At the normal boiling point of benzene,ΔH°_vap = 30.7 kJ/mol and ΔS°_vap = 86.9 J/(mol ∙ K).What is the normal boiling point of benzene?

Question

Quizplus · Accepted Answer

The normal boiling point of benzene can be calculated using the Clausius-Clapeyron equation, which relates the enthalpy of vaporization ($\Delta H_{vap}^\circ$) and the entropy of vaporization ($\Delta S_{vap}^\circ$) to the boiling point temperature ($T_b$) at 1 atm pressure. The equation in a simplified form for the boiling point is given by $T_b = \frac{\Delta H_{vap}^\circ}{\Delta S_{vap}^\circ}$.Given $\Delta H_{vap}^\circ = 30.7 \, \text{kJ/mol} = 30700 \, \text{J/mol}$ and $\Delta S_{vap}^\circ = 86.9 \, \text{J/(mol} \cdot \text{K)}$, we can calculate the boiling point as follows:$$T_b = \frac{30700 \, \text{J/mol}}{86.9 \, \text{J/(mol} \cdot \text{K)}} \approx 353 \, \text{K}$$Therefore, the normal boiling point of benzene is approximately 353 K.

At the Normal Boiling Point of Benzene,ΔH°vap = 30

At the Normal Boiling Point of Benzene,ΔH°_vap = 30