Question

(a) Using dissociation constants from Appendix \(D\), determine the value for the equilibrium constant for each of the following reactions. (Remember that when reactions are added, the corresponding equilibrium constants are multiplied.) (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?

Short answer

The equilibrium constant for Reaction (i) is K₅ = K₁ / K₂, and for Reaction (ii), it's K₆ = K₃ * K₄. Check if K₅ and K₆ are much greater than 1, and if they are, a single arrow can be used for the respective reaction(s).

Step-by-step solution

Find dissociation constants

Refer to Appendix D to find the dissociation constants (K) for the species involved in the reactions. For Reaction (i): HCO3⁻(aq) ↔ H⁺(aq) + CO3²⁻(aq) OH⁻(aq) ↔ H⁺(aq) + HO⁻(aq) For Reaction (ii): NH4⁺(aq) ↔ H⁺(aq) + NH3(aq) HCO3⁻(aq) ↔ H⁺(aq) + CO3²⁻(aq) You'll need to find: K₁ - dissociation constant for the reaction HCO3⁻(aq) K₂ - dissociation constant for the reaction OH⁻(aq) K₃ - dissociation constant for the reaction NH4⁺(aq) K₄ - dissociation constant for the reaction HCO3⁻(aq)

Determine the equilibrium constants

For each reaction, we will find the equilibrium constants using the relationship between constructed reactions and equilibrium constants. Reaction (i): HCO3⁻(aq) + OH⁻(aq) ↔ CO3²⁻(aq) + H₂O(l) This reaction can be constructed by combining the first two dissociation reactions (subtract the second dissociation reaction from the first one). Hence, the equilibrium constant for Reaction (i) will be: K₅ = K₁ / K₂ Reaction (ii): NH4⁺(aq) + CO3²⁻(aq) ↔ NH3(aq) + HCO3⁻(aq) This reaction can be constructed by combining the third and fourth dissociation reactions (add the third and fourth dissociation reactions). Hence, the equilibrium constant for Reaction (ii) will be: K₆ = K₃ * K₄

Decide if a single arrow is appropriate

We are asked to check if any of the equilibria can be written with a single arrow. A single arrow is appropriate when the forward reaction is appreciable (K much greater than 1) or when products escape from the system. Check if K₅ and K₆ are much greater than 1. If they are, then a single arrow can be used for the respective reaction(s). Based on your calculated values of K₅ and K₆, you can determine which reaction(s) may be represented with a single arrow.

Key concepts

Dissociation Constants

Understanding dissociation constants is essential when discussing chemical equilibrium in solutions. Dissociation constants, represented by the symbol 'K', are a measure of the extent to which a compound separates into its individual ions in a solution. For example, the dissociation of a weak acid HA in water can be represented like this:

\[ HA(aq) \rightleftharpoons H^+(aq) + A^-(aq) \]
The equilibrium constant for this dissociation \( K_a \) is calculated using the concentrations of the products and reactants at equilibrium:
\[ K_a = \frac{{[H^+][A^-]}}{{[HA]}} \]
A low value of K indicates that the compound does not dissociate significantly, remaining largely intact in solution. A high value of K indicates strong dissociation into ions. Understanding dissociation constants helps students calculate the equilibrium constant for reactions in a solution, as they are often the basic building blocks for constructing larger equilibrium reactions.

Chemical Equilibrium

In chemical reactions, chemical equilibrium is the state in which the concentrations of reactants and products do not change over time, meaning the forward and reverse reactions occur at the same rate. The point at which this occurs is crucial to understanding chemical behavior in reactions.

At equilibrium, the rate at which the reactants turn into products is equal to the rate at which the products revert back to reactants. The equilibrium constant (\( K \) or \( K_{eq} \) for reactions in aggregate) is a number that expresses the ratio of the concentration of the products to the concentration of the reactants, each raised to the power of their stoichiometric coefficients. It's important to note that only the species which change during the reaction appear in the expression for K.
\[ K_{eq} = \frac{{[Products]}}{{[Reactants]}} \]
If \( K_{eq} \) is greater than 1, the products are favored at equilibrium. If \( K_{eq} \) is less than 1, the reactants are favored. Chemical equilibrium is dynamic – although the concentrations remain constant, molecules are continually reacting and re-forming.

Reaction Quotient

The reaction quotient (\( Q \) closely resembles the equilibrium constant (\( K \) but is used to determine the direction in which a reaction needs to proceed to reach equilibrium. It is calculated in the exact same manner as \( K \) but using the concentrations of products and reactants at a particular moment in time, not necessarily at equilibrium.

\[ Q = \frac{{[Products]}}{{[Reactants]}} \]
If \( Q < K \) for a reaction at a given time, the forward reaction will be favored in order to reach equilibrium. If \( Q > K \) the reverse reaction is favored. When \( Q = K \) the system is at equilibrium and no net change occurs in the concentrations of reactants and products. This concept helps students predict the shift in the direction of the reaction and is crucial when working with chemical systems not at equilibrium.