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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

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in 0.1 \(\mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$\mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)$$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 M,\) the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\) , (a) What is the rate constant, \(k ?\) units of \(s^{-1}\) . (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to 0.100\(?\)

What is the molecularity of each of the following elementary reactions? Write the rate law for each. \(\begin{array}{l}{\text { (a) } \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g)} \\ {\text { (b) } \mathrm{OCl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HOCl}(a q)+\mathrm{OH}^{-}(a q)} \\\ {\text { (c) } \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}_{2}(g)}\end{array}\)

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(a l c),\) has an activation energy of 86.8 \(\mathrm{kJ} / \mathrm{mol}\) and a frequency factor of \(2.10 \times 10^{11} \mathrm{M}^{-1} \mathrm{s}^{-1}\) (a) Predict the rate constant for the reaction at \(35^{\circ} \mathrm{C} .\) (b) A g \(\mathrm{KOH}\) in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Similarly, 1.453 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reac-solution of \(\mathrm{KOH}\) in ethanol is made up by dissolving 0.335 g KOH in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Similarly, 1.453 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(35^{\circ} \mathrm{C} ?(\mathbf{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(50^{\circ} \mathrm{C}\) .

Consider the following reaction: $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in \(\mathrm{NO}\) . Write the rate law. (b) If the rate constant for this reaction at 1000 \(\mathrm{K}\) is \(6.0 \times 10^{4} M^{-2} \mathrm{s}^{-1}\) what is the reaction rate when \([\mathrm{NO}]=0.035 M\) and \(\left[\mathrm{H}_{2}\right]=0.015 M ?\) (c) What is the reaction rate at 1000 \(\mathrm{K}\) when the concentration of \(\mathrm{NO}\) is increased to 0.10 \(\mathrm{M}\)while the concentration of \(\mathrm{H}_{2}\) is 0.010\(M ?\) (d) What is the reaction rate at 1000 \(\mathrm{K}\) if \([\mathrm{NO}]\) is decreased to 0.010 \(\mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]\) is increased to 0.030 \(\mathrm{M}\) ?

(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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