Question

Calculate the lattice enthalpy (ΔHl° in kJ/mol) of the MF2 metal fluoride using the following data:


Bond dissociation enthalpy of F2(g) = 159 kJ/mol

Electron affinity of F = -328 kJ/mol

Enthalpy of sublimation of M = 112 kJ/mol

First ionization energy of M = 395 kJ/mol

Second ionization energy of M = 587 kJ/mol

Standard enthalpy of formation of MF2(s) = -377 kJ/mol

Answer 1

Lattice enthalpy of MF2=bond enthalpy of F2 + electron affinity of F +enthalpy of sublimation of M +first ionization energy of M + second ionization energy of M + enthalpy of formation of MF2.

$Lattice enthalpy= (159-328+112+395+587-377) kJ/mol$

Lattice enthalpy= 548 kJ/mol