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 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) shift left, b) shift left, c) shift right, d) shift right, e) no shift, f) shift left, g) no shift, h) shift right.

Step-by-step solution

Le Chatelier's Principle Review

Le Chatelier's Principle states that if a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the system will adjust to minimize the effect of that change. This principle will guide us in predicting the direction of shift for the given reactions.

Increase [Cd²⁺] (part a)

According to Le Chatelier's principle, increasing the concentration of [Cd²⁺] will result in the equilibrium shifting to the left to consume the excess [Cd²⁺] by forming more solid CdS .

Increase [S²⁻] (part b)

Similar to part a, an increase in [S²⁻] will cause the equilibrium to shift to the left. The system will try to decrease the concentration of [S²⁻] by forming more CdS solid.

Decrease [Cd²⁺] (part c)

When [Cd²⁺] decreases, the equilibrium shifts to the right. The system will produce more Cd²⁺ ions by dissolving more CdS , thus moving towards the production of ions.

Decrease [S²⁻] (part d)

A decrease in [S²⁻] shifts the equilibrium to the right. The system compensates by forming more [S²⁻] ions, thus dissolving more CdS into ions.

Add Solid CdS (part e)

Adding more CdS solid does not affect the equilibrium position, as the concentration of a solid does not appear in the equilibrium expression. Hence, there will be no shift in equilibrium.

Add Cd(NO₃)₂ (part f)

Adding solid Cd(NO₃)₂ increases [Cd²⁺] in solution. The equilibrium will shift to the left to reduce this increase, leading to the formation of more solid CdS .

Add NaNO₃ (part g)

Adding NaNO₃ has no effect on the equilibrium because neither Na⁺ nor NO₃⁻ ions participate in the equilibrium of CdS dissociation.

Add H⁺ Ions (part h)

Adding H⁺ ions will react with S²⁻ ions to form H₂S gas, which escapes the solution. This reaction decreases [S²⁻] , causing the equilibrium to shift to the right to produce more ions.

Key concepts

Equilibrium Shift

In any chemical reaction at equilibrium, the system maintains a delicate balance between the reactants and products. When changes occur, such as fluctuations in concentration, temperature, or pressure, the system experiences an 'equilibrium shift'. Le Chatelier's Principle provides guidance on how these shifts happen. For instance, consider the reaction involving cadmium sulfide \( \mathrm{CdS} \). If the concentration of \( \mathrm{Cd}^{2+} \) ions is increased, the equilibrium shifts to the left, meaning it favors the formation of more solid \( \mathrm{CdS} \) to counterbalance the increase.
However, if we decrease \( \mathrm{Cd}^{2+} \), the opposite happens, and the equilibrium shifts to the right, leading to more dissolution of \( \mathrm{CdS} \). Each of these shifts helps the system re-establish equilibrium in response to the added stress. Understanding these shifts forms a fundamental part of predicting how a reaction will respond to changes.

Chemical Equilibrium

Chemical equilibrium is a state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of the reactants and products remain constant over time. However, equilibrium does not mean that the reactants and products are in equal concentrations, but rather that their rates of conversion are steady.
Using the dissociation of cadmium sulfide as an example, the concentrations of \( \mathrm{Cd}^{2+} \) and \( \mathrm{S}^{2-} \) ions are stable at equilibrium unless a change is introduced. Similarly, the addition of a solid like \( \mathrm{Cd} \left( \mathrm{NO}_3 \right)_2 \), which increases \( \mathrm{Cd}^{2+} \) ions, disturbs this balance and results in an attempt to restore equilibrium by shifting.
The concept of chemical equilibrium is crucial for predicting the direction of changes in a reaction when subjected to various disturbances, providing insights into the system’s stability and response mechanisms.

Reaction Predictability

The predictability of a reaction at equilibrium under various stresses is rooted in the guidelines given by Le Chatelier's Principle. Through this principle, we can predict how a chemical system will react to maintain balance when subjected to changes. For instance, adding \( \mathrm{H}^{+} \) ions in the \( \mathrm{CdS} \) dissociation reaction results in a predictable shift. The \( \mathrm{H}^{+} \) ions react with \( \mathrm{S}^{2-} \), forming \( \mathrm{H}_2\mathrm{S} \), which leaves the solution as a gas. Consequently, this decrease in \( \mathrm{S}^{2-} \) ions causes the equilibrium to shift to the right.
By understanding such predictable shifts, you can anticipate the necessary adjustments to restore equilibrium or even control the direction of the product formation. Knowing how a system will respond to various interventions is key to fine-tuning conditions for desired outcomes in chemical processes.