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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 15.5 g of ethanol at 79.2 °C?

Answer :

Final answer:

The change in entropy (ΔS) of the system during the vaporization of 15.5 g of ethanol at 79.2 °C is 36.90 J/K.

Explanation:

The question is asking for the change in entropy (ΔS) of the system during the vaporization of ethanol. Entropy is a measure of the randomness or disorder within a system. The formula for calculating ΔS is ΔS = q_rev/T, where q_rev is the heat absorbed or released at constant temperature (in this case, the heat of vaporization), and T is the temperature in Kelvin.

First, we need to convert the given mass of ethanol (15.5 g) into moles using the molar mass (46.07 g/mol). So, the quantity of ethanol is 15.5 g / 46.07 g/mol = 0.3365 mol.

Next, convert the temperature from °C to K. To do this, add 273.15 to the Celsius temperature. So, T = 79.2 °C + 273.15 = 352.35 K.

The heat of vaporization for 1 mole of ethanol is given as 38.6 kJ. Therefore, the total heat for 0.3365 mol is 38.6 kJ/mol * 0.3365 mol = 13 kJ = 13000 J. We convert the energy to J (joules) because entropy is typically expressed in J/K/mol.

Finally, we can calculate ΔS = q_rev/T = 13000 J / 352.35 K = 36.90 J/K.

Learn more about Entropy here:

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