The normal boiling temperature of benzene, C₆H₆, is 353.3 K, whilst the triple point occurs at a temperature of 278.5 K and pressure of 4813 Pa. Determine the average value for the enthalpy of vaporization of benzene over this range. A) 33.3 kJ mol^-1 B) 4.01 kJ mol^-1 C) 30.7 kJ mol^-1 D) 57.1 kJ mol^-1

Question

Quizplus · Accepted Answer

The enthalpy of vaporization is given by the Clausius-Clapeyron equation:$$\ln\frac{P_2}{P_1} = \frac{\Delta H_{vap}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2} ight)$$where $P_1$ and $T_1$ are the pressure and temperature at the triple point, and $P_2$ and $T_2$ are the pressure and temperature at the normal boiling point.Substituting the given values into the equation, we get:$$\ln\frac{101325}{4813} = \frac{\Delta H_{vap}}{8.314} \left(\frac{1}{278.5} - \frac{1}{353.3} ight)$$Solving for $\Delta H_{vap}$, we get:$$\Delta H_{vap} = 33.3 ext{ kJ mol}^{-1}$$Therefore, the average value for the enthalpy of vaporization of benzene over this range is 33.3 kJ mol^-1.

The Normal Boiling Temperature of Benzene, C6H6, Is 353

The Normal Boiling Temperature of Benzene, C₆H₆, Is 353