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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.

An enthalpy of atomization is the enthalpy change for making one mole of separated gaseous chemical atoms from a specified starting chemical substance. For sodium, the relevant step is sublimation to monatomic gas; for diatomic chlorine, it is half the compatible molar bond-dissociation energy expressed as an enthalpy.

Equivalent chemical bonds in an intact molecule need not have equal successive bond-dissociation enthalpies. Once one chemical bond is removed, the next reactant has a different electronic quantum state and may relax to a different molecular geometry. The successive steps compare different species. This distinction also explains why a tabulated average chemical bond strength is not necessarily the cost of the first cleavage.^[ref]

For an invented species $A{B}_2$, stipulate the following gas-phase steps, with all enthalpies at the same temperature and the same standard states:The symbols stand for hypothetical species with consistently chosen electronic states. They are not values for water or another measured molecule. By Hess's law, the total enthalpy of atomization for $A{B}_2(g)\to A(g)+2B(g)$ is $700\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; the arithmetic mean per broken chemical bond is $350\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Neither individual step equals that mean. Nor must the second step be cheaper: here it is deliberately more expensive. An exact sequence sums exactly; substituting averages taken from other molecules generally gives only an estimate.