Ethanol has an enthalpy of vaporization of 42.3 kJ/mol.The compound has a vapor pressure of 1.00 atm at 78.3 °C.At what temperature is the vapor pressure equal to 0.800 atm? (R = 8.314 J/K ⋅ mol)
A) -83.8 °C
B) -24.4 °C
C) 62.6 °C
D) 73.0 °C
E) 78.0 °C

Question

Quizplus · Accepted Answer

First, we need to use the Clausius-Clapeyron equation to determine the temperature at which the vapor pressure is 0.800 atm:
ln(P₂/P₁) = -(ΔH_vap/R) × (1/T₂ - 1/T₁)
Since we know P₁ (1.00 atm) and T₁ (78.3°C + 273.15 = 351.45 K) and P₂ (0.800 atm), we can solve for T₂:
ln(0.800/1.00) = -(42.3 × 10³/R) × (1/T₂ - 1/351.45)
-0.2231 = -(42.3 × 10³/8.314) × (1/T₂ - 1/351.45)
-0.2231 = -5095.5 × (1/T₂ - 0.002839)
1/T₂ = 0.0135 + 1/351.45
1/T₂ = 0.016939
T₂ = 59.0°C + 273.15 = 332.15 K
Therefore, the temperature at which the vapor pressure of ethanol is 0.800 atm is 59.0°C, which is closest to answer choice D: 73.0°C.

Ethanol Has an Enthalpy of Vaporization of 42