HSC Exam Practice

Sample response. Dynamic equilibrium is the state of a reversible reaction in which the forward rate equals the reverse rate and both are non-zero. Macroscopic concentrations of all species remain constant, but at the molecular level both the forward and reverse reactions continue simultaneously at equal rates.

Marking notes. 1 mark for forward rate = reverse rate (both non-zero); 1 mark for constant macroscopic concentrations while molecular-level activity continues.

Sample response. Forward rate: as the reaction proceeds, reactants are consumed and their concentration decreases. This reduces the frequency of reactant–reactant collisions, reducing the frequency of effective forward collisions, so the forward rate decreases. Reverse rate: initially there are no products, so the reverse rate is zero. As products accumulate, the concentration of products increases, increasing the frequency of effective reverse collisions, so the reverse rate increases. When both rates equalise, dynamic equilibrium is established.

Marking notes. 1 mark for forward rate decrease linked to decreasing reactant concentration → fewer effective forward collisions; 1 mark for reverse rate starting at zero (no products); 1 mark for reverse rate increasing as products accumulate → more effective reverse collisions.

Sample response. The relationship is ΔH = E_a(forward) − E_a(reverse). For an exothermic forward reaction (ΔH < 0), products are lower in energy than reactants. The energy barrier from reactants to the transition state (E_a(forward)) is smaller than the barrier from products back to the transition state (E_a(reverse)). Therefore E_a(forward) < E_a(reverse) for any exothermic forward reaction.

Marking notes. 1 mark for stating ΔH = E_a(fwd) − E_a(rev); 1 mark for correctly deducing E_a(fwd) < E_a(rev) for exothermic reaction.

Sample response. A catalyst increases the rate of reaching equilibrium by providing an alternative reaction pathway with a lower activation energy, increasing the frequency of effective collisions in both forward and reverse directions. However, it has no effect on the equilibrium position or K_eq, because it lowers E_a by the same amount for both the forward and reverse reactions. The ratio of forward to reverse rates — which determines K_eq and equilibrium position — is unchanged. Rate is affected because more particles can cross the lowered barrier; equilibrium position is not affected because both directions benefit equally.

Marking notes. 1 mark for catalyst increasing rate (lower E_a → more effective collisions); 1 mark for no effect on equilibrium position / K_eq; 1 mark for explaining why: E_a lowered equally for both directions → ratio of rates unchanged.

Sample response. The forward reaction 2NO₂(g) → N₂O₄(g) is exothermic (ΔH = −57 kJ mol⁻¹), so E_a(forward) < E_a(reverse). Cooling decreases the average kinetic energy of all particles, reducing both forward and reverse rates. However, the reduction is proportionally greater for the reaction with the higher activation energy. Since E_a(reverse) > E_a(forward), the reverse rate (N₂O₄ → 2NO₂) decreases more than the forward rate. Forward rate > reverse rate → net forward reaction → more N₂O₄ forms. N₂O₄ is colourless and NO₂ is brown, so the mixture becomes paler.

Marking notes. 1 mark for identifying forward reaction as exothermic → E_a(fwd) < E_a(rev); 1 mark for cooling reducing reverse rate more than forward rate → net forward shift; 1 mark for linking increased N₂O₄ (colourless) to paler colour.

Sample response. Temperature is the only factor that changes K_eq. Temperature changes alter the Boltzmann distribution of particle energies, raising or lowering the fraction of particles with E ≥ E_a for forward and reverse reactions by different amounts (because E_a(fwd) ≠ E_a(rev)), thus changing the ratio of rates and hence K_eq. A catalyst lowers E_a equally for both directions — the ratio of rates is unchanged, so K_eq is unchanged. Concentration and pressure changes temporarily disturb the balance of rates but the rates re-equalise at the same ratio (same temperature), restoring the same K_eq.

Marking notes. 1 mark for identifying temperature as the only factor; 1 mark for a valid explanation of why other factors (at least one named) do not change K_eq.

(a) Sample response. Between t = 0 and point P, the reaction is proceeding forward from pure reactants. The concentration of CO and H₂ is highest at the start, maximising the frequency of reactant–reactant collisions and therefore the forward rate. Products (CH₄ and H₂O) are absent initially, so the reverse rate is zero. As products accumulate, the reverse rate rises, but it has not yet reached the forward rate — there is still a net forward conversion. Hence r₁ > r₂ throughout this interval.

Marking notes (a). 1 mark for high reactant concentration → high frequency of effective forward collisions → r₁ high; 1 mark for products absent/low → r₂ starts at zero and is still rising toward r₁.

(b) Sample response. Immediately after Q, the forward rate (r₁) spikes upward. Adding CO increases the concentration of CO in the vessel. This increases the frequency of collisions between CO and H₂ molecules per unit time. More of these collisions are effective (energy ≥ E_a), so the forward rate increases suddenly. The reverse rate is initially unchanged because the concentrations of products (CH₄ and H₂O) have not yet changed.

Marking notes (b). 1 mark for r₁ spikes upward; 1 mark for increased [CO] → increased collision frequency → more effective forward collisions.

(c) Sample response. The new equilibrium rate plateau after Q is higher than the plateau before P. After CO is added, higher concentrations of all species at the new equilibrium mean both forward and reverse reactions are occurring at a faster rate than before. The ratio of rates is the same (K_eq is unchanged at constant temperature), so both rates equalise at a higher common value. A higher absolute rate at equilibrium reflects the higher concentrations of all species in the vessel compared to before Q.

Marking notes (c). 1 mark for observing the new plateau is higher; 1 mark for explaining that higher species concentrations at the new equilibrium → higher absolute forward and reverse rates at the new common plateau (K_eq unchanged).

Sample response. Temperature and catalysts are both used to control industrial equilibria, but they serve fundamentally different purposes. Temperature changes K_eq by altering the ratio of forward to reverse rates, while catalysts only change the rate of reaching equilibrium without affecting K_eq at all. Understanding this distinction is essential for evaluating their relative importance.

Temperature is the only tool that can change the equilibrium yield (K_eq). For the Contact process (2SO₂ + O₂ ⇌ 2SO₃, ΔH = −196 kJ mol⁻¹), the forward reaction is strongly exothermic, so E_a(reverse) > E_a(forward). Decreasing temperature preferentially reduces the endothermic reverse rate more than the forward rate (because the reverse rate is more sensitive to the temperature drop — more particles fall below the higher E_a(reverse) threshold), shifting the equilibrium toward more SO₃ and increasing K_eq. Similarly for the Haber process (ΔH = −92 kJ mol⁻¹), lower temperatures favour NH₃ production. Temperature is therefore the more powerful tool for controlling equilibrium yield.

Catalysts, by contrast, cannot improve equilibrium yield. The V₂O₅ catalyst in the Contact process and the Fe catalyst in the Haber process both lower E_a by the same amount for both forward and reverse reactions. More particles can cross both barriers by the same factor, so the ratio of rates (and hence K_eq) is unchanged. Their industrial value lies in reaching the chosen equilibrium position quickly enough to be economically viable. At the temperature needed for good yield (low temperature), uncatalysed reactions are too slow. The catalyst enables an acceptable rate at that temperature. Raising temperature to increase rate without a catalyst would shift both exothermic equilibria left, destroying yield — the catalyst breaks this rate-yield compromise.

Overall judgement: temperature is the more important tool for controlling equilibrium yield because it is the only factor that changes K_eq. Catalysts are essential for industrial feasibility (rate) but provide no yield advantage. The two tools are complementary rather than competing — both are needed to optimise industrial equilibrium processes.

Marking notes. 1 mark — temperature changes K_eq; catalyst does not. 1 mark — correct E_a reasoning for why cooling shifts exothermic equilibrium right (reverse rate falls more; higher E_a reverse). 1 mark — named example with ΔH: Contact process or Haber process, correctly applied. 1 mark — catalyst lowers E_a equally both directions → K_eq unchanged. 1 mark — industrial role of catalyst explained (rate, not yield; enables low-temperature operation). 1 mark — reference to both named industrial processes, correctly linked to the respective catalysts (V₂O₅ / Fe). 1 mark — evaluative judgement that temperature is more important for yield; catalyst is more important for rate; both are needed; or equivalent evidence-based conclusion.