Calculate the standard entropy of vaporization of ethanol at its boiling point, 352 K.The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ.mol^-¹. A) +40.5 kJ.K^-¹.mol^-¹ B) +115 J.K^-¹.mol^-¹ C) -40.5 J.K^-¹.mol^-¹ D) +513 J.K^-¹.mol^-¹ E) -115 J.K^-¹.mol^-¹

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Quizplus · Accepted Answer

The standard entropy of vaporization ($\Delta S_{vap}$) can be calculated using the formula $\Delta S_{vap} = \frac{\Delta H_{vap}}{T}$, where $\Delta H_{vap}$ is the enthalpy of vaporization (40.5 kJ/mol) and $T$ is the temperature in Kelvin (352 K). Converting $\Delta H_{vap}$ to J/mol (1 kJ = 1000 J), we get 40500 J/mol. Thus, $\Delta S_{vap} = \frac{40500 J/mol}{352 K} \approx 115 J/(mol \cdot K)$.

Calculate the Standard Entropy of Vaporization of Ethanol at Its