Year 12 Chemistry Module 5 ⏱ ~35 min Lesson 2 of 18

Reversibility, Non-Equilibrium Systems and Entropy

In 1909 Fritz Haber recorded an ammonia yield of only about 6% from nitrogen and hydrogen, while a burning candle converts essentially all of its wax to carbon dioxide and water. Both reactions are spontaneous, yet only one stops part-way. The difference is thermodynamic, and it is what this lesson is about.

Today's hook, You cannot un-burn a log, but you can push ammonia back apart into nitrogen and hydrogen. Enthalpy, entropy and the size of the free energy change decide which reactions stop at equilibrium and which run to completion.

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Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

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Think First

In 1909, Fritz Haber's laboratory notebook recorded that N₂ + 3H₂ ⇌ 2NH₃ gave only 6% ammonia yield at 600°C and 200 atm, far less than thermodynamics seemed to allow. Compare this to methane combustion:

(1) Combustion of methane, CH₄ + 2O₂ → CO₂ + 2H₂O, ΔG = −818 kJ/mol
(2) Haber's 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.

Key Relationships, This Lesson

ΔG = ΔH − TΔS   (from Module 4, revisited here)

Spontaneous: ΔG < 0
Non-spontaneous: ΔG > 0
At equilibrium: ΔG = 0 (driving force exhausted)

Large negative ΔG° → products strongly favoured → reaction goes essentially to completion → treated as irreversible
Small negative ΔG° → products only slightly favoured → significant amounts of both present → reversible equilibrium

Note the wording. No reaction is truly irreversible, and ΔG° never becomes infinite. A large negative ΔG° makes the equilibrium position lie so far to the right that the reverse reaction is too slight to detect, so we treat the reaction as going to completion. “Irreversible” is a practical description of where the equilibrium sits, not a separate class of reaction.

Learning Intentions

Know

The thermodynamic spectrum from irreversible to reversible reactions
The connection between the magnitude of ΔG° and the position of equilibrium
Why at equilibrium, ΔG = 0

Understand

Why large negative ΔG° means the reaction goes essentially to completion
Why endothermic reactions can be spontaneous (entropy-driven)
Why combustion is a non-equilibrium system

Can Do

Classify reactions as reversible or effectively irreversible using ΔG and reasoning
Analyse non-equilibrium systems using both ΔH and ΔS components
Explain why photosynthesis requires continuous external energy input

Key Terms, scan these before reading

Reversible reaction
A reaction that proceeds in both forward and reverse directions, reaching dynamic equilibrium.

Irreversible reaction
A reaction that goes effectively to completion with no significant reverse reaction (e.g., combustion).

Entropy (S)
A measure of the disorder or dispersal of energy in a system.

Gibbs free energy (ΔG°)
A thermodynamic quantity combining enthalpy and entropy; ΔG° = ΔH° − TΔS°.

Non-equilibrium system
A reaction mixture where Q ≠ Keq and net reaction is still proceeding.

Spontaneous reaction
A reaction that proceeds without continuous energy input; has ΔG < 0.

Misconceptions to Fix

✗ 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.

Reversible vs irreversible

Core Content

Reversible vs Irreversible, The Thermodynamic View

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.

Interactive, Reversibility Classifier

Use the classifier. Which of these processes is reversible under the same conditions, so it can reach equilibrium?

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.

Must know

The ⇌ symbol in a chemical equation is a signal, it tells you the reaction reaches dynamic equilibrium, not that it always produces a 50/50 mixture. The position of equilibrium (how far right or left) depends on K_eq and temperature, which you will study in IQ3.

Common error

Students assume that because a reaction is written with ⇌, the concentrations of reactants and products must be similar at equilibrium. This is wrong, a reaction can be reversible (written ⇌) and still have K_eq = 10⁶, meaning the equilibrium position lies almost entirely on the products side. Reversibility refers to the existence of a meaningful reverse reaction, not to the ratio of products to reactants.

How the magnitude of ΔG° determines whether a reaction is irreversible or reversible and which side is favoured.

ΔG° magnitude determines reaction type, the spectrum from irreversible to reversible

What to write in your book

Reversibility is a spectrum, large negative ΔG° means irreversible; near-zero ΔG° means reversible equilibrium
The ⇌ symbol signals dynamic equilibrium, NOT a 50/50 mixture
Example: 2Mg + O₂ (ΔG° = −1138 kJ/mol) irreversible; N₂ + 3H₂ ⇌ 2NH₃ (ΔG° = −33 kJ/mol) reversible

Which of the following reactions is most likely to be IRREVERSIBLE?

Investigation, Testing Whether a Reaction Can Be Reversed

Syllabus: conduct a practical investigation to demonstrate reversible chemical reactions

We just saw the distinction between reversible and effectively irreversible reactions. That raises a question: how would you show which one you have, rather than being told? This card answers it → you drive one system backwards and forwards on demand, and try, and fail, to do the same to another.

Inquiry question. Can the products of a reaction be converted back into the reactants by changing the conditions, and what distinguishes a system where this is possible from one where it is not?

Safety

Teacher supervision and a written risk assessment are required. Safety glasses and gloves throughout. Iron(III) nitrate and potassium thiocyanate are irritants, thiocyanate must never be mixed with acid, it can release toxic HCN. Burning magnesium emits intense UV, never look directly at the flame, ignite with tongs at arm’s length over a heatproof mat, and never use water on a magnesium fire. Part B is a teacher demonstration in most schools. Dispose of iron/thiocyanate solutions in the labelled residues container.

Equipment. Part A: 0.002 mol L⁻¹ Fe(NO₃)₃, 0.002 mol L⁻¹ KSCN, solid KSCN or concentrated stock, solid Fe(NO₃)₃, 0.1 mol L⁻¹ NaOH, distilled water, 5 test tubes in a rack, dropper pipettes, white tile. Part B: magnesium ribbon (2 cm), tongs, Bunsen burner, heatproof mat, watch glass.

Part A, set up the system. Mix 25 mL of the dilute Fe(NO₃)₃ with 25 mL of the dilute KSCN. Record the colour. Divide the mixture between five test tubes, keeping tube 1 as an untouched control for comparison.

Tube 2: add a few crystals of KSCN. Record the colour change against the control.

Tube 3: add a few crystals of Fe(NO₃)₃. Record the colour change against the control.

Tube 4: add NaOH solution dropwise until a change is seen. This removes Fe³⁺ as insoluble Fe(OH)₃. Record the colour change and the precipitate.

The reversal test. To tube 4, add Fe(NO₃)₃ solution dropwise. Record whether the original colour returns. This is the step that decides the question, going one way is not evidence of reversibility, coming back is.

Part B, the comparison. Hold a 2 cm strip of magnesium ribbon in tongs and ignite it over a heatproof mat, looking away from the flame. Collect the white MgO ash on a watch glass.

Attempt to reverse it. Add water to the ash, then warm it. Record every attempt made to recover magnesium metal, and the outcome of each.

Results, record the colour or appearance before and after each change
Tube / part	Change made	Direction of shift	Reversible?
1	control, none	—	—
2	added KSCN
3	added Fe(NO₃)₃
4	added NaOH
5	Fe³⁺ restored to tube 4
B	burned Mg, then water/heat

Analysis.

Write the equation for the equilibrium in Part A and identify which species is responsible for the deep red colour.
For tubes 2, 3 and 4, explain each colour change in terms of the direction the equilibrium shifted, and why.
Tube 5 is the control that matters. What would it have meant if the colour had not returned?
In Part B, name every condition you changed and state why none of them recovered magnesium. Is the reaction reversible in principle, and is it reversible under the conditions available to you? These are different questions.

Conclusion. State which system you were able to drive in both directions and what evidence supports that. Then state what “irreversible” means operationally, on the evidence you collected, rather than as a definition you were given.

Why the control tube matters: colour judgements are unreliable from memory. Every observation is made side by side with tube 1, held against a white tile, or you will report shifts that are not there.

Common Error: Concluding that adding NaOH “made the reaction go backwards” and stopping there. NaOH removes Fe³⁺ from the system. The equilibrium responds to that removal, which is the point, and step 5 is what shows the response was a shift rather than the destruction of the system.

Reversibility is demonstrated by returning a system to its original state, not by observing one change. Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺: add either reactant, colour deepens; remove Fe³⁺ with OH⁻, colour fades; restore Fe³⁺, colour returns. Burning Mg gives MgO that no available condition converts back, so it is treated as irreversible.

Pause, write the highlighted method and its logic into your book.

Free energy and the equilibrium position

Gibbs Free Energy and Equilibrium

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	K_eq Magnitude
Very large negative (e.g. −500 kJ/mol)	Almost entirely products	K_eq >> 1 (e.g. 10⁸⁰)
Moderately negative (e.g. −20 kJ/mol)	Products favoured but reactants present	K_eq > 1 (e.g. 10³)
Near zero (e.g. ±5 kJ/mol)	Significant amounts of both	K_eq ≈ 1
Moderately positive (e.g. +20 kJ/mol)	Reactants favoured but products present	K_eq < 1 (e.g. 10⁻³)
Very large positive (e.g. +500 kJ/mol)	Almost entirely reactants	K_eq << 1

Extension

Extension, beyond the Module 5 dot points. The relationship $\Delta G° = -RT \ln K_{eq}$ is not examinable in this course and is introduced only for interest in L14. What you do need is the qualitative connection: large negative ΔG° → large K_eq → products strongly favoured. You do not need to calculate this in IQ1.

Insight

Equilibrium is a thermodynamic concept, it is the state of minimum free energy, regardless of kinetics. A catalyst can make you reach equilibrium faster, but it cannot change where that minimum is. This is why the iron catalyst in the Haber process does not improve yield, it only improves the rate of reaching equilibrium.

What to write in your book

At dynamic equilibrium: ΔG = 0, free energy minimum reached; no driving force in either direction
Large negative ΔG° → large K_eq → products heavily favoured
Near-zero ΔG° → K_eq ≈ 1 → both reactants and products present at equilibrium

At dynamic equilibrium, ΔG = 0 because the system has reached its minimum free energy and neither the forward nor the reverse reaction is spontaneous.

Deep dive, Extension. The qualitative link between the sign and magnitude of ΔG° and the position of equilibrium is core. Quantitative and standard-state Gibbs treatment, including calculating ΔG° or converting between ΔG and ΔG°, is beyond the Module 5 core and is not assessed here.

Non-equilibrium systems: combustion

Non-Equilibrium Systems, Combustion

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:

Enthalpy factor (ΔH < 0): the products (CO₂ and H₂O) have much lower enthalpy than the reactants, they are thermodynamically far more stable. A large amount of energy must be supplied to reverse the reaction.
Combined Gibbs effect: ΔG = −818 kJ/mol is an enormous negative value. The equilibrium lies so far to the right that the reverse reaction, CO₂ and water recombining to form methane and oxygen, is thermodynamically negligible.

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).

HSC answer tip

In HSC answers analysing non-equilibrium systems, address both ΔH and ΔS components of ΔG. "The combustion of methane is irreversible because the large negative enthalpy change and favourable entropy change combine to give a large negative ΔG, the products are overwhelmingly more thermodynamically stable than the reactants" is a full-mark answer.

Common error

Students often say combustion is irreversible "because it releases heat." Heat release alone (ΔH < 0) is not sufficient to make a reaction irreversible, there are many reversible exothermic reactions. It is the magnitude of ΔG (the combination of ΔH and TΔS) that determines irreversibility. Always invoke ΔG, not just ΔH.

What to write in your book

Combustion: non-equilibrium because ΔG = −818 kJ/mol (very large negative), both ΔH and TΔS terms reinforce each other
Always address BOTH ΔH and ΔS when explaining irreversibility in HSC answers
The reverse of combustion requires ΔG = +818 kJ/mol, essentially impossible spontaneously

Complete: Combustion is irreversible because the very large ___ ΔG° means the products are overwhelmingly more thermodynamically stable than the reactants.

Entropy-driven reactions

Entropy-Driven Reactions, When Endothermic Means Spontaneous

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:

Dissolution of ammonium nitrate: NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq), ΔH = +25.7 kJ/mol (solution gets cold), but ΔS is large and positive (ions dispersing into solution) → ΔG < 0 at room temperature → spontaneous.
Decomposition of CaCO₃ at high temperature: CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ/mol, but TΔS becomes very large at high temperatures (large ΔS due to production of CO₂ gas) → ΔG becomes negative above ~840°C → spontaneous at high temperature.

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.

Must know

For an endothermic reaction to be spontaneous, ΔS must be positive AND the temperature must be high enough that TΔS > ΔH. Always specify both conditions in HSC answers about entropy-driven reactions.

Common error

Students say endothermic reactions cannot be spontaneous. This is wrong, dissolution of ammonium nitrate and melting of ice (above 0°C) are both endothermic and spontaneous. The Gibbs equation ΔG = ΔH − TΔS always governs spontaneity, not ΔH alone.

Insight

Photosynthesis and respiration together form a beautiful non-equilibrium cycle, photosynthesis uses solar energy to create energy-rich glucose; respiration releases that energy by converting glucose back to CO₂ and water. Neither process is at equilibrium because both require open system conditions. Life itself is a sustained non-equilibrium thermodynamic system.

Gibbs free energy spontaneity conditions as a 2x2 matrix of ΔH vs ΔS

ΔG = ΔH − TΔS, four ΔH/ΔS combinations and their temperature-dependent spontaneity

What to write in your book

Endothermic reactions can be spontaneous IF: ΔS > 0 AND TΔS > ΔH
NH₄NO₃ dissolving: ΔH = +25.7 kJ/mol but large +ΔS from ion dispersal → ΔG < 0 → spontaneous
CaCO₃ decomposition: spontaneous above ~840°C because TΔS exceeds ΔH at high T
Photosynthesis: ΔG = +2870 kJ/mol, requires continuous solar energy (open system)

For the dissolution of NH₄NO₃ (ΔH = +25.7 kJ/mol) to be spontaneous at room temperature, which condition must be true?

Cross-lesson links: The ΔG = ΔH − TΔS analysis introduced here connects back to Module 4 and forward to L09 (Keq expressions) and L13 (temperature changes Keq). Fritz Haber's ammonia process introduced in Card 5 is analysed in full in L07 (industrial applications). The entropy-driven reactions in Card 4 underpin the temperature-dependence of Keq covered in L13.

Haber and a reversible reaction that changed history

Fritz Haber & the Reversible Reaction That Changed History

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.

Historical context

Fritz Haber is deeply controversial. His synthesis of ammonia saved billions of lives through fertilisers. He also pioneered the use of chlorine gas as a chemical weapon in World War I, supervising the first large-scale chemical warfare attack at Ypres in 1915, an event his wife Clara Immerwahr, also a chemist, opposed so strongly she took her own life the night Haber returned from Ypres. Chemistry's power to sustain life and end it simultaneously is rarely more starkly illustrated.

Module 5 anchor

The Haber process is the central real-world anchor for IQ1 and IQ2. This lesson introduces it as an example of reversibility and non-equilibrium analysis. It returns in L07 (full industrial analysis) and throughout IQ2 as the primary Le Chatelier's Principle application.

What to write in your book

N₂(g) + 3H₂(g) ⇌ 2NH₃(g): ΔH = −92 kJ/mol, ΔG° ≈ −33 kJ/mol at 25°C
ΔG° only moderately negative → reversible → 15–25% yield per pass industrially
Unconverted gases are recycled, reversibility is the industrial advantage
Haber process feeds ~4 billion people through nitrogen fertilisers

The Haber process achieves high industrial yield per pass because ΔG° = −33 kJ/mol is a very large negative value, meaning almost all N₂ and H₂ convert to NH₃.

✓

Worked examples and activities

✏️ Worked Examples

Worked Example 1, Classifying reactions as reversible or irreversible using ΔG

For each reaction, classify as reversible or irreversible, and justify using ΔG and thermodynamic reasoning.

(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.

Worked Example 2, Analysing a non-equilibrium system using ΔH and ΔS

The combustion of glucose: C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l), ΔH = −2803 kJ/mol.

(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.

Copy Into Your Books, Full Summary

Reversibility and ΔG°

Large negative ΔG° → irreversible; reaction goes to completion
Small negative or near-zero ΔG° → reversible; equilibrium with both present
At equilibrium: ΔG = 0 (free energy minimum reached)
ΔG° and K_eq are related: large negative ΔG° → large K_eq

Non-Equilibrium Systems

Combustion: large negative ΔG from both ΔH and TΔS terms → irreversible
Photosynthesis: ΔG = +2870 kJ/mol → requires solar energy input
Open systems cannot maintain dynamic equilibrium
Addressing BOTH ΔH and ΔS in HSC answers earns full marks

Entropy-Driven Spontaneous Reactions

Spontaneous despite ΔH > 0 if: ΔS > 0 AND TΔS > ΔH
Example: NH₄NO₃ dissolving, ΔH = +25.7 kJ/mol but ΔS is large positive
Example: CaCO₃ decomposition at high T, ΔS large positive due to CO₂ gas
Temperature must be high enough for TΔS to exceed ΔH

Haber Process, Key Values

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
ΔH = −92 kJ/mol (exothermic)
ΔG° ≈ −33 kJ/mol at 25°C (moderately negative → reversible)
Industrial yield per pass: 15–25%, recycling of unreacted gases needed

Activities

Reversible or Irreversible? Using ΔG°

For each reaction, classify as reversible or irreversible using ΔG reasoning. Then explain which side of equilibrium is favoured.

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

Sort the steps+7 XP

Put the conditions required to establish dynamic equilibrium in the correct logical order.

The forward and reverse rates become equal.
The reaction must be reversible (products can reform reactants).
Macroscopic properties (concentrations, pressure, colour) remain constant.
The system must be closed (matter cannot escape).
Dynamic equilibrium is established, both reactions continue at equal rates.

✓

Practice questions

Multiple Choice

+5 XP

A fresh set drawn from this lesson's question bank, with feedback shown immediately. +5 XP per correct · +25 XP all correct

✍️ Short Answer

UnderstandRecall(3 marks) Q6. 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.

ApplyApply(4 marks) Q7. 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)

EvaluateAnalyse(5 marks) Q8. 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)

Show all answers

MC Explanations

The quiz draws a fresh set from this lesson's bank and shuffles the options each time, so these notes are keyed to the idea being tested, not to a fixed option letter.

Magnesium combustion. 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. Saying the products are merely "kinetically stable" confuses thermodynamics with kinetics.

Ammonium nitrate dissolving. 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.

Haber process reversibility. A small ΔG° means the free energy minimum lies at an intermediate composition. Both N₂/H₂ and NH₃ are present at equilibrium. A catalyst does not determine reversibility, and the reaction is exothermic (ΔH = −92 kJ/mol), not endothermic.

ΔG at equilibrium. 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.

Calcium carbonate decomposition. 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.

Short Answer Model Answers

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].

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Extended Response

Analyse the statement: "The difference between combustion of methane and the Haber process is simply that one is exothermic and one is not." Using ΔG, ΔH, ΔS, and the concept of equilibrium, write a comprehensive response explaining why this statement is incorrect and what actually determines whether a reaction is irreversible or reaches dynamic equilibrium. (6 marks)

How did your thinking change?

Go back to your Think First response about Haber's 1909 result. Now that you've studied the relationship between ΔG° and reversibility:
• Was your prediction correct about which reaction goes to completion?
• Why did Haber only get 6% ammonia yield at 600°C, even though ΔG° = −33 kJ/mol is negative? Use the magnitude of ΔG° to explain why reversible reactions stop before completion.
• Can you state why ΔG = 0 at equilibrium and explain what "minimum free energy" means physically?

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