IB Chemistry SL & HL Topic 5 — The Rate of Chemical Change Paper 1 & 2 Core skill ~10 min read

Energy Profiles With and Without Catalysts

A catalyst does not shove the reactants over the hill. It shows them a lower path around it. Once you picture it that way, every catalyst question on the paper — including the tricky ones about ΔH and yield — answers itself.

📘 What you need to know

What the profile looks like

Draw both routes on the same axes and the picture is very clear. Two humps, two different heights, but they start and finish at exactly the same levels.

Two routes, one destination The catalyst lowers the hill, not the finishing level no catalyst with catalyst bigger Ea smaller Ea ΔHreactants products one ΔH arrow serves both routes, because both start and end togetherA catalyst changes how fast, never how far.
The single ΔH arrow is the whole point of this diagram. Both curves leave the same reactants line and land on the same products line, so the energy released is identical either way.
Say it carefully. The catalysed route has its own lower activation energy — the original pathway still exists with its original Ea. That is why the accepted wording is “provides an alternative pathway of lower activation energy” rather than “lowers the activation energy”.

How a catalyst actually pulls it off

The profile tells you what happens but not how. For a solid catalyst working on gases, the trick is that the reactants stick to the surface, and being stuck there weakens their bonds before they even meet each other.

A solid catalyst at work, in three stages Stick, react, leave — and the surface is ready again 1. STICK ON catalyst surface bonds weakened 2. REACT catalyst surface held in place, new bonds form 3. LEAVE catalyst unchanged product floats offThe surface holds the reactants still and weakens their bonds. Nothing is consumed, so one small piece of catalyst can process a huge amount.
Two jobs are being done at once here: the reactants are held in the right orientation, and their bonds are pre-weakened. Both make a successful collision far more likely.
This picture also explains why surface area matters so much for solid catalysts, and why they are usually spread thinly over a honeycomb support — more surface means more places for the reactants to stick.

Homogeneous and heterogeneous

The words look intimidating but the split is simple: is the catalyst in the same physical state as the reactants, or not?

TypeMeaningExamplePractical note
HomogeneousCatalyst is in the same phase as the reactantsAn acid catalyst in solution with dissolved reactantsMixes perfectly, but can be awkward to separate from the product afterwards
HeterogeneousCatalyst is in a different phase from the reactantsA solid metal catalyst with gaseous reactantsEasy to separate and reuse, so it suits continuous industrial processes

Where you meet them

Why catalysts matter beyond the exam

A lower activation energy means the reaction runs acceptably fast at a lower temperature. That has three knock-on effects industry cares about a great deal:

Rate versus yield, one more time. A catalyst gets you to equilibrium faster, but it speeds up the forward and reverse reactions by the same factor. The position of equilibrium, and therefore the yield, does not move. If a question asks how to increase the yield, a catalyst is the wrong answer.

Worked examples

WORKED EXAMPLE

Identifying the arrows on a profile

A profile shows an exothermic reaction with and without a catalyst. Arrow p runs from the reactants level to the higher peak, arrow q from the reactants level to the lower peak, and arrow r from the reactants level down to the products level. Which arrow is ΔH, and which is Ea for the catalysed reaction?

Step 1: ΔH connects reactants and products That is arrow r — it ignores both peaks. Step 2: the catalysed route is the lower peak Its Ea runs from reactants up to that lower peak, which is arrow q. ΔH = r, and Ea(catalysed) = q arrow p is the uncatalysed Ea — the biggest of the three
WORKED EXAMPLE

Working out both barriers

Without a catalyst, a reaction has Ea = 185 kJ mol−1. A catalyst lowers this to 95 kJ mol−1. The reaction has ΔH = −70 kJ mol−1. Find Ea for the reverse reaction on the catalysed route, and state the value of ΔH with the catalyst present.

Step 1: reverse barrier on the catalysed route Ea(reverse) = Ea(forward) − ΔH = 95 − (−70) = 165 kJ mol⁻¹ Step 2: ΔH with the catalyst still −70 kJ mol⁻¹ Ea(reverse, catalysed) = 165 kJ mol⁻¹; ΔH unchanged at −70 kJ mol⁻¹ the catalyst lowered both barriers by the same 90 kJ mol⁻¹, which is why ΔH cannot move
WORKED EXAMPLE

Explaining why ΔH does not change

Explain why adding a catalyst changes the rate of a reaction but not its enthalpy change. (2 marks)

Mark 1: what the catalyst does It provides an alternative pathway with a lower activation energy, so a greater proportion of collisions is successful and the rate increases. Mark 2: what it does not do ΔH depends only on the energies of the reactants and products, and the catalyst changes neither — it only alters the route between them. “only the route changes, not the start or the finish” is the sentence to remember

💡 Exam tip

⚠ Common mix-up

Up next: Maxwell-Boltzmann Distributions — the graph that shows exactly which particles can get over the hill, and the clearest way to explain both temperature and catalysts in one picture.

Want this explained one-to-one?

Book a free session with an experienced IB Chemistry tutor and get your trickiest topics made simple.

Book a Free Session →