9.6 Dissolution ΔG°

Free Energy of Dissolution

Pick a salt and slide T to break ΔG°soln into three factors — lattice cost, solvent cavity, and ion–dipole hydration. Watch the big ΔH terms nearly cancel to a small net ΔG°.

3-Factor ΔG°7 SaltsΔH − TΔSCancellation
Topic 9.6

Free Energy of Dissolution

Explain the relationship between the solubility of a salt and changes in the enthalpy and entropy that occur in the dissolution process.

EK 9.6.A.1 breaks dissolution into three contributions, and asks you to consider both the enthalpic and entropic side of each:

  1. Breaking the intermolecular interactions that hold the solid together. Endothermic (ΔH > 0), and entropy increases as the ordered lattice disperses (ΔS > 0).
  2. Reorganization of the solvent around the dissolved species. Endothermic to open a cavity, and the ordering of solvent molecules into hydration shells decreases entropy (ΔS < 0).
  3. Interaction of the dissolved species with the solvent. Exothermic (ΔH < 0), as ion–dipole or hydrogen-bonding interactions form.

The EK then makes an unusual admission that is itself examinable: predicting the total free energy of dissolution is challenging because of cancellations among these three factors. You are expected to be able to estimate the sign and relative magnitude of each contribution, and to recognize that the overall result is a near-cancellation.

That is why solubility is so hard to predict from first principles, and why dissolution can be exothermic (CaCl₂, used in hand warmers) or endothermic (NH₄NO₃, used in cold packs) with no simple rule.

Why endothermic dissolution happens at all. ΔG°soln = ΔH°soln − TΔS°soln. Even with ΔH°soln > 0, a sufficiently positive ΔS°soln — from the lattice breaking apart into freely moving ions — can make ΔG° negative. Dissolution of NH₄NO₃ is entropy-driven.

Temperature dependence follows. For an endothermic dissolution, raising the temperature makes the TΔS° term larger and increases solubility — which is why most salts dissolve better in hot water. For an exothermic dissolution, solubility decreases with temperature (Le Châtelier, 7.9). Gases in water are exothermic, which is why warm soda goes flat.

Connection to Ksp: ΔG°soln = −RT ln Ksp. A very small Ksp corresponds to a large positive ΔG°soln.

Key points

  • Dissolution has three contributions — lattice breaking, solvent reorganization, and solute–solvent interaction — and they nearly cancel.
  • Entropy can drive an endothermic dissolution to be favored.
  • Endothermic dissolution → solubility rises with temperature. Exothermic → solubility falls.
  • ΔG°_soln = −RT ln Ksp connects this topic directly to Unit 7.

Equations

  • not on the sheetThe general ΔG° relationship applied to dissolution.
  • not on the sheetLinks free energy to the solubility product.

Common mistakes

  • Assuming exothermic means soluble. Many salts dissolve endothermically.
  • Assuming dissolution always increases entropy. Highly charged small ions order water so strongly that ΔS can be negative.
  • Ignoring solvent reorganization. The CED names it as one of the three factors.
  • Predicting solubility from ΔH alone. The whole point of this topic is that you cannot.

Worked example

Dissolving NH₄NO₃ in water is endothermic (ΔH°_soln = +25.7 kJ/mol) yet the salt is highly soluble. (a) Explain using free energy. (b) Predict the effect of raising the temperature on its solubility. (c) Identify the three contributions to ΔH°_soln and give the sign of each.

(a) Why it dissolves. Favorability is determined by ΔG°soln = ΔH°soln − TΔS°soln, not by ΔH° alone. Dissolving NH₄NO₃ breaks a highly ordered ionic lattice into NH₄⁺ and NO₃⁻ ions that move freely throughout the solution, a large increase in the dispersal of matter and therefore a substantially positive ΔS°soln. At 298 K the −TΔS° term is negative and large enough to outweigh the +25.7 kJ/mol enthalpy cost, making ΔG°soln negative. The dissolution is entropy-driven.

(b) Effect of temperature. Solubility increases. Treating energy as a reactant for an endothermic process, raising the temperature shifts the dissolution equilibrium toward the dissolved ions (Le Châtelier). Equivalently, a larger T makes the −TΔS° term more negative, pushing ΔG°soln further below zero and raising Ksp. This is why cold packs work: the endothermic dissolution absorbs energy from the surroundings.

(c) The three contributions.

  • Breaking the ionic lattice: endothermic, ΔH > 0 — energy is required to overcome the Coulombic attractions between NH₄⁺ and NO₃⁻.
  • Reorganizing the solvent to make room for the ions: endothermic, ΔH > 0 — hydrogen bonds between water molecules must be disrupted.
  • Forming ion–dipole interactions between the ions and water: exothermic, ΔH < 0 — new attractive interactions release energy.

For NH₄NO₃ the first two costs slightly exceed the third release, giving a small net positive ΔH°soln. The near-cancellation among these three terms is exactly what EK 9.6.A.1 identifies as making dissolution predictions difficult.

Full notes for topic 9.6 →