Worked out Example

Find out the lattice energy of formation of ZnI₂ from Zn²⁺ and I^– with the help of concept of Born–Haber cycle.Useful information:

(i)	Zn (g) → Zn²⁺ (g) + 2e^–	ΔH = 2680 kJ
(ii)	Zn (s) → Zn (g)	ΔH = 116 kJ
(iii)	I₂ (g) → 2I (g)	ΔH = 144 kJ
(iv)	I₂(g) + 2e^– → 2I^– (g)	ΔH = –288 kJ
(v)	I₂(s) → I₂ (g)	ΔH = 38 kJ
(vi)	Zn²⁺ (g) + 2I^–(g) → ZnI₂ (s)	ΔH =?
(vii)	Zn (s) + I₂(s) → ZnI₂ (s)	ΔH = –204 kJ

Sol.

For solving out such type of problems one should write the series of reactions in the form of cycle which represents the correct order of the reactions taking place which eventually leads to products.

Here the major reaction is

Zn²⁺ (g) + 2I^– (g) → ZnI₂ (s) ΔH₆ = ?

We have to find out the lattice energy of ZnI₂

Now write the series of events that eventually lead to formation of ZnI₂

(i)	Zn (s) → Zn (g)	ΔH₁ =116 kJ
(ii)	Zn (g) → Zn²⁺ (g)	ΔH₂ = 2680 kJ
(iii)	I₂(s) → I₂ (g)	ΔH₃ = 38 kJ
(iv)	I₂(g) → 2I (g)	ΔH₄ = 144 kJ
(v)	I (g)+ e^– → I^– (g)	ΔH₅ = –288 kJ
(vi)	Zn²⁺ (g) + 2I^– (g) → ZnI₂ (s)	ΔH₆ =?

This can be given in the form of a table

From the Hess law we know that :
ΔH_f	=	ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ + ΔH₅ + ΔH₆
⇒ –204	=	116 + 2680 + 38 + 144 – 2(288) + x
–204	=	2978 – 576 + x
–204	=	2402 + x
x	=	–2606 kJ
Hence the lattice energy of formation of ZnI₂ (s) is –2606 kJ.