/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 \(\mathrm{NO}_{2}+\mathrm{O}_{3}... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\mathrm{NO}_{2}+\mathrm{O}_{3} \rightarrow \mathrm{NO}_{3}+\mathrm{O}_{2}\) \(\quad\) Slow \(\mathrm{NO}_{3}+\mathrm{NO}_{2} \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}\) \(\quad\) Fast A proposed reaction mechanism for the reaction of nitrogen dioxide and ozone is detailed above. Which of the following is the rate law for the reaction? (A) Rate \(=k\left[\mathrm{NO}_{2}\right]\left[\mathrm{O}_{3}\right]\) (B) Rate \(=k\left[\mathrm{NO}_{3}\right]\left[\mathrm{NO}_{2}\right]\) (C) Rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\left[\mathrm{O}_{3}\right]\) (D) Rate \(=k\left[\mathrm{NO}_{3}\right]\left[\mathrm{O}_{2}\right]\)

Short Answer

Expert verified
The rate law for the reaction is Rate = k[\(NO_{2}\)][\(O_{3}\)]. So, option (A) is correct.

Step by step solution

01

Identify the slow step

Firstly, it's important to recognise the slow step in the reaction mechanism provided. The mechanism consists of two steps, and it is given that the first step is slow while the second step is fast. So, the slow step is \(NO_{2} + O_{3} \rightarrow NO_{3} + O_{2}\).
02

Write down the rate law

The rate law of a reaction is determined by its slowest (rate-determining) step. Therefore, we should write the rate law based on the slow step. Since the reaction \(NO_{2} + O_{3} \rightarrow NO_{3} + O_{2}\) involves one molecule of \(NO_{2}\) and one of \(O_{3}\), the rate law should be: Rate = k[\(NO_{2}\)][\(O_{3}\)]. Here 'k' is the rate constant, [] denote the concentration of the species inside, and the superscripts are the stoichiometric coefficients from the balanced chemical equation for the slow step.

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Key Concepts

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

Rate Law
The rate law of a chemical reaction expresses how the rate depends on the concentration of reactants. In this case, we determine the rate law from the slowest step in the reaction mechanism, often called the rate-determining step. This step dictates the overall speed of the reaction, similar to how the slowest runner in a relay race decides the finishing time of the team.
In the provided exercise, the reaction of nitrogen dioxide (NO_2) and ozone (O_3) progresses through a slow step followed by a fast one. The slow step is crucial because it sets the stage for the rate law. According to the stoichiometry of this step, the rate law is expressed as:
\[\text{Rate} = k[\mathrm{NO}_2][\mathrm{O}_3]\]
Here, 'k' stands for the rate constant, which is specific to the reaction and temperature conditions. It has units that vary depending on the reaction order. The concentrations of reactants are shown in square brackets. This particular rate law highlights the dependence on the first powers of each reactant concentration, simplifying it as a first-order reaction with respect to each reactant.
Slow Step
In reaction mechanisms, the slow step acts as the bottleneck, determining the pace at which reactants convert into products. Identifying it within a mechanism is key to predicting reaction speeds accurately.
Our provided mechanism consists of two steps: the first involving NO_2 and O_3 transforming into NO_3 and O_2 is the slow step, while the subsequent reaction is fast. The slow step is critical because it limits the rate at which intermediates are produced, influencing the overall reaction rate.
By recognizing the slow step, chemists can derive the rate law directly related to the initial reactants ( NO_2 and O_3 ), leading to a straightforward depiction of how changes in concentration affect the reaction rate. It's crucial because focusing on the slow step omits the need for complex calculations involving intermediates.
Nitrogen Dioxide
Nitrogen dioxide (NO_2) is a brown gas notable for its significant role in atmospheric chemistry and pollution. As a reactive compound, it often acts as a precursor to various environmental effects, such as smog and acid rain.
In the discussed mechanism, NO_2 serves as a vital reactant in the initial slow step, reacting with O_3 to form NO_3 and O_2. Its concentration directly influences the rate of the reaction due to its presence in the rate law equation:
\[\text{Rate} = k[\mathrm{NO}_2][\mathrm{O}_3]\]
This highlights its critical role in controlling the rate at which products form. In broader terms, understanding NO_2's behavior at a molecular level can help predict its environmental impact, further emphasizing the importance of its role in both scientific and environmental studies.
Ozone
Ozone (O_3) is a triatomic molecule commonly associated with its protective role in Earth’s stratosphere. Yet, in the troposphere, it's a significant pollutant, posing health risks and participating in various chemical reactions.
In this reaction mechanism, O_3 pairs with nitrogen dioxide in the rate-determining slow step, forming oxygen and another nitrogen oxide. Its concentration is crucial as it's part of the rate law expression:
\[\text{Rate} = k[\mathrm{NO}_2][\mathrm{O}_3]\]
Ozone's involvement in the rate law underscores its importance in determining how quickly the reaction proceeds. Beyond this mechanism, understanding ozone's reactivity can also provide insights into discussions about air quality and environmental health. Overall, O_3's dual role in protection and pollution makes it a fascinating subject in both chemistry and environmental sciences.

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Most popular questions from this chapter

A sample of \(\mathrm{H}_{2} \mathrm{S}\) gas is placed in an evacuated, sealed container and heated until the following decomposition reaction occurs at \(1000 \mathrm{K} :\) \(2 \mathrm{H}_{2} \mathrm{S}(g) \rightarrow 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \qquad K_{\mathrm{c}}=1.0 \times 10^{-6}\) Which option best describes what will immediately occur to the reaction rates if the pressure on the system is increased after it has reached equilibrium? (A) The rate of both the forward and reverse reactions will increase. (B) The rate of the forward reaction will increase while the rate of the reverse reaction decreases. (C) The rate of the forward reaction will decrease while the rate of the reverse reaction increases. (D) Neither the rate of the forward nor reverse reactions will change.

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A sample of \(\mathrm{H}_{2} \mathrm{S}\) gas is placed in an evacuated, sealed container and heated until the following decomposition reaction occurs at \(1000 \mathrm{K} :\) $$2 \mathrm{H}_{2} \mathrm{S}(g) \rightarrow 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \qquad K_{\mathrm{c}}=1.0 \times 10^{-6}$$ As the reaction progresses at a constant temperature of 1000 K, how does the value for the Gibbs free energy constant for the reaction change? (A) It stays constant. (B) It increases exponentially. (C) It increases linearly. (D) It decreases exponentially.

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