Chemistry • Year 11 • Module 4 • Lesson 13

Gibbs Free Energy & Spontaneity

Two extended Band 5–6 tasks: data-driven multi-criteria evaluation and source critique. Demonstrate synthesis, evaluation, and evidence-based reasoning with the ΔG framework.

Master • Band 5–6 • Extended Response

Question 1 — The Haber process temperature paradox: data-driven evaluation

8 marks • ~20 minutes • Band 5–6

Stimulus: Haber process ΔG° data

The Haber process N₂(g) + 3H₂(g) → 2NH₃(g) has ΔH° = −92.4 kJ mol⁻¹ and ΔS° = −198.9 J K⁻¹ mol⁻¹. The table below shows ΔG° at various temperatures.

Temperature (K)	ΔG° (kJ mol⁻¹)	Spontaneous?	Practical rate?
298 (25°C)	−33.1	Yes	Essentially zero
465 (192°C)	≈ 0	Equilibrium	Very slow
673 (400°C)	+41.8	No	Acceptable (with Fe catalyst)
773 (500°C)	+61.4	No	Good (with Fe catalyst)

Additional context: The industrial compromise uses T = 400–500°C, P = 150–300 atm, iron catalyst, and continuous removal of NH₃. The Haber process produces ∼175 million tonnes of ammonia annually, supplying nitrogen for ∼50% of global food production.

Using the data above and your understanding of Gibbs free energy, evaluate the claim:

“The industrial Haber process operates at 400–500°C because this is the temperature range at which ΔG° is most negative, making the reaction most spontaneous.”

In your response you must:

State whether the claim is correct or incorrect.
Calculate ΔG° at 298 K to verify whether the Haber process is spontaneous at room temperature, showing unit conversions.
Calculate T_cross and identify the exact ΔH/ΔS case (A, B, C, or D) this reaction represents.
Explain why the industrial operating temperature is a kinetic compromise, not a thermodynamic optimum, using the data table as evidence.
State two other industrial modifications used in the Haber process and explain how each compensates for the thermodynamic or kinetic disadvantages at 400–500°C.

Question 2 — Source critique: a student’s chemistry blog post

7 marks • ~18 minutes • Band 5–6

Source: “The Gibbs Guide” — student chemistry blog, published online 2025

“Gibbs free energy is the total energy available in a reaction. If ΔG is negative, the reaction is spontaneous and will release a lot of heat to the surroundings. The more negative ΔG is, the faster the reaction proceeds. For any reaction where both ΔH and ΔS are negative, like the Haber process, increasing the temperature is the best way to make ΔG more negative and drive the reaction forward faster. Temperature is the master variable: hotter always means more spontaneous and faster for any reaction where ΔH < 0.”

The passage contains three distinct scientific errors. For each error:

Identify and quote the specific phrase that is incorrect.
State the correct chemistry, with reference to the ΔG = ΔH − TΔS framework.
For the third error, suggest an experimental observation or calculation that would demonstrate the error to a fellow student.

Answer Key & Marking Guidelines

Question 1 — 8 marks

Mark allocation:

1 mark: Correctly identifies the claim as incorrect and states that 400–500°C is thermodynamically less favourable than lower temperatures for this reaction.
2 marks: Calculation at 298 K with both unit conversions shown. T = 298 K (correct); ΔS° = −198.9 ÷ 1000 = −0.1989 kJ K⁻¹ mol⁻¹; ΔG° = −92.4 − (298 × −0.1989) = −92.4 + 59.3 = −33.1 kJ mol⁻¹ (spontaneous at 25°C). Award 1 mark for conversions, 1 mark for correct answer and classification.
1 mark: T_cross = ΔH°/ΔS° = −92.4/(−0.1989) = 465 K (192°C); correctly identifies Case C (ΔH < 0, ΔS < 0, spontaneous at low T only).
2 marks: Clear explanation that at 400–500°C the reaction is thermodynamically non-spontaneous (ΔG° = +41.8 to +61.4 kJ mol⁻¹) but proceeds at a practical rate because the N≡N bond has a very high activation energy that requires high temperature to overcome. Distinguishes thermodynamic favourability from kinetic rate (1 mark each). Evidence from data table cited explicitly.
2 marks: Two of the following: (i) High pressure (150–300 atm) increases concentration and pushes equilibrium towards fewer moles of gas (NH₃), increasing yield despite ΔG > 0. (ii) Iron catalyst lowers activation energy so the required rate is achieved at a lower temperature (mitigating the thermodynamic penalty somewhat). (iii) Continuous removal of NH₃ drives the equilibrium forward (Le Chatelier), compensating for the non-spontaneous ΔG° by maintaining low Q. 1 mark per correct modification + explanation.

Band 5: Correct identification + calculation + T_cross + kinetic reasoning. Band 6: All criteria met with clear data citation, two modifications correctly explained with thermodynamic/kinetic distinction.

Question 2 — 7 marks

Error 1 (2 marks): “ΔG is the total energy available in a reaction.” This is incorrect. ΔG is the maximum useful (non-expansion) work available, or the free energy change. It is not the total energy — that would be ΔH. ΔG is the enthalpy change adjusted for the entropy contribution at a given temperature (ΔG = ΔH − TΔS).

Error 2 (2 marks): “The more negative ΔG is, the faster the reaction proceeds.” This is incorrect. ΔG is a thermodynamic quantity that tells you if a reaction is spontaneous, not how fast. Rate is governed by kinetics — specifically activation energy (E_a) and collision frequency — which are independent of ΔG. The Haber process has ΔG° = −33.1 kJ mol⁻¹ at 25°C (spontaneous) but the rate is essentially zero at room temperature because of the very high E_a for the N≡N bond.

Error 3 (3 marks): “Increasing the temperature is the best way to make ΔG more negative… for any reaction where ΔH < 0.” This is incorrect for reactions where both ΔH and ΔS are negative (Case C). In this case, ΔG = ΔH − T(−) = ΔH + T|ΔS|, so increasing T makes ΔG more positive (less spontaneous), not more negative. 1 mark for identifying the specific error; 1 mark for correct chemistry with equation; 1 mark for experimental demonstration: calculate ΔG° for the Haber process at 25°C (−33 kJ) and 500°C (+61 kJ) — the increasing value proves the claim is wrong. Alternatively, show the ΔG vs T graph with positive slope for Case C.