Born–Haber cycles, hydration and solution

What you’ll learn

• Apply a consistent lattice-energy sign convention.
• Construct enthalpy cycles for ionic solids and solutions.
• Explain charge-density effects and evidence for covalency.

Understand the idea

Pearson defines lattice energy here as the enthalpy change when one mole of an ionic solid forms from its gaseous ions; it is therefore exothermic. A Born–Haber cycle connects this step to atomisation, ionisation and electron-affinity changes using Hess's law. The first electron affinity is often exothermic, while adding an electron to an already negative ion requires energy.

Higher ionic charges and smaller radii generally make lattice formation and hydration more exothermic. Dissolving a lattice first requires separation of its ions and then releases energy when water hydrates them. Compare experimental and ideal ionic lattice energies to assess covalent character: a small highly charged cation can polarise a large anion. Solubility depends on the total entropy balance as well as enthalpy of solution.

Key formulae

$$\mathrm{M^+(g)+X^-(g)\longrightarrow MX(s)}\quad\Delta H_{\mathrm{latt}}<0$$

$$\Delta H_f^\circ=\Delta H_{\mathrm{atomisation}}+\mathrm{IE}+\mathrm{EA}+\Delta H_{\mathrm{latt}}$$

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

$$\Delta H_{\mathrm{hyd}}:\ \mathrm{M^{z+}(g)\longrightarrow M^{z+}(aq)}$$

Worked example

For a model ionic solid MX, $\Delta H_f^\circ=-410\,\mathrm{kJ\,mol^{-1}}$. The total atomisation enthalpy is $+220$, ionisation energy $+500$ and electron affinity $-350$, all in $\mathrm{kJ\,mol^{-1}}$. Find the lattice formation enthalpy.

1. The gaseous-ion route contributes $220+500-350=370\,\mathrm{kJ\,mol^{-1}}$ before lattice formation.
2. Hess's law gives $-410=370+\Delta H_{\mathrm{latt}}$.
3. Subtract $370$ from both sides.

Answer: $\Delta H_{\mathrm{latt}}=-780\,\mathrm{kJ\,mol^{-1}}$.