Answer :
Final answer:
The change in entropy (ΔSsys) for the vaporization of 8.00 g of ethanol at 78.2 °C is calculated by first working out the number of moles of ethanol, then applying the formula for calculating the entropy of vaporization, and finally multiplying by the number of moles. The calculated ΔSsys is approximately 19.1 J/K.
Explanation:
To calculate the change in entropy for the vaporization of ethanol (C₂H5OH), we need to first calculate the moles of ethanol existing in 8.00 grams. Using the molecular weight (46.07 g/mol), we find that there are approximately 0.1736 moles.
Next, we apply the formula for calculating the entropy of vaporization ΔSvap = ΔHvap/T where ΔHvap is the enthalpy change of vaporization and T is the absolute temperature in Kelvin. We are given that ΔHvap is 38.6 kJ/mol and the temperature is 78.2°C, which we convert to Kelvin (78.2°C + 273.15 = 351.35 K).
Plugging in our values, we find ΔSvap = 38.6 kJ/mol / 351.35 K = 109.9 J/mol·K (note: to make units consistent, we converted kJ to J by multiplying by 1000).
Finally, to find ΔSsys for the process, we multiply ΔSvap by the number of moles. ΔSsys = 0.1736 moles * 109.9 J/(mol·K) = 19.1 J/K.
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