The enthaplpy changes state for the following processes are listed below:
`Cl_(2)(g)=2Cl(g)` : `242.3KJmol^(-1)`
`I_(2)(g)=2I(g)` , `151.0KJ mol^(-1)`
`ICl(g)=I(g)+Cl(g)` : `211.3KJ mol^(-1)`
`I_(2)(s)=l_(2)(g)` , `62.76KJ mol^(-1)`
Given that the standard states for iodine chlorine are `I_(2)(s)` and `Cl_(2)(g)` , the standard enthalpy of formation for `ICl(g)` is:

Options

A. `-14.6 kJ mol^(-1)`
B. `-16.8 kJ mol^(-1)`
C. `+16.8 kJ mol^(-1)`
D. `+244.8 kJ mol^(-1)`

Answer

C

Explanation

`1//2I_2(s) + 1//2Cl_2(g) to I Cl (g) , DeltaH_f = ?`

`2Cl(g) + 2I(g) to 2I Cl(g), Delta H = - 422.6 kJ mol^(-1)`
`I_2(s) + Cl_2(g) to 2I Cl(g)`,
`DeltaH = - 422.6 + 151 + 62.76 + 242.3 = 33.46 kJ`
So heat of formation of `I Cl = (33.46)/2 = 16.73 kJ mol^(-1)`.