Reaction Mechanisms & Catalysis

You’ll be able to

Describe a reaction as a sequence of elementary steps and identify the rate-determining step
Distinguish an intermediate from a catalyst in a mechanism
Explain how a catalyst speeds a reaction by lowering the activation energy

Reactions happen in steps

Most reactions do not occur in a single collision. The overall equation is really a sum of elementary steps — the actual molecular events. This sequence is the reaction mechanism. Add the elementary steps together and they must reproduce the overall balanced equation, with any species made and then consumed canceling out.

The rate-determining step sets the pace

A mechanism moves only as fast as its slowest elementary step — the rate-determining step (RDS), the bottleneck. Because an elementary step reacts exactly as written, its rate law can be taken straight from its molecularity (its own coefficients). So the overall rate law reflects the RDS, which is why experimental orders need not match the overall equation.

Intermediates vs. catalysts

Two species look similar but are opposites in timing. An intermediate is produced in an early step and consumed later — it does not appear in the overall equation and starts at zero. A catalyst is added at the start (consumed early) and regenerated later — it appears among the reactants of an early step and is remade, so it too cancels from the overall equation but is present from the beginning.

Spotting each species in a mechanism

Intermediate: appears first as a product, then a reactant. Catalyst: appears first as a reactant, then a product.

Both cancel out of the overall equation — the order in which they appear (made-then-used vs. used-then-remade) is what tells them apart.

Worked example

A proposed two-step mechanism is: Step 1 (fast): NO + NO ⇌ N₂O₂ Step 2 (slow): N₂O₂ + O₂ → 2 NO₂ Identify the overall reaction, the intermediate, and the rate-determining step.

1.Add the steps: NO + NO + N₂O₂ + O₂ → N₂O₂ + 2 NO₂.
2.Cancel species on both sides: N₂O₂ appears as a product in Step 1 and a reactant in Step 2, so it cancels.
3.Overall equation: 2 NO + O₂ → 2 NO₂.
4.N₂O₂ is made first and then consumed → it is an intermediate. Nothing is present at the start and regenerated, so there is no catalyst here.
5.The slow step (Step 2) is the rate-determining step; it sets the overall rate.

Answer: Overall: 2 NO + O₂ → 2 NO₂; intermediate = N₂O₂; RDS = Step 2 (the slow step)

Checkpoint

In the mechanism Step 1: A + B → C (C then reacts in Step 2: C + A → D), the species C is best described as:

How a catalyst actually works

A catalyst provides an alternative pathway with a lower activation energy. On a reaction-energy diagram, it lowers the height of the Eₐ barrier (often via a new intermediate) without touching the energies of reactants or products — so ΔH is unchanged. A lower Eₐ means a larger fraction of collisions succeed, speeding both forward and reverse reactions equally; the catalyst is regenerated, so it is never used up.

Watch out

A catalyst changes the kinetics, not the thermodynamics. It lowers Eₐ and speeds the approach to equilibrium, but it does not change ΔH, the position of equilibrium, or K. Anything that alters ΔH or the amount of product is not a catalytic effect.

Checkpoint

A catalyst increases the rate of a reaction by:

On the exam

To recover the RDS from data: if the experimental rate law is rate = k[A]²[B], the slow step must consume 2 A and 1 B (or an equivalent set once an intermediate is expressed via a fast equilibrium). Mechanism, RDS, and rate law must all agree.

Answer the 2 checkpoints as you read.