HSCScienceExam practice
Direction

Chemistry  ·  Year 12  ·  Module 5  ·  Lesson 6

HSC Exam Practice

Le Chatelier's Principle: Pressure, Volume & Catalysts

9 questions / 3 sections / 35 marks total
Section 1

Short answer

1.Short answer

1.1

Define Le Chatelier's Principle and state the condition under which a pressure change has no effect on a gas-phase equilibrium.

3marks Band 3
1.2

Identify the direction of equilibrium shift, if any, when the volume of the container for N2(g) + 3H2(g) ⇌ 2NH3(g) is halved. Justify your answer by counting gas moles on each side.

3marks Band 3
1.3

Explain why adding argon (an inert gas) to the equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g) at constant volume does not shift the equilibrium position.

2marks Band 3
1.4

Describe the effect of adding a vanadium pentoxide (V2O5) catalyst to the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g) on: (a) the equilibrium position; (b) the value of Keq; (c) the activation energies of the forward and reverse reactions.

3marks Band 3–4
1.5

Distinguish between the effect of increasing temperature and increasing pressure on the value of Keq for an exothermic gas-phase equilibrium where the forward reaction produces fewer moles of gas.

3marks Band 4
1.6

Account for the two-stage colour change observed when a syringe containing 2NO2(g) ⇌ N2O4(g) at equilibrium is rapidly compressed. Include the immediate observation and the observation after re-equilibration.

4marks Band 4
Section 2

Data response

2.Data response — Contact process yield data

2.1

The graph below shows the equilibrium % yield of SO3 in the Contact process (2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −196 kJ/mol) as a function of pressure at three temperatures.

0 20 40 60 80 100 Equilibrium % yield of SO₃ 1 2 5 10 20 50 Pressure (atm) 400°C 500°C 600°C
Figure 2.1. Equilibrium % yield of SO3 vs pressure at 400°C, 500°C and 600°C for the Contact process. Data stylised after Atkins & de Paula (2010), Physical Chemistry, 9th ed.

(a) Describe the trend in SO3 yield as pressure increases at 500°C. 2 marks

(b) Explain, with reference to gas mole counts and Le Chatelier's Principle, why increasing pressure increases the yield of SO3. 3 marks

(c) At a given pressure, the yield of SO3 decreases as temperature increases from 400°C to 600°C. Using the sign of ΔH and Le Chatelier's Principle, account for this observation. State whether Keq increases or decreases as temperature rises. 3 marks

8marks Band 4–5

3.Source critique — catalyst and equilibrium

3.1

A study notes website states: “Adding a platinum catalyst to the equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g) shifts the equilibrium to the right, producing more SO3. This is why the Contact process uses a catalyst to improve yield.”

Identify the flaw in this statement and write the correct scientific explanation of the catalyst's role.

4marks Band 4–5
Section 3

Extended response

4.Extended response

4.1

Analyse the role of pressure, temperature, and a catalyst in determining the yield and rate of ammonia production in the Haber process (N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ/mol). In your response, distinguish between factors that change the equilibrium position and factors that change Keq, and assess the conditions used at Incitec Pivot's Australian plant (~200 atm, 400–500°C, iron catalyst).

7marks Band 5–6

Chemistry · Year 12 · Module 5 · Lesson 6

Answer Key & Marking Guidelines

1.1

Section 1 · Short answer · 3 marks · Band 3

Sample response. Le Chatelier's Principle states that if a system at equilibrium is subjected to a disturbance (change in concentration, pressure, or temperature), the system will shift in the direction that partially opposes the disturbance and re-establishes equilibrium. A pressure change has no effect on equilibrium position when the number of moles of gas is equal on both sides of the balanced equation.

Marking notes. 1 mark for defining LCP (system shifts to oppose disturbance / re-establish equilibrium); 1 mark for the condition of no pressure effect (equal moles of gas on both sides); 1 mark for specifying only gaseous species are counted (or an equivalent qualification such as “solids and liquids are excluded from the mole count”). Accept any scientifically correct phrasing.

1.2

Section 1 · Short answer · 3 marks · Band 3

Sample response. Halving the volume doubles the pressure. N2(g) + 3H2(g) ⇌ 2NH3(g): left side has 1 + 3 = 4 moles of gas; right side has 2 moles of gas. By Le Chatelier's Principle, increasing pressure shifts equilibrium toward the side with fewer moles of gas — the right side. Equilibrium shifts right; more NH3 is produced. Keq is unchanged.

Marking notes. 1 mark for correct mole count (4 left, 2 right); 1 mark for direction of shift (right / toward fewer gas moles); 1 mark for linking to Le Chatelier's Principle or mechanism (shift reduces pressure). Deduct 1 if student states Keq changes.

1.3

Section 1 · Short answer · 2 marks · Band 3

Sample response. At constant volume, adding argon does not change the partial pressures of SO2, O2, or SO3. Each reacting gas still exerts the same pressure it did before the argon was added. Because equilibrium is determined by the partial pressures (or concentrations) of the reacting species, and these are unchanged, there is no driving force to shift the equilibrium position.

Marking notes. 1 mark for stating partial pressures of reacting gases are unchanged at constant V; 1 mark for concluding no shift occurs (or: the driving force / Q vs Keq comparison is unchanged). Accept “concentrations are unchanged” as equivalent to partial pressures unchanged.

1.4

Section 1 · Short answer · 3 marks · Band 3–4

Sample response. (a) Equilibrium position: no change — the catalyst does not shift equilibrium. (b) Keq: no change — Keq depends only on temperature, which is unchanged. (c) Activation energies: the catalyst lowers the activation energy of both the forward and reverse reactions equally by providing an alternative reaction pathway; neither direction is selectively favoured.

Marking notes. 1 mark for (a) — no change to equilibrium position; 1 mark for (b) — no change to Keq; 1 mark for (c) — both Ea values lowered equally (do not award this mark if student states only forward Ea is lowered).

1.5

Section 1 · Short answer · 3 marks · Band 4

Sample response. Increasing temperature changes Keq: for an exothermic forward reaction, increasing temperature shifts equilibrium left (favouring the endothermic reverse), so Keq decreases. Increasing pressure does not change Keq; it shifts the equilibrium position toward fewer gas moles (the product side, in this case), increasing yield at the same Keq value. Temperature is the only variable that changes Keq; pressure changes only the equilibrium position.

Marking notes. 1 mark for temperature — changes Keq (decreases for exothermic forward with rising T); 1 mark for pressure — does not change Keq; 1 mark for correct direction of pressure shift (toward product / fewer gas moles) and the distinction that this is a position change not a Keq change.

1.6

Section 1 · Short answer · 4 marks · Band 4

Sample response. Immediate observation (Stage 1): the colour darkens rapidly. Compression halves the volume, instantaneously doubling the concentration of both NO2 (brown) and N2O4 (colourless). The higher [NO2] per unit volume produces a darker brown colour. No equilibrium shift has occurred yet.

After re-equilibration (Stage 2): the colour pales from the momentarily darkened state (though remains darker than the original, pre-compression baseline). By Le Chatelier's Principle, increasing pressure shifts equilibrium toward fewer moles of gas. 2NO2(g) (2 mol left) ⇌ N2O4(g) (1 mol right) — equilibrium shifts right, converting brown NO2 to colourless N2O4. The concentration of NO2 decreases from its instantaneous peak, so the colour pales. Keq is unchanged.

Marking notes. Stage 1: 1 mark for ‘colour darkens immediately’; 1 mark for explanation (concentration/density increase, not equilibrium shift). Stage 2: 1 mark for ‘colour pales from peak’ and LCP + mole count (2 NO2 left, 1 N2O4 right, shifts right); 1 mark for linking NO2 consumption (right shift) to reduced colour intensity.

2.1

Section 2 · Data response · 8 marks · Band 4–5

Sample response (a). As pressure increases from 1 atm to 50 atm at 500°C, the equilibrium % yield of SO3 increases, from approximately 30% at 1 atm to approximately 87% at 50 atm. The relationship is non-linear: larger gains in yield occur at lower pressures, with diminishing returns as pressure increases further. The trend is consistent with Le Chatelier's Principle prediction for a reaction that produces fewer moles of gas. 1 mark for direction (increases with pressure); 1 mark for supporting data values from the graph.

Sample response (b). 2SO2(g) + O2(g) ⇌ 2SO3(g): left side 2 + 1 = 3 mol gas; right side 2 mol gas. Le Chatelier's Principle: increasing pressure shifts equilibrium toward fewer moles of gas — the right side (products). More SO3 is formed at new equilibrium, so % yield increases. 1 mark for mole count (3 left, 2 right); 1 mark for applying LCP (shift right); 1 mark for linking to increased SO3 yield.

Sample response (c). The forward reaction is exothermic (ΔH = −196 kJ/mol). By Le Chatelier's Principle, increasing temperature favours the endothermic direction (reverse) — equilibrium shifts left, so less SO3 is produced and the % yield decreases. Keq decreases as temperature rises because the ratio of equilibrium product concentrations to reactant concentrations is smaller at higher temperatures. 1 mark for identifying forward reaction is exothermic; 1 mark for LCP — rise in T shifts equilibrium left / toward endothermic direction; 1 mark for stating Keq decreases as temperature increases.

3.1

Section 2 · Source critique · 4 marks · Band 4–5

Sample response. The statement is incorrect in two ways. First, adding a catalyst does not shift the equilibrium position. A catalyst lowers the activation energy of both the forward and reverse reactions equally, so both rates increase by the same factor and no net shift occurs. The equilibrium concentrations of SO2, O2, and SO3 are unchanged. Second, the catalyst does not improve the yield of SO3. Yield at equilibrium is determined by Keq, which depends only on temperature. The correct role of the V2O5 catalyst in the Contact process is to allow equilibrium to be reached rapidly at the industrially chosen temperature (∼450°C), without changing the yield. Without the catalyst, the reaction rate at 450°C would be too slow for commercial production.

Marking notes. 1 mark for identifying the flaw: catalyst does not shift equilibrium to the right; 1 mark for mechanism: lowers Ea equally for both directions, so no net shift; 1 mark for identifying the second flaw: catalyst does not improve yield / Keq; 1 mark for correct role of catalyst: speeds attainment of equilibrium, not yield change.

4.1

Section 3 · Extended response · 7 marks · Band 5–6

Sample response. The Haber process N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ/mol, is the industrial synthesis of ammonia and is operated at Incitec Pivot's Gibson Island plant at approximately 200 atm and 400–500°C with an iron catalyst. Each operating condition involves a trade-off between yield and rate.

Pressure (changes equilibrium position, not Keq): The left side has 4 moles of gas (1 N2 + 3 H2); the right has 2 (2 NH3). By Le Chatelier's Principle, increasing pressure shifts equilibrium right toward fewer gas moles — increasing yield. 200 atm is chosen rather than higher pressures because the engineering and capital cost of building vessels that safely contain >200 atm exceeds the additional revenue from the yield gain. This is an economic constraint: Le Chatelier says “higher is better”, but engineering says “200 atm is the practical optimum.” Pressure does not change Keq — it changes only the equilibrium position.

Temperature (changes Keq): The forward reaction is exothermic. Lower temperatures shift equilibrium right (more NH3, higher Keq) and higher temperatures shift left (less NH3, lower Keq). Temperature is the only factor that changes Keq. At 300°C the yield is very high but the reaction rate is unacceptably slow — even with catalyst. At 400–500°C, the rate is commercially viable. This is the rate–yield compromise: higher temperature sacrifices yield (lower Keq) to gain a usable rate.

Catalyst (changes neither equilibrium position nor Keq): The iron catalyst lowers the activation energy of both the forward and reverse reactions equally. It does not preferentially accelerate the forward reaction, does not shift the equilibrium position, and does not change Keq. Its sole industrial function is to allow equilibrium to be attained in a residence time compatible with continuous flow operation at 400–500°C. Without the catalyst, the rate would be too slow for profitability.

Assessment of Incitec Pivot conditions: The chosen conditions (200 atm, 400–500°C, Fe catalyst) represent the engineering and economic optimum: high enough pressure to achieve meaningful yield without unsustainable infrastructure cost; high enough temperature to achieve commercial rate despite the yield compromise; and a catalyst to accelerate equilibrium attainment. Neither the catalyst nor the pressure changes Keq; temperature is the only lever for Keq.

Marking notes. 1 mark — gas mole count and LCP applied correctly to pressure (4 left, 2 right, shifts right); 1 mark — pressure increases yield but does not change Keq; 1 mark — identifies rate–yield trade-off for temperature (lower T = higher yield but slower rate); 1 mark — states temperature is the only factor that changes Keq (and states direction: exothermic, higher T lowers Keq); 1 mark — correctly describes catalyst role: lowers Ea equally for both directions, no change to position or Keq; 1 mark — uses Incitec Pivot / Australian industrial context with at least one specific value (200 atm or 400–500°C); 1 mark — reaches an explicit evaluative judgement about why the chosen conditions represent a commercially rational compromise.