Entropy and Gibbs Free Energy

Enthalpy alone is not enough to decide whether a chemical change will actually happen. Some endothermic reactions — ammonium nitrate dissolving in water, the thermal decomposition of limestone above 900 °C — proceed without external intervention, in apparent defiance of any simple "exothermic-is-favourable" rule. The deeper principle is the second law of thermodynamics: the entropy of an isolated system tends to increase, where entropy quantifies the number of microscopic configurations consistent with a given macroscopic state — equivalently, the positional and energetic probability distribution of the system's particles. For chemists working at constant temperature and pressure, the practical statement of the second law is encoded in Gibbs free energy: ΔG = ΔH − TΔS. A reaction is thermodynamically spontaneous (feasible) exactly when ΔG < 0. The four sign combinations of ΔH and ΔS give a simple decision rule: (ΔH−, ΔS+) is always spontaneous; (ΔH+, ΔS−) is never spontaneous; the two mixed cases swap sign at a crossover temperature T = ΔH/ΔS that you can compute from tabulated data. This lesson develops both halves of that equation in quantitative detail.

Spec mapping (AQA 7405): This lesson maps to §3.1.8 (A2 thermodynamics — entropy, free-energy change and feasibility). It builds directly on lesson 0 (enthalpy and calorimetry — the source of ΔH input data), lesson 3 (Born-Haber cycles, which provide ΔH routes for ionic compounds), and lesson 4 (enthalpy of solution and the entropy of dissolution, where ΔS first appears qualitatively); it feeds forward into lesson 6 (free-energy applications — industrial feasibility of extraction processes), and connects synoptically to §3.1.6 / §3.1.10 (equilibria — at equilibrium ΔG = 0, and the equilibrium constant K is related to ΔG° by ΔG° = −RT ln K). Refer to the official AQA specification document for the exact wording of each section.

Assessment objectives: AO1 recall — definition of entropy in microscopic / statistical terms (a measure of the number of accessible microstates, not a vague "disorder"); definition of Gibbs free energy as G = H − TS with ΔG = ΔH − TΔS; units (S and ΔS in J K⁻¹ mol⁻¹, ΔH and ΔG in kJ mol⁻¹); qualitative statement of the second law. AO2 application — compute ΔS_rxn = ΣS(products) − ΣS(reactants) from a data table; compute ΔG at a specified temperature with explicit unit reconciliation (either divide ΔS by 1000 to convert J → kJ, or multiply ΔH by 1000 to convert kJ → J — the most common A-Level mark loss in thermodynamics is omitting this conversion). AO3 analysis — predict feasibility for each of the four ΔH/ΔS sign combinations; compute the crossover temperature T = ΔH/ΔS at which ΔG = 0; rationalise why an endothermic decomposition (e.g. CaCO₃) becomes spontaneous on heating, and why a spontaneous reaction (e.g. diamond → graphite) may proceed at negligible rate.

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Entropy: A Microscopic Quantity

Entropy, symbol S, is a state function: its value depends only on the present state of the system (temperature, pressure, composition, phase), not on the path by which the system arrived there. Its SI units are J K⁻¹ mol⁻¹ — joules per kelvin per mole. Note the joules (not kilojoules) — this is the single most important unit point in the topic.

A precise modern definition is statistical: S = k_B ln W, where W is the number of microstates (microscopic configurations of position and momentum) consistent with the macroscopic state, and k_B = 1.381 × 10⁻²³ J K⁻¹ is the Boltzmann constant. The everyday gloss is "entropy measures disorder", but a sharper way to think is "entropy measures the number of equally probable arrangements" — equivalently, the positional and energetic probability spread of the system's particles. A gas at 298 K has astronomically more accessible microstates than the same substance as a perfect crystal at 0 K, because molecules in a gas can occupy any of an enormous number of position and momentum configurations.

Factors that increase entropy

• Temperature. Higher T means more thermal energy distributed over the same set of energy levels, so more microstates are accessible. Entropy rises monotonically with T (slowly within a phase; sharply at phase transitions).
• State of matter. S(solid) < S(liquid) << S(gas). The gas phase is dramatically higher because translational degrees of freedom unlock once particles are no longer confined to lattice sites or to a condensed liquid volume. For water at 298 K: S°(H₂O, l) = 70 J K⁻¹ mol⁻¹; for steam at 373 K: S°(H₂O, g) = 189 J K⁻¹ mol⁻¹ — a jump of ~119 J K⁻¹ mol⁻¹ at the boiling point.
• Mixing. Combining two ideal gases or miscible liquids increases entropy because each component now has access to a larger volume.
• Dissolution. Dissolving a solid in a solvent typically increases entropy: the lattice is dismantled and the ions or molecules disperse through the solvent. (Caveat: for small, highly charged ions in water — Al³⁺, Mg²⁺ — the strong ordering of water in the hydration shell can produce a small negative ΔS for dissolution. Most A-Level examples sidestep this subtlety.)
• More moles of gas. Reactions that produce more moles of gas than they consume have positive ΔS; the converse have negative ΔS. This is the single best predictor at A-Level.
• Molecular complexity. Larger, more complex molecules have more vibrational and rotational modes and hence higher molar entropy than smaller molecules. Compare S°(Ar, g) = 155, S°(CH₄, g) = 186, S°(C₂H₆, g) = 230 J K⁻¹ mol⁻¹.

Standard molar entropies at 298 K

Substance	State	S° (J K⁻¹ mol⁻¹)
C (diamond)	s	2.4
C (graphite)	s	5.7
Na	s	51.0
NaCl	s	72.0
CaCO₃	s	92.9
CaO	s	39.7
H₂O	l	70.0
H₂O	g	189.0
Cl₂	g	223.0
O₂	g	205.0
N₂	g	191.6
H₂	g	130.6
NH₃	g	192.3
CO₂	g	214.0
CH₄	g	186.2
C₂H₅OH	l	160.7

Note that diamond — with its rigid, tetrahedrally-connected covalent lattice — has one of the lowest entropies of any known substance at room temperature. Graphite, with its weakly-bonded layers, has higher entropy. Both pale beside any gas.

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Computing the Entropy Change of a Reaction

For a reaction at constant temperature, the standard entropy change is the difference of standard molar entropies of products and reactants, weighted by stoichiometry:

ΔS°_rxn = Σ νᵢ S°(products) − Σ νᵢ S°(reactants)

where νᵢ is the stoichiometric coefficient of species i in the balanced equation. The data table above (or one provided in the question) supplies the molar entropies; substitute and subtract.

Worked Example 1: Combustion of methane

Reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

ΔS° = [S°(CO₂, g) + 2 S°(H₂O, l)] − [S°(CH₄, g) + 2 S°(O₂, g)] ΔS° = [214 + 2(70)] − [186 + 2(205)] ΔS° = [214 + 140] − [186 + 410] ΔS° = 354 − 596 = −242 J K⁻¹ mol⁻¹

The entropy decreases. The cue: three moles of gas on the left (1 CH₄ + 2 O₂), one mole of gas on the right (CO₂), plus two moles of liquid water — net loss of gas-phase species, so ΔS is negative. The combustion is feasible at 298 K only because the very large negative ΔH (about −890 kJ mol⁻¹) dominates the −TΔS penalty.

Worked Example 2: Thermal decomposition of calcium carbonate

Reaction: CaCO₃(s) → CaO(s) + CO₂(g)

ΔS° = [S°(CaO, s) + S°(CO₂, g)] − [S°(CaCO₃, s)] ΔS° = [39.7 + 214] − [92.9] ΔS° = 253.7 − 92.9 = +161 J K⁻¹ mol⁻¹

A large positive entropy change — a solid decomposes to a solid plus a mole of gas. This is the textbook example of an endothermic reaction (ΔH ≈ +178 kJ mol⁻¹) driven entropically at high temperature.

Worked Example 3: Boiling of water

Reaction: H₂O(l) → H₂O(g) at 373 K

ΔS° = S°(H₂O, g) − S°(H₂O, l) = 189 − 70 = +119 J K⁻¹ mol⁻¹

This is the entropy of vaporisation. At the boiling point and standard pressure, ΔG = 0, so ΔH = TΔS — from which the enthalpy of vaporisation is predicted as ΔH = 373 × 0.119 = +44.4 kJ mol⁻¹, in excellent agreement with the calorimetric value of +40.7 kJ mol⁻¹. (The small discrepancy reflects the fact that boiling at 1 atm is not quite at the standard state.) This is a beautifully self-consistent check of the framework.

Worked Example 4: Ammonia synthesis

Reaction: N₂(g) + 3H₂(g) → 2NH₃(g)

ΔS° = [2 × 192.3] − [191.6 + 3 × 130.6] ΔS° = 384.6 − [191.6 + 391.8] ΔS° = 384.6 − 583.4 = −198.8 J K⁻¹ mol⁻¹

Four moles of gas collapse to two — a large entropy penalty, which is exactly why the Haber process is operated under high pressure (to shift the equilibrium back towards ammonia) and at the lowest temperature consistent with a usable rate.

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Gibbs Free Energy and the Criterion for Spontaneity

The Gibbs free energy is defined as G = H − TS, and for a process at constant T and p:

ΔG = ΔH − TΔS

ΔG is the maximum non-expansion (useful) work obtainable from the process, and its sign tells us whether the process is spontaneous:

• ΔG < 0 — the reaction is spontaneous (thermodynamically feasible). The forward direction has a net thermodynamic driving force.
• ΔG > 0 — the reaction is non-spontaneous in the forward direction (the reverse is spontaneous).
• ΔG = 0 — the system is at equilibrium. Forward and reverse rates are equal; macroscopic composition is constant.

Where the unit care matters: in A-Level data tables, ΔH is in kJ mol⁻¹ and ΔS is in J K⁻¹ mol⁻¹. The factor of 1000 must be reconciled. Two equivalent routes:

• Convert ΔS from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ by dividing by 1000; then ΔG comes out in kJ mol⁻¹.
• Convert ΔH from kJ mol⁻¹ to J mol⁻¹ by multiplying by 1000; then ΔG comes out in J mol⁻¹.

CRITICAL: Forgetting this 1000-fold conversion is the single most common A-Level mark loss in thermodynamics. Always write the conversion line explicitly.

The four sign combinations

ΔH	ΔS	ΔG = ΔH − TΔS	Feasibility
− (exothermic)	+ (entropy rises)	Always negative	Always spontaneous (any T) — e.g. combustion of acetylene C₂H₂ + 5/2 O₂ → 2CO₂ + H₂O
− (exothermic)	− (entropy falls)	Negative at low T, positive at high T	Spontaneous at low T — e.g. freezing of water (T < 273 K), ammonia synthesis at moderate T
+ (endothermic)	+ (entropy rises)	Positive at low T, negative at high T	Spontaneous at high T — e.g. CaCO₃ decomposition above ~1100 K, NaCl dissolving (slightly endothermic but ΔS > 0)
+ (endothermic)	− (entropy falls)	Always positive	Never spontaneous (any T) — e.g. 2 CO₂(g) → 2 CO(g) + O₂(g) at ordinary conditions

The two mixed cases — exothermic with entropy decrease, and endothermic with entropy increase — both feature a crossover temperature at which ΔG passes through zero. Below or above that temperature, the sign of ΔG flips.

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Crossover Temperature

Set ΔG = 0 in ΔG = ΔH − TΔS and solve:

T_crossover = ΔH / ΔS

Both ΔH and ΔS must be in consistent units before substituting.

Worked Example 5: CaCO₃ decomposition

For CaCO₃(s) → CaO(s) + CO₂(g): ΔH° = +178 kJ mol⁻¹; ΔS° = +161 J K⁻¹ mol⁻¹.

Convert ΔH to J: ΔH = +178 000 J mol⁻¹. (Or convert ΔS to kJ: ΔS = +0.161 kJ K⁻¹ mol⁻¹. Either route gives the same answer.)

T_crossover = 178 000 / 161 = 1106 K (about 833 °C)

Below 1106 K, ΔG > 0 — limestone is stable, as observed at room temperature. Above 1106 K, ΔG < 0 — limestone decomposes spontaneously. Industrial lime kilns are operated above 900 °C (≈ 1170 K) precisely to push past the crossover, and the resulting calcium oxide (quicklime) is one of the highest-volume products of the inorganic chemical industry.

Worked Example 6: ΔG for ammonia synthesis at 298 K

For N₂(g) + 3H₂(g) → 2NH₃(g): ΔH° = −92 kJ mol⁻¹; ΔS° = −198.8 J K⁻¹ mol⁻¹.

Convert ΔS to kJ: ΔS = −0.1988 kJ K⁻¹ mol⁻¹.

ΔG = ΔH − TΔS = (−92) − (298)(−0.1988) = −92 + 59.2 = −32.8 kJ mol⁻¹

ΔG is negative — the synthesis is thermodynamically feasible at 298 K. Yet in practice the Haber process is run at about 723 K (450 °C). Why heat it, when room temperature looks favourable? Two reasons. First, kinetically: at 298 K the rate is effectively zero — the activation energy for breaking the very strong N≡N triple bond is huge. The catalyst (iron promoted with K₂O / Al₂O₃) lowers Ea, but only becomes effective at elevated temperatures. Second, the crossover temperature for ΔG = 0 is T = ΔH / ΔS = −92 / −0.1988 = 463 K (190 °C); above this temperature the reaction becomes thermodynamically unfavourable. The industrial compromise: run hot enough for a usable rate but apply high pressure (typically 200 atm) to shift the position of equilibrium back towards ammonia — Le Chatelier's principle exploiting the negative Δn_gas of the reaction.

Worked Example 7: Water boiling at 298 K and 373 K

For H₂O(l) → H₂O(g): ΔH° = +44.0 kJ mol⁻¹; ΔS° = +119 J K⁻¹ mol⁻¹ (= +0.119 kJ K⁻¹ mol⁻¹).

At 298 K: ΔG = 44.0 − (298)(0.119) = 44.0 − 35.5 = +8.5 kJ mol⁻¹ — positive, so liquid water is the stable phase. A puddle does not spontaneously vaporise into vapour at standard pressure and room temperature.

At 373 K (boiling point): ΔG = 44.0 − (373)(0.119) = 44.0 − 44.4 ≈ 0 kJ mol⁻¹ — water and steam are in equilibrium, exactly as expected. The crossover temperature T = 44.0 / 0.119 = 370 K is in excellent agreement with the observed boiling point of 373 K.

At 400 K: ΔG = 44.0 − (400)(0.119) = 44.0 − 47.6 = −3.6 kJ mol⁻¹ — vapour is now the stable phase. Above the boiling point, liquid water at standard pressure is metastable and will boil away.

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Spontaneity Is Not Speed

A spontaneous reaction (ΔG < 0) is one that can happen — it has a thermodynamic driving force. Whether it does happen, in any reasonable time, is a separate (kinetic) question governed by the activation energy and the rate constant.

The classic example is the conversion of diamond to graphite: at 298 K and 1 atm, ΔG ≈ −3 kJ mol⁻¹ — graphite is the thermodynamically stable allotrope of carbon. Yet diamonds are essentially eternal: the activation barrier for rearranging the rigid sp³ tetrahedral network into sp² sheets is enormous (around 540 kJ mol⁻¹), and at room temperature the Boltzmann factor exp(−Ea / RT) is vanishingly small. A diamond is thermodynamically unstable but kinetically stable.

The same logic applies to many "spontaneous" reactions: a mixture of hydrogen and oxygen at room temperature has a huge negative ΔG for forming water (−237 kJ mol⁻¹), yet sits indefinitely without reaction unless ignited. Striking a spark or adding a platinum catalyst lowers the activation barrier and the reaction proceeds explosively.

The lesson for A-Level work: always distinguish "ΔG < 0 means the reaction is feasible" from "the reaction will be observed". The former is thermodynamics; the latter requires kinetics — covered in §3.1.5 of the AQA specification and in this course's sibling kinetics lessons. Lesson 6 of this course (free-energy applications) develops the industrial trade-offs in detail.

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