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It takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol (molar mass: 46.07 g/mol). What will be the change in entropy of the system ([tex]\Delta S_{\text{sys}}[/tex]) for the vaporization of 8.00 g of ethanol at 79.4 °C?

Answer :

Final answer:

To calculate the change in entropy for vaporization of 8.00 g of ethanol, convert the mass to moles, calculate the heat required using the molar heat of vaporization, convert that heat to joules, and apply it to the formula ∆Ssys = qrev / T with temperature in kelvins.

Explanation:

The change in entropy (∆Ssys) for the vaporization of 8.00 g of ethanol at 79.4 °C can be calculated using the provided heat of vaporization and assuming that the process occurs at a constant temperature. You can use the formula ∆Ssys = qrev / T, where qrev is the heat required for the reversible process (in joules or J), and T is the absolute temperature (in kelvins or K).

First, determine how many moles of ethanol are in 8.00 g using its molecular weight (46.07 g/mol). Then calculate the amount of heat energy required for vaporizing this much ethanol by scaling the given heat per mole (38.6 kJ/mol) to the number of moles you have. Convert this energy to joules (since 1 kJ = 1000 J) and plug it into the formula with the temperature (79.4 °C converted to kelvins).

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