Free energy and spontaneity: study guide

Key ideas

Free energy combines enthalpy and entropy into one test for spontaneity. A process is spontaneous at a given temperature when ΔG is negative, and ΔG° is directly tied to the equilibrium constant.

• ΔG = ΔH − TΔS, with T in kelvin. ΔG < 0 means spontaneous, ΔG > 0 non-spontaneous, and ΔG = 0 equilibrium.
• The four sign combinations: ΔH < 0 and ΔS > 0 is always spontaneous, ΔH > 0 and ΔS < 0 never, and the other two depend on temperature.
• Where the sign changes, T = ΔH ÷ ΔS. Keep ΔH in joules or ΔS in kJ so the units match.
• ΔG° = −RT ln K, so a negative ΔG° means K > 1 and the products are favored at equilibrium.

Key equations

• ΔG = ΔH − TΔS
• ΔG° = −RT ln K
• ΔS_surr = −ΔH ÷ T
• ΔS_total = ΔS_sys + ΔS_surr
• T_switch = ΔH ÷ ΔS

Worked example

A process has ΔH = +40 kJ/mol and ΔS = +100 J/(mol·K). At 298 K, ΔG = 40 000 − 298 × 100 = +10 200 J/mol, so it is not spontaneous. It becomes spontaneous above T = 40 000 ÷ 100 = 400 K. For a reaction with ΔG° = −10.0 kJ/mol at 298 K, ln K = 10 000 ÷ (8.314 × 298) = 4.04, so K ≈ 57.

Common mistakes

• Mixing kJ for ΔH with J for ΔS in the same formula.
• Using °C instead of kelvin for T.
• Reading ΔG° < 0 as meaning the reaction is fast. Free energy says nothing about rate.

Practice

Test yourself with practice questions on this topic: Free energy and spontaneity