Chemistry • Year 11 • Module 4 • Lesson 12
Calculating ΔS° & Standard Entropy
Apply the ΔS° formula to real reaction data, interpret quantitative entropy changes at the molecular level, and use the Australian industrial context (Haber process; Boral Marulan cement).
Reference: Standard Entropy Data Table
Use the values below for all calculations in this worksheet. Data table — not assessed separately
| Substance | S° (J K⁻¹ mol⁻¹) | Substance | S° (J K⁻¹ mol⁻¹) |
|---|---|---|---|
| H&sub2;(g) | 130.7 | CO&sub2;(g) | 213.8 |
| O&sub2;(g) | 205.2 | CO(g) | 197.7 |
| N&sub2;(g) | 191.6 | CH&sub4;(g) | 186.3 |
| NH&sub3;(g) | 192.4 | H&sub2;O(g) | 188.7 |
| CaCO&sub3;(s) | 92.9 | H&sub2;O(l) | 69.9 |
| CaO(s) | 39.8 | C&sub2;H&sub4;(g) | 219.6 |
| C(graphite) | 5.7 | C&sub2;H&sub6;(g) | 229.6 |
| NaCl(s) | 72.1 | Na+(aq) | 59.0 |
| Cl−(aq) | 56.5 | HCl(g) | 186.9 |
1. Interpreting S° patterns across reactions
The table below shows ΔS° values for four industrially relevant reactions. 8 marks
| Reaction | Equation | Δngas | ΔS° (J K⁻¹ mol⁻¹) |
|---|---|---|---|
| Haber process | N&sub2;(g) + 3H&sub2;(g) → 2NH&sub3;(g) | −2 | −198.9 |
| Limestone decomposition | CaCO&sub3;(s) → CaO(s) + CO&sub2;(g) | +1 | +160.7 |
| Methane combustion (steam) | CH&sub4;(g) + 2O&sub2;(g) → CO&sub2;(g) + 2H&sub2;O(g) | 0 | −5.1 |
| Water vaporisation | H&sub2;O(l) → H&sub2;O(g) | +1 | +118.8 |
1.1 The Haber process has a strongly negative ΔS°. Explain at the molecular level why combining N&sub2; and 3H&sub2; to form 2NH&sub3; reduces entropy, even though all species are gases. 2 marks
1.2 The Boral Marulan cement works (NSW) decomposes limestone: CaCO&sub3;(s) → CaO(s) + CO&sub2;(g). Verify ΔS° = +160.7 J K⁻¹ mol⁻¹ using the data table, showing full working. Then explain in molecular terms why ΔS° is positive. 3 marks
1.3 Compare the magnitude of ΔS° for limestone decomposition (+160.7) versus water vaporisation (+118.8). Both produce one mole of gas from zero. Suggest why limestone decomposition gives a larger ΔS° despite the same Δngas. 2 marks
1.4 Methane combustion to steam has Δngas = 0 yet ΔS° = −5.1 J K⁻¹ mol⁻¹. Explain why ΔS° ≈ 0 in this case, and why the small negative value is consistent with detailed molecular analysis. 1 mark
2. Graph interpretation — S° accumulation across phases
The bar chart below shows the standard entropy values for water in three phases and for selected elements and compounds. Use it to answer the sub-questions. 6 marks
2.1 Describe the trend in S° for water as it moves from solid to liquid to gas. Estimate the magnitude of the increase at each phase transition from the bar chart. 2 marks
2.2 The bar chart shows that N&sub2;(g) and NH&sub3;(g) have almost identical S° values (~191–192 J K⁻¹ mol⁻¹) yet NH&sub3; has one more atom. Explain why the difference is small despite NH&sub3; being more complex. 2 marks
2.3 C(graphite) has the lowest S° of any substance shown (5.7 J K⁻¹ mol⁻¹), yet it is a solid element at 298 K. Explain this extremely low value in terms of its crystal structure. 2 marks
3. Cause-and-effect chain — interpreting the negative ΔS° of the Haber process
The Haber process (N&sub2; + 3H&sub2; → 2NH&sub3;) has ΔS° = −198.9 J K⁻¹ mol⁻¹. Trace the causal chain from the change in moles of gas to the sign and magnitude of ΔS°, then to what this means for the number of accessible microstates in the system. 5 marks
4 mol gas (N&sub2; + 3H&sub2;) react to form 2 mol gas (2NH&sub3;)
Effect on Δngas
Effect on number of microstates
Sign of ΔS° and why the magnitude is large
ΔS° = −198.9 J K⁻¹ mol⁻¹ (consistent with qualitative prediction)
Hints: Δngas = products − reactants = ___; fewer gas molecules means ___; gas particles have far more ___ than other phases.
Stuck? Re-read the qualitative entropy rules from Lesson 11 (Card 2: Δn(gas) rule) and the quantitative calculation in Card 3 of this lesson.4. Predict and justify
Answer in 3–5 sentences using precise chemical reasoning. 4 marks
Q4. A student dissolves MgSO&sub4;(s) in water. She predicts ΔS° > 0 because “solids always increase in entropy when dissolved.” Her teacher points out that some ionic salts dissolve with ΔS° < 0. Predict whether ΔS° is positive or negative for MgSO&sub4; dissolving, and justify your prediction by discussing what happens to the hydration shells of Mg2+ ions compared to the dissolution of NaCl. 4 marks
Q1.1 — Haber: why ΔS° is strongly negative
Four moles of gas (1 N&sub2; + 3 H&sub2;) are converted into two moles of gas (2 NH&sub3;), so Δngas = −2. Halving the moles of gas dramatically reduces the number of accessible microstates (translational, rotational, vibrational modes). Fewer independent gas molecules means far fewer ways to distribute energy, so entropy decreases significantly. [2 marks: 1 for Δngas = −2 / fewer gas molecules; 1 for linking to fewer microstates / freedom of movement.]
Q1.2 — Limestone decomposition verification
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 ✓
Positive because: one mole of CO&sub2; gas is produced from a solid; gas particles have far more accessible microstates (freedom of translation, rotation, vibration) than solid lattice particles, so the system’s disorder increases substantially. [1 for working; 1 for correct answer; 1 for molecular explanation.]
Q1.3 — Limestone vs water vaporisation magnitude
In the limestone reaction, the reactant is a solid (CaCO&sub3;, S° = 92.9 J K⁻¹ mol⁻¹) with very little entropy. When a mole of CO&sub2; gas is produced, the “entropy starting point” is low, so the gain is large. In water vaporisation, the reactant is liquid H&sub2;O (S° = 69.9 J K⁻¹ mol⁻¹), which already has more entropy than a solid lattice; the increment from liquid to gas is smaller in relative terms. Accept also: CO&sub2; is a larger, more complex molecule than H&sub2;O, with more vibrational modes, so its S° (213.8) is higher than H&sub2;O(g) (188.7). [2 marks]
Q1.4 — Methane combustion, Δngas = 0
When Δngas = 0, the main entropy driver (change in moles of gas) is absent, so ΔS° ≈ 0. The small negative value (−5.1) reflects that CO&sub2; and 2H&sub2;O(g) have slightly lower combined entropy than CH&sub4; and 2O&sub2; — the individual S° values do not cancel exactly at constant Δngas. [1 mark]
Q2.1 — Water S° trend
S° increases in each step: solid (~41) → liquid (69.9) → gas (188.7). The solid-to-liquid increase is approximately +29 J K⁻¹ mol⁻¹ (melting adds positional freedom as the rigid lattice breaks down). The liquid-to-gas increase is approximately +119 J K⁻¹ mol⁻¹ (vaporisation is far larger because gaseous molecules gain full translational freedom in three dimensions). [2 marks: 1 for trend; 1 for magnitudes with ~correct estimates]
Q2.2 — N&sub2; vs NH&sub3; S° nearly equal
Although NH&sub3; has more atoms (and therefore more vibrational modes than N&sub2;), N&sub2; has a higher molar mass per mole of molecules (28 g mol⁻¹ vs 17 g mol⁻¹), which increases translational entropy. The opposing effects — greater molecular complexity in NH&sub3; vs greater molar mass in N&sub2; — approximately cancel, giving very similar S° values. [2 marks]
Q2.3 — C(graphite) very low S°
Graphite consists of planar sheets of carbon atoms covalently bonded in a hexagonal lattice. Atoms are strongly held in fixed positions with very limited ability to vibrate or move. At 298 K the number of accessible microstates is extremely small — only a fraction of thermal energy is enough to excite vibrations. This highly ordered, rigid structure gives C(graphite) the lowest S° of any common solid (5.7 J K⁻¹ mol⁻¹). [2 marks: 1 for ordered rigid lattice; 1 for linking to few microstates]
Q3 — Cause-and-effect chain (sample)
Effect 1 (Δngas): Δngas = 2 − (1 + 3) = −2. The number of moles of gas decreases by 2. [1 mark]
Effect 2 (microstates): Fewer moles of gas particles means fewer possible positions, speeds, and energy distributions — the number of accessible microstates decreases significantly. Gas particles have far more translational freedom than molecules in other phases, so losing 2 moles of gas particles is a large reduction in microstates. [1 mark]
Effect 3 (sign and magnitude of ΔS°): Because microstates decrease, ΔS° is negative. The magnitude is large (−198.9 J K⁻¹ mol⁻¹) because Δngas = −2 is a large change — two moles of translational freedom are lost. This is consistent with the quantitative calculation: ΣS°(products) − ΣS°(reactants) = 384.8 − 583.7 = −198.9 J K⁻¹ mol⁻¹. [1 mark for correct sign; 1 mark for magnitude explanation]
[1 mark per valid step; total 5 marks available across all three effect boxes and linking explanation]
Q4 — MgSO&sub4; dissolution prediction
ΔS° is likely negative or close to zero for MgSO&sub4;(aq). [1] Mg2+ is a small, highly charged ion that strongly attracts surrounding water molecules, ordering them into a tight hydration shell — this “freezing” of water molecules decreases the entropy of the solvent significantly. [1] By contrast, Na+ has a lower charge-to-radius ratio and a weaker hydration shell; NaCl dissolution has a modest positive ΔS° (+43.4 J K⁻¹ mol⁻¹) because the entropy gained from dispersing the ions outweighs the ordering of solvent. [1] For Mg2+, the water-ordering effect dominates, so overall entropy decreases. Experimentally, ΔS° for MgSO&sub4;(aq) is approximately −123 J K⁻¹ mol⁻¹. [1 for correct prediction with justification; accept ~negative with clear hydration reasoning]