Question

Using a Born-Haber cycle, calculate the lattice energy for lithium fluoride, LiF(s), given the following data: Sublimation energy for Li(s) = 166 kJ ⁄ mol first ionization energy for Li(g) = 520 kJ ⁄ mol bond energy for F2(g) = 158 kJ ⁄ mol–1 electron affinity for F(g) = –328 kJ ⁄ mol–1 enthalpy of formation of LiF(s) = –617 kJ ⁄ mol

Answer 1

Given:

Sublimation energy of Li(s) = 166 kJmol⁻¹

First ionization energy for Li(g) = 520 kJmol⁻¹

Bond energy of F₂(g) = 158 kJmol⁻¹

Electron affinity for F(g) = -328 kJmol⁻¹

Enthalpy of formation of LiF(s) = -617 kJmol⁻¹

To find:

Lattice energy of lithium fluoride LiF(s)

Solution:

To calculate lattice energy using Born-Haber cycle we have-

     $\Delta _fH_{LiF}= \Delta _{sub}H_{Li}+ \Delta H_{F-F} + \Delta _iH_{Li} +\Delta _{eg}H_F + \Delta _LH_{LiF}$

here, $\Delta _fH_{LiF}$ = Enthalpy of formation of LiF

        $\Delta _{sub}H_{Li}$ = Enthalpy of sublimation of Li

        $\Delta H_{F-F}$ = Bond energy of F-F

        $\Delta _iH_{Li}$ = First ionization energy of Li

        $\Delta _{eg}H_F$ = Electron affinity of F

        $\Delta _LH_{LiF}$ = Lattice energy of LiF

therefore,

  $\Delta _LH_{LiF} =\Delta _fH_{LiF}-( \Delta _{sub}H_{Li}+ \Delta H_{F-F} + \Delta _iH_{Li} +\Delta _{eg}H_F)$

  $\Delta _LH_{LiF} =-617-( 166+ 158 + 520 -328)$

  $\Delta _LH_{LiF} = -1133 kJmol^-^1$

Hence, Lattice energy of lithium fluoride is -1133kJmol⁻¹.