Using data from Appendix 4, calculate \(\Delta H^{\circ}, \Delta G^{\circ}\), and \(K_{\mathrm{p}}\) (at \(298 \mathrm{~K}\) ) for the production of ozone from oxygen: $$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$ At \(30 \mathrm{~km}\) above the surface of the earth, the temperature is about \(230 . \mathrm{K}\), and the partial pressure of oxygen is about \(1.0 \times 10^{-3}\) atm. Estimate the partial pressure of ozone in equilibrium with oxygen at \(30 \mathrm{~km}\) above the earth's surface. Is it reasonable to assume that the equilibrium between oxygen and ozone is maintained under these conditions? Explain.

In Appendix 4, we find the standard enthalpy \(\Delta H^{\circ}_{f}\) and standard Gibbs free energy \(\Delta G^{\circ}_{f}\) of formation for \(\mathrm{O}_2(g)\) and \(\mathrm{O}_3(g)\). \(\Delta H^{\circ}_{f} (\mathrm{O}_{2}(g))=0\) and \(\Delta G^{\circ}_{f} (\mathrm{O}_{2}(g)) = 0\), since O2 is the elemental form. \(\Delta H^{\circ}_{f} (\mathrm{O}_{3}(g)) = 142.2 \mathrm{~kJ/mol}\) and \(\Delta G^{\circ}_{([\mathrm{O}_3(g)]) = 163.3 \mathrm{~kJ/mol}\) Now we can use these values to find the standard enthalpy change \(\Delta H^{\circ}\) and the standard Gibbs free energy change \(\Delta G^{\circ}\) for the given reaction: $$\Delta H^{\circ} = \sum \Delta H^{\circ}_{f}(\mathrm{products}) - \sum \Delta H^{\circ}_{f}(\mathrm{reactants})$$ $$\Delta H^{\circ} = 2 \times \Delta H^{\circ}_{f}(\mathrm{O}_{3}(g)) - 3 \times \Delta H^{\circ}_{f}(\mathrm{O}_{2}(g))$$ $$\Delta H^{\circ} = 2 \times 142.2 - 3 \times 0 = 284.4 \mathrm{~kJ/mol}$$ Similarly, $$\Delta G^{\circ} = \sum \Delta G^{\circ}_{f}(\mathrm{products}) - \sum \Delta G^{\circ}_{f}(\mathrm{reactants})$$ $$\Delta G^{\circ} = 2 \times \Delta G^{\circ}_{f}(\mathrm{O}_{3}(g)) - 3 \times \Delta G^{\circ}_{f}(\mathrm{O}_{2}(g))$$ $$\Delta G^{\circ} = 2 \times 163.3 - 3 \times 0 = 326.6 \mathrm{~kJ/mol}$$

Now that we have the standard Gibbs free energy change, we can calculate the equilibrium constant \(K_{\mathrm{p}}\) at \(298 \mathrm{~K}\) using the equation: $$\Delta G^{\circ} = -RT \ln K_{\mathrm{p}}$$ Rearrange the equation to solve for \(K_\mathrm{p}\): $$K_{\mathrm{p}} = e^{\frac{-\Delta G^{\circ}}{RT}}$$ $$K_{\mathrm{p}} = e^{\frac{-(326.6 \times 10^3 \mathrm{~J/mol})}{(8.314 \mathrm{~J/mol\cdot K})(298 \mathrm{~K})}}$$ $$K_{\mathrm{p}} = 3.2 \times 10^{-36}$$

Assuming the system is in equilibrium, we can now find the partial pressure of ozone using \(K_\mathrm{p}\) and the given conditions: \(K_{\mathrm{p}} = \frac{P_{\mathrm{O}_3}^2}{P_{\mathrm{O}_2}^3}\) Given partial pressure of O2: \(P_{\mathrm{O}_2} = 1.0 \times 10^{-3} \mathrm{~atm}\) The partial pressure of ozone can be calculated as: $$P_{\mathrm{O}_3} = \sqrt[2]{K_{\mathrm{p}}P_{\mathrm{O}_2}^3}$$ $$P_{\mathrm{O}_3} = \sqrt[2]{(3.2 \times 10^{-36})(1.0 \times 10^{-3 \cdot 3})}$$ $$P_{\mathrm{O}_3} \approx 4.7 \times 10^{-20} \mathrm{~atm}$$

Since the equilibrium constant is very small, the forward reaction (formation of ozone) is highly unfavorable under these conditions. The partial pressure of ozone calculated above is extremely low compared to the partial pressure of oxygen. Thus, it is reasonable to assume that the equilibrium between oxygen and ozone is not maintained at \(30 \mathrm{~km}\) above the earth's surface, and the concentration of ozone would be minimal.