HSC Exam Practice

Sample response. Gibbs free energy (ΔG) is the thermodynamic potential that determines whether a reaction is spontaneous at constant temperature and pressure. It is defined by ΔG = ΔH − TΔS, where ΔH is the enthalpy change, T is the absolute temperature in Kelvin, and ΔS is the entropy change.

Marking notes. 1 mark for a correct definition (thermodynamic potential / free energy that predicts spontaneity at constant T and P). 1 mark for stating ΔG = ΔH − TΔS with correct symbols.

Sample response. (i) ΔG < 0 — spontaneous (thermodynamically favourable, forward reaction proceeds without continuous energy input). (ii) ΔG > 0 — non-spontaneous (thermodynamically unfavourable, reverse reaction favoured). (iii) ΔG = 0 — equilibrium (forward and reverse reactions proceed at equal rates; no net change in composition).

Marking notes. 1 mark each for the correct ΔG value/sign and a correct statement of what it means (3 marks total).

Sample response. Temperature must be in Kelvin because the Gibbs equation uses absolute temperature (0 K = absolute zero); using °C gives an arbitrary zero point and incorrect TΔS values. ΔS must be in kJ K⁻¹ mol⁻¹ (not J) so that the units of TΔS (K × kJ K⁻¹ mol⁻¹) match the units of ΔH (kJ mol⁻¹); using J inflates TΔS by a factor of 1000 and gives a ΔG value that is 1000× too large.

Marking notes. 1 mark for Kelvin (absolute scale needed). 1 mark for unit consistency — TΔS and ΔH must both be in kJ mol⁻¹; failure to convert J → kJ gives 1000× error.

Sample response. A spontaneous reaction has ΔG < 0, meaning it is thermodynamically favourable — the forward reaction can proceed without a continuous energy input. A fast reaction is one with a low activation energy (E_a) and high collision frequency, giving a rapid rate. These properties are independent: a reaction can be spontaneous but have a very high E_a, making it extremely slow. Example: the conversion of diamond to graphite has ΔG < 0 at room temperature and pressure (spontaneous) yet proceeds so slowly it is undetectable on human timescales, because the E_a for breaking the C—C lattice bonds is enormous.

Marking notes. 1 mark for correct ΔG criterion for spontaneity. 1 mark for distinguishing kinetics (E_a, rate) from thermodynamics (ΔG). 1 mark for a named example with both spontaneous and slow features identified.

Sample response. For ΔH > 0 and ΔS > 0, ΔG = ΔH − TΔS. At low T the positive ΔH dominates, making ΔG > 0 (non-spontaneous). As T increases, the TΔS term grows; when TΔS = ΔH the reaction reaches the crossover temperature, T_cross = ΔH/ΔS, where ΔG = 0 (equilibrium). Above T_cross, TΔS > ΔH, so ΔG < 0 and the reaction becomes spontaneous.

Marking notes. 1 mark for identifying that at low T the reaction is non-spontaneous (ΔG > 0). 1 mark for explaining the mechanism (TΔS grows until it exceeds ΔH). 1 mark for naming the crossover temperature and its formula T_cross = ΔH/ΔS.

Sample response. Case A: ΔH < 0, ΔS > 0 — always spontaneous at all temperatures; ΔG = (−) − T(+) = always negative. Case B: ΔH > 0, ΔS < 0 — never spontaneous at any temperature; ΔG = (+) − T(−) = always positive. Case C: ΔH < 0, ΔS < 0 — spontaneous at low temperatures only; above T_cross = ΔH/ΔS, ΔG becomes positive. Case D: ΔH > 0, ΔS > 0 — spontaneous at high temperatures only; above T_cross = ΔH/ΔS, ΔG becomes negative.

Marking notes. 1 mark per case: correct signs and correct spontaneity classification (always / never / temperature-dependent with correct T direction). Award half marks if signs correct but spontaneity wrong.

(a) 2 marks. From the graph: approximately 1100–1115 K (accept ±50 K). This temperature is called the crossover temperature (T_cross). 1 mark for reading ~1100–1115 K from graph; 1 mark for naming crossover temperature.

(b) 3 marks. Convert: ΔS° = 160.6 ÷ 1000 = 0.1606 kJ K⁻¹ mol⁻¹; T = 900 + 273 = 1173 K. ΔG° = 178 − (1173 × 0.1606) = 178 − 188.4 = −10.4 kJ mol⁻¹. Reaction is spontaneous at 900°C. Mark allocation: 1 mark for both conversions; 1 mark for correct calculation; 1 mark for correct sign interpretation.

(c) 3 marks. 1 mark: ΔH > 0 and ΔS > 0 — this is Case D. 1 mark: Below T_cross, ΔH (positive) dominates and ΔG > 0 (reaction non-spontaneous). 1 mark: Above T_cross, the TΔS term exceeds ΔH, making ΔG < 0 (spontaneous). Boral must exceed ~835°C to drive CaO production thermodynamically.

(a) 3 marks. T = 298 K (correct). ΔS° = −120.7 ÷ 1000 = −0.1207 kJ K⁻¹ mol⁻¹. ΔG° = −137 − (298 × −0.1207) = −137 − (−35.97) = −137 + 35.97 = −101.0 kJ mol⁻¹. Spontaneous at 25°C. 1 mark conversions; 1 mark calculation; 1 mark spontaneity.

(b) 2 marks. T_cross = ΔH°/ΔS° = −137/(−0.1207) = 1135 K (862°C). Above this temperature, ΔG becomes positive and the reaction is no-longer spontaneous. 1 mark for value (accept 1130–1140 K); 1 mark for correct statement of what happens to spontaneity.

(c) 1 mark. ΔG° indicates thermodynamic favourability but gives no information about the rate of reaction. Even with ΔG° = −101 kJ mol⁻¹, the reaction may require a catalyst or high temperature to proceed at an industrially useful rate, because rate depends on activation energy (E_a), not ΔG.

Sample response. Gibbs free energy is a powerful but incomplete tool for industrial process design. Its main strength is predicting thermodynamic feasibility: a negative ΔG° tells chemists that the reaction can proceed in the forward direction without a continuous energy input, while a positive value indicates it cannot proceed under those conditions. For the Haber process (N₂ + 3H₂ → 2NH₃; ΔH° = −92.4 kJ mol⁻¹, ΔS° = −199 J K⁻¹ mol⁻¹), ΔG calculation at 25°C gives −33.1 kJ mol⁻¹, confirming thermodynamic spontaneity. The crossover temperature T_cross = 465 K (192°C) reveals that above this temperature the reaction becomes thermodynamically non-spontaneous, which is vital information for reactor design. However, ΔG does not give information about kinetics — the rate of reaction. At 25°C the Haber process is spontaneous but essentially does not occur because the activation energy for breaking the N≡N triple bond is prohibitively high. Industrial plants therefore run at 400–500°C (well above T_cross, where ΔG° = +42 to +62 kJ mol⁻¹) to obtain a practical rate. This demonstrates that ΔG and reaction rate are fundamentally independent: the selection of operating temperature is a kinetic decision, not a thermodynamic one. Chemists compensate by using an iron catalyst (lowers E_a), high pressure (drives equilibrium toward fewer moles of gas), and continuous NH₃ removal (Le Chatelier). In summary, ΔG is essential for understanding if and when a reaction is thermodynamically favourable and how temperature and entropy interact — but it must be paired with kinetic analysis and process engineering for full industrial usefulness.

Marking notes. 1 mark — defines ΔG° and its criterion for spontaneity. 1 mark — correctly calculates or quotes ΔG° at 25°C for the Haber process with reference to ΔH/ΔS signs (Case C). 1 mark — identifies T_cross (~465 K) and explains that above this temperature the Haber reaction is thermodynamically non-spontaneous. 1 mark — identifies the key limitation: ΔG says nothing about kinetics or rate. 1 mark — explains the Haber paradox: industrial temperature (400–500°C) is chosen for kinetics, not because ΔG is favourable. 1 mark — names at least two process modifications (catalyst, pressure, or removal of product) and links each to its thermodynamic or kinetic purpose. 1 mark — reaches an explicit evaluative conclusion: ΔG is necessary but insufficient as a standalone tool; must be paired with kinetics.