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It takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol (molecular weight: 46.07 g/mol). What will be [tex]\Delta S_{sys}[/tex] for the vaporization of 21.5 g of ethanol at 79.6 °C?

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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