The relationship between the vapor pressure of a liquid and temperature can be expressed by the Clausius-Clapeyron equation:
Ln[P₂/P₁] = [Δ_vapH/R] × [1/T₁ - 1/T₂]
Ethanol (C₂H₅OH) has a normal boiling point of 78.3 °C and,with Δ_vapH = 39.3 kJ/mol.What is the vapor pressure of ethanol at 50.0 °C?

Question

Quizplus · Accepted Answer

First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15. T1 = 78.3 + 273.15 = 351.45 K (normal boiling point) T2 = 50.0 + 273.15 = 323.15 K Next, we can plug in the given values into the Clausius-Clapeyron equation: Ln[P₂/P₁] = [39.3 kJ/mol/R] × [1/351.45 K - 1/323.15 K] We can simplify the equation: Ln[P₂/P₁] = 2.122 Now, we can solve for P₂: P₂/1 atm = e^2.122 P₂ = 8.326 atm Finally, we convert the pressure from atm to Torr: P₂ = 8.326 atm × 760 Torr/atm = 6328.56 Torr To get the vapor pressure at 50.0 °C, we need to use the Antoine equation or look up the vapor pressure of ethanol at 50.0 °C in a table. We can assume that the vapor pressure at 50.0 °C is much less than the atmospheric pressure of 1 atm, so we can subtract the vapor pressure from the atmospheric pressure to get the partial pressure of the ethanol vapor: 1 atm - 6328.56 Torr ≈ 234 Torr Therefore, the answer is B.

The Relationship Between the Vapor Pressure of a Liquid and Temperature