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At \(25^{\circ} \mathrm{C}, \mathrm{K}_{\mathrm{C}}\) for the equilibrium \(2 \mathrm{O}_{3}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{O}_{2}(\mathrm{~g})\) is \(5.5 \times 10^{57}\). This explains (1) the oxidizing power of ozone (2) the absence of detectable amounts of ozone in the lower atmospherc (3) the formation of ozonc laycr in the upper atmosphcre (4) the polluting nature of ozonc

Short Answer

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The high \(\text{K}_\text{C}\) explains the absence of detectable amounts of ozone in the lower atmosphere.

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01

Understand the Equilibrium Constant

The given problem provides the value of the equilibrium constant, \(\text{K}_\text{C} = 5.5 \times 10^{57}\), for the reaction \(2 \text{O}_3(\text{g}) \rightleftharpoons 3 \text{O}_2(\text{g})\). This large value indicates that at equilibrium, the concentration of \(\text{O}_2\) is much higher than that of \(\text{O}_3\).
02

Analyze the Implication of a High \(\text{K}_\text{C}\)

A very high equilibrium constant suggests that the reaction strongly favors the production of \(\text{O}_2\) over \(\text{O}_3\). This means that ozone (\(\text{O}_3\)) is highly reactive and tends to decompose into oxygen (\(\text{O}_2\)).
03

Explain the Context of Each Option

Evaluate each given option: \(1.\) Ozone's high oxidizing power is due to its tendency to react and decompose readily, as indicated by the high \(\text{K}_\text{C}\). \(2.\) The absence of detectable amounts of ozone in the lower atmosphere can also be explained by the reaction favoring the production of \(\text{O}_2\). \(3.\) The formation of an ozone layer in the upper atmosphere may involve factors other than just the equilibrium constant. \(4.\) The polluting nature of ozone relates to its reactivity and potential to form harmful compounds, again tied to its high tendency to decompose.
04

Determine the Correct Option

The given \(\text{K}_\text{C}\) value best explains the absence of detectable amounts of ozone in the lower atmosphere. Ozone decomposes readily into oxygen, making it difficult to detect significant amounts.

Key Concepts

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

Equilibrium Constant
In chemical reactions, the equilibrium constant \(\text{K}_\text{C}\) expresses the ratio of the concentration of products to reactants at equilibrium. For the reaction \(2 \text{O}_3(\text{g}) \rightleftharpoons 3 \text{O}_2(\text{g})\), a high \(\text{K}_\text{C}\) value of \(5.5 \times 10^{57}\) means that oxygen (\text{O}_2) is heavily favored over ozone (\text{O}_3) when the system is at equilibrium. This tells us that the reaction progresses significantly towards the products, leaving very little reactant. A high equilibrium constant often indicates a reaction that goes nearly to completion. This is particularly important in understanding the behavior and presence of ozone in various environments.
Ozone Decomposition
Ozone (\text{O}_3) is known for its instability and tendency to decompose into oxygen (\text{O}_2). This process is summarized by the reaction \(2 \text{O}_3(\text{g}) \rightleftharpoons 3 \text{O}_2(\text{g})\). The high \(\text{K}_\text{C}\) value for this reaction suggests that ozone readily breaks down into oxygen, showcasing its high chemical reactivity. This rapid decomposition explains why ozone is not commonly found in large amounts in the lower atmosphere, as it quickly converts through this reaction. The high reactivity and decomposition of ozone also underpin its powerful oxidizing ability, driving various chemical processes.
Ozone in Atmosphere
The ozone present in the atmosphere is crucial for absorbing harmful ultraviolet radiation from the sun. The concentration of ozone varies with altitude. In the lower atmosphere, the high equilibrium constant \(\text{K}_\text{C}\) for its decomposition reaction means less ozone is present; it mostly exists as oxygen. However, in the upper atmosphere, known as the stratosphere, ozone forms a protective layer. Here, the conditions—such as the presence of UV light—promote the formation and maintenance of ozone, balancing its creation and decomposition differently than in the lower atmosphere. This layer plays a vital role in protecting living organisms from UV radiation.
Chemical Reactivity
Chemical reactivity refers to how readily a substance undergoes a chemical reaction. Ozone is a highly reactive molecule due to its ease of decomposition into oxygen. This reactivity can explain why it acts as a powerful oxidizing agent, capable of reacting with a variety of other substances. This characteristic also contributes to its pollutant nature, as reactions involving ozone can produce harmful compounds and contribute to air pollution issues like smog. The equilibrium constant \(\text{K}_\text{C}\) reflects its tendency to change states easily, underpinning its reactive nature. Understanding ozone's chemical reactivity helps explain its diverse roles from an oxidizer to an atmospheric pollutant.

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Most popular questions from this chapter

\(1.1\) mole of \(\Lambda\) is mixed with \(2.2\) molc of \(B\) and the mixture is then kept in \(1-\mathrm{L}\) flask till the cquilibrium is attained \(\Lambda+2 \mathrm{~B} \rightleftharpoons 2 \mathrm{C}+\mathrm{D} \cdot \Lambda \mathrm{t}\) the equilibrium \(0.2\) mole of \(C\) are formed. The equilibrium constant of the reaction is (1) \(0.001\) (2) \(0.222\) (3) \(0.003\) (4) \(0.004\)

At a given temperature, the \(\mathrm{K}_{\mathrm{c}}\) for the reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) is \(2.4 \times 10^{-3} .\) At the same temperature, the \(K_{\mathrm{C}}\) for the reaction \(\mathrm{PCl}_{3}(\mathrm{~g})+\) \(\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{~g})\) is (1) \(2.4 \times 10^{-3}\) (2) \(2.4 \times 10^{-3}\) (3) \(4.2 \times 10^{2}\) (4) \(4.8 \times 10^{-2}\)

At \(1000^{\circ} \mathrm{C}\), the equilibrium constant for the reaction of the system \(2 \mathrm{II}_{2}(\mathrm{~g}) \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{II}_{2} \mathrm{O}(\mathrm{g})\) is very largc. This implics that (1) \(\mathrm{II}_{2} \mathrm{O}(\mathrm{g})\) is unstable at \(1000^{\circ} \mathrm{C}\) (2) \(\mathrm{II}_{2}(\mathrm{~g})\) is unstable at \(1000^{\circ} \mathrm{C}\) (3) \(\mathrm{II}_{2}\) and \(\mathrm{O}_{2}\) have very little tendency to combinc at \(1000^{\circ} \mathrm{C}\) (4) \(\mathrm{II}_{2} \mathrm{O}(\mathrm{g})\) has very little tendency to decompose into \(\mathrm{II}_{2}(\mathrm{~g})\) and \(\mathrm{O}_{2}(\mathrm{~g})\) at \(1000^{\circ} \mathrm{C}\)

Which oxide of nitrogen is the most stable? (1) \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{O}(\mathrm{g}) \mathrm{k}=6.7 \times 10^{6} \mathrm{~mol}\) litre (2) \(2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \mathrm{k}=2.2 \times 10^{30} \mathrm{~mol}\) litre (3) \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \mathrm{k}=1.2 \times 10^{34}\) mol litre (4) \(2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \mathrm{k}=3.5 \times 10^{23}\) mol litre

The decreasing order of acidic nature of perchloric acid, sulphuric acid and sodium bisulphate is (1) sulphuric acid \(>\) sodium bisulphate \(>\) perchloric acid (2) sodium bisulphate \(>\) sulphuric acid \(>\) perchloric acid (3) perchloric acid > sodium bisulphate \(>\) sulphuric acid (4) perchloric acid > sulphuric acid \(>\) sodium bisulphate

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