THIS QUESTION IS BASED ON HABER’S BORN CYCLE
OCCURRING RXN WILL BE
ca(s) + cl2(g) __ Cacl2 now deltaHf = – 795 kj/mol
total ionization enthalpy = 589.5 + 1145 = 1734.5 kj/mol
enthalpy of sublimation = 121 kj/mol
total bond dissociation enthalpy = 242.7 kj/mol
we have two cl atoms, so electron gain enthalpy for two atoms is = 2*(-349) = -698 kj/mol as electron gain enthalpy = -(electron affinity)
so using the haber born cycle
enthalpy of formation = enthalpy of sublimation + enthalpy of ionization + enthalpy of bond dissociation + enthalpy of electron gain + lattice enthalpy
-795 = 1734.5+121+242.7-698+net enthalpy
therefore lattice enthalpy = -795 – 1400.2 = – 2195.2 kj/mol