Question

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.

Short answer

In the given reaction mechanism, the catalyst is nitric oxide (NO) and the intermediate species is nitrogen dioxide (NO2). The ratio of the rate constants for the catalyzed reaction to the uncatalyzed reaction at 25°C is approximately 2.33.

Step-by-step solution

Identify the catalyst and intermediate species

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.

Calculate the rate constant ratio

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: \( ratio = e^{\frac{14.0 kJ/mol - 11.9 kJ/mol}{8.314 \times 10^{-3} kJ/mol\cdot K \times 298 K}} = e^{2.1/2.48}\)

Calculate the final ratio

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.

Key concepts

Catalyst

Imagine cooking with the help of a blender. The blender speeds up the process of making a smoothie, but it remains unchanged afterwards. In chemical kinetics, a catalyst behaves similarly. It is a substance that accelerates a chemical reaction without being consumed or altered permanently.
In the ozone destruction example, nitric oxide ( ext{NO}) serves as the catalyst. It's involved in both the slow and fast steps of the reaction mechanism but doesn't appear in the overall balanced equation. This means it makes reactions happen more quickly without being used up, just like a blender is still ready for the next task after making a smoothie.

Intermediate Species

An intermediate species is like a temporary helper in a chemical reaction. It is produced in one step of a reaction mechanism and consumed in another, making it a short-lived participant that does not appear in the overall equation.
In our ozone destruction scenario, nitrogen dioxide ( ext{NO}_2) acts as an intermediate. It is created in the slow step when ext{O}_3 reacts with ext{NO}, and then used in the fast step when it reacts with ext{O}. By the time the final products are formed, ext{NO}_2 has "come and gone," thus it's not present in the final equation. This lifecycle of creating and consuming intermediates is crucial for understanding complex reactions.

Activation Energy

Activation energy is like the initial push you need to start a cranking engine on a cold winter's day. It's the minimum amount of energy required for reactants to transform into products in a chemical reaction.
In our discussion about ozone, the activation energy for the uncatalyzed reaction ext{O}_3 + ext{O} ightarrow 2 ext{O}_2 is 14.0 kJ. However, when a catalyst is used, this activation energy is lowered to 11.9 kJ.
Lowering the activation energy makes the reaction proceed faster because it's easier to "start the engine" of the reaction. This demonstrates one significant role of catalysts - they effectively decrease the energy barrier, facilitating quicker chemical transformations.

Arrhenius Equation

The Arrhenius equation is a handy tool that helps us understand how reaction rates depend on temperature and activation energy. It is given by the formula: \[ k = A imes e^{-\frac{E_a}{RT}} \] where:

• \(k\) is the rate constant, indicating how fast a reaction proceeds
• \(A\) is the frequency factor, a constant specific to each reaction
• \(E_a\) is the activation energy
• \(R\) is the gas constant, approximately 8.314 J/mol·K
• \(T\) is the temperature in Kelvin

In our ozone example, we used the Arrhenius equation to compare rate constants for catalyzed and uncatalyzed reactions. By knowing the activation energies and temperature, we determined that the catalyzed reaction is 2.33 times faster at 25°C. The lower the activation energy in the equation, the larger the rate constant becomes, thus speeding up the reaction.