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.

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

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

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 \(p \Delta V\): \[p \Delta V = \Delta H_{\mathrm{vap}}^{\circ} - (q_p - p \Delta V)\] \[p \Delta V = 69.0 \mathrm{kJ/mol} - (69.0 \mathrm{kJ/mol} - p \Delta V)\] \[p \Delta V = 0\]

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

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. \(\Delta E = 69.0 \mathrm{kJ/mol}\)