If the enthalpy of sublimation of Li is 155 kj $$mol^{-1}$$, enthalpy of dissociation of $$F_{2}$$ is 150 kj $$mol^{-1}$$, ionization enthalpy of Li is 520 kj $$mol^{-1}$$, electron gain enthalpy of F is - 313 kj $$mol^{-1}$$, standard enthalpy of formation of LiF is - 594 kj $$mol^{-1}$$. The magnitude of lattice enthalpy of LiF is _________ kJ $$mol_{-1}$$. (Nearest Integer)

We use the Born-Haber cycle to determine the lattice enthalpy of LiF.

The formation of LiF from its elements is broken down into several processes: sublimation of Li(s) to Li(g) with $$\Delta H_{\text{sub}} = +155$$ kJ/mol; dissociation of $$\frac{1}{2}F_2(g) \rightarrow F(g)$$ with $$\frac{1}{2}\Delta H_{\text{diss}} = \frac{150}{2} = +75$$ kJ/mol; ionization of Li(g) to Li⁺(g) with $$IE = +520$$ kJ/mol; electron gain by F(g) to F⁻(g) with $$\Delta H_{eg} = -313$$ kJ/mol; and finally lattice formation Li⁺(g) + F⁻(g) → LiF(s) with enthalpy $$U$$ to be determined.

By Hess’s law, the standard enthalpy of formation $$\Delta H_f$$ equals the sum of these individual steps, so that $$\Delta H_f = \Delta H_{\text{sub}} + \frac{1}{2}\Delta H_{\text{diss}} + IE + \Delta H_{eg} + U$$.

Substituting the known values yields $$-594 = 155 + 75 + 520 + (-313) + U$$, which simplifies to $$-594 = 155 + 75 + 520 - 313 + U = 437 + U$$ and hence $$U = -594 - 437 = -1031 \text{ kJ/mol}$$.

The magnitude of the lattice enthalpy of LiF is $$|U| = \mathbf{1031}$$ kJ/mol.