One pathway for the destruction of ozone in the upper atmosphere is $$\begin{array}{l}\mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad \text { Slow } \\\\\mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) \quad \text { Fast } \\ \text { Overall reaction: } \mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\end{array}$$ a. Which species is a catalyst? b. Which species is an intermediate? c. The activation energy \(E_{\mathrm{a}}\) 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 by the presence of \(\mathrm{NO}\) is \(11.9 \mathrm{~kJ} .\) 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. d. One of the concerns about the use of Freons is that they will migrate to the upper atmosphere, where chlorine atoms can be generated by the reaction $$\mathrm{CCl}_{2} \mathrm{~F}_{2} \stackrel{\mathrm{hr}}{\longrightarrow} \mathrm{CF}_{2} \mathrm{Cl}+\mathrm{Cl}$$ Freon- 12 Chlorine atoms also can act as a catalyst for the destruction of ozone. The first step of a proposed mechanism for chlorinecatalyzed ozone destruction is $$\mathrm{Cl}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g)$$ Slow Assuming a two-step mechanism, propose the second step in the mechanism and give the overall balanced equation. e. The activation energy for Cl-catalyzed destruction of ozone is \(2.1 \mathrm{~kJ} / \mathrm{mol}\). Estimate the efficiency with which \(\mathrm{Cl}\) atoms destroy ozone as compared with NO molecules at \(25^{\circ} \mathrm{C}\). Assume that the frequency factor \(A\) is the same for each catalyzed reaction and assume similar rate laws for each catalyzed reaction.

Step by step solution

01

a. Identifying catalyst

In a reaction mechanism, a catalyst is a substance that is involved in the elementary steps of the reaction but is not used up. It is the substance that appears in the initial steps and is regenerated in the final steps. In this mechanism, \(\mathrm{NO}\) is involved in both steps 1 and 2 but is not present in the overall reaction. Therefore, \(\mathrm{NO}\) is the catalyst.

02

b. Identifying intermediate

An intermediate is a species that is formed and consumed during the reaction mechanism but does not appear in the overall reaction. In this mechanism, \(\mathrm{NO}_2\) is formed in the first step and consumed in the second step. Therefore, \(\mathrm{NO}_2\) is the intermediate.

03

c. Calculating rate constant ratio

The ratio of the rate constant for the catalyzed reaction to the uncatalyzed reaction can be calculated using the Arrhenius equation: \[\frac{k_{\text{catalyzed}}}{k_{\text{uncatalyzed}}} = \frac{e^{-E_{a,\text{catalyzed}}/RT}}{e^{-E_{a,\text{uncatalyzed}}/RT}}\] Given, \(E_{a,\text{catalyzed}} = 11.9\mathrm{~kJ}, E_{a,\text{uncatalyzed}} = 14.0\mathrm{~kJ}, T= 25^{\circ} \mathrm{C} = 298\mathrm{~K}.\) We know that \(R = 8.314\mathrm{~J~mol^{-1}~K^{-1}}\). So: \[\frac{k_{\text{catalyzed}}}{k_{\text{uncatalyzed}}} = \frac{e^{-(11.9\times10^3)/(8.314\times298)}}{e^{-(14.0\times10^3)/(8.314\times298)}} \approx 5.48\] Hence, the ratio of rate constant for catalyzed reaction to that of the uncatalyzed reaction at \(25^{\circ}\mathrm{C}\) is approximately 5.48.

04

d. Proposing the second step and overall balanced equation

Considering a two-step mechanism involving \(\mathrm{Cl}\) as a catalyst, the first step is already given: \[\mathrm{Cl}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \text{ (slow)}\] To regenerate the catalyst (\(\mathrm{Cl}\)) and balance the overall reaction, we can propose the second step to be: \[\mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \text{ (fast)}\] Adding these two steps and canceling the intermediate (\(\mathrm{ClO}\)), we obtain the overall balanced equation: \[\mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2\mathrm{O}_{2}(g)\]

05

e. Estimating efficiency of Cl compared to NO

Given the activation energy for the Cl-catalyzed destruction of ozone is \(2.1\mathrm{~kJ/mol}\). We can calculate the ratio of rate constants using Arrhenius equation with the same temperature: \[\frac{k_\text{Cl-catalyzed}}{k_\text{NO-catalyzed}} = \frac{e^{-(2.1\times10^3)/(8.314\times298)}}{e^{-(11.9\times10^3)/(8.314\times298)}} \approx 60.90\] The efficiency of \(\mathrm{Cl}\) atoms to destroy ozone is approximately 60.9 times more than that of \(\mathrm{NO}\) molecules at \(25^{\circ}\mathrm{C}\).

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