The enthalpy and entropy of vaporization of carbon disulfide are 26.74 kJ/mol and 83.7 J/mol . K, respectively.What is the normal boiling point of carbon disulfide in$\degree$C?

Question

Quizplus · Accepted Answer

The normal boiling point can be found using the Clausius-Clapeyron equation, which relates the enthalpy of vaporization ($\Delta H_{vap}$) and the entropy of vaporization ($\Delta S_{vap}$) to the boiling point at standard pressure. The equation in its simplified form for boiling point is $T_b = \frac{\Delta H_{vap}}{\Delta S_{vap}}$. Plugging in the given values, $T_b = \frac{26740 \, \text{J/mol}}{83.7 \, \text{J/mol K}} = 319.5 \, \text{K}$. To convert to degrees Celsius, subtract 273.15 from the Kelvin temperature, resulting in $46.3^\circ C$.

The Enthalpy and Entropy of Vaporization of Carbon Disulfide Are °\degree°

The Enthalpy and Entropy of Vaporization of Carbon Disulfide Are $\degree$