Explain the Born-Haber Cycle.

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.
$Na^{+}Cl_{(s)}^{-} \rightarrow Na_{(g)}^{+} + Cl_{(g)}^{-}; \Delta_{\text{lattice}}H^{\ominus} = +788 \ kJ \ mol^{-1}$
The different steps of the formation of $NaCl$ and their related enthalpies can be explained by the Born-Haber cycle as follows:
$(1)$ Sublimation of sodium: $Na_{(s)} \rightarrow Na_{(g)}; \Delta_{\text{sub}}H^{\ominus} = 108.4 \ kJ \ mol^{-1}$
$(2)$ Ionization enthalpy: $Na_{(g)} \rightarrow Na_{(g)}^{+} + e_{(g)}^{-}; \Delta_{\text{i}}H^{\ominus} = 496 \ kJ \ mol^{-1}$
$(3)$ Dissociation of chlorine: $\frac{1}{2} Cl_{2(g)} \rightarrow Cl_{(g)}; \frac{1}{2} \Delta_{\text{bond}}H^{\ominus} = 121 \ kJ \ mol^{-1}$
$(4)$ Electron gain enthalpy: $Cl_{(g)} + e^{-} \rightarrow Cl_{(g)}^{-}; \Delta_{\text{eg}}H^{\ominus} = -348.6 \ kJ \ mol^{-1}$
$(5)$ Lattice enthalpy: $Na_{(g)}^{+} + Cl_{(g)}^{-} \rightarrow Na^{+}Cl_{(s)}^{-}; \Delta_{U}H^{\ominus} = ?$
$(6)$ Enthalpy of formation of $NaCl$: $Na_{(s)} + \frac{1}{2} Cl_{2(g)} \rightarrow NaCl_{(s)}; \Delta_{\text{f}}H^{\ominus} = -411.2 \ kJ \ mol^{-1}$
Applying Hess's law,we get:
$\Delta_{\text{f}}H^{\ominus} = \Delta_{\text{sub}}H^{\ominus} + \Delta_{\text{i}}H^{\ominus} + \frac{1}{2} \Delta_{\text{bond}}H^{\ominus} + \Delta_{\text{eg}}H^{\ominus} + \Delta_{U}H^{\ominus}$
Rearranging for lattice enthalpy $(\Delta_{\text{lattice}}H^{\ominus} = -\Delta_{U}H^{\ominus})$:
$\Delta_{\text{lattice}}H^{\ominus} = \Delta_{\text{f}}H^{\ominus} - (\Delta_{\text{sub}}H^{\ominus} + \Delta_{\text{i}}H^{\ominus} + \frac{1}{2} \Delta_{\text{bond}}H^{\ominus} + \Delta_{\text{eg}}H^{\ominus})$
$= -411.2 - (108.4 + 496 + 121 - 348.6) = -411.2 - 376.8 = -788 \ kJ \ mol^{-1}$ (for formation from ions).
Thus,the lattice enthalpy for dissociation is $+788 \ kJ \ mol^{-1}$.