Entropy & Gibbs Free Energy

You’ll be able to

Predict the sign of ΔS from changes in phase, moles of gas, and disorder
Combine ΔH and ΔS into ΔG = ΔH − TΔS to judge spontaneity
Analyze how the signs of ΔH and ΔS set the temperature range of spontaneity

Entropy measures dispersal

Enthalpy is not the whole story — some endothermic changes happen spontaneously (ice melting, salt dissolving). The missing ingredient is entropy (S), a measure of how spread out a system's energy and matter are. More accessible arrangements means higher entropy. ΔS > 0 means the system became more disordered; ΔS < 0 means it became more ordered.

Reading the sign of ΔS

A few reliable cues: entropy rises going solid → liquid → gas, when a solid dissolves, and when a reaction produces more moles of gas than it consumes. It falls in the reverse cases. When gas moles change, that term usually dominates — count gas particles on each side first.

Gibbs free energy

ΔG = ΔH − T·ΔS

T is absolute temperature in kelvin. Watch units: ΔH is usually kJ while ΔS is usually J·K⁻¹ — convert ΔS to kJ·K⁻¹ (divide by 1000) before subtracting.

Gibbs free energy decides spontaneity

ΔG combines the two drivers — energy release (ΔH) and entropy increase (ΔS) — into one verdict. ΔG < 0: the process is spontaneous (thermodynamically favored). ΔG > 0: nonspontaneous (the reverse is favored). ΔG = 0: the system is at equilibrium. Temperature scales the entropy term, so it can tip the balance.

Worked example

For a reaction at 298 K, ΔH = −92.2 kJ·mol⁻¹ and ΔS = −198.7 J·mol⁻¹·K⁻¹. Calculate ΔG and state whether it is spontaneous.

1.Convert ΔS to kJ so the units match ΔH: −198.7 J·mol⁻¹·K⁻¹ ÷ 1000 = −0.1987 kJ·mol⁻¹·K⁻¹.
2.Compute the entropy term: T·ΔS = 298 K × (−0.1987 kJ·mol⁻¹·K⁻¹) = −59.2 kJ·mol⁻¹.
3.Apply ΔG = ΔH − T·ΔS = −92.2 − (−59.2) = −92.2 + 59.2.
4.ΔG = −33.0 kJ·mol⁻¹.

Answer: ΔG = −33.0 kJ·mol⁻¹ < 0, so the reaction is spontaneous at 298 K

Watch out

The single most common ΔG mistake is a units mismatch: ΔH in kJ, ΔS in J. Convert ΔS to kJ·K⁻¹ (divide by 1000) before you multiply by T, or your answer will be off by a factor of a thousand.

Checkpoint

Which process has a positive ΔS (an increase in entropy)?

Checkpoint

Which combination of signs makes a reaction spontaneous at ALL temperatures?

Checkpoint

A reaction is exothermic (ΔH < 0) with a decrease in entropy (ΔS < 0). Under what conditions is it spontaneous?

Checkpoint

A reaction has ΔH = +40.0 kJ·mol⁻¹ and ΔS = +100.0 J·mol⁻¹·K⁻¹. Above what temperature does it become spontaneous?

On the exam

Memorize the four-case table: (−, +) spontaneous at all T; (+, −) never; (−, −) spontaneous at low T; (+, +) spontaneous at high T. When ΔH and ΔS pull the same direction, temperature is irrelevant; when they conflict, T > ΔH/ΔS marks the switch point.

Answer the 4 checkpoints as you read.