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Chapter 21: Problem 64

The equilibrium constant \(\left(K_{P}\right)\) for the reaction \(2 \mathrm{CO}(g)\) \(+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{CO}_{2}(g)\) is \(1.4 \times 10^{90}\) at \(25^{\circ} \mathrm{C}\). Given this enormous value, why doesn't \(\mathrm{CO}\) convert totally to \(\mathrm{CO}_{2}\) in the troposphere?

Short Answer

Expert verified
Kinetic barriers and specific atmospheric conditions prevent complete conversion of CO to CO2 despite the large K_P value.

Step by step solution

01

Understand the Equilibrium Constant

The equilibrium constant \(K_P\) is a measure of the tendency of a chemical reaction to proceed to completion. A very large \(K_P\) value, such as \(1.4 \times 10^{90}\), suggests that the reaction strongly favors the formation of products, in this case, \(\mathrm{CO}_2\).
02

Identify the Reaction Context

The reaction in question occurs in the troposphere, the lowest part of Earth's atmosphere. Here, conditions such as temperature, pressure, and the presence of catalysts greatly influence reaction rates and equilibria.
03

Consider Reaction Kinetics

Despite the large \(K_P\), \(\mathrm{CO}\) might not completely convert to \(\mathrm{CO}_2\) quickly due to kinetic factors. Slow reaction rates, lack of sufficient activation energy, or absence of catalysts can all result in incomplete conversion.
04

Consider Environmental Conditions

Environmental factors such as temperature, pressure, and the presence of other gases (like nitrogen or water vapor) can inhibit the reaction from reaching full completion, even with a favorable equilibrium constant.
05

Conclusion

The high \(K_P\) suggests thermodynamic favorability for the formation of \(\mathrm{CO}_2\), but practical conditions such as slow kinetics and specific atmospheric composition prevent immediate and complete conversion of \(\mathrm{CO}\) to \(\mathrm{CO}_2\) in the troposphere.

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Key Concepts

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

Reaction Kinetics
Reaction kinetics is the study of the rates at which chemical reactions occur. While thermodynamics tells us whether a reaction can occur, kinetics tells us how fast it will happen.
In the context of our exercise, the reaction of carbon monoxide (\(\mathrm{CO}\)) with oxygen (\(\mathrm{O}_2\)) to form carbon dioxide (\(\mathrm{CO}_2\)) might be favored, but it doesn't necessarily happen instantaneously. Reaction rates can be affected by several factors:
  • **Concentration of Reactants:** Higher concentrations often lead to more collisions and faster reactions.
  • **Temperature:** Generally, increasing the temperature speeds up reactions.
  • **Presence of Catalysts:** Catalysts lower the activation energy, speeding up reactions without being consumed.
Despite the large equilibrium constant indicating thermodynamic favorability, if the kinetics are slow, the reaction won't proceed quickly. This is why \(\mathrm{CO}\) is not completely converted in the troposphere.
Atmospheric Chemistry
Atmospheric chemistry involves the study of chemical processes within Earth's atmosphere. The troposphere, where the reaction of \(\mathrm{CO}\) and \(\mathrm{O}_2\) occurs, is the lowest layer, directly influencing weather and climate.
In this layer:
  • **Composition:** The air consists mainly of nitrogen, oxygen, water vapor, carbon dioxide, and trace gases.
  • **Temperature and Pressure:** Both decrease with altitude, influencing how and if reactions occur.
  • **Pollutants and Natural Compounds:** Various pollutants, like \(\mathrm{NO}_x\) and \(\mathrm{SO}_x\), and natural compounds can participate in or inhibit certain reactions.
The complex interplay of these factors affects the rate and extent of chemical reactions, including the conversion of \(\mathrm{CO}\) to \(\mathrm{CO}_2\). Constraints like low temperatures or competing reactions can prevent immediate completion.
Reaction Equilibrium
Reaction equilibrium is a state where the forward and reverse reactions occur at the same rate, leading to stable concentrations of reactants and products. In our reaction, the equilibrium constant \(K_P\) is enormously high, suggesting that almost all \(\mathrm{CO}\) should transform into \(\mathrm{CO}_2\).
This doesn't mean the reaction completes in real-world conditions immediately:
  • **Dynamic Nature:** The system is in constant flux as molecules continue to react forward and back.
  • **Thermodynamic Favorability:** A high \(K_P\) points towards product formation but doesn't dictate how quickly equilibrium is reached.
  • **External Factors:** Environmental conditions can shift the equilibrium, either favoring or retarding the completion of product formation.
Overall, while equilibrium favors \(\mathrm{CO}_2\), actual atmospheric conditions can delay this conversion.
Activation Energy
Activation energy is the energy barrier that must be overcome for a reaction to occur. It determines the feasibility of a reaction under given conditions.
For the conversion of \(\mathrm{CO}\) to \(\mathrm{CO}_2\):
  • **Energy Requirement:** A minimum energy must exist to break bonds in the reactants, allowing new bonds to form in products.
  • **Factors Affecting Activation Energy:** Temperature and pressure can help in surpassing this energy barrier. However, if these factors aren't optimal, the reaction may progress slowly.
  • **Role of Catalysts:** While not altering the \(K_P\), catalysts can lower activation energy to speed up the reaction.
The abundant \(K_P\) doesn't guarantee fast reaction rates if the activation energy remains unsatisfied under present atmospheric conditions.

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Most popular questions from this chapter

As stated in the chapter, carbon monoxide has a much higher affinity for hemoglobin than oxygen does. (a) Write the equilibrium constant expression \(\left(K_{\mathrm{c}}\right)\) for the following process: $$ \mathrm{CO}(g)+\mathrm{HbO}_{2}(a q) \rightleftarrows \mathrm{O}_{2}(g)+\mathrm{HbCO}(a q) $$ where \(\mathrm{HbO}_{2}\) and \(\mathrm{HbCO}\) are oxygenated hemoglobin and carboxyhemoglobin, respectively. (b) The composition of a breath of air inhaled by a person smoking a cigarette is \(1.9 \times 10^{-6} \mathrm{~mol} / \mathrm{L} \mathrm{CO}\) and \(8.6 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\) \(\mathrm{O}_{2} .\) Calculate the ratio of \([\mathrm{HbCO}]\) to \(\left[\mathrm{HbO}_{2}\right]\), given that $$ K_{\mathrm{c}} \text { is } 212 \text { at } 37^{\circ} \mathrm{C} $$

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