Calculate the enthalpy of vaporisation of `I_(2)` if the sublimation energy and enthalpy of fusion of `I_(2)` is 57.3 kJ `mol^(-1)` and 15.5 kJ `mol^(-1)` respectively

Options

A. `-72.8" kJ "mol^(-1)`
B. `72.8" kJ "mol^(-1)`
C. `-41.8" kJ "mol^(-1)`
D. `+41.8" kJ "mol^(-1)`

Answer

D

Explanation

(i) `I_(2)(s) to I_(2)(g),DeltaH_(1)=+57.3" kJ "mol^(-1)`
(ii) `I_(2)(s) to I_(2)(l),DeltaH_(2)=+15.5" kJ "mol^(-1)`
(iii) `I_(2)(l)toI_(2)(g),DeltaH=?`
Equation (iii) can be achieved by =(i)-(ii)
`DeltaH=57.3-15.5=+41.8kJ" "mol^(-1)`