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

Beginner2023-03-06Added 2 answers

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

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