Problem 51 - General Chemistry

Question

White phosphorus, \(\mathrm{P}_{4}\), is normally a white, waxy solid melting at \(44^{\circ} \mathrm{C}\) to a colorless liquid. The liquid has a vapor pressure of \(400.0 \mathrm{mmHg}\) at \(251.0^{\circ} \mathrm{C}\) and \(760.0 \mathrm{mmHg}\) at \(280.0^{\circ} \mathrm{C}\). What is the heat of vaporization of this substance?

Step-by-Step Solution

Answer

The heat of vaporization for white phosphorus is approximately 50.36 kJ/mol.

Step 1: Understand the Clausius-Clapeyron Equation

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

Step 2: Convert Temperatures to Kelvin

The temperatures must be converted from Celsius to Kelvin since the Clausius-Clapeyron equation requires all temperatures to be in the absolute scale. To convert a temperature from Celsius to Kelvin, you use the formula: \[T(\text{K}) = T(\degree\text{C}) + 273.15\]So, \(251^{\circ}\text{C} = 524.15\) K and \(280^{\circ}\text{C} = 553.15\) K.

Step 3: Plug Values into the Clausius-Clapeyron Equation

Substitute the given values into the equation: \(P_1 = 400.0\) mmHg, \(P_2 = 760.0\) mmHg, \(T_1 = 524.15\) K, and \(T_2 = 553.15\) K. This gives:\[\ln \left( \frac{760.0}{400.0} \right) = -\frac{\Delta H_{vap}}{8.314} \left( \frac{1}{553.15} - \frac{1}{524.15} \right)\]

Step 4: Calculate the Natural Logarithm Ratio

Calculate \(\ln \left( \frac{760.0}{400.0} \right)\) using a calculator:\[\ln \left( \frac{760.0}{400.0} \right) \approx 0.606\]

Step 5: Calculate the Inverse Temperatures Difference

Calculate \(\left( \frac{1}{553.15} - \frac{1}{524.15} \right)\):\[\frac{1}{553.15} \approx 0.001808 \text{ K}^{-1}, \quad \frac{1}{524.15} \approx 0.001908 \text{ K}^{-1}\]\[0.001908 - 0.001808 = 0.0001 \text{ K}^{-1}\]

Step 6: Solve for the Heat of Vaporization

Rearrange the equation to solve for \(\Delta H_{vap}\):\[0.606 = -\frac{\Delta H_{vap}}{8.314} \times 0.0001\]Multiply both sides by \(8.314\) and divide by \(0.0001\) to isolate \(\Delta H_{vap}\):\[\Delta H_{vap} = -0.606 \times \frac{8.314}{0.0001} = 50355.84 \, \text{J/mol} = 50.36 \, \text{kJ/mol}\]

Step 7: Finalize the Answer

The heat of vaporization for white phosphorus \(\mathrm{P}_4\) is approximately \(50.36 \, \text{kJ/mol}\). This is a measure of the energy required to vaporize one mole of \(\mathrm{P}_4\) from a liquid to a gas at its boiling point.

Key Concepts

Heat of Vaporization

The heat of vaporization is a key concept in thermodynamics, referring to the energy required to transform a given quantity of a substance from a liquid into a gas at constant temperature and pressure. This energy is typically expressed in joules per mole (J/mol) or kilojoules per mole (kJ/mol). Understanding this concept helps explain how substances change state.

When a liquid absorbs heat, its molecules gain enough energy to overcome intermolecular forces, allowing them to escape into the gas phase. The heat of vaporization quantifies this energy requirement, indicating how much heat is needed to convert a specific amount of a liquid into vapor.

• Relation to Pressure: The stronger the intermolecular forces, the higher the heat of vaporization. Liquids with high vapor pressures generally have lower heats of vaporization.
• Temperature Dependency: Heat of vaporization can vary with temperature, but for small temperature ranges, it's often treated as constant.
• Applications: Calculating the heat of vaporization is crucial in fields like chemical engineering and meteorology, aiding in the design of processes like distillation and understanding weather patterns.

Vapor Pressure

Vapor pressure is a measure of a liquid's volatility and represents the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. It indicates how easily a liquid can evaporate, which is critical for understanding phase transitions and equilibria.

Vapor pressure increases with temperature, as the kinetic energy of the molecules increases, allowing more molecules to escape the liquid surface into the gaseous phase. The Clausius-Clapeyron equation links vapor pressure and temperature, helping to calculate changes in vapor pressure with temperature variations.

• Impact of Temperature: Warmer liquids have higher vapor pressures due to increased molecular motion.
• Substance Characteristics: Different substances have varying vapor pressures based on molecular structure and intermolecular forces.
• Practical Implications: Vapor pressure is used in various applications, such as predicting boiling points and refining chemical mixtures.

Temperature Conversion

Temperature conversion is crucial for thermodynamics problems because different equations, like the Clausius-Clapeyron equation, require temperature inputs in specific units, typically Kelvin. Understanding how to convert temperatures from Celsius to Kelvin is fundamental.

To convert Celsius to Kelvin, simply add 273.15 to the Celsius value. This conversion allows temperatures to be on an absolute scale, which is necessary for thermodynamic calculations to ensure all values are positive and proportional to the energy within the system.

• Formula Use: The formula for conversion is straightforward: \[ T(K) = T(^\circ C) + 273.15 \]
• Importance: Kelvin is the SI unit for temperature measurement, making it indispensable in scientific calculations.
• Applications: Accurate temperature conversions are vital in experimental settings, ensuring consistency and accuracy in measurements across different conditions.