Question

One pathway for the destruction of ozone in the upper

atmosphere is

$$\begin{gathered} O3\left(g\right)+NO\left(g\right)NO2\left(g\right)+O2\left(g\right)Slow \\ NO2\left(g\right)+O\left(g\right)NO\left(g\right)+O2\left(g\right)Fast \end{gathered}$$

Overallreaction:$O3\left(g\right)+O\left(g\right)\to 2O2\left(g\right)$

a. Which species is a catalyst?

b. Which species is an intermediate?

c. Ea for the uncatalyzed reaction

$O3\left(g\right)+O\left(g\right)\to 2O2\left(g\right)$

is 14.0 kJ. Ea for the same reaction when catalyzed

is 11.9 kJ. What is the ratio of the rate constant for

catalyzed reaction to that of the uncatalyzed

reaction at 25⁰C? Assume the frequency factor A is

it the same for each reaction.

Short answer

1. NO is a catalyst.
2. NO₂ is a reaction intermediate.
3. The ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at 25⁰C is 2.33.

Step-by-step solution

Explain the catalyst.

The NO molecule is a catalyst because it contributes to the production of NO₂. It does not consumed in the reaction.

 Explain the intermediate

NO₂ molecule is an intermediate due to the slow reaction rate. These intermediates are the high-energy activated compounds that exist at the highest activation energy during the reaction.

Explain the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at 258o C

Given data -

Activation energy on catalyst surface $\left(E_1\right)$= 11.9 kJ/mol

Activation energy on without catalyst surface $\left(E_2\right)$= 14 kJ/mol

Temperature $\left(T\right)$= 298K

Gas constant, R = 8.314 J/mol K

The pre-exponential factor $\left(A\right)$is the same for each reaction.

The Arrhenius equation is given as

$\ln k=\ln A-\frac{E}{RT}$

On catalyst surface with rate constant $\left(k_1\right)$,

role="math" localid="1663763958674" $In{k}_1=\ln A-\frac{E_1}{RT}....\left(1\right)$

Without catalyst surface with rate constant $\left(k_2\right)$,

$\ln k_2=\ln A-\frac{E_2}{RT}....\left(2\right)$

Subtracting (2) From (1), we get

$$\begin{gathered} \backslash \mathrm{begingatheredln}{k}_1=\ln A-\frac{E_1}{RT}........\left(1\right) \\ \ln k_2=\ln A-\frac{E_2}{RT}.......\left(2\right) \\ \ln \left(\frac{k_1}{k_2}\right)=\frac{1}{8.314\times \left(298\right)}\left(14-11.9\right)\left({10}^3\right) \end{gathered}$$

$$\begin{gathered} \ln \left(\frac{k_1}{k_2}\right)=\frac{1}{2477.572}\left(2.1\right)\left({10}^3\right) \\ \ln \left(\frac{k_1}{k_2}\right)=\left(0.0008486\right)\left({10}^3\right) \\ \ln \left(\frac{k_1}{k_2}\right)=0.8486 \\ \frac{k_1}{k_2}=e^{0.8486} \\ \frac{k_1}{k_2}=2.33 \end{gathered}$$

Therefore, the ratio of the rate constants for the catalyzed reaction to the uncatalyzed reaction is 2.33 at 25⁰C.