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Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: $$ \begin{gathered} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{gathered} $$ (a) What is the overall equation for this process? (b) What is the catalyst in the reaction? How do you know? (c) What is the intermediate in the reaction? How do you distinguish it from the catalyst?

Short Answer

Expert verified
The overall equation for the ozone destruction process is \(O_{3}(g) + O(g) \longrightarrow 2O_{2}(g)\). The catalyst in this reaction is the Cl (Chlorine) atom, as it participates in the reaction steps but remains chemically unchanged at the end. The intermediate is ClO (Chlorine monoxide), as it forms during the reaction and gets consumed later in another step. The main difference between the catalyst and the intermediate is their roles in the reaction: the catalyst speeds up the reaction, while the intermediate is a temporary species that forms and gets consumed during the reaction process.

Step by step solution

01

Determine the overall equation

To find the overall equation for this two-step mechanism, we can add the equations of both steps and then simplify by cancelling out common species that appear on both sides of the reaction. Step 1: \(Cl(g) + O_{3}(g) \longrightarrow ClO(g) + O_{2}(g) \) Step 2: \(ClO(g) + O(g) \longrightarrow Cl(g) + O_{2}(g) \) Adding the equations together, we get: \(Cl(g) + O_{3}(g) + ClO(g) + O(g) \longrightarrow ClO(g) + O_{2}(g) + Cl(g) + O_{2}(g) \) Now, we can simplify by cancelling out the common species (Cl and ClO) that appear on both sides of the reaction: \(O_{3}(g) + O(g) \longrightarrow 2O_{2}(g) \) So, the overall equation for this process is: \(O_{3}(g) + O(g) \longrightarrow 2O_{2}(g) \)
02

Identify the catalyst

A catalyst is a substance that takes part in a reaction and speeds up the rate of the reaction but remains chemically unchanged at the end of the reaction. If we look at the two-step mechanism and the overall equation, we can see that the Cl (Chlorine) atom is participating in both steps of the reaction, but it is not consumed in the overall reaction. Therefore, Cl (Chlorine) is the catalyst in this reaction.
03

Identify the intermediate

An intermediate is a species that is formed during a reaction but does not appear in the overall equation as it gets consumed during the reaction. In the two-step mechanism, we can see that ClO (Chlorine monoxide) is formed in the first step and then consumed in the second step. It does not appear in the overall equation. Therefore, ClO (Chlorine monoxide) is the intermediate in this reaction.
04

Distinguish between the catalyst and the intermediate

The main difference between the catalyst and the intermediate in this reaction is their roles and behavior during the reaction: 1. Catalyst (Cl) - It participates in the reaction steps but remains chemically unchanged at the end of the reaction. It helps to speed up the reaction by lowering the activation energy. 2. Intermediate (ClO) - It is formed during one of the reaction steps and is consumed later in another step. It does not appear in the overall equation as it is not a part of the initial reactants or final products. In this ozone destruction reaction, the catalyst (Cl - Chlorine) is responsible for speeding up the reaction, while the intermediate (ClO - Chlorine monoxide) is a temporary species that forms and later gets consumed during the reaction process.

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Most popular questions from this chapter

The following mechanism has been proposed for the gas-phase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{aligned} &\mathrm{H}_{2}(\mathrm{~g})+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ &\mathrm{HI}(g)+\mathrm{ICl}(g) \rightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{aligned} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) Write rate laws for each elementary reaction in the mechanism. (d) If the first step is slow and the second one is fast, what rate law do you expect to be observed for the overall reaction?

The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows: $$ \begin{array}{ll} \hline \text { Temperature (K) } & \text { Rate Constant (s }^{-1} \text { ) } \\\ \hline 300 & 3.2 \times 10^{-11} \\ 320 & 1.0 \times 10^{-9} \\ 340 & 3.0 \times 10^{-8} \\ 355 & 2.4 \times 10^{-7} \\ \hline \end{array} $$ From these data, calculate the activation energy in units of \(\mathrm{kJ} / \mathrm{mol}\).

Cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) reacts with itself to form dicyclopentadiene \(\left(\mathrm{C}_{10} \mathrm{H}_{12}\right) .\) A \(0.0400 \mathrm{M}\) solution of \(\mathrm{C}_{5} \mathrm{H}_{6}\) was monitored as a function of time as the reaction \(2 \mathrm{C}_{5} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12}\) proceeded. The following data were collected: $$ \begin{array}{rl} \text { Time (s) } & {\left[\mathrm{C}_{5} \mathrm{H}_{6}\right](M)} \\ \hline 0.0 & 0.0400 \\ 50.0 & 0.0300 \\ 100.0 & 0.0240 \\ 150.0 & 0.0200 \\ 200.0 & 0.0174 \end{array} $$ Plot \(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, \(\ln \left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, and \(1 /\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time. What is the order of the reaction? What is the value of the rate constant?

You have studied the gas-phase oxidation of \(\mathrm{HBr}\) by \(\mathrm{O}_{2}\) : $$ 4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g) $$ You find the reaction to be first order with respect to \(\mathrm{HBr}\) and first order with respect to \(\mathrm{O}_{2}\). You propose the following mechanism: $$ \begin{aligned} \mathrm{HBr}(g)+\mathrm{O}_{2}(g) & \rightarrow \mathrm{HOOBr}(g) \\ \mathrm{HOOBr}(g)+\mathrm{HBr}(g) & \longrightarrow 2 \mathrm{HOBr}(g) \\ \mathrm{HOBr}(g)+\mathrm{HBr}(g) \longrightarrow & \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g) \end{aligned} $$ (a) Indicate how the elementary reactions add to give the overall reaction. (Hint: You will need to multiply the coefficients of one of the equations by 2.) (b) Based on the rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

(a) Define the following symbols that are encountered in rate equations: \([\mathrm{A}]_{0}, t_{1 / 2}[\mathrm{~A}]_{t}, k .(\mathrm{b})\) What quantity, when graphed versus time, will yield a straight line for a firstorder reaction?

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