Every car built since the 1970s carries a thin layer of platinum and palladium in its exhaust pipe. Hundreds of thousands of kilometres of toxic gases pass through it — and the platinum is still there, completely unchanged. How can something that is never used up help a reaction happen faster? And if it doesn't alter the products or reactants, what exactly does it change?
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A catalytic converter on a modern car contains a thin honeycomb of platinum and palladium. Every day, highly toxic gases (carbon monoxide, unburnt hydrocarbons, nitrogen oxides) flow through it at high temperature. They are converted to carbon dioxide, water, and nitrogen — far less harmful products. After 200,000 km, the platinum is still there, completely intact.
Two common claims about catalysts:
Which claim is correct? Which is false? Commit to an answer now, before any theory is shown.
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
📚 Core Content
A catalyst provides an alternative reaction pathway with a lower activation energy, increasing the proportion of collisions that have enough energy to react — without being consumed in the process.
A catalyst works by providing an alternative mechanism — a different sequence of bond-breaking and bond-forming steps, each of which requires less energy at each stage. On an energy profile diagram, the catalysed pathway shows a lower peak (lower Ea) but the same starting point (reactants) and the same ending point (products).
What a catalyst does and does not change:
| Property | With catalyst | Without catalyst |
|---|---|---|
| Ea (forward) | Lower | Higher |
| Ea (reverse) | Lower (same reduction) | Higher |
| ΔH | Unchanged | Same |
| Reactants | Unchanged | Same |
| Products | Unchanged | Same |
| Reaction rate | Faster | Slower |
| Catalyst consumed? | No — regenerated | Not applicable |
Catalysts are classified by whether they are in the same physical state (phase) as the reactants — homogeneous — or a different phase — heterogeneous.
| Catalyst | Type | Reaction | Phase of catalyst → reactants |
|---|---|---|---|
| Pt, Pd, Rh | Heterogeneous | Catalytic converter: CO + hydrocarbons + NOx → CO₂, N₂, H₂O | Solid → gas |
| Fe | Heterogeneous | Haber process: N₂(g) + 3H₂(g) → 2NH₃(g) | Solid → gas |
| H⁺(aq) | Homogeneous | Esterification of CH₃COOH(aq) + C₂H₅OH(aq) | Aqueous → aqueous |
| MnO₄⁻(aq) | Homogeneous | Oxidation reactions in aqueous solution | Aqueous → aqueous |
A single diagram can show both pathways — two peaks at different heights, but the same reactant level, the same product level, and the same ΔH arrow.
Rules for drawing a correct catalysed comparison
The catalytic converter in a modern car contains a ceramic honeycomb coated with platinum, palladium, and rhodium. Exhaust gases (CO, unburnt hydrocarbons, NOx) pass over the solid metal surface at high temperature. The solid catalyst provides a surface for the gases to adsorb onto, lowering the Ea for reactions that convert these toxic species to CO₂, H₂O, and N₂. The platinum is heterogeneous — solid phase, gaseous reactants. After 200,000 km, the platinum is still there: not consumed, not altered. ΔH for each reaction is unchanged by the catalyst — the converter does not affect how much energy the exhaust reactions release. This concept appears again in Short Answer Q3 below.
Misconception 1: "A catalyst changes ΔH — it makes the reaction release more energy."
Why students think this: Faster reaction = more energy released, so the catalyst must be adding something. The logic seems intuitive.
What is actually true: A catalyst only lowers Ea. ΔH depends on the energy of reactants vs products — and since the catalyst leaves both unchanged, ΔH is identical with or without the catalyst. The reaction releases the same total energy; it just releases it faster.
Misconception 2: "A catalyst provides energy to the reactants so they can react."
Why students think this: "Lowering the energy barrier" sounds like the catalyst is supplying energy to help molecules get over it.
What is actually true: The catalyst provides an alternative mechanism with a lower barrier. Molecules still need thermal energy from collisions to reach the transition state — but that threshold is now lower, so a greater proportion of collisions succeed. The catalyst does not supply energy; it lowers the requirement.
Misconception 3: "Spontaneous reactions don't need a catalyst — they already happen."
Why students think this: "Spontaneous" is interpreted as "fast" or "already occurring."
What is actually true: Spontaneous means ΔG < 0 — thermodynamically favourable. It says nothing about rate. Diamond → graphite is spontaneous but needs geological timescales because Ea is enormous. Catalysts are used to make spontaneous reactions occur at a useful rate — they don't change spontaneity, they change speed.
🔬 Worked Examples
A reaction has an uncatalysed Ea of 180 kJ mol⁻¹ and ΔH = −75 kJ mol⁻¹. In the presence of a catalyst, Ea drops to 95 kJ mol⁻¹. (a) What is Ea of the reverse uncatalysed reaction? (b) What is ΔH for the catalysed reaction? (c) What is Ea of the catalysed reverse reaction?
The Haber process reacts N₂(g) and H₂(g) over an iron (Fe) catalyst to produce NH₃(g). In a second reaction, H⁺(aq) ions catalyse the reaction between CH₃COOH(aq) and C₂H₅OH(aq) to form an ester. Identify whether each catalyst is homogeneous or heterogeneous, justify your answer, and state what effect each catalyst has on ΔH for its reaction.
🧪 Activities
a Error 1 (marked ①): Ea(uncatalysed) is drawn from the x-axis to the peak. What is wrong and what should it show?
b Error 2 (marked ②): The product level for the catalysed pathway is drawn lower than the product level for the uncatalysed pathway. Why is this wrong?
c Error 3 (marked ③): The ΔH arrow is drawn from the reactant level to the peak. What is wrong and where should ΔH be drawn?
d Error 4 (marked ④): Ea(catalysed) is drawn from the product level upward to the catalysed peak. What is wrong?
Type your full Spot + Fix responses before revealing answers:
Answer in your workbook, then reveal to check.
1 "A catalyst increases ΔH for the reaction, making it release more energy per mole."
2 "Platinum in a catalytic converter is a heterogeneous catalyst because it is a solid and the exhaust gases are in the gas phase — they are in different physical states."
3 "Adding a catalyst to an endothermic reaction will make it become exothermic, since less energy is needed to start the reaction."
4 "A catalyst lowers the activation energy of both the forward and reverse reactions by the same amount."
5 "H⁺(aq) is a heterogeneous catalyst in an aqueous esterification reaction because it is an ion, not a molecule."
6 "A catalyst is not consumed in the overall reaction — it is regenerated at the end of the mechanism and does not appear in the balanced overall equation as a reactant."
Type your True/False responses and corrections below:
Answer in your workbook.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Wrong: A catalyst changes the enthalpy change (ΔH) of a reaction.
Right: A catalyst provides an alternative reaction pathway with lower activation energy but does not change the reactant or product energy levels. ΔH remains unchanged because it depends only on the difference between product and reactant enthalpies, not the pathway taken.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. A reaction has the following energy data: Ea(forward, uncatalysed) = 210 kJ mol⁻¹; ΔH = +55 kJ mol⁻¹. A catalyst reduces Ea by 80 kJ mol⁻¹.
(a) Is this reaction exothermic or endothermic? (1 mark)
(b) Calculate Ea(reverse, uncatalysed). (1 mark)
(c) Calculate Ea(forward, catalysed) and Ea(reverse, catalysed). (2 marks)
(d) What is ΔH for the catalysed reaction? Explain. (1 mark)
5 MARKS
Type your answer — show all working:
Show full working in your workbook.
7. The industrial production of sulfuric acid uses vanadium(V) oxide (V₂O₅) as a catalyst in the Contact process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). V₂O₅ is a solid.
(a) Classify V₂O₅ as a homogeneous or heterogeneous catalyst. Justify your answer by referring to physical states. (2 marks)
(b) Explain, using the concept of activation energy, how V₂O₅ increases the rate of production of SO₃. (2 marks)
(c) A student suggests that using a catalyst will produce more SO₃ at equilibrium. Is this correct? Explain. (1 mark)
5 MARKS
Type your full answer:
Answer in your workbook.
8. A catalytic converter in a car converts toxic exhaust gases to less harmful products. The converter contains platinum (Pt), palladium (Pd), and rhodium (Rh) embedded on a ceramic honeycomb surface.
(a) Identify the physical state of the catalyst and the physical state of the exhaust gases. Hence classify the catalytic converter as using a homogeneous or heterogeneous catalyst. (2 marks)
(b) The uncatalysed combustion of CO has Ea = 232 kJ mol⁻¹. The catalytic converter reduces this to 75 kJ mol⁻¹. Use the concept of activation energy to explain why the catalytic converter is effective at converting CO at typical exhaust temperatures (400–600°C), but a car without a converter rarely oxidises CO completely at these temperatures. (3 marks)
(c) Explain why the catalytic converter does not alter the total energy released by the combustion reactions in the exhaust. (1 mark)
6 MARKS
Type your answer below:
Answer in your workbook.
Go back to your Think First response. You were asked which of two claims about catalysts was correct:
Type your reflection below:
Write your reflection in your book.
Error 1: Ea(uncatalysed) drawn from the x-axis. Fix: draw from the reactant energy level to the peak. Ea = height of the peak above the reactant level, not above zero.
Error 2: Catalysed product level drawn lower than uncatalysed. Fix: both pathways must end at exactly the same product energy level. A catalyst does not change ΔH or the identity of products.
Error 3: ΔH arrow goes from reactant level to the peak. Fix: ΔH arrow must go from reactant level to product level. The peak is the transition state, not the product.
Error 4: Ea(catalysed) measured from the product level upward. Fix: for the forward reaction's Ea, the arrow must start at the reactant level and end at the catalysed peak. From the product level would give Ea of the reverse catalysed reaction.
1. FALSE — Catalyst does not change ΔH; total energy per mole is the same with or without catalyst.
2. TRUE — Correctly classifies Pt as heterogeneous; gives correct reasoning (solid vs gas phase).
3. FALSE — ΔH is unchanged; endothermic stays endothermic. Catalyst lowers Ea only.
4. TRUE — Catalyst lowers the peak by the same amount regardless of direction, so both Ea(forward) and Ea(reverse) decrease equally.
5. FALSE — H⁺(aq) is homogeneous (same aqueous phase as reactants); catalyst type is about phase, not ionic/molecular nature.
6. TRUE — Correctly defines catalyst: regenerated, not consumed, not in overall balanced equation as reactant.
1. C — Catalyst lowers Ea; ΔH is unchanged. Options A and D are false (Ea does not increase; ΔH does not change). Option B falsely states ΔH becomes more negative.
2. B — Ea(reverse) = Ea(forward) − ΔH = 145 − (−60) = 145 + 60 = 205 kJ mol⁻¹. For an exothermic reaction, products are lower, so the reverse reaction must climb higher from the product level to the same peak.
3. B — Heterogeneous classification requires different physical states: Pt is solid, exhaust gases are gaseous. Option C (not consumed) describes the definition of a catalyst generally, but does not explain why it is heterogeneous specifically.
4. A — Ea = difference between reactant level and transition state peak. Option B describes Ea of the reverse reaction. Option C is the common error of measuring from the x-axis. Option D confuses Ea with ΔH.
5. D — The correct explanation: ΔH depends on reactant and product energies, which are unchanged by a catalyst. Whether a reaction is exothermic or endothermic is independent of Ea. A catalyst can lower Ea for an endothermic reaction — it remains endothermic.
Q6 (5 marks):
(a) ΔH = +55 kJ mol⁻¹ → positive → endothermic [1].
(b) Ea(reverse, uncat) = 210 − (+55) = 155 kJ mol⁻¹ [1]. (Products are 55 kJ mol⁻¹ higher than reactants; the reverse reaction starts from a higher level and needs less energy to reach the same peak.)
(c) Ea(forward, cat) = 210 − 80 = 130 kJ mol⁻¹ [1]; Ea(reverse, cat) = 130 − 55 = 75 kJ mol⁻¹ [1]. (The same reduction of 80 kJ mol⁻¹ applies to the peak; the reverse Ea is measured from the product level to the new lower peak.)
(d) ΔH(catalysed) = +55 kJ mol⁻¹ — unchanged. A catalyst does not change ΔH because reactants and products are identical to the uncatalysed reaction; only the pathway (and hence Ea) is altered [1].
Q7 (5 marks):
(a) V₂O₅ is a solid; SO₂ and O₂ are gases [½]. Solid ≠ gas → different physical states → heterogeneous catalyst [1 + ½].
(b) V₂O₅ provides an alternative reaction mechanism with a lower activation energy than the uncatalysed pathway [1]. At operating temperature, a greater proportion of SO₂ and O₂ molecules now possess energy ≥ Ea(catalysed), so the frequency of successful collisions increases, increasing the rate of SO₃ production [1].
(c) No — the student is incorrect [½]. A catalyst changes the rate at which equilibrium is reached but does not change the position of equilibrium (the ratio of products to reactants at equilibrium). The equilibrium amount of SO₃ depends on ΔG and temperature, not on whether a catalyst is present [½].
Q8 (6 marks):
(a) Catalyst (Pt, Pd, Rh) = solid. Exhaust gases (CO, hydrocarbons, NOx) = gas [½]. Solid ≠ gas → heterogeneous catalyst [1 + ½].
(b) Without converter: Ea = 232 kJ mol⁻¹. At 400–600°C, a relatively small proportion of CO molecules possess ≥ 232 kJ mol⁻¹ of kinetic energy, so the uncatalysed oxidation rate is low [1]. With converter: the Pt surface provides an alternative mechanism (adsorption, surface reaction, desorption) with Ea = 75 kJ mol⁻¹ [1]. At the same exhaust temperature, a much greater proportion of molecules now have sufficient energy — so the rate of CO oxidation increases dramatically and conversion is essentially complete before the gas exits the converter [1].
(c) ΔH depends on the energies of the reactants and products, which are identical with or without the catalyst [½]. The catalyst only alters the pathway (Ea), not the starting or ending energy levels — so total energy released is unchanged [½].
Put your knowledge of Activation Energy, Catalysts & Energy Diagrams to the test. Answer correctly to deal damage — get it wrong and the boss hits back. Pool: lessons 1–5.
Tick when you've finished all activities and checked your answers.