Question

The molar heats of fusion and sublimation of molecular iodine are \(15.27 \mathrm{~kJ} / \mathrm{mol}\) and \(62.30 \mathrm{~kJ} / \mathrm{mol}\), respectively. Estimate the molar heat of vaporization of liquid iodine.

Short answer

The molar heat of vaporization of liquid iodine is approximately 47.03 kJ/mol

Step-by-step solution

Understand the problem

The problem involves the molar heat of fusion (\( \Delta H_{fus} \)), the molar heat of sublimation (\( \Delta H_{sub} \)), and the molar heat of vaporization (\( \Delta H_{vap} \)). The molar heat of fusion and sublimation values are given, while the molar heat of vaporization value is what we need to find.

Apply the relationship between heats

Knowing that the molar heat of sublimation equals the molar heat of fusion plus molar heat of vaporization (\( \Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap} \)), we can isolate the molar heat of vaporization (\( \Delta H_{vap} \)) by subtracting the molar heat of fusion (\( \Delta H_{fus} \)) from both sides. Hence, \( \Delta H_{vap} = \Delta H_{sub} - \Delta H_{fus} \).

Substitute the given values

Substitute \( \Delta H_{sub} = 62.30 \mathrm{~kJ} / \mathrm{mol}\) and \( \Delta H_{fus} = 15.27 \mathrm{~kJ} / \mathrm{mol}\) into the above equation to calculate the value of \( \Delta H_{vap} \).

Compute for \( \Delta H_{vap} \)

Perform the subtraction to find the value of \( \Delta H_{vap} \).