One mechanism for the destruction of ozone in the upper atmosphere is $$ \begin{array}{ll} \mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & \text { Slov } \\ \mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \text { Fast } \\ \hline \end{array} $$ Overall reaction \(\mathrm{O}_{3}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{~g})\) a. Which species is a catalyst? b. Which species is an intermediate? c. \(E_{\mathrm{a}}\) for the uncatalyzed reaction $$ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2} $$ is \(14.0 \mathrm{~kJ} . E_{\mathrm{a}}\) for the same reaction when catalyzed 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.

Step by step solution

01

Identifying the catalyst

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the overall reaction. In the given mechanism, we observe that NO is present in both elementary reactions as both a reactant and a product, and doesn't appear in the overall reaction. Thus, NO is the catalyst for this mechanism.

02

Identifying the intermediate

An intermediate species is formed and consumed in the course of a reaction mechanism, and does not appear in the overall reaction. In the given mechanism, we observe that NO2 is formed in the first elementary reaction and consumed in the second elementary reaction, and it doesn't appear in the overall reaction. Thus, NO2 is an intermediate species.

03

Applying the Arrhenius equation to find the rate constant ratio

The Arrhenius equation is used to relate the rate constant k of a reaction to the activation energy Ea, the temperature T, and the frequency factor A: \(k=A \cdot e^{-\frac{Ea}{RT}}\) where R is the gas constant (8.314 J/mol·K in this case). We are given the activation energies Ea for both the catalyzed (11.9 kJ/mol) and uncatalyzed (14.0 kJ/mol) reactions, and the temperature T = 298 K = 25°C. We need to determine the ratio of the rate constants for the catalyzed (k1) and uncatalyzed (k2) reactions. The ratio can be calculated as: \(\frac{k1}{k2} = \frac{A \cdot e^{-\frac{Ea_{1}}{RT}}}{A \cdot e^{-\frac{Ea_{2}}{RT}}}\) Since the frequency factors A are the same for both reactions, they will cancel out, leaving only the exponential terms: \(\frac{k1}{k2} = \frac{e^{-\frac{Ea_{1}}{RT}}}{e^{-\frac{Ea_{2}}{RT}}}\) Now, we can plug in the given values for Ea and T, and calculate the rate constant ratio.

04

Calculate the rate constant ratio

With the values given: \(Ea_{1} = 11.9 \times 10^{3} J/mol\) \(Ea_{2} = 14.0 \times 10^{3} J/mol\) \(T = 298 K\) \(R = 8.314 J/mol·K\) Applying these into the equation: \(\frac{k1}{k2} = \frac{e^{-\frac{11.9 \times 10^{3}}{8.314 \times 298}}}{e^{-\frac{14.0 \times 10^{3}}{8.314 \times 298}}}\) Computing this gives: \(\frac{k1}{k2} \approx 4.04\) So, the rate constant for the catalyzed reaction is roughly 4.04 times greater than that for the uncatalyzed reaction at 25°C.

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