Answer :
Final answer:
To calculate the change in entropy for the vaporization of 0.50 mole of ethanol, convert the boiling point to Kelvin, use the molar heat of vaporization, and adjust for the number of moles. The value of ΔvapS for the vaporization of 0.50 mole of ethanol is 0.0559 kJ/K.
Explanation:
To calculate the change in entropy (ΔvapS) during the vaporization of 0.50 mole of ethanol, we can use the formula ΔvapS = ΔvapH / Tb, where ΔvapH is the molar heat of vaporization and Tb is the boiling point of ethanol in Kelvin. The molar heat of vaporization of ethanol is given as 39.3 kJ/mol and the boiling point is 78.3°C (351.45 K when converted to Kelvin).
Firstly, we calculate the entropy change for one mole of ethanol and then adjust for 0.50 mole:
ΔvapS (for 1 mole) = 39.3 kJ/mol / 351.45 K = 0.1118 kJ/(mol·K)
ΔvapS (for 0.50 mole) = 0.1118 kJ/(mol·K) × 0.50 mol = 0.0559 kJ/K
Therefore, the value of ΔvapS for the vaporization of 0.50 mole of ethanol is 0.0559 kJ/K. This represents the increase in disorder as ethanol transitions from a liquid to a gaseous state at its boiling point.
