Question

Carbon tetrachloride melts at \(250 .\) K. The vapor pressure of the liquid is 10,539 Pa at \(290 .\) K and 74,518 Pa at 340. K. The vapor pressure of the solid is \(270 .\) Pa at \(232 \mathrm{K}\) and 1092 Pa at \(250 .\) K. a. Calculate \(\Delta H_{\text {vaporization}}\) and \(\Delta H_{\text {sublimation}}\). b. Calculate \(\Delta H_{\text {fusion}}\). c. Calculate the normal boiling point and \(\Delta S_{\text {vaporization}}\) at the boiling point. d. Calculate the triple point pressure and temperature.

Short answer

a. \(\Delta H_{vaporization} ≈ 33,650\) J/mol and \(\Delta H_{sublimation} ≈ 56,210\) J/mol b. \(\Delta H_{fusion} ≈ 22,560\) J/mol c. Normal boiling point ≈ 351.9 K, \(\Delta S_{vaporization} ≈ 95.55\) J/(mol K) d. Triple point pressure ≈ 1020 Pa and triple point temperature ≈ 249 K

Step-by-step solution

a. Calculation of \(\Delta H_{\text {vaporization}}\) and \(\Delta H_{\text {sublimation}}\)

To calculate \(\Delta H_{vaporization}\), we will use the Clausius-Clapeyron equation and the given data for the vapor pressure of liquid carbon tetrachloride: \[ \ln \left( \frac{74,518\, \text{Pa}}{10,539\, \text{Pa}} \right) = - \frac{\Delta H_{vaporization}}{R} \left( \frac{1}{340 \, \text{K}} - \frac{1}{290 \, \text{K}} \right) \] Solving for \(\Delta H_{vaporization}\), we get: \(\Delta H_{vaporization} ≈ 33,650\) J/mol Next, we will now calculate \(\Delta H_{sublimation}\) using the given data for the vapor pressure of solid carbon tetrachloride: \[ \ln \left( \frac{1092 \, \text{Pa}}{270 \, \text{Pa}} \right) = - \frac{\Delta H_{sublimation}}{R} \left( \frac{1}{250 \, \text{K}} - \frac{1}{232 \, \text{K}} \right) \] Solving for \(\Delta H_{sublimation}\), we get: \(\Delta H_{sublimation} ≈ 56,210\) J/mol

b. Calculation of \(\Delta H_{\text {fusion}}\)

Now, we will calculate \(\Delta H_{fusion}\) using the relationship between the enthalpies: \(\Delta H_{fusion} = \Delta H_{sublimation} - \Delta H_{vaporization}\) \(\Delta H_{fusion} ≈ 56,210 \, \text{J/mol} - 33,650 \, \text{J/mol} = 22,560 \, \text{J/mol}\)

c. Calculation of normal boiling point and \(\Delta S_{\text {vaporization}}\)

For the normal boiling point, we will set the vapor pressure of the liquid equal to 1 atm, and solve for the temperature using the Clausius-Clapeyron equation: \[ \ln \left( \frac{101,325 \, \text{Pa}}{10,539 \, \text{Pa}} \right) = - \frac{33,650 \, \text{J/mol}}{R} \left( \frac{1}{T_{bp}} - \frac{1}{290 \, \text{K}} \right) \] Solving for the boiling point temperature, we get: \(T_{bp} ≈ 351.9\) K Next, we can use the definition of entropy change for the vaporization process while considering the normal boiling point temperature: \[ \Delta S_{vaporization} = \frac{\Delta H_{vaporization}}{T_{bp}} \] \(\Delta S_{vaporization} ≈ \frac{33,650 \, \text{J/mol}}{351.9 \, \text{K}} = 95.55 \, \text{J/(mol K)}\)

d. Calculation of triple point pressure and temperature

We can apply the Clausius-Clapeyron equation between the solid and liquid phases at the triple point: \[ \ln \left( \frac{P_{triple}}{1092 \, \text{Pa}} \right) = - \frac{\Delta H_{fusion}}{R} \left( \frac{1}{T_{triple}} - \frac{1}{250 \, \text{K}} \right) \] To eliminate \(P_{triple}\), we use the Clausius-Clapeyron equation between liquid and solid states at the triple point: \[ \ln \left( \frac{P_{triple}}{270 \, \text{Pa}} \right) = - \frac{\Delta H_{sublimation}}{R} \left( \frac{1}{T_{triple}} - \frac{1}{232 \, \text{K}} \right) \] Divide both equations and solve for \(T_{triple}\), we get: \(T_{triple} ≈ 249\) K Now, we can plug this temperature back into either equation to solve for \(P_{triple}\). Using the second equation, we get: \(P_{triple} ≈ 1020\) Pa The results are: a. \(\Delta H_{vaporization} ≈ 33,650\) J/mol and \(\Delta H_{sublimation} ≈ 56,210\) J/mol b. \(\Delta H_{fusion} ≈ 22,560\) J/mol c. Normal boiling point ≈ 351.9 K, \(\Delta S_{vaporization} ≈ 95.55\) J/(mol K) d. Triple point pressure ≈ 1020 Pa and triple point temperature ≈ 249 K

Key concepts

Vapor Pressure

Vapor pressure refers to the pressure created by the vapor formed above a liquid (or solid) in a closed container. It's an essential concept in understanding how substances transition between different phases, like liquid to gas.

When a liquid is in a container, molecules continuously escape into the vapor phase. At equilibrium, the rate of evaporation equals the rate of condensation. This equilibrium pressure is the vapor pressure.

Important characteristics of vapor pressure include:

• It increases with temperature, meaning more molecules have enough energy to escape into the vapor phase.
• Each substance has a unique vapor pressure at a given temperature, influenced by its molecular properties.

This concept is crucial for calculating phase change enthalpies, like enthalpy of vaporization, using equations such as Clausius-Clapeyron.

Enthalpy of Vaporization

Enthalpy of vaporization, (\( \Delta H_{vaporization} \), is the energy required to convert a liquid into a gas at a constant pressure. It is a key thermodynamic quantity to understand how substances absorb energy during phase transitions from liquid to vapor.

Considerations for enthalpy of vaporization include:

• It's typically measured in joules per mole (J/mol).
• The value depends on temperature and pressure, often determined by using the Clausius-Clapeyron equation.

In context, the enthalpy of vaporization for carbon tetrachloride was calculated using the pressures at two different temperatures. By applying the Clausius-Clapeyron equation, one determines the change in enthalpy necessary to vaporize the liquid.

Enthalpy of Sublimation

Enthalpy of sublimation, (\(\Delta H_{sublimation} \)), is the energy required to change a solid directly into a gas without passing through the liquid state. Understanding this concept helps explain how substances can skip melting under specific conditions.

Important facts about enthalpy of sublimation include:

• Commonly expressed in J/mol, it provides a measure of the energy needed for phase change from solid to gas.
• The Clausius-Clapeyron equation is used to determine this value by considering the vapor pressures at two temperatures.

Carbon tetrachloride's enthalpy of sublimation was calculated using vapor pressure data for the solid. This information is key for understanding energy changes at different temperature stages.

Enthalpy of Fusion

Enthalpy of fusion, (\( \Delta H_{fusion} \), is the energy required to change a solid into a liquid. Unlike sublimation, fusion involves only passing through one phase change and is crucial for understanding melting phenomena.

Attributes of enthalpy of fusion include:

• It's expressed in J/mol and denotes the energy needed to convert solids to liquids at their melting point.
• Calculated as the difference between enthalpy of sublimation and vaporization, reflecting the energy balance during melting.

In the case of carbon tetrachloride, this was determined by simply subtracting the enthalpy of vaporization from the enthalpy of sublimation, providing insights into its melting dynamics.