1. Example Calculate the reticular energy for LiF, knowing: Li (s) + ½ F 2(g) LiF (s) ΔH° - 594.1 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 -594.1 kJ) Born Haber Cyle, applying Hess’es Law

2. Li + (g) + F - (g) Li (g) + F(g) Li (s) +1/2 F 2 (g) LiF (s)  H° 5 = -1017 kJ  H°overall = -594.1 kJ  H° 1 = 155.2 kJ  H° 2 = 75.3 kJ  H° 3 = 520 kJ  H° 4 = -328 kJ

3. Steps: 1.- Li (s) Li (g) ΔH 1 = 155.2 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) = - 594.1 kJ formation energy -594.1 = 155.2 + 75.3 + 520 + (– 328) + U U = - 1017 k J

4. 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 828 610 LiBr 787 550 LiI 732 450 NaCl 788 801 NaBr 736 750 NaI 686 662 KCl 699 772 KBr 689 735 KI 632 680 MgCl 2 2527 714 MgO3890 2800