Problem 40 - Chemistry and Chemical Reactivity

Question

Consider the following equilibrium: \(\operatorname{COBr}_{2}(g) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=0.190\) at \(73^{\circ} \mathrm{C}\) (a) \(A\) 0.50 mol sample of \(\operatorname{COBr}_{2}\) is transferred to a 9.50-L. flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species. (b) The volume of the container is decreased to 4.5 L and the system allowed to return to equilibrium. Calculate the new equilibrium concentrations. (Hint: The calculation will be easier if you view this as a new problem with 0.5 mol of \(\mathrm{COBr}_{2}\) transferred to a 4.5 -L flask. (c) What is the effect of decreasing the container volume from 9.50 L. to 4.50 L?

Step-by-Step Solution

Answer

(a) \([\operatorname{COBr}_2] = 0.0295 \, \text{M}\), \([\mathrm{CO}] = 0.0231 \, \text{M}\), \([\mathrm{Br}_2] = 0.0231 \, \text{M}\). (b) Reduce the initial concentration and solve again for the new equilibrium. (c) Decreasing volume increases pressure, shifting equilibrium to \(\operatorname{COBr}_2\).

Step 1: Define Initial Conditions

We start with 0.50 moles of \( \operatorname{COBr}_2 \) in a 9.50 L flask. The initial concentration \( [\operatorname{COBr}_2]_0 \) is calculated as \( \frac{0.50 \, \text{mol}}{9.50 \, \text{L}} = 0.0526 \, \text{M} \). Initially, \( [\operatorname{CO}] \) and \( [\mathrm{Br}_2] \) are both 0 M.

Step 2: Set Up the Equilibrium Expression

The equilibrium expression for the reaction \( \operatorname{COBr}_2(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{Br}_2(g) \) is given by \( K_c = \frac{[\mathrm{CO}][\mathrm{Br}_2]}{[\operatorname{COBr}_2]} \). Since \( K_c = 0.190 \), we can use this to find the equilibrium concentrations.

Step 3: Use an ICE Table

Create an ICE table: - Initial concentrations: \([\operatorname{COBr}_2] = 0.0526 \, \text{M}, [\mathrm{CO}] = 0, [\mathrm{Br}_2] = 0\) - Change: \([-x, +x, +x]\) - Equilibrium concentrations: \( [\operatorname{COBr}_2] = 0.0526 - x \), \([\mathrm{CO}] = x \), \([\mathrm{Br}_2] = x \).

Step 4: Solve for Equilibrium Concentration

Substitute the ICE table values into the equilibrium expression: \[ K_c = \frac{x \times x}{0.0526 - x} = \frac{x^2}{0.0526 - x} = 0.190 \]Solve the quadratic equation to find \( x \). Simplifying gives:\[ x^2 = 0.190(0.0526 - x) \], which leads to the quadratic equation \( x^2 + 0.190x - 0.00999 = 0 \).Solving this, we find \( x = 0.0231 \).

Step 5: Calculate Equilibrium Concentrations

With \( x = 0.0231 \), the equilibrium concentrations are:- \([\operatorname{COBr}_2] = 0.0526 - 0.0231 = 0.0295 \, \text{M}\)- \([\mathrm{CO}] = 0.0231 \, \text{M}\)- \([\mathrm{Br}_2] = 0.0231 \, \text{M}\).

Step 6: Adjust for New Volume in Part (b)

For part (b), recalculate using 0.50 mol of \( \operatorname{COBr}_2 \) in a 4.5-L flask, giving an initial concentration of \( \frac{0.50}{4.5} = 0.111 \, \text{M} \). Repeat steps 2 through 5 with the new initial concentration.

Step 7: Calculate New Equilibrium Concentrations

Repeat ICE and solve for equilibrium with the initial concentration of \( 0.111 \, \text{M} \). The equilibrium expression and resultant concentrations will be different due to the initial value change.

Step 8: Determine Effect of Volume Change for Part (c)

Decreasing the volume increases the pressure, which shifts the equilibrium to the side with fewer moles of gas. In this case, it shifts toward reactants (\(\operatorname{COBr}_2\)), increasing its concentration and decreasing \([\mathrm{CO}]\) and \([\mathrm{Br}_2]\).

Key Concepts

Understanding the Equilibrium Expression

In chemical reactions that reach equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products remain constant over time. The equilibrium expression, also known as the equilibrium constant expression, helps describe this balance. For the reaction \(\operatorname{COBr}_2(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{Br}_2(g)\), the expression is based on concentrations:

• \(K_c = \frac{[\mathrm{CO}][\mathrm{Br}_2]}{[\operatorname{COBr}_2]}\)

The constant \(K_c\) reflects the ratio of product concentrations to reactant concentrations at equilibrium. When given \(K_c = 0.190\), it tells us how far the equilibrium lies towards the products or reactants. A lower \(K_c\) indicates a reaction that's less product-favored at equilibrium.
Understanding this is key to predicting how the system reacts to changes, such as pressure adjustments or concentration shifts.

Using the ICE Table Method

An ICE table is a handy tool for organizing information about a chemical equilibrium, which stands for Initial, Change, and Equilibrium. For our reaction, we begin with the initial concentrations and track how they change until equilibrium is reached.

• **Initial:** Start with known concentrations, e.g., \([\operatorname{COBr}_2] = 0.0526 \, \text{M}\)
• **Change:** Represents how concentrations shift to reach equilibrium, usually denoted as \(-x, +x, +x\).
• **Equilibrium:** Final concentrations expressed in terms of \(x\), based on initial values and changes.

By solving the equilibrium expression \(K_c = \frac{x^2}{0.0526 - x} = 0.190\), we can find \(x\) and thus the equilibrium concentrations. The use of an ICE table simplifies these processes, giving a clear step-by-step path to solving equilibrium problems.

Determining Equilibrium Concentrations

Calculating equilibrium concentrations involves substituting values back into your expressions from the ICE table. After solving the quadratic equation, let's say \(x = 0.0231\). Now:

• \([\operatorname{COBr}_2] = 0.0526 - x = 0.0295 \, \text{M}\)
• \([\mathrm{CO}] = x = 0.0231 \, \text{M}\)
• \([\mathrm{Br}_2] = x = 0.0231 \, \text{M}\)

Subsequent calculations, like those in different volumes, follow similar logic. Begin with the new initial concentration, use the ICE table, and solve for \(x\) using the quadratic equation derived from the equilibrium expression. This process enables accurate predictions of system behavior under various conditions.

Applying Le Chatelier's Principle

Le Chatelier's Principle is crucial in understanding how a reaction at equilibrium responds to external changes. When you decrease the volume of a gas reaction, the pressure increases, causing the equilibrium to shift to reduce that pressure.

• In our system, decreasing the volume shifts the equilibrium towards fewer moles of gas: \(\operatorname{COBr}_2\).
• This results in increased \([\operatorname{COBr}_2]\) and decreased \([\mathrm{CO}]\) and \([\mathrm{Br}_2]\) concentrations, aligning with what's predicted by Le Chatelier's Principle.

Understanding this principle allows you to predict qualitatively what will happen before even performing calculations. It's a valuable tool for anticipating the effects of changes such as temperature, pressure, and concentration adjustments on equilibrium systems.