Write the Born-Haber cycle for NaCl.

Explanation

Consider the energy changes involvd in the formation of sodium chloride crystal from metallic sodium and chlorine gas and calculate lattice energy of `NaCl_((s))`. The reaction involves the following steps.
(a) Sublimation of metallic sodium
`Na_((s)) rarr Na_((g)) , Delta_("sub")H^(@) = 108.4 kJ mol^(-1)`
(b) Ionisation of Na atom.
`Na_((s)) rarr Na_((g))^(+) + e^(-) , Delta_(1)H^(@) = 495.6 kJ mol^(-1)`
(c ) Dissociation of `Cl_(2)`
`Cl_(2(g)) rarr 2Cl_((g)), Delta_(a)H^(@) = 121 kJ mol^(-1)`
`Delta_(a)H^(@)` = enthalpy of atomisation (or) dissociation.
(d) Conversion of Cl(g) to `Cl^(-) (g)`
`Cl(g) + e^(-) rarr Cl^(-)(g) Delta_(c )H^(@) = -348.6 kJ mol^(-1)`
`Delta_(c )H^(@)` is electron gain enthalpy.
(e) Combination of `Na^(+)` (g) and `Cl^(-)` (g) ions to form one mole of NaCl(s).
`Na^(+)(g) + Cl^(-)(g) rarr NaCl(g) , Delta_(L)H^(@) = ?`
Applying Hess's law we get
`Delta_(f)H^(@) = underset("sub")(Delta H^(@)) + Delta underset(i)(H^(@)) + (1)/(2)underset("diss")(Delta H^(@)) + Delta underset(e)(H^(@)) + Delta underset(L)(H^(@))`
`Delta_(L)H^(@) = -411.2 - 108.4 - 121 - 495.6 +348.6 = -787.6 kJ mol^(-1)`
Thus Lattice enthalpy of `NaCl_((s))`. is `-787.6 kJ mol^(-1)`