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The decomposition of ozone in the stratosphere can occur by the following two- step mechanism: Step I: \(\mathrm{Br}+\mathrm{O}_{3} \rightarrow \mathrm{BrO}+\mathrm{O}_{2}\) Step II: \(\mathrm{BrO}+\mathrm{O} \rightarrow \mathrm{Br}+\mathrm{O}_{2}\) Which species is an intermediate in this mechanism? a. \(\mathrm{BrO}\) b. \(\mathrm{Br}\) c. \(\mathrm{O}_{3}\) d. \(\mathrm{O}\)

Short Answer

Expert verified
The intermediate is \(\mathrm{BrO}\).

Step by step solution

01

Identifying Reaction Mechanism Components

In a chemical mechanism, an intermediate is a species that is formed in one step of the mechanism and consumed in another. It does not appear in the overall balanced equation for the reaction.
02

Analyzing Step I

In Step I of the mechanism, \(\mathrm{Br} + \mathrm{O}_{3} \rightarrow \mathrm{BrO} + \mathrm{O}_{2}\), bromine monoxide \(\mathrm{BrO}\) is produced as a product.
03

Analyzing Step II

In Step II, \(\mathrm{BrO} + \mathrm{O} \rightarrow \mathrm{Br} + \mathrm{O}_{2}\), bromine monoxide \(\mathrm{BrO}\) is consumed as a reactant, producing bromine \(\mathrm{Br}\) and oxygen \(\mathrm{O}_{2}\).
04

Determining the Intermediate

Since \(\mathrm{BrO}\) is produced in Step I and consumed in Step II, and does not appear in the overall reaction, it fits the definition of an intermediate.

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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 crucial chemical process, particularly in the stratosphere where it plays a significant role in atmospheric chemistry. This natural process involves the breakdown of ozone molecules, transforming them into oxygen molecules. The decomposition of ozone typically occurs via a series of reactions rather than a single-step reaction, which is more complex but essential to understand. The ozone layer acts as a shield, protecting life on Earth from harmful ultraviolet (UV) radiation. However, substances such as bromine and chlorine, through various mechanisms, can lead to the decomposition of ozone. In the mechanism provided, ozone (O_3) decomposes in the presence of bromine (Br), starting a sequence of reactions that produces oxygen (O_2). Key points to understand while studying ozone decomposition:
  • Ozone molecules absorb UV light, which can initiate their breakdown.
  • Decomposition doesn't happen all at once; instead, it involves a series of steps where different species interact.
  • The role of catalysts like bromine is to speed up the reaction without being consumed permanently in the overall process.
Understanding each part of this sequence helps in grasping the larger picture of how substances in the atmosphere influence ozone levels.
Chemical Mechanisms
Chemical mechanisms are essentially a detailed sequence of steps that describe how a reaction occurs at a molecular level. In the case of ozone decomposition, these mechanisms outline the interactions between various molecules and radicals. When analyzing such mechanisms:
  • Each step has reactants and products, and collectively, they explain how the initial reactants are transformed into final products.
  • These steps are governed by rules of chemical kinetics and dynamics, including how energy is transferred and how molecular structures change.
  • Species that appear and disappear in different steps are key to understanding the overall mechanism.
Mechanisms are broken into individual reactions, each with their own rate, highlighting the intricate nature of how substances transform. For instance, in the ozone decomposition mechanism, the first step forms BrO, which is not seen in the final equation as it gets used up in the subsequent step. This sequential transformation is what helps us comprehend complex chemical changes occurring in environments like the stratosphere.
Intermediate Species Identification
Identifying intermediate species is a crucial skill in chemistry, especially when dissecting complex reactions like ozone decomposition. An intermediate is a transient species produced in one step and consumed in another, making them elusive in the overall balanced equation. In the decomposition mechanism of ozone with bromine, bromine monoxide (BrO) serves as an intermediate. Here’s what makes understanding intermediates essential:
  • They help in visualizing and verifying each stage of a multi-step reaction.
  • Intermediates provide insights into the dynamics and efficiencies of chemical reactions.
  • Studying intermediates helps chemists develop catalysts and other means to control reactions.
In practical terms, identifying intermediates means looking for species that appear in the equations of initial steps but disappear in the overall reaction. These fleeting species are usually not stable enough to be isolated but are critical for the progression of the reaction. For students, mastering this concept involves recognizing the formation and consumption patterns in chemical equations.

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Most popular questions from this chapter

The rate constant for the reaction, \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) is \(3.0 \times 10^{-5} \mathrm{~s}^{-1}\). If the rate is \(2.40 \times 10^{-5}\) mol litre \(^{-1}\) \(\mathrm{s}^{-1}\), then the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (in mol litre \(^{-1}\) ) is a. \(1.4\) b. \(1.2\) c. \(0.04\) d. \(0.8\)

The bromination of acetone that occurs in acid solution is represented by this equation. \(\mathrm{CH}_{3} \mathrm{COCH}_{3}(\mathrm{aq})+\mathrm{Br}_{2}\) (aq) \(\rightarrow\) \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{Br}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Br}(\mathrm{aq})\) These kinetic data were obtained from given reaction concentrations. Initial concentrations, (M) \(\begin{array}{lll}{\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]} & {\left[\mathrm{Br}_{2}\right]} & {\left[\mathrm{H}^{+}\right]} \\ 0.30 & 0.05 & 0.05 \\ 0.30 & 0.10 & 0.05 \\\ 0.30 & 0.10 & 0.10 \\ 0.40 & 0.05 & 0.20 \\ \text { Initial rate, disappearance of } & \end{array}\) disappearance of \(\mathrm{Br}_{2}, \mathrm{Ms}^{-1}\) \(5.7 \times 10^{-5}\) \(5.7 \times 10^{-5}\) \(1.2 \times 10^{-4}\) \(3.1 \times 10^{-4}\) Based on these data, the rate equation is: a. Rate \(=\mathrm{k}\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{Br}_{2}\right]\left[\mathrm{H}^{+}\right]^{2}\) b. Rate \(=\mathrm{k}\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{Br}_{2}\right]\left[\mathrm{H}^{+}\right]\) c. Rate \(=\mathrm{k}\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{H}^{+}\right]\) d. Rate \(=\mathrm{k}\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]\left[\mathrm{Br}_{2}\right]\)

If \(60 \%\) of a first order reaction was completed in 60 minutes, \(50 \%\) of the same reaction would be completed in approximately a. 50 minutes b. 45 minutes c. 60 minutes d. 40 minutes \((\log 4=0.60, \log 5=0.69)\)

The data given below is for the reaction of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) to form \(\mathrm{NOCl}\) at \(295 \mathrm{~K}\) What is the rate law? $$ \begin{array}{lll} \hline\left[\mathrm{Cl}_{2}\right] & {[\mathrm{NO}]} & \begin{array}{l} \text { Initial rate } \\ \left(\mathrm{mol}^{-1} \mathrm{~s}^{-1}\right) \end{array} \\ \hline 0.05 & 0.05 & 1 \times 10^{-3} \\ 0.15 & 0.05 & 3 \times 10^{-3} \\ 0.05 & 0.15 & 9 \times 10^{-3} \\ \hline \end{array} $$ a. \(\mathrm{r}=\mathrm{k}[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right]\) b. \(\mathrm{r}=\mathrm{k}\left[\mathrm{Cl}_{2}\right]^{1}[\mathrm{NO}]^{2}\) c. \(\mathrm{r}=\mathrm{k}\left[\mathrm{Cl}_{2}\right]^{2}[\mathrm{NO}]\) d. \(\mathrm{r}=\mathrm{k}\left[\mathrm{Cl}_{2}\right]^{1}\)

Which of the following is/are experimentally determined? a. Rate law b. Order c. Molecularity d. Rate constant

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