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Question

The ΔHf 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 FF bond enthalpy is 158 kJ/mol. Calculate the absolute value of lattice energy (in kJ/mol) of KF(s)


Solution

Born-Haber cycle for the formation of KF is:

Now, using Hess's law :
ΔHf=ΔH1+ΔH2+ΔH3+ΔH4+ΔH5
563=88+419+79322+ΔH5
ΔH5=827 kJ/mol

Lattice energy of KF (s) is 827 kJ/mol

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