zatynlpv

2023-03-05

The $\Delta {H}_{f}$ of KF(s) is −563 kJ/mol. The ionisation energy of potassium K(g) is 419 kJ/mol and enthalpy of sublimation of potassium is 88 kJ/mol. The electron affinity of F(g) is −322 kJ/mol and F−F bond enthalpy is 158 kJ/mol. Calculate the absolute value of lattice energy (in kJ/mol) of KF(s)

GrierveAlierskgc

For the formation of KF, the Born-Haber cycle is:

Now, using Hess's law :
$\Delta {H}_{f}=\Delta {H}_{1}+\Delta {H}_{2}+\Delta {H}_{3}+\Delta {H}_{4}+\Delta {H}_{5}$
$-563=88+419+79-322+\Delta {H}_{5}$
$\Delta {H}_{5}=-827kJ/mol$
∴ Lattice energy of KF (s) is −827 kJ/mol

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