The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compoud present in its gaseous state, dissociates into its ions. It is impossible to determine it directly by experiment. Suggest and explain in indirect method to measure lattice enthalpy of NaCl(s).

Explanation

The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state. For the reaction
`Na^(+)Cl^(-)(S) rarr Na^(+)(g)+Cl^(-)(g),Delta_("lattice")H^(Θ) = +788 kJ mol^(-1)`
Since, it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a Born-Haber cycle.
Let us now calculate the enthalpy of `Na^(+)Cl^(-)` (s) by following steps given below
(i) `Na^(+)(s) rarr Na(g)`, Sublimation of sodium metal, `Delta_(sub)H^(Θ) = 108.4 kJ mol^(-1)`
(ii) `Na(g) rarr Na^(+)(g)+e^(-)(g)`, The ionisation of sodium atoms, ionisation enthalpy `Delta_(i)H^(Θ) = 496 kJ mol^(-1)`
(iii) `(1)/(2) Cl_(2)(g) rarr Cl(g)`, The dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy `(1)/(2)Delta_("bond") H^(Θ) = 121 kJ mol^(-1)`
(iv) `Cl(g)+e^(-)(g) rarr Cl^(-)(g)`, electron gained by chlorine atoms. The electron gain enthalpy, `Delta_(eg)H^(Θ) = -348.6 kJ mol^(-1)`

(v) `Na^(+)(g)+Cl^(-)(g) rarr Na^(+)Cl^(-)(s)`
The sequence of steps is shown in given figure and is shown as Born-Haber cycle. The importance of the cycle is that, the sum of the enthalpy changes round a cycle is zero.
Applying Hess,s law, we get
`Delta_("lattice")H^(Θ)= 411.2 + 108.4 + 121 + 496 - 348.6`
`Delta_("lattice") H^(Θ) = + 788kJ`