Question

For rubidium \(\Delta H_{\text {vap }}^{\circ}=69.0 \mathrm{kJ} / \mathrm{mol}\) at \(686^{\circ} \mathrm{C},\) 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 its boiling point and 1.00 atm pressure, the following values are calculated: - Entropy change \(\Delta S^\circ = 71.95 \frac{\mathrm{J}}{\mathrm{mol \cdot K}}\) - Heat q = \(69.0\times10^3\) J - Work w \(\approx 0\) - Change in internal energy \(\Delta E = 69.0\times10^3\) J

Step-by-step solution

Calculate the entropy change \(\Delta S^\circ\)

Enthalpy of vaporization \(\Delta H_{\text {vap }}^\circ\) and entropy change \(\Delta S^\circ\) are related by the following formula at boiling point: \[\Delta H_{\text {vap }}^\circ = T\Delta S^\circ\] First, we need to convert the boiling point temperature from Celsius to Kelvin: \[T = 686 ^\circ C + 273.15 K = 959.15 K\] Now, we can find the entropy change by rearranging the formula: \[\Delta S^\circ = \frac{\Delta H_{\text {vap }}^\circ}{T}\] Using the provided values of \(\Delta H_{\text {vap }}^\circ\) and \(T\): \[\Delta S^\circ = \frac{69.0 \mathrm{kJ/mol}}{959.15 \mathrm K}\] In order to have consistent units, we need to convert kJ to J: \[\Delta S^\circ = \frac{69.0\times10^3 \mathrm J/mol}{959.15 \mathrm K}\] \[\Delta S ^\circ = 71.95 \frac{\mathrm{J}}{\mathrm{mol \cdot K}}\] Now, let's find the heat q.

Calculate the heat q

The heat required for vaporization is equal to the enthalpy change for the process: \[q = \Delta H_{\text {vap }}^\circ = 69.0 \,\mathrm{kJ/mol}\] For 1.00 mole of rubidium: \[q = 69.0\times10^3 \, \mathrm J\] Now, let's find the work w.

Calculate the work w

For vaporization at constant pressure, the work done is given by the following formula: \[w = -P\Delta V\] However, since the volume change during vaporization is difficult to determine and the work done is negligible compared to the heat q, we can assume: \[w \approx 0\] Finally, let's find the change in internal energy \(\Delta E\).

Determine the change in internal energy \(\Delta E\)

We can use the first law of thermodynamics to calculate the change in internal energy: \[\Delta E = q + w\] Substituting the values of q and w calculated in steps 2 and 3: \[\Delta E = 69.0 \times 10^3 \, \mathrm J + 0\] \[\Delta E = 69.0 \times 10^3 \, \mathrm J\] Now we have calculated all the required values: - Entropy change \(\Delta S^\circ = 71.95 \frac{\mathrm{J}}{\mathrm{mol \cdot K}}\) - Heat q = \(69.0\times10^3\) J - Work w \(\approx 0\) - Change in internal energy \(\Delta E = 69.0\times10^3\) J

Key concepts

Entropy Change

Entropy is a measure of the disorder or randomness in a system. During the vaporization of a substance like rubidium, the entropy increases because the molecules move from a more ordered liquid phase to a disordered gaseous phase.

The formula that relates enthalpy of vaporization \(\Delta H_{\text{vap}}^\circ\) and entropy change \(\Delta S^\circ\) is given by:

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

At the boiling point, the temperature \(T\) is crucial because it determines how much disorder is introduced as heat energy is added.

For rubidium, the temperature is converted to Kelvin because thermodynamic calculations are generally performed in this unit (Kelvin). To find \(\Delta S^\circ\):

• Convert boiling point temperature from Celsius to Kelvin.
• Use the enthalpy of vaporization and convert it from kJ to J for consistency.
• Divide the enthalpy by temperature to get the change in entropy.

This calculation tells us that for rubidium's vaporization, entropy change \(\Delta S^\circ\) is approximately 71.95 \(\text{J/mol·K}\), signifying a sensible increase in disorder.

Enthalpy of Vaporization

The enthalpy of vaporization is the amount of energy required to convert a mole of a liquid into a vapor at constant pressure. It is an endothermic process, which means that it absorbs heat from the surroundings. This is why the enthalpy of vaporization is always a positive value.

In this exercise, for the vaporization of rubidium:

• The enthalpy of vaporization \(\Delta H_{\text{vap}}^\circ\) is provided as 69.0 kJ/mol.
• It reflects the energy needed to overcome intermolecular forces in the liquid state.
• The conversion to \(69.0 \times 10^3\) J/mol helps with uniformity in math operations.

This value indicates that rubidium requires this amount of energy for each mole to change from liquid to gas, illustrating how volatile rubidium is at its boiling point.

Internal Energy

Internal energy \(\Delta E\) of a system involves all types of energy contained within that system, including kinetic and potential energy of molecules. The change in internal energy \(\Delta E\) for a process can be deduced from the first law of thermodynamics:

• \(\Delta E = q + w\)

Here, \(q\) is the heat exchanged and \(w\) is the work done. During vaporization at constant pressure:

• The process absorbs heat, \(q\), equal to the enthalpy of vaporization.
• The work \(w\) is nearly zero because the expansion work is negligible.

Thus, for rubidium vaporization: \(\Delta E\) is mainly determined by the heat term, as work is nearly negligible. Thus, \(\Delta E\) equals 69.0 kJ (or \(69.0 \times 10^3\) J), which is the same as the heat input, indicating that all energy used goes into changing phase without significant expansion work.