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It takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol (MW: 46.07 g/mol). What will be [tex]\Delta S_{sys}[/tex] for the vaporization of 6.50 g of ethanol at 79.0 °C?

Answer :

Final answer:

The entropy change for the vaporization of 6.5g of ethanol, where the process requires 38.6 kJ of energy for each mole, is 0.1095 kJ/mol-K.

Explanation:

The question seeks to calculate the change the change in entropy (delta S) during the vaporization of 6.5g of ethanol, a process that requires 38.6 kJ of energy for each mole of ethanol. Given that the molar mass of ethanol is 46.07 g/mol, we can first find out how many moles are in 6.5g, then multiply that by the energy needed for each mole to vaporize.

To calculate moles, we employ the formula: moles = mass (g) / molar mass (g/mol). Substituting, we have: moles = 6.5g / 46.07 g/mol = 0.141 mol.

We know that one mole of ethanol needs 38.6 kJ of energy to be vaporized, therefore, delta H (enthalpy change) for vaporization would be 38.6 kJ/mol.

Then, we can calculate delta S (entropy change) using the formula: Delta S = Delta H / T, where 'T' represent temperature and is in Kelvins.

To convert temperature from Celsius to Kelvin, we use the formular: T (K) = T (°C) + 273.15.

Thus, T = 79 + 273.15 = 352.15K.

Finally, substituting, Delta S(sys) = 38.6 kJ/mol / 352.15K = 0.1095 kJ/mol-K.

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