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Chapter 7: Problem 6

For the following equilibrium, \(K_{c}=6.3 \times 10^{14}\) at \(1000 \mathrm{~K}\) \(\mathrm{NO}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(\mathrm{~g})\) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is \(K_{c}\), for the reverse reaction?

Short Answer

Expert verified
\( K_{c} \) for the reverse reaction is \( 1.59 \times 10^{-15} \).

Step by step solution

01

Understand the Equilibrium Constant

For any given chemical reaction at equilibrium, the equilibrium constant, \( K_c \), represents the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients.
02

Identify the Given Forward Reaction

Here, we are given the forward reaction: \( \mathrm{NO}(\mathrm{g}) + \mathrm{O}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{NO}_{2}(g) + \mathrm{O}_{2}(\mathrm{~g}) \). The equilibrium constant for this reaction is \( K_{c} = 6.3 \times 10^{14} \) at \( 1000 \mathrm{~K} \).
03

Write the Reverse Reaction

The reverse reaction is: \( \mathrm{NO}_{2}(g) + \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) + \mathrm{O}_{3}(\mathrm{~g}) \).
04

Determine the Relationship for the Reverse Reaction

The equilibrium constant for the reverse reaction, \( K_{c, ext{reverse}} \), is the inverse of the equilibrium constant of the forward reaction, \( K_{c} \). This is because the equilibrium constant is a ratio, and reversing the reaction inverts this ratio.
05

Calculate \( K_{c} \) for the Reverse Reaction

Since \( K_{c, ext{forward}} = 6.3 \times 10^{14} \), the equilibrium constant for the reverse reaction is: \[ K_{c, ext{reverse}} = \frac{1}{6.3 \times 10^{14}} \approx 1.59 \times 10^{-15} \]
06

Conclusion

Thus, the equilibrium constant \( K_{c} \) for the reverse reaction \( \mathrm{NO}_{2}(g) + \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) + \mathrm{O}_{3}(\mathrm{~g}) \) is approximately \( 1.59 \times 10^{-15} \).

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Key Concepts

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

Understanding Chemical Equilibrium
Chemical equilibrium is a fascinating concept that occurs when a reversible chemical reaction reaches a state where the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of reactants and products remain constant over time, even though they are by no means static.

Key points to remember about chemical equilibrium include:
  • The equilibrium state is dynamic, meaning reactions continue to occur but without changes in concentration.
  • The equilibrium can be disrupted by changes in temperature, pressure, or concentration, a principle known as Le Chatelier鈥檚 principle.
  • The equilibrium constant, represented by the symbol \( K_c \), provides a snapshot of the equilibrium position and helps determine whether the reaction favors products or reactants.
Chemical equilibrium is an essential concept in understanding how different chemical systems behave and is fundamental when studying reaction kinetics and thermodynamics.
The Nature of Bimolecular Reactions
Bimolecular reactions are a type of elementary chemical reaction that involve two reactant molecules coming together to form products. These reactions are crucial in many biological and industrial processes.

Here's what you need to know about bimolecular reactions:
  • They typically involve a collision between two reactant molecules, each contributing to the reaction's progress.
  • The rate of the reaction often depends on the concentration of both reactants, given mathematically as \( ext{Rate} = k[A][B] \), where \( k \) is the rate constant, and \([A]\) and \([B]\) are the concentrations of the reactants.
  • Bimolecular reactions can be part of a larger mechanism, where this step is one of several that lead to the final product.
Understanding bimolecular reactions is key to predicting how fast reactions proceed and what factors might enhance or inhibit their progress.
Analyzing the Reverse Reaction
When discussing reversible reactions, recognizing that reactions can proceed in both directions is vital. For the reverse reaction, the equilibrium constant \( K_{c, \text{reverse}} \) holds substantial importance.

Here's what makes reverse reactions unique:
  • The equilibrium constant for a reverse reaction is the reciprocal of the forward reaction's equilibrium constant: \( K_{c, \text{reverse}} = \frac{1}{K_{c, \text{forward}}} \).
  • The concept signifies that reversing the reaction flips the ratio of products to reactants.
  • Understanding the reverse reaction is critical in processes such as recycling of reactants, optimizing yields, and enhancing reaction efficiency in industrial settings.
By grasping the dynamics of the reverse reaction and its equilibrium constant, you can better predict and control various chemical reactions for practical applications.

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Most popular questions from this chapter

It has been found that the pH of a \(0.01 \mathrm{M}\) solution of an organic acid is \(4.15 .\) Calculate the concentration of the anion, the ionization constant of the acid and its \(\mathrm{p} K_{\mathrm{a}}\).

What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: \(\mathrm{HNO}_{2}, \mathrm{CN}, \mathrm{HClO}_{4} \mathrm{~F}^{-}, \mathrm{OH}, \mathrm{CO}_{3}^{2-}\), and \(\mathrm{S}^{2}\)

The ionization constant of acetic acid is \(1.74 \times 10^{-5} .\) Calculate the degree of dissociation of acetic acid in its \(0.05 \mathrm{M}\) solution. Calculate the concentration of acetate ion in the solution and its \(\mathrm{pH}\).

At \(473 \mathrm{~K}\), equilibrium constant \(K_{c}\) for decomposition of phosphorus pentachloride, \(\mathrm{PCl}_{5}\) is \(8.3 \times 10^{3}\). If decomposition is depicted as. \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \quad \Delta_{t} H^{\ominus}=124.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) a) write an expression for \(K\), for the reaction. b) what is the value of \(K_{c}\) for the reverse reaction at the same temperature? c) \(\quad\) what would be the effect on \(K_{c}\) if (i) more \(\mathrm{PCl}_{3}\) is added (ii) pressure is increased (iii) the temperature is increased ?

One mole of \(\mathrm{H}_{2} \mathrm{O}\) and one mole of \(\mathrm{CO}\) are taken in \(10 \mathrm{~L}\) vessel and heated to \(725 \mathrm{~K}\). At cquilibrium \(40 \%\) of water (by mass) reacts with \(\mathrm{CO}\) according to the equation, $$ \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) $$ Calculate the equilibrium constant for the reaction.
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