Problem 69 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understanding Reaction Equilibrium Constants

Before determining if a reaction is product- or reactant-favored, we need to understand that the equilibrium constant \( K_p \) reflects the ratio of the concentrations of products to reactants at equilibrium. A very large \( K_p \) indicates a product-favored reaction, while a very small \( K_p \) indicates a reactant-favored reaction.

Step 2: Evaluating Reaction (a)

For the reaction \( \mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}_{2}(\mathrm{g}) \), the equilibrium constant \( K_{p} = 1.2 \times 10^{45} \) is extremely large, suggesting the reaction strongly favors the formation of products, which in this case is \( \mathrm{CO}_{2}(\mathrm{g}) \).

Step 3: Evaluating Reaction (b)

For the decomposition of water \( \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \rightleftarrows \mathrm{H}_{2}(\mathrm{g}) + 1 / 2 \mathrm{O}_{2}(\mathrm{g}) \), the equilibrium constant \( K_{p} = 9.1 \times 10^{-41} \) is very small. This indicates that the reaction heavily favors the reactants, meaning it is very unlikely for \( \mathrm{H}_{2} \) and \( \mathrm{O}_{2} \) to form from \( \mathrm{H}_{2} \mathrm{O} \) under standard conditions.

Step 4: Evaluating Reaction (c)

For the reaction \( \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_2(\mathrm{g}) \rightleftarrows \mathrm{COCl}_2(\mathrm{g}) \), \( K_{p} = 6.5 \times 10^{11} \) is a large constant value. This signifies that the reaction is product-favored, meaning \( \mathrm{COCl}_2(\mathrm{g}) \) forms predominantly at equilibrium.

Key Concepts

Product-Favored Reactions

In the context of chemical reactions, understanding whether a reaction is product-favored at equilibrium is crucial. Product-favored reactions are characterized by a significant formation of products when the reaction reaches equilibrium. This typically occurs when the equilibrium constant, denoted as \( K_p \) or \( K_c \) depending on the phase of reactants and products, has a very large value. A large \( K_p \) value indicates that, at equilibrium, the concentration of products is much higher than that of reactants. This is why reactions with large \( K_p \) values are often referred to as being product-favored.

For instance, consider the reaction \( \mathrm{CO} + 1/2 \mathrm{O}_2 \rightleftarrows \mathrm{CO}_2 \) with a \( K_p \) of \( 1.2 \times 10^{45} \). Such an enormous value of \( K_p \) suggests that almost all reactants are converted into products, indicating a high favorability towards product formation.

Reactant-Favored Reactions

Reactant-favored reactions occur when the reverse is true: the equilibrium state of the reaction has a higher concentration of reactants than products. This happens when the equilibrium constant is very small, often close to zero. A small \( K_p \) implies that very little product is formed before the reaction reaches equilibrium, indicating that the reactants are strongly favored.

Take, for example, the decomposition of water, \( \mathrm{H}_2\mathrm{O}(g) \rightleftarrows \mathrm{H}_2(g) + 1/2 \mathrm{O}_2(g) \), with a \( K_p \) of \( 9.1 \times 10^{-41} \). This exceptionally small \( K_p \) value highlights that the reaction does not proceed much towards the formation of products, making it a reactant-favored reaction as the reactants dominate at equilibrium.

Chemical Equilibrium Analysis

Analyzing chemical equilibrium involves understanding the state where the forward and reverse reactions occur at the same rate, maintaining constant concentrations of reactants and products. The tool used to conduct this analysis is the equilibrium constant, \( K \). By evaluating the equilibrium constant, one can determine the extent of the reaction—whether it is product-favored or reactant-favored.

- **Equilibrium Constant \( K \):** Provides insight into the ratio of product concentration to reactant concentration at equilibrium.
- **Magnitude: Larger \( K \) values** signify product-favored equilibria, while **smaller values** indicate reactant-favored ones.
- **Applications:** Equilibrium analysis is essential for predicting the behavior of reactions under different conditions, allowing chemists to manipulate conditions to achieve desired equilibria.

Reaction Equilibrium

Reaction equilibrium refers to the balanced state of a reversible chemical reaction, where the rate of the forward reaction equals the rate of the reverse reaction. When a system is at equilibrium, there is no net change in the concentration of reactants and products over time, even though both reactions are still proceeding.

- **Dynamic Equilibrium:** Even though the concentrations are constant, molecules continue to react, indicating a dynamic balance.
- **Equilibrium Shift:** Changes in conditions such as temperature, pressure, or concentration can shift the equilibrium position, either favoring the products or the reactants depending on the Le Chatelier's principle.
- **Le Chatelier's Principle:** States that if a dynamic equilibrium is disturbed by changing the conditions, the system responds by counteracting the change, thereby restoring a new equilibrium.