Problem 126 - Chemistry: The Central Science

Question

(a) Using dissociation constants from Appendix D, determine the value for the equilibrium constant for each of the following reactions. (i) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (ii) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCO}_{3}^{-}(a q)\) (b) We usually use single arrows for reactions when the forward reaction is appreciable ( \(K\) much greater than 1 ) or when products escape from the system, so that equilibrium is never established. If we follow this convention, which of these equilibria might be written with a single arrow?

Step-by-Step Solution

Answer

(a) (i) For the reaction \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\), the equilibrium constant is \(K = \frac{K_{a1} K_{w}}{[\mathrm{H}^{+}]}\). (ii) For the reaction \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCO}_{3}^{-}(a q)\), the equilibrium constant is \(K = \frac{K_{b1}}{K_{a2}}\). (b) To determine if any of these equilibria should be written with a single arrow, compare the calculated values of K with the given conditions (K>>1 or products escape the system).

Step 1: Identify the Dissociation Constants

Using Appendix D, find the dissociation constants for the relevant species in the given reaction. - \(K_{a1}\) for \( \mathrm{HCO}_{3}^{-}(a q)\) - \(K_{w}\) for \( \mathrm{OH}^{-}(a q)\)

Step 2: Write the Dissociation Reactions for Each Species

Write down the dissociation reactions and their equilibrium constants. - \( \mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q)\) with \(K_{a1}\) - \( \mathrm{OH}^{-}(a q)+\mathrm{H}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(l) \) with \(K_{w}\)

Step 3: Find the Expressions for the Equilibrium Constant of Each Dissociation Reaction

Write down the expressions for the equilibrium constants of each dissociation reaction in terms of concentrations. - For \(K_{a1} = \frac{[\mathrm{H}^{+}][\mathrm{CO}_{3}^{2-}]}{[\mathrm{HCO}_{3}^{-}]}\) - For \(K_{w} = [\mathrm{H}^{+}][\mathrm{OH}^{-}]\)

Step 4: Write the Expression for the Equilibrium Constant of the Given Reaction

Write down the expression for the equilibrium constant of the given reaction in terms of concentrations. - \(K = \frac{[\mathrm{CO}_{3}^{2-}][\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{HCO}_{3}^{-}][\mathrm{OH}^{-}]}\)

Step 5: Determine the Equilibrium Constant (K) for the Given Reaction

Use the dissociation constants' expressions to determine the value of K for the given reaction. - Multiply both sides of the \(K_{a1}\) equation by both sides of the \(K_{w}\) equation, then divide both sides of the resulting equation by \([\mathrm{H}^{+}]\). - The result is: \(K = \frac{K_{a1} K_{w}}{[\mathrm{H}^{+}]} = \frac{[\mathrm{CO}_{3}^{2-}][\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{HCO}_{3}^{-}][\mathrm{OH}^{-}]}\). Since water is a pure liquid, its concentration is constant and will cancel out from the expression. - So, \(K = \frac{K_{a1} K_{w}}{[\mathrm{H}^{+}]} \) (a) (ii) Determine the equilibrium constant for the reaction: \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q)+\mathrm{HCO}_{3}^{-}(a q)\).

Step 1: Identify the Dissociation Constants

Using Appendix D, find the dissociation constants for the relevant species in the given reaction. - \(K_{b1}\) for \( \mathrm{NH}_{4}^{+}(a q)\) - \(K_{a2}\) for \( \mathrm{HCO}_{3}^{-}(a q)\)

Step 2: Write the Dissociation Reactions for Each Species

Write down the dissociation reactions and their equilibrium constants. - \( \mathrm{NH}_{4}^{+}(a q) \rightleftharpoons \mathrm{NH}_{3}(a q) + \mathrm{H}^{+}(a q)\) with \(K_{b1}\) - \( \mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q)\) with \(K_{a2}\)

Step 3: Find the Expressions for the Equilibrium Constant of Each Dissociation Reaction

Write down the expressions for the equilibrium constants of each dissociation reaction in terms of concentrations. - For \(K_{b1} = \frac{[\mathrm{NH}_{3}][\mathrm{H}^{+}]}{[\mathrm{NH}_{4}^{+}]}\) - For \(K_{a2} = \frac{[\mathrm{H}^{+}][\mathrm{CO}_{3}^{2-}]}{[\mathrm{HCO}_{3}^{-}]}\)

Step 4: Write the Expression for the Equilibrium Constant of the Given Reaction

Write down the expression for the equilibrium constant of the given reaction in terms of concentrations. - \(K = \frac{[\mathrm{NH}_{3}][\mathrm{HCO}_{3}^{-}]}{[\mathrm{NH}_{4}^{+}][\mathrm{CO}_{3}^{2-}]}\)

Step 5: Determine the Equilibrium Constant (K) for the Given Reaction

Use the dissociation constants' expressions to determine the value of K for the given reaction. - Divide both sides of the \(K_{b1}\) equation by both sides of the \(K_{a2}\) equation, then divide both sides of the resulting equation by \([\mathrm{H}^{+}]\). - The result is: \(K = \frac{K_{b1}}{K_{a2}} = \frac{[\mathrm{NH}_{3}][\mathrm{HCO}_{3}^{-}]}{[\mathrm{NH}_{4}^{+}][\mathrm{CO}_{3}^{2-}]}\). (b) If the forward reaction is appreciable (K>>1) or the products escape the system such that equilibrium is never established, the reactions could be written with a single arrow. To answer this question, we would need to analyze the values of K obtained in part (a). If any of those values fulfill the above conditions, the equilibrium could be written with a single arrow.

Key Concepts

Dissociation Constant

Understanding the dissociation constant (often denoted as K_a for acids and K_b for bases) is crucial for students delving into the world of chemical equilibrium. This numerical value provides insight into the degree at which an acid or base dissociates in solution to form ions. In simpler terms, a high dissociation constant indicates a strong acid or base that dissociates almost completely, while a low value suggests a weak acid or base that only partially dissociates.

When faced with the calculation of an equilibrium constant for a reaction, such as
\(\mathrm{HCO}_{3}^{-}(aq) + \mathrm{OH}^{-}(aq) \rightleftharpoons \mathrm{CO}_{3}^{2-}(aq) + \mathrm{H}_{2}O(l)\),
we start by finding the K_a for HCO₃^- and the K_w for OH^-, which is the ionic product of water. Remember, water's dissociation is unique since it self-dissociates into H⁺ and OH^- ions.

To make the concept graspable, visualize the dissociation like a dance: each pair (molecule) separates (dissociates) into the dance floor (solution) as ion-partners. The K_a or K_b is like the music's tempo – a fast tempo for a strong acid/base causes most dancers to rush onto the dance floor, representing nearly complete dissociation.

Chemical Equilibrium

Chemical equilibrium comes into play when the rate of the forward reaction equals the rate of the reverse reaction, meaning the concentrations of reactants and products remain constant over time. It’s a delicate balance between the two competing processes, not unlike a teeter-totter perfectly levelled with equal weights on both sides. This state doesn't imply that the reactants and products are in equal concentrations, but rather that their ratios don’t change.

For instance, in the reaction
\(\mathrm{NH}_{4}^{+}(aq) + \mathrm{CO}_{3}^{2-}(aq) \rightleftharpoons \mathrm{NH}_{3}(aq) + \mathrm{HCO}_{3}^{-}(aq)\),
equilibrium will be achieved when the rate at which ammonium ions and carbonate ions react to form ammonia and bicarbonate is equal to the rate at which the reverse reaction occurs. It's a dance where dancers exchange partners at a rate where the number of dancing couples and resting couples remain constant.

Each reaction has a unique equilibrium constant, K, reflective of those specific conditions at a given temperature, expressing the ratio of product concentrations to reactant concentrations, each raised to the power of its stoichiometric coefficient. This constant is profoundly important because it tels us the extent of a reaction and guides us about the achievable concentrations of reactants and products at equilibrium.

Reaction Quotient

Moving forward to the reaction quotient (Q), which is the mathematical twin of the equilibrium constant (K). While K is the ratio of the concentrations of the products to reactants at equilibrium, Q is that ratio at any point during a reaction. Think of Q as a snapshot of the system’s current state, while K is a portrait of the system at peace – in equilibrium. As the reaction progresses, Q changes and ultimately, it will be equal to K when the system is at equilibrium.

When comparing Q to K:

• If Q < K, the forward reaction is favored, and the reaction will proceed to form more products.
• If Q > K, the reverse reaction is preferred, and the reaction will shift toward the formation of more reactants.
• If Q = K, the system is already in a state of equilibrium.

Identifying and calculating Q is a dynamic way to predict the direction in which a reaction will proceed. In the context of our exercise, knowing how to manipulate the dissociation constants and understanding their relationship with the equilibrium constant is essential—it’s essentially piecing together the dance steps from the tempo to predict the dance's outcome before the music ends.