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

Step by step solution

01

Combine the given reactions to find the overall equation

We have the following two reactions: \( \begin{array}{c}{\mathrm{Cl}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g)} \\\ {\mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g)}\end{array} \) Add the reactions together to find the overall equation: \( \begin{array}{c} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ +\text{ } \mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \\ \hline \mathrm{O}_{3}(g) + \mathrm{O}(g) \longrightarrow 2\mathrm{O}_{2}(g) \end{array} \) The overall equation for the process is \(\mathrm{O}_{3}(g) + \mathrm{O}(g) \longrightarrow 2\mathrm{O}_{2}(g)\).
02

Identify the catalyst

A catalyst is a substance that participates in a reaction, but is not consumed during the process. In this case, we need to look for a substance that appears in both reactions but does not appear as a reactant or product in the overall equation. Looking at the two steps of the reaction, we can see that the substance "Cl" appears in both steps and is not present in the overall equation. Thus, the catalyst for this reaction is Chlorine (\(\mathrm{Cl}\)).
03

Identify the intermediate

An intermediate is a substance that is produced in one step of a reaction, but is consumed in a subsequent step. Again, it is not present as a reactant or product in the overall equation. Looking at the two steps of the reaction, we can see that the substance "ClO" appears in both steps and is not present in the overall equation. Thus, the intermediate for this reaction is Chlorine Monoxide (\(\mathrm{ClO}\)).
04

Results

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 Chlorine (\(\mathrm{Cl}\)). c) The intermediate in the reaction is Chlorine Monoxide (\(\mathrm{ClO}\)).

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Key Concepts

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

Catalytic Decomposition of Ozone
The upper atmosphere contains an essential layer of ozone (O₃) which protects life on Earth from harmful ultraviolet (UV) radiation. However, certain chemical processes can lead to the decomposition of this ozone, a phenomenon often referred to as ozone depletion. One such process involves a catalytic cycle where certain substances accelerate the breakdown of ozone without being consumed in the overall reaction.

Let's consider the simple yet illustrative two-step reaction provided in the exercise. Initially, a chlorine atom (Cl) reacts with ozone (O₃) to produce chlorine monoxide (ClO) and oxygen (O₂). Subsequently, the ClO reacts with a single oxygen atom (O) to regenerate the chlorine atom and form additional O₂. When these two reactions are combined to discern the overall effect on ozone, we find that the net result is simply the conversion of a molecule of O₃ and an atom of O into two molecules of O₂, with the chlorine acting as a catalyst.

This sequence exemplifies the catalytic decomposition of ozone—a single chlorine atom can destroy many ozone molecules due to its regeneration cycle and this detachment from the overall equation.
Chlorine as a Catalyst
In the context of ozone layer chemistry, chlorine is a notorious catalyst. A catalyst is a substance that speeds up a chemical reaction but remains unchanged at the end. It effectively lowers the energy barrier for the reaction, making it easier for other substances to react.

Chlorine atoms are particularly effective catalysts for ozone decomposition because they can react with ozone to initiate the breakdown, and yet chlorine itself is reformed by the end of the cycle, ready to commence the process anew. This cyclic nature allows for a single chlorine atom to destroy a vast number of ozone molecules over time.

The source of atmospheric chlorine is primarily from man-made chlorofluorocarbons (CFCs), once used in refrigeration and aerosol applications, which release chlorine atoms when broken down by UV light—a reminder of the profound impact human activity can have on environmentally critical chemistry.
Reaction Intermediates in Ozone Layer
During the catalytic decomposition of ozone, certain molecules are created and destroyed within the reaction cycle; these are known as intermediates. In the exercise, we identify chlorine monoxide (ClO) as an intermediate.

Intermediates such as ClO are crucial for the overall reaction to proceed. After chlorine (Cl) reacts with ozone (O₃) to form ClO, this molecule subsequently reacts with a single oxygen atom (O) to regenerate the chlorine atom and create more oxygen (O₂). Though vital for the transition from reactants to final products, intermediates like ClO do not appear in the overall reaction equation because they are not present at the start or finish—they exist transiently.

Understanding the role of intermediates is essential for comprehending complex reaction mechanisms in atmospheric chemistry, which often include several sequential steps involving numerous transitional species.

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Most popular questions from this chapter

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right),\) with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{s}^{-1}\) at \(100^{\circ} \mathrm{C} .\) In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{s}^{-1}\) at \(21^{\circ} \mathrm{C}\) . (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C},\) what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{Cas} \mathrm{com}-\) pared to that at \(21^{\circ} \mathrm{C} ?(\mathbf{d})\) On the basis of parts \((\mathrm{c})\) and \((\mathrm{d}),\) what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{nm}\). The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has an extinction coefficient of \(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(520 \mathrm{nm}\). (a) Calculate the initial concentration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction. (b) The absorbance falls to 0.250 at \(30.0 \mathrm{~min}\). Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

The gas-phase reaction of NO with \(\mathrm{F}_{2}\) to form \(\mathrm{NOF}\) and \(\mathrm{F}\) has an activation energy of \(E_{a}=6.3 \mathrm{kJ} / \mathrm{mol} .\) and a frequency factor of \(A=6.0 \times 10^{8} M^{-1} \mathrm{s}^{-1} .\) The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g)$$ (a) Calculate the rate constant at \(100^{\circ} \mathrm{C}\) . (b) Draw the Lewis structures for the NO and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule, (c) Predict the shape for the NOF molecule.Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.

Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a \(1 \times 10^{5}\) -fold increase in the reaction rate?

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\) . When \([\mathrm{NO}]=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M},\) the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\) . (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of 1.8\(?\)

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