Problem 2 - Chemical Principles in the Laboratory

Question

Given the following two reactions and their equilibrium constants: $$\begin{array}{ll}\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(\mathrm{aq})+4 \mathrm{H}_{2} \mathrm{O}(\ell) & K_{1}=5 \times 10^{12} \\\\\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}=-\mathrm{CuCO}_{3}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\ell) & K_{2}=7 \times 10^{9} \end{array}$$ a Evaluate the equilibrium constant for the reaction (see discussion): $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq})=\mathrm{CuCO}_{3}(\mathrm{s})+4 \mathrm{NH}_{3}(\mathrm{aq})$$ b. When \(1.0 \mathrm{M}\) concentrations of all dissolved species are present, which is more stable, \(\mathrm{CuCO}_{3}\) or \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} ?\) Explain your reasoning.

Step-by-Step Solution

Answer

The equilibrium favors \(\mathrm{Cu}\left(\mathrm{NH}_3\right)_4^{2+}\), making it more stable than \(\mathrm{CuCO}_3\).

Step 1: Understand the Reactions and Equilibrium Constants

We have two reactions, each with a given equilibrium constant. The key here is how these reactions relate to the one we need to evaluate. The first reaction involves the equilibrium between the copper-ammonia complex and the copper-water complex with an equilibrium constant of \(K_1 = 5 \times 10^{12}\). The second reaction involves the copper-water complex in equilibrium with solid copper carbonate, with \(K_2 = 7 \times 10^9\).

Step 2: Write the Target Reaction

The target reaction is composed of reversing Reaction 1 and combining with Reaction 2. The equation is:\[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4^{2+}(\mathrm{aq}) + \mathrm{CO}_3^{2-}(\mathrm{aq}) \rightleftharpoons \mathrm{CuCO}_3(\mathrm{s}) + 4 \mathrm{NH}_3(\mathrm{aq})\]

Step 3: Reverse Reaction 1 and Calculate New Equilibrium Constant

Reversing Reaction 1 gives:\[\mathrm{Cu}\left(\mathrm{NH}_3\right)_{4}^{2+}(\mathrm{aq}) + 4 \mathrm{H}_2\mathrm{O}(\ell) \rightleftharpoons \mathrm{Cu}\left(\mathrm{H}_2\mathrm{O}\right)_4^{2+}(\mathrm{aq}) + 4 \mathrm{NH}_3(\mathrm{aq})\]The new equilibrium constant is the reciprocal of \(K_1\), so \(K_{1, \text{new}} = \frac{1}{5 \times 10^{12}}\).

Step 4: Combine Reactions and Calculate Overall Equilibrium Constant

The target reaction involves combining the reversed Reaction 1 with Reaction 2:1. Reversed Reaction 1: Makes the copper ammonia complex react with carbonate.2. Reaction 2 keeps the formation of copper carbonate.The overall equilibrium constant \(K\) for the target reaction is:\[K = K_{1, \text{new}} \times K_2 = \frac{1}{5 \times 10^{12}} \times 7 \times 10^9\]Carry out the calculation to find \(K = 1.4 \times 10^{-3}\).

Step 5: Determine Stability of Complexes

With 1.0 M concentrations of all dissolved species, the small value of \(K = 1.4 \times 10^{-3}\) indicates that the reaction does not favor the products, i.e., copper carbonate and ammonia in equilibrium. This tells us that at equilibrium, more \(\mathrm{Cu}\left(\mathrm{NH}_3\right)_4^{2+}\) will remain than will be converted to \(\mathrm{CuCO}_3\). Therefore, \(\mathrm{Cu}\left(\mathrm{NH}_3\right)_4^{2+}\) is more stable.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating state in which a chemical reaction's forward and reverse processes are balanced. This means the rate at which the reactants transform into products is equal to the rate at which products revert back into reactants. At equilibrium, the concentrations of all species remain constant, even though reactions are still proceeding.
To describe this equilibrium, we use the equilibrium constant ( K ), a vital tool in understanding reaction tendencies. The equilibrium constant is specific for a given reaction at a particular temperature. It provides insight into the relative amounts of products and reactants when the reaction is at equilibrium.

• When K >> 1, products are favored, indicating a "forward" reaction tendency.
• Conversely, when K << 1, reactants are favored, suggesting reverse reactants prevail more.

In the exercise, the equilibrium constant highlights which copper complex is more stable at equilibrium conditions. It serves as a mathematical representation, aiding us in predicting the reaction's dynamics.

Stability of Complexes

The stability of a complex involves how strongly and how often a complex remains intact under given conditions. Complexes are formed from a central atom or ion bonded to surrounding molecules or ions, called ligands. The more stable a complex, the less likely it is to undergo changes or break apart.
In the provided reactions, two complexes are examined: \( \mathrm{Cu(NH_3)_4^{2+}} \) and \( \mathrm{CuCO_3} \). To determine which is more stable, we evaluate their behavior when starting at equal concentrations of 1.0 M. The smaller overall equilibrium constant (K = 1.4 \times 10^{-3}) for the target reaction indicates low product formation, meaning \( \mathrm{Cu(NH_3)_4^{2+}} \) remains largely unconverted.

• This result suggests that the ammonia-complex copper ion is more inherently more stable than the carbonate version.
• It resists transforming into another chemical form even when competing ligands, like carbonate, are available.

This information can guide us in predicting reaction outcomes in situations where different ligands compete for bonding with a central ion.

Reaction Reversibility

Reaction reversibility is an essential aspect of chemical equilibrium, describing how readily a chemical reaction can proceed in both directions. It means that during a reversible reaction, products can convert back into reactants, often illustrated by the double-headed arrow (e.g., \( \rightleftharpoons \)).
Reversible reactions involve reaching a dynamic equilibrium where the forward and backward reaction rates are equal, not necessarily in composition but in activity.

• In our exercise, the reaction involving \( \mathrm{Cu(NH_3)_4^{2+}} \) and \( \mathrm{CuCO_3} \) is reversible.
• Even though products form, the rate at which the reactants regenerate controls the stability and proportions of the complexes.

Understanding reversibility is crucial because it underpins the concept of equilibrium. It allows chemists to infer which conditions favor shifts in reactions, either toward making more products or preserving reactants, essential for controlling chemical processes in industry and research.