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 under equilibrium conditions, in $\circ$ C?

Question

Quizplus · Accepted Answer

The boiling point of ethanol can be calculated using the Clausius-Clapeyron equation:

ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)

where P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively, ΔHvap is the enthalpy of vaporization, R is the gas constant, and ln represents the natural logarithm.

Assuming standard pressure (1 atm) as P1 and using the given values of ΔHvap and entropy of vaporization (ΔSvap = ΔHvap/Tb), we can rearrange the equation to solve for Tb:

Tb = ΔHvap / (ΔSvap + R ln(P2/P1))

Rearranging again and converting units to SI:

Tb = (38.6 kJ/mol) / (109.8 J/mol*K + 8.314 J/mol*K * ln(101325/P1))

where P1 = 1 atm.

Using the boiling point at standard pressure as a first guess (78.5°C), we can iterate in small steps until the calculated vapor pressure matches the actual vapor pressure at the given boiling point. This yields a final answer of 78.5°C = 173.3°F. Therefore, the best choice is B.

The Enthalpy and Entropy of Vaporization of Ethanol Are 38 ∘\circ∘

The Enthalpy and Entropy of Vaporization of Ethanol Are 38 $\circ$