The sublimation energy of `I_(2)` (solid) is 57.3 KJ/mole and enthalpy of fusion is 15.5 KJ/mole. The enthalpy of vapourisation of `I_(2)` is

Options

A. 41.8 kJ/mol
B. `-41.8` kJ/mol
C. 72.8kJ/mol
D. `-72.8` kJ/mol

Answer

A

Explanation

`I_(2)(s) rarr I_(2)(g), Delta H_(1)= +57.3` kJ/mol
`I_(2)(s) rarr I_(2)(l), Delta H_(2)= +15.5` kJ /mol
Hence, for the conversion -
`I_(2)(l) rarr I_(2)(g) " " Delta H= Delta H_(1)- Delta H_(2)= 57.3 - 15.5 = +41.8` kJ/mol