Question
The molar heat of vaporization of ethanol is 39.3 kJ/mol and the boiling point of ethanol is $78.3^{\circ} \mathrm{C}$. Calculate $\Delta S$ for the vaporization of 0.50 mol ethanol.
Step 1
- Molar heat of vaporization (\( \Delta H_{\text{vap}} \)) = 39.3 kJ/mol - Boiling point of ethanol = \( 78.3^{\circ} \mathrm{C} \) - Amount of ethanol = 0.50 mol Show more…
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The molar heat of vaporization of ethanol is $39.3 \mathrm{kJ} / \mathrm{mol}$, and the boiling point of ethanol is $78.3^{\circ} \mathrm{C}$. Calculate $\Delta S$ for the vaporization of 0.50 mole of ethanol.
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