Answer :
The entropy change when 1.85 moles of ethanol vapor condense at its boiling point of 78°C and 1 atm is 0.203 kJ/K. This can be calculated using the equation ΔS = q/T, where q is the heat transferred and T is the temperature in Kelvin.
The entropy change when 1.85 moles of ethanol vapor condenses at its boiling point of 78°C and 1 atm pressure can be calculated using the equation:
ΔS = q/T
Where ΔS is the entropy change, q is the heat transferred, and T is the temperature in Kelvin.
78°C + 273.15 = 351.15 K
q = nΔH
q = 71.41 kJ
ΔS = 0.203 kJ/K
To find the entropy change when 1.85 moles of ethanol vapor condense at its boiling point of 78°C and 1 atm pressure, we can use the equation ΔS = q/T, where ΔS is the entropy change, q is the heat transferred, and T is the temperature in Kelvin. First, we convert the temperature from Celsius to Kelvin by adding 273.15. Next, we calculate the heat transferred using the heat of vaporization, which is given as 38.6 kJ/mol. By multiplying the number of moles (1.85) by the heat of vaporization, we find that q is equal to 71.41 kJ. Finally, we substitute the values into the equation and calculate the entropy change to be 0.203 kJ/K.
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