Chemistry • Year 12 • Module 5 • Lesson 8

Consolidation — LCP Mastery

Build HSC Band 5–6 extended-response technique on multi-variable Le Chatelier’s Principle problems, graph interpretation, and the rate–yield distinction in industrial chemical contexts.

Master • Extended Response • Band 5–6

1. Stimulus-based extended response — the Contact Process for SO3 production (Band 5–6)

8 marks   Band 5–6

Stimulus. The Contact Process is the industrial method used in Australia and globally to produce sulfuric acid via the equilibrium:

2SO2(g) + O2(g) ⇌ 2SO3(g)    ΔH = −197 kJ/mol

The table below shows equilibrium SO3 yield (%) under different conditions measured by CSIRO researchers investigating catalyst efficiency.

Temperature (°C)Pressure (atm)CatalystEquilibrium yield of SO3 (%)
4001V2O599.2
4501V2O597.5
5001V2O591.4
6001V2O573.6
4502V2O598.9
4501None97.5

Adapted from CSIRO process chemistry data (illustrative values consistent with published thermodynamic data for this equilibrium).

Q1. Analyse and evaluate the data in the table and your knowledge of Le Chatelier’s Principle to explain the industrial operating conditions chosen for the Contact Process. In your response you must:

  • Explain the trend in equilibrium yield as temperature increases, linking your explanation to ΔH and Le Chatelier’s Principle.
  • Explain what the data reveal about the role of the V2O5 catalyst, with reference to rate and yield.
  • Explain the effect of pressure on yield using gas mole counts, and account for why the industrial process operates at only 1–2 atm despite higher pressure giving better yield.
  • Evaluate the trade-off between rate and yield that determines the operating temperature of 450°C, referencing the rate/yield distinction from the lesson.
  • Reach a justified conclusion about which factor (temperature, pressure, or catalyst) has the greatest influence on equilibrium yield.
Structure your answer: temperature trend → catalyst role → pressure effect → 450°C trade-off → overall conclusion. Always state Keq effect alongside position shift.

2. Stimulus-based extended response — the refrigeration cycle and the rate/yield distinction (Band 5–6)

8 marks   Band 5–6

Stimulus. The ammonia refrigeration cycle, widely used in large commercial refrigeration systems across Australian food processing plants (including Australian Pork Limited facilities in rural NSW), exploits the following equilibrium-like processes. The key reversible step is:

NH3(g) ⇌ NH3(l)    ΔH = −23 kJ/mol

To cool a refrigerated space, liquid ammonia is allowed to expand rapidly into a low-pressure evaporator compartment. To return ammonia to liquid for the next cycle, it is compressed at high pressure in the condenser compartment at ambient temperature. A student reads that “increasing pressure in the condenser improves the conversion of NH3 gas to liquid and also speeds up the process — it’s better in every way.”

In a separate industrial context, the same student encounters the Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ/mol, operated at 400–500°C and 150–300 atm with an iron catalyst, and argues that the high temperature is chosen because “it improves both rate and yield simultaneously.”

Q2. Analyse and evaluate both student claims using Le Chatelier’s Principle, collision theory, and the rate–yield distinction. In your response you must:

  • Evaluate the first student claim about the refrigeration condenser — is it correct that increasing pressure “is better in every way”? Explain using Le Chatelier’s Principle and state the effect on Keq.
  • Identify and correct every error in the second claim about the Haber process high temperature, using the rate–yield distinction and the exothermic ΔH.
  • Use collision theory to explain why high temperature increases the rate of Haber process NH3 synthesis, even though it decreases the yield.
  • Explain why 400–500°C and an iron catalyst together represent the industrial compromise, referencing both rate and yield in your answer.
  • Justify why high pressure (150–300 atm) is used in the Haber process and explain whether this changes Keq.
Dissect each claim in order. For claim 1: pressure and the liquid→gas equilibrium — note that liquid has negligible volume compared to gas, so increasing pressure favours liquid. For claim 2: the two-part error is (a) high T decreases yield for exothermic reaction, (b) catalyst (not T) is the rate fix.
Answers — Do not peek before attempting

Q1 — Sample Band 6 response (8 marks), annotated

Temperature trend: As temperature increases from 400°C to 600°C, equilibrium yield falls from 99.2% to 73.6%. The forward reaction is exothermic (ΔH = −197 kJ/mol), so increasing temperature favours the endothermic reverse direction (Le Chatelier’s Principle), shifting equilibrium left and producing less SO3. Keq decreases with rising temperature for this reaction.

Marking criteria.

  • 1 mark — Identifies the decreasing trend in yield with increasing temperature and links it to ΔH < 0 (exothermic forward) and LCP (shift left at higher T).
  • 1 mark — States that Keq decreases with increasing temperature for an exothermic forward reaction.
  • 1 mark — Correctly identifies the catalyst’s role from the data: comparing rows at 450°C/1 atm with and without catalyst — yield is identical (97.5%), confirming V2O5 does not change equilibrium yield or Keq; it only speeds attainment of equilibrium (rate effect).
  • 1 mark — Explains the pressure effect: left side has 3 mol gas (2 SO2 + 1 O2), right side has 2 mol gas (2 SO3). Increasing pressure shifts equilibrium right (toward fewer gas moles), increasing yield from 97.5% to 98.9%. Keq is unchanged.
  • 1 mark — Accounts for why the industrial process uses only 1–2 atm despite higher pressure giving better yield: the equipment cost and safety risks of very high pressure for a relatively small yield gain (97.5% at 1 atm is already commercially viable) do not justify the engineering expense.
  • 1 mark — Evaluates the 450°C trade-off: at 400°C the yield is 99.2% but the rate is prohibitively slow (even with catalyst); at 600°C the rate is faster but yield drops to 73.6%. 450°C with V2O5 gives 97.5% yield at a commercially acceptable rate — the classic rate/yield compromise.
  • 1 mark — Explicitly uses the rate/yield distinction: rate = kinetics (speed to equilibrium, improved by T and catalyst); yield = thermodynamics (proportion of SO3 at equilibrium, worsened by high T for this exothermic reaction).
  • 1 mark — Reaches a justified conclusion: temperature has the greatest effect on equilibrium yield (data show a 25.6 percentage-point reduction from 400 to 600°C), larger than the 1.4-point gain from doubling pressure. The catalyst has no effect on yield at all. Therefore temperature is the dominant factor in determining equilibrium yield.

Q2 — Sample Band 6 response (8 marks), annotated

Claim 1 (refrigeration condenser): The first student is broadly correct that increasing pressure in the condenser improves the conversion of NH3 gas to liquid, but the claim that it is “better in every way” oversimplifies. For NH3(g) ⇌ NH3(l), ΔH = −23 kJ/mol, the liquid phase occupies negligible volume relative to the gas phase, so increasing pressure strongly favours the liquid (fewer“gas moles” on the liquid side in a volume sense). By LCP, increasing pressure shifts the equilibrium toward the liquid — consistent with more efficient condensation. Keq is unchanged (pressure does not change Keq). However, increasing pressure also increases equipment cost, compressor energy demands, and safety requirements — so “better in every way” ignores engineering constraints.

Claim 2 (Haber process temperature) — errors and corrections:

  • Error 1: “High temperature improves yield.” This is wrong. N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ/mol — the forward reaction is exothermic. Increasing temperature favours the endothermic reverse direction (LCP), shifts equilibrium left, decreases Keq, and reduces the equilibrium yield of NH3.
  • Error 2 (implied): that improving rate and yield always go together for temperature changes. For exothermic reactions, rate and yield are in direct conflict when temperature is the variable: higher T gives higher rate but lower yield.

Collision theory for rate: Increasing temperature increases the average kinetic energy of all particles. A greater proportion of collisions now have energy equal to or exceeding the activation energy Ea — collision frequency and the fraction of effective collisions both increase. This increases the rate of forward (and reverse) reactions, allowing equilibrium to be reached more quickly — even though the equilibrium position reached is less product-rich than at lower temperature.

Industrial compromise — 400–500°C + iron catalyst: At low temperatures, yield is high (Keq is large) but the rate of reaching equilibrium is prohibitively slow — economically unviable. The iron catalyst lowers the activation energy for both forward and reverse reactions equally, allowing commercially acceptable reaction rates at 400–500°C without the yield penalty that would come from relying on temperature alone to achieve adequate rate. The catalyst does not change equilibrium position or Keq; it is purely a kinetic tool.

High pressure (150–300 atm): Left side: 4 mol gas (N2 + 3H2). Right side: 2 mol gas (2NH3). Increasing pressure shifts equilibrium right (fewer gas moles), increasing yield by Le Chatelier’s Principle. Keq is unchanged (pressure does not change Keq). High pressure is used because it both shifts the position to give more NH3 and also increases the rate by increasing reactant concentrations (collision frequency), contributing to feasible industrial throughput.

Marking criteria.

  • 1 mark — Evaluates Claim 1: increasing pressure in the condenser does improve NH3(g)→NH3(l) conversion via LCP (liquid favoured at higher pressure); Keq is unchanged. Identifies that “better in every way” is an overstatement (engineering/cost constraints).
  • 1 mark — Identifies Error 1 in Claim 2: for an exothermic forward reaction, increasing temperature decreases yield (shifts equilibrium left, decreases Keq).
  • 1 mark — Identifies Error 2 in Claim 2 (implicit): rate and yield are governed by different factors for temperature; they are in conflict for exothermic reactions.
  • 1 mark — Uses collision theory to explain why high T increases rate: more particles with E ≥ Ea, more effective collisions per unit time, equilibrium reached faster.
  • 1 mark — Explains the catalyst’s role in the industrial compromise: lowers Ea for both directions equally, allowing acceptable rate at 400–500°C without the yield loss of high T.
  • 1 mark — Explicitly states that catalyst does not change equilibrium position or Keq — it is a kinetic tool, not a thermodynamic one.
  • 1 mark — Explains the high-pressure justification: 4 mol gas (left) vs 2 mol gas (right); pressure increase shifts equilibrium right; Keq unchanged.
  • 1 mark — Reaches a coherent synthesis: high temperature in the Haber process is necessary for acceptable rate (kinetics), but this comes at a cost to yield (thermodynamics); the iron catalyst and high pressure are the key levers that recover commercial viability, not temperature.

HSCScience • Chemistry • Year 12 • Module 5 • Lesson 8 • Worksheet 3 (Master)