Given the following data for bromine, $$ \begin{aligned} &\mathrm{Br}_{2}(l): S^{\circ}=152.2 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ &\mathrm{Br}_{2}(g): S^{\circ}=245.4 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \quad \Delta H_{\mathrm{f}}^{\circ}=30.91 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ estimate the normal boiling point of bromine. $$ \mathrm{Br}_{2}(l) \rightleftharpoons \mathrm{Br}_{2}(g) $$

The Clausius-Clapeyron equation relates the vapor pressure of a substance to its temperature, enthalpy of vaporization, and the value of the ideal gas constant. The equation can be written as: $$ \ln \frac{P_2}{P_1}= \frac{\Delta H_v}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) $$ Where \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\) respectively, \(\Delta H_v\) is the enthalpy of vaporization, and \(R\) is the ideal gas constant in appropriate units.

To find the enthalpy of vaporization (\(\Delta H_v\)), we can use the given enthalpy of formation value for gaseous bromine (\(\Delta H_{f}^{\circ}=30.91 \mathrm{~kJ} / \mathrm{mol}\)) since liquid and gaseous bromine are the only components in the equilibrium equation: $$ \Delta H_v = \Delta H_{f}^{\circ} = 30.91 \mathrm{~kJ} / \mathrm{mol} = 30,910 \mathrm{~J} / \mathrm{mol} $$

Now we need to relate the given entropy values of liquid and gaseous bromine to the boiling point. At the boiling point, the vapor pressure of liquid bromine equals the vapor pressure of gaseous bromine. Therefore, we can write the relation between \(\Delta S\) (change in entropy), \(T_{b}\) (boiling point), and \(\Delta H_v\) (enthalpy of vaporization) as: $$ \Delta S = \frac{\Delta H_v}{T_b} $$ Using the given entropy values for liquid (\(S^{\circ}_{l} = 152.2 \mathrm{~J} / \mathrm{mol}\cdot K\)) and gaseous bromine (\(S^{\circ}_{g} = 245.4 \mathrm{~J} / \mathrm{mol}\cdot K\)), we can calculate the change in entropy (\(\Delta S\)) as: $$ \Delta S = S^{\circ}_{g} - S^{\circ}_{l} = 245.4 - 152.2 = 93.2 \mathrm{~J} / \mathrm{mol}\cdot K $$

Now we can use the relation between \(\Delta S\), \(T_b\), and \(\Delta H_v\) to find the boiling point: $$ T_b = \frac{\Delta H_v}{\Delta S} = \frac{30,910 \mathrm{~J} / \mathrm{mol}}{93.2 \mathrm{~J} / \mathrm{mol}\cdot K} = 331.7 \mathrm{~K} $$ Thus, the estimated normal boiling point of bromine is approximately 331.7 K.