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Entropy & Gibbs Free Energy — Spontaneity & Temperature

You’ll be able to

Quantifying ΔS from standard entropies

Beyond the qualitative cues (solid → liquid → gas, dissolving, more moles of gas), entropy changes are computed from tabulated standard molar entropies S° the same way enthalpy is: ΔS°rxn = Σ S°(products) − Σ S°(reactants), weighting each by its coefficient. Unlike ΔH°f, the S° of an element is not zero — every substance has real, positive entropy. When the moles of gas change, that term dominates the sign, so count gas particles first as a quick prediction, then confirm with the numbers.

Entropy change of a reaction
ΔS°rxn = Σ S°(products) − Σ S°(reactants)
S° values are in J·mol⁻¹·K⁻¹ and are always positive (even for elements). More moles of gas on the product side → ΔS > 0.

Gibbs free energy weighs enthalpy against entropy

ΔG = ΔH − T·ΔS folds both drivers into one verdict: ΔG < 0 is spontaneous, ΔG > 0 is nonspontaneous, and ΔG = 0 is equilibrium. The entropy term carries a factor of T, so temperature can amplify or suppress it. The persistent trap is a units mismatch — ΔH comes in kJ, ΔS in J·K⁻¹ — so convert ΔS to kJ·K⁻¹ (divide by 1000) before multiplying by T.

Gibbs free energy
ΔG = ΔH − T·ΔS
T is absolute temperature (kelvin). Convert ΔS from J·K⁻¹ to kJ·K⁻¹ before subtracting. ΔG < 0 spontaneous, ΔG > 0 nonspontaneous, ΔG = 0 at equilibrium.
Worked example

For 2 H₂(g) + O₂(g) → 2 H₂O(g) at 298 K, ΔH° = −483.6 kJ. Using S° = 130.7 (H₂), 205.2 (O₂), 188.8 (H₂O, g) in J·mol⁻¹·K⁻¹, find ΔS° and ΔG°, and state whether it is spontaneous.

  1. 1.ΔS° = Σ S°(products) − Σ S°(reactants) = 2(188.8) − [2(130.7) + 205.2].
  2. 2.Products: 2(188.8) = 377.6. Reactants: 2(130.7) + 205.2 = 261.4 + 205.2 = 466.6.
  3. 3.ΔS° = 377.6 − 466.6 = −89.0 J·K⁻¹ (negative — 3 mol of gas collapse to 2 mol of gas).
  4. 4.Convert and form the entropy term: T·ΔS = 298 × (−0.0890 kJ·K⁻¹) = −26.5 kJ.
  5. 5.ΔG° = ΔH° − T·ΔS° = −483.6 − (−26.5) = −483.6 + 26.5 = −457.1 kJ.
Answer: ΔS° = −89.0 J·K⁻¹ and ΔG° = −457.1 kJ < 0, so the reaction is spontaneous at 298 K — the large negative ΔH overwhelms the unfavorable entropy loss.
Watch out

Note the sign gymnastics in ΔG = ΔH − TΔS when ΔS is negative: subtracting a negative T·ΔS adds a positive quantity, making ΔG less negative. A negative-entropy reaction is only spontaneous when ΔH is negative enough to pay for it.

The four sign cases and the crossover temperature

The signs of ΔH and ΔS sort every reaction into four cases: (−, +) spontaneous at all T; (+, −) spontaneous at no T; (−, −) spontaneous only at low T; (+, +) spontaneous only at high T. In the two conflicting cases the sign of ΔG flips at a crossover temperature found by setting ΔG = 0: T = ΔH / ΔS. Below it and above it, opposite verdicts hold — so identifying the case tells you which side of the crossover is spontaneous.

Crossover (equilibrium) temperature
ΔG = 0 ⟹ T = ΔH / ΔS
Use ΔH in joules to match ΔS in J·K⁻¹ (or ΔH in kJ with ΔS in kJ·K⁻¹). For (+, +) the reaction turns spontaneous above T; for (−, −) it turns nonspontaneous above T.
Worked example

Limestone decomposes by CaCO₃(s) → CaO(s) + CO₂(g), with ΔH° = +178.3 kJ and ΔS° = +160.6 J·K⁻¹. Find the crossover temperature and state the range in which the reaction is spontaneous.

  1. 1.Identify the case: ΔH > 0 and ΔS > 0 is the (+, +) case — spontaneous only at high temperature.
  2. 2.Set ΔG = 0 at the crossover: 0 = ΔH − TΔS, so T = ΔH / ΔS.
  3. 3.Match units by converting ΔH to joules: 178.3 kJ = 178300 J.
  4. 4.Divide: T = 178300 J ÷ 160.6 J·K⁻¹ = 1110 K (about 837 °C).
  5. 5.Because it is the (+, +) case, ΔG < 0 only when T exceeds this value.
Answer: T ≈ 1110 K (≈ 837 °C). The decomposition is spontaneous above ~1110 K and nonspontaneous below it — which is why lime kilns are run red-hot.
Checkpoint

What is the sign of ΔS for 2 NO₂(g) → N₂O₄(g)?

Checkpoint

A reaction has ΔH = +58.0 kJ and ΔS = +176 J·K⁻¹. At what temperature does ΔG = 0 (the crossover temperature)?

Checkpoint

A reaction is endothermic (ΔH > 0) with an increase in entropy (ΔS > 0). Under what conditions is it spontaneous?

On the exam

A reliable free-response routine: compute ΔH°rxn and ΔS°rxn from tables, decide the sign case, and if ΔH and ΔS conflict solve T = ΔH/ΔS for the crossover (keep units consistent — ΔH in J with ΔS in J·K⁻¹). Then answer the temperature question by naming which side of the crossover is spontaneous.

Answer the 3 checkpoints as you read.

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