Chemistry • Year 11 • Module 4 • Lesson 13

Gibbs Free Energy & Spontaneity

Apply ΔG calculations, read a ΔG vs T data graph, reason through real industrial contexts, and predict crossover temperatures.

Apply • Band 4–5 • Data & Reasoning

1. Interpret a ΔG° vs temperature graph — limestone decomposition at Boral cement works

Boral operates a large cement plant at Berrima, NSW, where limestone (CaCO3) is heated in rotary kilns to produce quicklime (CaO) and carbon dioxide:

CaCO3(s) → CaO(s) + CO2(g)    (ΔH° = +178 kJ mol−1, ΔS° = +160.6 J K−1 mol−1)

The graph below shows ΔG° for this reaction plotted against temperature. 9 marks

ΔG=0 Tₐ ≈ 1108 K ΔG > 0 (non-spontaneous) ΔG < 0 (spontaneous) +150 +100 +50 0 −50 −90 ΔG° (kJ mol⁻¹) 300 500 700 900 1100 1300 Temperature (K) →
Figure 1.1. ΔG° vs temperature for CaCO3(s) → CaO(s) + CO2(g). Adapted from thermodynamic data consistent with Boral cement production contexts, Berrima NSW.

1.1 Describe the trend in ΔG° as temperature increases from 300 K to 1400 K. 2 marks

1.2 From the graph, estimate the crossover temperature (where ΔG° = 0). Then verify this by calculating Tcross = ΔH° / ΔS° using the data given. Show your working. 3 marks

1.3 Explain why Boral must heat its kilns to temperatures above Tcross for the decomposition to be thermodynamically spontaneous. Identify the ΔH/ΔS sign combination for this reaction and state which case from the 2×2 table it represents. 4 marks

2. Interpret ΔG data for the Haber process at Port Kembla

BlueScope Steel’s Port Kembla plant uses high-temperature industrial chemistry. The table below shows ΔG° values for the Haber process N2(g) + 3H2(g) → 2NH3(g) at four operating temperatures. 7 marks

Temperature (°C) Temperature (K) ΔG° (kJ mol−1) Spontaneous?
25298−33.1
192465≈ 0
400673+41.8
500773+61.4

2.1 Complete the “Spontaneous?” column for all four temperatures. 2 marks

(Write your answers directly in the table above.)

2.2 At 400–500°C (the industrial operating range), ΔG° is positive. Account for why industrial plants nonetheless operate at these temperatures, despite the reaction being thermodynamically non-spontaneous. Refer to both thermodynamics and kinetics in your answer. 3 marks

2.3 The Haber process has ΔH° = −92.4 kJ mol−1 and ΔS° = −198.9 J K−1 mol−1. Identify the ΔH/ΔS case for this reaction and predict what happens to ΔG° as temperature increases. 2 marks

3. Cause-and-effect chain — increasing temperature in an endothermic, entropy-increasing reaction

Consider a reaction where ΔH > 0 and ΔS > 0. Trace the cause-and-effect chain when operating temperature is raised from below Tcross to above Tcross. Fill each effect box. 5 marks

Temperature is increased above Tcross
→
Effect on TΔS term
↓
TΔS term becomes large positive
→
Effect on sign of ΔG
↓
ΔG changes sign from + to −
→
What this means for spontaneity

Overall outcome (so…):

4. Predict and justify — iron ore smelting at Port Kembla

In the blast furnaces at BlueScope’s Port Kembla steelworks, iron oxide is reduced to iron at temperatures above 1200°C. One of the key reactions is:

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)    (ΔH° = −25 kJ mol−1, ΔS° = +15 J K−1 mol−1)

A student claims: “This reaction has ΔH < 0 and ΔS > 0, so it will be more spontaneous at very high temperatures like 1200°C.” 4 marks

4.1 Predict whether the student is correct. Then calculate ΔG° at 25°C (298 K) and at 1200°C (1473 K) to confirm your prediction. Show all unit conversions. 4 marks

5. Compare and contrast — Haber process vs limestone decomposition

Complete the comparison table using the ΔG framework. 8 marks

Feature Haber process (N2 + 3H2 → 2NH3) Limestone decomposition (CaCO3 → CaO + CO2)
Sign of ΔH°
Sign of ΔS°
ΔH/ΔS case (A/B/C/D)
Spontaneous at low T?
Effect of increasing T on ΔG
Why high T is used industrially
Australian industrial context
Approximate Tcross

Answer Key

Section 1 — ΔG vs T graph

1.1 (2 marks) 1 mark: ΔG° decreases linearly (becomes less positive / more negative) as temperature increases. 1 mark: This occurs because the TΔS term increases with T, subtracting more from ΔH (which is constant), so ΔG = ΔH − TΔS becomes progressively more negative.

1.2 (3 marks) Graph estimate: approximately 1100–1115 K (accept ±50 K). 1 mark. Calculation: Tcross = ΔH°/ΔS° = 178 000 J mol−1 ÷ 160.6 J K−1 mol−1 = 1108 K (note: both in J units, or both in kJ). 2 marks (1 conversion, 1 answer).

1.3 (4 marks) 1 mark: The reaction is endothermic (ΔH > 0) with positive entropy (ΔS > 0) — this is Case D. 1 mark: Below Tcross (~1108 K), the +ΔH dominates and ΔG > 0 (non-spontaneous). 1 mark: Above Tcross the large TΔS term exceeds ΔH, making ΔG < 0. 1 mark: Boral must exceed ~835°C to make the reaction thermodynamically spontaneous and drive CaO formation.

Section 2 — Haber process data table

2.1 (2 marks): 25°C: Yes (ΔG < 0); 192°C: Equilibrium (ΔG ≈ 0); 400°C: No (ΔG > 0); 500°C: No (ΔG > 0). 0.5 marks each row.

2.2 (3 marks) 1 mark: At 25°C the reaction is thermodynamically spontaneous but the activation energy for breaking the N≡N triple bond is so high that the rate is essentially zero. 1 mark: High temperature (400–500°C) increases collision energy so the reaction proceeds at a practical rate. 1 mark: The thermodynamically unfavourable regime is accepted because kinetics (rate) is the limiting factor, and NH3 is continuously removed to drive equilibrium forward (Le Chatelier). Both thermodynamics and kinetics must be distinguished.

2.3 (2 marks) 1 mark: ΔH < 0 and ΔS < 0 → Case C. 1 mark: As T increases, TΔS (which is negative × T) grows more negative, so ΔG = ΔH − TΔS = (−) − (−large) = (−) + (large) → ΔG becomes less negative then positive. The reaction becomes non-spontaneous above Tcross ≈ 465 K (192°C).

Section 3 — Cause-and-effect chain

Effect 1 (TΔS term): TΔS grows larger (more positive) because T is now large and ΔS > 0.

Effect 2 (sign of ΔG): ΔG = ΔH − TΔS = (+) − (large positive) = becomes negative.

Effect 3 (spontaneity): ΔG < 0 → the reaction becomes spontaneous above Tcross.

Overall: Raising T above Tcross for a ΔH > 0, ΔS > 0 reaction makes it thermodynamically spontaneous, because the entropy “pay-off” finally outweighs the unfavourable enthalpy. (5 marks: 1 per effect box + 1 overall.)

Section 4 — Port Kembla iron smelting

4.1 (4 marks) Student is correct — with ΔH < 0, ΔS > 0 (Case A) the reaction is always spontaneous and becomes more so at high T. Convert: ΔS° = +15 ÷ 1000 = +0.015 kJ K−1 mol−1. ΔG°(298 K) = −25 − (298 × 0.015) = −25 − 4.47 = −29.5 kJ mol−1 (spontaneous). ΔG°(1473 K) = −25 − (1473 × 0.015) = −25 − 22.1 = −47.1 kJ mol−1 (more negative, more spontaneous). Marks: 1 conversion, 1 calculation at 298 K, 1 calculation at 1473 K, 1 interpretation.

Section 5 — Compare and contrast (1 mark per row, 8 marks)

FeatureHaberLimestone
Sign of ΔH°Negative (−92.4 kJ/mol)Positive (+178 kJ/mol)
Sign of ΔS°Negative (−0.199 kJ/K/mol)Positive (+0.161 kJ/K/mol)
CaseCD
Spontaneous at low T?Yes (below ~465 K)No
Effect of increasing T on ΔGΔG increases (becomes less negative, then positive)ΔG decreases (becomes less positive, then negative)
Why high T usedKinetics — rate is the limiting factor, not thermodynamicsThermodynamics — must exceed Tcross to make reaction spontaneous
Australian industrial contextHaber process / Port Kembla fertiliser chainBoral Berrima cement works (also Boral Maldon)
Approximate Tcross~465 K (192°C)~1108 K (835°C)