{c}

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>.<Chemistry 2e by Paul Flowers and others/7.5 Strengths of Ionic and Covalent Bonds>[“The Born-Haber Cycle”, Figure 7.13 and Table 7.4]{ref}

For one <mole> of <sodium chloride> <formula units>, the overall formation reaction is
$$
\mathrm{Na}(s)+\tfrac12\mathrm{Cl}_2(g)\longrightarrow\mathrm{NaCl}(s),\qquad F=\Delta H_{\rm f}^{\circ}.
$$
The <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:
$$
\begin{aligned}
\mathrm{Na}(s)&\longrightarrow\mathrm{Na}(g) && S>0,\\
\tfrac12\mathrm{Cl}_2(g)&\longrightarrow\mathrm{Cl}(g) && B>0,\\
\mathrm{Na}(g)&\longrightarrow\mathrm{Na}^+(g)+e^-(g) && I>0,\\
\mathrm{Cl}(g)+e^-(g)&\longrightarrow\mathrm{Cl}^-(g) && A=\Delta H_{\rm attach}^{\circ}<0,\\
\mathrm{Na}^+(g)+\mathrm{Cl}^-(g)&\longrightarrow\mathrm{NaCl}(s) && L_{\rm form}<0.
\end{aligned}
$$
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 $\mathrm{Cl}_2(g)\to2\mathrm{Cl}(g)$, 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. Thus
$$
F=S+B+I+A+L_{\rm form},\qquad
L_{\rm diss}=-L_{\rm form}=S+B+I+A-F.
$$
The closed traversal returns from the <solid> to the elemental reference states, adding $-F$; therefore $S+B+I+A+L_{\rm form}-F=0$. Notice that the signed attachment term $A$ is negative. Substituting a positive spectroscopic <electron affinity> for $A$ would reverse the attachment contribution.

\Image[/sodium-chloride-born-haber-enthalpy-cycle.png]
{title=A closed <Born-Haber cycle> for <sodium chloride> using explicitly synthetic molar <enthalpies>: levels represent complete material inventories, and horizontal position is bookkeeping rather than a <reaction coordinate>}
{id=image-sodium-chloride-born-haber-enthalpy-cycle}

The diagram deliberately uses invented, internally consistent inputs in $\mathrm{kJ\,mol^{-1}}$: $S=100$, $B=120$, $I=500$, $A=-350$ and $F=-400$. Preparation of separated gaseous <ions> raises the <enthalpy> by $370$, so $L_{\rm form}=-770$ and $L_{\rm diss}=+770$. 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 $769\,\mathrm{kJ\,mol^{-1}}$ 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>.


 Born-Haber cycle (source code)