Explain the Lattice Enthalpy.

Lattice Enthalpy: The lattice enthalpy of an ionic compound is the enthalpy change that occurs when one mole of an ionic compound dissociates into its ions in the gaseous state.
$Na^{+}Cl^{-}_{(s)} \rightarrow Na^{+}_{(g)} + Cl^{-}_{(g)}; \Delta_{lattice} H = +788 \ kJ / mol$
Different steps of the formation of $NaCl$ and its related enthalpy can be explained by the Born-Haber cycle as follows:
$(1)$ $Na_{(s)} \rightarrow Na_{(g)}$; sublimation of sodium,$\Delta_{sub} H = 108.4 \ kJ \ mol^{-1}$
$(2)$ Ionization enthalpy: $Na_{(g)} \rightarrow Na^{+}_{(g)} + e^{-}_{(g)}; \Delta_{i} H = 496 \ kJ / mol$
$(3)$ Dissociation of chlorine: $\frac{1}{2} Cl_{2(g)} \rightarrow Cl_{(g)}; \frac{1}{2} \Delta_{bond} H = 121 \ kJ / mol$
$(4)$ Electron gain enthalpy: $Cl_{(g)} + e^{-} \rightarrow Cl^{-}_{(g)}; \Delta_{eg} H = -348.6 \ kJ / mol$
$(5)$ Lattice formation: $Na^{+}_{(g)} + Cl^{-}_{(g)} \rightarrow Na^{+}Cl^{-}_{(s)}; \Delta_{U} H = ?$
$(6)$ Enthalpy of formation of $NaCl$: $Na_{(s)} + \frac{1}{2} Cl_{2(g)} \rightarrow NaCl_{(s)}; \Delta_{f} H = -411.2 \ kJ / mol$
Applying Hess's law,we get:
$\Delta_{lattice} H = \Delta_{f} H - (\Delta_{sub} H + \frac{1}{2} \Delta_{bond} H + \Delta_{i} H + \Delta_{eg} H)$
$= -(-411.2) - (108.4 + 121 + 496 - 348.6) = +788 \ kJ / mol$
For $NaCl_{(s)} \rightarrow Na^{+}_{(g)} + Cl^{-}_{(g)}$,the internal energy is smaller by $2RT$ and is equal to $+783 \ kJ / mol$.
We use the lattice enthalpy to calculate the enthalpy of solution using the expression:
$\Delta_{sol} H = \Delta_{lattice} H + \Delta_{hyd} H$
For one mole of $NaCl_{(s)}$,lattice enthalpy $= +788 \ kJ / mol$ and $\Delta_{hyd} H = -784 \ kJ / mol$.
$\therefore \Delta_{sol} H = +788 - 784 = +4 \ kJ / mol$