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The rate law for this reaction $$ \mathrm{CO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{NO}(g) $$ is rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\). Suggest a plausible mechanism for the reaction, given that the unstable species \(\mathrm{NO}_{3}\) is an intermediate.

Short Answer

Expert verified
The suggested mechanism for the reaction is as follows: \n1. Slow: \(\mathrm{NO}_{2}(g) + \(\mathrm{NO}_{2}(g) \rightarrow \(\mathrm{NO}_{3}(g) + \(\mathrm{NO}(g)\) \n2. Fast: \(\mathrm{CO}(g) + \(\mathrm{NO}_{3}(g) \rightarrow \(\mathrm{CO}_{2}(g) + \(\mathrm{NO}_{2}(g)\)

Step by step solution

01

Identifying Elementary Steps

In the first step, it is inferred that two molecules of \(\mathrm{NO}_{2}(g)\) must come together to form \(\mathrm{NO}_{3}\) and \(\mathrm{NO}(g)\). This makes sense as \(\mathrm{NO}_{3}\) is an intermediate species and is not found in the final products. The elementary reaction for the first step can be represented as: \(\mathrm{NO}_{2}(g) + \(\mathrm{NO}_{2}(g) \rightarrow \(\mathrm{NO}_{3}(g) + \(\mathrm{NO}(g)
02

Completing the Reaction

The next step is needed to generate the final products. In this step, \(\mathrm{CO}(g)\) reacts with the \(\mathrm{NO}_{3}(g)\) intermediate to produce \(\mathrm{CO}_{2}(g)\) and \(\mathrm{NO}_{2}(g)\). The elementary reaction for the second step can be represented as: \(\mathrm{CO}(g) + \(\mathrm{NO}_{3}(g) \rightarrow \(\mathrm{CO}_{2}(g) + \(\mathrm{NO}_{2}(g)
03

Confirming the Mechanism

Now that both steps of the reaction mechanism have been identified, it is necessary to ensure they align with the given rate law, rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\). In order for this to be true, the first step must be the rate-determining step, meaning it is the slowest and thus controls the rate of the overall reaction. As the rate law is second order with respect to \(\mathrm{NO}_{2}\), and the reaction consumes two molecules of \(\mathrm{NO}_{2}\), this is indeed the rate-determining step.

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Key Concepts

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

Rate Law
Understanding the rate law is fundamental when studying the kinetics of a chemical reaction. A rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. Mathematically, it is represented by an equation showing the rate (usually denoted by the variable 'rate' or 'v') as proportional to the product of constant 'k' (the rate constant) and the concentrations of the reactants raised to some powers (the reaction orders).

For example, consider the rate law for the reaction between carbon monoxide, CO(g), and nitrogen dioxide, NO2(g), which is: rate = k[NO2]2. This indicates that the reaction rate is directly proportional to the square of the concentration of NO2, making it a second-order reaction with respect to NO2.

Here, 'k' is specific to the reaction and temperature-dependent, the square of [NO2] implies that if you double the concentration of NO2, the rate of the reaction will increase by a factor of four. This is key to predicting how changes in concentration affect the rate, which can be tested through experiments and used to confirm the proposed reaction mechanism.
Elementary Steps
Elementary steps are the basic, individual processes that occur during a chemical reaction. Each of these steps involves a direct interaction between molecules or atoms resulting in the formation or breaking of bonds. The overall reaction is the sum of these elementary steps. The mechanism that comprises the elementary steps gives us insight into the 'molecular-level' pathway of a complex reaction.

In our example, where CO(g) and NO2(g) react to form CO2(g) and NO(g), the mechanism involves two elementary steps. The first elementary step proposes the formation of an unstable intermediate, NO3, by the dimerization of two NO2 molecules. The occurrence of such an intermediate is consistent with the rate law, as the rate-determining step involves NO2 molecules interacting with each other. The second step involves CO(g) reacting with NO3, the intermediate, to form the final products, CO2 and NO. Since each step is considered elementary, all of them occur in a single-stage rather than through a series of events.
Rate-Determining Step
The rate-determining step in a reaction mechanism is the slowest step, which therefore controls the overall rate of the reaction. Think of it as the 'bottleneck' in the process. No matter how fast the other steps are, the overall reaction cannot proceed any faster than this slowest step.

In our reaction, the formation of NO3 from two NO2 molecules is the rate-determining step since it must be the slowest step to be consistent with the rate law given. It poignantly demonstrates that only the rate-determining step's reactant concentrations appear in the rate law. The subsequent steps are faster and do not influence the rate law directly. Identifying the rate-determining step is imperative when deducing the mechanism of a reaction. It aids in optimizing and controlling the conditions to improve reaction rates for practical applications, such as in industrial chemical processes.

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Most popular questions from this chapter

For the reaction \(X_{2}+Y+Z \longrightarrow X Y+X Z\) it is found that doubling the concentration of \(\mathrm{X}_{2}\) doubles the reaction rate, tripling the concentration of \(Y\) triples the rate, and doubling the concentration of \(Z\) has no effect. (a) What is the rate law for this reaction? (b) Why is it that the change in the concentration of \(Z\) has no effect on the rate? (c) Suggest a mechanism for the reaction that is consistent with the rate law.

The equation for the combustion of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) is $$ 2 \mathrm{C}_{2} \mathrm{H}_{6}+7 \mathrm{O}_{2} \longrightarrow 4 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O} $$ Explain why it is unlikely that this equation also represents the elementary step for the reaction.

The rate law for the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\). Which of these changes will change the value of \(k ?\) (a) The pressure of \(\mathrm{NO}_{2}\) is doubled. (b) The reaction is run in an organic solvent. (c) The volume of the container is doubled. (d) The temperature is decreased. (e) A catalyst is added to the container.

The following gas-phase reaction was studied at \(290^{\circ} \mathrm{C}\) by observing the change in pressure as a function of time in a constant-volume vessel: $$ \mathrm{ClCO}_{2} \mathrm{CCl}_{3}(g) \longrightarrow 2 \mathrm{COCl}_{2}(g) $$ Determine the order of the reaction and the rate constant based on the following data: $$ \begin{array}{rc} \text { Time }(\mathrm{s}) & \mathrm{P}(\mathrm{mmHg}) \\ \hline 0 & 15.76 \\ 181 & 18.88 \\ 513 & 22.79 \\ 1164 & 27.08 \end{array} $$ where \(P\) is the total pressure.

These data were collected for the reaction between hydrogen and nitric oxide at \(700^{\circ} \mathrm{C}\) : \(2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)\) $$ \begin{array}{cllc} \text { Experiment } & {\left[\mathrm{H}_{2}\right]} & {[\mathrm{NO}]} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1 & 0.010 & 0.025 & 2.4 \times 10^{-6} \\ 2 & 0.0050 & 0.025 & 1.2 \times 10^{-6} \\ 3 & 0.010 & 0.0125 & 0.60 \times 10^{-6} \end{array} $$ (a) Determine the order of the reaction. (b) Calculate the rate constant. (c) Suggest a plausible mechanism that is consistent with the rate law. (Hint: Assume the oxygen atom is the intermediate.)

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