Topic 11 · page 10 of 21

The rate-determining step and the rate equation

The slowest step in a mechanism controls the overall rate, so the species in the rate equation are the ones in that step. Use this to go from a mechanism to a rate equation, and back.

14 min read11.1(vii)11.7

Read first: Rate, rate equation and order, Orders from initial-rate data

The slowest step 11.1(vii)

Many reactions do not happen in one collision. They happen in a series of steps, and the balanced equation is the sum of those steps. The species formed in one step and used up in a later one are intermediates. They do not appear in the overall equation.

Think of a pipe with one narrow section. No matter how wide the rest is, the flow is set by the narrow part. In a reaction, the steps after the slow one cannot go faster than the intermediate arrives, and the steps before it just queue up.

A reaction profile with two humps. Enthalpy is on the vertical axis and reaction pathway on the horizontal axis. The reactants rise over a high first hump to an intermediate in a valley, then over a lower second hump down to the products, which are lower than the reactants. Arrows show a large activation energy for step 1 and a smaller one for step 2. Step 1 is labelled rate-determining step.
A two-step reaction. Step 1 has the larger activation energy, so it is the slow, rate-determining step.

The step with the largest activation energy is the slowest, and so is the rate-determining step.

From mechanism to rate equation 11.7

Worked exampleSlow step first

The reaction 2NO₂ + F₂ → 2NO₂F goes by two steps.

Step 1 (slow): NOX2+FX2→NOX2F+F\ce{NO2 + F2 -> NO2F + F}

Step 2 (fast): NOX2+F→NOX2F\ce{NO2 + F -> NO2F}

  1. The rate-determining step is step 1. It contains one NO₂ and one F₂.
  2. So rate = k[NO₂][F₂].
  3. F is an intermediate and does not appear in the rate equation.

If the slow step uses an intermediate, replace it by the substances that made it. A fast step before the slow one has time to settle, so the intermediate's concentration is proportional to the concentrations of what formed it.

Worked exampleSlow step second

The reaction 2NO + O₂ → 2NO₂ goes by two steps.

Step 1 (fast): 2 NO⇌NX2OX2\ce{2NO <=> N2O2}

Step 2 (slow): NX2OX2+OX2→2 NOX2\ce{N2O2 + O2 -> 2NO2}

  1. The slow step contains one N₂O₂ and one O₂, so the rate is proportional to [N₂O₂][O₂].
  2. N₂O₂ is an intermediate, so it cannot stay in the rate equation. It is made from two NO in the fast step, so [N₂O₂] is proportional to [NO]².
  3. So rate = k[NO]²[O₂], third order overall.

From rate equation to mechanism 11.7

The reasoning reverses. Each species in the rate equation is in the rate-determining step (or in a fast step before it), and its order is the number of particles of it that are involved.

Worked exampleTesting a mechanism against a rate equation

The reaction NO₂ + CO → NO + CO₂ has rate = k[NO₂]². Two mechanisms are proposed. Which is consistent?

Mechanism A, one step: NOX2+CO→NO+COX2\ce{NO2 + CO -> NO + CO2}

Mechanism B:

Step 1 (slow): 2 NOX2→NOX3+NO\ce{2NO2 -> NO3 + NO}

Step 2 (fast): NOX3+CO→NOX2+COX2\ce{NO3 + CO -> NO2 + CO2}

  1. In A the slow step is the only step. It would give rate = k[NO₂][CO], which does not match the rate equation.
  2. In B the slow step contains two NO₂ and no CO, so rate = k[NO₂]². This matches.
  3. Check that B adds up: 2NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂. Cancel NO₃ and one NO₂ from each side to get NO₂ + CO → NO + CO₂. So B is consistent.

Check yourself

  1. Define rate-determining step.
  2. Hydrogen peroxide decomposes in the presence of iodide ions by this mechanism. Step 1 (slow): H₂O₂ + I⁻ → H₂O + IO⁻. Step 2 (fast): H₂O₂ + IO⁻ → H₂O + O₂ + I⁻. Write the overall equation, give the rate equation and name the catalyst.
  3. A reaction between A, B and C has rate = k[A][B]. State what this tells you about the rate-determining step and where C reacts.
  4. For 2NO + 2H₂ → N₂ + 2H₂O the mechanism is: step 1 (fast) 2NO ⇌ N₂O₂; step 2 (slow) N₂O₂ + H₂ → N₂O + H₂O; step 3 (fast) N₂O + H₂ → N₂ + H₂O. Show that the steps add up to the overall equation and deduce the rate equation.
  5. Explain why a species can be in the balanced equation but not in the rate equation.
  6. In a two-step reaction, step 1 has an activation energy of 80 kJ mol⁻¹ and step 2 has 40 kJ mol⁻¹. Which is the rate-determining step, and why?

Answers cover these until you have tried the questions

  1. The slowest step in the mechanism of a reaction, which controls the overall rate.
  2. Adding the steps and cancelling IO⁻ and I⁻ gives 2H₂O₂ → 2H₂O + O₂ (overall). The slow step contains one H₂O₂ and one I⁻, so rate = k[H₂O₂][I⁻]. The catalyst is the iodide ion: it is used in step 1 and made again in step 2, so it is not in the overall equation.
  3. The rate-determining step contains one particle of A and one of B (or species made from them in an earlier fast step). C is zero order, so it is not in the slow step: it reacts in a fast step after it.
  4. Step 1: 2NO → N₂O₂. Step 2: N₂O₂ + H₂ → N₂O + H₂O. Step 3: N₂O + H₂ → N₂ + H₂O. Adding: 2NO + N₂O₂ + H₂ + N₂O + H₂ → N₂O₂ + N₂O + H₂O + N₂ + H₂O. Cancel N₂O₂ and N₂O: 2NO + 2H₂ → N₂ + 2H₂O. The slow step contains one N₂O₂ and one H₂. N₂O₂ is made from two NO, so [N₂O₂] ∝ [NO]², and rate = k[NO]²[H₂].
  5. It may react in a fast step after the rate-determining step. The rate is set only by the slow step, so changing the concentration of a species that is not in it does not change the rate, even though the species is used up overall.
  6. Step 1, because it has the higher activation energy, so fewer collisions succeed and it is the slowest step.