Free Energy of Dissolution

Full Notes

What Happens During Dissolution

When a salt dissolves in a solvent – typically water – several energetic changes take place. Combined, these determine whether the process is thermodynamically favorable. Let’s break it down into three key steps:

Breaking Ionic Bonds in the Solid

To dissolve a salt, the strong electrostatic forces holding the ions together in the solid crystal must be overcome. This requires energy, making this step endothermic (+ΔH). The stronger the ionic lattice, the more energy that is needed.

Separating Solvent Molecules

Water molecules are strongly hydrogen bonded to each other. In order to make space for ions to enter the solution, some of these interactions must be disrupted. This also requires energy and is endothermic (+ΔH).

Hydration (Solvation) of Ions

Once the ions are in the solution, they are surrounded by water molecules. The negative and positive ends of the polar water molecules orient around the ions, forming strong ion-dipole interactions. This releases energy, so this step is exothermic (-ΔH°).

The overall enthalpy change of dissolution (ΔH°) is the sum of these three effects. Whether it’s positive or negative depends on which step dominates.

Entropy Considerations (ΔS°)

When a crystalline solid dissolves into free ions the system becomes more disordered. The transition from a highly ordered solid to dispersed ions in solution usually results in an increase in entropy (+ΔS°).

A positive ΔS° helps drive the dissolution process forward because it contributes a negative term (−TΔS°) to the Gibbs free energy equation (see below), helping make ΔG° more negative.

Gibbs Free Energy and Solubility

Whether dissolution actually occurs depends on the Gibbs free energy change:

ΔG° = ΔH° − TΔS°

• If ΔG° < 0, dissolution is thermodynamically favored – the salt is soluble.
• If ΔG° > 0, dissolution is not favored – the salt is insoluble or only dissolves sparingly.

This equation tells us that both enthalpy (ΔH°) and entropy (ΔS°) contribute to solubility. Sometimes, a positive ΔH° (which resists dissolution) can be overcome by a large positive ΔS° (which promotes it), making the overall ΔG° negative. The opposite can also be true.

Why It’s Difficult to Predict Solubility

While it’s possible to estimate each individual energy contribution, predicting solubility from ΔH° and ΔS° can still be difficult. That’s because the energetic terms often cancel out. For example, breaking ionic bonds (endothermic) may be nearly offset by hydration energy (exothermic).

An increase in entropy usually helps make dissolution favorable, but its effect depends on temperature and how ordered the initial solid is compared to the dissolved state.

Examples:

• A salt with a positive ΔH° but a large positive ΔS° may still dissolve because the entropy gain makes ΔG° negative.
• However, a salt with a negative ΔH° (exothermic) but little or no entropy gain might still be insoluble.

Summary

• Dissolution involves breaking solute–solute and solvent–solvent interactions and forming solute–solvent (hydration) interactions.
• These processes contribute to the enthalpy change (ΔH°), while the increase in disorder of particles contributes to entropy change (ΔS°).
• We can use ΔG° = ΔH° − TΔS° to determine if dissolution is favored.
• A negative ΔG° means the salt is soluble; a positive ΔG° means it isn’t.
• Because of offsetting energy terms, predicting solubility purely from ΔH° or ΔS° is challenging, and ΔG° gives the full thermodynamic picture.