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

Answer

A

Explanation

`1/2 I_(2)+1/2 Cl_(2) rarr ICl, DeltaH=H_(R)-H_(P)`