Problem 97 - General Chemistry: Principles and Modern Applications

Question

In organic synthesis many reactions produce very little yield, that is \(K \ll 1 .\) Consider the following hypothetical reaction: \(\mathrm{A}(\mathrm{aq})+\mathrm{B}(\mathrm{aq}) \longrightarrow \mathrm{C}(\mathrm{aq}), K=1 \times 10^{-2}\) We can extract product, \(\mathrm{C}\), from the aqueous layer by adding an organic layer in which \(\mathrm{C}(\mathrm{aq}) \longrightarrow \mathrm{C}(\mathrm{or})\), \(K=15 .\) Given initial concentrations of \([\mathrm{A}]=0.1 \mathrm{M}\) \([\mathrm{B}]=0.1,\) and \([\mathrm{C}]=0.1,\) calculate how much \(\mathrm{C}\) will be found in the organic layer. If the organic layer was not present, how much C would be produced?

Step-by-Step Solution

Answer

If the organic layer is present, the final concentration of C in the organic layer is \(1.5 \times 10^{-3}\)M. Without the organic layer, the amount of C produced would be \(1 \times 10^{-4}\) M.

Step 1: Set up the first reaction

Here, we need to establish the equilibrium expression for the first reaction and also calculate the equilibrium concentrations of A, B, and C in the aqueous layer. The equilibrium expression for reaction A(aq) + B(aq) -> C(aq) is given as \(K = \frac{[C]}{[A]*[B]}\). Given that \(K = 1 \times 10^{-2}\) and the initial concentrations of A, B, and C are all 0.1M, plugging in these values we get \([C] = K \times [A] \times [B] = 1 \times 10^{-2} \times 0.1 \times 0.1 = 1 \times 10^{-4} M\). Thus, the equilibrium concentration of C in the aqueous layer is \(1 \times 10^{-4}\) M.

Step 2: Set up the second reaction

For the second reaction, where C(aq) -> C(or), an organic layer is added that causes C to transfer from the aqueous layer to the organic layer. The equilibrium expression for this reaction is given as \(K = \frac{[C_{or}]}{[C_{aq}]}\). Rearranging this expression gives \([C_{or}] = K \times [C_{aq}]\). The equilibrium constant for this reaction is given as 15. Therefore, plugging in the values we get \([C_{or}] = 15 \times [1 \times 10^{-4} M] = 1.5 \times 10^{-3} M\). Thus, the concentration of C in the organic layer is \(1.5 \times 10^{-3}\) M.

Step 3: Calculate the amount of C without the organic layer

Without the organic layer present, the distribution of C will only be determined by the first reaction (A(aq) + B(aq) -> C(aq)). Using the same process as in Step 1 and since the concentrations of A and B remain as 0.1M, the amount of C produced will be \([C] = K \times [A] \times [B] = 1 \times 10^{-2} \times 0.1 \times 0.1 = 1 \times 10^{-4} M\).

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is an essential concept in chemistry that measures the balance between reactants and products in a reversible chemical reaction at equilibrium. For the reaction \( \mathrm{A}(\mathrm{aq}) + \mathrm{B}(\mathrm{aq}) \rightarrow \mathrm{C}(\mathrm{aq}) \), the equilibrium constant is calculated as: \[ K = \frac{[C]}{[A][B]} \] This formula implies that at equilibrium, the product of the concentrations of the products, in this case \( [C] \), divided by the product of the concentrations of reactants \( [A][B] \), will equal a constant value that is characteristic of the system under specific conditions.

• **Balance**: An equilibrium constant much less than 1 (\( K \ll 1 \)) indicates that, at equilibrium, the reaction mixture contains more reactants than products. This is typical for reactions where the forward reaction is not favored.
• **Impact of \( K \)**: A small \( K \) value, like \( 1 \times 10^{-2} \), shows that the reaction favors the reactants, resulting in a low yield of product \( C \). Simply put, the higher the \( K \), the more the reaction favors the formation of products.
• **Condition-Dependent**: The constant depends on temperature and pressure; changing these conditions will change \( K \).

Reaction Yield

Reaction yield refers to the amount of product formed compared to the maximum possible amount. It indicates the efficiency of a reaction. In our example, the yield is low because the equilibrium constant \( K = 1 \times 10^{-2} \) signals that significantly more reactants remain than products at equilibrium. Here, we observe how equilibrium impacts reaction yield: - **Equilibrium Shift**: Since \( K \) is relatively low, adding more reactant or removing product could drive the reaction toward more product formation.- **Maximizing Yield**: Techniques like adding an organic layer can extract product \( C \), increasing the effective yield by altering the system conditions favorably.- **Practical Yield**: The actual yield in the lab often differs from theoretical predictions due to side reactions or incomplete conversions. The existence of side reactions or transfer processes, like extraction, is a pragmatic approach to optimize yield effectively. Ultimately, understanding yield helps chemists tailor reaction conditions to enhance product formation where needed.

Aqueous Solution Extraction

Aqueous solution extraction is a method used to separate components based on their distribution between two immiscible liquids, typically water (aqueous) and an organic solvent. This technique depends on the differing solubilities of substances in two phases. - **How It Works**: In a reaction like \( \mathrm{C}(\mathrm{aq}) \rightarrow \mathrm{C}(\mathrm{or}) \), \( C \) is more soluble in the organic phase than the aqueous phase. Thus, \( C \) naturally moves to the organic layer, which can be separated to remove \( C \) from the reaction environment. - **Equilibrium in Extraction**: The partition coefficient or the distribution coefficient is similar to an equilibrium constant in that it determines the ratio of concentrations in the organic layer to the aqueous layer: \[ K = \frac{[C_{or}]}{[C_{aq}]} \] In the given example, \( K = 15 \) indicates a significant preference for the product \( C \) to be in the organic phase, enhancing the yield of usable \( C \).- **Efficiency of Extraction**: The extent of extraction depends on the choice of solvent and the equilibrium constant for the transfer, which should ideally be much greater than 1 to ensure efficient separation. This method tactically shifts the chemical equilibrium and augments the reaction's output potential, making it a valuable tool in organic synthesis.