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A mechanism for a naturally occurring reaction that destroys ozone is: Step I: \(\mathrm{O}_{3}(\mathrm{~g})+\mathrm{HO}(\mathrm{g}) \rightarrow \mathrm{HO}_{2}(\mathrm{~g})+\mathrm{O}_{2}\) Step II: \(\mathrm{HO}_{2}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{HO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) Which species is an intermediate? a. \(\mathrm{O}\) b. \(\mathrm{O}_{3}\) c. HO d. \(\mathrm{HO}_{2}\)

Short Answer

Expert verified
The intermediate is HO2.

Step by step solution

01

Understand the Mechanism

Examine the given reaction mechanism consisting of two steps. In these steps, the transformation of reactants to products occurs through intermediates. One of these species appears in each reaction step but cancels out in the overall reaction.
02

Identify Intermediates

An intermediate is a species that is produced in one step of the reaction and consumed in a subsequent step. Here, \(nputs]O](mathrm{~g})\) is a reactant in Step I but appears as a product in Step II, and it cancels in the overall reaction. Similarly, \(nputs]O_{2}](mathrm{~g})\) is formed in Step I and used in Step II, indicating it does not appear in the final reaction equation.
03

Determine the Intermediate

Since \(HO_{2}(\mathrm{~g})\) is produced in Step I and completely used in Step II without appearing in the overall reaction, it acts as an intermediate. This confirms \(HO_{2}\) is the intermediate species.

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Key Concepts

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

Ozone Depletion
Ozone depletion is a critical environmental issue caused by various chemical reactions that occur in the stratosphere. Ozone, or \( \text{O}_3 \, \,properly\), helps absorb the sun's harmful ultraviolet rays, acting as Earth's protective shield. When this layer is depleted, increased UV radiation reaches the Earth's surface, which can lead to more cases of skin cancer, cataracts, and have harmful effects on animals and plants.
Many factors contribute to the destruction of the ozone layer, but certain chemical reactions involving naturally occurring and human-made substances are significant culprits. For instance, hydroxyl radicals (\( \text{HO} \)) and other reactive substances can break ozone molecules, initiating a chain of reactions that continue to deplete ozone. These reactions occur through mechanisms composed of several steps, often involving reaction intermediates that cycle the reactive species, thus amplifying the amount of ozone destroyed by these processes. Understanding these mechanisms is crucial in environmental chemistry and helps formulate policies and interventions to protect our ozone layer.
Intermediates in Reactions
In chemical reactions, especially those involved in complex mechanisms like ozone depletion, intermediates play a vital role. Intermediates are species that are formed in one step of a reaction mechanism and used up in a subsequent step without appearing in the overall reaction equation. Recognizing these intermediates is important as they help drive the process from reactants to products, even though they don't appear in the final balanced reaction.
For instance, in the provided exercise, \(\text{HO}_2(\mathrm{g})\) is an intermediate. It is generated as a product in the first reaction step and then consumed as a reactant in the next step. This means it facilitates the reaction but is not part of the final outcome, highlighting its temporary yet crucial role. By identifying and studying these intermediates, chemists can better understand and manipulate chemical reactions, including those significant to environmental and industrial processes.
Chemical Kinetics
Chemical kinetics is the study of the rates at which chemical reactions occur. It involves understanding how different factors such as temperature, pressure, concentration, and catalysts affect these rates. In the context of reaction mechanisms like ozone depletion, kinetics provides insight into how quickly these reactions progress and under what conditions they are most effective.
Reaction mechanisms often consist of several steps, each with its own rate, which collectively determine the overall reaction rate. The step with the slowest rate is typically known as the rate-determining step because it acts as a bottleneck for the process. By studying kinetics, researchers can identify these steps and optimize conditions to either speed up beneficial reactions or slow down harmful ones, like those leading to ozone depletion.
  • Understanding kinetics helps in controlling environmental impacts.
  • Kinetics provides the tools to develop efficient industrial chemical processes.
In environmental chemistry, kinetics is essential for modeling how pollutants affect natural systems and for designing strategies to mitigate their effects.

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Most popular questions from this chapter

In the following question two statements Assertion (A) and Reason (R) are given Mark. a. If \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\); b. If \(A\) and \(R\) both are correct but \(R\) is not the correct explanation of \(\mathrm{A}\); c. \(\mathrm{A}\) is true but \(\mathrm{R}\) is false; d. \(\mathrm{A}\) is false but \(\mathrm{R}\) is true, e. \(\mathrm{A}\) and \(\mathrm{R}\) both are false. (A): For the hydrogen halogen photochemical reaction, the quantum yield for the formation of \(\mathrm{HBr}\), is lower than that of \(\mathrm{HCl}\). (R): \(\mathrm{Br}+\mathrm{H}_{2} \rightarrow \mathrm{HBr}+\mathrm{H}\) has higher activation energy than \(\mathrm{Cl}+\mathrm{H}_{2} \rightarrow \mathrm{HCl}+\mathrm{H}\)

When the concentration of \(\mathrm{A}\) is doubled, the rate for the reaction: \(2 \mathrm{~A}+\mathrm{B} \rightarrow 2 \mathrm{C}\) quadruples. When the concentration of \(\mathrm{B}\) is doubled the rate remains the same. Which mechanism below is consistent with the experimental observations? a. Step I: \(2 \mathrm{~A} \rightleftharpoons \mathrm{D}\) (fast equilibrium) Step II: \(\mathrm{B}+\mathrm{D} \rightarrow \mathrm{E}\) (slow) Step III: \(\mathrm{E} \rightarrow 2 \mathrm{C}\) (fast) b. Step I: \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{D}\) (fast equilibrium) Step II: \(\mathrm{A}+\mathrm{D} \rightarrow 2 \mathrm{C}\) (slow) c. Step \(\mathrm{I}: \mathrm{A}+\mathrm{B} \rightarrow \mathrm{D}\) (slow) Step II: \(\mathrm{A}+\mathrm{D} \rightleftharpoons 2 \mathrm{C}\) (fast equilibrium) d. Step I: \(2 \mathrm{~A} \rightarrow \mathrm{D}\) (slow) Step II: \(\mathrm{B}+\mathrm{D} \rightarrow \mathrm{E}\) (fast) Step III: \(\mathrm{E} \rightarrow 2 \mathrm{C}\) (fast)

The rate constant of a reaction is \(1.5 \times 10^{7} \mathrm{~s}^{-1}\) at \(50^{\circ} \mathrm{C}\) and \(4.5 \times 10^{7} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). What is the value of activation energy? a. \(2.2 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\) b. \(2300 \mathrm{~J} \mathrm{~mol}^{-1}\) c. \(2.2 \times 10^{4} \mathrm{~J} \mathrm{~mol}^{-1}\) d. \(220 \mathrm{~J} \mathrm{~mol}^{-1}\)

Which of the following graphs for a first order reaction ( \(\mathrm{A} \rightarrow\) Products) would be straight line? a. Rate vs time b. Rate vs \([\mathrm{A}]\) c. Rate vs \(\log [\mathrm{A}]\) d. \(\log [\mathrm{A}]\) vs time

The following set of data was obtained by the method of initial rates for the reaction: $$ \begin{array}{r} \mathrm{BrO}_{3}^{-}(\mathrm{aq})+5 \mathrm{Br}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \\ 3 \mathrm{Br}_{2}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O} \text { (1) } \end{array} $$ Calculate the initial rate when \(\mathrm{BrO}_{3}^{-}\)is \(0.30 \mathrm{M}\), \(\mathrm{Br}\) is \(0.050 \mathrm{M}\) and \(\mathrm{H}^{+}\)is \(0.15 \mathrm{M}\). $$ \begin{array}{llll} \hline\left[\mathrm{BrO}_{3}^{-}\right], \mathrm{M} & {[\mathrm{Br}], \mathrm{M}} & {\left[\mathrm{H}^{+}\right], \mathrm{M}} & \text { Rate, } \mathrm{M} / \mathrm{s} \\ \hline 0.10 & 0.10 & 0.10 & 8.0 \times 10^{-4} \\ 0.20 & 0.10 & 0.10 & 1.6 \times 10^{-3} \\ 0.20 & 0.15 & 0.10 & 2.4 \times 10^{-3} \\ 0.10 & 0.10 & 0.25 & 5.0 \times 10^{-3} \\ \hline \end{array} $$ a. \(3.17 \times 10^{-4} \mathrm{M} / \mathrm{s}\) b. \(6.7 \times 10^{-3} \mathrm{M} / \mathrm{s}\) c. \(2.7 \times 10^{-3} \mathrm{M} / \mathrm{s}\) d. \(1.71 \times 10^{-3} \mathrm{M} / \mathrm{s}\)

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