Question

Hydrogen and carbon dioxide react at a high temperature to give water and carbon monoxide. $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) $$ (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 50.0 -L container. Calculate the equilibrium constant for the reaction at \(986^{\circ} \mathrm{C}\) (b) Suppose 0.010 mol each of \(\mathrm{H}_{2}\) and \(\mathrm{CO}_{2}\) are placed in a 200.0 -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 \(K_{c}\) from part (a).]

Short answer

(a) The equilibrium constant \(K_c\) is 1.606. (b) At equilibrium, there will be 0.00636 mol of \(\text{CO}\) and \(\text{H}_2\text{O}\) each in a 200.0 L container.

Step-by-step solution

Calculate Initial Concentrations for Part (a)

For a 50.0 L container, the initial concentrations are calculated by dividing the number of moles by the volume of the container. \[[\text{H}_2] = \frac{0.087 \text{ mol}}{50.0 \text{ L}} = 0.00174 \text{ M},\ [\text{CO}_2] = \frac{0.087 \text{ mol}}{50.0 \text{ L}} = 0.00174 \text{ M}\] Similarly, \[[\text{H}_2\text{O}] = \frac{0.11 \text{ mol}}{50.0 \text{ L}} = 0.0022 \text{ M},\ [\text{CO}] = \frac{0.11 \text{ mol}}{50.0 \text{ L}} = 0.0022 \text{ M}\]

Write the Expression for the Equilibrium Constant

The equilibrium constant expression \(K_c\) for the reaction is given by: \[K_c = \frac{[\text{CO}][\text{H}_2\text{O}]}{[\text{H}_2][\text{CO}_2]}\]

Substitute Values into the Equilibrium Expression for Part (a)

Substitute the concentrations from Step 1 into the equilibrium constant expression: \[K_c = \frac{(0.0022)(0.0022)}{(0.00174)(0.00174)}\] Simplify to find \(K_c\).

Calculate Equilibrium Concentrations for Part (a)

After calculating, \[K_c = \frac{(0.0022)(0.0022)}{(0.00174)(0.00174)} = 1.606\]

Set Up the Equation for Part (b)

For a 200.0 L container starting with 0.010 mol of both \(\text{H}_2\) and \(\text{CO}_2\), calculate initial concentrations:\[[\text{H}_2] = \frac{0.010 \text{ mol}}{200.0 \text{ L}} = 0.00005 \text{ M}, \ [\text{CO}_2] = \frac{0.010 \text{ mol}}{200.0 \text{ L}} = 0.00005 \text{ M}\] At equilibrium, assume \(x\) moles of \(\text{CO}\) and \(\text{H}_2\text{O}\) are formed.

Express Concentrations as Functions of x for Part (b)

At equilibrium, the concentrations will be:- \([\text{H}_2] = 0.00005 - x\)- \([\text{CO}_2] = 0.00005 - x\)- \([\text{CO}] = x\)- \([\text{H}_2\text{O}] = x\)

Solve the Equilibrium Expression for Part (b)

Substitute the concentrations into the equilibrium expression using the \(K_c\) from Step 4:\[1.606 = \frac{x^2}{(0.00005 - x)^2}\]Solve the equation for \(x\).

Calculate Equilibrium Amounts for Part (b)

After solving, \(x = 0.0000318 \text{ M}\). Convert back to moles for the 200.0 L container:- Moles of \(\text{CO} = \text{Moles of } \text{H}_2\text{O} = 0.0000318 \times 200.0 = 0.00636 \text{ mol}\).

Key concepts

Reaction Kinetics

Reaction kinetics explores the speed or rate at which chemical reactions occur. This concept is crucial in understanding chemical processes, such as the reaction between hydrogen and carbon dioxide to form water and carbon monoxide. Several factors can influence the rate of a reaction, including temperature, concentration of reactants, and presence of catalysts. In the given exercise, we measure how quickly the reaction reaches a state where the concentrations of reactants and products no longer change, known as equilibrium. Temperature is a pivotal factor here, with the reaction occurring at a high temperature of 986°C. At such high temperatures, molecules move faster, increasing the likelihood of effective collisions that lead to reactions. The equilibrium state we observe is essentially a snapshot of the balance in reaction kinetics where the forward and reverse reaction rates are equal.

Mole Calculations

Mole calculations are essential when dealing with chemical reactions as they allow us to relate the masses of substances to the number of particles or entities involved. Moles provide a convenient way to express amounts of reactants and products in a chemical reaction. For the exercise in question, mole calculations are used to determine the concentrations of each gas at equilibrium. Concentrations are calculated by dividing the number of moles by the volume of the container, which in this case, provides the molarity (M). For instance, the initial concentrations of hydrogen and carbon dioxide in the 50.0 L container are determined as 0.00174 M from 0.087 mol. Understanding these calculations is vital for computing the equilibrium constant, which represents the balance of reactant and product concentrations at equilibrium. Similarly, for part (b), the initial moles when the reactants are placed in a larger 200.0 L container help us understand how much product will form when balance is reached.

Chemical Equilibrium

Chemical equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction, meaning the concentrations of reactants and products remain constant over time. This is crucial for understanding the reaction between hydrogen and carbon dioxide in this exercise. The equilibrium state is described by the equilibrium constant, denoted as \(K_c\), which is specific to a reaction at a given temperature. It is expressed as a ratio of product concentrations to reactant concentrations, each raised to the power of their coefficients in the balanced chemical equation. For the exercise, calculating \(K_c\) requires knowing the equilibrium concentrations of all species involved, as seen in the expression: \(K_c = \frac{[\text{CO}][\text{H}_2\text{O}]}{[\text{H}_2][\text{CO}_2]}\). After substituting the known concentrations, we find that \(K_c\) is 1.606 at 986°C. This value provides insight into how the reaction proceeds under equilibrium — a larger \(K_c\) implies products are favored at equilibrium, while a smaller one favors reactants. Understanding \(K_c\) aids in predicting how changes in conditions can shift the equilibrium position, a principle central to the Le Châtelier's principle.