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.

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.

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.

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 \(\Delta E\).

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