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

Answer

B

Explanation

`(1)/(2)I_(2(g))+(1)/(2)Cl_(2(g))rarr ICl_((g)), Delta H_(f)(ICl) = ?`
`eq.(v)=(1)/(2)[eq(i)+eq(ii)+eq(iv)]-eq(iii)`
`Delta H=(1)/(2)[Delta H_(1)+Delta H_(2)+Delta H_(4)]-Delta H_(3)`