Answer :
Final answer:
The change in entropy for the vaporization of 0.30 mol of ethanol is calculated using the molar heat of vaporization and the boiling point. The result is ΔS = 33.54 J/K.
Explanation:
The question asks to calculate the change in entropy for the vaporization of 0.30 mol of ethanol, using the molar heat of vaporization (39.3 kJ/mol) and boiling point (78.3°C) of ethanol. The formula for calculating the change in entropy (ΔS) during vaporization is given by:
ΔS = ΔHvap / Tb
Where ΔHvap is the molar heat of vaporization and Tb is the boiling point in Kelvin. The change in entropy can be calculated using the following steps:
Finally, multiply by the number of moles: ΔS for 0.30 mol = 111.8 J/K·mol
0.30 mol = 33.54 J/K.
Therefore, the change in entropy for the vaporization of 0.30 mol of ethanol is 33.54 J/K.