This AP Chemistry lesson on reaction mechanism notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on reaction mechanism and check yourself with the quiz.
Quick answer: A reaction mechanism is a sequence of elementary steps whose sum is the overall equation. The slowest step sets the rate law. A catalyst opens an alternative mechanism with lower activation energy and is restored, whereas an intermediate is made and then used up. Energy profiles show peaks (transition states), valleys (intermediates), Ea and ΔH.
Level: AP Chemistry · Curriculum: AP Chem Unit 5
Reaction mechanism notes: the key ideas
What is a reaction mechanism?
Most reactions do not happen in one collision. A mechanism is the sequence of simple steps by which reactants become products. Each step is called an elementary step, and for an elementary step, and only for it, the rate law can be written straight from the coefficients.
An elementary step is a single molecular event. Its molecularity is the number of particles that collide in it: unimolecular (1), bimolecular (2), termolecular (3, rare). An intermediate is formed in one step and used up in a later one, so it is absent from the overall equation.
| Elementary step | Molecularity | Rate law |
|---|---|---|
| A → products | Unimolecular | rate = k[A] |
| A + B → products | Bimolecular | rate = k[A][B] |
| 2 A → products | Bimolecular | rate = k[A]² |
How does the slow step set the rate law?
In a chain of steps, the slowest one is the rate-determining step, like the narrowest door in a corridor. The rate of the overall reaction equals the rate of that step. A proposed mechanism is acceptable only if its steps add up to the overall equation and its predicted rate law matches experiment.
Kofi in Accra studies NO₂ + CO → NO + CO₂ and finds rate = k[NO₂]². A proposed mechanism is: step 1 (slow) 2 NO₂ → NO₃ + NO; step 2 (fast) NO₃ + CO → NO₂ + CO₂.
Add the steps: 2 NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂. Cancel NO₃ (intermediate) and one NO₂: NO₂ + CO → NO + CO₂ ✓.
The slow step is bimolecular in NO₂, so rate = k₁[NO₂]² — it matches the experiment. CO does not appear because it reacts after the slow step.
If the slow step contains an intermediate, replace it using a fast step that comes before it. When the first step is fast and reversible, it reaches equilibrium, so [intermediate] = K × [reactants].
For 2 NO + O₂ → 2 NO₂, step 1 is a fast equilibrium: 2 NO ⇌ N₂O₂ (equilibrium constant K). Step 2 is slow: N₂O₂ + O₂ → 2 NO₂ (k₂).
Rate = k₂[N₂O₂][O₂]. Since [N₂O₂] = K[NO]², rate = k₂K[NO]²[O₂] = k[NO]²[O₂], with k = k₂K. The rate law is third order overall.
What does an energy profile show?
An energy profile plots potential energy against progress of the reaction. Each peak is a transition state, each valley an intermediate. The activation energy Eₐ is the climb from the reactants to a peak. The overall enthalpy change ΔH is the difference between products and reactants. The step with the highest peak above its starting level is normally the slow one.
How does a catalyst speed up a reaction?
A catalyst provides a different mechanism with a lower activation energy and is restored at the end. It speeds up the forward and reverse reactions equally, so it does not change ΔH or the equilibrium position.
| Feature | Catalyst | Intermediate |
|---|---|---|
| First appears | Reactant side (consumed early) | Product side of an early step |
| Fate | Restored in a later step | Used up in a later step |
| In the overall equation? | No (cancels) | No (cancels) |
| Effect | Lowers Eₐ | None by itself |
A catalyst does not make a reaction “more favourable” and does not shift the equilibrium. It does not lower the energy of the products. Also remember that a catalyst is consumed in one step and re-formed in another, while an intermediate is formed and then consumed; both cancel in the overall equation, so you need the mechanism to tell them apart.
How are rate constant, temperature and activation energy linked?
k = Ae−Eₐ/RT, so ln k = ln A − Eₐ/(RT). For two temperatures: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂). Eₐ in J/mol, R = 8.314 J/(mol K), T in kelvin, A = frequency factor.
Emma in Leeds measures k = 1.0 × 10⁻³ s⁻¹ at 300 K and 4.0 × 10⁻³ s⁻¹ at 320 K. Find Eₐ.
ln 4.0 = 1.386; 1/300 − 1/320 = 2.083 × 10⁻⁴ K⁻¹.
Eₐ = 8.314 × 1.386/(2.083 × 10⁻⁴) = 5.53 × 10⁴ J/mol = 55 kJ/mol.
A catalyst that lowered Eₐ by 15 kJ/mol at 300 K would raise k by e15000/(8.314 × 300) = e6.01 ≈ 4 × 10², if A is unchanged.
To tell a catalyst from an intermediate, add the steps and cross out what appears on both sides. A species that is on the left of the first step it enters and returns on the right later is the catalyst; a species born first and spent later is the intermediate.
- Elementary step rate law = coefficients as exponents; overall reactions need experiments.
- The slowest step is rate-determining; the mechanism must add up to the overall equation.
- Catalyst: lowers Ea, restored, unchanged ΔH and equilibrium; intermediate: formed then consumed.
- Arrhenius: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂), with R = 8.314 J/(mol K).
- Raising temperature increases the fraction of collisions with energy ≥ Ea.
Mechanisms, catalysts and activation energy: frequently asked questions
How do you know which step is rate-determining?
It is the slowest step, usually the one with the largest activation energy. The experimental rate law helps: the slow step must produce the same concentration terms.
Can an intermediate appear in the rate law?
Not in the final rate law, because intermediates cannot be measured easily. Replace them using an earlier fast equilibrium so that only reactants and catalysts remain.
Does a catalyst change the equilibrium constant?
No. It lowers the activation energy in both directions by the same amount, so equilibrium is reached sooner but at the same position.
What is the difference between a transition state and an intermediate?
A transition state is the peak of the profile and cannot be isolated. An intermediate sits in a valley between two peaks and may live long enough to be detected.
Why does a small rise in temperature increase the rate so much?
The fraction of collisions with energy at least Ea grows exponentially with temperature. For many reactions, 10 °C raises the rate by roughly two or three times.
