Question

When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and \(1.50 \mathrm{~mol} \mathrm{H}_{2}\) are placed in a \(0.750-\mathrm{L}\) container at \(395^{\circ} \mathrm{C}\), the following equilibrium is achieved: \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) .\) If \(K_{c}=0.802\), what are the concentrations of each substance in the equilibrium mixture?

Short answer

The equilibrium concentrations of the substances are approximately: CO₂ (1.249 M), H₂ (1.249 M), CO (0.751 M), and H₂O (0.751 M).

Step-by-step solution

Write the balanced chemical equation and the equilibrium constant expression.

The balanced chemical equation is given: \[\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\] The equilibrium constant expression (Kc) for this reaction is: \[K_c = \frac{[\mathrm{CO}][\mathrm{H}_2\mathrm{O}]}{[\mathrm{CO}_2][\mathrm{H}_2]}\]

Calculate the initial concentrations of the substances.

Using the initial moles and volume of the container, let's calculate the initial concentrations of the reactants. Initial concentration of CO₂: \[\frac{1.5 \thinspace \text{mol}}{0.75 \thinspace \text{L}} = 2.0 \thinspace\text{M}\] Initial concentration of H₂: \[\frac{1.5 \thinspace \text{mol}}{0.75 \thinspace \text{L}} = 2.0 \thinspace\text{M}\] The container is empty initially for the products CO and H₂O, so their concentrations are 0 M.

Set up the expressions for equilibrium concentrations and equilibrium constant.

Let x be the change in concentration due to the reaction occurring. Thus, concentrations at equilibrium will be: CO₂: (2.0 - x) M H₂: (2.0 - x) M CO: x M H₂O: x M Now, substitute these concentrations into the equilibrium constant expression: \[0.802 = \frac{x^2}{(2.0 - x)^2}\]

Solve the equilibrium constant equation for x.

Solving this equation for x. First we can square both sides: \[0.802^2 = \frac{x^2}{(2.0 - x)^2}\] Rearrange this equation and simplify: \[(2.0-x)^2 = \frac{x^2}{0.802^2} \\ 2x^2 = (2.0 - x) ^ 2 \cdot 0.802^2 \\ x=0.751\] (rounded to three decimal places)

Calculate the equilibrium concentrations for all substances.

Now, let's substitute x back into the expressions for equilibrium concentrations: CO₂: (2.0 - 0.751) M ≈ 1.249 M H₂: (2.0 - 0.751) M ≈ 1.249 M CO: 0.751 M H₂O: 0.751 M The equilibrium concentrations of the substances are approximately: CO₂ (1.249 M), H₂ (1.249 M), CO (0.751 M), and H₂O (0.751 M).

Key concepts

Equilibrium Constant (Kc)

The Equilibrium Constant, denoted as \(K_c\), is vital in understanding chemical equilibrium. It portrays the ratio of product concentrations to reactant concentrations at equilibrium. This constant is specific to each reaction at a particular temperature.
For the given chemical reaction \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\), the expression for the equilibrium constant \(K_c\) is:

• \[K_c = \frac{[\mathrm{CO}][\mathrm{H}_2\mathrm{O}]}{[\mathrm{CO}_2][\mathrm{H}_2]}\]

This formula shows how the concentrations of products and reactants influence equilibrium. When \(K_c\) is greater than 1, products are favored. Conversely, when \(K_c\) is less than 1, reactants are favored at equilibrium.
Understanding \(K_c\) helps predict the direction and extent of a reaction. It guides whether a reaction proceeds more to form products or remains dominated by reactants.

Equilibrium Concentrations

Equilibrium concentrations are the concentrations of reactants and products when a reaction reaches a state of balance. At this point, the forward and reverse reaction rates are equal, and the concentrations of all substances remain constant over time.
To find equilibrium concentrations, we start with the initial concentrations of all species. For the problem provided, the initial concentration of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\) is both 2.0 M in a 0.750 L container. The products, \(\mathrm{CO}\) and \(\mathrm{H}_2\mathrm{O}\), initially have concentrations of 0 M.
During the reaction, the concentration of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\) decreases by a quantity \(x\), which equals the concentration of products formed, \(x\). Thus, the equilibrium concentrations are:

• \( \mathrm{CO}_2: (2.0 - x) \text{ M} \)
• \( \mathrm{H}_2: (2.0 - x) \text{ M} \)
• \( \mathrm{CO}: x \text{ M} \)
• \( \mathrm{H}_2\mathrm{O}: x \text{ M} \)

To find the value of \(x\), we substitute these expressions into the equilibrium constant equation and solve. In this scenario, the equilibrium concentrations are \(\mathrm{CO}_2\) and \(\mathrm{H}_2\) at approximately 1.249 M, and \(\mathrm{CO}\) and \(\mathrm{H}_2\mathrm{O}\) at approximately 0.751 M each.

Balanced Chemical Equation

A balanced chemical equation is a fundamental tool in chemistry, ensuring the conservation of mass in a reaction. It ensures that the number of atoms for each element is the same on both sides of the equation.
For the given reaction:

• \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}\mathrm{O}(g)\)

This equation is already balanced with one carbon, two oxygens, and two hydrogens on each side.
Balancing the equation is crucial as it helps in calculating the correct stoichiometric coefficients. These coefficients serve as a basis for determining the proportions of reactants and products involved in the reaction. When dealing with equilibrium problems, a balanced equation is the starting point to set up the equilibrium expression properly.
Whether performing calculations for \(K_c\) or predicting reaction yields, always ensure the chemical equation is balanced first. This foundational step avoids errors and ensures understanding of the reaction's dynamics.