Question

Calculate the lattice formation enthalpy in ${\text{kJ mol}}^{-1}$of $LiF$ given that
Enthalpy of sublimation of lithium = $155.3{\text{kJ mol}}^{-1}$
Dissociation enthalpy of half mole of $F_2=75.3\text{kJ}$
Ionization enthalpy of lithium $=520{\text{kJ mol}}^{-1}$
Electron gain enthalpy of 1 mol of $F\left(g\right)=-333\text{kJ}{\Delta }_f{H}_{\text{overall}}=-594{\text{kJ mol}}^{-1}$

• A. $-1011.6{\text{kJ mol}}^{-1}$
• B. $-1200.4{\text{kJ mol}}^{-1}$
• C. $-984.6{\text{kJ mol}}^{-1}$
• D. $-1488.6{\text{kJ mol}}^{-1}$

Answer 1

The correct option is A $-1011.6{\text{kJ mol}}^{-1}$

Given,
$Li\left(s\right)\to Li\left(g\right);\Delta H_{\text{Sub}}^o=+155.3{\text{kJ mol}}^{-1}Li\left(g\right)\to L{i}^{+};IE=520{\text{kJ mol}}^{-1}\frac{1}{2}{F}_2\left(g\right)\to F\left(g\right);\Delta H_{F-F}^o=75.3{\text{kJ mol}}^{-1}F\left(g\right)+e^{-1}\to F^{1-};\Delta H_{eg}=-141{\text{kJ mol}}^{-1}{\Delta }_f{H}_{\text{overall}}=-594{\text{kJ mol}}^{-1}ForLiF;\Delta H_f=\Delta H_{\text{Sub}}^o+\frac{1}{2}\Delta H_{F-F}+\Delta H_i+\Delta H_{eg}+\Delta H_{\text{Lattice}}-594.1=155.2+75.3+520-333+\Delta H_{\text{Lattice}}\Delta H_{\text{Lattice}}=-1011.6{\text{kJ mol}}^{-1}$
Hence, the lattice formation enthalpy of $LiF$ is given by $-1011.6{\text{kJ mol}}^{-1}$.