Short on time? This page is the recap. For explanations and worked examples, use the full study guide.
Recap
- The slowest elementary step sets the rate of a reaction.
- If the first step is slow, its rate law is the rate law of the reaction (Topic 5.8).
- If a later step is slow, its rate law usually contains an intermediate. You cannot set an intermediate’s concentration in an experiment, so it must not appear in the final rate law.
- When the step before the slow step is fast and reversible, assume it stays at equilibrium: its forward and reverse rates are equal. This is the pre-equilibrium approximation.
The recipe
- Write the slow step’s rate law from its molecularity.
- Find the intermediate in it.
- Fast step: forward rate = reverse rate. Solve for [intermediate].
- Substitute, then combine the constants into one k.
Key relationships
| Mechanism pattern | Intermediate | Rate law |
|---|---|---|
| A + B ⇌ I (fast); I + B → P (slow) | [I] = (k₁ / k₋₁)[A][B] | rate = (k₂k₁ / k₋₁)[A][B]² |
| X₂ ⇌ 2X (fast); X + Y → XY (slow) | [X] = (K₁[X₂])^½ | rate = k₂K₁^½[X₂]^½[Y] |
| A + Cat ⇌ ACat (fast); ACat + B → AB + Cat (slow) | [ACat] = K₁[A][Cat] | rate = k₂K₁[A][Cat][B] |
K₁ = k₁ / k₋₁ (forward over reverse).
Assumptions
- The fast step goes back and forth much faster than the slow step uses up the intermediate.
- Every step is elementary, so its rate law follows from its molecularity.
- Steps after the slow step do not affect the rate law.
Mistakes to avoid
- Leaving the intermediate in. The slow step’s rate law alone is not the answer.
- Using overall coefficients as orders.
- Flipping K₁. Forward constant on top.
- Losing the square root. X₂ ⇌ 2X gives [X] proportional to [X₂]^½.
- Claiming proof. A matching rate law means “consistent with”, not “proven”.
Quick self-check
- A fast step has k₁ = 6.0 × 10³ M⁻¹ s⁻¹ and k₋₁ = 2.0 × 10⁵ s⁻¹. What is K₁? (0.030 M⁻¹)
- Rate = k[X₂]^½. By what factor does the rate change if [X₂] is multiplied by 9? (3)
- Fast: M + N ⇌ MN. Slow: MN → P + Q. Rate law? (rate = k[M][N], with k = k₂k₁ / k₋₁)
Next: practice questions.