Chemistry • Year 12 • Module 5 • Lesson 7
Industrial Applications of Equilibrium & Collision Theory
Apply equilibrium and collision theory reasoning to real data, Australian industrial contexts, and yield–condition trade-off graphs.
1. Interpret a yield-vs-temperature graph — Haber process
The graph below shows the equilibrium percentage yield of ammonia (NH3) as a function of temperature at three different operating pressures for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). The vertical dashed line marks the typical industrial operating temperature. 9 marks
Adapted from representative Haber process equilibrium data. Values are approximate; used for analysis only.
1.1 Describe the general trend in NH3 yield as temperature increases from 200°C to 700°C, and explain this trend using Le Chatelier’s Principle. 3 marks
1.2 At 450°C, read the approximate NH3 yield at 200 atm from the graph. Explain why the industrial process is not run at 200°C (where yield is highest) despite the graph showing much higher yield at that temperature. 3 marks
1.3 The graph shows that increasing pressure from 100 atm to 300 atm increases yield at any given temperature. Using Le Chatelier’s Principle and collision theory, explain why higher pressure favours both yield AND rate simultaneously for this reaction. 3 marks
2. Interpret industrial process data — Contact process
A research group compared five sets of operating conditions for the Contact process (SO2 + ½O2 ⇌ SO3, ΔH = −99 kJ mol−1). The table shows the equilibrium yield of SO3 and the time taken to reach equilibrium under each set of conditions. 8 marks
| Condition set | Temperature (°C) | Pressure (atm) | Catalyst | Equilibrium yield of SO3 (%) | Time to equilibrium (min) |
|---|---|---|---|---|---|
| A | 350 | 1 | None | 95 | >2000 |
| B | 450 | 1 | V2O5 | 88 | 12 |
| C | 600 | 1 | V2O5 | 62 | 3 |
| D | 450 | 5 | V2O5 | 94 | 11 |
| E | 450 | 1 | None | 88 | 340 |
Hypothetical representative data for analysis.
2.1 Compare Condition Sets B and C. What is the effect of increasing temperature from 450°C to 600°C on (a) equilibrium yield and (b) time to equilibrium? Use the data to support your answer. 2 marks
2.2 Compare Condition Sets B and E. What is the effect of the V2O5 catalyst on (a) equilibrium yield and (b) time to equilibrium? Explain both findings using collision theory. 3 marks
2.3 A process engineer proposes using Condition Set A (350°C, no catalyst) because it gives the highest yield (95%). Evaluate this proposal with reference to both yield and rate. 3 marks
3. Australian industrial case study — Incitec Pivot Phosphate Hill
Incitec Pivot’s Phosphate Hill facility in Queensland produces ammonia via the Haber process for use in nitrogenous fertilisers, supplying the Australian agricultural sector. The plant uses synthesis gas (N2 and H2) derived from natural gas reforming, and operates at approximately 450°C and 200 atm with an iron catalyst. Annual production capacity is approximately 225,000 tonnes of ammonia. 8 marks
3.1 Using the Haber process equilibrium N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol−1, predict and explain what would happen to the equilibrium yield of NH3 if the temperature at Phosphate Hill were raised from 450°C to 600°C. Use Le Chatelier’s Principle in your answer. 3 marks
3.2 A media report claims: “The iron catalyst used at Phosphate Hill is the key to maximising the amount of ammonia produced.” Identify the scientific inaccuracy in this statement and write the corrected version. 2 marks
3.3 The Phosphate Hill plant recycles unreacted N2 and H2 back into the reactor after NH3 is removed by condensation. Explain, using Le Chatelier’s Principle, how continuously removing NH3 as it forms (and recycling the unconverted reactants) drives the overall conversion to well above the per-pass yield. 3 marks
4. Cause-and-effect chain — increasing temperature in the Haber process
The boxes on the left describe causes. Fill in the right-hand effect boxes. The final “overall outcome” box asks for the net industrial implication. 5 marks
Temperature increases from 450°C to 600°C in the Haber process reactor.
Average kinetic energy of all gas particles increases.
Reverse reaction (endothermic) has a higher Ea than the forward reaction.
Reverse rate increases MORE than the forward rate.
Overall industrial implication (so…)
Q1.1 — Graph trend and explanation
As temperature increases from 200°C to 700°C, the equilibrium yield of NH3 decreases (e.g. from approximately 52% at 200°C to approximately 5% at 700°C at 200 atm) [1]. The forward reaction (N2 + 3H2 → 2NH3) is exothermic (ΔH = −92 kJ mol−1) [1]. By Le Chatelier’s Principle, increasing temperature disturbs the equilibrium by adding thermal energy; the system shifts in the endothermic (reverse) direction to oppose this, decreasing [NH3] and reducing yield [1].
Q1.2 — Why not 200°C?
Approximate yield at 450°C and 200 atm: ~18–22% [1]. Although yield at 200°C is much higher (approximately 52% at 200 atm), the rate at 200°C is far too slow to be commercially viable [1] — the activation energy barrier means very few particles have sufficient energy for effective collisions, and even with a catalyst the time to reach equilibrium would be extremely long. The iron catalyst also becomes active only above approximately 300°C [1].
Q1.3 — Why higher pressure helps both yield and rate
Le Chatelier’s Principle: the Haber process has 4 mol gas on the left and 2 mol gas on the right. Increasing pressure causes the system to shift to the right (fewer gas moles), increasing equilibrium yield [1]. Collision theory: increasing pressure increases the concentration of all gas-phase species, increasing the frequency of effective collisions per unit time, therefore increasing reaction rate [1]. Because both effects work in the same direction, pressure is not a trade-off — it is limited only by engineering cost and safety [1].
Q2.1 — Temperature effect (B vs C)
(a) Equilibrium yield decreases from 88% (450°C) to 62% (600°C), a fall of 26 percentage points. (b) Time to equilibrium decreases from 12 min to 3 min — the reaction reaches equilibrium much faster at higher temperature. (1 mark per comparison with data support.)
Q2.2 — Catalyst effect (B vs E)
(a) The catalyst has NO effect on equilibrium yield: both B and E give 88% yield. (b) The V2O5 catalyst reduces time to equilibrium dramatically: 12 min (with catalyst, B) vs 340 min (without, E) [1]. Explanation: the catalyst provides an alternative reaction pathway with a lower activation energy (Ea). This increases the proportion of particles with sufficient energy for effective collisions, increasing both the forward and reverse rates equally [1]. Because Keq = ratefwd/raterev and both rates increase by the same factor, Keq and equilibrium yield are unchanged [1].
Q2.3 — Evaluating Condition Set A
The proposal should be rejected. Although Set A gives the highest yield (95%), the time to equilibrium is >2000 min (more than 33 hours) without a catalyst [1]. This makes commercial production rate far too slow: the amount of SO3 produced per day would be negligible despite the high equilibrium yield [1]. Industrial viability requires both an adequate yield AND an adequate rate. Set B (450°C, catalyst, 88% yield, 12 min) provides a commercially acceptable balance [1].
Q3.1 — Effect of raising temperature to 600°C
Equilibrium yield of NH3 would decrease [1]. The forward reaction is exothermic (ΔH = −92 kJ mol−1); by Le Chatelier’s Principle, increasing temperature from 450°C to 600°C shifts equilibrium to the left (endothermic direction) to oppose the added heat [1]. This means less NH3 at equilibrium and a lower Keq; annual output would fall significantly, reducing the plant’s commercial viability [1].
Q3.2 — Media report inaccuracy
Inaccuracy: the catalyst does NOT maximise the amount of NH3 produced (yield) [1]. Corrected statement: “The iron catalyst used at Phosphate Hill is the key to maximising the rate at which ammonia is produced at the industrial operating temperature of 450°C — it does not affect the equilibrium yield, which is determined by temperature and pressure.” [1]
Q3.3 — Recycling and Le Chatelier
Removing NH3 by condensation decreases the product concentration. By Le Chatelier’s Principle, the system shifts to the right (forward direction) to oppose the removal, producing more NH3 from the remaining N2 and H2 [1]. Returning unreacted N2 and H2 increases their concentration in the reactor feed, further shifting the equilibrium to the right [1]. By continuously removing product and returning reactants, the net forward reaction is sustained well beyond what would occur in a single closed pass, achieving >95% overall conversion despite only 15–25% conversion per pass [1].
Q4 — Cause-and-effect chain
Effect 1: Both forward and reverse reaction rates increase (more particles exceed Ea).
Effect 2: A greater proportion of particles now have enough kinetic energy to overcome Ea.
Effect 3: The proportion of reverse-reaction particles exceeding the higher reverse Ea increases more than those exceeding the lower forward Ea.
Effect 4: Equilibrium shifts to the left; [NH3] decreases; [N2] and [H2] increase; Keq decreases.
Overall industrial implication: Operating at 600°C would increase rate but significantly reduce the per-pass yield of NH3, reducing commercial output and profitability — the temperature compromise (450°C) already balances rate and yield; raising it further costs more in lost yield than it gains in speed.