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

To determine which species is a catalyst, we need to look for a substance that participates in the reaction mechanism but is not consumed overall. In other words, it must be present in the reactants and products side of the reaction mechanism but not in the overall reaction. By comparing the two given reactions and the overall reaction, we can conclude that the catalyst is NO(g). It takes part in the reaction but does not appear in the overall reaction.

02

Identifying the intermediate

To determine which species is an intermediate, we need to look for a substance that is produced in one of the reactions but is consumed in another reaction within the mechanism. In this case, the intermediate is NO2(g), as it is produced in the first reaction and consumed in the second reaction, but does not appear in the overall reaction.

03

Applying the Arrhenius equation

We are given the activation energies of the uncatalyzed and the catalyzed reactions. To find the ratio of the rate constants for the catalyzed (k_c) and the uncatalyzed reaction (k_u), we will use the Arrhenius equation: \[k = Ae^{-E_a / RT}\] where k is the rate constant, A is the frequency factor, E_a is the activation energy, R is the gas constant (8.314 J/mol·K), and T is the temperature in Kelvin. We know that the frequency factor (A) is the same for both the uncatalyzed and catalyzed reactions, and the temperature is given as 25°C (298K). Thus, we can write the following equations for the rate constants of the uncatalyzed and catalyzed reactions: \[k_u = Ae^{-E_{au}/RT}\] \[k_c = Ae^{-E_{ac}/RT}\]

04

Calculating the ratio of rate constants

To determine the ratio of the rate constants, we will divide the equation for k_c by the equation for k_u: \[\frac{k_c}{k_u} = \frac{Ae^{-E_{ac}/RT}}{Ae^{-E_{au}/RT}}\] As the frequency factors (A) are the same for both reactions, they cancel out each other, and this simplifies the equation as: \[\frac{k_c}{k_u} = e^{(E_{au}-E_{ac})/RT}\] Now, we can insert the given activation energies (E_{au} = 14.0 kJ/mol and E_{ac} = 11.9 kJ/mol) and the given temperature (T = 298K), and the gas constant (R = 8.314 J/mol·K = 0.008314 kJ/mol·K) into the equation: \[\frac{k_c}{k_u} = e^{(14.0-11.9)/(0.008314·298)}\]

05

Calculating the final result

By plugging the given values into the equation, we get: \[\frac{k_c}{k_u} = e^{2.1/(0.008314·298)} ≈ 8.29\] The ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at 25°C is approximately 8.29. This means that the rate of the catalyzed reaction is about 8.29 times that of the uncatalyzed reaction under the same conditions.

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