Chemistry • Year 11 • Module 4 • Lesson 12

Calculating ΔS° & Standard Entropy

Build HSC Band 5–6 extended-response technique on standard entropy calculation, molecular interpretation, and evaluation of industrial entropy trade-offs.

Master • Extended Response

Reference data (use for both questions)

SubstanceS° (J K⁻¹ mol⁻¹)SubstanceS° (J K⁻¹ mol⁻¹)
H&sub2;(g)130.7CO&sub2;(g)213.8
O&sub2;(g)205.2CO(g)197.7
N&sub2;(g)191.6CH&sub4;(g)186.3
NH&sub3;(g)192.4H&sub2;O(g)188.7
CaCO&sub3;(s)92.9H&sub2;O(l)69.9
CaO(s)39.8C&sub2;H&sub4;(g)219.6
C(graphite)5.7C&sub2;H&sub6;(g)229.6
NaCl(s)72.1Na+(aq)59.0
Cl(aq)56.5HCl(g)186.9

1. Data + scenario + multi-criteria evaluation (Band 5–6)

8 marks   Band 5–6

Scenario. The Orica Kooragang Island ammonia synthesis plant near Newcastle (NSW) operates the Haber process at approximately 450°C and 150–300 atm. Ammonia (NH&sub3;) is a key input for fertilisers supporting Australian agriculture. The chief chemical engineer states:

“The entropy change of the reaction is one of the key thermodynamic factors that drove us to high-pressure operation. The negative ΔS° means we need to compensate thermodynamically to achieve an acceptable yield.”

The table below compares the ΔS° of three related industrial reactions.

ReactionEquationΔS° (J K⁻¹ mol⁻¹)
Haber — NH&sub3; synthesisN&sub2;(g) + 3H&sub2;(g) → 2NH&sub3;(g)−198.9
Limestone decompositionCaCO&sub3;(s) → CaO(s) + CO&sub2;(g)+160.7
Ethene hydrogenationC&sub2;H&sub4;(g) + H&sub2;(g) → C&sub2;H&sub6;(g)−120.7

Q1. Using the data and scenario above, write an extended response that:

  • Verifies ΔS° for the Haber process using the reference data table above, showing all working.
  • Explains at a molecular level why the Haber process has a large negative ΔS°, using microstate reasoning.
  • Compares the sign and magnitude of ΔS° across the three reactions on at least two criteria (e.g. Δngas, phase of reactants/products, molecular complexity).
  • Evaluates the engineer’s claim: explain specifically how a negative ΔS° relates to pressure as an operating variable, referencing Le Chatelier’s principle and the effect on ΔG° = ΔH° − TΔS°.
  • Reaches a justified judgement about whether ΔS° alone determines whether a reaction is industrially viable.
Plan first: calculation → molecular interpretation → comparison table (2 criteria) → Le Chatelier link → judgement (is ΔS° the whole story?). Allow 15–20 min.

2. Multi-step calculation and interpretation (Band 5–6)

7 marks   Band 5–6

Context. Boral operates a lime kiln at Marulan (NSW) that decomposes limestone to quicklime: CaCO&sub3;(s) → CaO(s) + CO&sub2;(g). The kiln is fired at ~900°C (1173 K). The standard enthalpy for this reaction is ΔH° = +178 kJ mol⁻¹ (strongly endothermic).

(a) Calculate ΔS° for the limestone decomposition reaction, showing all working and stating units. 2 marks

(b) Convert your ΔS° answer to kJ K⁻¹ mol⁻¹. Then calculate ΔG° at 25°C (298 K) using ΔG° = ΔH° − TΔS°. Comment on whether the reaction is spontaneous at room temperature. 3 marks

(c) The kiln operates at ~1173 K. Without recalculating (use your values from (a) and the given ΔH°), predict whether ΔG° becomes negative at 1173 K and justify your prediction. State one industrial implication of your answer. 2 marks

For (b): convert ΔS° by ÷1000 first. For (c): at high T, the −TΔS° term becomes large and negative — what does that do to ΔG° = ΔH° − TΔS° when ΔS° is positive?
Answers — Do not peek before attempting

Q1 — Marking guide (8 marks)

Verification (2 marks):
ΣS°(products) = 2 × 192.4 = 384.8 J K⁻¹ mol⁻¹ [1 for scaling by coefficient 2]
ΣS°(reactants) = 191.6 + 3(130.7) = 191.6 + 392.1 = 583.7 J K⁻¹ mol⁻¹ [1 for S°[N&sub2;] and S°[H&sub2;] both non-zero, scaled correctly]
ΔS° = 384.8 − 583.7 = −198.9 J K⁻¹ mol⁻¹ ✓

Molecular explanation (2 marks):
4 moles of gas (N&sub2; + 3H&sub2;) condense to 2 moles of gas (2NH&sub3;); Δngas = −2 [1]. Fewer independent gas molecules means fewer accessible microstates (fewer ways to distribute translational, rotational and vibrational energy), so the system’s entropy decreases [1].

Comparison across 3 reactions on 2 criteria (2 marks):
Criterion 1 (Δngas): Haber Δngas = −2 (−198.9); ethene hydrogenation Δngas = −1 (−120.7); limestone decomposition Δngas = +1 (+160.7). Sign of Δngas correctly predicts sign of ΔS° in all three cases [1].
Criterion 2 (magnitude): larger |Δngas| → larger |ΔS°|; limestone starts from a solid (low S° reactant) which amplifies the gain when CO&sub2;(g) is released [1].

Evaluation of engineer’s claim (1 mark):
Negative ΔS° means the −TΔS° term in ΔG° = ΔH° − TΔS° is positive (subtracting a negative number); at lower pressure the equilibrium favours the higher-entropy reactant side. By Le Chatelier’s principle, increasing pressure shifts equilibrium towards fewer moles of gas (products), compensating for the entropy-driven thermodynamic disadvantage [1].

Justified judgement (1 mark):
ΔS° alone does not determine industrial viability. ΔG° = ΔH° − TΔS° requires both ΔH° and ΔS° plus temperature; kinetics, pressure, catalyst availability and economic factors also determine viability. The Haber process is viable despite negative ΔS° because ΔH° is sufficiently negative and conditions can be optimised [1].

Q2(a) — ΔS° for limestone decomposition (2 marks)

ΣS°(products) = S°[CaO(s)] + S°[CO&sub2;(g)] = 39.8 + 213.8 = 253.6 J K⁻¹ mol⁻¹
ΣS°(reactants) = S°[CaCO&sub3;(s)] = 92.9 J K⁻¹ mol⁻¹
ΔS° = 253.6 − 92.9 = +160.7 J K⁻¹ mol⁻¹ [1 for working; 1 for correct value + units]

Q2(b) — ΔG° at 298 K (3 marks)

Convert: ΔS° = +160.7 ÷ 1000 = +0.1607 kJ K⁻¹ mol⁻¹ [1]
ΔG° = ΔH° − TΔS° = +178 − (298 × 0.1607) = +178 − 47.9 = +130.1 kJ mol⁻¹ [1]
ΔG° > 0 at 298 K: reaction is non-spontaneous at room temperature [1]. The large positive ΔH° dominates at low temperature; the favourable entropy term (+TΔS°) is insufficient at 298 K.

Q2(c) — Spontaneity at 1173 K (2 marks)

ΔG° will become negative at 1173 K [1]. Since ΔS° > 0, the term −TΔS° = −1173 × 0.1607 = −188.7 kJ mol⁻¹. Then ΔG° = +178 − 188.7 = −10.7 kJ mol⁻¹ < 0 — spontaneous above approximately T = ΔH° / ΔS° = 178/0.1607 ≈ 1108 K. Industrial implication: the Boral kiln must be operated above ~835°C (1108 K) to thermodynamically favour decomposition; below this temperature, the reaction does not proceed efficiently even with a catalyst [1 for correct prediction with TΔS reasoning; 1 for industrial implication].