Problem 85 - Chemistry The Central Science In Si Units

Question

When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and \(1.50 \mathrm{~mol} \mathrm{H}_{2}\) are placed in a 3.00-L container at \(395^{\circ} \mathrm{C}\), the following reaction occurs: \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\). If \(K_{c}=0.802,\) what are the concentrations of each substance in the equilibrium mixture?

Step-by-Step Solution

Answer

Equilibrium concentrations are: \([\mathrm{CO}_{2}] = 0.264 \text{ M}\), \([\mathrm{H}_{2}] = 0.264 \text{ M}\), \([\mathrm{CO}] = 0.236 \text{ M}\), \([\mathrm{H}_{2}\mathrm{O}] = 0.236 \text{ M}\).

Step 1: Write the equilibrium expression

For the reaction \( \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \), the equilibrium constant expression is: \[ K_c = \frac{[\mathrm{CO}][\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{CO}_{2}][\mathrm{H}_{2}]} \] Given \( K_c = 0.802 \).

Step 2: Calculate initial concentrations

The initial moles of \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \) are 1.50 mol each in a 3.00 L container. Therefore, \[ [\mathrm{CO}_{2}]_0 = [\mathrm{H}_{2}]_0 = \frac{1.50 \text{ mol}}{3.00 \text{ L}} = 0.50 \text{ M} \] Initially, there is no \( \mathrm{CO} \) or \( \mathrm{H}_{2}\mathrm{O} \).

Step 3: Define change in concentrations

Let \( x \) be the amount of \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \) that react to reach equilibrium. Then, the changes are: - \( [\mathrm{CO}_{2}] = 0.50 - x \) - \( [\mathrm{H}_{2}] = 0.50 - x \) - \( [\mathrm{CO}] = x \) - \( [\mathrm{H}_{2}\mathrm{O}] = x \).

Step 4: Set up the equilibrium expression

Substitute the equilibrium concentrations into the equilibrium expression:\[ K_c = \frac{x^2}{(0.50-x)^2} = 0.802 \]

Step 5: Solve the equilibrium equation

Take the square root of both sides to simplify solving for \( x \):\[ \frac{x}{0.50-x} = \sqrt{0.802} \approx 0.895 \]Solving for \( x \):\[ x = 0.895(0.50-x) \]Expanding and solving the equation:\[ x = 0.4475 - 0.895x \] \[ 1.895x = 0.4475 \] \[ x \approx 0.236 \].

Step 6: Calculate equilibrium concentrations

Substitute \( x = 0.236 \) back into the equilibrium concentrations:- \([\mathrm{CO}_{2}] = 0.50 - 0.236 = 0.264 \text{ M}\) - \([\mathrm{H}_{2}] = 0.50 - 0.236 = 0.264 \text{ M}\) - \([\mathrm{CO}] = 0.236 \text{ M}\) - \([\mathrm{H}_{2}\mathrm{O}] = 0.236 \text{ M} \).

Key Concepts

Equilibrium Constant

In chemical reactions, the concept of the equilibrium constant \( K_c \) is fundamental. It tells us how the concentrations of reactants and products relate when a reaction reaches equilibrium. For a reaction such as \( \mathrm{CO}_{2}(g) + \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H}_2\mathrm{O}(g) \), the equilibrium constant is expressed using the formula:

• \( K_c = \frac{[\mathrm{CO}][\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{CO}_{2}][\mathrm{H}_{2}]} \)

This equation shows that at equilibrium, the ratio of product concentrations to reactant concentrations equals the equilibrium constant, \( K_c \). It's a numeric value that remains unchanged as long as temperature does not vary, and it helps predict how much of each substance is present after a reaction has equilibrated.

Reaction Quotient

The reaction quotient, \( Q_c \), serves a vital role in predicting the direction a reaction will proceed before reaching equilibrium. By comparing \( Q_c \) to \( K_c \), one can determine if the reaction will move towards forming more products or reactants. When initially calculating \( Q_c \) using the same expression as \( K_c \), remember:

• If \( Q_c < K_c \), the reaction proceeds to the right, favoring product formation.
• If \( Q_c > K_c \), the reaction shifts left, favoring reactants.
• If \( Q_c = K_c \), the reaction is at equilibrium.

Understanding the reaction quotient helps to assess how close a reaction mixture is to equilibrium and predict concentration adjustments.

Concentration Calculation

Calculating concentrations involves determining initial values and how they change as the reaction approaches equilibrium. For example, with initial moles provided in a specific container volume, it is straightforward:

• Determine initial concentration: \([\mathrm{CO}_{2}]_0 = [\mathrm{H}_{2}]_0 = \frac{1.50 \text{ mol}}{3.00 \text{ L}} = 0.50 \text{ M}\)

As the reaction proceeds, reactant concentrations decrease while product concentrations grow. Use an unknown \( x \) to represent the change. Then substitute these changes in the equilibrium expression to solve for \( x \); note how changes reflect upon initial values and result in equilibrium concentrations.

Equilibrium Expression

The equilibrium expression is crucial in solving for unknown concentrations in a chemical reaction at equilibrium. Start by substituting the changes in concentrations into the equilibrium constant expression. For instance, for our system:

• \( K_c = \frac{x^2}{(0.50-x)^2} = 0.802 \)

Solving this expression provides the change in concentration, \( x \). Substitute \( x \) back to find each individual concentration of reactants and products, ultimately serving as a tool to understand the extent of the reaction and to confirm that the calculated concentrations respect the given \( K_c \) value.