Question

For rubidium \(\Delta H_{\mathrm{vap}}^{\circ}=69.0 \mathrm{kJ} / \mathrm{mol}\) \(686^{\circ} \mathrm{C},\) its boiling point. Calculate \(\Delta S^{\circ}, q, w,\) and \(\Delta E\) for the vaporization of 1.00 mole of rubidium at \(686^{\circ} \mathrm{C}\) and 1.00 atm pressure.

Short answer

For the vaporization of 1.00 mole of rubidium at \(686^{\circ} \mathrm{C}\) and 1.00 atm pressure, we have \(\Delta S^{\circ} = 0.072 \mathrm{kJ/mol \cdot K}\), \(q_p = 69.0 \mathrm{kJ/mol}\), \(p \Delta V = 0\), and \(\Delta E = 69.0 \mathrm{kJ/mol}\).

Step-by-step solution

Calculate \(\Delta S^{\circ}\)

Using equation 2, we can find \(\Delta S^{\circ}\) as follows: \[\Delta S^{\circ} = \frac{\Delta H_{\mathrm{vap}}^{\circ}}{T}\] First, we need to convert the temperature from Celsius to Kelvin: \(T = 686^{\circ} \mathrm{C} + 273.15 = 959.15 \mathrm{K}\) Now, we can calculate \(\Delta S^{\circ}\): \[\Delta S^{\circ} = \frac{69.0 \mathrm{kJ/mol}}{959.15 \mathrm{K}} = 0.072 \mathrm{kJ/mol \cdot K}\]

Calculate \(q_p\)

For the vaporization of rubidium at constant pressure, the heat flow (\(q_p\)) is equal to the enthalpy change (\(\Delta H^{\circ}\)): \[q_p = \Delta H_{\mathrm{vap}}^{\circ} = 69.0 \mathrm{kJ/mol}\]

Calculate \(p \Delta V\)

Now, we need to find the work done, \(p \Delta V\). Since we are given pressure, we can use equation 1: \[\Delta H_{\mathrm{vap}}^{\circ} = \Delta E + p \Delta V\] But we want to find \(p \Delta V\), so we need to rearrange the equation: \[p \Delta V = \Delta H_{\mathrm{vap}}^{\circ} - \Delta E\] We don't have \(\Delta E\) yet, but we can find it using equation 5: \[\Delta E = q + w\] Since \(w = p \Delta V\), we can substitute it into the equation: \[\Delta E = q_p - p \Delta V\] Now, we can substitute this expression for \(\Delta E\) into the equation for \(p \Delta V\): \[p \Delta V = \Delta H_{\mathrm{vap}}^{\circ} - (q_p - p \Delta V)\] \[p \Delta V = 69.0 \mathrm{kJ/mol} - (69.0 \mathrm{kJ/mol} - p \Delta V)\] \[p \Delta V = 0\]

Calculate \(\Delta E\)

Since \(p \Delta V = 0\), we can substitute this result into the expression for \(\Delta E\): \[\Delta E = q_p - p \Delta V = 69.0 \mathrm{kJ/mol} - 0 = 69.0 \mathrm{kJ/mol}\]

Summary

We have found the following results for the vaporization of 1.00 mole of rubidium at \(686^{\circ} \mathrm{C}\) and 1.00 atm pressure: 1. \(\Delta S^{\circ} = 0.072 \mathrm{kJ/mol \cdot K}\) 2. \(q_p = 69.0 \mathrm{kJ/mol}\) 3. \(p \Delta V = 0\) 4. \(\Delta E = 69.0 \mathrm{kJ/mol}\)

Key concepts

Enthalpy Change

Enthalpy change, denoted by \(\Delta H\), is a measure of the heat absorbed or released by a system at constant pressure. For the vaporization process of rubidium, its enthalpy change \(\Delta H_{\mathrm{vap}}^{\circ}\) is given as 69.0 kJ/mol. This value indicates the amount of energy required to convert one mole of liquid rubidium into its gaseous form at its boiling point, which is 686°C and 1.00 atm pressure.

When a substance undergoes a phase change under constant pressure, the heat transferred into the system (heat flow, denoted as \(q_p\)) is equal to the enthalpy change. Thus in this scenario, \(q_p = \Delta H_{\mathrm{vap}}^{\circ} = 69.0 \,\text{kJ/mol}\). Understanding enthalpy change is crucial because it describes the energy required for phase transitions, such as vaporization, melting, or sublimation.

Entropy Change

Entropy change, denoted by \(\Delta S^{\circ}\), measures the degree of disorder or randomness in a system. During vaporization, entropy typically increases because gas molecules have more freedom and occupy more space than liquid molecules. The formula used to calculate \(\Delta S^{\circ}\) in this case is:

• \(\Delta S^{\circ} = \frac{\Delta H_{\mathrm{vap}}^{\circ}}{T}\)

where \(T\) is the temperature in Kelvin.

For rubidium vaporization, the boiling point temperature was converted from Celsius to Kelvin: \(686^{\circ} \mathrm{C} + 273.15 = 959.15 \, \mathrm{K}\). Thus, the entropy change is:

• \(\Delta S^{\circ} = \frac{69.0 \, \mathrm{kJ/mol}}{959.15 \, \mathrm{K}} = 0.072 \, \mathrm{kJ/mol \cdot K}\)

The positive value of \(\Delta S^{\circ}\) reflects the increase in randomness as rubidium transitions from liquid to vapor.

Vaporization

Vaporization is the process where a liquid turns into a vapor. This can occur through boiling or evaporation. For instance, rubidium undergoes vaporization at its boiling point, which is 686°C. During this process, heat is absorbed, allowing the rubidium atoms to break free from the liquid and transition into a gaseous state.

Rubidium's enthalpy change for vaporization, \(\Delta H_{\mathrm{vap}}^{\circ} = 69.0 \, \mathrm{kJ/mol}\)), indicates the energy needed to change one mole of rubidium from liquid to gas at this temperature. In this transformation, key thermodynamic values include:

• Entropy change indicating increased disorder (\(\Delta S^{\circ}\))
• Energy related to system work (\(p \Delta V\))

In this scenario, the work performed (\(p \Delta V\)) is zero, because the volume change of the container under constant pressure doesn't lead to external pressure work.

Rubidium

Rubidium is a soft, silvery-white metal belonging to the alkali metal group in the periodic table. This element is highly reactive, especially with water, and is characterized by its low density and melting point. Rubidium's boiling point is 686°C, a crucial detail when considering its phase transitions.

In this exercise, rubidium was chosen to study the thermodynamics of vaporization. Its enthalpy of vaporization and associated entropy change showcase the energy dynamics involved in the transformation from liquid to gas. The values provided, such as \(\Delta H_{\mathrm{vap}}^{\circ} = 69.0 \, \mathrm{kJ/mol}\), are important for students to understand how specific substances behave under thermal transitions, illustrating their practical applications in scientific research and industrial processes. Rubidium, like other alkali metals, has unique properties that make it a fascinating subject of study in thermodynamics.