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Chapter 13: Problem 118

Explain why NO is a catalyst in the following two-step process that results in the depletion of ozone in the stratosphere: (1) \(\quad \mathrm{NO}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{O}(g)\) (2) \(\quad \mathrm{O}(g)+\mathrm{NO}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g)\) Overall: \(\quad \mathrm{O}(g)+\mathrm{O}_{3}(g) \rightarrow 2 \mathrm{O}_{2}(g)\)

Short Answer

Expert verified
Answer: NO is considered a catalyst in the given two-step process of ozone depletion because it helps increase the rate of the reaction without being consumed. It participates in the reaction temporarily and is regenerated at the end of the process, making it a catalyst in the overall depletion of ozone in the stratosphere.

Step by step solution

01

Understand the reactions

In the given two-step process, the first reaction involves nitrogen monoxide (NO) reacting with ozone (O3) to form nitrogen dioxide (NO2) and an oxygen atom (O). In the second reaction, the oxygen atom reacts with NO2 to form NO and an oxygen molecule (O2).
02

Identify the final product

Combining the two reactions, we get the overall reaction: O(g) + O3(g) 鈫� 2 O2(g) This reaction represents the depletion of ozone, where an oxygen atom and an ozone molecule combine to form two oxygen molecules.
03

Determine the role of NO

Let's examine the change in NO throughout the reactions. In the first reaction, NO is consumed, while NO2 is produced. In the second reaction, NO is produced and NO2 is consumed. This means that, after both reactions occur, the initial amount of NO is regenerated. Consequently, NO does not undergo any net change and is not consumed in the overall reaction.
04

Define a catalyst

A catalyst is a substance that increases the rate of a chemical reaction, without being consumed. It participates in the reaction temporarily, helping the molecules to react, but is regenerated at the end of the process.
05

Conclude the role of NO

Since NO is not consumed in the two-step process and is regenerated at the end of the reaction, we can conclude that NO acts as a catalyst in this process that results in the depletion of ozone in the stratosphere.

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Key Concepts

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

Catalyst
A catalyst is a fascinating component in chemistry. It speeds up chemical reactions without being consumed or permanently altered. How does it manage this? By providing an alternative reaction pathway with lower activation energy.
In other words, it helps the reactants interact more efficiently. Catalysts remain unchanged at the end of the reaction.
This makes them valuable in various processes as they can be used over and over again.
  • Increases reaction rate
  • Not consumed in the reaction
  • Regenerated and unchanged
Catalysts are essential in environmental chemistry too. For instance, they play a crucial role in reducing ozone depletion in the stratosphere.
Nitrogen Monoxide (NO)
Nitrogen Monoxide, also known as nitric oxide, is a simple yet crucial molecule in atmospheric chemistry. It has various roles but is particularly notable when discussing air pollution and ozone chemistry.
In the context of ozone depletion, NO acts as a catalyst. It participates in reactions without being consumed, facilitating the breakdown of ozone molecules.
  • Acts as a catalyst in ozone-depleting reactions
  • Regenerates itself after reactions
  • Contributes to pollution but also aids studies in atmospheric chemistry
Understanding NO's role is critical to grasping the mechanisms behind ozone layer depletion.
Stratosphere
The stratosphere is a layer of Earth's atmosphere situated above the troposphere, typically spanning heights from 10 km to 50 km. This region plays a critical role in life on Earth as it contains the ozone layer.
The ozone layer absorbs and scatters ultraviolet solar radiation. This shields living organisms on Earth from harmful UV rays. However, certain chemical reactions, like those involving nitrogen monoxide, can deplete ozone levels.
  • Second layer of Earth's atmosphere
  • Home to the protective ozone layer
  • Site of critical atmospheric reactions
Understanding the stratosphere is crucial for comprehending where and how ozone depletion occurs.
Chemical Reaction Mechanism
A chemical reaction mechanism details each step of a chemical process, illustrating how reactants transform into products. This concept is vital for understanding complex reactions like the two-step process involving nitrogen monoxide and ozone.
In the provided mechanism:
  • First, NO reacts with ozone to form NO2 and a free oxygen atom.
  • Then, this oxygen atom reacts with NO2 to regenerate NO and produce O2.
The beauty of the mechanism lies in its ability to reveal why NO serves as a catalyst, showing how it enters and exits the reactions unchanged. By focusing on individual steps, scientists can manipulate conditions to enhance desired reactions or inhibit harmful ones.

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Most popular questions from this chapter

Why are the units of the rate constants different for reactions of different order?

Bacterial Degradation of Ammonia Nitrosomonas bacteria convert ammonia into nitrite in the presence of oxygen by the following reaction: \(2 \mathrm{NH}_{3}(a q)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{2}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) A. How are the rates of formation of \(\mathrm{H}^{+}\) and \(\mathrm{NO}_{2}^{-}\) related to the rate of consumption of \(\mathrm{NH}_{3} ?\) b. How is the rate of formation of \(\mathrm{NO}_{2}^{-}\) related to the rate of consumption of \(\mathrm{O}_{2} ?\) c. How is the rate of consumption of \(\mathrm{NH}_{3}\) related to the rate of consumption of \(\mathrm{O}_{2} ?\)

Does a catalyst affect both the rate and the rate constant of a reaction?

The compound 1,1 -difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride: $$\mathrm{CH}_{3} \mathrm{CHF}_{2}(g) \rightarrow \mathrm{CH}_{2} \mathrm{CHF}(g)+\mathrm{HF}(g)$$ At \(460^{\circ} \mathrm{C}, k=5.8 \times 10^{-6} \mathrm{s}^{-1}\) and \(E_{\mathrm{a}}=265 \mathrm{kJ} / \mathrm{mol} .\) To what temperature would you have to raise the reaction to make it go four times as fast?

Because the units of concentration in the term \(\ln \left([\mathrm{X}] /[\mathrm{X}]_{0}\right)\) cancel out in the integrated rate law for first- order reactions (Equation 13.16 ), molar concentration can be replaced by any concentration term. With gases, for example, partial pressures may be used. The decomposition of phosphine gas \(\left(\mathrm{PH}_{3}\right)\) at \(600^{\circ} \mathrm{C}\) is first order in \(\mathrm{PH}_{3}\) with \(k=0.023 \mathrm{s}^{-1}\) $$4 \mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g)$$ If the initial partial pressure of \(\mathrm{PH}_{3}\) is 375 torr, what percent of \(\mathrm{PH}_{3}\) reacts in \(1 \mathrm{min}\) ?
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