1.

The sublimation energy ofI_(2_((s)))is is 57.3 kJ/mol and the enthalpy of fusion is 15.5 kJ/mol. The enthalpy of vaporisation of I_(2) is

Answer»

41.8 kJ/mol
`-41.8 kJ//mol`
`72.8 kJ//mol`
`-72.8 kJ//mol`

Solution :`I_(2)(s)rarrI_(2)(g), DeltaH_(1)=+57.3 kJ//mol`
`I_(2)(s)rarrI_(2)(l),DeltaH_(2)=+15.5 kJ//mol`
HENCE, for the CONVERSION -
`I_(2)(l)rarrI_(2)(g)""DeltaH=DeltaH_(1)-DeltaH_(2)`
`=57.3-15.5 =+41.8 kJ//mol`.


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