Question

Given the following equilibrium constants at \(427^{\circ} \mathrm{C}\), $$ \begin{array}{ll} \mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{1}=2 \times 10^{-25} \\ \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{2}=2 \times 10^{-5} \\ \mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{3}=5 \times 10^{-29} \\ \mathrm{NaO}_{2}(s) \rightleftharpoons \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{4}=3 \times 10^{-14} \end{array} $$ determine the values for the equilibrium constants for the following reactions. a. \(\mathrm{Na}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) b. \(\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l)\) c. \(2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)

Short answer

The equilibrium constants for the desired reactions are: a. \(K_a = 2.5 \times 10^{-4}\) b. \(K_b = 5 \times 10^{-9}\) c. \(K_c = 1 \times 10^{-8}\)

Step-by-step solution

Determine the relationship between the given reactions and the desired reaction

Subtract the first reaction from the third reaction: $(\mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2\mathrm{Na}(l)+\mathrm{O}_{2}(g)) - (\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g)) \\\ \Rightarrow (\mathrm{Na}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s))$

Multiply the equilibrium constants

To find the equilibrium constant for the desired reaction, we must divide the equilibrium constant of the third reaction by the equilibrium constant of the first reaction: \(K_a = \frac{K_{3}}{K_{1}} = \frac{5 \times 10^{-29}}{2 \times 10^{-25}} = 2.5 \times 10^{-4}\) b. Determine the equilibrium constant for \(\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l)\).

Determine the relationship between the given reactions and the desired reaction

Subtract the first reaction from the second reaction and add the result to the third reaction: $((\mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g)) - (\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g))) + (\mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g)) \\ \Rightarrow (\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l))$

Multiply the equilibrium constants

To find the equilibrium constant for the desired reaction, we must multiply the equilibrium constants of the second and third reactions and divide by the equilibrium constant of the first reaction: \(K_b = \frac{K_{2} \times K_{3}}{K_{1}} = \frac{(2 \times 10^{-5}) (5 \times 10^{-29})}{2 \times 10^{-25}} = 5 \times 10^{-9}\) c. Determine the equilibrium constant for \(2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\).

Determine the relationship between the given reactions and the desired reaction

Subtract the first reaction twice from the third reaction: $2((\mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g))) - 2(\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g)) + (\mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g)) \\ \Rightarrow (2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s))$

Multiply the equilibrium constants

To find the equilibrium constant for the desired reaction, we must square the equilibrium constant of the second reaction and multiply by the equilibrium constant of the third reaction, then divide by the square of the equilibrium constant of the first reaction: \(K_c = \frac{(K_{2}^2) K_{3}}{K_{1}^2} = \frac{(2 \times 10^{-5})^2 (5 \times 10^{-29})}{(2 \times 10^{-25})^2} = 1 \times 10^{-8}\)

Key concepts

Equilibrium Constant

Chemical equilibrium plays a central role in determining how reactions behave under different conditions. The equilibrium constant, denoted as \( K \), provides a numerical measure of the position of equilibrium for a given reaction. It is essentially the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their respective coefficients in the balanced equation. For gases, the equilibrium constant is often expressed in terms of partial pressures, while for solutions it can be described using concentrations in molarity.

• In general, a large \( K \) value indicates that the equilibrium position is towards the products, signifying a reaction that nearly "goes to completion."
• Conversely, a small \( K \) value means the equilibrium position favors the reactants, with little product formed.
• Equilibrium constants are temperature dependent and remain constant for a given reaction at a particular temperature.

For instance, in the original exercise, the different \( K \) values provided at \( 427^{\circ} \mathrm{C} \) illustrate how various sodium oxide-related equilibria vary in their tendencies to form products under these specific conditions.
Understanding equilibrium constants is crucial for manipulating reaction conditions effectively in industry and laboratory settings.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemistry that predicts how a system at equilibrium responds to disturbances. When a dynamic equilibrium undergoes a change in concentration, temperature, or pressure, the system adjusts itself to counteract the change and reestablish equilibrium.

Application of Le Chatelier's Principle:

• Concentration Changes: If the concentration of a reactant is increased, the equilibrium shifts to favor the formation of products to reduce the added reactant. Conversely, removing a product also shifts equilibrium toward the product side.
• Pressure Changes: For gas-phase equilibria, increasing the pressure shifts the equilibrium toward the side with fewer moles of gas. This shift occurs because reducing the total number of gas particles reduces the pressure.
• Temperature Changes: If a reaction is exothermic, increasing the temperature favors the reactants because the system shifts to absorb added heat. The opposite is true for endothermic reactions.

Le Chatelier’s Principle provides valuable insight into how conditions can be adjusted to drive reactions toward desired products, which is often exploited in industrial processes to optimize production yield.

Reaction Quotients

The reaction quotient, \( Q \), is a tool that helps predict the direction in which a reaction must proceed to reach equilibrium. It uses the same expression as the equilibrium constant \( K \) but with the current concentrations or pressures of reactants and products, rather than those at equilibrium.

Understanding Reaction Quotients:

• If \( Q < K \), the forward reaction is favored, meaning the reaction will proceed to form more products until equilibrium is achieved.
• If \( Q > K \), the reverse reaction is favored, indicating that the reaction will shift to form more reactants.
• When \( Q = K \), the system is already at equilibrium, and no net change occurs.

In the exercise at hand, understanding how to manipulate \( Q \) provides a framework for solving complex equilibrium problems by evaluating which direction a reaction mixture needs to adjust in order to achieve equilibrium. This understanding is crucial for designing experiments and for anticipating the effects of changes in concentration or pressure on a reaction system.