The molar heats of sublimation and fusion of iodine are 62.3kjmol and 15.3kjmol respectively calculate the molar heat of vaporization of liquid iodide?

Answer 1

It is a known fact : Molar heat of sublimation = molar heat of fusion + molar heat of vaporization so, molar heat of vaporization = molar heat of sublimation - molar heat of fusion Mv = 62.3 kJ/mol - 15.3 kJ/mol Mv = 47 kJ/mol.