More freedom → more entropy: S(gas) > S(liquid) > S(solid). Melting and vaporizing raise entropy.

Compare moles of gas on each side: more gas moles = higher entropy.

Dissolving a solid into ions raises entropy; precipitating lowers it.

ΔS° = ΣS°(products) − ΣS°(reactants), each S° × its coefficient. (Standard molar entropies are tabulated and nonzero, even for elements.)

ΔG° = ΣΔG°f(products) − ΣΔG°f(reactants). ΔG°f = 0 for an element in its standard state.

When ΔG°f isn't available, use ΔG° = ΔH° − TΔS° (T in K). Watch units: ΔH° in kJ, ΔS° in J/K, so divide ΔS° by 1000 first.

ΔG° < 0 spontaneous; ΔG° > 0 nonspontaneous; ΔG° = 0 at equilibrium.

ΔG° = −RT ln K (T in K, R = 8.314 J/(mol·K)). K > 1 ↔ ΔG° < 0 (product-favored).

ΔG = ΔG° + RT ln Q. ΔG < 0 → proceeds forward; ΔG > 0 → reverse; ΔG = 0 → at equilibrium.

ΔS_univ = ΔS_sys + ΔS_surr, with ΔS_surr = −ΔH_sys/T (const T, P). Spontaneous ↔ ΔS_univ > 0.

When ΔH° and ΔS° are both positive, the reaction turns spontaneous above T = ΔH°/ΔS° (where ΔG° changes sign).

At a boiling or freezing point the two phases are at equilibrium, so ΔG° = 0 and the transition temperature is T = ΔH°/ΔS° for that phase change.