Question

for propane, the $\delta h^\circ$ of vaporization is 22.9 kj/mol and the $\delta s^\circ$ of vaporization is 99.0 j·mol⁻¹k⁻¹. at 1.00 atm, what is the boiling point of propane, in k?

Explanation:

Step1: Recall the relationship at boiling point

At boiling point, the Gibbs free energy change \(\Delta G^{\circ}\) for the vaporization process is zero (equilibrium between liquid and gas phases). The formula relating \(\Delta G^{\circ}\), \(\Delta H^{\circ}\), \(\Delta S^{\circ}\), and temperature \(T\) is \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\). Setting \(\Delta G^{\circ} = 0\) (at boiling point), we get \(0=\Delta H^{\circ}-T\Delta S^{\circ}\), which can be rearranged to \(T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}\).

Step2: Convert units of \(\Delta H^{\circ}\)

Given \(\Delta H^{\circ}=22.9\space kJ/mol\). We need to convert this to \(J/mol\) to match the units of \(\Delta S^{\circ}\) (which is in \(J\cdot mol^{-1}K^{-1}\)). Since \(1\space kJ = 1000\space J\), \(\Delta H^{\circ}=22.9\times1000 = 22900\space J/mol\).

Step3: Calculate the boiling point \(T\)

Now we use the formula \(T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}\). Substituting \(\Delta H^{\circ}=22900\space J/mol\) and \(\Delta S^{\circ}=99.0\space J\cdot mol^{-1}K^{-1}\) into the formula:
\(T=\frac{22900\space J/mol}{99.0\space J\cdot mol^{-1}K^{-1}}\approx231.31\space K\) (we can also keep more precise calculation: \(\frac{22900}{99}\approx231.31\))

Answer:

The boiling point of propane is approximately \(\boldsymbol{231\space K}\) (or more precisely around \(231.3\space K\))