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Answer :
Final answer:
The entropy change for the vaporization of ethanol at 78.0 °C is calculated by dividing the enthalpy of vaporization (39.3 kJ/mol) by the temperature in Kelvin (351.15 K), resulting in an entropy change of 111.9 J/K·mol.
Explanation:
To calculate the entropy change (Δr S°) for the vaporization of ethanol, C2H5OH, at its normal boiling point, you can use the formula:
Δr S° = Δvap H° / T
Where Δvap H° is the enthalpy of vaporization and T is the temperature in Kelvin.
The enthalpy of vaporization of ethanol is 39.3 kJ/mol, and the temperature is 78.0 °C, which is 351.15 K (78.0 + 273.15).
Δr S° = 39.3 kJ/mol / 351.15 K = 0.1119 kJ/K·mol
To express this in standard entropy units, we convert kJ to J:
Δr S° = 0.1119 kJ/K·mol × 1000 J/kJ = 111.9 J/K·mol
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