O3(g) + O(g) -> 2 O2(g) The decomposition of O3(g) in the upper atmosphere is represented by the equ — Kinetics Chemistry Question

STEPS:

1. Understand the role of a catalyst in a mechanism:
A catalyst (in this case, $\text{NO}(g)$) is a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy barrier. At the molecular level, a catalyst is consumed in an early elementary step and regenerated in a later step, meaning it does not appear in the overall balanced equation.
2. Understand the role of reaction intermediates:
An intermediate (such as $\text{NO}_2(g)$) is a species that is produced in an early step and then consumed in a subsequent step. Like the catalyst, it cancels out when the elementary steps are combined and does not appear in the final overall equation.
3. Verify the overall reaction of the proposed mechanisms:
To be a valid mechanism, the sum of the individual elementary steps must add up exactly to the given overall reaction:
$$\text{O}_3(g) + \text{O}(g) \rightarrow 2\ \text{O}_2(g)$$
Let's sum the steps of Mechanism B:
* Step 1: $\text{O}_3(g) + \text{NO}(g) \rightarrow \text{NO}_2(g) + \text{O}_2(g)$
* Step 2: $\text{NO}_2(g) + \text{O}(g) \rightarrow \text{NO}(g) + \text{O}_2(g)$
* Sum:
$$\text{O}_3(g) + \cancel{\text{NO}(g)} + \cancel{\text{NO}_2(g)} + \text{O}(g) \rightarrow \cancel{\text{NO}_2(g)} + \text{O}_2(g) + \cancel{\text{NO}(g)} + \text{O}_2(g)$$
which simplifies perfectly to:
$$\text{O}_3(g) + \text{O}(g) \rightarrow 2\ \text{O}_2(g)$$
This confirms that Mechanism B is stoichiometrically consistent with the overall reaction.
4. Correlate the mechanism steps with the Potential Energy Diagram:
* The potential energy diagram shows that in the presence of $\text{NO}(g)$ (represented by the dashed line), the reaction pathway has two distinct peaks (activation energy barriers). This indicates that the catalyzed reaction occurs via a two-step mechanism.
* The first peak is higher than the second peak, which means the first elementary step has a larger activation energy ($E_a$) than the second step.
* Because a larger activation energy results in a slower reaction rate, Step 1 must be the slow (rate-determining) step, and Step 2 must be the fast step.
* Mechanism B correctly designates Step 1 as "slow" and Step 2 as "fast", which is fully consistent with the energy profile of the diagram.

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WHY_OTHERS_WRONG:

• Option A is incorrect: This option represents a single-step mechanism. The potential energy diagram in the presence of $\text{NO}$ shows two distinct transition states (two peaks), which is characteristic of a multi-step reaction rather than a single elementary step. Furthermore, it does not sum to the correct overall reaction.
• Option C is incorrect: In this mechanism, $\text{NO}_2$ is a reactant in the first step and $\text{NO}$ is a product in the first step. This would mean $\text{NO}_2$ is the catalyst and $\text{NO}$ is an intermediate. However, the problem specifies that the reaction is catalyzed in the presence of $\text{NO}(g)$, meaning $\text{NO}$ must be a reactant in the first step. Furthermore, the steps do not sum to yield the overall reaction.
• Option D is incorrect: Similar to Option C, this mechanism mistakenly places $\text{NO}_2$ as the starting reactant (catalyst) instead of $\text{NO}$. It also introduces $\text{NO}_3$ as an intermediate, which does not match the catalysis by nitric oxide ($\text{NO}$) described in the prompt and the potential energy diagram.