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Example Calculate the reticular energy for LiF, knowing:

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

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Example Calculate the reticular energy for LiF, knowing:

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  1. Born Haber Cyle, applying Hess’es Law Example Calculate the reticular energy for LiF, knowing: Li(s) + ½ F2(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)

  2. Li+(g) + F-(g) H°3= 520 kJ H°4= -328 kJ Li (g) + F(g) H°5 = -1017 kJ H°1= 155.2 kJ H°2= 75.3 kJ Li (s) +1/2 F2(g) LiF (s) H°overall = -594.1 kJ

  3. Steps: 1.- Li(s) Li(g)ΔH1= 155.2 kJ sublimationn 2.-½( F2(g) 2 F(g)) ΔH2= ½(150.6) kJ dissociation 3.-Li(g) Li+(g) + 1e- ΔH3= 520kJ ionization energy 4.-F(g) + 1e- F-(g)ΔH4= -328kJ electronic affinity 5.- F-(g) + Li+(g) - LiF(s) = U reticular energy Li(s) + ½ F2(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. Compound Network Energy (kJ/mol) Melting Point (0C) 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 MgCl2 2527 714 MgO 3890 2800

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