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Answer :
Final answer:
The change in entropy (ΔS) for the vaporization of 0.50 mole of ethanol at its boiling point is 55.9 J/K.
Explanation:
To calculate the change in entropy (ΔS) during the vaporization of ethanol, you need to use the formula:
ΔS = ΔH / T
where ΔH is the molar heat of vaporization, and T is the absolute temperature in Kelvin. For ethanol, ΔH is given as 39.3 kJ/mol, and the boiling point is 78.3°C, which is 351.45 K (78.3 + 273.15).
Since we have 0.50 mol of ethanol, the total heat absorbed during vaporization would be:
0.50 mol × 39.3 kJ/mol = 19.65 kJ
To find the change in entropy, we then divide this heat by the absolute temperature:
ΔS = 19.65 kJ / 351.45 K = 0.0559 kJ/K
Converting kJ to J, we get:
ΔS = 55.9 J/K for 0.50 mol of ethanol.
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