Fritz Haber's discovery that nitrogen and hydrogen gas could be converted to ammonia — reversibly — changed the course of human history. The reaction that feeds half the world's population works precisely because it doesn't go all the way to completion.
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You know from Module 4 that Gibbs free energy (ΔG) determines whether a reaction is spontaneous. Consider two reactions:
(1) Combustion of methane — CH₄ + 2O₂ → CO₂ + 2H₂O, ΔG = −818 kJ/mol
(2) Formation of ammonia — N₂ + 3H₂ ⇌ 2NH₃, ΔG° = −33 kJ/mol
Both have negative ΔG values — both are spontaneous in the forward direction. But one reaches dynamic equilibrium with significant amounts of reactants remaining, while the other goes essentially to completion. Before reading on — which one goes to completion, and why do you think the magnitude of ΔG matters? Write your prediction.
📚 Core Content
Wrong: Entropy always increases in every chemical reaction.
Right: The Second Law states that total entropy of the universe increases for spontaneous processes, but the system alone can have ΔS < 0. Endothermic reactions with negative ΔS can still be spontaneous if TΔS is outweighed by a large negative ΔH, or at high temperature if ΔS is positive.
Whether a reaction is reversible or irreversible is not a binary switch — it is a spectrum determined by how strongly thermodynamics favours the products over the reactants.
An irreversible reaction is one in which the forward reaction is so thermodynamically favoured (large negative ΔG°) that the reverse reaction is negligible under the same conditions. The system effectively goes to completion — almost all reactants are converted to products. Written with →. Examples: combustion of hydrocarbons; neutralisation of strong acid with strong base.
A reversible reaction is one in which both forward and reverse reactions are thermodynamically accessible — neither direction is overwhelmingly favoured. The system reaches dynamic equilibrium with measurable amounts of both reactants and products present. Written with ⇌. Examples: Haber process (N₂ + 3H₂ ⇌ 2NH₃); decomposition of N₂O₄; formation of HI from H₂ and I₂.
The magnitude of ΔG° is the quantitative indicator: reactions with ΔG° << −100 kJ/mol are effectively irreversible; reactions with |ΔG°| < 50 kJ/mol often show meaningful equilibrium mixtures.
ΔG° magnitude determines reaction type — the spectrum from irreversible to reversible
Gibbs free energy is the driving force for a reaction — and equilibrium is the point where that driving force is exhausted: the system has found the lowest accessible free energy state.
From Module 4, you know that ΔG = ΔH − TΔS. A reaction is spontaneous when ΔG < 0 and non-spontaneous when ΔG > 0. At dynamic equilibrium, the free energy of the system is at its minimum — ΔG = 0. Neither forward nor reverse is spontaneous; the system has no driving force for net change in either direction.
For an irreversible reaction, the free energy minimum is reached only when essentially all reactants have become products. For a reversible reaction, the minimum is at an intermediate composition — with both reactants and products present.
| ΔG° Value | Position of Equilibrium | Keq Magnitude |
|---|---|---|
| Very large negative (e.g. −500 kJ/mol) | Almost entirely products | Keq >> 1 (e.g. 10⁸⁰) |
| Moderately negative (e.g. −20 kJ/mol) | Products favoured but reactants present | Keq > 1 (e.g. 10³) |
| Near zero (e.g. ±5 kJ/mol) | Significant amounts of both | Keq ≈ 1 |
| Moderately positive (e.g. +20 kJ/mol) | Reactants favoured but products present | Keq < 1 (e.g. 10⁻³) |
| Very large positive (e.g. +500 kJ/mol) | Almost entirely reactants | Keq << 1 |
Combustion is the archetypal non-equilibrium system — the products are so thermodynamically stable that the reverse reaction is effectively impossible under normal conditions.
Consider the combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), ΔH = −890 kJ/mol, ΔG = −818 kJ/mol. This reaction is non-equilibrium for two reinforcing thermodynamic reasons:
This is why you cannot un-burn a log: the products are in a far lower energy state and the entropic conditions make the reverse reaction essentially impossible without a massive external energy input (like photosynthesis, which uses solar energy).
Some reactions absorb heat and are still spontaneous — because the entropy gain is large enough to overcome the enthalpy cost, making ΔG negative despite positive ΔH.
From Module 4, ΔG = ΔH − TΔS. For ΔG to be negative when ΔH is positive (endothermic), the term TΔS must be larger than ΔH — the entropy gain must be large enough and/or the temperature high enough.
Examples of spontaneous endothermic reactions:
Photosynthesis — non-equilibrium endothermic example: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂, ΔG = +2870 kJ/mol. This is highly non-spontaneous — it requires continuous solar energy input and occurs in an open system. Without continuous energy input, the reverse reaction (respiration/combustion) is spontaneous.
ΔG = ΔH − TΔS — four ΔH/ΔS combinations and their temperature-dependent spontaneity
The Haber process works because nitrogen and hydrogen don't combust — they form ammonia reversibly, meaning the reaction can be controlled, optimised, and run continuously without the products escaping the system.
In the early 20th century, the world faced a crisis: agricultural soil was being depleted of nitrogen faster than natural processes could replenish it. Fritz Haber discovered that N₂ and H₂ could be combined to form ammonia:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ/mol, ΔG° ≈ −33 kJ/mol at 25°C
The reaction is reversible and reaches dynamic equilibrium. The challenge: because ΔG° is only moderately negative, not all N₂ and H₂ convert to NH₃ at equilibrium — industrial yields are typically only 15–25% per pass. The strength: because the reaction is reversible in a closed system, unconverted N₂ and H₂ can be recycled through the reactor multiple times, dramatically increasing overall yield.
Today, ammonia produced by the Haber process is the basis of nitrogen fertilisers that feed approximately half the world's population — an estimated 4 billion people.
✏️ Worked Examples
(a) 2H₂(g) + O₂(g) → 2H₂O(g), ΔG = −457 kJ/mol per mol H₂O
(b) CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq), ΔG° = +27 kJ/mol
(c) N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔG° = −33 kJ/mol
ΔG = −457 kJ/mol — a very large negative value. The products (water) are overwhelmingly more thermodynamically stable than the reactants. The reverse reaction (splitting water spontaneously) has ΔG = +457 kJ/mol — highly non-spontaneous. This reaction goes to completion.
→ Irreversible.
ΔG° = +27 kJ/mol — positive, meaning the forward reaction as written is non-spontaneous under standard conditions. However, the reverse reaction (ΔG° = −27 kJ/mol) is spontaneous. In reality, acetic acid partially dissociates — both forward and reverse reactions occur, reaching dynamic equilibrium. The positive ΔG° means the equilibrium lies on the reactants side (more acetic acid than ions at equilibrium — weak acid).
→ Reversible equilibrium — equilibrium position favours reactants.
ΔG° = −33 kJ/mol — moderately negative. The forward reaction is spontaneous but not overwhelmingly so. The reverse reaction (ΔG° = +33 kJ/mol) is non-spontaneous but not impossible. Both reactions are thermodynamically accessible, and the system reaches dynamic equilibrium with measurable amounts of both N₂/H₂ and NH₃ present.
→ Reversible equilibrium — equilibrium position favours products moderately.
Summary: (a) Irreversible — very large negative ΔG means only products remain at equilibrium. (b) Reversible — small ΔG°; equilibrium favours reactants (weak acid partially dissociates). (c) Reversible — moderately negative ΔG°; equilibrium has significant amounts of both reactants and products.
(a) Predict the sign of ΔS and explain. (b) Use ΔG = ΔH − TΔS to explain why this reaction is irreversible. (c) Explain why photosynthesis (the reverse) requires continuous external energy and is not at equilibrium.
Products include 6 mol CO₂(g) — gas molecules have much higher entropy than solids. The conversion of solid glucose to CO₂ gas and liquid water represents a net increase in disorder. ΔS > 0 (positive entropy change).
ΔG = ΔH − TΔS = (−2803) − T(positive value). Both terms make ΔG more negative — ΔH contributes a large negative value and −TΔS contributes an additional negative value. ΔG is very large and negative at any temperature.
The reverse reaction would require ΔG = +2803 kJ/mol — completely non-spontaneous without energy input. The reaction goes to completion and cannot reverse spontaneously.
Photosynthesis is the reverse of combustion — it has ΔG = +2870 kJ/mol — extremely non-spontaneous. Solar energy (absorbed by chlorophyll) is the external energy input that drives this thermodynamically unfavourable reaction forward.
Because it requires continuous energy input and occurs in an open system (leaves exchange CO₂, O₂, and water with the atmosphere), it cannot reach dynamic equilibrium — it is a sustained non-equilibrium process.
Summary: (a) ΔS > 0 — solid glucose converts to gases and liquid; overall disorder increases. (b) Both ΔH and the TΔS term make ΔG extremely large and negative — products are overwhelmingly more stable; reverse reaction has ΔG = +2803 kJ/mol and is essentially impossible spontaneously. (c) Photosynthesis has ΔG = +2870 kJ/mol — continuous solar energy input is required; it occurs in an open system and cannot reach dynamic equilibrium.
🧪 Activities
| Reaction | ΔG° (kJ/mol) | Reversible or Irreversible? | Equilibrium favours… |
|---|---|---|---|
| 2Mg(s) + O₂(g) → 2MgO(s) | −1138 | Your answer | Your answer |
| H₂(g) + I₂(g) ⇌ 2HI(g) | −3.5 | Your answer | Your answer |
| PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) | +21 | Your answer | Your answer |
| 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) (Contact process) | −141 | Your answer | Your answer |
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
❓ Multiple Choice
1. Which of the following best explains why the combustion of magnesium (2Mg + O₂ → 2MgO) is irreversible?
2. The dissolution of ammonium nitrate (ΔH = +25.7 kJ/mol) is spontaneous at room temperature. Which explanation is correct?
3. N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔG° = −33 kJ/mol. Why is this reaction described as reversible rather than going to completion?
4. At dynamic equilibrium, the value of ΔG for the system is:
5. A student states: "The decomposition of CaCO₃(s) → CaO(s) + CO₂(g) is endothermic (ΔH = +178 kJ/mol), therefore it can never be spontaneous." Evaluate this claim.
✍️ Short Answer
6. Explain what it means for a reaction to be "at equilibrium" in terms of Gibbs free energy. Why is ΔG = 0 at equilibrium? Use the concept of driving force to explain. 3 MARKS
7. The following reactions both have ΔH < 0 (exothermic), but one is reversible and one is irreversible:
Reaction 1: H₂(g) + F₂(g) → 2HF(g), ΔG° = −543 kJ/mol
Reaction 2: 2NO₂(g) ⇌ N₂O₄(g), ΔG° = −4.7 kJ/mol
(a) Classify each as reversible or irreversible and justify using ΔG°. (2 marks)
(b) For the reversible reaction, describe what happens macroscopically and microscopically after equilibrium is established in a sealed flask. (2 marks)
4 MARKS
8. Haber Process Analysis: Fritz Haber's process N₂(g) + 3H₂(g) ⇌ 2NH₃(g) has ΔG° = −33 kJ/mol and ΔH = −92 kJ/mol.
(a) Explain, using Gibbs free energy, why the Haber process reaches equilibrium rather than going to completion. (2 marks)
(b) Industrial plants typically operate at high temperature (400–500°C). Considering ΔG = ΔH − TΔS, explain how increasing temperature affects the spontaneity of the forward reaction. Use the signs of ΔH and ΔS for this reaction. (3 marks)
5 MARKS
Go back to your Think First response. Now that you've studied the relationship between ΔG° and reversibility:
2Mg + O₂, ΔG° = −1138: Irreversible — enormously negative ΔG° means MgO is overwhelmingly more thermodynamically stable than Mg and O₂. Products almost entirely favoured.
H₂ + I₂, ΔG° = −3.5: Reversible — ΔG° is near zero; both H₂/I₂ and HI are thermodynamically accessible; equilibrium has significant amounts of all three species. Products slightly favoured (Keq slightly > 1).
PCl₅, ΔG° = +21: Reversible — positive ΔG° means reactants (PCl₅) are slightly favoured under standard conditions; equilibrium lies on the reactants side (Keq < 1). But both directions are accessible.
2SO₂ + O₂, ΔG° = −141: Moderately reversible (borderline, but reversible equilibrium) — ΔG° is substantially negative; SO₃ is strongly favoured but some SO₂ and O₂ remain at equilibrium. Written with ⇌ in industry.
Student A error: Says "releases heat" — but the pack gets cold, so the dissolution absorbs heat (ΔH = +25.7 kJ/mol, endothermic). The student confused the sign of ΔH. Correction: The dissolution is spontaneous not because it releases heat but because ΔS is large and positive (ions dispersing into solution increases disorder significantly) — TΔS exceeds ΔH, making ΔG < 0.
Student B error: Claims combustion of petrol is reversible because "all reactions can theoretically go in both directions." While thermodynamically true at an extreme level, combustion of petrol has ΔG° << −1000 kJ/mol — the reverse reaction is so thermodynamically unfavourable it is effectively impossible under ordinary conditions. Correction: Combustion is irreversible because the magnitude of ΔG is so large that the reverse reaction has an enormous positive ΔG and essentially cannot occur.
Student C error: Confuses reversible (⇌) with equal concentrations. The ⇌ symbol indicates a reaction that reaches dynamic equilibrium — not that concentrations are equal. Correction: For N₂ + 3H₂ ⇌ 2NH₃, the equilibrium position favours NH₃ (Keq > 1 at 25°C), so [NH₃] > [N₂] and [H₂] at equilibrium. Equal concentrations would only occur if Keq ≈ 1.
1. B — Irreversibility is a thermodynamic property — the very large negative ΔG° for Mg combustion means MgO is so much more stable that the reverse reaction would require enormous positive ΔG input. Option A confuses thermodynamics with kinetics.
2. B — ΔG = ΔH − TΔS. When ΔH is positive but ΔS is large and positive, TΔS can exceed ΔH, making ΔG negative. The dispersion of NH₄⁺ and NO₃⁻ ions into solution represents a very large entropy increase — the thermodynamic driving force.
3. C — A small ΔG° means the free energy minimum lies at an intermediate composition. Both N₂/H₂ and NH₃ are present at equilibrium. Option A is wrong — catalysts don't determine reversibility. Option B is wrong — the reaction is exothermic (ΔH = −92 kJ/mol).
4. D — At dynamic equilibrium, the system has reached its minimum Gibbs free energy — the driving force for any further change in either direction is zero (ΔG = 0). ΔG° is the standard free energy change and is generally non-zero; ΔG (the actual driving force) is zero only at equilibrium.
5. A — ΔG = ΔH − TΔS. For CaCO₃ decomposition, producing CO₂(g) from a solid creates a large positive ΔS. As T increases, TΔS increases. At ~840°C, TΔS = ΔH → ΔG = 0. Above this temperature, TΔS > ΔH → ΔG < 0 → spontaneous. Endothermic reactions can be spontaneous at high enough temperatures when ΔS is positive.
Q6 (3 marks): At equilibrium, the system has reached its minimum Gibbs free energy — the free energy can no longer decrease in either the forward or reverse direction [1]. ΔG = 0 at equilibrium because the driving force for the forward reaction exactly equals the driving force for the reverse reaction — neither direction is thermodynamically favoured [1]. Before equilibrium, ΔG < 0 for the forward direction, providing the thermodynamic "push" for the net forward reaction; at equilibrium, this push is exhausted — no net conversion occurs in either direction [1].
Q7 (4 marks): (a) Reaction 1, ΔG° = −543 kJ/mol: Irreversible — the very large negative ΔG° means HF is overwhelmingly more thermodynamically stable than H₂ and F₂; the reverse reaction (ΔG° = +543 kJ/mol) is essentially impossible; reaction goes to completion [1]. Reaction 2, ΔG° = −4.7 kJ/mol: Reversible — ΔG° is near zero; both 2NO₂ and N₂O₄ are thermodynamically accessible; equilibrium is established with both present [1]. (b) Macroscopically: the concentrations of NO₂ and N₂O₄ remain constant — the flask appears static with a stable colour [1]. Microscopically: both forward (2NO₂ → N₂O₄) and reverse (N₂O₄ → 2NO₂) reactions continue simultaneously at equal, non-zero rates — molecules are constantly converting between the two species [1].
Q8 (5 marks): (a) ΔG° = −33 kJ/mol is only moderately negative — neither the forward nor the reverse reaction is overwhelmingly favoured. Both N₂/H₂ and NH₃ are thermodynamically accessible [1]. The free energy minimum is at an intermediate composition with measurable amounts of all three gases present at equilibrium — the reaction does not need to proceed to completion to reach this minimum [1]. (b) For N₂ + 3H₂ → 2NH₃: 4 mol gas → 2 mol gas — a decrease in gas moles means a decrease in disorder → ΔS < 0 (negative) [1]. ΔG = ΔH − TΔS = (−92) − T(negative) = −92 + T|ΔS| [1]. As T increases, the positive T|ΔS| term grows, making ΔG less negative (smaller driving force for the forward reaction) and eventually positive (forward reaction non-spontaneous) — higher temperature actually reduces the yield of NH₃. This is the industrial compromise: temperature must be high enough for acceptable rate but not so high that yield collapses [1].
Answer questions on Reversibility, Non-Equilibrium Systems & Entropy before your opponents cross the line. Fast answers = faster car. Pool: lessons 1–2.
Tick when you've finished all activities and checked your answers.