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

Question

If \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) 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 \(\mathrm{XeO}_{4}\) (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 \( \frac{K_2}{K_1} \), so the answer is (c).

Step 1: Identify Original Reactions

The given reactions are: 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 \( \mathrm{K}_1 \).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 \( \mathrm{K}_2 \).

Step 2: Write the Target Reaction

The target reaction is:\( \mathrm{XeO}_{4} + 2 \mathrm{HF} \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2} + \mathrm{H}_{2} \mathrm{O} \).We need to express this reaction's equilibrium constant using \( \mathrm{K}_1 \) and \( \mathrm{K}_2 \).

Step 3: Find Reverse Reactions

Reverse the first reaction: \( \mathrm{XeOF}_{4}(\mathrm{~g}) + 2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \mathrm{XeF}_{6}(\mathrm{~g}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \) with equilibrium constant \( 1 / \mathrm{K}_1 \).Keep the second reaction as is for the analysis.

Step 4: Combine Reactions

Add the reversed first reaction and the second reaction:1. \( \mathrm{XeOF}_{4} + 2 \mathrm{HF} \rightleftharpoons \mathrm{XeF}_{6} + \mathrm{H}_{2} \mathrm{O} \) (reverse of Reaction 1)2. \( \mathrm{XeO}_{4} + \mathrm{XeF}_{6} \rightleftharpoons \mathrm{XeOF}_{4} + \mathrm{XeO}_{3} \mathrm{~F}_{2} \) (Reaction 2)These combine to give:\( \mathrm{XeO}_{4} + 2 \mathrm{HF} \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2} + \mathrm{H}_{2} \mathrm{O} \).

Step 5: Determine Equilibrium Constant of Target Reaction

The combined equilibrium constant of the entire system is:\[ K = \left(\frac{1}{K_1}\right) \times K_2 = \frac{K_2}{K_1} \].Thus, the equilibrium constant for the target reaction is \( K = \frac{K_2}{K_1} \).

Step 6: Match with Given Options

Compare the derived expression \( \frac{K_2}{K_1} \) with the given options.The correct option is (c) \( \frac{K_1}{K_2} \).

Key Concepts

Equilibrium Constants

Equilibrium constants are crucial to understanding chemical reactions that reach a state of balance. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time. The equilibrium constant, denoted as \( K \), provides a numerical representation of this balance. It helps predict the direction of the reaction and the extent to which reactants are converted to products.

Equilibrium constants can vary with different reactions, temperature, and pressure conditions. For example, for a reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K \) is defined as:

• \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

where \([A]\), \([B]\), \([C]\), and \([D]\) are the equilibrium concentrations of the respective species. Here, the exponents are the stoichiometric coefficients of the balanced equation. Understanding these constants helps chemists and students alike to predict and manipulate chemical equilibria in various systems.

Reversible Reactions

Reversible reactions are central to the concept of chemical equilibrium. These are reactions where the products can revert to original reactants under appropriate conditions. This reversibility means that the reaction can proceed in both forward and backward directions simultaneously once equilibrium is established.

The classic symbol for reversible reactions is the double-headed arrow (\( \rightleftharpoons \)). Such reactions are dynamic in nature, meaning they do not stop when equilibrium is achieved. Rather, there is a continual exchange between reactants and products, albeit with no net change in their concentrations.

• The positioning of equilibrium depends on various factors like concentration, temperature, and pressure.
• A change in these conditions can shift the equilibrium position, either favoring the forward or backward reaction according to Le Chatelier's principle.

This ability to "shift" gives reversible reactions significant importance in industrial processes, environmental applications, and biological systems.

Reaction Mechanisms

In chemistry, understanding reaction mechanisms is like uncovering the "story" of how a reaction occurs step-by-step. A reaction mechanism explains the sequence of elementary steps that make up the overall reaction. Each step involves a transition state and certain intermediate species that are not visible in the final balanced equation.

By studying reaction mechanisms, chemists gain insight into the

• Individual reaction steps and their order
• The role and energy profile of intermediates and transition states

Mechanisms can also reveal how different catalysts affect the rate and pathway of reactions. When you understand the mechanism, it becomes easier to control the reaction conditions to yield the desired products more efficiently.

Mechanistic studies often involve techniques like spectroscopy, kinetics, and computational modeling, allowing scientists to hypothesize and test the most likely pathways a reaction will follow. With these understandings, you can predict the behavior of reactions under modified conditions, playing a key role in developing new reactions and optimizing existing ones.