Answer :
The result will be the change in entropy of the system (ΔSsys) for the vaporization is 0.051 kJ/K.
To calculate the change in entropy (ΔSsys) for the vaporization of 21.5 g of ethanol at 79.6 °C, we need to use the equation:
ΔSsys = qrev / T
Where:
ΔSsys is the change in entropy of the system
qrev is the heat transferred in a reversible process
T is the temperature in Kelvin
First, let's convert the mass of ethanol from grams to moles:
moles of ethanol = mass / molar mass
moles of ethanol = 21.5 g / 46.07 g/mol
moles of ethanol = 0.467 mol
Now, let's calculate the heat transferred (qrev) for the vaporization of 0.467 mol of ethanol:
qrev = energy required per mole * number of moles
qrev = 38.6 kJ/mol * 0.467 mol
qrev = 18.02 kJ
Since the temperature given is in Celsius, we need to convert it to Kelvin:
T = 79.6 °C + 273.15
T = 352.75 K
Finally, we can calculate the change in entropy:
ΔSsys = qrev / T
ΔSsys = 18.02 kJ / 352.75 K
ΔSsys ≈ 0.051 kJ/K
Learn more about the change in entropy: https://brainly.com/question/28244712
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