HSC Exam Practice

Sample response. An industrial compromise is the set of operating conditions (temperature, pressure, catalyst) chosen to balance the conflicting demands of equilibrium yield and reaction rate in a way that makes the process economically viable, accepting that neither yield nor rate is fully optimised.

Marking notes. 1 mark for identifying the balance between yield and rate; 1 mark for linking to economic/commercial viability or identifying that neither is maximised alone.

Sample response. A catalyst lowers the activation energy (E_a) for both the forward and reverse reactions by the same amount. Because both rates increase by the same factor, the ratio of forward to reverse rates (which determines K_eq) is unchanged, so the equilibrium position and yield are unaffected. However, because more particles now have sufficient energy to overcome the lower E_a, both rates increase substantially, so equilibrium is reached much faster.

Marking notes. 1 mark for stating that E_a is lowered equally for both directions; 1 mark for explaining that equal rate changes leave K_eq and equilibrium yield unchanged; 1 mark for explaining that lower E_a means more particles exceed it, increasing rate and reaching equilibrium faster.

Sample response. Contact process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g); catalyst is vanadium(V) oxide (V₂O₅). The operating pressure is kept low because the equilibrium yield at 1–2 atm is already very high (~97–99%); the additional yield gain from increasing to very high pressures (like the Haber process) is too small to justify the enormous engineering cost and safety risks of high-pressure equipment.

Marking notes. 1 mark for correct equation or identification of SO₂ + O₂ → SO₃; 1 mark for naming V₂O₅; 1 mark for explaining that the already-high equilibrium yield at low pressure does not economically justify high-pressure engineering (or: the LCP gain from higher pressure is marginal given the already-favourable equilibrium).

Sample response. Effect on rate: increasing temperature increases the average kinetic energy of particles. A greater proportion of particles now exceed the activation energy (E_a), leading to more frequent effective collisions; the rate increases. Effect on equilibrium yield: the Haber process forward reaction is exothermic (ΔH = −92 kJ mol⁻¹). By Le Chatelier’s Principle, increasing temperature causes the system to shift in the endothermic (reverse) direction to oppose the disturbance; K_eq decreases and equilibrium yield of NH₃ decreases. These two effects are in direct opposition, creating the yield–rate trade-off.

Marking notes. 1 mark for correct collision theory explanation of rate increase (must mention E_a or proportion of particles with sufficient energy); 1 mark for correct LCP explanation of yield decrease with temperature for exothermic forward reaction; 1 mark for explicitly stating the two effects are in opposition (trade-off).

Sample response. The unreacted N₂ and H₂ gases are separated from the NH₃ product by condensation (NH₃ liquefies at higher temperatures than N₂ and H₂) and then recycled back into the reactor. Over many passes, essentially all the reactants are eventually converted to NH₃, giving >95% overall conversion despite only 15–25% per pass.

Marking notes. 1 mark for identifying the recycling/return of unreacted gases; 1 mark for explaining the separation mechanism (condensation of NH₃) or stating that multiple passes accumulate to high overall conversion.

Sample response. Removing NH₃ decreases the concentration of the product. By Le Chatelier’s Principle, the system shifts in the forward direction (to the right) to oppose the disturbance, producing more NH₃ from the remaining N₂ and H₂ until a new equilibrium is reached.

Marking notes. 1 mark for stating the equilibrium shifts to the right / forward direction; 1 mark for a correct LCP explanation (system opposes the decrease in product concentration by producing more product).

Part (a) — 1 mark. Approximately 22–26% (accept 20–28%); the graph shows the curve at 200 atm yields approximately 24%.

Part (b) — 3 marks. Le Chatelier’s Principle: the reaction N₂ + 3H₂ ⇌ 2NH₃ has 4 mol gas on the left and 2 mol on the right. Increasing pressure shifts equilibrium to the right (fewer gas moles), increasing yield [1]. The graph flattening above ~300 atm indicates diminishing returns — each additional 100 atm increase adds progressively less yield [1]. Engineering cost for high-pressure vessels, pumps, and seals increases dramatically with pressure; the small extra yield (e.g. from ~36% at 300 atm to ~47% at 400 atm) does not justify the proportionally much larger increase in capital cost, energy use, and safety risk [1].

Part (c) — 3 marks. The student’s claim is incorrect [1]. K_eq is defined as the ratio of equilibrium concentrations at a given temperature; it changes only with temperature, not pressure [1]. The graph shows that as pressure increases, the equilibrium position shifts (more NH₃ at equilibrium in proportion), but this is a shift in the equilibrium position — the equilibrium constant K_eq itself is unchanged. The higher yield at higher pressure means the same K_eq is satisfied at different concentration ratios when the total pressure changes [1].

Part (a) — 3 marks. ICE table: N₂ starts at 25.0 mol; 18.0 mol NH₃ produced requires 18.0/2 = 9.0 mol N₂ consumed [1] and 3 × 9.0 = 27.0 mol H₂ consumed [1]. Remaining: N₂ = 25.0 − 9.0 = 16.0 mol; H₂ = 75.0 − 27.0 = 48.0 mol [1].

Part (b) — 2 marks. Maximum NH₃ possible: limited by N₂ (25.0 mol); max NH₃ = 2 × 25.0 = 50.0 mol [1]. Percentage yield = (18.0 / 50.0) × 100 = 36.0% [1]. (Accept 18.0/50.0 = 36.0%; full marks only if working shown.)

Part (c) — 2 marks. Assumption: NH₃ and the unreacted N₂/H₂ must have sufficiently different boiling points (or condensation temperatures) for complete separation by condensation at the operating pressure [1]. Limitation: any inert gas impurities (e.g. Ar from the N₂ feed) will accumulate in the recycled gas stream and must be purged periodically, reducing efficiency; or: energy cost of compression/recycling must be factored into commercial viability [1].

Sample response. The statement is incorrect: the conditions chosen for the Haber process do not maximise yield but rather represent a carefully negotiated compromise between yield, rate, and cost. At thermodynamically ideal conditions (low temperature, very high pressure), yield would be maximised — but commercial production would be impossible. The forward reaction (N₂ + 3H₂ → 2NH₃, ΔH = −92 kJ mol⁻¹) is exothermic; Le Chatelier’s Principle predicts that lower temperatures increase K_eq and yield. However, at low temperatures (< 300°C), the collision frequency of particles with energy exceeding E_a is too low for any practical rate — equilibrium would take years to reach. The industrial temperature of 400–500°C is a compromise: it allows the iron catalyst to function (active above ~300°C) while providing a commercially acceptable rate, despite giving a per-pass yield of only 15–25%. High pressure (150–300 atm) is used because it is the one variable where yield and rate both improve (LCP: 4 mol gas → 2 mol gas, shift right; rate increases via higher concentration). It is limited only by engineering cost, not chemistry. The iron catalyst lowers E_a equally for both reactions, increasing rate without changing K_eq or yield — it is the element that makes the industrial compromise feasible at 400–500°C. Recycling compensates for the low per-pass yield by returning unconverted N₂ and H₂, achieving >95% overall conversion. The conditions are therefore not chemically optimal (room temperature with extreme pressure would give higher yield) but are economically optimal: a compromise that maximises profit, not yield.

Marking notes.

• 1 mark — States and explains the trade-off for temperature: lower T gives higher yield (LCP, exothermic) but slower rate (collision theory, fewer particles exceed E_a).
• 1 mark — States and explains the compromise at 400–500°C with reference to catalyst activation temperature.
• 1 mark — Correctly explains that pressure improves both yield (LCP: 4 mol → 2 mol) AND rate (collision theory: higher concentration), and is limited only by engineering cost.
• 1 mark — Correctly explains the catalyst’s role: lowers E_a for both reactions equally; increases rate without changing K_eq or yield.
• 1 mark — Explains how recycling compensates for low per-pass yield to achieve high overall conversion.
• 1 mark — Uses both LCP and collision theory correctly in the response.
• 1 mark — Reaches an explicit, justified evaluative conclusion: conditions are NOT chemically optimal (yield not maximised) but are the best commercial/economic compromise.
• 1 mark — Quality: structured, precise language, correct use of K_eq, E_a, ΔH notation, evidence-based argumentation throughout.