Problem 27 - Chemistry The Central Science

Question

The following equilibria were attained at \(823 \mathrm{K} :\) $$\operatorname{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67$$ $$\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) \quad K_{c}=490$$ Based on these equilibria, calculate the equilibrium constant for \(\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at 823 \(\mathrm{K}\) .

Step-by-Step Solution

Answer

The equilibrium constant for the reaction H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g) at 823 K is approximately 7.31.

Step 1: Balanced equations for the given equilibria

The balanced equations for the given equilibria are already provided: 1. CoO(s) + H₂(g) ⇌ Co(s) + H₂O(g) (Kc = 67) 2. CoO(s) + CO(g) ⇌ Co(s) + CO₂(g) (Kc = 490)

Step 2: Determine the relationship between the given reactions and the desired reaction

We want to find the equilibrium constant for the reaction: H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g) Notice that we can obtain the desired reaction by: 1. Reversing reaction 1: Co(s) + H₂O(g) ⇌ CoO(s) + H₂(g) 2. Adding the reversed reaction 1 and reaction 2: (Co(s) + H₂O(g) ⇌ CoO(s) + H₂(g)) + (CoO(s) + CO(g) ⇌ Co(s) + CO₂(g)) By doing this, the Co(s) and CoO(s) species cancel out, resulting in the desired reaction: H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g)

Step 3: Calculate the equilibrium constant for the desired reaction

Since we want the equilibrium constant (Kc) for the desired reaction, we can make use of the fact that equilibrium constants multiply when reactions are combined: 1. Reversed reaction 1: Kc' = 1 / Kc₁ = 1 / 67 2. Reaction 2: Kc₂ = 490 Since we are combining these reactions to get the desired reaction, we can find the equilibrium constant Kc_desired by multiplying the constants for the individual reactions: Kc_desired = Kc' × Kc₂ = (1 / 67) × 490 = 490 / 67 ≈ 7.31 So, the equilibrium constant for the desired reaction H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g) at 823 K is approximately 7.31.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state where the concentrations of reactants and products have reached a balance, resulting in no net change over time. This occurs in a reversible reaction, where the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, the system is stable, meaning that the macroscopic properties like concentration, pressure, and temperature remain constant.

For students studying chemical equilibrium, it's important to understand that reactions don't stop when equilibrium is reached. Instead, they continue to occur at equal rates in both directions. This balance is often represented by an equilibrium constant, denoted by \( K_c \), which provides a ratio of product concentrations to reactant concentrations at equilibrium.

• Equilibrium is dynamic: reactions continue, but no net change occurs.
• Equilibrium constants are specific to a particular temperature.
• The value of \( K_c \) indicates the position of equilibrium. A large \( K_c \) favors products, whereas a small \( K_c \) favors reactants.

Understanding chemical equilibrium helps us predict how changes in conditions can affect the system's balance, a principle described by Le Chatelier's principle.

Reaction Reversibility

Reaction reversibility is a crucial concept in chemistry, where a chemical reaction can proceed in both forward and reverse directions. In reversible reactions, neither the reactants nor the products are completely consumed. Instead, they reach a state of balance or equilibrium. The double arrows (⇌) in a chemical equation symbolize this reversible nature.

For reactions to be reversible, they usually involve weak interactions, allowing them to readily change direction depending on the conditions. Not all reactions are reversible; some proceed to completion, going in only one direction. Factors that can affect reaction reversibility include:

• Change in temperature or pressure
• Concentration of reactants and products
• Presence of catalysts or inhibitors

Understanding reversibility is essential when studying chemical systems because it shows how equilibrium can shift when external conditions change. This knowledge is applied in real-world scenarios, such as chemical manufacturing processes, where controlling the direction of reactions can optimize yield and efficiency.

Equilibrium Reactions

Equilibrium reactions are reactions that occur in a closed system and can proceed in both forward and reverse directions at the same rates. This results in a state of equilibrium. For these reactions, we rely on the equilibrium constant \( K_c \) to quantify the concentration ratio of products and reactants.

In our given exercise, we dealt with equilibrium reactions to calculate the constant for a new reaction. By determining the relationship between two reactions and manipulating their equilibrium constants, we arrived at the desired \( K_c \) for the target reaction. Such manipulations emphasize that equilibrium reactions are interconnected and modifying one can influence another through a series of steps.

Key points to remember about equilibrium reactions include:

• These reactions are sensitive to initial conditions such as temperature and pressure.
• \( K_c \) values help predict how much of reactants are converted to products at equilibrium.
• Reaction conditions can shift the equilibrium position, impacting product yield in practice.

Recognizing and calculating equilibrium constants in reactions is a foundational skill in chemistry, aiding in the prediction and analysis of reaction behaviors.