Answer :
Final answer:
The entropy change for the vaporization of ethanol at its normal boiling point is calculated using the enthalpy of vaporization and the temperature in Kelvin, resulting in an entropy change of 111.9 J/K·mol.
Explanation:
To calculate the entropy change, Delta S°, for the vaporization of ethanol at its normal boiling point, we can use the formula Delta S° = Delta H°/vap / T, where Delta H°/vap is the enthalpy of vaporization and T is the temperature in Kelvin.
First, we need to convert the boiling point of ethanol from degrees Celsius to Kelvin. The boiling point of ethanol is 78.0°C, which is equivalent to 351.15 K (78 + 273.15).
Given that the enthalpy of vaporization, Delta H°/vap, for ethanol is 39.3 kJ/mol, we can now calculate the entropy change:
Delta S° = 39.3 kJ/mol / 351.15 K = 0.1119 kJ/K·mol
Since entropy is usually given in units of J/K·mol, we convert 0.1119 kJ/K·mol to 111.9 J/K·mol. Therefore, the entropy change for the vaporization of ethanol at its normal boiling point is 111.9 J/K·mol.
