The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K•mol. What is the approximate boiling point of 1-pentanol?

Question

Quizplus · Accepted Answer

The boiling point can be estimated using the Clausius-Clapeyron equation, which relates the heat of vaporization ($\Delta H_{vap}$) and the entropy of vaporization ($\Delta S_{vap}$) to the boiling point at atmospheric pressure. The equation is $\Delta H_{vap} = T_{b} \cdot \Delta S_{vap}$, where $T_{b}$ is the boiling point in Kelvin. Rearranging for $T_{b}$, we get $T_{b} = \frac{\Delta H_{vap}}{\Delta S_{vap}}$. Substituting the given values, $T_{b} = \frac{55.5 \times 10^3 \, \text{J/mol}}{148 \, \text{J/K} \cdot \text{mol}} \approx 375 \, K$. To convert to Celsius, subtract 273, resulting in approximately 102°C, which is closest to the option A (100°C).

The Heat of Vaporization of 1-Pentanol Is 55