The enthalpy and entropy of vaporization of ethanol are 38.6 kJ/mol and 109.8 J/mol · K, respectively. What is the boiling point of ethanol, in °C?

Question

Quizplus · Accepted Answer

To find the boiling point of ethanol, we use the Clausius-Clapeyron equation:
ln(P2/P1) = ΔvapH/R * (1/T1 - 1/T2)
where P1 and T1 are the known pressure and temperature (standard conditions), P2 is the unknown pressure at the boiling point, ΔvapH is the enthalpy of vaporization, R is the gas constant, and T2 is the unknown boiling point temperature.
Converting the given values to SI units, we have:
ΔvapH = 38.6 * 1000 J/mol
ΔvapS = 109.8 J/mol K
R = 8.314 J/mol K
P1 = 1 atm = 101325 Pa
T1 = 273 K
Substituting these values and solving for T2, we get:
T2 = ΔvapH/R * (1/T1 - ln(P2/P1)/ΔvapS)
At the boiling point, P2 = 1 atm, so:
T2 = ΔvapH/R * (1/T1 - ln(1/101325)/ΔvapS) = 351.5 K
Converting to Celsius, we get:
T2 = 78.5°C

The Enthalpy and Entropy of Vaporization of Ethanol Are 38