3: CO enters only in the fast step 2, so it doesn't appear in the rate law.
Answer: rate = k[NO₂]² (zero order in CO)
- Example 2
Choosing between mechanisms
The experimental rate law for 2A + B → C + D is rate = k[A][B]. Which mechanism is consistent? Mechanism 1: A + B → E (slow); E + A → C + D (fast). Mechanism 2: A + A → F (slow); F + B → C + D (fast).
Show the solutionHide the solution
- Step 1: Both mechanisms add up to 2A + B → C + D (E and F are intermediates that cancel).
- Step 2: Mechanism 1: slow first step A + B → E gives rate = k[A][B]. Matches.
- Step 3: Mechanism 2: slow first step A + A → F gives rate = k[A]². Doesn't match, because the experiment shows B in the rate law and only first order in A.
- Step 4: So only mechanism 1 is consistent.
Answer: Mechanism 1
- Example 3
Overall coefficients mislead (classic trap)
A student uses the overall equation for the NO₂ + CO reaction to predict rate = k[NO₂][CO]. The measured rate law is k[NO₂]². Explain the mismatch.
Show the solutionHide the solution
- Step 1: The overall equation isn't elementary, so its coefficients don't give the rate law.
- Step 2: The actual mechanism has a slow first step in which two NO₂ molecules collide, so the rate depends on [NO₂]².
- Step 3: CO reacts only in a later, fast step, so changing [CO] doesn't change the rate.
Answer: The rate law reflects the slow step (2NO₂ → NO₃ + NO), not the overall equation, so CO doesn't appear and NO₂ is second order.
Common mistakes
- Writing the rate law from the overall equation instead of from the slow step.
- Writing the rate law from a fast step.
- Leaving an intermediate in an overall rate law. If the slow step contains one, use the pre-equilibrium approach from topic 5.9.
- Claiming a mechanism is proven because its rate law matches the data.
On the exam
- A very common free-response task is: given a mechanism and an experimental rate law, decide whether the mechanism is consistent, and justify by writing the slow step's rate law and comparing.
- Expect questions on which species are zero order and why, based on where they enter the mechanism.
Connected topics
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Check yourself
4 questions on 5.8 Reaction Mechanism and Rate Law. Pick an answer to see if you got it, and why.
The overall reaction 2A + B → C has the experimentally determined rate law rate = k[A][B]. Which of the following mechanisms is consistent with this rate law?
The overall reaction A + B → C has the experimentally determined rate law rate = k[A]². Which of the following mechanisms is consistent with both the overall equation and the rate law?
The following mechanism is proposed for a reaction between NO₂ and CO at low temperature.
Step 1 (slow): NO₂(g) + NO₂(g) → NO₃(g) + NO(g)
Step 2 (fast): NO₃(g) + CO(g) → NO₂(g) + CO₂(g)
Proposed reaction mechanism
Which rate law is consistent with this mechanism?
The overall reaction 2NO₂Cl(g) → 2NO₂(g) + Cl₂(g) is thought to occur by the following mechanism.
Step 1 (slow): NO₂Cl(g) → NO₂(g) + Cl(g)
Step 2 (fast): NO₂Cl(g) + Cl(g) → NO₂(g) + Cl₂(g)
Proposed reaction mechanism
Which rate law is consistent with this mechanism?
0 of 4 answered