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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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slide1

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)

slide2

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

slide3

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

reticular energies for some ionic solids
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