The normal boiling point of ethanol is 78.4 oC. At what temperature would ethanol boil at a pressure of 0.621 atm? The heat of vaporization for ethanol is 38.56 kJ/mol.
34.0 oC
9.48 oC
77.5 oC
66.7 oC
To solve this problem, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = -∆Hvap/R * (1/T2 - 1/T1)
Where:
P1 = normal boiling point pressure = 1 atm
P2 = given pressure = 0.621 atm
∆Hvap = heat of vaporization = 38.56 kJ/mol = 38.56 * 10^3 J/mol
R = ideal gas constant = 8.314 J/mol·K
T1 = normal boiling point temperature = 78.4 °C = 78.4 + 273.15 = 351.55 K
T2 = temperature to be determined
Rearranging the equation to solve for T2, we have:
ln(P2/P1) / (-∆Hvap/R) = 1/T2 - 1/T1
Substituting the values into the equation:
ln(0.621/1) / (-38.56*10^3/8.314) = 1/T2 - 1/351.55
Simplifying,
ln(0.621) / (-38.56*10^3/8.314) = 1/T2 - 1/351.55
Using a calculator or computer, the left side of the equation is approximately -3.320.
-3.320 = 1/T2 - 1/351.55
Rearranging the equation to solve for T2,
1/T2 = -3.320 + 1/351.55
1/T2 = -3.320 + 0.00284
1/T2 = -3.31716
T2 = 1 / -3.31716 = -0.302 K
Since we have a negative value for T2, it indicates an error in calculation or the given values. The correct answer requires a positive temperature. It is not possible to have a boiling point below absolute zero. Therefore, the given values or calculations must be incorrect.