Short on time? This page is the recap. For explanations, the flow chart and worked examples, use the full study guide.
Recap
- In a mechanism, one step is usually much slower than the rest. This rate-limiting step sets the rate of the whole reaction.
- When the first step is the slow step, the rate law of the reaction is the rate law of that elementary step: each species in it is raised to its coefficient in the step.
- Reactants that react only in fast steps after the slow step do not appear in the rate law (zero order).
- Intermediates never appear in a rate law compared with experiment. Catalysts can, if they react in the slow step.
- A mechanism is consistent with the evidence if it adds up to the overall equation, has reasonable steps, and predicts the experimental rate law. Consistent is not the same as proven.
Key relationships
| Idea | What to write or do | Example (fictional) |
|---|---|---|
| Elementary step rate law | aA + bB → … gives rate = k[A]ᵃ[B]ᵇ | 2D → D₂ gives rate = k[D]² |
| Slow first step | Rate law of the reaction = rate law of step 1 | Step 1 (slow) X + Y₂ → … gives rate = k[X][Y₂] |
| Overall order | Sum of the exponents | k[A][B] is second order overall |
| Units of k (rate in M s⁻¹) | First order: s⁻¹; second order: M⁻¹ s⁻¹ | k[A][B]: M⁻¹ s⁻¹ |
| Effect of changing [A] by a factor f | Rate changes by fⁿ, where n is the order in A | Order 2, triple [A]: × 9 |
Assumptions behind the rule
- The steps are elementary, so the slow step’s coefficients show the particles colliding.
- The slow step is much slower than the steps after it, so the later steps do not hold up the reaction.
- The steps go one way only (single arrows). A fast reversible step before the slow step needs Topic 5.9.
- The slow step is the first step. If it is not, use the approximation in Topic 5.9.
Mistakes to avoid
- Using the overall coefficients for the rate law instead of the slow step’s.
- Including a reactant from a fast later step. It is zero order.
- Putting an intermediate in the final rate law.
- Leaving out a catalyst that reacts in the slow step.
- Saying a consistent mechanism is “proven”.
- Forgetting units of k, which depend on the overall order.
Quick self-check
- Step 1 (slow): J₂ + L → J₂L. Step 2 (fast): J₂L + L → J₂ + L₂. What is the rate law? (rate = k[J₂][L])
- In question 1, what role does J₂ play, and what happens to the rate if [L] is doubled? (J₂ is a catalyst; the rate doubles)
- Two mechanisms both predict the experimental rate law and both add up. Is either proven? (No. Both are consistent; more evidence, such as a detected intermediate, is needed.)
Next: practice questions.