Problem 101 - Chemistry The Molecular Science

Question

Two molecules of A react to form one molecule of \(\mathrm{B},\) as in the reaction $$ 2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons \mathrm{B}(\mathrm{g}) $$ Three experiments are done at different temperatures and equilibrium concentrations are measured. For each experiment, calculate the equilibrium constant, \(K_{\mathrm{c}^{*}}\) (a) \([\mathrm{A}]=0.74 \mathrm{~mol} / \mathrm{L},[\mathrm{B}]=0.74 \mathrm{~mol} / \mathrm{L}\) $$ \begin{array}{l} \text { (b) }[\mathrm{A}]=2.0 \mathrm{~mol} / \mathrm{L},[\mathrm{B}]=2.0 \mathrm{~mol} / \mathrm{L} \\ \text { (c) }[\mathrm{A}]=0.01 \mathrm{~mol} / \mathrm{L},[\mathrm{B}]=0.01 \mathrm{~mol} / \mathrm{L} \end{array} $$ What can you conclude about this statement: "If the concentrations of reactants and products are equal, then the equilibrium constant is always \(1.0 . "\)

Step-by-Step Solution

Answer

The equilibrium constant is not always 1.0 even when concentrations are equal.

Step 1: Understanding the Reaction

The equilibrium reaction given is \(2 \mathrm{~A} \rightleftharpoons \mathrm{B}\), where two molecules of A react to form one molecule of B. The equilibrium constant expression for this reaction, \(K_c\), is given by: \[ K_c = \frac{[\mathrm{B}]}{[\mathrm{A}]^2} \]where \([\mathrm{B}]\) and \([\mathrm{A}]\) are the equilibrium concentrations of B and A, respectively.

Step 2: Calculating Equilibrium Constant for Experiment (a)

Given: \([\mathrm{A}]=0.74 \,\mathrm{mol/L}\) and \([\mathrm{B}]=0.74 \,\mathrm{mol/L}\).Substitute these values into the equilibrium expression:\[ K_c = \frac{0.74}{(0.74)^2} = \frac{0.74}{0.5476} \approx 1.351 \]

Step 3: Calculating Equilibrium Constant for Experiment (b)

Given: \([\mathrm{A}]=2.0 \,\mathrm{mol/L}\) and \([\mathrm{B}]=2.0 \,\mathrm{mol/L}\).Substitute these values into the equilibrium expression:\[ K_c = \frac{2.0}{(2.0)^2} = \frac{2.0}{4.0} = 0.5 \]

Step 4: Calculating Equilibrium Constant for Experiment (c)

Given: \([\mathrm{A}]=0.01 \,\mathrm{mol/L}\) and \([\mathrm{B}]=0.01 \,\mathrm{mol/L}\).Substitute these values into the equilibrium expression:\[ K_c = \frac{0.01}{(0.01)^2} = \frac{0.01}{0.0001} = 100 \]

Step 5: Conclusion Based on Calculations

From the calculations, the value of \(K_c\) varies despite equal concentrations of reactants and products. For experiments (a), (b), and (c), the values were approximately 1.351, 0.5, and 100, respectively. This indicates that the equilibrium constant is not always 1.0 when the concentrations of reactants and products are equal.

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, leading to the concentrations of reactants and products remaining constant over time. This does not mean the amounts are equal, but that their proportions no longer change. For the reaction \(2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\), equilibrium is dynamic. Molecules of A are simultaneously turning into B and vice-versa at the same rate.

Equilibrium is affected by changes in temperature, concentration, and pressure. If one of these is altered, according to Le Chatelier's Principle, the system will shift to counteract the change and restore a new equilibrium. Understanding equilibrium is crucial for calculating equilibrium constants, as well as for predicting how changes will impact a reaction.

Reaction Quotients

The reaction quotient, \(Q_c\), like the equilibrium constant, \(K_c\), describes the ratio of the concentrations of products to reactants at any point during the reaction, not just at equilibrium. For our reaction \(2 \mathrm{~A} \rightleftharpoons \mathrm{B}\), the quotient is given by the same formula used for \(K_c\): \[ Q_c = \frac{[\mathrm{B}]}{[\mathrm{A}]^2} \]Evaluating \(Q_c\) helps determine if a reaction is at equilibrium (\(Q_c = K_c\)), or which direction it needs to shift to reach equilibrium:

• If \(Q_c < K_c\), the reaction will move forward, converting reactants to products.
• If \(Q_c > K_c\), the reaction will move in reverse, converting products to reactants to achieve equilibrium.

Concentration Calculations

Concentration calculations focus on determining the amounts of various substances in a reaction mixture, specifically at equilibrium. The given reaction \(2 \mathrm{~A} \rightleftharpoons \mathrm{B}\), requires that we understand how to substitute values into the equilibrium expression.

The equilibrium constant \(K_c\) is determined using the formula:\[ K_c = \frac{[\mathrm{B}]}{[\mathrm{A}]^2} \]where

• \([\mathrm{A}])\) is the concentration of reactant \(A\).
• \([\mathrm{B}])\) is the concentration of product \(B\).

Calculate these using the data provided, as illustrated in the example solution. This calculation reveals the substance ratio at equilibrium, shedding light on whether the forward or reverse reaction is favored. Understanding these calculations aids in predicting the behavior of a chemical system under various conditions.