Question

For the following reaction: $$ \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{3(\mathrm{~g})} \rightleftharpoons \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} $$ The value of \(K_{c}\) is \(8.2 \times 10^{4} .\) What will be the value of \(K_{c}\) for the reverse reaction? (a) \(8.2 \times 10^{4}\) (b) \(\frac{1}{8.2 \times 10^{4}}\) (c) \(\left(8.2 \times 10^{4}\right)^{2}\) (d) \(\sqrt{8.2 \times 10^{4}}\)

Short answer

\(\frac{1}{8.2 \times 10^{4}}\)

Step-by-step solution

Understand the Relationship Between Forward and Reverse Reaction Equilibria

For a chemical reaction at equilibrium, the equilibrium constant for the forward reaction is denoted as \(K_c\). For the reverse reaction, the equilibrium constant \(K_c'\) is the reciprocal of the equilibrium constant for the forward reaction. In mathematical terms, if the forward reaction has an equilibrium constant \(K_c\), then the reverse reaction's equilibrium constant \(K_c'\) can be represented as \(K_c' = \frac{1}{K_c}\).

Calculate the Equilibrium Constant for the Reverse Reaction

Given that the equilibrium constant for the forward reaction, \(K_c\), is \(8.2 \times 10^{4}\), then the equilibrium constant for the reverse reaction \(K_c'\) would be the reciprocal of this value. Therefore, \(K_c' = \frac{1}{8.2 \times 10^{4}}\).

Key concepts

Reversible Reactions

Chemical processes often involve reactions that can proceed in both forward and reverse directions. This type of chemical reaction is known as a reversible reaction. In these reactions, the reactants can convert to products, and simultaneously, the products can revert back to the original reactants. This dynamic process continues until a state of balance, or equilibrium, is reached.

At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, meaning no net change occurs in the concentration of reactants and products. However, it is crucial to understand that reactions at equilibrium are not static; they are dynamic, with constant conversion between reactants and products happening at equal rates. Familiarity with reversible reactions is essential for comprehending more complex chemical behaviors, such as the reactions occurring in biological systems or industrial processes.

Equilibrium Constant for Reverse Reaction

When considering reversible reactions, each direction of the reaction has its own equilibrium constant. The equilibrium constant for the forward reaction, commonly represented as \(K_c\), quantifies the ratio of the concentrations of products to reactants at equilibrium. Conversely, the equilibrium constant for the reverse reaction, denoted as \(K_c'\), measures the same ratio but with the positions of products and reactants switched.

In mathematical terms, \(K_c'\) is the reciprocal of \(K_c\), or \(K_c' = \frac{1}{K_c}\) when the reaction coefficients are equal to one. This reciprocal relationship provides a simple way to determine the equilibrium constant for a reverse reaction when the constant for the forward reaction is known. Understanding this reciprocal nature of equilibrium constants is fundamental in the study of chemical reactions and is instrumental in predicting the behavior of reversible reactions under varying conditions.

Equilibrium Constant Relationship

The relationship between the equilibrium constants for forward and reverse reactions is based on the principle of microscopic reversibility, which states that a system at equilibrium at a given temperature should have the reverse process occurring by the same mechanism as the forward process. This means that the equilibrium constants are intrinsically related.

For the given reaction \(\mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{3(\mathrm{~g})} \rightleftharpoons \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\), with a forward reaction equilibrium constant \(K_{c}\) of \(8.2 \times 10^{4}\), the equilibrium constant for the reverse reaction (\(K_{c}'\)) is, as previously established, the reciprocal. Therefore, knowing the relationship between \(K_c\) and \(K_c'\) allows us to solve for the latter, reinforcing the importance of these conceptual ties in various applications such as calculating concentrations, predicting the direction of reaction, and even in industrial design where chemical equilibria play a pivotal role in optimization of conditions for product yield.