The enthalpy change for the following process are listed below`:`
`(a)` `Cl_(2(g)) rarr 2Cl_((g)), DeltaH=242.3kJ mol^(-1)`
`(b)` `I_(2(g))rarr 2I_((g)), DeltaH=151.0kJ mol^(-1)`
`(c)` `ICl_((g)) rarr I_((g))+Cl_((g)), DeltaH=211.3kJ mol^(-1)`
`(d)` `I_(2(s)) rarr I_(2(g)), DeltaH=62.76kJ mol^(-1)`
If standard state of iodine and chloride are `I_(2(s))` and `Cl_(2(g))` , the standard enthalpy of formation for `ICl_((g))` is `:`

Options

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

Answer

C

Explanation

`I_(2(s))+Cl_(2(g))rarr 2ICl_((g)),`
`DeltaH=[e_(Cl_(2(g))+e_(I_(2(srarrg)))+e_(I_(2_((g)))))]-2xxe_(ICl)`
`=[242.3+62.76+151.0]-2xx211.3`
`=33.46`
`:. DeltaH//mol=(33.46)/(2)=16.73kJ mol^(-1)`