Gibbs Free Energy

Gibbs free energy brings together everything we have covered in this course — enthalpy and entropy — into a single quantity that tells us whether a reaction is thermodynamically feasible. This is one of the most important concepts in chemistry, and a frequent topic in A-Level exam questions. This lesson covers the Gibbs equation, the four cases of feasibility, finding transition temperatures, and worked calculations across a range of contexts.

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The Gibbs Free Energy Equation

ΔG = ΔH − TΔS

where:

• ΔG = Gibbs free energy change (kJ mol⁻¹)
• ΔH = enthalpy change (kJ mol⁻¹)
• T = temperature in kelvin (K)
• ΔS = entropy change (convert to kJ K⁻¹ mol⁻¹ to match ΔH units)

A reaction is thermodynamically feasible (spontaneous) when ΔG < 0 (negative).

A reaction is not feasible when ΔG > 0 (positive).

When ΔG = 0, the system is at equilibrium.

Critical reminder: ΔS is usually given in J K⁻¹ mol⁻¹. You MUST divide by 1000 to convert to kJ K⁻¹ mol⁻¹ before substituting into the equation. Forgetting this conversion is one of the most common errors in exam questions.

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The Four Combinations of ΔH and ΔS

The signs of ΔH and ΔS together determine whether a reaction is feasible and how temperature affects feasibility.

Case	ΔH	ΔS	ΔG = ΔH − TΔS	Feasibility
1	Negative	Positive	Always negative	Feasible at all temperatures
2	Positive	Negative	Always positive	Never feasible
3	Negative	Negative	Depends on T	Feasible at low temperatures
4	Positive	Positive	Depends on T	Feasible at high temperatures

Case 1: ΔH negative, ΔS positive

ΔG = (negative) − T(positive) = negative − positive = always negative

The reaction is feasible at all temperatures. Both the enthalpy and entropy changes favour the reaction.

Example: 2H₂O₂(l) → 2H₂O(l) + O₂(g), ΔH = −196 kJ mol⁻¹, ΔS = +125.6 J K⁻¹ mol⁻¹

Case 2: ΔH positive, ΔS negative

ΔG = (positive) − T(negative) = positive + positive = always positive

The reaction is never feasible at any temperature. Both terms oppose the reaction.

Example: 3O₂(g) → 2O₃(g). This reaction has positive ΔH and negative ΔS.

Case 3: ΔH negative, ΔS negative

ΔG = (negative) − T(negative) = negative + positive

At low temperatures, the negative ΔH dominates and ΔG is negative → feasible. At high temperatures, the TΔS term becomes large enough to make ΔG positive → not feasible.

Example: N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92 kJ mol⁻¹, ΔS = −198.8 J K⁻¹ mol⁻¹

Case 4: ΔH positive, ΔS positive

ΔG = (positive) − T(positive) = positive − positive

At low temperatures, the positive ΔH dominates and ΔG is positive → not feasible. At high temperatures, the TΔS term exceeds ΔH, making ΔG negative → feasible.

Example: CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹, ΔS = +160.4 J K⁻¹ mol⁻¹

graph LR
    subgraph "Case 3: ΔH < 0, ΔS < 0"
        A1["Low T: feasible"] --> A2["High T: not feasible"]
    end
    subgraph "Case 4: ΔH > 0, ΔS > 0"
        B1["Low T: not feasible"] --> B2["High T: feasible"]
    end

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Finding the Transition Temperature

For Cases 3 and 4, there is a specific temperature at which the reaction switches between feasible and not feasible. This is found by setting ΔG = 0:

ΔG = 0 when ΔH = TΔS

Therefore: T = ΔH / ΔS

(Use consistent units: ΔH in kJ mol⁻¹ and ΔS in kJ K⁻¹ mol⁻¹, or both in J.)

Worked Example 1: Thermal Decomposition of CaCO₃

CaCO₃(s) → CaO(s) + CO₂(g)

ΔH = +178 kJ mol⁻¹, ΔS = +160.4 J K⁻¹ mol⁻¹ = +0.1604 kJ K⁻¹ mol⁻¹

T = ΔH / ΔS = 178 / 0.1604 = 1110 K (approximately 837 °C)

Below this temperature, ΔG > 0 and the reaction is not feasible. Above this temperature, ΔG < 0 and the reaction is feasible.

In practice, limestone (CaCO₃) is heated to about 900–1000 °C in a lime kiln to produce quicklime (CaO), consistent with our calculation.

Worked Example 2: Synthesis of Ammonia

N₂(g) + 3H₂(g) → 2NH₃(g)

ΔH = −92 kJ mol⁻¹, ΔS = −198.8 J K⁻¹ mol⁻¹ = −0.1988 kJ K⁻¹ mol⁻¹

T = ΔH / ΔS = −92 / −0.1988 = 463 K (approximately 190 °C)

Below 463 K, ΔG < 0 — the reaction is feasible. Above 463 K, ΔG > 0 — the reaction is not feasible.

Yet the Haber process operates at 400–500 °C. Why? Because thermodynamic feasibility (ΔG < 0) does not guarantee a useful rate. At low temperatures the rate is too slow. The high temperature is used with a catalyst to achieve a reasonable rate, accepting a lower equilibrium yield.

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Worked ΔG Calculations

Worked Example 3: Is the Combustion of Ethanol Feasible at 298 K?

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

ΔH = −1367 kJ mol⁻¹, ΔS = −138.8 J K⁻¹ mol⁻¹ = −0.1388 kJ K⁻¹ mol⁻¹

ΔG = ΔH − TΔS = (−1367) − (298)(−0.1388) = −1367 + 41.4 = −1325.6 kJ mol⁻¹

ΔG is very negative — the reaction is strongly feasible. The small positive TΔS term barely affects the large exothermic enthalpy change.

Worked Example 4: Is the Dissolution of NH₄NO₃ Feasible at 298 K?

NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq)

ΔH = +25.7 kJ mol⁻¹, ΔS = +108.7 J K⁻¹ mol⁻¹ = +0.1087 kJ K⁻¹ mol⁻¹

ΔG = 25.7 − (298)(0.1087) = 25.7 − 32.4 = −6.7 kJ mol⁻¹

ΔG is negative despite ΔH being positive. The entropy increase drives the reaction at room temperature. This is why ammonium nitrate feels cold when it dissolves — the process is endothermic but thermodynamically feasible.

Worked Example 5: At What Temperature Does Fe₂O₃ Reduction Become Feasible?

Fe₂O₃(s) + 3C(s) → 2Fe(s) + 3CO(g)

ΔH = +490 kJ mol⁻¹, ΔS = +543 J K⁻¹ mol⁻¹ = +0.543 kJ K⁻¹ mol⁻¹

T = ΔH / ΔS = 490 / 0.543 = 902 K (approximately 629 °C)

Above about 900 K, the entropy term dominates and the reduction of iron oxide by carbon becomes feasible. In a blast furnace, temperatures exceed 1500 °C, well above this threshold.

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Limitations of Gibbs Free Energy

ΔG tells us whether a reaction can happen (thermodynamic feasibility), not whether it will happen quickly (kinetics):

• ΔG < 0 means the reaction is energetically downhill. But if the activation energy is very high, the rate may be negligibly slow. Diamond → graphite has ΔG < 0 at room temperature, but the rate is essentially zero without extreme conditions.
• ΔG > 0 means the reaction cannot occur spontaneously under those conditions, regardless of kinetics.
• The calculation assumes standard conditions (or the specified temperature). Changing concentrations or pressures affects the actual ΔG.

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Real-World Applications

• Industrial lime production: CaCO₃ → CaO + CO₂ requires temperatures above ~1110 K. Kilns operate at 900–1100 °C.
• Iron smelting: The reduction of iron ore in a blast furnace exploits the fact that ΔG becomes negative for carbon reduction above ~900 K.
• Biological processes: Many biochemical reactions in cells have ΔG > 0 on their own. They are coupled with the hydrolysis of ATP (ΔG = −30.5 kJ mol⁻¹) to make the overall ΔG negative, allowing unfavourable reactions to proceed.
• Fuel cells: The reaction 2H₂(g) + O₂(g) → 2H₂O(l) has ΔG = −474 kJ mol⁻¹, and fuel cells convert this free energy directly into electrical energy more efficiently than combustion engines.

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Common Mistakes

Mistake	Correction
Not converting ΔS from J to kJ	Divide ΔS by 1000 before using in ΔG = ΔH − TΔS
Using °C instead of K for temperature	Always convert: T(K) = T(°C) + 273
Saying ΔG < 0 means the reaction is fast	ΔG tells us about feasibility, not rate
Confusing the transition temperature with the actual operating temperature	T = ΔH/ΔS gives the boundary; industrial processes may operate well above or below
Forgetting that ΔG = 0 at equilibrium, not at the transition temperature only	At the transition temperature, the system is at the boundary between feasible and not feasible

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Summary

• ΔG = ΔH − TΔS determines feasibility: ΔG < 0 = feasible, ΔG > 0 = not feasible.
• Four cases based on the signs of ΔH and ΔS determine temperature dependence.
• Transition temperature: T = ΔH / ΔS (when ΔG = 0).
• Always convert ΔS to kJ K⁻¹ mol⁻¹ and temperature to kelvin.
• Feasibility ≠ rate: a thermodynamically feasible reaction may still be kinetically slow.

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A-Level Deep Dive: Gibbs Free Energy

Spec mapping

Edexcel 9CH0 specification Topic 13 introduces Gibbs free energy via ΔG° = ΔH° − TΔS° as the predictor of thermodynamic spontaneity at constant temperature and pressure (refer to the official specification document for exact wording). Candidates must compute ΔG°, predict whether a reaction is spontaneous (ΔG° < 0), and determine the temperature at which ΔG° = 0 — the "spontaneity threshold" above which an entropy-driven endothermic reaction becomes spontaneous, or below which an entropy-disfavoured exothermic reaction remains spontaneous. Gibbs questions appear on Paper 1 as multi-step calculations and on Paper 3 as synoptic items connecting thermodynamics to equilibrium constants (ΔG° = −RT ln K). Unit handling (J vs kJ; J K⁻¹ vs kJ K⁻¹) is critical and frequently tested.

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Worked example with full mark scheme

Question (8 marks):

For the thermal decomposition of magnesium carbonate: MgCO₃(s) → MgO(s) + CO₂(g).

ΔH° = +117 kJ mol⁻¹; ΔS° = +175 J K⁻¹ mol⁻¹.

(a) Calculate ΔG° at 298 K. (2) (b) Determine the temperature above which the reaction becomes thermodynamically spontaneous. (3) (c) Compare your answer to (b) with the experimental decomposition temperature of MgCO₃ (~350 °C ≈ 623 K) and account for any discrepancy. (3)

Solution with mark scheme:

(a) Convert ΔS° to kJ: ΔS° = 0.175 kJ K⁻¹ mol⁻¹.

ΔG° = ΔH° − TΔS° = 117 − 298 × 0.175 = 117 − 52.15 = +64.9 kJ mol⁻¹.

M1 — correct unit conversion or matching units. A1 — correct value with sign.

(b) Threshold: ΔG° = 0 ⟹ T = ΔH°/ΔS°.

T = 117 / 0.175 = 669 K (≈ 396 °C).

M1 — recognise ΔG° = 0 condition. M1 — substitute and divide. A1 — answer with units.

(c) B1 — calculated threshold (669 K) is higher than the observed decomposition temperature (623 K), a difference of ~46 K.

B1 — possible reasons: (i) the reaction does not need to be in equilibrium for decomposition to proceed if CO₂ is continuously removed (Le Chatelier); (ii) ΔH° and ΔS° have temperature dependence (assumed constant in the calculation); (iii) experimental "decomposition temperature" is the temperature of visible decomposition, not strict ΔG° = 0.

B1 — the small (~7%) discrepancy validates the thermodynamic prediction; the framework is robust within typical experimental uncertainties.

Total: 8 marks (M3 A2 B3).

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Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (6 marks):

Hydrogen iodide synthesis: H₂(g) + I₂(g) ⇌ 2HI(g). ΔH° = −10 kJ mol⁻¹; ΔS° = +22 J K⁻¹ mol⁻¹.

(a) Show that ΔG° at 298 K is negative and calculate its value. (2) (b) Use ΔG° = −RT ln K to estimate the equilibrium constant K at 298 K. (R = 8.31 J K⁻¹ mol⁻¹) (3) (c) Predict (with reasoning) whether K increases or decreases at 600 K. (1)