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The Born-Haber cycle for KCl is evaluated with the following
data: $Δ_{f} H for KCl = - 436.7 KJ / mole$
$Δ_{sub} H for K = 89.2 KJ / mole$
$Δ_{vapourisation} H for K = 419.0 KJ / mole$
$Δ_{bond} H for {Cl}_{2} = 243 kJ / mole$ The magnitude of lattice enthalpy of KCl in KJ/mole is _____

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Detailed Solution

By Born Haber cycle

$Δ_{f}^{H} KCl = + {ΔH}_{sub} + IP + \frac{D}{2} - EA - U$

$- U_{KCl} = Δ_{f} H KCl - {ΔH}_{sub} - IP - \frac{D}{2} + EA$

$= - 436.7 - 89.2 - 425 - \frac{243}{2} + 355.5 = - 719.8 kJ$

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