Topic 9.6 Notes – Free Energy of Dissolution

1. What Free Energy of Dissolution Is

When a salt dissolves, we ask one core question: Is the process thermodynamically favorable? That answer comes from Gibbs free energy.

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

• ΔG° < 0 → spontaneous → the substance dissolves (to some extent)
• ΔG° > 0 → nonspontaneous → essentially insoluble
• ΔH° = heat absorbed or released
• ΔS° = change in disorder
• T = temperature in Kelvin

Solubility depends on the combined effects of ΔH° and ΔS°. Large positive and negative contributions often nearly cancel, which is why predicting solubility can feel tricky.

2. The Three Energy Changes in Dissolution

When an ionic solid dissolves in water, three things happen at the same time. Seeing all three together helps you understand where ΔH° and ΔS° come from. The diagram shows $\ce{Na^{+}}$ and $\ce{Cl^{-}}$ ions surrounded by water molecules in hydration shells.

Hydration shells around Na⁺ and Cl⁻ in water

a. Breaking the Lattice

Ions are pulled apart from the crystal.

• ΔH°: positive (energy required)
• ΔS°: positive (ordered solid → dispersed particles)

Stronger lattices mean:

• Smaller ions
• Higher charges like $\ce{Mg^{2+}}$ vs $\ce{Na^{+}}$
• Strong electrostatic attraction

These salts are harder to dissolve.

b. Reorganizing the Solvent

Water molecules must separate to make room for ions.

• ΔH°: slightly positive
• ΔS°: negative (water becomes more ordered)

This step slightly opposes dissolution.

c. Hydration of Ions

Water surrounds ions with ion-dipole forces. Notice in the diagram that the oxygen ends of water point toward $\ce{Na^{+}}$, and the hydrogen ends point toward $\ce{Cl^{-}}$.

• ΔH°: negative (energy released)
• ΔS°: usually negative (structured hydration shells)

Hydration is stronger for:

• Smaller ions
• Higher charges

Notice something important. Small, highly charged ions increase both lattice energy and hydration energy. That competition is why predictions are not obvious.

Overall:

$$\Delta G^\circ_{\text{diss}} = \text{(lattice)} + \text{(solvent reorg)} + \text{(hydration)}$$

3. Enthalpy of Dissolution

For ionic solids, we often simplify enthalpy to:

$$\Delta H^\circ_{\text{diss}} = \Delta H^\circ_{\text{lattice}} + \Delta H^\circ_{\text{hydration}}$$

• Lattice energy → always positive
• Hydration energy → always negative

Two outcomes:

Exothermic Dissolution (ΔH° < 0)

Hydration outweighs lattice energy.
Solution warms.
Solubility often decreases as temperature increases.

Endothermic Dissolution (ΔH° > 0)

Lattice outweighs hydration.
Solution cools.
Solubility often increases as temperature increases.

Enthalpy alone does not decide solubility. Entropy and temperature can flip the outcome.

4. Entropy in Dissolution

Entropy measures overall disorder.

Two competing effects:

• Breaking a crystal → increases disorder
• Organizing water around ions → decreases disorder

Trends:

• Small, highly charged ions → strong hydration shells → more negative ΔS°
• Larger, lower-charge ions → smaller entropy decrease

For many salts, ΔS° is small. That means temperature becomes important because of the $T\Delta S^\circ$ term.

5. How ΔH°, ΔS°, and Temperature Control Solubility

The signs of ΔH° and ΔS° determine temperature behavior.

On tests, you’re often asked to justify why solubility increases or decreases with temperature. Tie it directly to the sign of ΔH° and how that affects ΔG° as T changes.

6. Why Predicting Solubility Is Hard

Here’s the part students underestimate.

• Lattice and hydration energies are both large and opposite in sign.
• Small changes in ion size or charge can flip ΔH°.
• Entropy effects may partially cancel too.
• Temperature multiplies ΔS°, so its impact grows at higher T.

On an FRQ, you usually aren’t calculating ΔG°. You’re explaining which factors increase or decrease it and comparing magnitudes qualitatively. If you clearly connect lattice strength, hydration strength, entropy change, and temperature to ΔG°, you earn the points.