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//2I_(2)(s)+1//2Cl_(2)(g)rarrICl(g), DeltaH_(f)=?`
`{:(Cl_(2)(g)rarr2Cl(g),,DeltaH=242.3 kJmol^(-1)),(I_(2)(s)rarrI_(2)(g),,DeltaH=62.76 kJ mol^(-1)),(I_(2)(g)rarr2I(g),,DeltaH=151.0 kJ mol^(-1)):}`
`2Cl(g)+2I(g)rarr2ICl(g), DeltaH=-422.6 kJ mol^(-1)`
`I_(2)(s)+Cl_(2)(g)rarr2ICl(g),`
`thereforeH=-422.6 +151+62.76 +242.3=33.46 kJ`
So heat of formation of `ICl=(33.46)/(2)=16.73 kJ mol^(-1)`.