HSC Exam Practice

Sample response. The K_eq expression for aA + bB ⇌ cC + dD is: K_eq = [C]^c[D]^d / ([A]^a[B]^b). Products appear in the numerator and reactants in the denominator. Each molar concentration is raised to the power of its stoichiometric coefficient in the balanced equation as written. Gas-phase (g) and aqueous (aq) species are included. Pure solids (s) and pure liquids (l) (including water when acting as solvent) are excluded because their thermodynamic activity is defined as 1.

Marking notes. 1 mark — correct general form (products numerator, reactants denominator, concentrations). 1 mark — stoichiometric coefficients as exponents. 1 mark — correct exclusion rule: pure solids (s) and pure liquids (l) excluded because activity = 1; (g) and (aq) included.

Sample response. K_eq = [CO₂]. CaCO₃(s) and CaO(s) are both pure solids — their activities are defined as 1 and they are excluded from the K_eq expression. Only CO₂(g) is included as it is a gas-phase species whose concentration is variable.

Marking notes. 1 mark — correct expression K_eq = [CO₂]. 1 mark — correct justification: CaCO₃ and CaO are pure solids excluded because activity = 1.

Sample response.

(a) K_eq = [SO₃]² / ([SO₂]²[O₂]). No exclusions — all species are gases.

(b) K_eq = [CH₃COO⁻][H⁺] / [CH₃COOH]. No exclusions required for the species shown; water is the solvent (pure liquid, activity = 1) but does not appear in the equation here as a product or reactant to exclude.

(c) K_eq = [H₂O]⁴ / [H₂]⁴. Excluded: Fe₂O₃(s) and Fe(s) — both pure solids, activity = 1. Note: H₂O is a gas product (g) in this reaction, not the solvent, so it is included.

Marking notes. 2 marks per part: 1 mark for correct expression, 1 mark for correct exclusion statement (or “no exclusions” if correct). For (c), must identify both solids as excluded AND note H₂O(g) is included for full marks.

Sample response. K_eq is the numerical value of the ratio of equilibrium concentrations raised to stoichiometric powers — it is a thermodynamic constant that depends only on temperature. The equilibrium position describes the relative amounts of reactants and products actually present at equilibrium. A concentration change (e.g. adding more reactant) disturbs the equilibrium position — the system shifts to restore the equilibrium ratio, reaching a new position where [products]/[reactants]^coeff again equals K_eq. K_eq is unchanged because no new temperature was applied; the equilibrium position changed (shifted right), but the value of the constant did not.

Marking notes. 1 mark — defines K_eq as temperature-dependent constant (ratio of concentrations at equilibrium). 1 mark — defines equilibrium position as relative amounts of reactants/products. 1 mark — explains that concentration change shifts position but does not change K_eq; K_eq changes only with temperature.

Sample response.

(a) Reverse: K_eq(reverse) = 1 / K_eq(forward) = 1 / 977 = 1.02 × 10⁻³ ≈ 1.0 × 10⁻³.

(b) Halved coefficients: K_eq(new) = (K_eqoriginal)^1/2 = (977)^0.5 = 31.3. So K_eq for ½N₂ + 3/2H₂ ⇌ NH₃ = 31.3 at 300°C.

Marking notes. (a) 1 mark for reciprocal rule stated; 1 mark for correct calculation. (b) 1 mark for multiplier rule (raise to power ½); 1 mark for correct calculation (accept 31 to 32).

Sample response (a). 400°C: K_eq = 10⁸ — strongly products-favoured (almost entirely SO₃). 450°C: K_eq = 1.8 × 10⁶ — strongly products-favoured. 500°C: K_eq = 6.0 × 10⁴ — products-favoured (both present but mostly SO₃). 600°C: K_eq = 250 — products still favoured but less strongly; significant amounts of both present. [2 marks: 1 for interpreting top two rows correctly; 1 for interpreting lower two rows with distinction between them.]

Sample response (b). K_eq decreases as temperature increases. The forward reaction is exothermic, so increasing temperature is equivalent to adding “heat” as a product. By Le Chatelier’s principle the equilibrium shifts left (toward SO₂ and O₂), reducing [SO₃]/[SO₂]²[O₂] at the new equilibrium — hence K_eq is smaller at higher temperatures. [1 mark trend; 1 mark mechanism.]

Sample response (c). At 400°C, K_eq is larger (better equilibrium yield), but the catalyst (V₂O₅) becomes less active at lower temperatures, so the rate of reaching equilibrium is much slower — commercially impractical throughput. At 450°C, K_eq = 1.8 × 10⁶ still gives ~98% yield while the catalyst remains sufficiently active. This is the optimal trade-off between thermodynamic yield (K_eq) and kinetic practicality. [1 mark for kinetics/catalyst reasoning; 1 mark for explicit trade-off conclusion referencing both K_eq and rate.]

Sample response (a). K_eq = [NO₂]² / ([NO]²[O₂]) = 1.0 × 10¹². [1 mark for correct expression with exponents.]

Sample response (b). Given: N₂(g) + O₂(g) ⇌ 2NO(g), K_eq = 1.0 × 10⁻³⁰. Reaction 2 is NO(g) ⇌ ½N₂(g) + ½O₂(g), which is the reverse of the given reaction with all coefficients halved (divided by 2).

Step 1: reverse: K_eq(reversed) = 1 / (1.0 × 10⁻³⁰) = 1.0 × 10³⁰.

Step 2: halve coefficients (multiply by ½): K_eq(Reaction 2) = (1.0 × 10³⁰)^0.5 = 1.0 × 10¹⁵. [1 mark reversal; 1 mark ½-power rule; 1 mark correct final value.]

Sample response (c). K_eq = 1.0 × 10¹⁵ for NO(g) ⇌ ½N₂ + ½O₂ is extremely large (much greater than 10³), meaning NO is strongly thermodynamically unstable at 25°C — at equilibrium virtually all NO decomposes. NO persists in the atmosphere at 25°C because the decomposition is kinetically very slow (high activation energy), not because K_eq favours NO. [1 mark for correct interpretation linking large K_eq to thermodynamic instability of NO.]

Sample response. The student’s statement contains two distinct errors.

Error 1 — K_eq is not a percentage. K_eq is the ratio of equilibrium concentrations raised to stoichiometric powers: K_eq = [products]^stoich / [reactants]^stoich. A value of K_eq = 0.01 does not mean 1% of reactants have converted. The actual percentage conversion depends on initial concentrations, temperature, and (for gases) pressure as well as K_eq. For example, for the Haber process N₂ + 3H₂ ⇌ 2NH₃ with K_eq = 0.013 at 500°C, the actual per-pass conversion at 200 atm is approximately 15–25% — not 1.3%. K_eq = 0.01 indicates the equilibrium position favours reactants (K_eq < 1), but the numerical value is not a direct percentage. [2 marks: 1 for identifying the error; 1 for correct explanation with example.]

Error 2 — Adding reactant does not change K_eq. The student claims that adding more reactant increases K_eq. This is incorrect. K_eq is a thermodynamic constant that depends only on temperature. Adding more reactant is a concentration change. By Le Chatelier’s principle, the system responds by shifting the equilibrium position to the right (more product forms), until the ratio [products]^stoich/[reactants]^stoich again equals K_eq. The value of K_eq is the same before and after the addition — the system simply establishes equilibrium at different concentrations that still satisfy K_eq. For example, if additional N₂ is added to the Haber process at 500°C, more NH₃ forms and [N₂] is partially restored until K_eq = [NH₃]²/([N₂][H₂]³) = 0.013 is re-established. K_eq = 0.013 throughout. [3 marks: 1 for identifying the error; 1 for explaining K_eq depends only on temperature; 1 for correct mechanism using Le Chatelier’s principle with named example.]

Correct framing. K_eq is a dimensionless ratio encoding the thermodynamic balance between products and reactants at a specific temperature. It tells us the direction of the equilibrium and, combined with initial concentrations, allows calculation of equilibrium concentrations — but its numerical value is not itself a conversion percentage, and only temperature changes it. [2 marks: 1 for correct definition/framing; 1 for overall evaluative judgement that explicitly refutes both errors and gives the defensible alternative.]

Marking criteria:

• 1 mark — Identifies Error 1: K_eq is not a percentage conversion.
• 1 mark — Explains correct meaning of K_eq and gives a worked example (Haber or equivalent) showing percentage depends on other factors.
• 1 mark — Identifies Error 2: adding reactant does not change K_eq.
• 1 mark — Explains that K_eq depends only on temperature; concentration changes shift position not K_eq.
• 1 mark — Explains the Le Chatelier shift mechanism (equilibrium shifts right, new concentrations satisfy the same K_eq value).
• 1 mark — Uses at least one named equilibrium from the lesson (Haber, Contact Process, CaCO₃, acetic acid dissociation, or similar) as a concrete example for either error.
• 1 mark — Reaches an explicit, coherent evaluative judgement that refutes both errors and provides the correct framing of what K_eq represents.