Problem 53 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(K_{1}\) and \(K_{2}\) are the respective equilibrium constants for the two reactions, \(\mathrm{XeF}_{6}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\mathrm{XeOF}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})\) \(\mathrm{XeO}_{4}(\mathrm{~g})+\mathrm{XeF}_{6}(\mathrm{~g}) \rightleftharpoons \mathrm{XeOF}_{4}(\mathrm{~g})+\mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})\) Then equilibrium constant of the reaction \(\mathrm{XeO}_{4}(\mathrm{~g})+\) \(2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) will be (a) \(\mathrm{K}_{1} /\left(\mathrm{K}_{2}\right)^{2}\) (b) \(\mathrm{K}_{1} \cdot \mathrm{K}_{2}\) (c) \(\mathrm{K}_{1} / \mathrm{K}_{2}\) (d) \(\mathrm{K}_{2} / \mathrm{K}_{\mathrm{t}}\)

Step-by-Step Solution

Answer

The equilibrium constant is \( \frac{K_2}{K_1} \), corresponding to option (c).

Step 1: Identify the Given Reactions and Constants

We are given two reactions with their respective equilibrium constants: 1. \( \mathrm{XeF}_{6}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\mathrm{XeOF}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) \) with equilibrium constant \( K_1 \).2. \( \mathrm{XeO}_{4}(\mathrm{~g})+\mathrm{XeF}_{6}(\mathrm{~g}) \rightleftharpoons \mathrm{XeOF}_{4}(\mathrm{~g})+\mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g}) \) with equilibrium constant \( K_2 \).

Step 2: Express Desired Reaction Using Given Reactions

The desired reaction is \( \mathrm{XeO}_{4}(\mathrm{~g})+ 2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \). Notice that the forward direction is the product of reaction 1 in reverse and reaction 2 as is.

Step 3: Determine the Equilibrium Constant for the Desired Reaction

Use the mathematical manipulation of equilibrium constants:- Reverse reaction 1 and find the expression for the equilibrium constant in reverse, which is \( \frac{1}{K_1} \).- Use reaction 2 as is, keeping the equilibrium constant as \( K_2 \).Therefore, the equilibrium constant for the combined reaction is the product of the inverted equilibrium constant of reaction 1 and the equilibrium constant of reaction 2: \( \frac{1}{K_1} \times K_2 = \frac{K_2}{K_1} \).

Step 4: Choose the Correct Option

From the given multiple-choice answers, the correct representation of the equilibrium constant for the desired reaction is option (c) \( \frac{K_1}{K_2} \).

Key Concepts

Equilibrium Constant

The concept of the equilibrium constant encapsulates the balance in reversible reactions. It describes the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their respective coefficients in the balanced equation. For a general reaction:\[aA + bB \rightleftharpoons cC + dD\]The equilibrium constant, denoted as \( K \), is defined as:\[K = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]Understanding the equilibrium constant is essential because:

• It provides insight into the position of equilibrium—whether products or reactants are favored.
• The magnitude of \( K \) tells us about the extent of the reaction at equilibrium.

When \( K > 1 \), the products are predominant; when \( K < 1 \), reactants are predominant at equilibrium. In our specific problem, manipulating the equilibrium constants \( K_1 \) and \( K_2 \) helps derive the equilibrium constant for the desired reaction.

Reaction Manipulation

Manipulating reactions to obtain a desired reaction is a powerful tool in chemistry. By reversing or combining given reactions, we can derive new reactions and calculate the corresponding equilibrium constants. Key aspects of reaction manipulation include:- **Reversing a Reaction:** When a reaction is reversed, the equilibrium constant becomes its reciprocal. For instance, if a forward reaction has an equilibrium constant \( K \), reversing it results in: \[ \frac{1}{K} \]- **Combining Reactions:** When reactions are added together, their equilibrium constants are multiplied. This is crucial when you need to construct a new reaction from known ones, as seen in the exercise.In the exercise, two reactions are given. The manipulation involves reversing one reaction and keeping the other to find a new equilibrium constant.

Reversible Reactions

Reversible reactions are chemical processes where the reactants form products, which can then react to form the original reactants. This back-and-forth nature means that these reactions can reach a state of equilibrium, where the rates of the forward and backward reactions are equal.- **Key Characteristics:**

• These reactions do not go to completion. Instead, they reach a state where the concentrations of reactants and products remain constant over time.
• The existence of a dynamic equilibrium where the forward reaction rate equals the backward reaction rate.

In our exercise, both reactions are reversible, each associated with an equilibrium constant \( K \). Understanding how to manipulate these to deduce the equilibrium expression for a desired reaction is critical in solving such problems.