Born–Haber cycle

A Born-Haber cycle uses Hess's law to infer a lattice energy from measurable enthalpy differences and isolated-species data. The motivating difficulty is that separating a mole of an ionic crystal into mutually separated gaseous ions is not a convenient direct calorimetry experiment. Instead, compare two accounting routes between the same elemental reference states and the same solid. Because enthalpy is a state function, the signed differences close exactly when all inputs describe compatible states. These routes do not assert a reaction mechanism or a temporal sequence of reaction intermediates.^[ref]

For one mole of sodium chloride formula units, the overall formation reaction isThe stoichiometry requires half a mole of diatomic chlorine molecules, supplying one mole of chlorine chemical atoms. A half coefficient counts an amount of substance; it does not describe half of an individual molecule. Write each auxiliary chemical reaction with its own sign:Here $S$ is the sublimation enthalpy, $B$ the enthalpy of atomization of chlorine per mole of chemical atoms, and $I$ the molar ionization enthalpy represented by the ionization energy input in elementary treatments. If the bond-dissociation energy table gives the compatible molar enthalpy for ${\mathrm{Cl}}_2(g)\to 2\mathrm{Cl}(g)$, use half that number for $B$. Both ions in the last step are gaseous, and no aqueous hydration of ions belongs in that step.

Adding these equations cancels the gaseous chemical atoms, the gaseous ions and the free electron, leaving precisely the overall formation reaction. ThusThe closed traversal returns from the solid to the elemental reference states, adding $-F$; therefore $S+B+I+A+L_{\mathrm{form}}-F=0$. Notice that the signed attachment term $A$ is negative. Substituting a positive spectroscopic electron affinity for $A$ would reverse the attachment contribution.

The diagram deliberately uses invented, internally consistent inputs in $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$: $S=100$, $B=120$, $I=500$, $A=-350$ and $F=-400$. Preparation of separated gaseous ions raises the enthalpy by $370$, so $L_{\mathrm{form}}=-770$ and $L_{\mathrm{diss}}=+770$. The closure check is $100+120+500-350-770+400=0$. These numbers are not measured sodium chloride data and do not replace the source's $769\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ example in lattice energy. Every level retains one sodium nucleus, one chlorine nucleus and zero total electric charge; the ionization level explicitly retains the free electron.