If the enthalpy of sublimation of Li is 155 kJ mol−1, enthalpy of dissociation of F2 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 ____

If the enthalpy of sublimation of Li is 155 kJ mol−1 enthalpy of dissociation of F2 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

Q. If the enthalpy of sublimation of Li is 155 kJ mol⁻¹, enthalpy of dissociation of F₂ 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)

Correct Answer: 1031

Explanation

This problem is solved using the Born–Haber cycle for the formation of ionic compound LiF.

The steps involved in the formation of LiF from its elements are:

Sublimation of lithium:

Li(s) → Li(g) ΔH = +155 kJ mol⁻¹

Dissociation of fluorine molecule:

½F₂(g) → F(g) ΔH = +75 kJ mol⁻¹

Ionization of lithium atom:

Li(g) → Li⁺(g) + e⁻ ΔH = +520 kJ mol⁻¹

Electron gain by fluorine atom:

F(g) + e⁻ → F⁻(g) ΔH = −313 kJ mol⁻¹

Lattice formation:

Li⁺(g) + F⁻(g) → LiF(s) ΔH = −U

According to Born–Haber cycle:

−594 = 155 + 75 + 520 − 313 − U

Simplifying:

−594 = 437 − U

U = 437 + 594

U = 1031 kJ mol⁻¹

Hence, the magnitude of lattice enthalpy of LiF is 1031 kJ mol⁻¹.

Explanation

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