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The decomposition of ozone in the upper atmosphere is facilitated by NO. The overall reaction and the rate law are $$ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g) \quad \text { Rate }=k\left[\mathrm{O}_{3}\right][\mathrm{NO}] $$ Write a mechanism that is consistent with the rate law.

Short Answer

Expert verified
The mechanism is: \(\mathrm{O}_3 + \mathrm{NO} \rightarrow \mathrm{NO}_2 + \mathrm{O}_2\) (slow), followed by \(\mathrm{NO}_2 + \mathrm{O} \rightarrow \mathrm{NO} + \mathrm{O}_2\) (fast).

Step by step solution

01

Identify the Given Information

We are provided with the overall reaction \( \mathrm{O}_3(g) + \mathrm{O}(g) \rightarrow 2\mathrm{O}_2(g) \) and the rate law \( \text{Rate} = k[\mathrm{O}_3][\mathrm{NO}] \). Our goal is to determine a mechanism that explains this rate law.
02

Determine the Key Components of the Mechanism

From the rate law, the presence of \([\mathrm{NO}]\) in the rate equation suggests that NO might act as a catalyst. Since it does not appear in the overall reaction, it must facilitate the reaction without being consumed. The mechanism should include NO interacting with Ozone (\(\mathrm{O}_3\)) or O (\(\mathrm{O}\)).
03

Propose Elementary Steps According to the Rate Law

Since the rate law is first order with respect to both \(\mathrm{O}_3\) and \(\mathrm{NO}\), the slow (rate-determining) step must involve both of these species. We propose the following mechanism:1. \(\mathrm{O}_3 + \mathrm{NO} \rightarrow \mathrm{NO}_2 + \mathrm{O}_2\) (slow)2. \(\mathrm{NO}_2 + \mathrm{O} \rightarrow \mathrm{NO} + \mathrm{O}_2\) (fast)Combined:\(\mathrm{O}_3 + \mathrm{O} + \mathrm{NO} \rightarrow 2\mathrm{O}_2 + \mathrm{NO}\) which simplifies to the overall reaction.
04

Confirm Consistency with the Overall Reaction and Rate Law

The proposed mechanism is consistent with the provided overall reaction, as \(\mathrm{NO}\) is regenerated, acting as a catalyst. In the first step, \(\mathrm{O}_3\) and \(\mathrm{NO}\) are involved, which matches the terms in the rate law \(\text{Rate} = k[\mathrm{O}_3][\mathrm{NO}] \). The second step ensures that \(\mathrm{O}\) is consumed and \(\mathrm{NO}\) is regenerated, resulting in the formation of two oxygen molecules.

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Key Concepts

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

Ozone Decomposition
Ozone decomposition is a critical process in the atmosphere that involves the breakdown of ozone molecules into oxygen gas. Ozone, which is a molecule made of three oxygen atoms (\(\mathrm{O}_3\)), plays a vital role in protecting the Earth from harmful ultraviolet radiation. However, its decomposition is equally important for maintaining the balance in the ozone layer.
During ozone decomposition, an individual ozone molecule reacts with an oxygen atom (\(\mathrm{O}\)) to produce two molecules of oxygen gas (\(\mathrm{O}_2\)).
  • The overall reaction is: \(\mathrm{O}_3(g) + \mathrm{O}(g) \rightarrow 2\mathrm{O}_2(g)\).
  • This process is crucial as it helps in the regeneration of dioxygen, which is necessary to sustain life on Earth.
In the context of catalytic processes, ozone decomposition can be facilitated by specific chemicals like nitrogen oxides. This aspect of the decomposition is particularly interesting when considering how certain compounds interact with ozone, speeding up its breakdown by acting as catalysts.
Rate Law
A rate law expresses the relationship between the concentration of reactants and the rate of a chemical reaction. In the case of ozone decomposition, understanding the rate law helps us predict how quickly ozone is converted into oxygen gas.
The rate law for the ozone decomposition reaction is given by:
  • \(\text{Rate} = k\left[\mathrm{O}_3\right]\left[\mathrm{NO}\right]\).
  • This indicates that the reaction rate depends on the concentration of ozone and nitrogen monoxide (NO).
The form of the rate law is crucial for determining the mechanism of the reaction. It suggests that both ozone and NO are involved in the rate-determining step, meaning that this step is the slowest and thus controls the overall reaction rate.
By proposing a mechanism consistent with this rate law, we can understand better how constituents like NO interact with ozone during the decomposition process.
Catalysis in Chemistry
Catalysis in chemistry involves the acceleration of a chemical reaction by a substance called a catalyst, which itself remains unchanged at the end of the reaction. In the decomposition of ozone, NO acts as a catalyst.
Catalysts play an essential role because they reduce the activation energy required for a reaction, allowing it to proceed faster or at a lower temperature. The presence of a catalyst like NO in the ozone decomposition enables the reaction to occur more readily.
  • In the proposed mechanism, NO combines with \(\mathrm{O}_3\) to form \(\mathrm{NO}_2\) and \(\mathrm{O}_2\) (a slow step).
  • Then \(\mathrm{NO}_2\) reacts with \(\mathrm{O}\) to regenerate NO and produce another \(\mathrm{O}_2\) (a fast step).
This cycle highlights how NO helps in decomposing ozone while being continuously regenerated and not consumed, therefore it is categorized as a true catalyst. Understanding these mechanisms sheds light on the interaction of compounds in atmospheric chemistry and the pivotal role catalysts play in natural processes.

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Most popular questions from this chapter

The mechanism for catalytic destruction of ozone by chlorine radicals is: (1) \(2\left[\mathrm{Cl}+\mathrm{O}_{3} \longrightarrow \mathrm{O}_{2}+\mathrm{ClO}\right]\) Faster (2) \(2 \mathrm{ClO} \longrightarrow \mathrm{Cl}_{2} \mathrm{O}_{2} \quad\) Slower, Rate-determining step (3) \(\mathrm{Cl}_{2} \mathrm{O}_{2}+h v \longrightarrow 2 \mathrm{Cl}+\mathrm{O}_{2}\) Faster \(2 \mathrm{O}_{3}+h v \longrightarrow 3 \mathrm{O}_{2}\) Overall (a) Write the rate law for the rate-determining step. (b) The rate constant for this reaction at \(190 \mathrm{~K}\) is \(7.2 \times 10^{-13} \mathrm{~cm}^{3} /\) molecule \(\cdot \mathrm{s} .\) Calculate the rate of reaction for Step 2 when the concentration of \(\mathrm{ClO}\) is \(2.4 \times 10^{9}\) molecules \(/ \mathrm{cm}^{3}\). (c) Calculate the rate of ozone loss, which is determined by the rate- determining step.

Bromomethane is converted to methanol in an alkaline solution. The reaction is first order in each reactant. $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$(a) Write the rate law. (b) How does the reaction rate change if the OH concentration is decreased by a factor of \(5 ?\) (c) What is the change in rate if the concentrations of both reactants are doubled?

A \(0.500 \mathrm{~L}\) reaction vessel equipped with a movable piston is filled completely with a \(3.00 \%\) aqueous solution of hydrogen peroxide. The \(\mathrm{H}_{2} \mathrm{O}_{2}\) decomposes to water and \(\mathrm{O}_{2}\) gas in a first-order reaction that has a half-life of \(10.7 \mathrm{~h}\). As the reaction proceeds, the gas formed pushes the piston against a constant external atmospheric pressure of \(738 \mathrm{~mm} \mathrm{Hg} .\) Calculate the \(\mathrm{PV}\) work done (in joules) after a reaction time of \(4.02 \mathrm{~h}\). (You may assume that the density of the solution is \(1.00 \mathrm{~g} / \mathrm{mL}\) and that the temperature of the system is maintained at \(20^{\circ} \mathrm{C}\).)

At high substrate concentrations, the rate of product formation is zeroth order in S. (a) By what factor does the rate of an enzyme-catalyzed reaction change when the substrate concentration is changed from \(2.8 \times 10^{-3} \mathrm{M}\) to \(4.8 \times 10^{-3} \mathrm{M} ?\) (b) Why is the enzyme-catalyzed reaction zero order in \(\mathrm{S}\) at high substrate concentrations?

What fraction of the molecules in a gas at \(300 \mathrm{~K}\) collide with an energy equal to or greater than \(E_{a}\) when \(E_{a}\) equals \(50 \mathrm{~kJ} / \mathrm{mol} ?\) What is the value of this fraction when \(E_{a}\) is \(100 \mathrm{~kJ} / \mathrm{mol} ?\)

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