Problem 85 - Chemistry The Molecular Science

Question

At \(20 .{ }^{\circ} \mathrm{C}\) the vapor pressure of white phosphorus is \(0.0254 \mathrm{mmHg}\); at \(40 .{ }^{\circ} \mathrm{C}\) it is \(0.133 \mathrm{mmHg}\). Use the Clausius-Clapeyron equation to estimate the heat of sublimation (J/mol) of white phosphorus.

Step-by-Step Solution

Answer

The heat of sublimation of white phosphorus is approximately 63,123 J/mol.

Step 1: Understand the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation relates the change in vapor pressure with temperature to the enthalpy change of the phase transition. It is given as \( \ln \left( \frac{P_2}{P_1} \right) = \frac{\Delta H_{sub}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \), where \( P_1 \) and \( P_2 \) are the vapor pressures at temperatures \( T_1 \) and \( T_2 \) respectively, \( R \) is the gas constant, and \( \Delta H_{sub} \) is the heat of sublimation.

Step 2: Convert Temperature to Kelvin

Convert the given temperatures from Celsius to Kelvin using the formula \( T(K) = T(°C) + 273.15 \). For \( 20^{\circ}C \), \( T_1 = 293.15 \) K; for \( 40^{\circ}C \), \( T_2 = 313.15 \) K.

Step 3: Identify Given Vapor Pressures

Assign the given vapor pressures to appropriate variables: \( P_1 = 0.0254 \) mmHg at \( T_1 = 293.15 \) K and \( P_2 = 0.133 \) mmHg at \( T_2 = 313.15 \) K.

Step 4: Substitute Values into the Clausius-Clapeyron Equation

Substitute \( P_1 \), \( P_2 \), \( T_1 \), and \( T_2 \) into the equation \( \ln \left( \frac{P_2}{P_1} \right) = \frac{\Delta H_{sub}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \). This becomes \( \ln \left( \frac{0.133}{0.0254} \right) = \frac{\Delta H_{sub}}{8.314} \left( \frac{1}{293.15} - \frac{1}{313.15} \right) \).

Step 5: Solve for the Heat of Sublimation

Calculate \( \ln \left( \frac{P_2}{P_1} \right) \) which is \( \ln(5.2362) \approx 1.6553 \). Evaluate \( \frac{1}{293.15} - \frac{1}{313.15} \approx -0.0002183 \). Substitute into the equation and solve for \( \Delta H_{sub} \): \( 1.6553 = \frac{\Delta H_{sub}}{8.314} (-0.0002183) \). Rearranging gives \( \Delta H_{sub} = \frac{1.6553 \times 8.314}{-0.0002183} \approx 63123 \) J/mol.

Key Concepts

Heat of Sublimation

The heat of sublimation refers to the energy required to transform a substance from a solid to a gas without passing through a liquid state. This process occurs when the molecules of the solid gain enough energy to overcome intermolecular forces.
It's an important concept in thermodynamics, as it helps determine how much energy is needed during this phase change.

• The Clausius-Clapeyron equation is frequently used to calculate the heat of sublimation by relating vapor pressure and temperature changes.
• It ties together the physical states of matter, temperature changes, and energy exchange, making it valuable for various applications.
• The heat of sublimation is often expressed in J/mol, indicating the energy needed per mole of substance.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature. It's a crucial factor in understanding phase transitions such as sublimation.
Higher vapor pressure implies that a substance more readily changes from a solid to gas.

• Substances with high vapor pressures at a particular temperature transition to the gas phase easily.
• This concept is significant for understanding weather patterns and various industrial applications.
• In the exercise, vapor pressure readings at two temperatures help in finding the heat of sublimation using the Clausius-Clapeyron equation.

Temperature Conversion

Converting temperatures is essential to solving thermodynamic problems accurately. In many cases, such as with the Clausius-Clapeyron equation, temperature must be expressed in Kelvin.
The conversion from Celsius to Kelvin is straightforward: just add 273.15.

• Kelvin is the SI unit of temperature and is necessary for calculative consistency in physics and chemistry.
• This conversion ensures all temperature measurements are absolute, simplifying calculations.
• In the exercise, converting 20°C to 293.15 K and 40°C to 313.15 K allowed for the correct application of the Clausius-Clapeyron equation.

Enthalpy Change

Enthalpy change, a key concept in thermodynamics, represents the heat change at constant pressure during a reaction or phase change. It helps us understand energy transfer essential for processes such as sublimation.
For sublimation, the enthalpy change is the heat of sublimation.

• The Clausius-Clapeyron equation bridges this concept by relating enthalpy change with observable parameters like vapor pressure and temperature.
• This concept is crucial in both natural phenomena and technological processes, including refrigeration and chemical manufacturing.
• Understanding enthalpy changes can predict whether a process will release or absorb heat, providing valuable insights for engineers and scientists.