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, \(\mathrm{XeF}_{6}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{XeOF}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})\) \(\mathrm{XeO}_{4}(\mathrm{~g})+\mathrm{XeF}_{6}(\mathrm{~g}) \rightleftharpoons \mathrm{XeOF}_{4}(\mathrm{~g})+\mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})\) 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) \(\mathrm{K}_{2} / \mathrm{K}_{1}\)

Step-by-Step Solution

Answer

Short answer goes here based on solution steps.

Step 1: Understand the Given Reactions

First, let's list the given reactions:1. \( \mathrm{XeF}_{6}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{XeOF}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) \) 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 \( K_{2} \).

Key Concepts

Equilibrium Constants

Equilibrium constants, often represented by the symbol \( K \), play a crucial role in understanding chemical equilibria. These constants provide valuable information about the concentration of reactants and products at equilibrium. Each chemical reaction has its own unique equilibrium constant based on the temperature and pressures of the involved species.

Here's what you should remember about equilibrium constants:

• An equilibrium constant is expressed as the ratio of the concentration (or pressures, for gaseous reactions) of the products over the reactants, each raised to the power of their stoichiometric coefficients from the balanced equation.
• If \( K = 1 \), the reaction is perfectly balanced, with neither side being favored.
• If \( K > 1 \), the reaction favors the formation of products.
• If \( K < 1 \), the reaction favors the formation of reactants.

For the reactions in the exercise, understanding these relationships helps to determine the overall equilibrium constant of a third derived reaction.

Reactions and Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Accurate stoichiometric calculations ensure that the balance of atoms is maintained before and after a reaction, essential for correctly determining the equilibrium constant.

In the context of the given reactions, stoichiometry plays a critical role:

• Each reactant and product has a coefficient that indicates the number of molecules involved in the reaction, affecting the calculation of the equilibrium constant.
• The overall stoichiometry of a derived reaction could alter the ratio of reactants to products, which in turn affects its equilibrium expression.
• By analyzing the stoichiometric ratios of the initial reactions, we can deduce the equilibrium constant for a new reaction by combining and manipulating these expressions.

For our task, understanding how coefficients affect reactions can guide us through finding relationships between \(K_1\) and \(K_2\) for the desired outcome.

Gaseous Reactions

Gaseous reactions involve reactants and products primarily in the gas phase. These reactions are often influenced by pressure and temperature, which can significantly affect their equilibrium constants.

When dealing with gaseous reactions, specific considerations must be taken into account:

• Reactions involving gases are often expressed in terms of partial pressures. The equilibrium expressions will involve the partial pressures of the gaseous species.
• Changes in pressure can shift the position of equilibrium. This is governed by Le Chatelier's Principle, which states that a system at equilibrium will adjust to counteract changes.
• Exploring the implications of pressure variations helps us understand how equilibrium constants like \(K_1\) and \(K_2\) relate when formulating new reaction equations.

In scenarios like the one in the exercise, understanding the behavior of gaseous species under different conditions is essential to correctly identify the relationship and outcome of new reactions formed from existing ones.