Problem 96 - Chemistry The Molecular Science

Question

These amounts of \(\mathrm{CO}(\mathrm{g}), \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{CO}_{2}(\mathrm{~g})\), and \(\mathrm{H}_{2}(\mathrm{~g})\) are introduced into a \(10.00-\mathrm{L}\) flask. The flask is sealed and heated to a very high temperature. \begin{tabular}{lcccc} \hline & \(n_{\mathrm{CO}}(\mathrm{mol})\) & \(n_{\mathrm{H}_{2} \mathrm{O}}(\mathrm{mol})\) & \(n_{\mathrm{CO}_{2}}(\mathrm{~mol})\) & \(n_{\mathrm{H}_{2}}(\mathrm{~mol})\) \\ \hline Case a & 1.0 & 0.10 & 0.10 & 0.10 \\ Case b & \(10 .\) & 1.0 & 1.0 & 1.0 \\ Case c & \(10 .\) & \(10 .\) & 1.0 & 1.0 \\ Case d & 5.62 & 0.381 & 1.75 & 1.75 \\ \hline \end{tabular} The equilibrium constant for the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) has the value \(K_{\mathrm{c}}=4.00\) at this temperature. For which cases will the concentration of CO increase as equilibrium is attained, and in which cases will the concentration of CO decrease?

Step-by-Step Solution

Answer

In Cases c and d, the concentration of CO increases; in Cases a and b, it decreases.

Step 1: Write the Expression for Kc

The reaction is \( \mathrm{CO(g)} + \mathrm{H_2O(g)} \rightleftharpoons \mathrm{CO_2(g)} + \mathrm{H_2(g)} \). The equilibrium constant expression \( K_c \) is given by \( K_c = \frac{[\mathrm{CO_2}][\mathrm{H_2}]}{[\mathrm{CO}][\mathrm{H_2O}]} \). Here, the brackets denote concentration in moles per liter.

Step 2: Calculate Initial Concentrations

For each case, calculate the initial concentrations using \( [X] = \frac{n_X}{V} \), where \( n_X \) is the number of moles and \( V = 10.00 \text{ L} \). For example, in Case a: \( [\mathrm{CO}] = \frac{1.0}{10.0} = 0.1 \text{ M} \), \( [\mathrm{H_2O}] = \frac{0.10}{10.0} = 0.01 \text{ M} \), and similarly for \( \mathrm{CO_2} \) and \( \mathrm{H_2} \).

Step 3: Calculate the Reaction Quotient (Qc)

For each case, use the initial concentrations to calculate \( Q_c \) using \( Q_c = \frac{[\mathrm{CO_2}][\mathrm{H_2}]}{[\mathrm{CO}][\mathrm{H_2O}]} \). Cover all cases, for example, in Case a: \( Q_c = \frac{0.01 \times 0.01}{0.1 \times 0.01} = 0.01 \).

Step 4: Compare Qc to Kc

Determine whether \( Q_c \) is greater than, less than, or equal to \( K_c = 4.0 \). If \( Q_c < K_c \), the concentration of \( \mathrm{CO} \) will decrease as equilibrium is reached. If \( Q_c > K_c \), the concentration of \( \mathrm{CO} \) will increase. Continue this analysis for all cases.

Step 5: Draw Conclusion from Qc vs Kc Analysis

Summarize findings for each case: - **Case a**: \( Q_c = 0.01 < K_c \), \( \mathrm{CO} \) decreases. - **Case b**: \( Q_c = 1 < K_c \), \( \mathrm{CO} \) decreases. - **Case c**: \( Q_c = 10 > K_c \), \( \mathrm{CO} \) increases. - **Case d**: \( Q_c = 8 < K_c \), \( \mathrm{CO} \) increases.

Key Concepts

Equilibrium Constant

In a chemical reaction at equilibrium, the equilibrium constant, denoted as \( K_{c} \), is a numerical value that expresses the ratio of product concentrations to reactant concentrations. For a reaction \( \mathrm{CO(g)} + \mathrm{H_2O(g)} \rightleftharpoons \mathrm{CO_2(g)} + \mathrm{H_2(g)} \), the equilibrium constant is given by \( K_{c} = \frac{[\mathrm{CO_{2}}][\mathrm{H_{2}}]}{[\mathrm{CO}][\mathrm{H_{2}O}]} \). This particular reaction illustrates how \( K_{c} \) provides insight into the position of equilibrium. - If \( K_{c} \) is greater than 1, the products are favored at equilibrium.- If \( K_{c} \) is less than 1, the reactants are favored.In our exercise, the equilibrium constant is \( K_{c} = 4.0 \), indicating that at equilibrium, the product side is more favored than the reactant side. Understanding \( K_{c} \) is crucial as it helps predict the behavior of a reaction when disturbed and points out the direction in which the reaction will shift to re-establish equilibrium.

Reaction Quotient

The reaction quotient, denoted as \( Q_{c} \), serves as a snapshot of a reaction's condition at any given moment—before or while approaching equilibrium. Similar to \( K_{c} \), \( Q_{c} \) is calculated using the concentration of the products over the reactants: \[ Q_{c} = \frac{[\mathrm{CO_{2}}][\mathrm{H_{2}}]}{[\mathrm{CO}][\mathrm{H_{2}O}]} \]. By comparing \( Q_{c} \) to \( K_{c} \), one can deduce the directional shift required to reach equilibrium:- If \( Q_{c} < K_{c} \), the system will shift toward the products to reach equilibrium, meaning more CO will be consumed.- If \( Q_{c} > K_{c} \), the system will shift toward the reactants, causing more CO to be produced.- If \( Q_{c} = K_{c} \), the reaction is at equilibrium.In our problem, calculating \( Q_{c} \) for each case using initial concentrations and comparing it to \( K_{c} = 4.0 \) helps determine which direction the reaction will proceed, thereby predicting concentration changes in CO.

Concentration Calculations

Calculating concentrations is essential when dealing with chemical equilibria. Concentration, symbolized by square brackets such as \([X]\), represents the amount of a substance in a given volume, typically expressed in molarity (mol/L). To compute initial concentrations in the provided flask, use the formula \([X] = \frac{n_{X}}{V}\), where \(n_{X}\) is the number of moles of component \(X\) and \(V\) is the volume of the flask, which is 10.0 L in this exercise.Example calculations are as follows:- For Case a, the concentration of CO is computed as \([\mathrm{CO}] = \frac{1.0 \text{ mol}}{10.0 \text{ L}} = 0.1 \text{ M}\).- Similarly, perform these calculations for all reactants and products in each case to obtain their initial concentrations.These calculated concentrations are crucial to determine the reaction quotient \( Q_{c} \). By using these values, you can assess how far a system is from equilibrium and predict the changes necessary to reach a state where the rates of the forward and reverse reactions are equal, underlining the importance of accurate concentration calculations in chemical equilibrium studies.