Chemistry • Year 12 • Module 5 • Lesson 2

Reversibility, Non-Equilibrium Systems & Entropy

Apply ΔG reasoning to real data, trace cause-and-effect in non-equilibrium systems, and interpret a ΔG vs composition graph for the Haber process.

Apply · Data & Reasoning

1. Interpret data — classifying reactions by ΔG° and predicting equilibrium positions

The table below gives ΔG° values and ΔH values for six reactions. Use this data to answer the questions. 8 marks

Reaction	ΔH (kJ/mol)	ΔG° (kJ/mol)	Classification	Equilibrium favours
A: 2Mg(s) + O₂(g) → 2MgO(s)	−1204	−1138	Your answer	Your answer
B: H₂(g) + I₂(g) ⇌ 2HI(g)	−9	−3.5	Your answer	Your answer
C: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)	−92	−33	Your answer	Your answer
D: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)	+88	+21	Your answer	Your answer
E: NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq)	+25.7	−17	Your answer	Your answer
F: CaCO₃(s) → CaO(s) + CO₂(g)	+178	+130 at 25°C; −1 at ~850°C	Your answer (temperature-dependent)	Your answer

1.1 Identify which reactions in the table are irreversible and explain the thermodynamic criterion you used to classify them. 2 marks

1.2 Reaction E (dissolution of ammonium nitrate) has a positive ΔH yet a negative ΔG°. Using ΔG = ΔH − TΔS, explain how this is possible. 3 marks

1.3 Reaction F (decomposition of CaCO₃) has a positive ΔG° at 25°C but a negative ΔG° near 850°C. Explain this temperature dependence using ΔG = ΔH − TΔS and the sign of ΔS for this reaction. 3 marks

Stuck? Revisit Cards 2 and 4 in Lesson 2 and the ΔH/ΔS spontaneity table (mind map SVG).

2. Graph interpretation — Gibbs free energy along the reaction coordinate

The graph below shows how the Gibbs free energy (G) of a closed reaction system changes as the composition shifts from pure reactants (left) to pure products (right) for the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The curve is schematic but quantitatively consistent with known data. 7 marks

Schematic Gibbs free energy profile for N₂ + 3H₂ ⇌ 2NH₃ at 25°C. Adapted from thermodynamic principles; ΔG° = −33 kJ/mol at 298 K.

2.1 At the point labelled “Equilibrium (min G)”, what is the value of ΔG? Explain what this value tells you about the driving forces acting on the system at this point. 2 marks

2.2 The equilibrium minimum is located closer to the reactants side than the products side. What does this tell you about the industrial yield of NH₃ per pass through the reactor? 2 marks

2.3 For an irreversible reaction such as combustion of methane (ΔG = −818 kJ/mol), describe how the G vs composition curve would differ from the Haber process curve above. 3 marks

Stuck? Revisit Card 2 (‘Gibbs Free Energy and Equilibrium’) in Lesson 2 and the callout on equilibrium being a free energy minimum.

3. Cause-and-effect chain — why photosynthesis is a non-equilibrium system

The boxes below show a cause-and-effect chain for photosynthesis as a non-equilibrium system. The causes (shaded) are given; fill in each effect box. Finish with the “Overall outcome” line. 5 marks

Cause: Photosynthesis requires the conversion of CO₂ + H₂O → C₆H₁₂O₆ + O₂, which has ΔG = +2870 kJ/mol.

↓

Effect 1:

Cause: Solar energy (photons) is absorbed continuously by chlorophyll in the leaf.

↓

Effect 2:

Cause: Leaves exchange CO₂, O₂, and water vapour with the atmosphere through stomata.

↓

Effect 3:

Cause: At night, energy input stops; the reverse reaction (respiration, ΔG = −2870 kJ/mol) is spontaneous.

↓

Effect 4:

Overall outcome (so…): Photosynthesis is a non-equilibrium process because…

Stuck? Revisit Card 4 (entropy-driven reactions) and the ‘Insight’ callout about photosynthesis and respiration in Lesson 2.

4. Australian case study — reversible reactions at Incitec Pivot, Gibson Island

Incitec Pivot operates Australia’s largest ammonia plant at Gibson Island, Brisbane. The plant uses the Haber process (N₂ + 3H₂ ⇌ 2NH₃, ΔG° = −33 kJ/mol) and the Contact process for sulfuric acid (2SO₂ + O₂ ⇌ 2SO₃, ΔG° = −141 kJ/mol). The plant uses recycling loops to boost overall yield even though equilibrium conversion per pass is low. 5 marks

4.1 Explain why the ΔG° values for both reactions classify them as reversible equilibrium reactions rather than reactions that go to completion. 2 marks

4.2 Incitec Pivot achieves a far higher overall yield of NH₃ than a single equilibrium pass would give. Using the concept of a closed system and dynamic equilibrium, explain how recycling unconverted N₂ and H₂ back through the reactor improves total yield without changing the position of equilibrium or the value of K. 3 marks

Stuck? Revisit Card 5 (Haber process) in Lesson 2.

5. Compare and contrast — reversible vs irreversible reactions

Complete the two-column table below. For each feature, describe how it applies to a reversible reaction (e.g. H₂ + I₂ ⇌ 2HI) and to an irreversible reaction (e.g. combustion of a eucalyptus forest in a bushfire). 8 marks (1 per row)

Feature	Reversible (H₂ + I₂ ⇌ 2HI)	Irreversible (bushfire combustion)
Magnitude of ΔG°
Can be conducted in a sealed flask?
Arrow notation used
Value of ΔG at equilibrium
Presence of reactants at “completion”
Sign of ΔS for the reaction
System type (open/closed)
Role of a catalyst

Stuck? Use Cards 1–3 in Lesson 2, the combustion callout and the worked examples.

Answers — Do not peek before attempting

Q1 — Data table classification

Irreversible: Reaction A (2Mg + O₂, ΔG° = −1138 kJ/mol) — the enormous negative ΔG° places it well past the −100 kJ/mol threshold; products are overwhelmingly more stable. All other reactions are reversible (moderate or near-zero ΔG°).

Criterion used (1 mark): A reaction is classified as irreversible when |ΔG°| >> 100 kJ/mol, meaning the equilibrium lies almost entirely on the product side with negligible reverse reaction under the same conditions. A reaction is classified as reversible when |ΔG°| is small enough (< ~100 kJ/mol, or ΔG° near zero or positive) that both forward and reverse reactions are thermodynamically accessible.

Q1.2 — NH₄NO₃ dissolution: positive ΔH, negative ΔG°

Using ΔG = ΔH − TΔS: when NH₄NO₃ dissolves, the solid lattice breaks apart and NH₄⁺ and NO₃⁻ ions disperse throughout the solvent—a very large increase in disorder, so ΔS is large and positive [1]. Although ΔH = +25.7 kJ/mol is positive (endothermic), the TΔS term is larger at room temperature (TΔS > ΔH) [1], so ΔG = ΔH − TΔS < 0 [1]. The reaction is entropy-driven—spontaneous because the entropy gain overcomes the enthalpy cost.

Q1.3 — CaCO₃ decomposition: temperature dependence

CaCO₃(s) → CaO(s) + CO₂(g): ΔH = +178 kJ/mol (endothermic); CO₂(g) is produced from a solid, so ΔS is large and positive [1]. Using ΔG = ΔH − TΔS: at 25°C (298 K), the TΔS term is small and ΔG remains positive (non-spontaneous) [1]. As temperature increases, TΔS grows. At ~850°C, TΔS ≈ ΔH so ΔG ≈ 0; above this temperature TΔS > ΔH and ΔG < 0—the reaction becomes spontaneous [1].

Q2.1 — ΔG at the equilibrium minimum

ΔG = 0 at the equilibrium minimum [1]. This means neither the forward nor the reverse reaction has a thermodynamic driving force—the system has reached its lowest possible Gibbs free energy and there is no tendency for net change in either direction [1].

Q2.2 — Equilibrium position and industrial yield

The equilibrium minimum lying close to the reactants side (around 15–25% conversion) indicates that the equilibrium position strongly favours the reactant side relative to pure products [1]. In industrial terms, only about 15–25% of N₂ and H₂ is converted to NH₃ per pass through the reactor—a relatively low single-pass yield. This is why the Haber process uses recycling loops to return unreacted gases [1].

Q2.3 — G vs composition for an irreversible reaction

For combustion of methane (ΔG = −818 kJ/mol), the curve would drop steeply from the reactant side and reach its minimum only when the system has converted almost all reactants to products—the minimum would be located very close to the right-hand (pure products) end [1]. The difference in G between reactants and products would be enormous (∼818 kJ/mol per mole), compared to only 33 kJ/mol for the Haber process [1]. There would be essentially no region where the curve “curves back” to favour any significant amount of reactants at equilibrium—the equilibrium mixture is almost purely products [1].

Q3 — Cause-and-effect chain (sample answers)

Effect 1: The forward reaction (photosynthesis) is highly non-spontaneous (ΔG >> 0) and cannot proceed without a large external energy input; left alone, the reaction would not occur at all.

Effect 2: Absorbed solar energy provides the thermodynamic input needed to drive the non-spontaneous forward reaction; the light energy is converted into chemical potential energy stored in glucose molecules.

Effect 3: The leaf is an open system—CO₂ enters and O₂ leaves continuously through stomata, so the system can never reach a closed-system dynamic equilibrium; the concentrations of reactants and products are continuously replenished and depleted.

Effect 4: Without photon input, the non-spontaneous photosynthesis reaction stops; the spontaneous reverse reaction (cellular respiration and decomposition, ΔG = −2870 kJ/mol) dominates, converting glucose back to CO₂ and H₂O.

Overall outcome: Photosynthesis is a non-equilibrium process because it requires continuous external energy input (solar radiation), occurs in an open system that cannot attain dynamic equilibrium, and has a large positive ΔG that makes the forward reaction non-spontaneous without that energy input. [1 mark]

Q4.1 — Incitec Pivot: ΔG° and reversibility

Both ΔG° values (−33 and −141 kJ/mol) are negative but not very large in magnitude. This means that, while the forward direction is thermodynamically favoured, the reverse direction is also thermodynamically accessible—the free energy minimum lies at an intermediate composition with measurable amounts of both reactants and products present. Neither reaction goes “to completion” as a reaction with |ΔG°| >> 100 kJ/mol would.

Q4.2 — How recycling improves yield without changing K

At equilibrium in a single pass, the system reaches a fixed composition determined by K [1]. By removing NH₃ from the exit stream (product extraction) and returning the remaining N₂ and H₂ to the reactor inlet, the plant is effectively running the same closed-system equilibrium multiple times [1]. Each pass converts 15–25% of the recirculating gas to NH₃; after multiple cycles, the cumulative yield can exceed 95%—all without changing K, ΔG°, or the position of equilibrium for any single pass [1].

Q5 — Compare and contrast table (sample answers)

Reversible (H₂ + I₂ ⇌ 2HI) | Irreversible (bushfire combustion):

|ΔG°|: Small/moderate (~3.5 kJ/mol) | Very large (~hundreds–thousands kJ/mol)
Sealed flask possible?: Yes, equilibrium reached | No, open system; gases escape
Arrow: Double arrow ⇌ | Single arrow →
ΔG at equilibrium: ΔG = 0 at equilibrium | Not applicable (no equilibrium reached)
Reactants at completion: Yes, measurable amounts of H₂, I₂ remain | No, essentially all reactants consumed
ΔS: Near zero (similar moles of gas, 2 mol → 2 mol) | Large positive (solid biomass → CO₂ gas + H₂O)
System type: Closed (sealed container) | Open (products disperse into atmosphere)
Catalyst role: Speeds up approach to same equilibrium; does not change yield | Speeds up combustion (ignition); does not change products or irreversibility

Award 1 mark per row where both columns are correctly and meaningfully contrasted (max 8).