Unit 5 · Kinetics

AP Chemistry Reaction Mechanisms

Reading a mechanism, finding the intermediates and catalysts, and deriving the rate law, including the fast-first-step case, in four worked examples.

A mechanism question gives you a list of elementary steps and asks one of a few things: what is the overall reaction, which species are intermediates or catalysts, what rate law the mechanism predicts, or whether it is consistent with a measured rate law. The reasoning is the same each time. The Unit 5 review outlines the ideas; this page works through the algebra and the bookkeeping, and the rate law page covers how the observed rate law is found from data in the first place.

How a Reaction Mechanism Works and What to Look For

• Elementary steps: each step is a single molecular event, so its rate law uses the step's own coefficients as exponents, for example rate = k[A][B] for A + B → products.
• Molecularity: the number of particles colliding in a step. Unimolecular and bimolecular steps are common; termolecular steps are rare.
• The steps must sum to the overall equation. Add them and cancel species that appear on both sides.
• Intermediates are made in one step and used in a later one. Catalysts are used in an early step and regenerated later.
• The rate-determining step is the slowest step and sets the rate. The overall rate law cannot contain an intermediate, because intermediates do not appear in the overall reaction.

The diagram shows the case that needs the most care: a fast reversible first step followed by a slow second step. When the slow step contains an intermediate, the equilibrium of the fast step is used to replace it, the pre-equilibrium approximation described in the Unit 5 review.

Worked Example 1: Adding Steps to Get the Overall Reaction and a Rate Law From a Slow First Step

A proposed mechanism for NO₂ + CO → NO + CO₂ is:

Step 1 (slow): NO₂ + NO₂ → NO₃ + NO
Step 2 (fast): NO₃ + CO → NO₂ + CO₂

Confirm that the steps give the overall reaction, name the intermediate, and write the rate law.

Add the steps: 2NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂. Cancel one NO₂ and the NO₃ from both sides to get NO₂ + CO → NO + CO₂, as required.
Intermediate: NO₃, which is produced in step 1 and consumed in step 2.
Rate law: the first step is slow, so it sets the rate, and it contains only reactants: rate = k[NO₂]².

CO does not appear in the rate law even though it is a reactant, because it reacts only in the fast step after the bottleneck. The overall equation has a coefficient of 1 for NO₂, yet the rate is second order in it, which is why the order cannot be read from the balanced equation.

Worked Example 2: Telling a Catalyst From an Intermediate

In the upper atmosphere, chlorine atoms destroy ozone by this mechanism:

Step 1: Cl + O₃ → ClO + O₂
Step 2: ClO + O → Cl + O₂

Find the overall reaction and classify Cl and ClO.

Overall: Cl + O₃ + ClO + O → ClO + O₂ + Cl + O₂. Cancel Cl and ClO to get O₃ + O → 2O₂.
Cl is a catalyst: it is a reactant in step 1 and is regenerated as a product of step 2.
ClO is an intermediate: it is a product of step 1 and a reactant in step 2.

A quick test: the species that appears first as a reactant and reappears at the end is a catalyst, and the one that appears first as a product is an intermediate. A catalyst lowers the energy barrier without being used up; its effect on the energy profile is covered in activation energy.

Worked Example 3: Deriving the Rate Law When the First Step Is a Fast Equilibrium

A mechanism for 2NO + O₂ → 2NO₂ is:

Step 1 (fast, reversible): 2NO ⇌ N₂O₂
Step 2 (slow): N₂O₂ + O₂ → 2NO₂

Check the overall reaction, then find the rate law the mechanism predicts.

Overall: 2NO + N₂O₂ + O₂ → N₂O₂ + 2NO₂, which reduces to 2NO + O₂ → 2NO₂.
Slow step: rate = k₂[N₂O₂][O₂]. N₂O₂ is an intermediate, so it cannot stay in the final rate law.
Fast equilibrium: because step 1 is fast and reversible, it stays near equilibrium, so K = [N₂O₂]/[NO]² and [N₂O₂] = K[NO]².
Substitute: rate = k₂K[NO]²[O₂] = k[NO]²[O₂], where k = k₂K.

The result is third order overall, second order in NO and first order in O₂, even though no single step involves three particles colliding. This is the main reason the pre-equilibrium method is useful: it explains rate laws whose orders match the overall equation without requiring a termolecular step.

Worked Example 4: Testing Candidate Mechanisms Against an Observed Rate Law

The reaction 2NO + Cl₂ → 2NOCl has the experimental rate law rate = k[NO]²[Cl₂]. Which of these mechanisms is consistent with it?

Mechanism A: one termolecular step, 2NO + Cl₂ → 2NOCl.
Mechanism B: NO + Cl₂ ⇌ NOCl₂ (fast); NOCl₂ + NO → 2NOCl (slow).
Mechanism C: NO + Cl₂ → NOCl + Cl (slow); Cl + NO → NOCl (fast).

A: rate = k[NO]²[Cl₂]. It matches, but a termolecular step is very unlikely, so it is a poor candidate.
B: slow step rate = k₂[NOCl₂][NO]. From the fast step, [NOCl₂] = K[NO][Cl₂], so rate = k₂K[NO]²[Cl₂]. It matches.
C: the slow step is first, so rate = k[NO][Cl₂]. This is first order in NO, so it does not match.

Mechanism B is the best supported, and C is ruled out. Keep in mind that agreement does not prove B: A also agrees, and other mechanisms could as well. A mismatch eliminates a mechanism, but a match only keeps it as a possibility. Each step must also be a reasonable elementary step, with an energy profile like the ones in reading reaction energy profiles.

Five Reaction Mechanism Practice Questions With Answers

1. Write the rate law for the elementary step O₃ + NO → O₂ + NO₂ and give its molecularity.
   Show answer

   Rate = k[O₃][NO]. It is bimolecular, because two particles collide.
2. Step 1 (slow): A + B → C. Step 2 (fast): C + B → D. Give the overall reaction, the intermediate, and the rate law.
   Show answer

   Overall: A + 2B → D. Intermediate: C. The slow step is first, so rate = k[A][B].
3. Step 1 (slow): H₂O₂ + I⁻ → H₂O + IO⁻. Step 2 (fast): H₂O₂ + IO⁻ → H₂O + O₂ + I⁻. Identify the catalyst and the intermediate, and write the rate law.
   Show answer

   I⁻ is the catalyst (used in step 1, regenerated in step 2) and IO⁻ is the intermediate. Overall: 2H₂O₂ → 2H₂O + O₂. Rate = k[H₂O₂][I⁻].
4. Step 1 (fast, reversible): A + B ⇌ C. Step 2 (slow): C + B → D. Find the rate law.
   Show answer

   Slow step: rate = k₂[C][B]. From the fast step, [C] = K[A][B], so rate = k₂K[A][B]² = k[A][B]².
5. Can the order of an overall reaction be read from its balanced equation? Explain.
   Show answer

   No. The exponents match the coefficients only for a single elementary step. For an overall reaction the order is found from experimental data, and it reflects the rate-determining step and any pre-equilibrium.

Common Reaction Mechanism Mistakes

• Writing the rate law from the overall equation. Use the rate-determining step, not the overall coefficients.
• Leaving an intermediate in the rate law. If the slow step contains one, replace it using the earlier fast equilibrium.
• Mixing up catalysts and intermediates. A catalyst is used first and regenerated; an intermediate is formed first and used up.
• Skipping the sum check. The steps must add to the overall equation after canceling species that appear on both sides.
• Assuming the slow step is first. Read which step is labeled slow before deciding how to write the rate law.
• Saying a matching rate law proves a mechanism. It only shows that the mechanism is consistent with the data.

Mechanism problems combine algebra, reasoning, and vocabulary, so a few practiced patterns go a long way. To see how a stronger Unit 5 would shift your overall estimate, the AP Chem Score Calculator turns section scores into a 1–5 range.

Related Resources

• AP Chem Score Calculator
• Unit 5 Review: Kinetics
• AP Chemistry Rate Law
• AP Chemistry Activation Energy
• AP Chemistry Catalysts
• AP Chemistry Equilibrium Constant Kc and Kp
• AP Chemistry Le Chatelier's Principle
• AP Chemistry Balancing Equations
• AP Chemistry Study Guide

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