HSC Exam Practice

Sample response. Dynamic equilibrium is a state in a reversible reaction (in a closed system) where the forward and reverse reactions are both occurring simultaneously at equal, non-zero rates, so macroscopic properties (concentrations, colour, pressure) remain constant. Static equilibrium is a state where a reaction has completely stopped — the forward rate reaches zero and the reverse rate is also zero; it occurs in irreversible reactions when reactants are fully consumed.

Marking notes. 1 mark for dynamic equilibrium: forward = reverse rate, both non-zero, macroscopic properties constant. 1 mark for static equilibrium: reactions stopped, rates = 0, irreversible. 1 mark for explicit contrast (dynamic = ongoing activity, static = no activity).

Sample response. Concentrations are constant because the rate of production of each species exactly equals its rate of consumption — at the molecular level, both forward and reverse reactions are continuously occurring, and their effects cancel out. Concentrations are not necessarily equal because their values at equilibrium are determined by K_eq (the ratio of product to reactant concentrations raised to stoichiometric powers at equilibrium), which can be any value from very small (reactant-favoured) to very large (product-favoured). “Equal rates” does not require “equal concentrations.”

Marking notes. 1 mark for constant because rate of production = rate of consumption for each species. 1 mark for K_eq determines the actual values (not necessarily equal). 1 mark for explicitly distinguishing “equal rates” from “equal concentrations.”

Sample response. A catalyst provides an alternative reaction pathway with a lower activation energy. It lowers E_a by the same amount for both the forward and reverse reactions. Therefore, both rates increase by the same factor — the ratio of forward to reverse rate (which determines K_eq) is unchanged. The equilibrium position does not shift, and K_eq is unchanged. The only effect of the catalyst is to shorten the time required to reach equilibrium; the yield at equilibrium is identical to that without the catalyst.

Marking notes. 1 mark for lower E_a (alternative pathway). 1 mark for E_a lowered equally for both forward and reverse reactions. 1 mark for equilibrium position unchanged (no shift). 1 mark for K_eq unchanged / only time to equilibrium is reduced.

Sample response. In a saturated NaCl solution with excess solid, the amount of dissolved NaCl appears constant. The NaCl(s) ⇄ Na⁺(aq) + Cl⁻(aq) equilibrium appears static. However, when a crystal of NaCl labelled with radioactive ²⁴Na is added to the saturated solution, radioactive Na⁺ ions gradually appear in the solution even though the total [NaCl(aq)] does not change. This proves that Na⁺ and Cl⁻ ions are continuously leaving the crystal surface (dissolution) and returning to it (recrystallisation) at equal rates — molecular activity is ongoing, confirming dynamic (not static) equilibrium.

Marking notes. 1 mark for identifying the apparently unchanging macroscopic state. 1 mark for the radioactive tracer (²⁴Na) experiment — radioactive ions appear in solution. 1 mark for the conclusion: ongoing ion exchange at equal rates proves dynamic equilibrium (not static).

Sample response. Both experiments contain exactly the same total atomic composition: 2 mol of H atoms and 2 mol of I atoms in 1 L at the same temperature. Equilibrium is a thermodynamic minimum free energy state that depends only on the total composition and temperature — not on the direction from which it is approached. Starting with H₂ + I₂ (Q = 0), the net forward reaction proceeds until Q = K_eq. Starting with 2HI (Q = ∞), the net reverse reaction proceeds until Q = K_eq. Both converge on the same equilibrium concentrations because K_eq at that temperature is fixed.

Marking notes. 1 mark for identical total atomic composition. 1 mark for thermodynamic/K_eq argument: equilibrium position depends only on T and composition. 1 mark for describing the direction of approach from each starting point (one forward, one reverse) reaching the same K_eq.

Sample response. On a concentration-vs-time graph, dynamic equilibrium is established at the point where all concentration curves simultaneously become horizontal (flat) and cease to change. This point is not necessarily where curves cross because the crossing point represents the moment when two concentrations happen to have equal values — which only coincides with equilibrium when K_eq ≈ 1 (equal concentrations at equilibrium). For most reactions, K_eq ≠ 1: if K_eq > 1, equilibrium is product-favoured and the product curve may never even drop to cross the reactant curve; if K_eq < 1, equilibrium is reactant-favoured.

Marking notes. 1 mark for equilibrium = all curves simultaneously horizontal. 1 mark for crossing point = equal concentrations, not necessarily equilibrium. 1 mark for when they coincide: only when K_eq ≈ 1.

Sample response. At t₁, [NO₂] spikes suddenly upward then decreases toward a new equilibrium, while [N₂O₄] rises to a higher value. This is consistent with the injection of additional NO₂ gas into the vessel. The sudden increase in [NO₂] then triggers a net forward reaction (2NO₂ → N₂O₄), consuming NO₂ and producing more N₂O₄, until a new equilibrium is established.

Marking notes. 1 mark for identifying “additional NO₂ was added (injected).” 1 mark for justification: [NO₂] spikes up (sudden increase = addition, not production), followed by gradual decrease as equilibrium is restored.

Sample response. The forward reaction (2NO₂ → N₂O₄) is exothermic (ΔH = −57 kJ mol⁻¹). Increasing temperature increases the average kinetic energy of all particles — both forward and reverse reaction rates increase. However, the reverse reaction (N₂O₄ → 2NO₂) has a higher activation energy (E_a) than the forward reaction (since the forward is exothermic). The proportional increase in rate is greater for the reaction with higher E_a, so the reverse rate increases more than the forward rate. Le Chatelier’s Principle: the system opposes the temperature increase by favouring the endothermic reverse reaction, consuming N₂O₄ and producing NO₂. Therefore [N₂O₄] decreases.

Marking notes. 1 mark for identifying forward reaction is exothermic, so reverse is endothermic / E_a(reverse) > E_a(forward). 1 mark for reverse rate increases more than forward rate at higher T (proportional E_a argument). 1 mark for Le Chatelier: shift left (reverse favoured) → [N₂O₄] decreases.

Sample response. Yes, K_eq changes between the equilibrium after t₁ and the equilibrium after t₂. Temperature is the only factor that changes K_eq. At t₁, additional NO₂ was added at constant temperature — K_eq was unchanged (same T). At t₂, temperature was increased — this changes K_eq. Since the reaction is exothermic, increasing temperature shifts equilibrium left (toward NO₂) and K_eq decreases (ratio of [N₂O₄]/[NO₂]² is smaller at higher T).

Marking notes. 1 mark for K_eq changes at t₂ because temperature changed (temperature is the only factor that changes K_eq). 1 mark for correct direction: K_eq decreases because exothermic reaction, higher T favours reverse reaction.

Sample response. The source contains two errors. Error 1: “the catalyst shifts the equilibrium to the right.” A platinum (or V₂O₅) catalyst does not shift the equilibrium position. It lowers the activation energy equally for both the forward (2SO₂ + O₂ → 2SO₃) and reverse (2SO₃ → 2SO₂ + O₂) reactions. Because both rates increase by the same factor, the ratio of forward to reverse rate — and therefore the equilibrium position — is unchanged. Error 2: “raising the equilibrium constant K_eq.” K_eq depends only on temperature. Adding a catalyst at constant temperature cannot change K_eq. The yield of SO₃ at equilibrium is therefore unchanged by the catalyst. The catalyst’s only role is to reduce the time required to reach equilibrium, which is commercially important (the industrial process operates faster) but does not increase yield.

Marking notes. 1 mark for identifying Error 1: catalyst does not shift equilibrium. 1 mark for explaining why: lowers E_a equally for both reactions, so ratio of rates unchanged. 1 mark for identifying Error 2: K_eq depends only on temperature; catalyst does not change K_eq. 1 mark for correct statement: catalyst only speeds up reaching equilibrium; yield unchanged.

Sample response. The claim is incorrect. “Dynamic equilibrium” by definition is not an endpoint at which molecular activity ceases — the word “dynamic” means that activity is continuous at the molecular level. At dynamic equilibrium, both the forward and reverse reactions continue to occur simultaneously at equal, non-zero rates. The apparent constancy of macroscopic properties (concentration, colour, pressure) arises not because nothing is happening, but because the rate at which each species is produced equals the rate at which it is consumed — the effects cancel at the macroscopic level while molecular activity continues.

The saturated NaCl solution provides direct experimental evidence. When solid NaCl is added to water until no more dissolves, the system appears static: the dissolved NaCl concentration is constant and the crystal size appears unchanged. The claim would predict that nothing is happening. However, when a crystal of NaCl labelled with radioactive ²⁴Na is added to this apparently static system, radioactive Na⁺ ions gradually appear in solution even though the total [NaCl(aq)] does not increase. This is only possible if ion exchange is continuously occurring: Na⁺ and Cl⁻ ions leave the crystal surface (dissolution, forward) and return from solution to the crystal (recrystallisation, reverse) at equal rates. This is conclusive evidence of ongoing molecular activity in an apparently static system — directly refuting the claim.

The claim conflates dynamic equilibrium with static equilibrium. Static equilibrium does occur in irreversible reactions (e.g. complete combustion of Mg) where the forward rate reaches zero and no reverse reaction is possible. Dynamic equilibrium describes reversible reactions in closed systems; here the system reaches a thermodynamic minimum free energy state with both reactions still proceeding. Describing dynamic equilibrium as a “stable endpoint where no further molecular activity occurs” would only be accurate for static equilibrium, not dynamic equilibrium.

Marking criteria (6 marks):

1 mark — evaluative judgement: the claim is incorrect (and a reason is given).

1 mark — explains significance of “dynamic”: ongoing molecular activity, both forward and reverse reactions continue simultaneously at equal, non-zero rates.

1 mark — explains why concentrations appear constant: rate of production = rate of consumption for each species (not because reactions stopped).

1 mark — names the NaCl / radioactive tracer experiment as experimental evidence, or another valid experimental example of ongoing molecular activity at equilibrium.

1 mark — correctly interprets the experimental evidence: radioactive ions appear in solution despite constant [NaCl(aq)], proving ion exchange is ongoing.

1 mark — distinguishes dynamic equilibrium from static equilibrium: the claim describes static equilibrium; dynamic equilibrium requires a reversible reaction in a closed system with continuous molecular activity.