Problem 10 - Chemistry and Chemical Reactivity

Question

Hydrogen and carbon dioxide react at a high temperature to give water and carbon monoxide. $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) $$ (a) Laboratory measurements at \(986^{\circ} \mathrm{C}\) show that there are 0.11 mol each of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) vapor and 0.087 mol each of \(\mathrm{H}_{2}\) and \(\mathrm{CO}_{2}\) at equilibrium in a 1.0-L container. Calculate the equilibrium constant for the reaction at \(986^{\circ} \mathrm{C}\) (b) Suppose 0.050 mol each of \(\mathrm{H}_{2}\) and \(\mathrm{CO}_{2}\) are placed in a \(2.0-\mathrm{L}\) container. When equilibrium is achieved at \(986^{\circ} \mathrm{C},\) what amounts of \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}),\) in moles, would be present? [Use the value of \(K_{\mathrm{c}}\) from part \((a) .]\)

Step-by-Step Solution

Answer

(a) \(K_c = 1.60\). (b) \([\text{CO}], [\text{H}_2\text{O}] = 0.0111\, \text{mol}\.\)

Step 1: Write the Equilibrium Expression

The equilibrium constant expression for the reaction \( ext{H}_2(g) + \text{CO}_2(g) \rightleftharpoons \text{H}_2\text{O}(g) + \text{CO}(g)\) is given by \(K_c = \frac{[\text{H}_2\text{O}][\text{CO}]}{[\text{H}_2][\text{CO}_2]}\). Each concentration here is in moles per liter.

Step 2: Identify Equilibrium Concentrations

At equilibrium at \(986^{\circ} \text{C}\) in a 1.0 L container, we are given: 0.11 mol of \(\text{CO}\), 0.11 mol of \(\text{H}_2\text{O}\), 0.087 mol of \(\text{H}_2\), and 0.087 mol of \(\text{CO}_2\). Since the container is 1.0 L, these quantities are also their concentrations.

Step 3: Calculate Equilibrium Constant \(K_c\)

Substitute the given equilibrium concentrations into the \(K_c\) expression: \(K_c = \frac{(0.11)(0.11)}{(0.087)(0.087)}\), which simplifies to \(K_c = \frac{0.0121}{0.007569}\). Calculate the value of \(K_c\).

Step 4: Initial Moles in New Scenario

For part (b), calculate the initial moles per liter of \(\text{H}_2\) and \(\text{CO}_2\) in the 2.0 L container. Initially, \(\text{H}_2 = \text{CO}_2 = \frac{0.050}{2.0} = 0.025\, \text{mol/L}\).

Step 5: Set Up Equilibrium Changes

Let the change in moles for \(\text{H}_2\) and \(\text{CO}_2\) be \(-x\) and for \(\text{H}_2\text{O}\) and \(\text{CO}\) be \(+x\). At equilibrium, \([\text{H}_2] = [\text{CO}_2] = 0.025 - x\) and \([\text{H}_2\text{O}] = [\text{CO}] = x\).

Step 6: Apply \(K_c\) Expression

Substitute into the \(K_c\) expression: \(K_c = \frac{x^2}{(0.025-x)^2}\). Replace \(K_c\) with the calculated value from part (a) and solve for \(x\), the equilibrium concentrations of \(\text{H}_2\text{O}\) and \(\text{CO}\).

Step 7: Solve for \(x\)

Simplify and solve the equation \(\frac{x^2}{(0.025-x)^2} = K_c\), where \(K_c\) is from part (a). Solve the equation to find \(x\), which represents the moles per liter of both \(\text{CO}\) and \(\text{H}_2\text{O}\) at equilibrium in the 2.0 L container.

Step 8: Calculate the Moles

Multiply \(x\) by the volume of the container (2.0 L) to find the number of moles of \(\text{CO}\) and \(\text{H}_2\text{O}\) at equilibrium.

Key Concepts

Equilibrium Constant

Chemical equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction. This balance allows specific amounts of reactants and products to coexist. The equilibrium constant, denoted as \(K_c\), quantifies this balance at a given temperature. In this specific reaction involving hydrogen and carbon dioxide forming water and carbon monoxide, the equilibrium constant is expressed as a ratio of the product concentrations to the reactant concentrations: \(K_c = \frac{[\text{H}_2\text{O}][\text{CO}]}{[\text{H}_2][\text{CO}_2]}\). Each concentration is measured in moles per liter (mol/L).

The value of \(K_c\) provides insight into the position of equilibrium. A \(K_c\) greater than one indicates products are favored, while a \(K_c\) less than one suggests that reactants are predominant. However, it is important to note that \(K_c\) is temperature-dependent and will vary if the reaction's temperature changes.

Reaction Stoichiometry

Stoichiometry concerns the quantitative analysis of reactants and products in a chemical reaction. In our reaction of interest, the stoichiometry is one-to-one for all species involved: - 1 mol of hydrogen \([\text{H}_2]\)- reacts with 1 mol of carbon dioxide \([\text{CO}_2]\)
- to produce 1 mol of water \([\text{H}_2\text{O}]\) and 1 mol of carbon monoxide \([\text{CO}]\).
This balanced relationship is essential for calculating changes in concentrations during the reaction, especially when determining the equilibrium position. If any change in the quantity of one species occurs, stoichiometry dictates that equivalent changes will happen for others according to the coefficients in the balanced equation.

Gas Laws

The reaction takes place with all components in the gaseous state. Therefore, the behavior of these gases might be influenced by gas laws, particularly if the volume or temperature changes. Although we assume constant conditions here, knowing the basics is useful:
- **Ideal Gas Law:** \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is moles, \(R\) is the gas constant, and \(T\) is temperature (in Kelvin). This relates the physical properties of a gas and can be helpful when calculating initial concentrations if volumes change.- **Avogadro's Law:** At the same temperature and pressure, equal volumes of gases contain an equal number of molecules. This aids in comparing the amounts in different gaseous states.These laws provide foundational support and ensure that reactions can be controlled or predicted if conditions change.

Mole Calculations

Calculating moles involves determining how much of each reactant and product is present. For our system in a 1.0 L container, the amount of moles is numerically equal to the concentration in molarity (mol/L). Given the equilibrium concentrations of hydrogen and carbon dioxide as 0.087 mol each, and 0.11 mol each for water and carbon monoxide, we applied these values in the equilibrium constant expression.

For other scenarios, initial moles must be adjusted for container volume to find concentrations, as shown in part (b) of the problem. For example, starting with 0.050 moles in a 2.0 L container equates to concentrations of 0.025 mol/L. As reactions progress, changes occur (denoted as \(-x\) or \(+x\), where \(x\) is the change in moles).
Moles directly influence concentration and reactant-product balance within the equilibrium framework, making precise mole calculations crucial for determining how the system will behave.