AP Chemistry · Lesson 3

Mechanisms, energy profiles and catalysts

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A mechanism is a series of elementary steps. The steps must add up to the overall equation. Unlike an overall equation, an elementary step's rate law can be written straight from its coefficients: for A + 2B → C as one step, rate = k[A][B]².

An intermediate is made in one step and used up in a later step. A catalyst is used in an early step and made again later. Neither appears in the overall equation.

The slowest step is the rate-determining step. If the slow step comes first, the predicted rate law is that step's rate law. If a fast, reversible step comes before the slow step, the slow step's rate law may contain an intermediate; replace the intermediate using the fast equilibrium, so the final rate law has only reactants (and any catalyst).

On a reaction energy profile, each peak is a transition state and each dip between peaks is an intermediate. The activation energy of a step is the height of its peak above the point where that step starts. ΔH is products minus reactants. A catalyst gives a new path with a lower activation energy. It speeds the forward and reverse reactions and does not change ΔH.

Worked example Easy

A proposed mechanism for the reaction X₂ + 2Y → 2XY is: Step 1 (slow): X₂ + Y → XY + X. Step 2 (fast): X + Y → XY.

Which rate law does this mechanism predict?

  1. rate = k[X₂][Y]²
  2. rate = k[X₂][Y]Answer
  3. rate = k[X][Y]
  4. rate = k[X₂]

How to solve it

  1. The slow step is first, so the rate law comes from step 1.
  2. Step 1 is elementary: X₂ + Y means rate = k[X₂][Y].
  3. Check: adding the steps gives X₂ + 2Y → 2XY, and X cancels as an intermediate.

Why each choice is right or wrong

  • A. Incorrect. This rate law copies the overall equation's coefficients; the slow step has only one Y, so the predicted rate law is first order in Y.
  • B. Correct. Step 1 is slow and comes first, so its rate law, written from its coefficients, sets the rate: rate = k[X₂][Y].
  • C. Incorrect. This is the rate law of step 2, the fast step, and it contains X, an intermediate that is not a reactant.
  • D. Incorrect. This leaves out Y, but Y is one of the two particles that collide in the slow first step.

Trap Intermediate or catalyst?

Look at where a species appears first. An intermediate is made first (a product of an early step) and used later. A catalyst is used first (a reactant of an early step) and made again later. Both cancel out of the overall equation.

Explain Explain it

Task verbs: explain, evaluate

To evaluate a mechanism, check two things: (1) the steps add up to the overall equation, and (2) the rate law the mechanism predicts matches the rate law found by experiment. A mechanism that passes both tests is consistent with the data; it is not proved, because another mechanism might also pass.

Worked example Medium

A proposed mechanism for the reaction A + 2B → AB₂ is: Step 1 (fast, reversible): A + B ⇌ AB. Step 2 (slow): AB + B → AB₂.

Which rate law does this mechanism predict?

  1. rate = k[AB][B]
  2. rate = k[A][B]
  3. rate = k[A][B]²Answer
  4. rate = k[A]²[B]

How to solve it

  1. The slow step gives rate = k₂[AB][B], but AB is an intermediate.
  2. Step 1 is a fast equilibrium: [AB] = K[A][B].
  3. Substitute: rate = k₂K[A][B][B] = k[A][B]².

Why each choice is right or wrong

  • A. Incorrect. This is the slow step's rate law, but AB is an intermediate; it must be replaced using the fast equilibrium of step 1.
  • B. Incorrect. This is the rate law of step 1, the fast step, which does not limit the overall rate.
  • C. Correct. Rate = k₂[AB][B], and the fast equilibrium gives [AB] = K[A][B], so rate = k[A][B]².
  • D. Incorrect. This makes A second order, but only one A enters the mechanism, in step 1, so A must be first order.

Explain Your turn

A small amount of a catalyst is added to a reaction mixture, and the reaction speeds up. Explain, in terms of a reaction energy profile, how the catalyst speeds up the reaction, and say whether it changes the ΔH of the reaction.

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Compare with a model answer

A full answer does this

  • Says the catalyst gives a different pathway with a lower activation energy
  • Connects the lower activation energy to a larger fraction of collisions having enough energy to react
  • Says ΔH does not change

Model answer

The catalyst gives the reaction a different pathway, with a lower peak on the energy profile, so the activation energy is lower. With a lower barrier, a larger fraction of the colliding particles have at least the needed energy, so more collisions succeed each second and the rate rises. ΔH does not change, because the reactants and products sit at the same energies with or without the catalyst.

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