Question

The vapor pressure of \(\mathrm{SiCl}_{4}\) is \(100 \mathrm{~mm} \mathrm{Hg}\) at \(5.4^{\circ} \mathrm{C}\), and the normal boiling point is \(57.7{ }^{\circ} \mathrm{C}\). What is \(\Delta H_{\mathrm{vap}}\) for \(\mathrm{SiCl}_{4}\) in \(\mathrm{kJ} / \mathrm{mol}\) ?

Short answer

The heat of vaporization, \(\Delta H_{\text{vap}}\), is \(112.47 \text{kJ/mol}\).

Step-by-step solution

Understand the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is used to relate the vapor pressure of a substance to its temperature and its heat of vaporization. It is given by: \[ \ln \left( \frac{P_2}{P_1} \right) = \frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] where \( P_1 \) and \( P_2 \) are the vapor pressures at temperatures \( T_1 \) and \( T_2 \) respectively, \( \Delta H_{\text{vap}} \) is the enthalpy of vaporization, and \( R \) is the gas constant \( 8.314 \text{ J/mol·K} \).

Gather Given Information

We are given: \( P_1 = 100 \text{ mm Hg} \) at \( T_1 = 5.4^{\circ} \text{C} = 278.4 \text{ K} \) (since \( 5.4 + 273 \)). Normal boiling point indicates vapor pressure \( P_2 = 760 \text{ mm Hg} \) at \( T_2 = 57.7^{\circ} \text{C} = 330.7 \text{ K} \).

Substitute Values in Clausius-Clapeyron Equation

Substitute the known values into the Clausius-Clapeyron equation. \[ \ln \left( \frac{760}{100} \right) = \frac{\Delta H_{\text{vap}}}{8.314} \left( \frac{1}{278.4} - \frac{1}{330.7} \right) \]

Calculate the Natural Logarithm and Temperatures

Calculate \( \ln(7.6) \), which is approximately \( 2.028 \). Find the difference in the reciprocals of the temperatures: \[ \frac{1}{278.4} - \frac{1}{330.7} \approx 0.000452 - 0.000302 = 0.000150 \text{ K}^{-1} \]

Solve for \( \Delta H_{\text{vap}} \)

Re-arrange to solve for \( \Delta H_{\text{vap}} \): \[ \Delta H_{\text{vap}} = \frac{2.028 \times 8.314}{0.000150} \] Calculate the result: \[ \Delta H_{\text{vap}} \approx 112,467 \text{ J/mol} = 112.47 \text{ kJ/mol} \]

Conclusion

The heat of vaporization, \( \Delta H_{\text{vap}} \), for \( \text{SiCl}_4 \) is \( 112.47 \text{ kJ/mol} \).

Key concepts

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase. It increases with temperature since more molecules have enough energy to escape into the vapor phase. For example, in the Clausius-Clapeyron exercise provided, the vapor pressure of silicon tetrachloride (\( \text{SiCl}_4 \)) is given as \(100 \text{ mm Hg}\) at \(5.4^{\circ} \text{C}\). At its boiling point, the vapor pressure reaches \(760 \text{ mm Hg}\), matching atmospheric pressure.

This is a key concept in understanding boiling and evaporation processes:

• Higher vapor pressure means a substance evaporates more easily.
• It depends on temperature; as temperature increases, vapor pressure rises.

Understanding vapor pressure helps in applications like predicting weather patterns and designing distillation processes.

Enthalpy of Vaporization

The enthalpy of vaporization \(\Delta H_{\text{vap}}\) is the energy required to convert one mole of a liquid into vapor without a change in temperature. In the exercise, using the Clausius-Clapeyron equation gave us \(\Delta H_{\text{vap}}\) for \(\text{SiCl}_4\) as \(112.47 \text{ kJ/mol}\).

This energy reflects the strength of intermolecular forces in the liquid:

• Larger values indicate stronger forces needing more energy to vaporize.
• It's important in processes like distillation, where separation of components based on boiling points occurs.

Knowing the enthalpy of vaporization helps in understanding heat transfer and energy efficiency in industrial processes.

Boiling Point

The boiling point is the temperature at which a liquid's vapor pressure equals the surrounding atmospheric pressure, causing it to convert into vapor. For \(\text{SiCl}_4\), the boiling point is given as \(57.7^{\circ} \text{C}\) in the exercise. At this temperature, the vapor pressure is \(760 \text{ mm Hg}\).

• At high altitudes, atmospheric pressure is lower, so boiling points are lower.
• This concept explains why cooking times vary at different altitudes.

Boiling points are crucial in applications such as chemical manufacturing and culinary arts, where precise temperature control is essential.

Temperature Conversion

Temperature conversion is essential in scientific calculations, especially when using equations like Clausius-Clapeyron, which require temperature in Kelvin. In the exercise, the temperatures \(5.4^{\circ} \text{C}\) and \(57.7^{\circ} \text{C}\) were converted to Kelvin as \(278.4 \text{ K}\) and \(330.7 \text{ K}\), respectively.

Here's how to convert:

• Add \(273.15\) to a Celsius temperature to get Kelvin.
• Subtract \(273.15\) to convert from Kelvin to Celsius.

Temperature conversion is fundamental in chemistry, physics, and engineering, ensuring consistency in data when using different scales.