U7.14 · Free Energy of DissolutionPractice · 溶解的自由能

ΔG = ΔH − TΔS. Ions leaving an ordered crystal and spreading through the solution increase the entropy, and at room temperature TΔS exceeds the ΔH absorbed, so ΔG < 0 and dissolving is thermodynamically favorable. The cooling is the result of the endothermic dissolving. Physical changes follow thermodynamics too.

ΔG° = −RT ln Ksp = −8.314 × 298 × ln(5.0 × 10⁻¹³) = −2478 × (−28.32) = 70175 J/mol ≈ +70.2 kJ/mol. Ksp is far below 1, so ΔG° is positive and dissolving is unfavorable.

ΔG° = −RT ln K. For K < 1, ln K < 0 and ΔG° > 0: under standard conditions (each ion at 1 M) dissolving is unfavorable. K > 1 corresponds to ΔG° < 0, and K = 1 to ΔG° = 0.

Ions leaving an ordered lattice for the solution gain many more ways to move, raising the entropy; but each ion attracts and orders the water molecules around it, lowering the entropy. For small, highly charged ions such as Mg²⁺ and Al³⁺, this ordering is so strong that ΔS for dissolving can even be negative.

ln Ksp = −ΔG° ÷ RT = −45000 ÷ (8.314 × 298) = −18.16, so Ksp = e^−18.16 = 1.3 × 10⁻⁸. Convert ΔG° to J/mol so its units match R.