Answer :
Final answer:
The value of ΔSvap for the vaporization of 0.50 mole of ethanol is 0.0559 kJ/K, calculated by dividing the molar heat of vaporization by the boiling temperature in Kelvin and then multiplying by the number of moles.
Explanation:
To calculate the value of ΔSvap for the vaporization of 0.50 mole of ethanol, we use the given molar heat of vaporization (ΔHvap) of ethanol and its boiling point. The molar heat of vaporization is 39.3 kJ/mol, and the boiling point is 78.3°C (or 351.45 K, converting to Kelvin by adding 273.15).
ΔSvap is calculated using the formula:
ΔSvap = ΔHvap / Tboiling
For the calculation of the entropy change of vaporization (ΔSvap) for 0.50 mole of ethanol:
ΔSvap = (39.3 kJ/mol) / (351.45 K) = 0.1118 kJ/K·mol
However, since we have only 0.50 moles, we multiply this value by the number of moles:
ΔSvap (for 0.50 moles) = 0.1118 kJ/K·mol × 0.50 moles = 0.0559 kJ/K
Thus, the value of ΔSvap for the vaporization of 0.50 mole of ethanol is 0.0559 kJ/K.