Problem 91 - A Complete Resource Book in Chemistry for JEE Main

Question

In which of the following reactions, the concentration of product is higher than the concentration of reactant at equilibrium? = equilibrium constant) (a) \(\mathrm{A} \rightleftharpoons \mathrm{B} ; \mathrm{K}=0.001\) (b) \(\mathrm{M} \rightleftharpoons \mathrm{N} ; \mathrm{K}=10\) (c) \(\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}=0.005\) (d) \(\mathrm{R} \longrightarrow \mathrm{P} ; \mathrm{K}=0.01\)

Step-by-Step Solution

Answer

Reaction (b) with \( K=10 \) has a higher concentration of product than reactant at equilibrium.

Step 1: Understanding Equilibrium Constant

The equilibrium constant \( K \) indicates the ratio of the concentration of products to the concentration of reactants at equilibrium. A large \( K \) (much greater than 1) means products are favored. Conversely, a small \( K \) (much less than 1) means reactants are favored.

Step 2: Analyze Each Reaction

Evaluate each given reaction: - Reaction (a): \( K = 0.001 \), very small, favoring reactants.- Reaction (b): \( K = 10 \), greater than 1, favoring products.- Reaction (c): \( K = 0.005 \), very small, favoring reactants.- Reaction (d): \( K = 0.01 \), small, favoring reactants.

Step 3: Determine which Reaction Favors Products

Only reaction (b) has an equilibrium constant \( K \) greater than 1, indicating that the concentration of the product \( \mathrm{N} \) is higher than the concentration of the reactant \( \mathrm{M} \) at equilibrium.

Key Concepts

Reaction Favorability

Reactions at equilibrium can lean more towards either products or reactants, and determining which direction they favor is key. The equilibrium constant, often denoted as \( K \), is the value that reveals this tendency. - **Equilibrium Constant:** It is calculated as the ratio of the concentration of products to the concentration of reactants. - **Favored Side:** A large equilibrium constant (\( K \gg 1 \)) means products are highly favored, pushing the reaction towards product formation. However, a small equilibrium constant (\( K \ll 1 \)) highlights that reactants are predominant.By evaluating \( K \), you can predict the reaction's outcome and determine which species will be more prevalent once the system reaches equilibrium. If \( K \) is closer to 1, neither side is strongly favored, indicating a balanced system between reactants and products.

Concentration of Products

The concentration of products at equilibrium is directly influenced by the equilibrium constant \( K \). Let's understand how this plays a role:- **Understanding \( K \):** For a reaction with \( K > 1 \), it means that at equilibrium, the concentration of products will be higher than that of reactants. This implies that the products are more stable under the given conditions and are thus favored.- **Practical Implication:** For example, in reaction \( \mathrm{M} \rightleftharpoons \mathrm{N} \) where \( K = 10 \), products \( \mathrm{N} \) predominate. If you were observing this system in a lab, you'd find more of \( \mathrm{N} \) than \( \mathrm{M} \) present once equilibrium is established.Understanding this helps in making predictions about the amounts and behavior of chemicals in a reaction, especially in industries where product yield is critical.

Equilibrium Analysis

The process of equilibrium analysis is crucial to understanding the dynamics of a chemical reaction:- **Analyzing Reactions:** Equilibrium analysis involves determining whether a reaction is more likely to produce reactants or products. This is achieved by looking at the value of the equilibrium constant \( K \).- **Case Evaluation:** When examining reactions such as our examples, each \( K \) value provides a snapshot of the equilibrium state. For reaction \( \mathrm{X} \rightleftharpoons \mathrm{Y} \) with \( K = 0.005 \), reactants distinctly dominate, revealing that the system has little interest in staying as product \( \mathrm{Y} \).- **Carbon Copy for Reaction Tuning:** This approach allows chemists to adjust conditions, like temperature or pressure, to sway the balance of \( K \) if possible, thus promoting the desired outcome of more products or reactants based on their application needs.Equilibrium analysis is therefore a powerful tool in both academic and practical chemistry, providing insights into optimizing reactions and industrial processes.