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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 the change in entropy of the system ([tex]\Delta S_{\text{sys}}[/tex]) for the vaporization of 14.0 g of ethanol at 80.0 °C?

Answer :

Final answer:

The change in entropy (ΔSsys) for the vaporization of 14.0 g of ethanol at 80.0 °C is approximately 33.20 J/K, calculated by determining the moles of ethanol, finding the heat added during vaporization, and dividing by the temperature in Kelvin.

Explanation:

The student asked about calculating the change in entropy (ΔSsys) for the vaporization of 14.0 g of ethanol at 80.0 °C, given that it takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol. First, we should calculate the number of moles of ethanol in 14.0 g using the molecular weight (46.07 g/mol). The number of moles of ethanol (n) is:

n = mass (g) / MW (g/mol) = 14.0 g / 46.07 g/mol ≈ 0.304 mol

Next, we use the formula ΔS = qrev / T to calculate the entropy change, where qrev is the heat added in a reversible process (in this case, vaporization) and T is the temperature in Kelvin. Knowing that it takes 38.6 kJ to vaporize 1 mol, we find the total energy required for 0.304 mol:

qrev = 38.6 kJ/mol × 0.304 mol ≈ 11.73 kJ

Since entropy is in J/K, we convert kJ to J:

qrev = 11.73 kJ × 1000 J/kJ ≈ 11730 J

Now, convert the temperature to Kelvin:

T = 80.0 °C + 273.15 = 353.15 K

Finally, calculate the entropy change:

ΔSsys = qrev / T = 11730 J / 353.15 K ≈ 33.20 J/K

Hence, the change in entropy for the vaporization of 14.0 g of ethanol at 80.0 °C is approximately 33.20 J/K.

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