Using the two-point form of the Clausius-Clapeyron equation, calculate the vapor pressure of methanol at 12.0°C given its normal boiling point is 64.6°C and its heat of vaporization is 35.2 kJ/mol.

• A. 0.12 atm
• B. 1.00 atm
• C. 0.35 atm
• D. 0.75 atm

Answer: (C)

Verified step by step guidance

1. Identify the two temperatures involved: the normal boiling point (64.6°C) and the temperature at which you want to find the vapor pressure (12.0°C). Convert these temperatures from Celsius to Kelvin by adding 273.15.
2. Use the two-point form of the Clausius-Clapeyron equation: \( \ln \left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \), where \( P_1 \) is the vapor pressure at the normal boiling point (1.00 atm), \( P_2 \) is the vapor pressure at 12.0°C, \( \Delta H_{vap} \) is the heat of vaporization, and \( R \) is the ideal gas constant (8.314 J/mol·K).
3. Substitute the known values into the equation: \( \Delta H_{vap} = 35200 \) J/mol, \( R = 8.314 \) J/mol·K, \( T_1 = 337.75 \) K, and \( T_2 = 285.15 \) K.
4. Rearrange the equation to solve for \( P_2 \): \( P_2 = P_1 \times \exp \left( -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \right) \).
5. Calculate the expression inside the exponential function and then compute \( P_2 \) to find the vapor pressure of methanol at 12.0°C.