Question

for a reaction where $\delta h^\circ = +573\\ \text{kj/mol}$ and $\delta s^\circ = +1.64\\ \text{kj} \cdot \text{mol}^{-1} \cdot \text{k}^{-1}$, at what temperature, in k, does $\text{x}(\text{l}) \
ightarrow \text{x}(\text{g})$ occur spontaneously?

Explanation:

Step1: Recall the spontaneity condition

For a reaction to be spontaneous, the Gibbs free energy change \(\Delta G^{\circ}\) must be less than zero. The formula relating \(\Delta G^{\circ}\), \(\Delta H^{\circ}\), \(\Delta S^{\circ}\), and temperature \(T\) is \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\). We set \(\Delta G^{\circ}<0\):
\(\Delta H^{\circ}-T\Delta S^{\circ}<0\)

Step2: Rearrange the inequality to solve for \(T\)

Starting with \(\Delta H^{\circ}-T\Delta S^{\circ}<0\), we can rearrange it to isolate \(T\). First, add \(T\Delta S^{\circ}\) to both sides:
\(\Delta H^{\circ}Then, divide both sides by \(\Delta S^{\circ}\) (since \(\Delta S^{\circ}\) is positive, the inequality sign remains the same):
\(T > \frac{\Delta H^{\circ}}{\Delta S^{\circ}}\)

Step3: Substitute the given values

We are given \(\Delta H^{\circ}= + 573\space kJ/mol\) and \(\Delta S^{\circ}=+1.64\space kJ\cdot mol^{-1}\cdot K^{-1}\). Substitute these values into the formula for \(T\):
\(T>\frac{573\space kJ/mol}{1.64\space kJ\cdot mol^{-1}\cdot K^{-1}}\)

Step4: Calculate the temperature

Perform the division:
\(\frac{573}{1.64}\approx349.39\space K\)
Since \(T\) must be greater than this value for the reaction to be spontaneous, the temperature at which the reaction occurs spontaneously is above approximately \(349\space K\) (more precisely, above \(\frac{573}{1.64}\space K\)). If we want the temperature at which the reaction just becomes spontaneous (the transition temperature), we set \(\Delta G^{\circ} = 0\), so \(T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}\), and calculating that gives \(T=\frac{573}{1.64}\approx349\space K\) (to a reasonable number of significant figures).

Answer:

The reaction occurs spontaneously at temperatures greater than \(\boldsymbol{\approx 349\space K}\) (the transition temperature is approximately \(349\space K\)). If we calculate the exact value from \(\frac{573}{1.64}\), we get \(T\approx349.39\space K\), so the temperature at which it becomes spontaneous (and above) is approximately \(\boldsymbol{349\space K}\) (or more precisely, \(\boldsymbol{\frac{573}{1.64}\space K\approx349\space K}\)).