Given data
First Ionization energy of Na: 496
kJ/mol                  Â
Na(g) ------->Na+ + e-
Electron affinity of F: –328
kJ/mol                             Â
F(g) + e- -------------> 2F(g)
Enthalpy of atomization Na: 108 kJ/mol Na(s) ------------->
Na(g)
bond energy of fluorine gas: 159
kJ/mol                  Â
F2(g)------------------> 2F(g)
Enthalpy of formation of NaF(s) = –574
kJ/mol         Na(s)
+ 1/2F2 ---------> NaF
Using Born Haber cycle
Enthalpy of formation of NaF(s) = Enthalpy of
atomization Na + First Ionization energy of Na + bond energy of
fluorine gas + Electron affinity of F + Lattice energy of NaF
-574 = 108 + 496 + 159 -328 + Lattice
energy     (all the energies are in
kj/mol)
Lattice energy = -574 -108-496-159+328 = -1009 kj/mol
