Example Calculate the reticular energy for LiF, knowing: Li (s) + ½ F 2(g) LiF (s) ΔH° kJ Applying Born Haber’s cycle, the formation of LiF is carried about through 5 steps, remembering that the sum of the enthalpy changes of all the steps is equal to the change of tnthalpy for the global reaction (in this case kJ) Born Haber Cyle, applying Hess’es Law

Li + (g) + F - (g) Li (g) + F(g) Li (s) +1/2 F 2 (g) LiF (s)  H° 5 = kJ  H°overall = kJ  H° 1 = kJ  H° 2 = 75.3 kJ  H° 3 = 520 kJ  H° 4 = -328 kJ

Steps: 1.- Li (s) Li (g) ΔH 1 = kJ sublimationn 2.-½( F 2(g) 2 F (g) ) ΔH 2 = ½(150.6) kJ dissociation 3.-Li (g) Li+ (g) + 1e- ΔH 3 = 520kJ ionization energy 4.-F (g) + 1e- F- (g) ΔH 4 = -328kJ electronic affinity 5.- F- (g) + Li+ (g) -  LiF (s) = U reticular energy Li(s) + ½ F 2(g) LiF (s) = kJ formation energy = (– 328) + U U = k J

Reticular Energies for some ionic solids The high values of the reticular energy explains why the ionic compounds are very stables solids, with a very high melting point. CompoundNetwork Energy (kJ/mol) Melting Point ( 0 C) LiCl LiBr LiI NaCl NaBr NaI KCl KBr KI MgCl MgO