Problem 96 - Chemistry The Central Science

Question

The following equilibria were measured at 823 K: \begin{equation} \begin{aligned} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) & \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{aligned} \end{equation} (a) Use these equilibria to calculate the equilibrium constant, \(K_{c},\) for the reaction \(\operatorname{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)\) \(+\mathrm{CO}_{2}(g)\) at 823 \(\mathrm{K}\) . (b) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{K} ?\) (c) If you were to place 5.00 \(\mathrm{g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of 250 \(\mathrm{mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of 1.00 atm and a temperature of \(298 \mathrm{K},\) what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). (d) If the reaction vessel from part (c) is heated to 823 \(\mathrm{K}\) and allowed to come to equilibrium, how much \(\mathrm{CoO}(s)\) remains?

Step-by-Step Solution

Answer

The equilibrium constant for the reaction \(CoO(s) + CO(g) \rightleftharpoons Co(s) + CO₂(g)\) is 0.00209. Carbon monoxide is a stronger reducing agent than hydrogen at 823 K. The concentration of CO gas is 0.0408 M. At equilibrium, there are 4.88 g of CoO remaining in the reaction vessel.

Step 1: Write down the target reaction

Combine the given reactions in order to obtain the desired reaction: CoO(s) + CO(g) ↔ Co(s) + CO₂(g)

Step 2: Invert and combine given reactions

Invert the first given reaction, then add both given reactions together: Co(s) + H₂O(g) ↔ CoO(s) + H₂(g) (Inverted, \(K_c=1/67\)) H₂(g) + CO₂(g) ↔ CO(g) + H₂O(g) (\(K_c=0.14\)) Upon adding these reactions, we obtain the desired reaction and can find the new equilibrium constant.

Step 3: Calculate the new equilibrium constant

When we invert a reaction, we take the inverse of the equilibrium constant, and when we add two reactions, we multiply their constants. Therefore, \(K_c = \frac{1}{67} \times 0.14 = 0.00209\) The equilibrium constant for the reaction \(CoO(s) + CO(g) \rightleftharpoons Co(s) + CO₂(g)\) is 0.00209. #b- Strength of CO as a reducing agent#

Step 4: Compare the equilibria for CO and H₂

Carbon monoxide is a stronger reducing agent if its reaction equilibrium constant is smaller than that of hydrogen gas in the same reaction system. Referring to the first given reaction, which involves hydrogen gas, and comparing it with our calculated value for carbon monoxide: \(K_c (H₂) = 67\) \(K_c (CO) = 0.00209\)

Step 5: Determine the strength of CO and H₂

The equilibrium constant for CO is smaller than that of H₂, so carbon monoxide is a stronger reducing agent than hydrogen at 823 K. #c- Find the concentration of CO gas#

Step 6: Use the ideal gas law

We will use the ideal gas law to find the concentration of CO gas: \(PV = nRT\) Where P is the pressure (1 atm), V is the volume (0.25 L), n is the number of moles of CO, R is the ideal gas constant (0.0821 L atm/(mol K)), and T is the temperature (298 K).

Step 7: Calculate the concentration of CO gas

Rearrange the ideal gas law to solve for the number of moles (n): \(n = \frac{PV}{RT}\) \(n = \frac{1 \times 0.25}{0.0821 \times 298} = 0.0102 \, \text{mol}\) Now, calculate the concentration by dividing the moles by the volume: \([\text{CO}] = \frac{0.0102}{0.25} = 0.0408 \, \text{M}\) The concentration of CO gas is 0.0408 M. #d- CoO remaining after equilibrium#

Step 8: Calculate moles at equilibrium

At equilibrium, the number of moles of CO and CO₂ are equal, as indicated by the stoichiometry of the target reaction. And, since the equilibrium constant is small, we assume that the change in moles of CO and CO₂ is negligible, keeping the concentration of CO constant. Let x be the moles of CO that react with CoO. \(K_c = \frac{[\text{CO}_2(g)]}{[\text{CO}(g)]} = \frac{x}{0.0408 - x}\) Solve for x: \(x = 0.00209 (0.0408 - x)\)

Step 9: Calculate remaining CoO

Notice that there are 5 g of CoO initially. The molar mass of CoO is approximately 75 g/mol. Initial moles of CoO: \(\frac{5}{75} = 0.067 \, \text{mol}\) When x moles of CO react, the same amount of CoO gets consumed, leaving 0.067 - x moles of CoO. Solve for x and find the remaining moles: \(x = 0.00197 \, \text{mol}\) Remaining CoO: \(0.067 - 0.00197 = 0.065 \, \text{mol}\)

Step 10: Calculate the remaining mass of CoO

Finally, convert moles of CoO remaining to grams: Remaining mass of CoO: \(0.065 \times 75 = 4.88 \, \text{g}\) At equilibrium, there are 4.88 g of CoO remaining in the reaction vessel.

Key Concepts

Equilibrium Constant

The equilibrium constant (\(K_c\)) is a ratio that gives us a snapshot of the relative concentrations of the products to the reactants of a chemical reaction at equilibrium. It can tell us the direction in which the reaction favors or the extent to which reactants get converted to products under a certain condition of temperature.

In the context of the exercise, it is important to note that the equilibrium constant changes when the balanced chemical equation changes. When calculating the equilibrium constant for a new reaction derived from a combination of other reactions, as done in the exercise, one must remember to invert the value of \(K_c\) when the direction of the original equation is reversed and to multiply the constants when reactions are added. This method is essential to find the equilibrium constant for a reaction not provided directly in the problem statement, ensuring a deep understanding.

Le Chatelier's Principle

Le Chatelier's principle is the go-to guide when predicting the effects of changes in concentration, temperature, or pressure on a system at equilibrium. This principle states that if an external change is imposed on a system at equilibrium, the system adjusts to partially counteract the impact of the change and re-establishes equilibrium.

When applying this principle, for example, an increase in the concentration of one of the reactants will typically shift the equilibrium position toward the formation of products, as the system strives to reduce the effect of the change introduced. By using Le Chatelier's principle, students can gain insights into how dynamic chemical equilibria respond to external disturbances in real-world applications.

Ideal Gas Law

The ideal gas law, expressed as \(PV = nRT\), relates the pressure (\(P\)), volume (\(V\)), number of moles (\(n\)), temperature (\(T\)), and the ideal gas constant (\(R\)) for a hypothetical ideal gas. By rearranging this equation, we can solve for various parameters such as the number of moles or concentration of a gaseous reactant or product under certain conditions.

To illustrate, the exercise demonstrates how to calculate the concentration of CO gas given the pressure, volume, and temperature. This calculation is crucial for performing further calculations on the chemical equilibrium involving gases in the system. Understanding the ideal gas law is vital for students, especially when working with gas-phase equilibrium reactions or real-world scenarios involving gases.

Reducing Agent Strength

Reducing agent strength is assessed by how well a substance can bring about the reduction of others, essentially by donating electrons in a chemical reaction. A stronger reducing agent readily loses electrons to reduce its counterpart, the oxidizing agent. The reduction potential of a reducing agent is inversely related to its equilibrium constant; a smaller \(K_c\) value signifies a more robust reducing agent in a given reaction.

In the exercise, when comparing the equilibrium constants of H\textsubscript{2} and CO, the smaller \(K_c\) for CO's reaction shows that CO is a stronger reducing agent than H\textsubscript{2} at 823 K. This realization is fundamental when dealing with redox reactions, as the ability to reduce and the strength of reducing agents are critical factors in understanding and predicting the outcome of these reactions.