HSC Exam Practice

Sample response. The reaction quotient Q is the ratio of product concentrations raised to their stoichiometric powers to reactant concentrations raised to their stoichiometric powers, calculated using the concentrations present at any given moment, whether or not the system is at equilibrium. K_eq uses the same algebraic expression but is evaluated using concentrations measured specifically at equilibrium, and is constant at a given temperature. The key difference is which concentrations are used: Q uses current (instantaneous) concentrations; K_eq uses equilibrium concentrations.

Marking notes. 1 mark: Q defined as same ratio as K_eq using current (not necessarily equilibrium) concentrations. 1 mark: K_eq defined as the same expression but using equilibrium concentrations and noted as constant at fixed temperature. 1 mark: explicitly distinguishes Q from K_eq by the type of concentrations used.

Sample response. (i) Q < K_eq: too few products relative to equilibrium → shift right (forward reaction favoured, more products form). (ii) Q > K_eq: too many products relative to equilibrium → shift left (reverse reaction favoured, reactants reform). (iii) Q = K_eq: system is at equilibrium → no net shift (forward and reverse rates are equal).

Marking notes. 1 mark per correct relationship with correct direction (3 marks total). “Shift right” and “shift left” must be stated; direction alone without comparison earns 0 for that entry.

Sample response. Q = [NH₃]² / ([N₂][H₂]³) = (0.150)² / (0.200)(0.300)³ = 0.02250 / (0.200 × 0.02700) = 0.02250 / 0.005400 = 4.17. Q = 4.17 > K_eq = 0.500 → system shifts left (reverse direction); NH₃ decomposes to N₂ and H₂.

Marking notes. 1 mark: correct Q expression written. 1 mark: correct numerical answer (Q = 4.17). 1 mark: correct comparison (Q > K_eq) and correct direction (left/reverse).

Sample response. Products appear in the numerator of the Q expression (raised to their stoichiometric powers). When a product is added to a system at equilibrium, its concentration increases, so the numerator of Q increases. The denominator (reactant concentrations) is momentarily unchanged, so Q = numerator/denominator increases above its previous value, which was equal to K_eq. Therefore Q > K_eq after the addition, and the system shifts left to reduce the product concentration and restore Q = K_eq.

Marking notes. 1 mark: products in numerator of Q stated. 1 mark: adding product increases numerator → Q increases. 1 mark: Q now exceeds K_eq → system shifts left.

Sample response. For a dissolution/precipitation equilibrium such as Ca₃(PO₄)₂(s) ⇌ 3Ca²⁺(aq) + 2PO₄³⁻(aq), K_sp is the solubility product constant (K_eq for that system), evaluated at equilibrium. Q (called the ion activity product, IAP, in solution chemistry) is the same expression but calculated from the actual current ion concentrations in solution. If Q (IAP) > K_sp, the solution is supersaturated — the ion concentrations exceed what can be sustained at equilibrium — and a precipitate will form (ions combine and leave solution) until Q = K_sp. If Q < K_sp, the solution is unsaturated and no precipitate forms.

Marking notes. 1 mark: Q (IAP) defined as same expression as K_sp using current ion concentrations. 1 mark: K_sp is the equilibrium value; Q > K_sp triggers precipitation. 1 mark: correct use of “supersaturated” in context (Q > K_sp → solution is supersaturated → precipitate forms).

Sample response. When the reaction begins from pure reactants, no products are present so [products] = 0 and the numerator of Q = 0; therefore Q = 0. Because Q = 0 < any positive K_eq, the reaction shifts right (forward direction). As the reaction proceeds, product concentrations increase (numerator of Q rises) and reactant concentrations decrease (denominator falls), so Q increases progressively. This continues until Q equals K_eq, at which point equilibrium is established and no further net change occurs.

Marking notes. 1 mark: initial Q = 0 because [products] = 0 (numerator = 0). 1 mark: Q increases as the forward reaction proceeds (products build up, reactants decrease). 1 mark: Q reaches K_eq at equilibrium.

Sample response (a) — Q between t = 0 and t = 3 (2 marks). Q rises from 0 at t = 0, increasing steeply at first then levelling off as it approaches K_eq = 54.3 by t = 3. This occurs because [H₂] and [I₂] are decreasing (denominator of Q decreases) while [HI] is increasing (numerator of Q increases), so Q increases toward K_eq. [1 describe shape: Q rises from 0, levels off at K_eq; 1 explain in terms of [HI] increase / [H₂] and [I₂] decrease]

Sample response (b) — disturbance at t = 5 (3 marks). The instantaneous drop in Q indicates that the Q expression decreased suddenly. The most likely cause is that a reactant was removed from the mixture—specifically H₂ or I₂ was removed (or both were diluted). In the Q expression Q = [HI]²/([H₂][I₂]), removing H₂ or I₂ decreases the denominator, but since we are starting from equilibrium and Q drops (not rises), the disturbance must have increased the denominator or decreased the numerator. Actually, removing HI (a product) would decrease the numerator → Q drops. Alternatively, adding a reactant increases the denominator → Q drops. The graph is consistent with either adding H₂/I₂ or removing HI. [1 identify Q drops = too few products; 1 identify the change: adding reactant increases denominator OR removing HI decreases numerator; 1 explain mechanism using Q expression structure]

Note: Accept either “reactant added” or “product removed” as the cause of the drop—both are consistent with Q < K_eq outcome shown in the graph.

Sample response (c) — Q rising after disturbance (2 marks). After the disturbance, Q < K_eq, so the system shifts right: H₂ and I₂ are consumed (denominator decreases) and HI is produced (numerator increases). Both changes cause Q to increase progressively until Q = K_eq = 54.3 at the new equilibrium. [1 shift right stated with correct reasoning from Q < K_eq; 1 correct concentration changes and their effects on Q numerator/denominator]

Sample response (a) — concentrations and Q (3 marks).
[PCl₃] = 0.500/2.00 = 0.250 mol/L; [Cl₂] = 0.300/2.00 = 0.150 mol/L; [PCl₅] = 1.20/2.00 = 0.600 mol/L.
Q = [PCl₅] / ([PCl₃][Cl₂]) = 0.600 / (0.250 × 0.150) = 0.600 / 0.03750 = 16.0.
[1 correct concentrations; 1 correct Q expression; 1 correct answer Q = 16.0]

Sample response (b) — direction (2 marks). Q = 16.0 < K_eq = 65.0 → system shifts right (forward direction): PCl₃ and Cl₂ combine to form more PCl₅ until Q increases to 65.0. [1 Q < K_eq stated with values; 1 direction right with chemical interpretation]

Sample response (c) — assumption and limitation (1 mark). The calculation assumes that all species behave as ideal gases (or that activity equals concentration in mol/L). At high pressures or with polar molecules such as PCl₃, intermolecular interactions may cause activity to deviate from concentration, so the actual Q (based on activities) may differ from the calculated Q (based on concentrations). Accept also: temperature assumed constant; system assumed perfectly sealed (no leakage). [1 mark for any valid assumption stated with a chemically sensible limitation]

Sample response. The claim is incorrect. While Le Chatelier’s Principle (LCP) correctly predicts the qualitative direction of an equilibrium shift in response to a stress, Q provides quantitative information that LCP cannot. LCP is a qualitative rule: it says “add a reactant, system shifts right” without any numerical basis. Q is the mathematical mechanism that explains why this happens and allows quantitative confirmation. When N₂ is added to the Haber process equilibrium (N₂ + 3H₂ ⇌ 2NH₃), LCP qualitatively predicts a shift right. Q shows the mechanism: [N₂] enters the denominator of Q; increasing [N₂] decreases Q below K_eq; the system shifts right to restore Q = K_eq. LCP and Q agree on direction. However, Q provides what LCP cannot: (i) a quantitative check on whether a given mixture is at equilibrium by calculating Q and comparing to K_eq—LCP gives no way to determine this without already knowing the equilibrium concentrations; (ii) a prediction for mixtures not produced by a simple stress on a known equilibrium—any arbitrary mixture of concentrations can be evaluated using Q, but LCP requires a pre-existing equilibrium to apply its “oppose the stress” logic; (iii) the basis for determining Q = K_sp in precipitation reactions—LCP cannot determine whether a specific mixture of Ca²⁺ and PO₄³⁻ ions in a Murray–Darling water sample will precipitate calcium phosphate; Q (IAP) compared to K_sp can. For these reasons, Q and LCP are complementary: LCP provides an intuitive qualitative rule; Q provides the quantitative framework that gives LCP its mechanistic justification and extends predictive power to cases where LCP fails. The claim is therefore rejected.

Marking criteria.

• 1 mark — States an overall evaluative judgement: the claim is wrong/overstated (LCP and Q are complementary, not redundant).
• 1 mark — Correctly describes LCP as a qualitative rule (direction only, no numerical basis).
• 1 mark — Correctly describes Q as providing quantitative information using current concentrations compared to K_eq.
• 1 mark — Uses a named chemical example (Haber process / PCl₃+Cl₂ / H₂+I₂ or other valid system) to show Q and LCP agree on direction.
• 1 mark — Identifies at least one specific situation where Q provides information LCP cannot: determining whether an arbitrary mixture is at equilibrium; applying Q to precipitation (K_sp / IAP); or checking a non-obvious mixture not derived from a simple stress.
• 1 mark — Explains the mechanism: product/reactant appears in numerator/denominator of Q; the change in concentration causes Q to deviate from K_eq; this determines direction.
• 1 mark — Reaches a justified conclusion that rejects the claim and frames Q as extending LCP rather than being redundant, using precise terminology (Q, K_eq, current concentrations, equilibrium).