If the sublimation energy and enthalpy of fusion of I2 are 57.3 kJ/mole and 15.5 kJ/mole respectively then the enthalpy of vaporisation of I2 will be

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-72.8 kJ/mole

72.8 kJ/mole

-41.8 kJ/mole

41.8 kJ/mole

answer is D.

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Detailed Solution

∆Hsub = ∆Hfusion +∆Hvaporisation∆Hvaporisation = 57.3 - 15.5 =41.8 kJ/mole

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