Question

The equilibrium constant for the reaction: \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) is \(K_{1}\) and the equilibrium constant for the reaction: \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) is \(K_{2}\), both at the same temperature. \(K_{1}\) and \(K_{2}\) are related as (a) \(K_{1}=\left(\frac{1}{K_{2}}\right)^{2}\) (b) \(K_{1}=K_{2}^{2}\) (c) \(K_{2}=\left(\frac{1}{K_{1}}\right)^{2}\) (d) \(K_{2}=K_{1}^{2}\)

Short answer

The correct relationship between the equilibrium constants is (a) \(K_{1}=\left(\frac{1}{K_{2}}\right)^{2}\).

Step-by-step solution

Write Down the Given Equations

Note the two given chemical equations and their equilibrium constants, 1) For the reaction \(\mathrm{N}_2(\mathrm{~g}) + \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) the equilibrium constant is \(K_{1}\) and 2) For the reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) the equilibrium constant is \(K_{2}\).

Invert the Second Reaction

To find the relationship between \(K_{1}\) and \(K_{2}\), invert the second reaction to obtain \(\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})\). The equilibrium constant for the inverted reaction becomes \(\frac{1}{K_{2}}\).

Balance the Inverted Reaction

To match the first equation, we need to multiply the inverted second reaction by 2, obtaining \(\mathrm{~N}_{2}(\mathrm{~g}) + \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\). Since equilibrium constants are multiplicative for reactions that are added or multiplied, the inverse squared is the equilibrium constant for the multiplied inverse reaction: \(\left(\frac{1}{K_{2}}\right)^2\).

Identify the Correct Relation

Compare the final balanced inverted reaction from Step 3 with the first given reaction. It is identical, which means their equilibrium constants must be equal. Therefore, we have \(K_{1} = \left(\frac{1}{K_{2}}\right)^2\) or \(K_{1} = \frac{1}{K_{2}^2}\).

Key concepts

Equilibrium Constant Relationship

Understanding the relationship between equilibrium constants of reversed or manipulated chemical reactions is fundamental in the study of chemical equilibria. An equilibrium constant, denoted as K, is a quantitative measure of the position of equilibrium in a chemical reaction. It is defined for the balance between the concentrations of products and reactants at equilibrium.
When we reverse a reaction, the equilibrium constant for the reversed reaction is the reciprocal of the original one. If we denote the equilibrium constant of the original reaction as K, then for the reversed reaction, it is \( \frac{1}{K} \). Furthermore, if the reaction is multiplied by a coefficient, the equilibrium constant is raised to the power of that coefficient.
For instance, when comparing \( \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g}) \) with a constant \( K_{1} \) and \( \mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2}\mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2}\mathrm{O}_{2}(\mathrm{~g}) \) with a constant \( K_{2} \), inverting the second reaction and then squaring it to match the first reaction leads us to a mathematical relationship: \( K_{1} = \left( \frac{1}{K_{2}} \right)^2 \). This demonstrates that the equilibrium constants of related chemical reactions are mathematically linked and can be determined through the operations applied to the reactions.

Chemical Reaction Equilibrium

Chemical reaction equilibrium occurs when the rates of the forward and reverse reactions are equal, leading to no net change in the amounts of reactants and products. It is a dynamic state where reactants are continuously converted to products and products are converted back into reactants at the same rate. The equilibrium position of a reaction is described using an equilibrium constant, K, which expresses the ratio of the concentration of products to the concentration of reactants, each raised to the power of their respective coefficients in the balanced chemical equation.
Consider the equilibrium expression for the reaction \( \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g}) \), which has an equilibrium constant \( K_{1} \). The value of this constant is determined by the concentrations of \( \mathrm{N}_{2} \), \( \mathrm{O}_{2} \), and \( \mathrm{NO} \) when the reaction is at equilibrium. If K is high, the equilibrium favors the products, and if K is low, it favors the reactants.

Le Chatelier's Principle

Le Chatelier's principle provides insight into how a system at equilibrium responds to changes in concentration, temperature, or pressure. It states that if an external stress is applied to a system at equilibrium, the system will adjust in such a way as to minimize the stress.

1. Adding more reactants or products will shift the equilibrium to the side that counteracts the change.
2. Changing the temperature can favor either the exothermic or endothermic direction, depending on the heat change of the reaction.
3. Varying the pressure by changing the volume of the system will shift the equilibrium in the direction that involves fewer moles of gas.

Consider the equilibrium between \( \mathrm{N}_{2} \), \( \mathrm{O}_{2} \), and \( \mathrm{NO} \). If the concentration of \( \mathrm{NO} \) is increased, Le Chatelier's principle predicts that the equilibrium will shift to the left, producing more of the reactants to lower the concentration of products. These adjustments are crucial for chemists to control reactions to obtain desired products in optimal quantities.