Enthalpies of solution and hydration

Enthalpy change of hydration (ΔHhyd) is the enthalpy change when one mole of gaseous ions is dissolved in water to give aqueous ions:

$\text{Na}^+(g) + aq \rightarrow \text{Na}^+(aq) \qquad \Delta H_{hyd}$

It is exothermic — the δ− oxygen atoms of water are attracted to a cation (and δ+ hydrogens to an anion), releasing energy as the hydration shell forms.

Enthalpy change of solution (ΔHsol) is the enthalpy change when one mole of solute dissolves completely in water (to a solution dilute enough that further water causes no further enthalpy change):

$\text{NaCl}(s) + aq \rightarrow \text{Na}^+(aq) + \text{Cl}^-(aq) \qquad \Delta H_{sol}$

ΔHsol can be exothermic or endothermic, depending on the balance between energy needed to break the lattice and energy released on hydration.

Dissolving an ionic solid can be split into two steps:

1. Break the lattice into gaseous ions — this is the reverse of lattice energy, so it costs −ΔHlatt (a positive amount, since ΔHlatt is negative).
2. Hydrate the gaseous ions — releasing ΣΔHhyd.

Adding the two steps gives:

$\Delta H_{sol} = \sum \Delta H_{hyd} - \Delta H_{latt}$

Worked (NaCl). ΔHlatt = −788; ΔHhyd(Na⁺) = −406; ΔHhyd(Cl⁻) = −378 (kJ mol⁻¹): $\Delta H_{sol} = (-406 - 378) - (-788) = -784 + 788 = +4\,\text{kJ mol}^{-1}$

So dissolving NaCl is very slightly endothermic — yet it dissolves readily (see the next section). You can rearrange the same cycle to find any one term — e.g. a hydration enthalpy from ΔHsol and ΔHlatt.

Hydration enthalpy depends on charge density (≈ charge ÷ radius):

$\Delta H_{hyd} \propto \frac{\text{charge}}{r}$

So ΔHhyd is more exothermic for ions that are smaller and more highly charged — they attract the polar water molecules more strongly. For example:

• $\text{Mg}^{2+}$ is far more exothermic than $\text{Na}^+$ (higher charge, smaller).
• $\text{Na}^+$ is more exothermic than $\text{K}^+$ (smaller).

Why endothermic salts still dissolve. Dissolving usually produces a large positive entropy change (an ordered solid becomes dispersed aqueous ions). Spontaneity depends on $\Delta G = \Delta H - T\Delta S$, not ΔH alone, so a small positive ΔHsol (like NaCl's +4 kJ mol⁻¹) is easily outweighed by the favourable $-T\Delta S$ term, giving ΔG < 0.