Reversibility, Non-Equilibrium Systems & Entropy

Q1 — Sample Band 6 response (8 marks), annotated

The student’s misunderstanding arises from conflating ΔG° with ΔG (the actual free energy change at a specific composition). ΔG° = −33 kJ/mol tells us that, under standard conditions (1 mol L⊃−¹ of all species at 25°C), the forward reaction is spontaneous—but it does not tell us the reaction will go to completion. ΔG° quantifies the thermodynamic driving force from the standard state, not from any mixture of reactants and products. [1 — correctly distinguishes ΔG° from ΔG]

The reason the system does not go to completion is the concept of entropy of mixing. Any mixture of reactants and products has higher entropy than either pure extreme, because mixing increases disorder. This means the Gibbs free energy of an intermediate composition is lower than both pure reactants and pure products—the system always “falls into” an intermediate minimum free energy state rather than converting completely to products. This is the thermodynamic reason all reactions with moderate |ΔG°| reach equilibrium with both species present. [2 — entropy of mixing explanation, 1 for concept and 1 for linking to lower G at intermediate composition]

The data confirm this: at equilibrium (t = 20–30), [N₂] = 0.530 mol L⁻¹—53% of the initial 1.000 mol L⁻¹ N₂ remains unreacted. Similarly, [H₂] = 1.590 mol L⁻¹ (53% remains). Only 0.940 mol L⁻¹ NH₃ has formed from a possible 2.000 mol L⁻¹—a 47% conversion. Significant amounts of both reactants are present at equilibrium. [1 — correct use of data to quantify unreacted N₂/H₂]

At equilibrium, ΔG = 0 (not ΔG°). This is the crucial distinction. ΔG = 0 means the free energy of the system is at its minimum—neither the forward nor the reverse reaction has a thermodynamic driving force at that composition. ΔG° = −33 kJ/mol is the driving force from the standard state (1 mol L⁻¹ of all species) towards equilibrium; it is non-zero because the standard state is not the equilibrium state. Once the system reaches equilibrium, ΔG = 0 and no further net change occurs in either direction. [2 — ΔG = 0 at equilibrium and its distinction from ΔG°]

If ΔG° were −500 kJ/mol rather than −33 kJ/mol, the Gibbs free energy minimum would shift almost entirely to the products side. The entropy of mixing contribution would be negligible compared to the enormous thermodynamic preference for products. Essentially all reactants would be converted—the reaction would behave as irreversible, and the data table would show [N₂] and [H₂] approaching zero at equilibrium. [1 — prediction applied correctly]

Marking criteria.

• 1 mark — Correctly states that ΔG° quantifies the standard-state driving force and does not predict completion; distinguishes ΔG° from ΔG.
• 1 mark — Names and correctly explains the entropy of mixing concept.
• 1 mark — Explains that intermediate mixtures have higher entropy (lower G) than either pure extreme, so the G minimum lies at an intermediate composition for moderate ΔG°.
• 1 mark — Uses the data table to quantify unreacted reactants at equilibrium (e.g. 53% of N₂ remains, 47% conversion of NH₃).
• 1 mark — States ΔG = 0 at equilibrium and explains this as the minimum free energy state with no thermodynamic driving force.
• 1 mark — Correctly distinguishes ΔG = 0 (equilibrium condition) from ΔG° ≠ 0 (standard-state driving force).
• 1 mark — Predicts (correctly) that ΔG° = −500 kJ/mol would shift the equilibrium almost entirely to products, approaching irreversible behaviour.
• 1 mark (bonus/Band 6 discriminator) — Explicitly connects all three points: entropy of mixing + magnitude of ΔG° + meaning of ΔG = 0, in a coherent argument that fully corrects the student’s misunderstanding without oversimplification.

Q2 — Sample Band 6 response (7 marks), annotated

The blog claim contains one defensible observation but is fundamentally incorrect in its core argument.

Defensible element: It is true that combustion of petrol in a car engine occurs in an open system (exhaust gases escape), and that open systems cannot reach closed-system dynamic equilibrium. [1 — concedes one correct element]

What is incorrect: The claim implies that the open system is the reason combustion is irreversible—suggesting that in a sealed flask, equilibrium with measurable petrol would reform. This is thermodynamically wrong. Irreversibility is determined by the magnitude of ΔG, not by the system type. [1 — correctly identifies the flaw in logic]

Using ΔG = ΔH − TΔS for combustion of octane (a major petrol component, C₈H₁₈): ΔH ≈ −5471 kJ/mol (strongly exothermic; products CO₂ and H₂O are far more thermodynamically stable than octane and O₂). ΔS is large and positive (solid/liquid octane and gas O₂ → CO₂(g) + H₂O(g); large increase in gas moles and disorder). Therefore ΔG = ΔH − TΔS is the sum of a very large negative enthalpy term and an additional negative −TΔS term—both terms drive ΔG to be extremely large and negative. This makes combustion “doubly irreversible”: the thermodynamic preference for products is reinforced by both enthalpy and entropy simultaneously. [2 — 1 for correct signs and 1 for “doubly irreversible” argument]

The sealed-flask scenario: Even in a perfectly sealed flask, the equilibrium constant K_eq would be astronomically large (related to ΔG° by the equation ΔG° = −RT ln K; for ΔG° ≈ −5000 kJ/mol, K_eq ≈ 10⁸⁷⁵). The amount of petrol that would reform at equilibrium is so vanishingly small as to be physically meaningless—no measurable petrol would ever appear. The claim that “equilibrium with reformed petrol” would exist in a sealed flask is therefore incorrect. [1 — sealed flask reasoning with K_eq argument]

Australian examples: Combustion of eucalypt oils during a Black Summer bushfire in south-east Australia (e.g. the 2019–2020 fires burning more than 18 million hectares) has ΔG ≈ −thousands of kJ/mol—irreversible; the burnt landscape cannot spontaneously regenerate by reversing the reaction. In contrast, N₂O₄(g) ⇌ 2NO₂(g) (the brown-gas equilibrium visible in sealed ampoules of nitrogen dioxide) has ΔG° = +4.7 kJ/mol at 25°C—a small value indicating a genuinely reversible equilibrium where both colourless N₂O₄ and brown NO₂ are measurably present. [1 — named Australian irreversible example and named reversible example with thermodynamic contrast]

Defensible reformulation: “Combustion of petrol is irreversible not primarily because it occurs in an open system, but because both ΔH and ΔS make ΔG extremely negative. The free energy minimum lies almost entirely at the products side, so the reverse reaction (reformation of petrol from CO₂ and H₂O) has a ΔG of thousands of kJ/mol in the positive direction—essentially impossible under normal conditions regardless of system type.” [1 — chemically and thermodynamically defensible reformulation]

Marking criteria.

• 1 mark — Identifies the one defensible element (open system cannot reach closed-system dynamic equilibrium; this part is correct in itself).
• 1 mark — Correctly identifies the flaw: the claim implies system type determines irreversibility; in fact it is the magnitude of ΔG.
• 1 mark — Correctly applies ΔG = ΔH − TΔS to combustion: states ΔH << 0 and ΔS > 0, so both terms make ΔG very large and negative.
• 1 mark — Makes the “doubly irreversible” argument: both ΔH and TΔS reinforce irreversibility simultaneously (unlike reactions where only one term favours the forward direction).
• 1 mark — Addresses the sealed-flask assertion: even in a sealed flask, K_eq is astronomically large so no measurable petrol would reform; the claim is incorrect.
• 1 mark — Names at least one real-world Australian irreversible combustion example (e.g. bushfire) and one reversible reaction (e.g. N₂O₄/NO₂, Haber process, H₂/I₂/HI) and uses both to illustrate the thermodynamic spectrum.
• 1 mark — Produces a chemically and thermodynamically defensible reformulation of the claim that correctly centres irreversibility on the magnitude of ΔG rather than system type.