c. A possible mechanism for this reaction involves two steps: NO2(g) + NO2(g) - NO3(g) + NO(g) slow NO3(g) + CO(g) → NO2(g) + CO2(g) fast Is this proposed mechanism consistent with the experimentally determined rate law? Explain. d. What is the reaction intermediate in the proposed mechanism?

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Chapter12: Chemical Kinetics
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Problem 117CP: You are studying the kinetics of the reaction H2(g) + F2(g) 2HF(g) and you wish to determine a...
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### Chemical Reaction Mechanism and Rate Law Analysis

#### c. Analysis of a Proposed Mechanism
A possible mechanism for the given reaction consists of two steps:

\[ \text{NO}_2(\text{g}) + \text{NO}_2(\text{g}) \rightarrow \text{NO}_3(\text{g}) + \text{NO}(\text{g}) \quad \text{[slow]} \]

\[ \text{NO}_3(\text{g}) + \text{CO}(\text{g}) \rightarrow \text{NO}_2(\text{g}) + \text{CO}_2(\text{g}) \quad \text{[fast]} \]

**Question:** Is this proposed mechanism consistent with the experimentally determined rate law? Explain.

#### d. Identifying the Reaction Intermediate
**Question:** What is the reaction intermediate in the proposed mechanism?

---

### Explanation

**Section c:** The proposed mechanism comprises two steps: the first step is slow, involving the reaction of two nitrogen dioxide \((\text{NO}_2)\) molecules to form nitrogen trioxide \((\text{NO}_3})\) and nitrogen monoxide \((\text{NO})\). The second, faster step involves the reaction of \(\text{NO}_3\) with carbon monoxide \((\text{CO})\) to produce \(\text{NO}_2}\) and carbon dioxide \((\text{CO}_2})\).

To determine consistency with the experimentally determined rate law, one needs to:

- Derive the rate law from the slow step (since it is the rate-determining step).
- Compare this derived rate law with the experimentally determined rate law to check for consistency.

**Section d:** The reaction intermediate is a molecule that is produced in one step of the reaction mechanism and consumed in a subsequent step. In this proposed mechanism, \(\text{NO}_3}\) is produced in the first step and consumed in the second step, thus it is the reaction intermediate.

---

Understanding and analyzing the proposed mechanism helps in verifying its consistency with the experimental observations and further elucidates the steps involved in the chemical reaction.
Transcribed Image Text:### Chemical Reaction Mechanism and Rate Law Analysis #### c. Analysis of a Proposed Mechanism A possible mechanism for the given reaction consists of two steps: \[ \text{NO}_2(\text{g}) + \text{NO}_2(\text{g}) \rightarrow \text{NO}_3(\text{g}) + \text{NO}(\text{g}) \quad \text{[slow]} \] \[ \text{NO}_3(\text{g}) + \text{CO}(\text{g}) \rightarrow \text{NO}_2(\text{g}) + \text{CO}_2(\text{g}) \quad \text{[fast]} \] **Question:** Is this proposed mechanism consistent with the experimentally determined rate law? Explain. #### d. Identifying the Reaction Intermediate **Question:** What is the reaction intermediate in the proposed mechanism? --- ### Explanation **Section c:** The proposed mechanism comprises two steps: the first step is slow, involving the reaction of two nitrogen dioxide \((\text{NO}_2)\) molecules to form nitrogen trioxide \((\text{NO}_3})\) and nitrogen monoxide \((\text{NO})\). The second, faster step involves the reaction of \(\text{NO}_3\) with carbon monoxide \((\text{CO})\) to produce \(\text{NO}_2}\) and carbon dioxide \((\text{CO}_2})\). To determine consistency with the experimentally determined rate law, one needs to: - Derive the rate law from the slow step (since it is the rate-determining step). - Compare this derived rate law with the experimentally determined rate law to check for consistency. **Section d:** The reaction intermediate is a molecule that is produced in one step of the reaction mechanism and consumed in a subsequent step. In this proposed mechanism, \(\text{NO}_3}\) is produced in the first step and consumed in the second step, thus it is the reaction intermediate. --- Understanding and analyzing the proposed mechanism helps in verifying its consistency with the experimental observations and further elucidates the steps involved in the chemical reaction.
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