Using data from Appendix 4, calculate \(\Delta H^{\circ}, \Delta G^{\circ}\), and \(K(\) 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.

From Appendix 4, we can find the standard enthalpies of formation \(\Delta H^{\circ}_\text{f}\) and standard Gibbs free energies of formation \(\Delta G^{\circ}_\text{f}\) for both oxygen (O2) and ozone (O3) at 298 K: Oxygen: \(\Delta H^{\circ}_\text{f}(O_2) = 0\ \text{kJ/mol}\) and \(\Delta G^{\circ}_\text{f}(O_2) = 0\ \text{kJ/mol}\) Ozone: \(\Delta H^{\circ}_\text{f}(O_3) = 142.7\ \text{kJ/mol}\) and \(\Delta G^{\circ}_\text{f}(O_3) = 163.3\ \text{kJ/mol}\) Now, let's calculate the \(\Delta H^{\circ}\) for the reaction: $$\Delta H^{\circ} = \Delta H^{\circ}_\text{f}(\text{products}) - \Delta H^{\circ}_\text{f}(\text{reactants})$$ $$\Delta H^{\circ} = [2 \cdot \Delta H^{\circ}_\text{f}(O_3)] - [3 \cdot \Delta H^{\circ}_\text{f}(O_2)]$$ $$\Delta H^{\circ} = 2 \cdot 142.7\ \text{kJ/mol} - 3 \cdot 0\ \text{kJ/mol} = 285.4\ \text{kJ/mol}$$ Next, let's calculate the \(\Delta G^{\circ}\) for the reaction: $$\Delta G^{\circ} = \Delta G^{\circ}_\text{f}(\text{products}) - \Delta G^{\circ}_\text{f}(\text{reactants})$$ $$\Delta G^{\circ} = [2 \cdot \Delta G^{\circ}_\text{f}(O_3)] - [3 \cdot \Delta G^{\circ}_\text{f}(O_2)]$$ $$\Delta G^{\circ} = 2 \cdot 163.3\ \text{kJ/mol} - 3 \cdot 0\ \text{kJ/mol} = 326.6\ \text{kJ/mol}$$

Using the calculated value of \(\Delta G^{\circ}\), we can determine the equilibrium constant \(K\) at 298 K using the equation: $$\Delta G^{\circ} = -RT \ln K$$ $$K = \text{e}^{-\frac{\Delta G^{\circ}}{RT}}$$ Where \(R = 8.314\ \text{J/mol}\cdot\text{K}\), \(T = 298\ \text{K}\), and \(\Delta G^{\circ} = 326.6\ \times 10^3\ \text{J/mol}\). $$K = \text{e}^{-\frac{326.6 \times 10^3}{8.314 \cdot 298}} \approx 3.53 \times 10^{-53}$$

To do this, let's first calculate the equilibrium constant \(K' \) at 230 K. We can use the van't Hoff equation for this: $$\ln \frac{K'}{K} = -\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T'} - \frac{1}{T}\right)$$ Where \(T' = 230\ \text{K}\). $$\ln \frac{K'}{K} = -\frac{285.4 \times 10^3}{8.314}\left(\frac{1}{230} - \frac{1}{298}\right)$$ $$\Rightarrow K' = K \text{e}^{-\frac{285.4 \times 10^3}{8.314}\left(\frac{1}{230} - \frac{1}{298}\right)} \approx 8.53 \times 10^{-53}$$ Now, let's write the equilibrium expression for the reaction: $$K' = \frac{[\text{O}_3]^2}{[\text{O}_2]^3}$$ At 30 km above the earth's surface, the partial pressure of oxygen is about \(1.0 \times 10^{-3}\ atm\). We can represent the partial pressure of ozone as \(x\ atm\). Assuming the ideal gas law, we have: $$K' = \frac{x^2}{(1.0 \times 10^{-3})^3}$$ Now, let's solve for \(x\): $$x^2 = K' \cdot (1.0 \times 10^{-3})^3$$ $$x = \sqrt{8.53 \times 10^{-53} \cdot (1.0 \times 10^{-3})^3} \approx 9.12 \times 10^{-25}\ \text{atm}$$

The calculated partial pressure of ozone at 30 km above the earth's surface is extremely low (\(9.12 \times 10^{-25}\ \text{atm}\)), compared to the partial pressure of oxygen (\(1.0 \times 10^{-3}\ \text{atm}\)). Given this imbalance and the extremely small value of \(K'\), it is not reasonable to assume that the equilibrium between oxygen and ozone is maintained under these conditions. Instead, since ozone is such a minor species in the atmosphere, we can conclude that other factors, such as photochemical reactions and transport processes, play a more significant role in the behavior of ozone in the stratosphere.