The bond dissociation energies for `Cl_(2)`, `I_(2)` and `IC l` are `242.3`, `151.0` and `211.3 kJ//"mole"` respectively. The enthalpy of sublimation of iodine is `62.8 kJ //"mole"`. What is the standard enthalpy of formation of `ICI(g)` nearly equal to

Options

A. `- 211.3 kJ //`mole
B. `- 14.6kJ //`mole
C. 16.8 kJ`//` mole
D. `33.5 kJ//` mole

Answer

C

Explanation

We aim `: (1)/(2) I_(2)(s) + (1)/(2) Cl_(2) (g) rarr ICI (g) `
We are given `:`
(i) `Cl_(2)(g)rarr 2Cl(g), DeltaH=242.3 kJ`
(ii) `I_(2)(g) rarr 2I(g),DeltaH = 151 kJ`
(iii) `ICI (g) rarr (i) + Cl(g), DeltaH =211.3 kJ`
(iv) `I_(2)(s) rarrI_(2)(g), DeltaH =62.8 kJ`
`(1)/(2) (ii) (1)/(2) (i)- (iii) +(1)/(2) ` gives the required equation with
`DeltaH =(1)/(2) (151) +(1)/(2) (292.3) -211.3 + (1)/(2) (62.8) = 16.75kJ`