The enthalpy changes for the following process are listed below :
`Cl_(2)(g)=2Cl(g)," "242.3" kJ"mol^(-1)`
`I_(2)(g)=2I(g)," "151.0" kJ"mol^(-1)`
`ICl(g)=2I(g)+Cl(g)," "211.3" kJ"mol^(-1)`
`I_(2)(s)=I_(2)(g)," "62.76" kJ"mol^(-1)`
Given that standard states for iodine and chlorine are `I_(2)(s)` and `Cl_(2)(g)`, the standerd 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)/(2) I_(2(s))+(1)/(2) Cl_(2) rarr ICl_((g))`
`DeltaH =[(1)/(2) Delta H_(I_(2(s))rarrI_(2(g)))+(1)/(2) mu_(1-1)+(1)/(2) mu Cl-Cl]-[mu_(I)-Cl]`
`=((1)/(2)xx62.76+(1)/(2)xx151.0+(1)/(2)xx242.3)-(211.3)`
`=228.03-211.13`
`Delta H =16.73`