Answer :
The entropy change (∆Ssys) for the vaporization of 14.00 g of ethanol at 78.2°C is -360.9 J/K.
Given data:
- Heat of vaporization (∆Hvap) of ethanol = 38.6 kJ/mol
- Molecular weight (MW) of ethanol = 46.07 g/mol
- Mass of ethanol (m) = 14.00 g
- Temperature (T) = 78.2°C = 78.2 + 273.15 K = 351.35 K
First, calculate the moles of ethanol:
n = m/MW = 14.00 g / 46.07 g/mol = 0.304 mol
Next, calculate the entropy change using the formula:
∆Ssys = ∆Hvap / T
Now, convert the heat of vaporization from kJ to J:
∆Hvap = 38.6 kJ × 1000 J/kJ = 38600 J
Plug in the values:
∆Ssys = 38600 J / 351.35 K = -109.8 J/K
The entropy change (∆Ssys) is negative because vaporization leads to a decrease in entropy. To convert the value to kilojoules per mole per kelvin (kJ/mol*K), divide by the number of moles:
∆Ssys = -109.8 J/K ÷ 0.304 mol = -360.9 J/mol*K
Therefore, the final entropy change (∆Ssys) for the vaporization of 14.00 g of ethanol at 78.2°C is -360.9 J/K.
