/*! 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 108 Ozone in the upper atmosphere ca... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
(a) The overall equation for this process is: \(O_{3}(g) + O(g) \longrightarrow 2 O_{2}(g)\) (b) The catalyst in this reaction is Cl(g). (c) The intermediate in this reaction is ClO(g).

Step by step solution

01

Write down both reactions

Write down the two reactions in their respective steps: 1) Cl(g) + O3(g) → ClO(g) + O2(g) 2) ClO(g) + O(g) → Cl(g) + O2(g)
02

Add the reactions and cancel common species

Add the two reactions and cancel out the common species on both sides of the reaction: 1) Cl(g) + O3(g) → ClO(g) + O2(g) 2) ClO(g) + O(g) → Cl(g) + O2(g) --------------------------------- Overall: O3(g) + O(g) → 2 O2(g) Hence, the overall equation for this process is: \(O_{3}(g) + O(g) \longrightarrow 2 O_{2}(g)\) (b) Catalysts speed up reactions but are not consumed in the process; they appear in the initial and final stages of the reaction but do not appear in the overall equation.
03

Identify the catalyst

From the reactions above, we can see that Cl(g) is present in the reactants of the first step and in the products of the second step. Thus, Cl(g) is the catalyst in this reaction. (c) Intermediates are produced in one step of a reaction and then consumed in a later step; they do not appear in the overall equation.
04

Identify the intermediate

From the reactions above, we can identify that ClO(g) is produced in the first step and then consumed in the second step. Therefore, ClO(g) is the intermediate in this reaction.

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Most popular questions from this chapter

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\), what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part (a)? (c) How do the half-lives of first-order and second-order reactions differ?

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2}\). The reaction proceeds according to: $$ \begin{aligned} &\mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ &2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{aligned} $$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3}\). (b) \(\mathrm{Do}\) we consider \(\mathrm{NO}_{2}\) a catalyst or an intermediate in this reaction? (c) Is this an example of homogeneous catalysis or heterogeneous catalysis?

For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (d) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{NH}_{3}(g)\)

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at \(330 \mathrm{~K}_{2} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(\)alc \()\), is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is \(0.0477 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 M\), the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

(a) What is meant by the term reaction rate? (b) Name three factors that can affect the rate of a chemical reaction. (c) Is the rate of disappearance of reactants always the same as the rate of appearance of products?

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