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

The reaction is:
`(1)/(2)I_(2)(s)+(1)/(2)Cl_(2)(g)toICl(g)`
`Delta_(f)H_(I-Cl)=[(1)/(2)DeltaH_(I_(2)(s)toI_(2)(g))+(1)/(2)DeltaH_(I-I)+(1)/(2)DeltaH_(Cl-Cl)]-[DeltaH_(I-Cl)]`
`=[(1)/(2)xx62.76+(1)/(2)xx151+(1)/(2)xx242.3]-[211.3]`
`=16.73kJ" "mol^(-1)`