Questions: The normal boiling point of liquid methyl acetate is 331 K. Assuming that its molar heat of vaporization is constant at 30.7 kJ / mol, the boiling point of CH3COOCH3 when the external pressure is 0.738 atm is

Solution Steps

The Clausius-Clapeyron equation relates the change in vapor pressure with temperature for a substance undergoing a phase change. The equation is given by:

\[ \ln \left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

where:

• \( P_1 \) and \( P_2 \) are the vapor pressures at temperatures \( T_1 \) and \( T_2 \) respectively,
• \( \Delta H_{vap} \) is the molar heat of vaporization,
• \( R \) is the universal gas constant (8.314 J/mol·K),
• \( T_1 \) and \( T_2 \) are the temperatures in Kelvin.

From the problem, we have:

• Normal boiling point \( T_1 = 331 \, \text{K} \) at \( P_1 = 1 \, \text{atm} \),
• \( P_2 = 0.738 \, \text{atm} \),
• \( \Delta H_{vap} = 30.7 \, \text{kJ/mol} = 30700 \, \text{J/mol} \).

We need to find \( T_2 \), the boiling point at \( P_2 \). Rearrange the Clausius-Clapeyron equation to solve for \( T_2 \):

\[ \ln \left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

\[ \frac{1}{T_2} = \frac{1}{T_1} + \frac{R}{\Delta H_{vap}} \ln \left( \frac{P_1}{P_2} \right) \]

Substitute the known values into the equation:

\[ \frac{1}{T_2} = \frac{1}{331} + \frac{8.314}{30700} \ln \left( \frac{1}{0.738} \right) \]

Calculate the natural logarithm and the fractions:

\[ \ln \left( \frac{1}{0.738} \right) \approx 0.3045 \]

\[ \frac{8.314}{30700} \approx 0.0002708 \]

\[ \frac{1}{T_2} = \frac{1}{331} + 0.0002708 \times 0.3045 \]

\[ \frac{1}{331} \approx 0.0030205 \]

\[ 0.0002708 \times 0.3045 \approx 0.0000825 \]

\[ \frac{1}{T_2} = 0.0030205 + 0.0000825 = 0.0031030 \]

Finally, solve for \( T_2 \):

\[ T_2 = \frac{1}{0.0031030} \approx 322.3 \, \text{K} \]

Final Answer