1.

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

Answer»

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

Solution :According to Hess's law, total HEAT changes during a chemical reaction are independent of path of reaction.
Given, `I_(2)(s) to I_(2)(g),DeltaH_(1)=57.3kJ//mol`. ..(i)
`I_(2)(s)toI_(2)(L),DeltaH_(2)=+15.5kJ//mol`. . . (ii)
REQUIRED EQUATION `I_(2)(l)toI_(2)(g)`,
Substract Eq. (ii) from eq. (i)
`therefore I_(2)(l) to I_(2)(g),Deltah=57.3+(-15.5)=+41.8kJ//mol`


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