HSC Exam Practice

Sample response. Le Chatelier’s Principle states that when a disturbance is applied to a system at equilibrium, the system will shift in the direction that partially counteracts the disturbance and re-establishes equilibrium. It applies only to systems at equilibrium because only at equilibrium are forward and reverse rates equal — any net reaction (rate imbalance) is the mechanism by which the system counteracts the disturbance.

Marking notes. 1 mark for a correct definition (disturbance → shift to counteract); 1 mark for linking the principle’s applicability to the concept of equal forward and reverse rates at equilibrium.

Sample response. Option (b) — increasing temperature from 450°C to 600°C. Only temperature changes K_eq. For this exothermic forward reaction (ΔH = −197 kJ/mol), increasing temperature shifts equilibrium left (endothermic direction), and K_eq decreases. Options (a) pressure, (c) catalyst, and (d) concentration all shift the equilibrium position but leave K_eq unchanged.

Marking notes. 1 mark for correctly selecting (b) only; 1 mark for justification linking temperature change to K_eq change and explaining why the others do not alter K_eq.

Sample response. Removing H₂O(g) decreases its concentration — removing a product. LCP: the system shifts to the right (forward reaction) to partially counteract the decrease by producing more H₂O and consuming CO and H₂. Collision theory: removing H₂O decreases the concentration of products, so the reverse reaction frequency decreases; forward rate > reverse rate → net forward reaction until rates re-equalise. K_eq is unchanged: removing H₂O is a concentration change, and only temperature changes K_eq.

Marking notes. 1 mark for identifying direction of shift (right) and linking to removal of product via LCP; 1 mark for collision theory (reverse rate decreases, forward rate dominates → forward shift); 1 mark for stating K_eq is unchanged with justification (concentration change only).

Sample response. Adding a catalyst to a system already at equilibrium has no effect on the equilibrium position or on K_eq. The catalyst lowers the activation energy equally for both forward and reverse reactions, so both rates increase by the same factor — they remain equal. On a concentration–time graph, all concentrations remain flat (no change is observed) because the system was already at equilibrium and no net shift occurs.

Marking notes. 1 mark for stating no change in equilibrium position or K_eq, with reason (E_a lowered equally both directions); 1 mark for correctly describing the graph signature (all concentration lines remain flat — no observable change).

Sample response. Rate = the speed at which the system reaches equilibrium (kinetics: governed by temperature, catalyst, concentration, and surface area). Yield = the proportion of reactants converted to products at equilibrium (thermodynamics: governed by K_eq, which depends only on temperature). For the Haber process (ΔH = −92 kJ/mol, exothermic forward): increasing temperature increases the average kinetic energy of particles, raising collision frequency and the proportion of effective collisions — rate increases. However, for this exothermic forward reaction, increasing temperature also shifts equilibrium left (Le Chatelier, endothermic direction favoured) and decreases K_eq — meaning less NH₃ is present at equilibrium — yield decreases. Rate and yield therefore respond in opposite ways to temperature for this exothermic reaction.

Marking notes. 1 mark for correctly defining rate (speed to equilibrium, kinetics) vs yield (proportion at equilibrium, thermodynamics/K_eq); 1 mark for explaining why temperature increases rate (kinetic energy, collision frequency, effective collisions); 1 mark for explaining why temperature decreases yield for this exothermic reaction (ΔH < 0 → LCP shifts left at higher T → K_eq decreases → less NH₃ at equilibrium).

Sample response. N₂O₄(g) ⇌ 2NO₂(g): left = 1 mol gas, right = 2 mol gas.

Disturbance 1 — increasing pressure: LCP shifts toward fewer gas moles → shift LEFT (toward 1 mol N₂O₄). K_eq unchanged.

Disturbance 2 — removing NO₂: Removing a product → shift RIGHT to produce more NO₂. K_eq unchanged.

Combined: Pressure shift LEFT vs NO₂ removal shift RIGHT — effects oppose each other. The overall direction depends on the relative magnitudes, but K_eq is unchanged in either case (neither pressure nor concentration changes K_eq; only temperature does).

Marking notes. 1 mark for correct direction for pressure increase (left, fewer gas moles, with mole count); 1 mark for correct direction for NO₂ removal (right, product removed); 1 mark for correctly identifying the effects as opposing; 1 mark for stating K_eq is unchanged for both disturbances (neither is a temperature change).

Part (a) — 2 marks. The disturbance at t₁ is a temperature increase. Evidence: after t₁, all concentrations change gradually (no sudden jump), with [NH₃] falling and [N₂], [H₂] rising. The gradual drift (not an instantaneous jump) is the diagnostic signature of a temperature change. A concentration or pressure change would cause a sudden instantaneous step in one or all species.

Marking notes (a). 1 mark for correct identification (temperature increase); 1 mark for citing the graph evidence (gradual change, no sudden jump; NH₃ falls while N₂, H₂ rise → consistent with exothermic forward → shift left at higher T).

Part (b) — 3 marks. The disturbance at t₂ is addition of N₂ gas. Evidence: [N₂] shows a sudden upward jump at t₂ while [H₂] and [NH₃] are momentarily unchanged — this is the signature of adding a single species. After the jump, [N₂] gradually decreases, [H₂] decreases, and [NH₃] increases: equilibrium shifts RIGHT. LCP: adding a reactant (N₂) disturbs equilibrium; the system shifts right to partially consume the added N₂. K_eq is unchanged: N₂ addition is a concentration change, and only temperature changes K_eq. (K_eq in fact decreased at t₁ due to the temperature increase, and remains at this lower value throughout t₂.)

Marking notes (b). 1 mark for correct identification (N₂ added) and graph evidence (N₂ jumps suddenly, others unchanged); 1 mark for direction of shift (right) with LCP justification; 1 mark for stating K_eq unchanged for the t₂ disturbance (concentration change, not temperature).

Part (a) — 2 marks. At every temperature in the table, increasing pressure increases NH₃ yield (e.g., at 400°C: 36% → 47% → 57%). LCP: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) — left side has 4 mol gas, right side has 2 mol gas. Increasing pressure shifts equilibrium toward the side with fewer moles of gas (right), producing more NH₃. K_eq is unchanged by pressure.

Part (b) — 2 marks. At 300°C the equilibrium yield is higher (e.g., 52% at 100 atm vs 36% at 400°C), but the rate of reaching equilibrium is extremely slow at this low temperature — the reaction would take impractically long times to reach equilibrium even with the iron catalyst [1]. The industrial compromise at 400–500°C with the iron catalyst gives a commercially acceptable rate of production (reaches equilibrium in minutes, not days) at a yield that is still economically viable [1].

Marking notes. Part (a): 1 mark for describing the trend with data evidence; 1 mark for LCP explanation with mole count (4 mol → 2 mol, shift right). Part (b): 1 mark for identifying that 300°C is too slow kinetically (rate issue); 1 mark for the rate/yield trade-off at 400–500°C + catalyst as the compromise (not exceeding acceptable time for equilibrium).

Sample Band 6 response.

Disturbance 1 — Increasing temperature: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −197 kJ/mol. The forward reaction is exothermic. Increasing temperature adds heat to the system; LCP shifts equilibrium in the endothermic direction (reverse/left) to counteract the heat input. Products SO₃ are consumed; reactants SO₂ and O₂ increase. Collision theory: increasing temperature increases average kinetic energy of all particles. Because the reverse (endothermic) reaction has a higher activation energy E_a than the forward (exothermic) reaction, a proportionally greater fraction of particles now exceed the reverse E_a → reverse rate increases more than forward rate → net reverse shift. K_eq decreases (higher T for exothermic forward → smaller K_eq). Graph signature: no sudden jump; all concentrations drift gradually — [SO₃] falls, [SO₂] and [O₂] rise — until new equilibrium is established at lower [SO₃].

Disturbance 2 — Decreasing volume (increasing pressure): Left side: 2 + 1 = 3 mol gas. Right side: 2 mol gas. Decreasing volume increases the concentration of all gas species simultaneously. LCP: increasing pressure shifts equilibrium toward the side with fewer moles of gas → toward the right (2 mol SO₃ < 3 mol reactants). Graph signature: all gas concentrations jump upward simultaneously and proportionally (instantaneous, not gradual) as volume halves → then system re-equilibrates rightward from this higher baseline → [SO₃] rises further; [SO₂] and [O₂] fall back toward new values. K_eq is unchanged (pressure does not change K_eq).

Opposing effects and net result: Temperature increase shifts LEFT; pressure increase (volume decrease) shifts RIGHT → effects OPPOSE each other on equilibrium position. K_eq unambiguously decreases (only temperature changes K_eq; pressure does not). For equilibrium position: the net direction depends on magnitudes. A very large temperature increase will dominate and produce a net left shift (lower [SO₃]); a very large pressure increase may produce a net right shift. Without quantitative data, the net position direction is uncertain, but K_eq decreasing is certain.

Marking criteria.

• 1 mark — Identifies temperature increase shifts equilibrium left (endothermic reverse favoured, exothermic forward ΔH < 0) via LCP.
• 1 mark — States K_eq decreases due to temperature increase (only factor that changes K_eq).
• 1 mark — Provides collision theory for temperature: reverse (endothermic) reaction has higher E_a; greater fraction of particles exceed higher E_a at higher T → reverse rate increases more → net left shift.
• 1 mark — Correctly describes graph signature for temperature: gradual drift, no sudden jump; [SO₃] falls, [SO₂] and [O₂] rise.
• 1 mark — Identifies volume decrease shifts equilibrium right (toward fewer gas moles: 3 mol left vs 2 mol right) via LCP, with correct mole count.
• 1 mark — States K_eq unchanged by pressure/volume change, with reason.
• 1 mark — Correctly describes graph signature for pressure increase: all gas concentrations jump simultaneously (proportional to volume compression) before re-equilibration; distinct from gradual temperature signature.
• 1 mark — Correctly identifies that temperature and pressure effects on equilibrium position OPPOSE each other.
• 1 mark — States that K_eq decrease is unambiguous (temperature effect), but net position direction is uncertain without quantitative data (or correctly predicts which dominates with justification).
• 1 mark — Response is logically structured, uses correct chemistry terminology throughout (LCP, collision theory, E_a, K_eq, equilibrium position), and reaches a clear evaluative conclusion about both K_eq and position.