Chemistry • Year 12 • Module 5 • Lesson 3
Collision Theory Applied to Equilibrium
Build Band 5–6 extended-response technique — synthesise collision theory, activation energy diagrams, real data, and industrial context into evaluative, evidence-based responses.
1. Data + scenario — evaluate the claim about a catalyst and an industrial equilibrium (Band 5–6)
8 marks Band 5–6
Scenario. A process engineer at Incitec Pivot’s Gibson Island ammonia plant (Brisbane, Queensland) presents the following data to management from a pilot trial comparing two catalyst systems for the Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol−1.
| Catalyst system | Temperature (°C) | Pressure (MPa) | Time to equilibrium (min) | % NH3 at equilibrium | Keq |
|---|---|---|---|---|---|
| Standard Fe catalyst | 450 | 20 | 24 | 18% | 6.0 × 10−2 |
| Promoted Fe–Ru catalyst | 350 | 20 | 19 | 28% | 1.8 × 10−1 |
Engineer’s claim: “The promoted Fe–Ru catalyst is better because it is more active — it lowers Ea more than the standard Fe catalyst, so it gives a higher equilibrium yield of NH3 and a faster rate. We should switch entirely to Fe–Ru.”
Q1. Evaluate the engineer’s claim using the data and your understanding of collision theory and activation energy. In your response you must:
- Define what a catalyst does in terms of activation energy for a reversible reaction.
- Assess whether the difference in % NH3 and Keq between the two systems can be explained by the catalyst alone. Justify using the data.
- Identify the most likely reason for the difference in equilibrium yield and Keq between the two catalyst systems.
- Evaluate whether the engineer’s conclusion (“better catalyst → higher yield”) is scientifically valid. Give an evidence-based judgement.
2. Source critique — identify the scientific flaw (Band 5–6)
7 marks Band 5–6
“Catalysts are the chemist’s secret weapon for improving reaction outcomes. By lowering the activation energy of the forward reaction more than the reverse, a catalyst tips the balance toward product formation, allowing industrial reactions like the Haber process to produce more ammonia than would otherwise be possible at the same temperature. This is why refineries invest so heavily in better catalysts — a more active catalyst means a higher equilibrium yield.”
— From a fictional chemistry blog post.
Q2. This passage contains a significant scientific error that would cost marks in an HSC extended response. Identify the specific flaw, explain the correct chemistry using activation energy concepts, and describe how you would design an experiment to demonstrate the correct behaviour. 7 marks
Q1 — Marking criteria (8 marks)
Definition of catalyst action (1 mark): A catalyst provides an alternative lower-energy reaction pathway, lowering Ea by the same amount for both the forward and reverse reactions. It is not consumed in the process.
Assess difference in % NH3 and Keq (2 marks): The two catalyst systems give different % NH3 (18% vs 28%) and different Keq (6.0 × 10−2 vs 1.8 × 10−1). A catalyst cannot change Keq because it lowers Ea equally for both directions — the ratio of forward to reverse rates is unchanged. The data therefore show that the difference in equilibrium yield is NOT caused by the catalyst. [1 mark for identifying that catalyst cannot change Keq; 1 mark for citing the different Keq values as evidence the catalyst is not the cause.]
Identify the true reason (2 marks): The data table shows that the two trials are run at different temperatures (450°C vs 350°C), not just different catalysts. Temperature is the only factor that changes Keq. For an exothermic forward reaction (ΔH = −92 kJ mol−1), decreasing temperature shifts equilibrium to the right, increasing Keq and % NH3. The higher % NH3 at 350°C is entirely attributable to the lower temperature, not the Fe–Ru catalyst. [1 mark for identifying temperature difference as the confounding variable; 1 mark for linking lower T to higher Keq for exothermic reaction.]
Evaluate the engineer’s conclusion (2 marks): The engineer’s claim is scientifically invalid. The claim conflates two separate effects: catalyst activity (which affects rate) and temperature (which affects equilibrium yield and Keq). The data do not isolate the effect of the catalyst because temperature is not controlled. A valid comparison would require both catalysts to be tested at the same temperature. The Fe–Ru system may indeed reach equilibrium faster at 350°C (a real advantage), but this is a rate benefit, not a yield benefit. [1 mark for correctly rejecting the claim with reasoning; 1 mark for identifying the uncontrolled variable (temperature) as the flaw in the engineer’s logic.]
Evidence-based judgement (1 mark): A strong response explicitly states: “The engineer’s claim is incorrect because catalysts cannot increase equilibrium yield; the higher yield observed at 350°C is due to the lower operating temperature favouring the exothermic forward reaction, not the catalyst.”
Q2 — Source critique marking criteria (7 marks)
Identify the specific flaw (2 marks): The passage claims that a catalyst lowers the activation energy of the forward reaction more than the reverse (“tips the balance toward product formation”). This is incorrect. A catalyst lowers Ea by the same amount for both the forward and reverse reactions. It does not preferentially favour either direction. [1 mark for identifying the incorrect claim about asymmetric Ea reduction; 1 mark for stating the correct principle: catalyst lowers Ea equally for both directions.]
Correct chemistry (3 marks): (a) A catalyst provides an alternative reaction mechanism with a lower activation energy barrier. On an energy diagram, the transition state peak is lowered by the same value x for both the forward and reverse pathways. (b) Because both barriers are lowered equally, both the forward and reverse rates increase by the same factor. The ratio of forward rate to reverse rate — which determines Keq and equilibrium position — is unchanged. (c) Therefore, a catalyst does not change the equilibrium position, does not change Keq, and does not increase the yield of any product at equilibrium. Keq depends only on temperature. The only industrial advantage of a catalyst is reaching equilibrium more quickly at a given temperature. [3 marks: 1 for equal Ea lowering shown on energy diagram; 1 for both rates increasing by same factor → ratio unchanged; 1 for explicit statement that Keq depends only on temperature.]
Experimental demonstration (2 marks): Set up two identical sealed vessels containing the same Haber process mixture (N2, H2) at the same temperature (e.g. 450°C) and pressure (e.g. 20 MPa). Add the catalyst to one vessel (experimental) and none to the other (control). Monitor NH3 concentration over time using gas chromatography or spectroscopy. Record: (1) time to reach equilibrium — predicted to be shorter with catalyst; (2) [NH3] at equilibrium — predicted to be identical in both vessels. If the equilibrium [NH3] is the same in both vessels, this falsifies the claim that a better catalyst gives a higher yield. [1 mark for controlled experimental design with identical temperature and pressure; 1 mark for measuring both rate to equilibrium and final concentration, with correct prediction that final concentrations will be equal.]