One mechanism for the destruction of ozone in the upper atmosphere is $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad \text { Slow }$$ $$\frac{\mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g)}{\mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)}\quad \text { Fast }$$ Overall reactiona. Which species is a catalyst? b. Which species is an intermediate? c. \(E_{\mathrm{a}}\) for the uncatalyzed reaction$$\mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)$$is \(14.0 \mathrm{kJ} . E_{\mathrm{a}}\) for the same reaction when catalyzed is 11.9 kJ. What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C} ?\) Assume that the frequency factor \(A\) is the same for each reaction.

We are given two reactions, and we need to identify the catalyst and the intermediate species. The catalyst is a species that appears on both sides of the reaction mechanism. It participates in the reaction but is not consumed. The intermediate species is formed and consumed during the overall reaction, so it does not appear in the final balanced equation of the process. Slow step: O3(g) + NO(g) -> NO2(g) + O2(g) Fast step: NO2(g) + O(g) -> NO(g) + O2(g) Catalyst: Nitric oxide (NO) is present in both the slow and fast steps, but not in the overall reaction, so it is the catalyst. Intermediate: Nitrogen dioxide (NO2) is also present in both the slow and fast steps, but not in the overall reaction, so it is the intermediate species.

We are given the activation energies (Ea) for the uncatalyzed reaction and the catalyzed reaction: Ea(uncatalyzed) = 14.0 kJ Ea(catalyzed) = 11.9 kJ And we are given the temperature: T = 25°C = 298 K To find the ratio of the rate constants (k) for the catalyzed and the uncatalyzed reactions, we use the Arrhenius equation: \(k = A \cdot e^{-\frac{E_a}{RT}}\) where A is the frequency factor, R is the gas constant (8.314 J/mol-K), and T is the absolute temperature in kelvin. Since we are given the assumption that A is the same for both reactions, it will cancel out in the ratio we need to calculate: ratio: \(k_c : k_u = \frac{Ae^{-\frac{E_{a_c}}{RT}}}{Ae^{-\frac{E_{a_u}}{RT}}} = e^{\frac{E_{a_u} - E_{a_c}}{RT}}\) Now, we can plug in the given activation energies and temperature to calculate the ratio: \( ratio = e^{\frac{14.0 kJ/mol - 11.9 kJ/mol}{8.314 \times 10^{-3} kJ/mol\cdot K \times 298 K}} = e^{2.1/2.48}\)

After calculating the value inside the exponential function: ratio = \(e^{0.846}\) Now, we can use a calculator to find the approximate value for the ratio of rate constants: ratio ≈ 2.33 So, the rate constant for the catalyzed reaction is approximately 2.33 times greater than the rate constant for the uncatalyzed reaction at the given temperature.