Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. Using the Clausius-Clapeyron equation, what is the vapor pressure of ethanol at 15 °C?

• A. 1.02 atm
• B. 0.12 atm
• C. 0.45 atm
• D. 0.78 atm

Answer: (C)

Verified step by step guidance

1. Start by understanding the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at two different temperatures. The equation is: ${\ln }_P=\frac{-\mathrm{\Delta H}/R}{(\frac{1}{T_2}-\frac{1}{T_1})}$, where $\mathrm{\Delta H}$ is the heat of vaporization, $R$ is the universal gas constant, and $T_1$ and $T_2$ are the temperatures in Kelvin.
2. Convert the temperatures from Celsius to Kelvin. The normal boiling point of ethanol is 78.4 °C, which converts to $351.4$ K. The temperature at which you want to find the vapor pressure is 15 °C, which converts to $288.15$ K.
3. Use the Clausius-Clapeyron equation to find the natural logarithm of the ratio of the vapor pressures at the two temperatures. Plug in the values: $\mathrm{\Delta H}$ = 38.56 kJ/mol (convert to J/mol by multiplying by 1000), $R$ = 8.314 J/(mol·K), $T_1$ = 351.4 K, and $T_2$ = 288.15 K.
4. Calculate the difference in the reciprocals of the temperatures: $\frac{1}{288.15}$ - $\frac{1}{351.4}$. This will be used in the Clausius-Clapeyron equation.
5. Solve for the vapor pressure at 15 °C by rearranging the Clausius-Clapeyron equation to find $P_2$, using the known vapor pressure at the boiling point (1 atm) and the calculated value from the previous steps.