HSC Exam Practice

Sample response. Entropy (S) is a measure of the disorder or dispersal of energy in a system. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), 4 mol of gas (1 + 3) are converted to 2 mol of gas, so the number of gas molecules decreases, leading to a decrease in disorder. Therefore ΔS < 0 (negative).

Marking notes. 1 mark for an acceptable definition of entropy (disorder / dispersal of energy); 1 mark for correctly identifying ΔS < 0 and linking to the decrease in moles of gas (4 mol → 2 mol).

Sample response. A reversible reaction is one where |ΔG°| is relatively small (typically < ~100 kJ/mol), so both forward and reverse reactions are thermodynamically accessible; the system reaches dynamic equilibrium with measurable amounts of both reactants and products present at apparent completion. An irreversible reaction has a very large negative ΔG° (e.g. << −100 kJ/mol), meaning the products are overwhelmingly more thermodynamically stable; the reaction goes to effective completion with essentially no reactants remaining at equilibrium.

Marking notes. 1 mark for reversible: small |ΔG°|, both directions accessible; 1 mark for irreversible: very large negative ΔG°; 1 mark for distinguishing the presence of reactants at completion (measurable for reversible; negligible for irreversible).

Sample response. At dynamic equilibrium the system has reached its minimum Gibbs free energy—the point where neither the forward nor the reverse reaction has a net thermodynamic driving force. ΔG = 0 at this specific composition because the free energy of the mixture cannot decrease further in either direction. ΔG° (the standard free energy change) is generally non-zero because it is measured at the standard state (1 mol L⊃−¹ of all species, 25°C), which is almost never the equilibrium composition—ΔG° describes the driving force from the standard state towards equilibrium, not the driving force at equilibrium itself.

Marking notes. 1 mark for ΔG = 0 meaning the free energy minimum has been reached / no net driving force in either direction; 1 mark for ΔG° measured at the standard state (not the equilibrium state); 1 mark for explicitly contrasting the two (ΔG = 0 at equilibrium; ΔG° measures the drive from standard state towards equilibrium and is non-zero for most reactions).

Sample response. Enthalpy (ΔH): Combustion of eucalyptus releases a very large amount of heat (ΔH << 0, highly exothermic). The products (CO₂(g) and H₂O(g)) are far more thermodynamically stable than the cellulose, oils, and oxygen reactants—a large contribution to a very negative ΔG. Entropy (ΔS): Combustion converts solid biomass to CO₂(g) and H₂O(g)—a large increase in the number of gas molecules and therefore a large positive ΔS. Since ΔG = ΔH − TΔS, the negative ΔH and the additional negative −TΔS (because ΔS > 0) together make ΔG extremely large and negative. Both effects reinforce each other, making the reverse reaction (reconstructing biomass from combustion gases) have a ΔG of thousands of kJ/mol in the positive direction—thermodynamically impossible without continuous massive energy input.

Marking notes. 1 mark for ΔH << 0 (exothermic) and explanation linking to thermodynamic stability of products; 1 mark for ΔS > 0 and explanation linking to production of gas molecules from solid biomass; 1 mark for correctly combining both using ΔG = ΔH − TΔS to show ΔG is very large and negative; 1 mark for stating that the reverse reaction is essentially impossible (large positive ΔG for the reverse) and therefore the process is irreversible/non-equilibrium.

Sample response. Although ΔH = +25.7 kJ/mol (endothermic, absorbs heat), the dissolution involves breaking the solid ionic lattice and dispersing NH₄⁺ and NO₃⁻ ions throughout the solvent—a large increase in disorder, so ΔS is large and positive. Applying ΔG = ΔH − TΔS at room temperature (298 K): the TΔS term exceeds ΔH, so ΔG < 0. The entropy gain outweighs the enthalpy cost, making the reaction spontaneous.

Marking notes. 1 mark for correctly identifying ΔS > 0 (large entropy increase from ion dispersal); 1 mark for applying ΔG = ΔH − TΔS and showing/stating that TΔS > ΔH; 1 mark for conclusion that ΔG < 0 (spontaneous) driven by entropy increase overcoming enthalpy cost.

Sample response. In both reversible and irreversible reactions, a catalyst increases the rate at which the reaction proceeds by providing an alternative reaction pathway with a lower activation energy. In a reversible reaction, the catalyst accelerates both the forward and reverse reactions equally, so the system reaches dynamic equilibrium faster—but the position of equilibrium (the ratio of products to reactants) and the value of K_eq are unchanged. In an irreversible reaction, the catalyst speeds the rate of product formation but does not change the thermodynamic outcome (still goes to completion); K_eq is not altered in either case.

Marking notes. 1 mark for stating that catalysts lower activation energy / speed up the approach to equilibrium for reversible reactions; 1 mark for stating that the position of equilibrium and K_eq are not changed by a catalyst; 1 mark for applying the same reasoning to irreversible reactions (speeds rate; outcome still goes to completion; thermodynamics unchanged).

Part (a) — 2 marks. Between T1 and T2 (the 60°C phase), [N₂O₄] decreases from approximately 0.80 mol L⁻¹ to approximately 0.55 mol L⁻¹ [1], while [NO₂] increases from approximately 0.40 mol L⁻¹ to approximately 0.90 mol L⁻¹ [1]. Both concentrations then reach new constant values at the higher temperature, indicating a new equilibrium position has been established.

Part (b) — 3 marks. The reaction N₂O₄ ⇌ 2NO₂ is endothermic (ΔH = +57.2 kJ/mol > 0). Since the reaction produces 2 mol of gas from 1 mol of gas, ΔS > 0 (entropy increases). Applying ΔG = ΔH − TΔS: when temperature rises, the TΔS term becomes larger in magnitude, making ΔG more negative for the forward (endothermic) reaction. This means the forward reaction becomes more thermodynamically favoured at higher temperature [1]; the free energy minimum shifts to a composition with more NO₂ and less N₂O₄ [1]; K_eq increases (consistent with the graph: K rises from 0.14 to 0.87 mol L⁻¹ between 25°C and 60°C) [1].

Part (c) — 2 marks. In the sealed bottle, neither N₂O₄ nor NO₂ can escape to the surroundings, so the closed system allows the concentrations to reach constant values where the forward and reverse reaction rates are equal—dynamic equilibrium [1]. In an open bottle, NO₂(g) would escape into the atmosphere, continuously removing product and preventing the reverse reaction from ever becoming equal to the forward reaction; the system would be driven continuously forward and could not reach equilibrium [1].

Part (a) — 2 marks. ΔG° = ΔH° − TΔS° = +178 kJ/mol − (298 K × 0.165 kJ mol⁻¹ K⁻¹) = +178 − 49.17 = +128.8 kJ/mol (accept +129 kJ/mol) [1]. Since ΔG° > 0, the reaction is non-spontaneous at 25°C [1].

Part (b) — 2 marks. At the crossover temperature, ΔG° = 0: ΔH° = TΔS°; T = ΔH°/ΔS° = 178 000 J/mol ÷ 165 J mol⁻¹ K⁻¹ = 1079 K = 806°C (accept 800–815°C depending on rounding) [1]. Above this temperature the reaction is spontaneous (ΔG < 0) [1].

Part (c) — 1 mark. The assumption that ΔH° and ΔS° are independent of temperature (constant) is a limitation: at very high temperatures (1450°C), the actual values of ΔH and ΔS for the reaction may differ significantly from the 25°C standard values due to heat capacity differences between reactants and products (Kirchhoff’s law). The prediction is an approximation [1]. Accept also: phase changes at intermediate temperatures not accounted for; the equation gives ΔG° (standard state), not necessarily ΔG under actual industrial concentrations.

Sample response. The claim is incorrect. The double arrow (⇌) notation indicates that a reaction reaches dynamic equilibrium—it does not imply equal concentrations of reactants and products at that equilibrium. The position of equilibrium is determined by ΔG° and can range from products strongly favoured to reactants strongly favoured.

The relationship between ΔG° and the equilibrium position is qualitative but consistent: a large negative ΔG° means the equilibrium lies far towards products (large K_eq); a near-zero ΔG° means similar amounts of both are present at equilibrium (K_eq ≈ 1); a positive ΔG° means the equilibrium favours reactants (K_eq < 1). Equal concentrations would only occur if K_eq is exactly 1, which is uncommon and requires ΔG° ≈ 0.

The deeper reason all reactions with moderate ΔG° reach equilibrium with both species present is the concept of entropy of mixing. Any mixture of reactants and products has higher entropy than either pure extreme (because mixing itself increases disorder). This entropy of mixing contribution lowers the Gibbs free energy of intermediate compositions below that of both pure reactants and pure products. Even when ΔG° is negative (products thermodynamically favoured), the G vs composition curve dips to a minimum at an intermediate point—the equilibrium position—rather than at pure products. The system “falls into” this free energy minimum and stops there: ΔG = 0.

Consider two named reactions: (1) H₂(g) + I₂(g) ⇌ 2HI(g), ΔG° = −3.5 kJ/mol. The very small ΔG° means the free energy minimum is close to the centre of the composition range. At equilibrium, significant amounts of H₂, I₂, and HI are all present—concentrations are not equal, but all species are measurable. (2) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔG° = −141 kJ/mol. Here the equilibrium lies strongly towards products; at equilibrium, [SO₃] >> [SO₂] and [O₂]—far from equal concentrations. Both reactions are written with ⇌ and both reach dynamic equilibrium, but their equilibrium compositions are very different because their ΔG° values differ by over 100 kJ/mol.

In summary, the double arrow notation tells us only that both forward and reverse reactions are thermodynamically accessible and that the system reaches a free energy minimum at an intermediate composition. The actual concentrations at equilibrium depend on K_eq, which is set by ΔG°. Equal concentrations at equilibrium would be a special case requiring K_eq = 1 and would not be the general expectation for any reaction with a non-zero ΔG°.

Marking notes.

• 1 mark — States explicitly that the claim is incorrect and that ⇌ does not mean equal concentrations; equilibrium position depends on K_eq and ΔG°.
• 1 mark — Explains the qualitative relationship between ΔG° and K_eq: large negative ΔG° → K >> 1 (products favoured); near-zero ΔG° → K ≈ 1; positive ΔG° → K < 1 (reactants favoured).
• 1 mark — Identifies ΔG = 0 at equilibrium (the free energy minimum) and distinguishes this from ΔG° ≠ 0.
• 1 mark — Names and correctly applies the concept of entropy of mixing: intermediate compositions have higher entropy and lower G than either pure extreme.
• 1 mark — Connects entropy of mixing to the existence of a free energy minimum at an intermediate composition on the G vs composition curve.
• 1 mark — Names first reaction (H₂ + I₂ ⇌ 2HI) with ΔG° ≈ −3.5 kJ/mol and correctly describes the near-central equilibrium position with all three species measurably present.
• 1 mark — Names second reaction (e.g. 2SO₂ + O₂ ⇌ 2SO₃) with a larger negative ΔG° and correctly describes the products-favoured equilibrium position; makes the contrast with the first reaction explicit.
• 1 mark — States the condition under which equal concentrations would be observed (K_eq = 1, ΔG° ≈ 0) and notes this is a special case.
• 1 mark (Band 6 discriminator) — Constructs a coherent, structured argument that moves from claim rejection → ΔG°/K_eq relationship → entropy of mixing mechanism → G minimum concept → named examples, demonstrating synthesis rather than a list of separate points.