Problem 71 - Chemistry: The Central Science

Question

When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and \(1.50 \mathrm{~mol} \mathrm{H}_{2}\) are placed in a 0.750-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?

Step-by-Step Solution

Answer

The equilibrium concentrations are approximately [CO] = 1.42 M, [H2O] = 1.42 M, [CO2] = 0.58 M, and [H2] = 0.58 M.

Step 1: Write the balanced chemical equation

We are given the balanced chemical equation: CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

Step 2: Calculate the initial concentrations

To calculate the initial concentrations, we will use moles and volume. The initial moles of CO2 and H2 are given as 1.50 mol each, and the volume of the container is 0.750 L. Initial concentration of CO2 = moles of CO2 / volume = 1.50 mol / 0.750 L ≈ 2.00 M Initial concentration of H2 = moles of H2 / volume = 1.50 mol / 0.750 L ≈ 2.00 M Initially, there are no products in the container, so the initial concentrations of CO and H2O are 0 M.

Step 3: Write the equilibrium constant expression

The equation for Kc is given by: Kc = [CO][H2O] / [CO2][H2] where Kc = 0.802 and [CO], [H2O], [CO2], and [H2] are the equilibrium concentrations.

Step 4: Express equilibrium concentrations in terms of a variable

Let x represent the change in the concentration of CO and H2O (mol/L), which will increase at equilibrium. At the same time, the concentrations of CO2 and H2 will decrease by x (mol/L). Then, the equilibrium concentrations are: [CO] = x [H2O] = x [CO2] = 2.00 - x [H2] = 2.00 - x

Step 5: Substitute equilibrium concentrations into the Kc expression

Plug in the equilibrium concentrations in terms of x into the Kc expression: 0.802 = (x)(x) / ((2.00 - x)(2.00 - x))

Step 6: Solve for x

This is a quadratic equation, and we can solve for x by rearranging terms: x^2 = 0.802(4 - 4x + x^2) Now, assuming a small x (as Kc is not too much larger than 1), we can simplify and solve for x: x ≈ √(0.802(4)) x ≈ 1.42 M

Step 7: Calculate the equilibrium concentrations

Now we know the value of x, we can find the equilibrium concentrations: [CO] = 1.42 M [H2O] = 1.42 M [CO2] = 2.00 - 1.42 ≈ 0.58 M [H2] = 2.00 - 1.42 ≈ 0.58 M Therefore, the concentrations at equilibrium are approximately [CO] = 1.42 M, [H2O] = 1.42 M, [CO2] = 0.58 M, and [H2] = 0.58 M.

Key Concepts

Equilibrium Constant Expression

Understanding the equilibrium constant expression involves analyzing a chemical system where the forward and reverse reactions occur at the same rate. This expression depends on the balanced chemical equation of the reaction at equilibrium.

For a general reaction where reactants A and B form products C and D, represented as \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant expression for this reaction, designated \(K_c\), is written as \[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \] where the square brackets represent molar concentrations of the reactants and products at equilibrium, and the lowercase letters represent the stoichiometric coefficients from the balanced equation.

In the provided exercise, the equilibrium constant expression is derived from the balanced equation \(\mathrm{CO}_2(g) + \mathrm{H}_2(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H}_2\mathrm{O}(g)\), and is given by \[ K_c = \frac{[CO][H_2O]}{[CO_2][H_2]} \]. Through this exercise, students learn how to apply this expression by substituting equilibrium concentrations and solving for the unknown.

Balanced Chemical Equation

A balanced chemical equation plays a pivotal role in determining the stoichiometry of the reactants and products involved in the chemical reaction. It ensures the law of conservation of mass, meaning that the number of atoms for each element must be the same on both sides of the equation.

In the practice problem, the chemical equation \(\mathrm{CO}_2(g) + \mathrm{H}_2(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H}_2\mathrm{O}(g)\) is already balanced, with one mole of each reactant generating one mole of each product. When solving problems related to chemical equilibrium, always start with a balanced chemical equation. From this, one can deduce the stoichiometry required to express the changes in concentration of substances as the system reaches equilibrium, referred to as the reaction quotient. This sets the foundation for subsequent calculations, including determining the equilibrium constant and solving for unknown concentrations.

Molar Concentration

Molar concentration, often represented with square brackets \([]\), is a measure of the amount of a substance (in moles) present in a given volume (in liters) of solution. It is expressed as moles per liter (mol/L) and is a key concept when dealing with solutions and reactions in chemistry.

In the context of the equilibrium problem at hand, initial molar concentrations of reactants \(\mathrm{CO}_2\) and \(\mathrm{H}_2\) are calculated using the formula: \[ \text{Concentration} = \frac{\text{Moles of substance}}{\text{Volume of solution}} \]. As the reaction progresses and reaches equilibrium, the molar concentration of the reactants decrease, while those of the products increase. This change is quantitatively represented by using a variable (like 'x' in the example) to indicate the change in molar concentrations as the system shifts to achieve equilibrium.

The concept of molar concentration is fundamental for students to grasp as it allows them to quantify the amounts of reactants and products in a chemical reaction and understand the quantitative aspects of chemical equilibrium.