Octane has a vapor pressure of 40. torr at 45.1°C and 400. torr at 104.0°C. What is its heat of vaporization?
A)39.0 kJ/mol
B)46.0 kJ/mol
C)590 kJ/mol
D)710 kJ/mol
E)None of these choices are correct.

Question

Quizplus · Accepted Answer

The Clausius-Clapeyron equation can be used to calculate the heat of vaporization ($ \Delta H_{vap} $) from the vapor pressure at two different temperatures. The equation in its linear form is:$$ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) $$Given:- $ P_1 = 40 $ torr = $ 40/760 $ atm (since 1 atm = 760 torr)- $ P_2 = 400 $ torr = $ 400/760 $ atm- $ T_1 = 45.1°C = 318.25 K $ (since $ T(K) = T(°C) + 273.15 $)- $ T_2 = 104.0°C = 377.15 K $Rearranging the equation to solve for $ \Delta H_{vap} $ gives:$$ \Delta H_{vap} = -R \left(\frac{\ln(P_2/P_1)}{(1/T_2 - 1/T_1)}\right) $$Using $ R = 8.314 \, J/(mol·K) $ and converting the final answer to kJ/mol:$$ \Delta H_{vap} = -8.314 \left(\frac{\ln(400/760 / (40/760))}{(1/377.15 - 1/318.25)}\right) \times 10^{-3} $$$$ \Delta H_{vap} \approx 39.0 \, kJ/mol $$Therefore, the correct answer is 39.0 kJ/mol, which corresponds to choice A.

Octane Has a Vapor Pressure of 40