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Question

Calculate the lattice formation enthalpy in kJ mol1of LiF given that
Enthalpy of sublimation of lithium = 155.3 kJ mol1
Dissociation enthalpy of half mole of F2=75.3 kJ
Ionization enthalpy of lithium =520 kJ mol1
Electron gain enthalpy of 1 mol of F(g) =333 kJΔfHoverall=594 kJ mol1

A
1011.6 kJ mol1
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B
1200.4 kJ mol1
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C
984.6 kJ mol1
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D
1488.6 kJ mol1
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Solution

The correct option is A 1011.6 kJ mol1
Given,
Li(s)Li(g) ; ΔHoSub=+155.3 kJ mol1Li (g)Li+ ; IE=520 kJ mol112F2 (g)F (g) ; ΔHoFF=75.3 kJ mol1F(g) +e1F1; ΔHeg=141 kJ mol1ΔfHoverall=594 kJ mol1For LiF; ΔHf=ΔHoSub+12ΔHFF+ΔHi+ΔHeg+ΔHLattice594.1=155.2+75.3+520333+ΔHLatticeΔHLattice=1011.6 kJ mol1
Hence, the lattice formation enthalpy of LiF is given by 1011.6 kJ mol1.

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