J.R. S. answered 06/11/18
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I2(g) + Cl2(g) ==> 2ICl(g) ∆H = -25.6 kJ
∆Hf ICl - ∆Hf I2 + ∆HfCl2 = -25.6 kJ
∆Hf ICI - 0 + 0 = -25.6 kJ
∆Hf ICl = -25.6 kJ/2