Activation Energy, Catalysts & Energy Diagrams
In 1975, the U.S. Environmental Protection Agency mandated catalytic converters on all new passenger vehicles. Ford and General Motors engineers specified a platinum–palladium–rhodium washcoat just 40 µm thick deposited on a ceramic honeycomb, presenting 2·0 m² of catalyst surface per gram. At exhaust temperatures of 400–600°C, this surface reduces CO oxidation activation energy from ~140 kJ mol⁻¹ to ~75 kJ mol⁻¹ — achieving over 99% conversion efficiency. The ΔH of combustion is unchanged; only Ea is lowered.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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:
- A catalyst makes a reaction release more energy — it improves the energetics of the reaction.
- A catalyst lowers the amount of energy the reactants need to possess before they can successfully react.
Which claim is correct? Which is false? Commit to an answer now, before any theory is shown.
Key Facts
- Ea = minimum energy for a collision to result in reaction
- A catalyst lowers Ea — it does NOT change ΔH, reactants, or products
- Homogeneous catalyst: same phase as reactants; heterogeneous: different phase
Concepts
- Why Ea and ΔH are independent quantities (kinetics vs thermodynamics)
- How a catalyst provides an alternative mechanism with lower energy steps
- Why a catalyst lowers Ea of the reverse reaction equally
Skills
- Draw and interpret catalysed vs uncatalysed energy profile diagrams
- Calculate Ea(reverse) from Ea(forward) and ΔH
- Classify a catalyst as homogeneous or heterogeneous with justification
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 |
A catalyst provides an alternative reaction pathway with a lower activation energy (Ea), increasing the proportion of successful collisions — without being consumed. On an energy profile diagram the catalysed pathway shows a lower peak; ΔH is unchanged.
Pause — copy the highlighted definition into your book before moving on.
Quick check: A catalyst is added to an exothermic reaction. Which of the following is correct?
We just saw that catalysts lower Ea without changing ΔH. That raises a question: are all catalysts the same, or do they work by fundamentally different mechanisms? This card answers it → by distinguishing homogeneous (same phase) from heterogeneous (different phase) catalysts.
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 |
Homogeneous catalysts are in the same phase as the reactants (e.g. H⁺(aq) in an aqueous reaction); heterogeneous catalysts are in a different phase (e.g. solid Pt catalysing a gas-phase reaction). Heterogeneous catalysts work by adsorbing reactants onto their surface.
Add the highlighted point to your notes before the check below.
Explain it: Vanadium(V) oxide (V₂O₅) is used as a catalyst in the Contact process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). V₂O₅ is a solid. In two sentences, classify V₂O₅ as homogeneous or heterogeneous and justify your answer.
We just saw that homo- and heterogeneous catalysts both lower Ea by different mechanisms. That raises a question: how do we show both the catalysed and uncatalysed pathways on a single energy profile diagram? This card answers it → with two peaks at different heights sharing the same reactant and product levels.
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:
- Draw both curves starting from the same reactant energy level
- Draw both curves ending at the same product energy level
- Uncatalysed curve: taller peak (higher Ea)
- Catalysed curve: lower peak (lower Ea), sometimes shows multiple smaller humps (multi-step mechanism)
- Label Ea(uncatalysed) and Ea(catalysed) from the reactant level to each respective peak
- Draw the ΔH arrow from the reactant level to the product level — identical for both pathways
A catalysed energy profile diagram shows two pathways: uncatalysed (higher peak, larger Ea) and catalysed (lower peak, smaller Ea). Both share identical reactant and product levels and the same ΔH — only activation energy differs.
Pause — write the highlighted definition into your book.
Fill in the blanks: Complete the sentences about catalyst diagrams.
Both the catalysed and uncatalysed curves must start at the same ___ level and end at the same ___ level, because a catalyst does not change ___. The only difference is the height of the ___, which is lower for the ___ pathway.
Worked examples · reveal as you go
Interpreting a catalysed energy profile diagram. 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?
GIVEN: Ea(forward, uncatalysed) = 180 kJ mol⁻¹ | ΔH = −75 kJ mol⁻¹ | Ea(forward, catalysed) = 95 kJ mol⁻¹
FIND: (a) Ea(reverse, uncat) | (b) ΔH(catalysed) | (c) Ea(reverse, catalysed)
Ea(reverse) = Ea(forward) − ΔH = 180 − (−75) = 180 + 75 = 255 kJ mol⁻¹
ΔH(catalysed) = −75 kJ mol⁻¹ (unchanged)
Ea(reverse, cat) = Ea(forward, cat) − ΔH = 95 − (−75) = 95 + 75 = 170 kJ mol⁻¹
Final Answers: (a) 255 kJ mol⁻¹ | (b) −75 kJ mol⁻¹ (unchanged) | (c) 170 kJ mol⁻¹
Identifying homogeneous vs heterogeneous catalysts. 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.
GIVEN: Fe catalyst, N₂(g) + H₂(g) reactants | H⁺(aq) catalyst, CH₃COOH(aq) + C₂H₅OH(aq) reactants
FIND: Catalyst type for each; effect on ΔH
Fe is a solid; N₂ and H₂ are gases. Solid ≠ gas → different phases → heterogeneous catalyst.
H⁺ is an aqueous ion; both reactants are in aqueous solution. Aqueous = aqueous → same phase → homogeneous catalyst.
Neither catalyst changes ΔH. For the Haber process, ΔH = −92 kJ mol⁻¹ with or without the Fe catalyst. For esterification, ΔH is unchanged by H⁺. Catalysts only lower Ea; they never alter the enthalpy of products or reactants.
Put the steps for reading activation energy from an energy profile diagram in the correct order.
- Measure the energy difference from the reactants level to the peak (transition state).
- Identify the reactants energy level on the y-axis.
- Record this as Ea — always a positive value.
- Identify the transition state at the highest point of the curve.
- Note that ΔH is measured separately: products level minus reactants level.
Key Relationships — This Lesson
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Common errors · the 3 traps that cost marks
"A catalyst changes ΔH — it makes the reaction release more energy"
Students write "using a catalyst releases more energy per mole" or "ΔH becomes more negative with a catalyst." The logic seems intuitive: faster reaction = more energy.
Fix: 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.
"Ea(catalysed) is measured from the x-axis / product level"
Students draw the Ea arrow for the catalysed pathway starting from the x-axis (zero) or from the product level rather than from the reactant level.
Fix: Ea for the forward reaction is always measured from the reactant energy level to the transition state peak — regardless of whether the pathway is catalysed or uncatalysed. Measuring from the product level gives Ea of the reverse reaction, not the forward reaction.
"The catalysed pathway ends at a lower product energy level"
Students draw the catalysed curve ending at a lower product energy level, implying the catalyst changes where the products end up energetically.
Fix: Both curves must end at exactly the same product energy level. The products are chemically identical in both pathways — same species, same bonds, same enthalpy. Only the height of the transition state peak changes. If your diagram shows different product levels, it implies ΔH changes, which is wrong.
Quick-fire practice · 5 reps +2 XP per reveal
A student drew an energy profile diagram with Ea(uncatalysed) drawn from the x-axis to the peak, and the catalysed product level lower than the uncatalysed product level. Identify both errors.
Error 2: A catalyst does not change the products or their energy. Both catalysed and uncatalysed pathways must end at the same product energy level. Different product levels would imply ΔH changes — it does not.
A reaction has Ea(forward, uncatalysed) = 210 kJ mol⁻¹ and ΔH = +55 kJ mol⁻¹. A catalyst reduces Ea by 80 kJ mol⁻¹. Calculate: (a) Ea(reverse, uncatalysed), (b) Ea(forward, catalysed), (c) Ea(reverse, catalysed).
(b) Ea(forward, cat) = 210 − 80 = 130 kJ mol⁻¹.
(c) Ea(reverse, cat) = 130 − 55 = 75 kJ mol⁻¹.
Classify each catalyst as homogeneous or heterogeneous and justify: (a) MnO₄⁻(aq) in an aqueous oxidation reaction; (b) Pt(s) in the conversion of SO₂(g) to SO₃(g); (c) concentrated H₂SO₄(l) in the dehydration of C₂H₅OH(l).
(b) Heterogeneous — Pt is solid; SO₂ and O₂ are gaseous. Solid ≠ gas → different phases.
(c) Homogeneous — H₂SO₄(l) is liquid; C₂H₅OH(l) is also liquid. Same phase.
A student argues: "Adding a catalyst to an endothermic reaction makes it exothermic, because the catalyst lowers Ea and the products are now at lower energy." Is this correct? Explain why or why not.
The uncatalysed oxidation of CO has Ea = 232 kJ mol⁻¹. The catalytic converter (Pt surface) reduces this to 75 kJ mol⁻¹. ΔH = −283 kJ mol⁻¹. Using the concept of Ea, explain why the converter is effective at exhaust temperatures (400–600°C) while the uncatalysed reaction is not.
With converter: the Pt surface provides an alternative mechanism (adsorption → surface reaction → desorption) with Ea = 75 kJ mol⁻¹. At the same exhaust temperature, a much greater proportion of CO molecules now have sufficient energy to react → rate increases dramatically → essentially complete conversion before gas exits the converter. ΔH remains −283 kJ mol⁻¹ in both cases.
Go back to your Think First response. Now that you've studied catalysts and energy diagrams, revisit the 1975 EPA catalytic converter mandate — Ford and GM's platinum–palladium system dropping CO oxidation Ea from ~140 kJ mol⁻¹ to ~75 kJ mol⁻¹:
- Claim 1 — FALSE: "A catalyst makes a reaction release more energy." The 1975 catalytic converter does not change ΔH for CO oxidation. The total energy released per mole is identical — because the reactants (CO + ½O₂) and products (CO₂) are unchanged, and ΔH depends only on their relative energies.
- Claim 2 — CORRECT: "A catalyst lowers the amount of energy reactants need to possess before they can react." The platinum surface drops Ea from ~140 to ~75 kJ mol⁻¹ — at exhaust temperatures (400–600°C), a far greater proportion of CO molecules now have sufficient energy to react. Hence 99% conversion efficiency, not ~10%.
- Draw a labelled energy profile diagram for the catalysed and uncatalysed CO oxidation pathways. Mark Ea(uncatalysed) = 140 kJ mol⁻¹, Ea(catalysed) = 75 kJ mol⁻¹, and show that ΔH is identical for both.
Pick your answer, then rate your confidence — that tells the system what to drill next.
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.
Wrong: Ea(catalysed) is measured from the x-axis or from the product level.
Right: Ea for the forward reaction is always measured from the reactant energy level to the transition state peak. Measuring from the product level gives Ea of the reverse reaction.
Wrong: The catalysed pathway ends at a lower product level because Ea is lower.
Right: Both pathways end at the same product level. The catalyst changes only the peak height, not the endpoint of the reaction.
Q6. 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
Q7. 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
Q8. 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
Show comprehensive answers ▼
Activity 1 — Diagram Errors
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.
Multiple Choice
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.
Short Answer Model Answers
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].
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