Chemistry • Year 11 • Module 4 • Lesson 5

Activation Energy, Catalysts & Energy Diagrams

Apply your understanding of E_a, ΔH and catalyst classification to real energy data, a graph-reading task, and a cause-and-effect scenario.

Apply · Data & Reasoning

1. Interpret energy data — catalysed vs uncatalysed reactions

The table below shows energy data for three industrial reactions, with and without catalyst. Use it to answer questions 1.1–1.4. 9 marks

Reaction	E_a (uncatalysed) kJ mol⁻¹	E_a (catalysed) kJ mol⁻¹	Catalyst used	ΔH kJ mol⁻¹
P — Haber process (NH₃ synthesis)	335	150	Fe(s) with K₂O/Al₂O₃ promoters	−92
Q — Decomposition of H₂O₂	75	29	MnO₂(s)	−196
R — Contact process (SO₃ synthesis)	272	80	V₂O₅(s)	−197

Note: reactants in all three reactions are gases or aqueous solutions at reaction temperature.

1.1 For each reaction (P, Q, R), classify the catalyst as homogeneous or heterogeneous. Justify each classification by referring to the physical states of the catalyst and reactants. 3 marks

1.2 Calculate E_a(reverse, uncatalysed) for Reaction P (Haber process). Show your working. 2 marks

1.3 A student says: "Because the Haber process catalyst reduces E_a from 335 to 150 kJ mol⁻¹, the reaction now releases more energy — ΔH becomes more negative." Is the student correct? Explain why or why not, using data from the table. 2 marks

1.4 Identify which reaction has the greatest reduction in E_a due to the catalyst, and explain what this tells you about the effectiveness of that catalyst. 2 marks

Stuck? Revisit lesson § Homogeneous vs Heterogeneous and the formula E_a(reverse) = E_a(forward) − ΔH.

2. Read an energy profile graph — Australian catalytic converters

The graph below shows potential energy versus reaction coordinate for the catalytic oxidation of carbon monoxide (CO) in a car's catalytic converter. The uncatalysed and catalysed pathways are both shown. Use the graph to answer questions 2.1–2.4. 9 marks

Figure 2.1. Energy profile for CO + ½O₂ → CO₂ in a catalytic converter. Adapted from standard kinetics data; E_a values representative of published Pt-surface catalysis literature.

2.1 From the graph, read and record: (a) E_a(uncatalysed); (b) E_a(catalysed); (c) ΔH for the reaction; (d) the reduction in E_a due to the Pt/Pd/Rh catalyst. 2 marks

2.2 Describe the shape difference between the uncatalysed and catalysed curves on the graph. What does this difference represent chemically? 2 marks

2.3 Use the graph to explain why a car without a catalytic converter rarely fully oxidises CO at exhaust temperatures of 400–600°C, but a car with a converter achieves near-complete conversion of CO to CO₂ at the same temperature. Refer to E_a values in your answer. 3 marks

2.4 Classify the Pt/Pd/Rh catalyst in a catalytic converter as homogeneous or heterogeneous. Justify your answer using physical states. 2 marks

Stuck? Revisit lesson § How Catalysts Work and § Homogeneous vs Heterogeneous.

3. Cause-and-effect chain — Incitec Pivot and the Haber process

Incitec Pivot (Brisbane) uses the Haber process to manufacture ammonia fertiliser at scale. An iron catalyst at ~450°C and ~200 atm is central to the process. Complete the chain below by filling in each effect box. 5 marks

CAUSE: Iron (Fe) catalyst is added to the Haber process reactor.

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Effect 1: The activation energy for the reaction N₂(g) + 3H₂(g) → 2NH₃(g) is _______________________

Effect 1 leads to: a lower E_a for both forward and reverse reactions.

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Effect 2: A greater proportion of N₂ and H₂ molecules now have sufficient energy to _______________________

Effect 2 leads to: more successful collisions per second.

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Effect 3: The rate of NH₃ production _______________________ compared to without catalyst.

Effect 3 leads to: faster NH₃ production, allowing a commercially viable plant.

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Effect 4: The ΔH for the Haber process reaction (with catalyst) is _______________________ kJ mol⁻¹ and is therefore _______________________.

Overall outcome (so what?): The Fe catalyst at Incitec Pivot enables the Haber process to run at a commercially viable rate, but the _______________________ of the reaction is unchanged — the same amount of energy is released per mole of NH₃ produced with or without the catalyst.

Stuck? Revisit lesson § How Catalysts Work and the catalyst summary table (what changes / what doesn't).

4. Compare and contrast — homogeneous vs heterogeneous catalysts

Complete the table by filling in each empty cell. Use the lesson examples (catalytic converter, Haber process, esterification, H₂O₂ decomposition). 10 marks

Feature	Homogeneous catalyst	Heterogeneous catalyst
Definition
Physical state relationship	Same phase as reactants
Named example		Pt(s) in catalytic converter; gases (g) as reactants
How it works (mechanism)	Forms an intermediate in solution with the reactants
Ease of separation from products		Easier — physically distinct from gaseous/liquid products
Does it change ΔH?

Stuck? Revisit lesson § Homogeneous vs Heterogeneous card and the comparison SVG.

Answers — Do not peek before attempting

Q1.1 — Catalyst classification

P (Haber — Fe(s), gases): Fe is solid; N₂ and H₂ are gases. Solid ≠ gas → heterogeneous.

Q (H₂O₂ decomposition — MnO₂(s), H₂O₂(aq)): MnO₂ is solid; H₂O₂ is aqueous. Solid ≠ aqueous → heterogeneous.

R (Contact process — V₂O₅(s), SO₂(g) + O₂(g)): V₂O₅ is solid; reactants are gases. Solid ≠ gas → heterogeneous. (1 mark per correct classification with justification; 3 marks total.)

Q1.2 — E_a(reverse, uncatalysed) for Haber process

E_a(reverse) = E_a(forward) − ΔH = 335 − (−92) = 335 + 92 = 427 kJ mol⁻¹. [1 mark formula/setup, 1 mark correct answer with units.] The reverse reaction (decomposition of NH₃) climbs from the product energy level to the same uncatalysed peak — since products are 92 kJ mol⁻¹ below reactants, the reverse pathway must climb 335 + 92 = 427 kJ mol⁻¹.

Q1.3 — Does the catalyst change ΔH?

The student is incorrect. [1 mark] ΔH = −92 kJ mol⁻¹ with or without the catalyst — this is confirmed by the table showing ΔH unchanged. A catalyst lowers E_a only; it does not alter the identity or energy of reactants or products. ΔH is a thermodynamic quantity determined solely by the energy difference between reactants and products, which is pathway-independent. [1 mark explanation citing unchanged ΔH.]

Q1.4 — Greatest reduction in E_a

Reaction R (Contact process, V₂O₅): reduction = 272 − 80 = 192 kJ mol⁻¹, the largest of the three. [1 mark] This indicates V₂O₅ provides an alternative mechanism that is significantly more energetically accessible than the uncatalysed pathway — it lowers the energy barrier by the greatest margin, making it a highly effective industrial catalyst for this reaction. [1 mark interpretation.]

Q2.1 — Graph readings

(a) E_a(uncatalysed) = 232 kJ mol⁻¹. (b) E_a(catalysed) = 75 kJ mol⁻¹. (c) ΔH = −283 kJ mol⁻¹. (d) Reduction = 232 − 75 = 157 kJ mol⁻¹.

Q2.2 — Shape difference and chemical meaning

The uncatalysed curve has a taller, higher peak; the catalysed curve has a lower peak. Both curves start at the same reactant energy level and end at the same product energy level — the ΔH arrow is identical for both. Chemically, the lower peak means the Pt/Pd/Rh surface provides an alternative mechanism in which the highest-energy intermediate is 157 kJ mol⁻¹ lower in energy — so far more CO molecules at exhaust temperatures possess enough energy to reach this lower transition state and proceed to CO₂.

Q2.3 — Why the converter works at exhaust temperatures

Without the converter, E_a = 232 kJ mol⁻¹. At 400–600°C, only a very small fraction of CO molecules have kinetic energy ≥ 232 kJ mol⁻¹, so the rate of uncatalysed oxidation is very slow and most CO exits unconverted. [1 mark] With the Pt/Pd/Rh surface, E_a = 75 kJ mol⁻¹. At the same temperature, a much larger fraction of CO molecules now possess ≥ 75 kJ mol⁻¹. [1 mark] The catalyst achieves this by providing an alternative mechanism (adsorption of CO and O₂ onto the Pt surface, surface reaction, desorption of CO₂), so the gas-phase reaction effectively occurs on the surface at much lower energy cost — conversion is essentially complete before the gas exits. [1 mark]

Q2.4 — Classification of Pt/Pd/Rh catalyst

Pt/Pd/Rh = solid; exhaust gases (CO, hydrocarbons, NO_x) = gaseous. Solid ≠ gas → heterogeneous catalyst. Reactions occur at the solid metal surface where gases adsorb, react, and desorb as product molecules.

Q3 — Cause-and-effect chain answers

Effect 1: lowered (from ~335 kJ mol⁻¹ to ~150 kJ mol⁻¹).

Effect 2: reach the transition state energy and react successfully to form NH₃.

Effect 3: increases significantly (approximately doubles at typical operating temperature).

Effect 4: −92 kJ mol⁻¹ — exothermic (unchanged by the catalyst).

Overall outcome: ΔH (enthalpy change) of the reaction is unchanged.

Q4 — Compare and contrast table

Definition row: Homogeneous — catalyst is in the same phase as the reactants. Heterogeneous — catalyst is in a different phase from the reactants.

Physical state relationship: Heterogeneous = different phase from reactants.

Named example: Homogeneous — H⁺(aq) in esterification of CH₃COOH(aq) + C₂H₅OH(aq); or MnO₄⁻(aq) in aqueous oxidation.

How it works: Heterogeneous — reactants adsorb onto the solid catalyst surface, bond-weakening and reaction occur on the surface, products desorb.

Ease of separation: Homogeneous — harder, same phase as products; requires distillation or selective extraction.

Does it change ΔH? Neither homogeneous nor heterogeneous catalysts change ΔH — both rows: No.