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.
