HSC Exam Practice

Sample response. The equilibrium concentration is the concentration of a species after the system has reached chemical equilibrium — the point at which the rates of the forward and reverse reactions are equal and concentrations remain constant. The initial concentration is the concentration of each species before any reaction has occurred, placed in the I (Initial) row of the ICE table. The equilibrium concentration equals the initial concentration plus the net change (the Change row entry), and is recorded in the E (Equilibrium) row.

Marking notes. 1 mark — correctly defines equilibrium concentration (constant, forward = reverse rate, or equivalent). 1 mark — correctly distinguishes from initial concentration by identifying the Change row or stating that initial concentration precedes any reaction.

Sample response. H₂O(l) and Fe₂O₃(s) are excluded. H₂O(l) is a pure liquid — its concentration is constant (approximately 55.5 mol L⁻¹) and does not change meaningfully during the reaction; its activity is defined as 1 and is incorporated into the numerical value of Keq. Fe₂O₃(s) is a pure solid, similarly excluded for the same reason. PCl₃(g) and HCl(aq) are gases and aqueous solutes, respectively; their concentrations change during the reaction and they are included in the Keq expression.

Marking notes. 1 mark — correctly identifies H₂O(l) as excluded with correct reason (pure liquid / constant concentration / activity = 1). 1 mark — correctly identifies Fe₂O₃(s) as excluded with correct reason (pure solid). 1 mark — correctly identifies PCl₃(g) and HCl(aq) as included (or implicitly by contrast). Do not award if a student also incorrectly excludes PCl₃ or HCl.

Sample response. Error: the student did not apply stoichiometric coefficients to the Change row. For 2SO₂ + O₂ ⇌ 2SO₃, the coefficients are 2, 1 and 2. The correct Change row is: SO₂: −2x; O₂: −x; SO₃: +2x. (x is defined as the decrease in [O₂].)

Marking notes. 1 mark — identifies the error (ignoring/not using stoichiometric coefficients or equivalent). 1 mark — writes the correct Change row with all three entries correct.

Sample response. The verification step involves substituting the E-row equilibrium concentrations back into the Keq expression and checking that the result matches the given Keq (within approximately ±2% for rounding). Its purpose is to detect arithmetic errors, stoichiometric ratio errors, and invalid use of the simplifying assumption that would otherwise go unnoticed. If the result differs from the given Keq by more than ~5%, the student should (1) check the Change row for correct stoichiometric ratios; (2) check whether the simplifying assumption was invalid and, if so, discard that answer and solve the full quadratic; and (3) recheck all algebraic steps.

Marking notes. 1 mark — states the purpose of verification (confirm answer is consistent with given Keq / detect errors). 1 mark — describes what to do (identify and locate the error, e.g. check stoichiometry or assumption validity). 1 mark — mentions the specific actions: check Change row stoichiometry AND/OR re-solve using the quadratic formula.

Sample response. The simplifying assumption is valid when x is very small relative to the initial concentration, specifically when (x / [initial]) × 100 < 5%. This can be pre-checked using the threshold Keq / [initial] × 100 < 5%; if this pre-check passes, x is likely to be small enough. After solving, the post-check (x / [initial] × 100) must confirm the assumption.

Marking notes. 1 mark — states the condition (x / [initial] < 5% or Keq / [initial] < 5%). 1 mark — specifies the numerical threshold (5%) and indicates both a pre-check and/or post-check is required.

Sample response. Q (the reaction quotient) has the same mathematical form as Keq but is calculated using concentrations at any point in time, not necessarily at equilibrium. Keq is the value of Q specifically at equilibrium at a given temperature, and is constant for a given reaction at that temperature. If Q < Keq, the numerator (products) is too small and the reaction proceeds forward to increase [products] and decrease [reactants] until Q = Keq. If Q > Keq, the numerator is too large and the reaction proceeds in reverse until Q = Keq. If Q = Keq, the system is already at equilibrium.

Marking notes. 1 mark — correctly distinguishes Q (any point) from Keq (at equilibrium). 1 mark — Q < Keq = forward shift; Q > Keq = reverse shift. 1 mark — Q = Keq means system is at equilibrium / no net change.

Part (a) — 2 marks. Sample response. As initial [H₂CO₃] increases, the equilibrium [H⁺] also increases, but not proportionally — the graph shows a curve of diminishing gradient (square-root shape), indicating that [H⁺] grows more slowly than [H₂CO₃]₀.

1 mark — identifies that [H⁺] increases with increasing [H₂CO₃]₀. 1 mark — identifies the non-linear / square-root / diminishing-gradient nature of the increase (must be specific about shape).

Part (b) — 5 marks. ICE table setup and Keq calculation.

ICE table: Initial: [H₂CO₃] = 0.0340; [H⁺] = 0; [HCO₃⁻] = 0. Change: H₂CO₃: −x; H⁺: +x; HCO₃⁻: +x. Equilibrium: H₂CO₃: 0.0340 − x; H⁺: x; HCO₃⁻: x. Since [H⁺]_eq = 1.21 × 10⁻⁴ mol L⁻¹, x = 1.21 × 10⁻⁴. 5% check: x/[initial] = 1.21×10⁻⁴/0.0340 = 0.356% < 5% — assumption valid. [H₂CO₃]_eq = 0.0340 − 1.21×10⁻⁴ ≈ 0.03388 mol L⁻¹. Keq = x² / (0.0340 − x) = (1.21×10⁻⁴)² / 0.03388 = 1.464×10⁻⁸ / 0.03388 = 4.32×10⁻⁷ ≈ 4.3×10⁻⁷. Consistent with given Keq.

Marking: 1 mark ICE table correct; 1 mark x identified from given [H⁺]_eq; 1 mark 5% check valid with working; 1 mark correct Keq expression substituted; 1 mark correct numerical value consistent with 4.3×10⁻⁷.

Part (c) — 2 marks. Sample response. Because the simplifying assumption is valid (x << [initial]), the ICE table gives [H⁺]_eq = x ≈ √(K_eq × [H₂CO₃]₀). This is a square-root relationship between [H⁺] and [H₂CO₃]₀, which produces a curve of the shape observed in Figure 2.1 — [H⁺] increases as [H₂CO₃]₀ increases, but at a decreasing rate (proportional to the square root, not linearly).

1 mark — derives or states the square-root relationship [H⁺] = √(Keq × [initial]). 1 mark — explicitly links this to the diminishing-gradient curve shape in Figure 2.1.

Part (a) — 2 marks. Check: Keq/[H₂] = 54.3/0.500 = 108.6%. This is much greater than 5%, so the simplifying assumption is NOT valid and a different approach is needed. However, for H₂ + I₂ ⇌ 2HI with equal initial concentrations of H₂ and I₂, the Keq expression reduces to (2x)²/([initial]−x)² = Keq, which can be solved by taking the square root of both sides without needing the quadratic formula. Taking √Keq = √54.3 = 7.369 and 2x/(0.500−x) = 7.369 is a linear equation in x. So the simplifying assumption is not needed; the equation can be solved exactly by rearranging linearly. 1 mark for attempting the 5% check; 1 mark for identifying that the square-root shortcut applies here (or for correctly noting assumption invalid and proposing the square-root method).

Part (b) — 3 marks. ICE: H₂: I; I₂: Initial=0.500; HI: 0. Change: H₂: −x; I₂: −x; HI: +2x. Equilibrium: H₂: 0.500−x; I₂: 0.500−x; HI: 2x. Keq = (2x)²/((0.500−x)²) = 54.3. Take square root: 2x/(0.500−x) = 7.369. 2x = 7.369(0.500−x) = 3.685 − 7.369x. 9.369x = 3.685. x = 0.3933 mol L⁻¹. [H₂]_eq = [I₂]_eq = 0.500 − 0.3933 = 0.107 mol L⁻¹. [HI]_eq = 2(0.3933) = 0.787 mol L⁻¹. 1 mark ICE table; 1 mark solving for x; 1 mark correct E-row concentrations.

Part (c) — 2 marks. Keq = (0.787)² / ((0.107)²) = 0.6194 / 0.01145 = 54.1 ≈ 54.3. Consistent with given Keq (within rounding). 1 mark for correct substitution; 1 mark for correct statement of consistency.

Sample response. The claim contains a partial truth and a significant error. The simplifying assumption is valid and time-saving in a specific and mathematically defined set of circumstances: when the ratio Keq/[initial] × 100 is less than 5%, x is small relative to the initial concentration and omitting it introduces less than 5% error, which is within acceptable rounding for HSC. For example, for H₂CO₃ dissociation with Keq = 4.3×10⁻⁷ and [initial] = 0.0340 mol L⁻¹, the ratio is approximately 0.0013%, far below the threshold — the assumption is entirely justified and produces [H⁺]_eq = √(4.3×10⁻⁷ × 0.0340) = 1.21×10⁻⁴ mol L⁻¹ with no measurable error. However, the claim that the simplifying assumption is always acceptable and that the quadratic formula is never needed is incorrect. When Keq is not much smaller than [initial] — specifically when Keq/[initial] × 100 ≥ 5% — the assumption introduces significant error. For example, for PCl₅ decomposition with Keq = 0.041 and [PCl₅]₀ = 0.500 mol L⁻¹, the ratio is 8.2% > 5%. Applying the assumption anyway gives x = √(0.041 × 0.500) = 0.143 mol L⁻¹ and [PCl₅]_eq = 0.357 mol L⁻¹. Substituting back: Keq = (0.143)²/0.357 = 0.0573, a 40% discrepancy from 0.041 — the assumption has failed. The correct approach uses the quadratic: x² + 0.041x − 0.0205 = 0, giving x = 0.124 mol L⁻¹ and verification Keq = (0.124)²/0.376 = 0.0409 ≈ 0.041. The claim is therefore partially correct (assumption saves time when valid) but incorrect in claiming it is always acceptable and that the quadratic is never needed — HSC questions routinely require the quadratic when Keq/[initial] exceeds 5%.

Marking notes. 1 mark — correctly identifies the condition for the assumption being valid (5% rule / Keq << [initial] with the threshold stated). 1 mark — uses a correct numerical example showing the assumption IS valid (any valid example with Keq/[initial] < 5%, showing the calculation). 1 mark — identifies the condition under which the claim is WRONG (Keq/[initial] ≥ 5%). 1 mark — uses a correct numerical example where the assumption fails, calculating the incorrect result from the assumption. 1 mark — calculates the correct result using the quadratic formula for the same example. 1 mark — verifies the quadratic result by substitution or equivalent. 1 mark — reaches an explicit evaluative judgement rejecting the claim by stating the conditions under which the assumption is valid vs invalid, without overgeneralising in either direction.