Born-Haber cycles and lattice enthalpyAQA A-Level Chemistry: Revision notes

The strength of ionic bonding in a solid is measured by its lattice enthalpy. It can be defined in two opposite ways, so always say which one you mean.

• Enthalpy of lattice formation (ΔH⊖LE): the enthalpy change when one mole of an ionic solid is formed from its gaseous ions, e.g. Na⁺(g) + Cl⁻(g) → NaCl(s). It is always exothermic.
• Enthalpy of lattice dissociation: the enthalpy change when one mole of an ionic solid is broken up into its gaseous ions, e.g. NaCl(s) → Na⁺(g) + Cl⁻(g). It is always endothermic, and equal in size to lattice formation but opposite in sign.

Lattice enthalpy cannot be measured directly, so it is found indirectly using a Born–Haber cycle.

Each step of a Born–Haber cycle has its own definition. All refer to standard conditions and one mole.

• Enthalpy of formation (ΔHf): one mole of a compound is formed from its elements in their standard states.
• Enthalpy of atomisation (ΔHat): one mole of gaseous atoms is formed from an element in its standard state, e.g. ½Cl₂(g) → Cl(g). For a diatomic gas this is half the bond dissociation enthalpy.
• First ionisation energy: one mole of gaseous atoms loses one mole of electrons to form one mole of gaseous 1+ ions. Always endothermic. Later ionisation energies are larger still.
• First electron affinity: one mole of gaseous atoms gains one mole of electrons to form one mole of gaseous 1− ions, e.g. Cl(g) + e⁻ → Cl⁻(g). Usually exothermic.
• Second electron affinity: one mole of gaseous 1− ions gains one mole of electrons, e.g. O⁻(g) + e⁻ → O²⁻(g). It is endothermic because the electron is added to a negative ion and is repelled.
• Bond dissociation enthalpy: one mole of covalent bonds is broken in gaseous molecules.

A Born–Haber cycle is an application of Hess's law. The route from the elements to the ionic solid in one step (the enthalpy of formation) equals the sum of the steps that go via gaseous atoms and gaseous ions.

For an ionic compound MX built up in this order:

1. atomise the metal: M(s) → M(g)
2. atomise the non-metal: ½X₂(g) → X(g)
3. ionise the metal: M(g) → M⁺(g) + e⁻ (use second ionisation energy as well for a 2+ ion)
4. add electron(s) to X: X(g) + e⁻ → X⁻(g)
5. form the lattice: M⁺(g) + X⁻(g) → MX(s)

So ΔHf = ΔHat(M) + ΔHat(X) + IE + EA + ΔH(lattice formation). For a compound such as MgCl₂ the chlorine terms must be doubled, because the formula contains two chlorine atoms.

Rearrange the Hess's law equation for whichever term is unknown.

Worked example. Find the enthalpy of lattice formation of potassium chloride. ΔHf(KCl) = −437; ΔHat(K) = +89; IE1(K) = +419; ΔHat(Cl) = +122; EA(Cl) = −349 (all kJ mol⁻¹).

−437 = 89 + 419 + 122 − 349 + ΔHlattice −437 = 281 + ΔHlattice ΔHlattice formation = −437 − 281 = −718 kJ mol⁻¹

The same equation can find any other term. For MgO, for example, the unknown could be the second electron affinity of oxygen, found by rearranging once the lattice enthalpy is known. Always check signs, and give the answer a sign and units.

In the perfect ionic model the compound is treated as a lattice of perfectly spherical ions with charge evenly distributed, held together only by electrostatic attraction. A theoretical lattice enthalpy can be calculated from this model using the charges and radii of the ions.

Lattice enthalpy is more exothermic for ions with higher charge and smaller radius, which give a stronger attraction. For example MgO is much more exothermic than NaCl, and BaO is less exothermic than CaO.

Compare the lattice enthalpy from the Born–Haber cycle (the experimental value, which reflects the real bonding) with the perfect ionic model value.

• If the two values are close, the compound is almost purely ionic. NaCl is an example.
• If the Born–Haber value is more exothermic than the theoretical value, there is extra stabilisation from covalent character. Silver iodide is an example.

Covalent character arises because a small, highly charged cation polarises a large anion, distorting its electron cloud towards the cation so that electron density lies between the nuclei. Large anions such as I⁻ are the most easily polarised. The ions are then no longer perfect spheres, so the model fails.