Question

For mercury, the enthalpy of vaporization is \(58.51 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of vaporization is \(92.92 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). What is the normal boiling point of mercury?

Short answer

The normal boiling point of mercury is approximately \(629.6 \ \text{K}\).

Step-by-step solution

Simplify the Clausius-Clapeyron equation

Since \(\ln{1} = 0\), we can simplify the Clausius-Clapeyron equation as follows: \[0 = \frac{-\Delta H_\text{vap}}{R} (\frac{1}{T_\text{b}} - \frac{1}{T_1})\]

Use \(\Delta S_\text{vap}\) to find the relationship between \(\Delta H_\text{vap}\) and \(T_\text{b}\)

We have the relationship: \[\Delta S_\text{vap} = \frac{\Delta H_\text{vap}}{T_\text{b}}\] Rearranging for \(T_\text{b}\): \[T_\text{b} = \frac{\Delta H_\text{vap}}{\Delta S_\text{vap}}\]

Plug in the values of enthalpy and entropy of vaporization for mercury

We are given the values: \[\Delta H_\text{vap} = 58.51 \ \text{kJ/mol}\] \[\Delta S_\text{vap} = 92.92 \ \text{J/K} \cdot \text{mol}\] First, let's convert the enthalpy of vaporization to J/mol: \[\Delta H_\text{vap} = 58.51 \ \text{kJ/mol} \times \frac{1000 \ \text{J}}{1 \ \text{kJ}} = 58510 \ \text{J/mol}\] Now we can plug the values into the equation: \[T_\text{b} = \frac{58510 \ \text{J/mol}}{92.92 \ \text{J/K} \cdot \text{mol}}\]

Calculate the normal boiling point of mercury

Dividing the enthalpy by the entropy: \[T_\text{b} = \frac{58510 \ \text{J/mol}}{92.92 \ \text{J/K} \cdot \text{mol}} = 629.6 \ \text{K}\] Therefore, the normal boiling point of mercury is approximately \(629.6 \ \text{K}\).