Question

Cadmium sulfide dissociates slightly in an aqueous solution as follows: $$ \mathrm{CdS}(s) \rightleftarrows \mathrm{Cd}^{2+}(a q)+\mathrm{S}^{2-}(a q) $$ Predict the direction of equilibrium shift for each of the following stresses: (a) increase \(\left[\mathrm{Cd}^{2+}\right]\) (b) increase \(\left[\mathrm{S}^{2-}\right]\) (c) decrease \(\left[\mathrm{Cd}^{2+}\right]\) (d) decrease \(\left[\mathrm{S}^{2-}\right]\) (e) add solid \(\mathrm{CdS}\) (f) add solid \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) (g) add solid \(\mathrm{NaNO}_{3}\) (h) add \(\mathrm{H}^{+}\)

Short answer

(a) Left, (b) Left, (c) Right, (d) Right, (e) No change, (f) Left, (g) No change, (h) Right.

Step-by-step solution

Understanding Le Châtelier's Principle

Le Châtelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This can involve shifts in the direction of the reaction to offset changes in concentration, pressure, or temperature.

Analyzing stress (a) - Increase [Cd²⁺]

An increase in the concentration of \([\mathrm{Cd}^{2+}]\) will cause the equilibrium to shift towards the left, favoring the formation of solid \(\mathrm{CdS}\), to reduce the concentration of \(\mathrm{Cd}^{2+}\).

Analyzing stress (b) - Increase [S²⁻]

Increasing \([\mathrm{S}^{2-}]\) leads to a similar effect as with \([\mathrm{Cd}^{2+}]\). The equilibrium will shift to the left, promoting the formation of \(\mathrm{CdS}(s)\) to decrease \([\mathrm{S}^{2-}]\).

Analyzing stress (c) - Decrease [Cd²⁺]

A decrease in \([\mathrm{Cd}^{2+}]\) results in a rightward shift in equilibrium, promoting dissociation of \(\mathrm{CdS}\) to produce more \(\mathrm{Cd}^{2+}\) and \(\mathrm{S}^{2-}\).

Analyzing stress (d) - Decrease [S²⁻]

By decreasing \([\mathrm{S}^{2-}]\), the equilibrium shifts to the right, leading to more dissociation of \(\mathrm{CdS}(s)\) to counter the decrease in \(\mathrm{S}^{2-}\).

Analyzing stress (e) - Adding solid CdS

Adding more \(\mathrm{CdS}(s)\) does not affect the equilibrium position as it is a pure solid and its concentration is unchanged in the equilibrium expression.

Analyzing stress (f) - Adding solid Cd(NO₃)₂

Adding \(\mathrm{Cd(NO}_3\mathrm{)_2}\) increases the concentration of \(\mathrm{Cd}^{2+}\) due to its dissociation in water, resulting in a leftward shift in equilibrium, promoting the formation of \(\mathrm{CdS}\).

Analyzing stress (g) - Adding solid NaNO₃

Adding \(\mathrm{NaNO}_3\) has no impact on the equilibrium since neither \(\mathrm{Na}^+\) nor \(\mathrm{NO}_3^-\) are involved in the equilibrium reaction.

Analyzing stress (h) - Adding H⁺

Adding \(\mathrm{H}^{+}\) ions will react with \(\mathrm{S}^{2-}\) ions to form \(\mathrm{HS}^{-}\) or \(\mathrm{H}_2\mathrm{S}\), thereby reducing \([\mathrm{S}^{2-}]\) and shifting the equilibrium to the right to replace the lost \([\mathrm{S}^{2-}]\).

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs in a closed system when the rates of the forward and reverse reactions are equal, leading to constant concentrations of reactants and products. While the concentrations stay constant, it does not mean the amounts of reactants and products are the same on both sides. It simply means that the reactions are occurring at such a rate that their concentrations no longer change over time. A state of balance is achieved, and externally, the reaction appears static. This is a dynamic but balanced situation.

In the case of the cadmium sulfide (\(\mathrm{CdS}\)) dissociation, this state of equilibrium can be pushed to favor either the solid cadmium sulfide or its ionized components (\(\mathrm{Cd}^{2+}\) and \(\mathrm{S}^{2-}\)), based on changes to the system, such as adding or removing reactants or products. These shifts indicate the system's adaptive nature in response to external influences.

Equilibrium Constant

The equilibrium constant (\(K_{eq}\)) is an essential concept in understanding chemical equilibria. It quantifies the ratio of product concentrations to reactant concentrations, each raised to the power of their respective coefficients at equilibrium. For a general reaction \(aA + bB \rightleftarrows cC + dD\), the equilibrium constant expression is:

\[\begin{align*}K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}\end{align*}\]

In this formula, square brackets represent the concentrations, and the letters denote stoichiometric coefficients and chemical species.When \(K_{eq}\) is large, the equilibrium lies towards products, indicating more products than reactants at equilibrium. When it is small, the equilibrium favors reactants. In the case of cadmium sulfide dissociation, solid \(\mathrm{CdS}\) does not appear in the expression for \(K_{eq}\), since pure solids and liquids are omitted. Only the dissolved \(\mathrm{Cd}^{2+}\) and \(\mathrm{S}^{2-}\) concentrations are considered, demonstrating how the equilibrium constant only reflects changes in aqueous or gaseous species.

Dissociation Reaction

A dissociation reaction involves a compound breaking apart into two or more components. In this context, the dissociation of cadmium sulfide (\(\mathrm{CdS}\)) in water results in the production of cadmium ions (\(\mathrm{Cd}^{2+}\)) and sulfide ions (\(\mathrm{S}^{2-}\)). This is represented by the equation:

\[\mathrm{CdS}(s) \rightleftarrows \mathrm{Cd}^{2+}(aq) + \mathrm{S}^{2-}(aq)\]

The dissociation reaction is a reversible process that can reach equilibrium. In this state, the forward rate (dissolution of \(\mathrm{CdS}\)) equals the reverse rate (reformation of \(\mathrm{CdS}\)), maintaining dynamic balance.Le Châtelier's Principle plays a critical role in understanding how such reactions behave when system changes occur. For instance, adding more \(\mathrm{Cd}^{2+}\) or \(\mathrm{S}^{2-}\) will shift the equilibrium towards solid \(\mathrm{CdS}\) formation, as the system strives to reduce the concentration of these increased ions by recombining them back into the solid state. Conversely, removing any of these ions will shift the equilibrium toward more dissociation of \(\mathrm{CdS}\). This exemplifies the flexibility of equilibrium systems in their response to external stress.