5.11 Catalysis

Catalyst Cycle Tracker

Watch catalysts bind reactants in Step 1 and release products in Step 2 across 3 cycles. Counters prove Δ[catalyst] = 0 while intermediate cancels overall.

3 Real CyclesConsumed & RegeneratedIntermediateLowers Eₐ
Topic 5.11

Catalysis

Explain the relationship between the effect of a catalyst on a reaction and changes in the reaction mechanism.

For a catalyst to increase the rate of a reaction, its addition must increase the number of effective collisions and/or provide a reaction path with a lower activation energy relative to the original reaction coordinate.

The key insight: a catalyst does not lower the activation energy of the existing pathway. It opens an entirely new mechanism with a lower barrier.

EK 5.11.A.2 describes catalyst bookkeeping: in a mechanism containing a catalyst, the net concentration of the catalyst is constant. However, the catalyst is frequently consumed in the rate-determining step and regenerated in a later step. That means a catalyst can legitimately appear in the experimental rate law — a fact that surprises many students.

The CED names three catalytic modes:

  • Binding catalysis: the catalyst binds the reactant(s), which are then oriented more favorably or react with lower activation energy. A new intermediate forms in which the catalyst is bound to the reactant. Many enzymes function this way.
  • Covalent/acid–base catalysis: covalent bonding between catalyst and reactant. Acid–base catalysis is the example — a reactant or intermediate gains or loses a proton, introducing new intermediates and new elementary reactions.
  • Surface catalysis: a reactant or intermediate adsorbs onto a solid surface, which weakens bonds and holds reactants in favorable orientations. Catalytic converters work this way.

What a catalyst does NOT change: ΔH, ΔG, ΔS, K, or the position of equilibrium. It lowers Ea equally for the forward and reverse directions, so both rates increase by the same factor — the system reaches the same equilibrium, just sooner.

Key points

  • A catalyst provides a new pathway with lower Ea; it does not lower the barrier of the old one.
  • Forward and reverse rates increase equally, so K and the equilibrium position are unchanged.
  • Net catalyst concentration is constant, but a catalyst may appear in the rate law.
  • Consumed early and regenerated later distinguishes a catalyst from an intermediate.

Common mistakes

  • "A catalyst shifts the equilibrium toward products." It does not. This is one of the most heavily penalized misconceptions in the course.
  • "A catalyst lowers the activation energy of the reaction." Imprecise — it provides an alternative pathway with a lower Ea.
  • "A catalyst changes ΔH." Never. Reactant and product energies are fixed.
  • "A catalyst cannot appear in the rate law." False — it can, if it participates in or before the RDS.

Worked example

The decomposition 2 H₂O₂ → 2 H₂O + O₂ is catalyzed by iodide ion: Step 1: H₂O₂ + I⁻ → H₂O + IO⁻ Step 2: H₂O₂ + IO⁻ → H₂O + O₂ + I⁻ (a) Show that I⁻ is a catalyst and identify the intermediate. (b) Explain why the catalyzed reaction is faster. (c) State the effect on the amount of O₂ ultimately produced.

(a) Summing: 2 H₂O₂ + I⁻ + IO⁻ → 2 H₂O + O₂ + IO⁻ + I⁻. Both I⁻ and IO⁻ cancel, giving 2 H₂O₂ → 2 H₂O + O₂ ✓

I⁻ is a catalyst: it is consumed in step 1 and regenerated in step 2, so its net concentration is unchanged.
IO⁻ is an intermediate: it is produced in step 1 and consumed in step 2.

(b) Iodide provides an alternative mechanism whose highest activation barrier is lower than the barrier of the uncatalyzed single-step decomposition. At a given temperature, a lower Ea means a larger fraction of collisions on the Maxwell–Boltzmann distribution has sufficient energy to reach the transition state, so more collisions are effective per second and the rate increases.

(c) No effect on the amount of O₂ produced. A catalyst lowers the barrier equally in both directions and does not change the energies of reactants or products. ΔH, ΔG, and K are all unchanged, so the final amount of O₂ is identical — it is simply produced sooner.

Full notes for topic 5.11 →