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Chapter 14: Problem 108

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: \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2\mathrm{O}_{2}(g)\) (b) The catalyst in the reaction is \(\mathrm{Cl}(g)\) (chlorine). (c) The intermediate in the reaction is \(\mathrm{ClO}(g)\) (chlorine monoxide).

Step by step solution

01

Add both reactions together to find the overall equation

Add both reactions together to determine the overall equation. You can sum up the reactions horizontally by adding the reactants and the products separately on both sides of the equation. $$ \begin{aligned} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) & \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) & \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \\ \hline \mathrm{Cl}(g)+\mathrm{O}_{3}(g)+\mathrm{O}(g)+\mathrm{ClO}(g) & \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ Now, eliminate the common terms on the left and right sides of the arrow, which are Cl and ClO: $$ \begin{aligned} \mathrm{O}_{3}(g)+\mathrm{O}(g) & \longrightarrow \mathrm{O}_{2}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ Combine the two O₂(g) on the right side: $$ \begin{aligned} \mathrm{O}_{3}(g)+\mathrm{O}(g) & \longrightarrow 2\mathrm{O}_{2}(g) \end{aligned} $$ This equation shows the overall process. (a) The overall equation for this process is: $$ \begin{aligned} \mathrm{O}_{3}(g)+\mathrm{O}(g) & \longrightarrow 2\mathrm{O}_{2}(g) \end{aligned} $$
02

Identify the catalyst

The catalyst will be present on both sides of the overall equation because it is not consumed in the reaction. - In this case, \(\mathrm{Cl}(g)\) and \(\mathrm{ClO}(g)\) are present in both sides of the overall equation, leading us to believe that they are the catalysts. (b) The catalyst in the reaction is \(\mathrm{Cl}(g)\) (chlorine).
03

Determine the intermediate in the reaction

The intermediate is a substance that appears in both reactions but not in the overall equation. - In this case, \(\mathrm{ClO}(g)\) is present in both reactions but not in the overall equation, so it is the intermediate. (c) The intermediate in the reaction is \(\mathrm{ClO}(g)\) (chlorine monoxide).

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Most popular questions from this chapter

Based on their activation energies and energy changes and assuming that all collision factors are the same, rank the following reactions from slowest to fastest. (a) $E_{a}=75 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-20 \mathrm{~kJ} / \mathrm{mol}$ (b) $E_{a}=100 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=+30 \mathrm{~kJ} / \mathrm{mol}$ (c) $E_{a}=85 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-50 \mathrm{~kJ} / \mathrm{mol}$

Many primary amines, \(\mathrm{RNH}_{2}\), where \(\mathrm{R}\) is a carboncontaining fragment such as $\mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2},$ and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a C atom for "R"). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

The reaction between ethyl bromide $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\( and hydroxide ion in ethyl alcohol at \)330 \mathrm{~K}$, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c),$ 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 \mathrm{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?

In a hydrocarbon solution, the gold compound $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\( decomposes into ethane \)\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ and a different gold compound, \(\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3} .\) The following mechanism has been proposed for the decomposition of $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}:$ Step 1: $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \underset{k-1}{\stackrel{k_{1}}{\rightleftharpoons}}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3} \quad$ (fast) Step 2: $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{\mathrm{k}_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au} \quad$ (slow) Step 3: $\left(\mathrm{CH}_{3}\right) \mathrm{Au}+\mathrm{PH}_{3} \stackrel{k_{3}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3} \quad$ (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary steps? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (f) What would be the effect on the reaction rate of adding \(\mathrm{PH}_{3}\) to the solution of $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} ?$

(a) A certain first-order reaction has a rate constant of $2.75 \times 10^{-2} \mathrm{~s}^{-1}\( at \)20^{\circ} \mathrm{C}\(. What is the value of \)k$ at \(60^{\circ} \mathrm{C}\) if $E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{b})\( Another first-order reaction also has a rate constant of \)2.75 \times 10^{-2} \mathrm{~s}^{-1}\( at \)20^{\circ} \mathrm{C}$. What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if $E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{c})$ What assumptions do you need to make in order to calculate answers for parts (a) and (b)?
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