Writing Keq Expressions

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

The K_eq expression for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is K_eq = [SO₃]² / ([SO₂]²[O₂]). All three species are gases (g) — all are included; no exclusions. [1 — correct expression including exponents and inclusion rationale]

At 400°C, K_eq = 1.0 × 10⁸ — much greater than 10³, so the equilibrium strongly favours products (SO₃), and ~99.6% of SO₂ is converted. At 550°C, K_eq = 3.2 × 10³ — still products-favoured, but only ~80% yield; both reactants and products are present in meaningful amounts. [1 — interpretation of K_eq at 400°C; 1 — interpretation at 550°C with reference to data]

K_eq decreases as temperature rises because the forward reaction is exothermic (ΔH = −197 kJ mol⁻¹). Increasing temperature adds energy equivalent to adding a product, shifting equilibrium left via Le Chatelier’s principle and decreasing the ratio [products]/[reactants] at equilibrium — hence K_eq decreases. More precisely, rising T increases the Gibbs free energy change ΔG°, making K_eq = e^−ΔG°/RT smaller. [1 — exothermic mechanism linked to K_eq decrease]

Consequences of a lower operating temperature (e.g. 400°C): (1) K_eq = 1.0 × 10⁸ gives ~99.6% yield — excellent from an equilibrium perspective; (2) however, lower temperatures slow reaction kinetics, meaning the catalyst (V₂O₅) is less active and equilibrium is reached more slowly, reducing production rate per unit time. [1 — two consequences at lower T, one K_eq and one outside K_eq]

Consequences of a higher operating temperature (e.g. 550°C): (1) K_eq = 3.2 × 10³ gives only ~80% yield — significantly less SO₃ at equilibrium; (2) but higher temperature accelerates reaction kinetics, so equilibrium is achieved faster and throughput is higher. [1 — two consequences at higher T]

At ~450°C, K_eq = 1.8 × 10⁶ gives ~98% yield — still an extremely high conversion — while the catalyst remains sufficiently active for commercially viable production rates. This represents the optimum trade-off: yield is only marginally below the 400°C maximum while kinetics are practical. The data confirm that 400°C is theoretically superior on K_eq grounds but commercially inferior due to kinetics. [1 — justified conclusion using data; 1 — explicit trade-off reasoning linking K_eq and kinetics]

Marking criteria:

• 1 mark — Correct K_eq expression with stoichiometric exponents; states all are gas-phase species.
• 1 mark — Correctly interprets K_eq at 400°C: much greater than 1, strongly products-favoured, ~99.6% yield from data.
• 1 mark — Correctly interprets K_eq at 550°C: still products-favoured but significantly reduced, ~80% yield, both species present.
• 1 mark — Explains K_eq decrease with temperature using exothermic nature of reaction.
• 1 mark — Two consequences of lower temperature: better K_eq/yield AND slower kinetics (or catalyst activity).
• 1 mark — Two consequences of higher temperature: worse K_eq/yield AND faster kinetics.
• 1 mark — Justified conclusion: 450°C is optimal trade-off, using data values.
• 1 mark — Explicit identification of the K_eq-vs-kinetics trade-off as the reason 400°C is not used despite superior K_eq.

Q2 — Sample Band 6 response (7 marks), annotated

What is defensible in the article: It is correct that K_eq for CaCO₃ dissolution increases with temperature — dissolution is endothermic, so higher temperatures do shift K_sp to a higher value, increasing solubility slightly. It is also correct that coral dissolution is accelerating. [1 — identifies the defensible element]

Error 1 — conflating equilibrium position with K_eq: The article claims that CO₂ emissions cause a “permanently higher K_eq”. This is incorrect. Ocean acidification is a concentration change (H⁺ from dissolved CO₂ removes CO₃²⁻), not a temperature change. Concentration changes shift the equilibrium position (more CaCO₃ dissolves) but do not alter K_sp. K_sp depends only on temperature. [1 — identifies error: CO₂ / acidification changes position not K_sp; 1 — explains mechanism: CO₃²⁻ removal shifts position right, K_sp unchanged]

Error 2 — “damage to K_eq is irreversible”: K_sp is not “damaged” by CO₂ emissions and is not irreversible in the chemical sense. If CO₂ concentrations in seawater decreased (e.g. by reducing emissions or by carbon capture), [H⁺] would fall, [CO₃²⁻] would recover, the dissolution equilibrium would shift back left (toward CaCO₃ formation), and coral calcification could resume. The K_sp value itself is a temperature-dependent thermodynamic constant — it is neither damaged nor made irreversible by atmospheric CO₂. [1 — refutes irreversibility; 1 — explains that reducing CO₂ restores equilibrium position]

Correct K_sp expression: K_sp = [Ca²⁺][CO₃²⁻] — CaCO₃ is excluded as a pure solid (activity = 1). The K_sp = 3.4 × 10⁻⁹ at 25°C — much less than 1 — indicates CaCO₃ is sparingly soluble; ocean acidification shifts this equilibrium to the right by removing CO₃²⁻, but K_sp remains constant unless temperature changes. [1 — correct expression and K_sp interpretation]

Defensible reformulation: “Ocean acidification caused by rising CO₂ reduces [CO₃²⁻] in seawater, shifting the CaCO₃ dissolution equilibrium to the right (more coral dissolves). This is a shift in equilibrium position — K_sp itself has not changed and would not change unless ocean temperature changes. Reducing atmospheric CO₂ would allow [CO₃²⁻] to recover, shifting the equilibrium back toward CaCO₃ formation — the process is chemically reversible at the level of equilibrium position, though geological and biological damage on human timescales may not be.” [1 — defensible reformulation distinguishing position from K_sp]

Marking criteria:

• 1 mark — Identifies the one defensible element: K_sp does increase slightly with temperature (dissolution is endothermic) and coral dissolution is indeed accelerating.
• 1 mark — Correctly identifies Error 1: ocean acidification / CO₂ changes equilibrium position, not K_sp.
• 1 mark — Explains the mechanism: CO₃²⁻ is removed by H⁺, shifting dissolution equilibrium right; K_sp depends only on temperature.
• 1 mark — Correctly identifies Error 2: the claim that K_sp is “irreversibly damaged” is wrong; K_sp is a thermodynamic constant.
• 1 mark — Explains reversibility: reducing CO₂ → H⁺ falls → CO₃²⁻ recovers → equilibrium shifts left (back toward calcification).
• 1 mark — Writes the correct K_sp expression ([Ca²⁺][CO₃²⁻]) and explains CaCO₃ exclusion (pure solid, activity = 1).
• 1 mark — Provides a scientifically defensible reformulation that correctly distinguishes equilibrium position from K_sp and avoids claiming irreversibility at the K_sp level.