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The reaction of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{CO}(\mathrm{g})\) is thought to occur in two steps: Step 1: Slow \(\mathrm{NO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{g})\) Step 2: Fast \(\mathrm{NO}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})\) (a) Show that the elementary steps add up to give the overall, stoichiometric equation. (b) What is the molecularity of each step? (c) For this mechanism to be consistent with kinetic data, what must be the experimental rate equation? (d) Identify any intermediates in this reaction.

Short Answer

Expert verified
(a) \( \mathrm{NO}_{2} + \mathrm{CO} \rightarrow \mathrm{NO} + \mathrm{CO}_{2} \); (b) Both steps are bimolecular; (c) Rate = \( k[\mathrm{NO}_{2}]^2 \); (d) Intermediate: \( \mathrm{NO}_{3} \).

Step by step solution

01

Combining Elementary Steps

First, we need to combine the two given steps to verify if they produce the overall reaction. Add together the reactants and products from each step.Step 1: \( \mathrm{NO}_{2} + \mathrm{NO}_{2} \rightarrow \mathrm{NO} + \mathrm{NO}_{3} \)Step 2: \( \mathrm{NO}_{3} + \mathrm{CO} \rightarrow \mathrm{NO}_{2} + \mathrm{CO}_{2} \)Adding both:Reactants: \( \mathrm{NO}_{2} + \mathrm{NO}_{2} + \mathrm{NO}_{3} + \mathrm{CO} \)Products: \( \mathrm{NO} + \mathrm{NO}_{3} + \mathrm{NO}_{2} + \mathrm{CO}_{2} \) Cancel \( \mathrm{NO}_{3} \) from both sides, leaving:\( 2\mathrm{NO}_{2} + \mathrm{CO} \rightarrow \mathrm{NO} + \mathrm{NO}_{2} + \mathrm{CO}_{2} \)Further simplify by canceling \( \mathrm{NO}_{2} \) on both sides:\( \mathrm{NO}_{2} + \mathrm{CO} \rightarrow \mathrm{NO} + \mathrm{CO}_{2} \)So, the simplified equation is \( \mathrm{NO}_{2} + \mathrm{CO} \rightarrow \mathrm{NO} + \mathrm{CO}_{2} \). This verifies the stoichiometric equation.
02

Determining Molecularity

Molecularity refers to the number of molecules coming together in a step of a reaction. For Step 1: \( \mathrm{NO}_{2} + \mathrm{NO}_{2} \rightarrow \mathrm{NO} + \mathrm{NO}_{3} \), two \( \mathrm{NO}_{2} \) molecules are involved, so it is bimolecular.For Step 2: \( \mathrm{NO}_{3} + \mathrm{CO} \rightarrow \mathrm{NO}_{2} + \mathrm{CO}_{2} \), one \( \mathrm{NO}_{3} \) and one \( \mathrm{CO} \) are involved, making it also bimolecular.
03

Deriving the Rate Equation

For kinetic data consistency, the rate law must reflect the reaction mechanism and include the slow step. The given mechanism indicates Step 1 is slow, and determines the rate law.For step 1: \( \mathrm{NO}_{2} + \mathrm{NO}_{2} \rightarrow \mathrm{NO} + \mathrm{NO}_{3} \), the rate law is based on the molecularity giving the rate as\( ext{Rate} = k [\mathrm{NO}_{2}]^2 \).
04

Identifying Reaction Intermediates

A reaction intermediate is a species that is produced in one step of a reaction mechanism and consumed in another. In this reaction, \( \mathrm{NO}_{3} \) is created in Step 1 and consumed in Step 2. Therefore, \( \mathrm{NO}_{3} \) is the intermediate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elementary Steps
An elementary step is a single reaction event that occurs as part of a larger reaction mechanism. These steps break down complex reactions into simpler events that show how molecules rearrange to form products. Each elementary step provides insight into the actual molecular interactions during the reaction process.

In the given reaction between \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{CO}(\mathrm{g})\), the mechanism is composed of two elementary steps. The first step involves two \(\mathrm{NO}_{2}\) molecules producing \(\mathrm{NO}\) and \(\mathrm{NO}_{3}\). The second step occurs when \(\mathrm{NO}_{3}\) reacts with \(\mathrm{CO}\) to produce \(\mathrm{NO}_{2}\) and \(\mathrm{CO}_{2}\).

Elementary steps must collectively summarize or simplify to represent the overall stoichiometric reaction, which is \(\mathrm{NO}_{2} + \mathrm{CO} \rightarrow \mathrm{NO} + \mathrm{CO}_{2}\).
Molecularity
Molecularity refers to the number of reactant molecules participating in an elementary step. It provides a direct count of molecules that collide to bring about the reaction.

  • **Unimolecular:** Involves one molecule decomposing or rearranging.
  • **Bimolecular:** Requires two molecules to interact or collide.
  • **Termolecular:** Involves three molecules; rare due to the improbability of three molecules colliding simultaneously.


In the problem, both steps are **bimolecular**:
- Step 1: Involves collision between two \(\mathrm{NO}_{2}\) molecules.
- Step 2: Involves interaction between \(\mathrm{NO}_{3}\) and \(\mathrm{CO}\).

Each step's molecularity reflects the complexity and likelihood of the process.
Rate Law
The rate law shows how the rate of reaction depends on the concentrations of the reactants. For reactions involving multiple steps, the rate law is often determined by the slowest step, also known as the rate-determining step.

In this particular mechanism, Step 1 is designated as the slow step. The slow step predominantly affects the overall reaction rate. Thus, the rate equation is derived from this step's reactants.

For Step 1: \(\mathrm{NO}_{2} + \mathrm{NO}_{2} \rightarrow \mathrm{NO} + \mathrm{NO}_{3}\)
The rate law based on the molecularity (bimolecular) is:
\[\text{Rate} = k [\mathrm{NO}_{2}]^2\]

This equation demonstrates that the reaction rate is proportional to the square of the concentration of \(\mathrm{NO}_{2}\). Understanding the rate law helps predict how changing concentrations can affect reaction speed.
Reaction Intermediates
Reaction intermediates are species that are formed in one elementary step and consumed in another during the reaction mechanism. They are transient, meaning they do not appear in the overall reaction equation.

In the given mechanism, \(\mathrm{NO}_{3}\) plays the role of an intermediate. It is created in Step 1 during the reaction \(\mathrm{NO}_{2} + \mathrm{NO}_{2} \rightarrow \mathrm{NO} + \mathrm{NO}_{3}\) and then immediately used up in Step 2, where it reacts with \(\mathrm{CO}\) to form products.

Intermediates are crucial for understanding the step-by-step progression of chemical transformations. Identifying these helps scientists infer the details of the molecular path taken by reactions.

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Most popular questions from this chapter

The following statements relate to the reaction for the formation of HI: $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{HI}(\mathrm{g}) \quad \text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]$$ Determine which of the following statements are true. If a statement is false, indicate why it is incorrect. (a) The reaction must occur in a single step. (b) This is a second-order reaction overall. (c) Raising the temperature will cause the value of \(k\) to decrease. (d) Raising the temperature lowers the activation energy for this reaction. (e) If the concentrations of both reactants are doubled, the rate will double. (f) Adding a catalyst in the reaction will cause the initial rate to increase.

Using the rate equation "Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}],\) " define the order of the reaction with respect to \(\mathrm{A}\) and \(\mathrm{B}\). What is the total order of the reaction?

What is the rate law for each of the following elementary reactions? (a) \(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(g) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\) (b) \(\mathrm{Cl}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{H}(\mathrm{g})\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(\mathrm{aq}) \rightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\)

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is a first-order reaction: $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ The rate constant for the reaction is \(2.8 \times 10^{-3} \mathrm{min}^{-1}\) at \(600 \mathrm{K}\). If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(1.24 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) how long will it take for the concentration to drop to \(0.31 \times 10^{-3} \mathrm{mol} / \mathrm{L} ?\)

We know that the decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ with a half-life of 245 minutes at \(600 \mathrm{K}\). If you begin with a partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) of \(25 \mathrm{mm}\) Hg in a 1.0-L. flask, what is the partial pressure of each reactant and product after 245 minutes? What is the partial pressure of each reactant after 12 hours?

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