HSC Exam Practice

Sample response. The Third Law of Thermodynamics states that the entropy of a perfect crystal at absolute zero (0 K) is exactly zero. This provides an absolute (physics-based) reference point for entropy measurement. Because entropy starts at a defined zero, the entropy accumulated by any substance as it is heated from 0 K to 298 K can be measured and recorded as an absolute S° value for every substance.

Marking notes. 1 mark for correctly stating the Third Law (perfect crystal, 0 K, S = 0). 1 mark for identifying that it provides an absolute reference point. 1 mark for explaining that this reference enables tabulation of absolute S° values (entropy accumulated from 0 K to 298 K).

Sample response. ΔH°_f[O&sub2;(g)] = 0 by human convention: we arbitrarily choose elements in their standard state as the reference point for enthalpy, so there is no “formation” reaction and ΔH°_f = 0. This is not a physical measurement, merely a defined reference. By contrast, S°[O&sub2;(g)] = 205.2 J K⁻¹ mol⁻¹ (non-zero) because entropy has an absolute reference at 0 K (Third Law). At 298 K, O&sub2; has accumulated real, measurable entropy through vibrational, rotational and translational motion. Setting S°(elements) = 0 would contradict the Third Law.

Marking notes. 1 mark for explaining ΔH°_f[element] = 0 is a convention/defined reference. 1 mark for explaining S°[O&sub2;] is non-zero because entropy is an absolute measurement from 0 K (Third Law). 1 mark for stating the correct direction: S°[O&sub2;] = 205.2 J K⁻¹ mol⁻¹ (or equivalent non-zero value) with correct justification.

Sample response. ΔS°_rxn = ΣS°(products) − ΣS°(reactants), where each S° is multiplied by its stoichiometric coefficient. The unit is J K⁻¹ mol⁻¹.

Marking notes. 1 mark for correct formula (products minus reactants). 1 mark for correct unit (J K⁻¹ mol⁻¹).

Sample response. In the liquid phase, H&sub2;O molecules are hydrogen-bonded in a partially ordered arrangement with limited translational freedom. In the gas phase, molecules move freely in three dimensions with large translational freedom; they also have more accessible rotational and vibrational energy levels. The number of microstates (ways to distribute energy among molecules) is far larger in the gas phase. Because entropy is proportional to the number of accessible microstates, S°(g) >> S°(l).

Marking notes. 1 mark for identifying that gas molecules have more translational (and rotational/vibrational) freedom. 1 mark for linking this to more accessible microstates. 1 mark for explicitly connecting more microstates to higher S°.

Sample response. ΣS°(products) = 2 × S°[NH&sub3;(g)] = 2 × 192.4 = 384.8 J K⁻¹ mol⁻¹ [scaling by coefficient 2].
ΣS°(reactants) = 1 × S°[N&sub2;(g)] + 3 × S°[H&sub2;(g)] = 191.6 + 3(130.7) = 191.6 + 392.1 = 583.7 J K⁻¹ mol⁻¹.
ΔS° = 384.8 − 583.7 = −198.9 J K⁻¹ mol⁻¹.
Conversion: −198.9 ÷ 1000 = −0.1989 kJ K⁻¹ mol⁻¹.

Marking notes. 1 mark for correctly scaling NH&sub3; by 2 (and not setting S°[N&sub2;] or S°[H&sub2;] = 0). 1 mark for correct ΣS°(reactants) = 583.7. 1 mark for correct ΔS° = −198.9 J K⁻¹ mol⁻¹ with unit. 1 mark for correct conversion to −0.1989 kJ K⁻¹ mol⁻¹.

Sample response. The qualitative method (L11) predicts only the sign of ΔS based on Δn_gas and phase-change rules; the quantitative method (L12) gives the exact value in J K⁻¹ mol⁻¹. In practice the sign should always agree: if Δn_gas > 0 then ΔS° > 0, etc. Conflicts arise when Δn_gas = 0 (the qualitative method predicts ΔS ≈ 0, but the precise S° values of individual species may give a small positive or negative result), or in dissolution reactions where the complex interplay between lattice entropy, ion dispersion and solvent ordering cannot be captured by simple gas-counting rules.

Marking notes. 1 mark for identifying the key difference (sign only vs exact value). 1 mark for stating both methods should agree in sign for simple reactions. 1 mark for identifying a valid scenario where they may conflict (Δn_gas = 0, or dissolution, or when phase changes and molecular complexity interact in unexpected ways).

Sample response. The Haber process has the largest magnitude ΔS° (−198.9 J K⁻¹ mol⁻¹). Δn_gas = 2 − 4 = −2: four moles of gas are converted to two moles of gas. Halving the number of independent gas molecules dramatically reduces accessible microstates, giving the largest entropy decrease.

Marking notes. 1 mark for identifying Haber with correct ΔS° value. 1 mark for Δn_gas = −2 and linking magnitude to large decrease in moles of gas.

Sample response. When Δn_gas = 0, the dominant entropy driver (change in moles of gas) is absent, so ΔS° ≈ 0. The small negative value reflects that the S° values of CO&sub2;(g) + 2H&sub2;O(g) are not identical to those of CH&sub4;(g) + 2O&sub2;(g): the individual molecular entropies differ due to differences in molecular mass, complexity and bond types, giving a net small negative result even with no change in the number of gas moles.

Marking notes. 1 mark for identifying Δn_gas = 0 as the reason ΔS° ≈ 0. 1 mark for explaining small non-zero value (differences in individual S° values due to molecular properties). Accept any equivalent scientific reasoning.

Sample response. Both reactions have negative ΔS°. Haber has Δn_gas = −2 (4 mol gas → 2 mol), while ethene hydrogenation has Δn_gas = −1 (2 mol gas → 1 mol). A larger reduction in moles of gas means a larger reduction in accessible microstates, so Haber has a greater magnitude (−198.9 vs −120.7 J K⁻¹ mol⁻¹). The ratio (≈1.65) is broadly consistent with the ratio of Δn_gas magnitudes (2:1), though not exact due to differences in individual molecular entropies.

Marking notes. 1 mark for correctly comparing Δn_gas (−2 vs −1). 1 mark for explaining why larger Δn_gas magnitude gives larger |ΔS°|.

Sample response. ΣS°(products) = 39.8 + 213.8 = 253.6 J K⁻¹ mol⁻¹
ΣS°(reactants) = 92.9 J K⁻¹ mol⁻¹
ΔS° = 253.6 − 92.9 = +160.7 J K⁻¹ mol⁻¹

Marking notes. 1 mark for correct working (both S° values used, products − reactants). 1 mark for correct final answer +160.7 J K⁻¹ mol⁻¹ with unit.

Sample response. Convert: ΔS° = +0.1607 kJ K⁻¹ mol⁻¹
ΔG° = ΔH° − TΔS° = +178 − (298 × 0.1607) = +178 − 47.9 = +130.1 kJ mol⁻¹
ΔG° > 0 at 298 K: the reaction is non-spontaneous at room temperature. The large endothermic ΔH° dominates; the favourable entropy term is insufficient at 25°C.

Marking notes. 1 mark for correct ΔS° conversion and substitution into Gibbs equation. 1 mark for correct ΔG° = +130.1 kJ mol⁻¹ (accept ±2 kJ). 1 mark for correct conclusion (non-spontaneous) with brief explanation linking ΔH >> TΔS at 298 K.

Sample response. The assumption is that ΔH° and ΔS° are essentially temperature-independent (i.e. their values at 25°C apply at 1100°C). In reality, both quantities vary with temperature through heat capacity effects (Kirchhoff’s law and the analogous entropy expression), so the 25°C values introduce error when applied to ~1100°C. Accept also: assumption that no additional phase changes occur between 298 K and 1373 K; or that activity coefficients are unity (pure solids and ideal gas).

Marking notes. 1 mark for any valid, clearly stated assumption about applying 298 K data to kiln conditions.

Sample response. The student has made two connected errors. First, they have applied the convention ΔH°_f(elements) = 0 incorrectly to standard entropy, assuming S°[N&sub2;] = S°[H&sub2;] = 0. This is a fundamental conceptual error: ΔH°_f(elements) = 0 is an arbitrary human-defined reference for enthalpy, with no physical meaning for entropy. Standard entropy is an absolute quantity governed by the Third Law of Thermodynamics, which states that entropy is zero only for a perfect crystal at absolute zero (0 K). At 298 K, N&sub2; and H&sub2; have accumulated substantial entropy (191.6 and 130.7 J K⁻¹ mol⁻¹ respectively) through heating, and these non-zero values must be used in the calculation. Second, because the student set the reactant entropies to zero, they calculated ΔS° = +384.8 J K⁻¹ mol⁻¹, obtaining the wrong sign as well as the wrong magnitude. The correct calculation is: ΣS°(products) = 2 × 192.4 = 384.8 J K⁻¹ mol⁻¹; ΣS°(reactants) = 191.6 + 3(130.7) = 583.7 J K⁻¹ mol⁻¹; ΔS° = 384.8 − 583.7 = −198.9 J K⁻¹ mol⁻¹. The negative sign is physically correct: four moles of gas become two moles, reducing accessible microstates and therefore entropy.

Marking notes. 1 mark for identifying that setting S°(elements) = 0 is the core error. 1 mark for correctly explaining the distinction: ΔH°_f(elements) = 0 is a convention; S°(elements) ≠ 0 is a physical measurement. 1 mark for invoking the Third Law (S = 0 at 0 K) as the reason entropy is absolute. 1 mark for correct ΣS°(reactants) = 583.7 with correct scaling. 1 mark for correct ΔS° = −198.9 J K⁻¹ mol⁻¹. 1 mark for explaining why the sign is negative (fewer moles of gas, fewer microstates).

Sample response. The statement is false. Spontaneity is determined by the sign of Gibbs free energy: ΔG° = ΔH° − TΔS°. A reaction is spontaneous when ΔG° < 0. A negative ΔS° contributes a positive term to ΔG° (since −T × negative = positive), but this does not necessarily make ΔG° positive if ΔH° is sufficiently negative.

Consider the Haber process: N&sub2;(g) + 3H&sub2;(g) → 2NH&sub3;(g). ΔS° = −198.9 J K⁻¹ mol⁻¹ (negative, as four moles of gas become two). Yet the Haber process is thermodynamically favourable (ΔG° < 0) at low temperatures because ΔH° = −92 kJ mol⁻¹ (exothermic). The large negative ΔH° more than compensates the ΔG° penalty from negative ΔS° at temperatures below approximately 460 K. The Orica Kooragang Island plant deliberately operates at moderate temperatures for this reason.

Similarly, the formation of liquid water from gases: 2H&sub2;(g) + O&sub2;(g) → 2H&sub2;O(l) has ΔS° ≈ −327 J K⁻¹ mol⁻¹ (strongly negative) yet ΔG° < 0 at 298 K because ΔH° = −572 kJ mol⁻¹ is overwhelmingly exothermic. This reaction is highly spontaneous. In contrast, if ΔH° > 0 and ΔS° < 0, the reaction is never spontaneous at any temperature (ΔG° = positive + positive > 0 always). The statement conflates a single contributor (ΔS°) with the full thermodynamic criterion (ΔG°). The correct generalisation is: spontaneity requires ΔG° < 0, which depends on the relative magnitudes of ΔH°, T and ΔS°. A negative ΔS° reduces the tendency toward spontaneity but does not preclude it when a sufficiently exothermic ΔH° dominates.

Marking notes. 1 mark for correctly stating the spontaneity criterion (ΔG° < 0). 1 mark for writing ΔG° = ΔH° − TΔS° and identifying the sign contribution of each term. 1 mark for a correct named example of a reaction with ΔS° < 0 that is spontaneous (Haber, water formation, or equivalent). 1 mark for calculating or explaining why ΔH° dominates in that example. 1 mark for correctly explaining the temperature dependence (as T increases, −TΔS° term becomes more positive, eventually making ΔG° > 0). 1 mark for identifying the only case where the statement is necessarily true: ΔH° > 0 AND ΔS° < 0 (never spontaneous at any T). 1 mark for overall evaluative judgement that clearly rejects the statement with a precise qualification. 1 mark for coherent, well-structured scientific argument throughout.