It takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol (MW: 46.07 g/mol). What will be ΔSsys for the vaporization of 8.00 g of ethanol at 79.6 °C

Answer:

[tex]\Delta S_{sys}=0.020kJ=20J[/tex]

Explanation:

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In this case, given the required energy to vaporize 1.00 mol of ethanol as the enthalpy of vaporization:

[tex]\Delta H_{vap}=\frac{38.6kJ}{1.00mol}=38.6kJ/mol[/tex]

We can compute the entropy of the system for the vaporization of 8.00 g of ethanol, by first computing the moles:

[tex]n_{et}=8.00g*\frac{1mol}{46.07g} =0.174mol[/tex]

And then setting up the following expression:

[tex]\Delta S_{sys}=\frac{n_{et}*\Delta H_{vap}}{T}[/tex]

Whereas the temperature is in kelvins; thus, we obtain:

[tex]\Delta S_{sys}=\frac{0.174mol*38.6\frac{kJ}{mol} }{79.6+273.15K}\\\\\Delta S_{sys}=0.020kJ=20J[/tex]

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The entropy of the system ([tex]\Delta S_{sys}[/tex]) for the vaporization of 8.00 grams of ethanol at 79.6 °C is 0.0190 kilojoules.

Given the following data:

Conversion:

Temperature = [tex]79.6 + 273 = 352.6 \;K[/tex]

To find the entropy of the system ([tex]\Delta S_{sys}[/tex]) for the vaporization of 8.00 grams of ethanol at 79.6 °C:

First of all, we would determine the heat of vaporization.

[tex]Heat\; of \;vaporization = \frac{Energy}{moles} \\\\Heat\; of \;vaporization = \frac{38.6}{1}[/tex]

Heat of vaporization = 38.6 kJ/mol.

Next, we would determine the number of moles in 8.00 grams of ethanol:

[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{8}{46.07}\\\\Number\;of\;moles = 0.1737 \;moles[/tex]

Mathematically, the entropy of a system ([tex]\Delta S_{sys}[/tex]) is given by the formula:

[tex]\Delta S_{sys} = \frac{n\Delta H }{T}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\Delta S_{sys} = \frac{0.1737 \times 38.6 }{352.6}\\\\\Delta S_{sys} = \frac{6.7048 }{352.6}\\\\\Delta S_{sys} = 0.0190 \;kJ[/tex]

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