Using the following data, calculate the standard heat of formation of ICL(g)in kJ/mol:(a)Cl2(g)→2Cl(g)∆H°=242.3kJ;(b)I2(g)→2I(g)∆H°=151.0kJ;(c)ICL(g)→I(g)+CL(g)∆H°=211.3kJ;(d)I2(s)→I2(g)∆H°=62.8kJ

Answer 1

Born Haber Cycle

I₂(s) ==> I₂(g) ... 62.8

I₂(g) ==> 2I(g) ... 151/2 = 75.5

Cl₂(g) ==> 2Cl(g) ... 242.3/2 = 121.7

ICl(g) ==> I(g) + Cl(g) ... -211.3

∑ = 62.8 + 75.5 + 121.7 -211.3

∆H_formation = 48.7 kJ/mol

Answer 2

(a) Cl₂ (g)→2Cl (g) ΔH_a=242.3KJ/mol

(b) I₂ (g)→2I (g) ΔH_b=151.0KJ/mol

(c) ICl (g)→I (g)+Cl (g) ΔH_c=211.3KJ/mol

(d) I₂ (s)→I₂ (g) ΔH_d=62.8KJ/mol

a+b+d-2c, Cl₂ (g)+I₂ (s)→2ICl (g)

ΔH=ΔH_a+ΔH_b+ΔH_d-2ΔH_c=242.3+151.0+62.8-2*211.3=33.5 kJ/mol