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Chapter 14: Problem 113

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: $$ \begin{array}{l} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{array} $$ (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
(a) The overall equation for this process is: \(O_3 (g) + O (g) \longrightarrow 2 O_2 (g)\). (b) The catalyst in the reaction is Chlorine (Cl). We know this because it is present in both steps of the mechanism but not consumed in the overall reaction. (c) The intermediate in the reaction is Chlorine monoxide (ClO). We distinguish it from the catalyst because it is formed and consumed within the reaction steps playing the role of a bridge, while the catalyst is not consumed during the whole process.

Step by step solution

01

Combine the two equations

To find the overall equation, we need to add the two given equations. Firstly, let's write them down: 1) \(Cl (g) + O_3 (g) \longrightarrow ClO (g) + O_2 (g)\) 2) \(ClO (g) + O (g) \longrightarrow Cl (g) + O_2 (g)\) Now, let's sum the equations: \(Cl (g) + O_3 (g) + ClO (g) + O (g) \longrightarrow ClO (g) + O_2 (g) + Cl (g) + O_2 (g)\)
02

Simplify the overall equation

Now, let's eliminate equal substances from both sides of the combined equation: - \(ClO (g)\) appears on both sides, so we can eliminate it. - Also, \(Cl (g)\) appears on both sides, so we can eliminate it too. Now, the simplified overall equation is: \(O_3 (g) + O (g) \longrightarrow 2 O_2 (g)\) (a) So, the overall equation for this process is: \(O_3 (g) + O (g) \longrightarrow 2 O_2 (g)\).
03

Identify the catalyst

A catalyst is a substance that is not consumed in the overall reaction and increases the reaction rate. From the overall equation, we can see that Chlorine (Cl) appears in both steps but is not part of the overall reaction. (b) Therefore, the catalyst in the reaction is Chlorine (Cl). We know this because it is present in both steps of the mechanism, but it is not consumed in the overall reaction.
04

Identify the intermediate

An intermediate is a substance that is formed in one step and consumed in another step. In this mechanism, Chlorine monoxide (ClO) is formed in the first step and consumed in the second step. (c) So, the intermediate in the reaction is Chlorine monoxide (ClO). We distinguish it from the catalyst because it is formed and consumed within the reaction steps playing the role of a bridge, while the catalyst is not consumed during the whole process.

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Most popular questions from this chapter

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the \(\mathrm{CO}_{2}\) accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to \(10^{7} \mathrm{CO}_{2}\) molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

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

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{~s}^{-1}\). Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from \(1.00 \mathrm{~L}\) of \(0.600 \mathrm{M} \mathrm{N}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of \(20.0 \mathrm{hr}\) if the gas is collected in a \(10.0-\mathrm{L}\) container. (Assume that the products do not dissolve in chloroform.)

(a) If you were going to build a system to check the effectiveness of automobile catalytic converters on cars, what substances would you want to look for in the car exhaust? (b) Automobile catalytic converters have to work at high temperatures, as hot exhaust gases stream through them. In what ways could this be an advantage? In what ways a disadvantage? (c) Why is the rate of flow of exhaust gases over a catalytic converter important?

(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 \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?(\mathrm{c})\) What assumptions do you need to make in order to calculate answers for parts (a) and (b)?
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