Calculate the entropy of vaporization of ethanol, C₂H₅OH, at 298 K from the following data. The constant-pressure molar heat capacities of ethanol as liquid and vapour are 111.46 and 65.44 J K^-1 mol^-1 respectively. The enthalpy of vaporization of ethanol at its boiling temperature, 352 K, is 38.56 kJ mol^-1.

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The entropy of vaporization ($\Delta S_{vap}$) can be calculated using the Clausius-Clapeyron equation and considering the change in heat capacity ($\Delta C_p$) between the liquid and vapor phases. First, calculate the entropy of vaporization at the boiling point using $\Delta S_{vap} = \Delta H_{vap} / T_{b}$, where $\Delta H_{vap}$ is the enthalpy of vaporization and $T_{b}$ is the boiling temperature. Then, adjust this value to the desired temperature (298 K) considering the difference in heat capacities.Given:- $\Delta H_{vap} = 38.56 \, \text{kJ mol}^{-1} = 38560 \, \text{J mol}^{-1}$- Boiling temperature, $T_{b} = 352 \, \text{K}$- Heat capacities: $C_{p,\text{liquid}} = 111.46 \, \text{J K}^{-1} \text{mol}^{-1}$, $C_{p,\text{vapor}} = 65.44 \, \text{J K}^{-1} \text{mol}^{-1}$- $\Delta C_p = C_{p,\text{vapor}} - C_{p,\text{liquid}} = 65.44 - 111.46 = -46.02 \, \text{J K}^{-1} \text{mol}^{-1}$First, calculate $\Delta S_{vap}$ at $T_{b}$:$$\Delta S_{vap} (T_{b}) = \frac{\Delta H_{vap}}{T_{b}} = \frac{38560}{352} \approx 109.55 \, \text{J K}^{-1} \text{mol}^{-1}$$Then, adjust $\Delta S_{vap}$ to 298 K considering $\Delta C_p$:$$\Delta S_{vap} (298 \, \text{K}) = \Delta S_{vap} (T_{b}) + \Delta C_p \ln\left(\frac{298}{T_{b}}\right)$$$$\Delta S_{vap} (298 \, \text{K}) = 109.55 - 46.02 \ln\left(\frac{298}{352}\right)$$$$\Delta S_{vap} (298 \, \text{K}) \approx 109.55 - (-7.66) \approx 117.21 \, \text{J K}^{-1} \text{mol}^{-1}$$Therefore, the correct answer is +117.21 J K$^{-1}$ mol$^{-1}$, which corresponds to choice A.

Calculate the Entropy of Vaporization of Ethanol, C2H5OH, at 298

Calculate the Entropy of Vaporization of Ethanol, C₂H₅OH, at 298