The standard free energies of formation and the standard enthalpies of formation at \(298 \mathrm{K}\) for diffuoroacetylene \(\left(\mathrm{C}_{2} \mathrm{F}_{2}\right)\) and hexafluorobenzene \(\left(\mathrm{C}_{6} \mathrm{F}_{6}\right)\) are $$\begin{array}{ccc} & \Delta G_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & \Delta H_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) \\ \hline \mathrm{C}_{2} \mathrm{F}_{2}(g) & 191.2 & 241.3 \\\ \mathrm{C}_{6} \mathrm{F}_{6}(g) & 78.2 & 132.8 \end{array}$$ For the following reaction: $$\mathrm{C}_{6} \mathrm{F}_{6}(g) \rightleftharpoons 3 \mathrm{C}_{2} \mathrm{F}_{2}(g)$$ a. calculate \(\Delta S^{\circ}\) at \(298 \mathrm{K}\). b. calculate \(K\) at 298 K. c. estimate \(K\) at \(3000 .\) K, assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

a. Calculate \(\Delta S^{\circ}\) at 298 K: By using the relation \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), we can calculate \(\Delta S^{\circ}\) as follows: \(\Delta G_{rxn}^{\circ} = \Delta H_{rxn}^{\circ} - 298 \Delta S_{rxn}^{\circ}\) Rearranging to solve for \(\Delta S_{rxn}^{\circ}\) gives: \(\Delta S_{rxn}^{\circ} = \frac{\Delta H_{rxn}^{\circ} - \Delta G_{rxn}^{\circ}}{298}\) Substituting the given values: \(\Delta S_{rxn}^{\circ} = \frac{(3\times241.3 - 132.8) - (3\times191.2 - 78.2)}{298}\) \(\Delta S_{rxn}^{\circ} = 297.6 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\) b. Calculate \(K\) at 298 K: Using the relation between Gibbs Free Energy and the equilibrium constant, the equation is: \(K = e^{-(\Delta G_{rxn}^{\circ})/RT}\) Substituting values: \(K = e^{-(3\times191.2 - 78.2)/(8.314\times10^{-3}\times298)}\) \(K \approx 9.54\times 10^{-7}\) c. Estimate \(K\) at 3000 K: Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, we can use the van't Hoff equation: \(\ln\frac{K_2}{K_1} = \frac{-\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\) We want to find \(K_2\), given \(K_1\) at 298 K and \(T_2 = 3000\, K\). Rearrange the equation: \(K_2 = K_1 \times e^{\frac{-\Delta H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)}\) Substituting values: \(K_2 = 9.54\times 10^{-7} \times e^{\frac{-(3\times241.3 - 132.8)}{8.314} \left(\frac{1}{3000} - \frac{1}{298}\right)}\) \(K_2 \approx 0.490\)