The molar heats of sublimation and fusion of iodine are 62.3 kJ/mol and15.3 kJ/mol, respectively. Calculate the molar heat of vaporization of liquid iodine.
A) -77.6 kJ/mol
B) 47.0 kJ/mol
C) -47.0 kJ/mol
D) 77.6 kJ/mol
E) 4.07 kJ/mol

Question

Quizplus · Accepted Answer

The molar heat of vaporization of liquid iodine can be calculated using Hess's Law, which states that the total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in. The molar heat of sublimation is the energy required to convert solid iodine directly into gaseous iodine. This process can be thought of as occurring in two steps: solid iodine melting to liquid iodine (fusion), and then the liquid iodine vaporizing to gaseous iodine (vaporization). Therefore, the molar heat of sublimation is equal to the sum of the molar heat of fusion and the molar heat of vaporization. Mathematically, this can be represented as:Heat of sublimation = Heat of fusion + Heat of vaporizationRearranging to solve for the heat of vaporization gives:Heat of vaporization = Heat of sublimation - Heat of fusionSubstituting the given values:Heat of vaporization = 62.3 kJ/mol - 15.3 kJ/mol = 47.0 kJ/mol

The Molar Heats of Sublimation and Fusion of Iodine Are