Problem 65 - Chemistry and Chemical Reactivity

Question

Characterize each of the following as product-or reactantfavored. (a) \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=1.2 \times 10^{45}\) (b) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftarrows \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=9.1 \times 10^{-41}\) (c) \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows \mathrm{COCl}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=6.5 \times 10^{11}\)

Step-by-Step Solution

Answer

(a) Product-favored, (b) Reactant-favored, (c) Product-favored.

Step 1: Understanding Equilibrium Constants

Equilibrium constant \( K_p \) is a measure of the extent of a reaction at equilibrium. A large \( K_p \) value (\( K_p >> 1 \)) indicates that the equilibrium lies far to the right, favoring product formation. Conversely, a small \( K_p \) value (\( K_p << 1 \)) indicates that the equilibrium lies far to the left, favoring reactant formation.

Step 2: Analyze Reaction (a)

For the reaction \( \mathrm{CO} (\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2} (\mathrm{g}) \rightleftarrows \mathrm{CO}_{2} (\mathrm{g}) \), the equilibrium constant \( K_p = 1.2 \times 10^{45} \). This extremely large \( K_p \) indicates that the reaction is heavily product-favored, meaning that at equilibrium, the concentration of \( \mathrm{CO}_{2} \) will be much greater than the concentrations of \( \mathrm{CO} \) and \( \mathrm{O}_{2} \).

Step 3: Analyze Reaction (b)

For the reaction \( \mathrm{H}_{2} \mathrm{O} (\mathrm{g}) \rightleftarrows \mathrm{H}_{2} (\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2} (\mathrm{g}) \), the equilibrium constant \( K_p = 9.1 \times 10^{-41} \). This very small \( K_p \) suggests that the reaction is strongly reactant-favored, meaning that at equilibrium, the concentration of \( \mathrm{H}_{2} \mathrm{O} \) will remain much higher than the concentrations of \( \mathrm{H}_{2} \) and \( \mathrm{O}_{2} \).

Step 4: Analyze Reaction (c)

For the reaction \( \mathrm{CO} (\mathrm{g}) + \mathrm{Cl}_{2} (\mathrm{g}) \rightleftarrows \mathrm{COCl}_{2} (\mathrm{g}) \), the equilibrium constant \( K_p = 6.5 \times 10^{11} \). This large \( K_p \) value indicates that the reaction is product-favored, indicating that the concentration of \( \mathrm{COCl}_{2} \) will be significantly higher at equilibrium than the concentrations of \( \mathrm{CO} \) and \( \mathrm{Cl}_{2} \).

Key Concepts

Equilibrium Constants

Equilibrium constants, often denoted as \( K_p \) for reactions involving gases, play a crucial role in understanding chemical equilibria. Simply put, an equilibrium constant is a numerical value that indicates the ratio of the concentrations of products to reactants at equilibrium. This constant helps predict the direction in which a reaction proceeds.

• If \( K_p \) is much greater than 1 (\( K_p >> 1 \)), it indicates that the products are favored at equilibrium. The reaction mixture will predominantly consist of products.
• If \( K_p \) is much less than 1 (\( K_p << 1 \)), it suggests that the reactants are favored—thus, the concentration of reactants will be higher compared to that of the products at equilibrium.

The value of \( K_p \) depends solely on the nature of the reactants and products as well as the temperature at which the reaction takes place. It remains unaffected by the initial concentrations of reactants and products. Understanding the value of \( K_p \) is essential for predicting the behavior of the system under different conditions.

Product-Favored Reactions

Product-favored reactions are characterized by a large equilibrium constant. In such reactions, as the system reaches equilibrium, the majority of reactants have transformed into products. This means that the concentration of products will be significantly higher than that of reactants at equilibrium.

Consider the reaction: \( \mathrm{CO} (\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2} (\mathrm{g}) \rightleftarrows \mathrm{CO}_{2} (\mathrm{g}) \), with an equilibrium constant \( K_p = 1.2 \times 10^{45} \). Such an extremely high \( K_p \) informs us that the reaction is heavily product-favored.

• The formation of \( \mathrm{CO}_{2} \) is essentially complete, resulting in very low amounts of \( \mathrm{CO} \) and \( \mathrm{O}_{2} \) remaining in the reaction mixture.
• Larger values of \( K_p \) indicate that the chemical equilibrium drastically leans towards the production side, often perceived as very efficient reactions towards product formation.

Understanding when a reaction is product-favored allows chemists to manipulate reaction conditions to enhance product yield, by potentially adjusting concentrations, pressure, or temperature.

Reactant-Favored Reactions

In contrast to product-favored reactions, reactant-favored reactions exhibit very small equilibrium constants. These reactions do not proceed extensively towards the formation of products; instead, the reactants remain in higher concentrations within the reaction mixture upon reaching equilibrium.

A great example is the reaction: \( \mathrm{H}_{2} \mathrm{O} (\mathrm{g}) \rightleftarrows \mathrm{H}_{2} (\mathrm{g}) + \frac{1}{2} \mathrm{O}_{2} (\mathrm{g}) \), with a tiny \( K_p = 9.1 \times 10^{-41} \). This small \( K_p \) indicates that the reaction is strongly reactant-favored.

• The concentration of \( \mathrm{H}_{2} \mathrm{O} \) will stay much larger than \( \mathrm{H}_{2} \) or \( \mathrm{O}_{2} \) at equilibrium, meaning very little of these gases is produced.
• Reactions with small \( K_p \) are less spontaneous towards forming products and often require different conditions, like the use of catalysts, to proceed further.

By recognizing reactant-favored reactions, we can better approach optimizing chemical processes to favor product formation under specific circumstances.