Chemistry • Year 12 • Module 5 • Lesson 7
Industrial Applications of Equilibrium & Collision Theory
Synthesise data, evaluate competing industrial strategies, and construct Band 6 extended responses integrating equilibrium, collision theory and economic reasoning.
Question 1 — Data, scenario and multi-criteria evaluation
8 marks
Scenario. Orica’s Kooragang Island facility near Newcastle, NSW, produces nitric acid (HNO3) for use in the manufacture of explosives and ammonium nitrate fertilisers. The first step in the process is the oxidation of ammonia (itself produced via the Haber process): 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g), ΔH = −906 kJ mol−1.
An engineering team at the facility is evaluating four candidate operating conditions for this oxidation step. The table below summarises the equilibrium yield of NO and the estimated time to reach operating equilibrium at each set of conditions. A platinum–rhodium alloy (Pt-Rh) catalyst is used in all four conditions.
| Condition | Temperature (°C) | Pressure (atm) | Equilibrium yield of NO (%) | Time to equilibrium (s) |
|---|---|---|---|---|
| W | 600 | 1 | 96 | 0.3 |
| X | 800 | 1 | 89 | 0.04 |
| Y | 900 | 8 | 82 | 0.02 |
| Z | 400 | 1 | 99 | 28 |
Note: For this reaction, more moles of gas are on the product side (4 NO + 6 H2O = 10 mol) than on the reactant side (4 NH3 + 5 O2 = 9 mol). Hypothetical representative data.
Extended response prompt: Evaluate which operating condition (W, X, Y or Z) is most suitable for industrial use at the Kooragang Island facility. In your response you must:
- Define the term industrial compromise and explain why it is necessary for this reaction.
- Compare ALL four conditions on the basis of yield AND rate, supported by data from the table.
- Use collision theory to explain the temperature effect on the rate of the reaction.
- Use Le Chatelier’s Principle to explain the effect of temperature on yield for this exothermic reaction.
- State your recommended condition with an evidence-based justification, referring to both chemical and economic considerations.
Question 2 — Evaluate and compare: Haber vs Contact process strategies
8 marks
Scenario. Wesfarmers CSBP operates an ammonia and ammonium sulfate plant at Kwinana, Western Australia. The facility produces ammonia via the Haber process and uses ammonia as a feedstock to make ammonium sulfate fertiliser. A nearby facility (a different company) uses the Contact process to produce sulfuric acid, which is also used in fertiliser production.
A Year 12 student writes in an assignment: “The Haber process and the Contact process both face the same yield-vs-rate trade-off because they are both exothermic forward reactions. This means they are operated under identical conditions to achieve the same type of industrial compromise.”
The student’s claim contains both a valid generalisation and a significant error. The data below are provided for reference.
| Feature | Haber process | Contact process |
|---|---|---|
| Reaction | N2(g) + 3H2(g) ⇌ 2NH3(g) | 2SO2(g) + O2(g) ⇌ 2SO3(g) |
| ΔH (kJ mol−1) | −92 | −197 |
| Gas mol: left → right | 4 → 2 | 3 → 2 |
| Catalyst | Iron (Fe) with K2O / Al2O3 promoters | Vanadium(V) oxide (V2O5) |
| Operating temperature | 400–500°C | 450–550°C |
| Operating pressure | 150–300 atm | 1–2 atm |
| Per-pass yield | ~15–25% | ~97–99% |
| Recycling used? | Yes | No (yield too high to need it) |
Extended response prompt: Evaluate the student’s claim. In your response you must:
- Identify the valid generalisation in the student’s claim and explain why it is correct.
- Identify and correct the significant error, with specific reference to the data table.
- Compare the two processes on at least THREE of the following criteria: pressure strategy, per-pass yield, recycling need, temperature range, catalyst choice.
- Use Le Chatelier’s Principle to explain why the Contact process operates at much lower pressure than the Haber process despite both involving gas-phase reactions with fewer product gas moles.
- Reach an evidence-based conclusion about whether the industrial compromises for the two processes are “the same”.
Q1 — Marking notes (8 marks)
Mark 1 — Definition of industrial compromise: The set of conditions (temperature, pressure, catalyst) chosen to balance equilibrium yield and reaction rate in a commercially viable way, accepting that neither factor can be fully optimised simultaneously.
Mark 2 — Data-referenced comparison of all four conditions: Award if student systematically addresses each condition. W = high yield (96%) and fast (0.3 s) — strong candidate. X = slightly lower yield (89%) but extremely fast (0.04 s). Y = lowest yield (82%) at high pressure, very fast. Z = highest yield (99%) but extremely slow (28 s) — commercially impractical. Comparison must reference at least two conditions with data.
Mark 3 — Collision theory explanation of temperature→rate: Higher temperature increases average kinetic energy of particles; a greater proportion exceed Ea; collision frequency with sufficient energy increases; rate increases. Award if mechanism (not just “more energy”) is stated.
Mark 4 — LCP explanation of temperature→yield: The forward reaction is exothermic (ΔH = −906 kJ mol−1). Le Chatelier’s Principle — increasing temperature opposes equilibrium by shifting in the endothermic (reverse) direction, decreasing [NO] and yield. Keq decreases with temperature for exothermic reactions.
Mark 5 — Pressure analysis (Y vs W/X/Z): More gas moles on product side (10 > 9), so LCP predicts increasing pressure shifts equilibrium LEFT, reducing yield. Award for noting that high pressure would actually hurt yield for this reaction (unlike the Haber process) — Y (8 atm) has the lowest yield, consistent with this. Low pressure (1 atm) is preferred for yield.
Mark 6 — Recommended condition with justification: Condition W (600°C, 1 atm) is the best balance — 96% yield with 0.3 s equilibration time (versus Z’s 28 s). Accept X if student justifies the faster rate compensates for the 7% yield reduction, with economic reasoning about throughput.
Mark 7 — Economic/engineering dimension: High throughput matters: even a lower yield per pass may give higher total output if equilibrium is reached in milliseconds. Z’s theoretical 99% yield is economically useless if equilibration is 1000× slower. Engineering cost of 8 atm (Y) does not justify the yield penalty.
Mark 8 — Quality of overall synthesis: Award for a structured response that explicitly states and justifies the recommendation, addresses the trade-off between yield and rate using both the data and chemical principles, and reaches a clear evidence-based conclusion.
Q2 — Marking notes (8 marks)
Mark 1 — Valid generalisation identified and explained: Correct: both the Haber process and the Contact process involve exothermic forward reactions, so both face the same type of yield–rate trade-off (lower temperature favours yield; higher temperature favours rate). The student is correct that this is a shared thermodynamic characteristic.
Mark 2 — Error identified and corrected: The error is that the two processes do NOT operate under identical conditions — they differ substantially in pressure strategy (Haber: 150–300 atm; Contact: 1–2 atm), per-pass yield (Haber: 15–25%; Contact: 97–99%), and recycling requirement (Haber: essential; Contact: not needed). Awarding requires specific data comparison from the table.
Mark 3 — Comparison on ≥3 criteria:
- Pressure: Haber uses 150–300 atm (high, improves yield); Contact uses 1–2 atm (low, avoids yield penalty from LCP).
- Per-pass yield: Haber ~15–25% vs Contact ~97–99%.
- Recycling: Haber requires recycling to achieve overall >95% conversion; Contact does not.
- Temperature range: similar (450–550°C for both) — both face the same catalyst activation temperature constraint.
- Catalyst: different choices (Fe-based vs V2O5) reflecting different chemical systems.
Award 1 mark for each clearly stated, data-referenced criterion comparison (max 3 marks for this component).
Mark 4 — LCP explanation for Contact process low pressure: Although both reactions have fewer gas moles on the product side (Contact: 3 mol left → 2 mol right), the Contact process per-pass yield is already ~97–99% at 1–2 atm. Increasing pressure would improve yield only marginally, and the engineering cost of high-pressure equipment (as required by Haber at 200 atm) does not justify the trivial additional yield gain. Accept also: LCP shows the direction of improvement from higher pressure but the economic cost is unjustified given the already-high equilibrium yield.
Mark 5 — Evidence-based conclusion: The student’s claim is partially correct (both face a yield–rate trade-off due to exothermic forward reactions) but substantially wrong in concluding that the compromises are identical. The Haber and Contact processes require significantly different industrial strategies primarily because the Haber process has a very unfavourable equilibrium position at industrial temperatures (requiring high pressure and recycling to compensate for low per-pass yield), whereas the Contact process has a highly favourable equilibrium position at 1–2 atm (no recycling needed, no high-pressure equipment required).
Mark 6 (quality/structure): Award for coherent structure, accurate use of the data table, correct LCP and collision theory terminology, and a judgement that goes beyond restating the prompt.