Problem 107 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constants for the reactions, \(\mathrm{XeF}_{6}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{XeOF}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})\) (i) and \(\mathrm{XeO}_{4}(\mathrm{~s})+\mathrm{XeF}_{6}(\mathrm{~g}) \rightleftharpoons \mathrm{XeOF}_{4}(\mathrm{~g})+\) \(\mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g}) \ldots\) (ii) are \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) respectively. The equilibrium constant (K) for the reaction \(\mathrm{XeO}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \mathrm{is}\) (a) \(\mathrm{K} / \mathrm{K}\) (b) \(\mathrm{K}_{2} / \mathrm{K}_{1}\) (c) \(\mathrm{K}_{1} / \mathrm{K}_{2}^{-}\) (d) \(\mathrm{K}_{1}^{2} / \mathrm{K}_{2}\)

Step-by-Step Solution

Answer

The answer is (b) \(K_2 / K_1\).

Step 1: Understand the Reactions

First, analyze the given reactions. We have:1. Reaction (i): \( \mathrm{XeF}_{6}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{XeOF}_{4}(\mathrm{g})+2 \mathrm{HF}(\mathrm{g}), \) with an equilibrium constant \( K_1. \)2. Reaction (ii): \( \mathrm{XeO}_{4}(\mathrm{s})+\mathrm{XeF}_{6}(\mathrm{g}) \rightleftharpoons \mathrm{XeOF}_{4}(\mathrm{g})+\mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{g}), \) with an equilibrium constant \( K_2. \)

Step 2: Write the Target Reaction

Write the reaction for which we are asked to find the equilibrium constant:\[ \mathrm{XeO}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \]

Step 3: Reversing Reaction (i)

Reverse Reaction (i) as follows:\[ \mathrm{XeOF}_{4}(\mathrm{g})+2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \mathrm{XeF}_{6}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \]The equilibrium constant for the reversed reaction is \( \frac{1}{K_1}. \)

Step 4: Combine the Reactions

Combine the reversed Reaction (i) with Reaction (ii):- The first part is already flipped to provide the right half of our target reaction.- Reaction (ii) provides the left half of our target reaction.Add their constants appropriately to get:\( K = \frac{1}{K_1} \cdot K_2. \)

Step 5: Conclusion on Equilibrium Constant

Therefore, the equilibrium constant \( K \) for the target reaction \( \mathrm{XeO}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \) is given by \( \frac{K_2}{K_1}. \) So the correct option is (b).

Key Concepts

Equilibrium Constant

In the world of chemistry, understanding chemical equilibrium is essential. When a reaction reaches balance, the rates of the forward and reverse reactions are equal. This is when we say the system is at equilibrium. At this point, the concentrations of reactants and products remain constant over time. The equilibrium constant, symbolized as \( K \), is a number that relates the concentrations of reactants and products at this point of balance.

The equilibrium constant for a given reaction is determined by the ratio of the concentrations of the products to the concentrations of the reactants. It's important to remember, each concentration is raised to the power of its coefficient from the balanced equation. The equilibrium constant can provide valuable insight into the reaction direction and extent:

• **If \( K \) is much greater than 1:** The products are favored; the forward reaction predominates.
• **If \( K \) is much less than 1:** The reactants are favored; the reverse reaction predominates.

Using the equilibrium constants for individual reactions, we can deduce the constant for a complex reaction. For instance, if we know \( K_1 \) and \( K_2 \) for two parts of a reaction, we can calculate the equilibrium constant for the overall reaction by combining them appropriately.

Reversible Reactions

Reversible reactions are fascinating because they do not go to completion. Instead, they occur in both forward and reverse directions. Many chemical reactions in our environment and bodies are reversible. They reach a state where the forward reaction rate equals the reverse reaction rate, achieving chemical equilibrium.

In reversible reactions, you'll often see the double arrow symbol \( \rightleftharpoons \), indicating that both directions of the reaction are happening simultaneously. At equilibrium, the concentrations of reactants and products remain constant, but they are not necessarily equal.

• **Dynamic Nature:** Even at equilibrium, particles continue to react, but the system as a whole remains stable.
• **Equilibrium Position:** The ratio of products to reactants isn't always the same; it's influenced by several factors, including temperature, pressure, and concentrations.
• **Le Chatelier's Principle:** This principle states that if a system at equilibrium is disturbed, the system will adjust to minimize that disturbance and regain equilibrium.

Understanding the nature of reversible reactions helps us manipulate reaction conditions to our advantage, such as increasing product yield or reforming reactants.

Reaction Quotients

The reaction quotient, denoted by \( Q \), is very similar to the equilibrium constant \( K \). However, \( Q \) can be calculated at any point during a reaction, not just at equilibrium. It's an important diagnostic tool in chemistry to predict the direction in which a reaction will proceed to reach equilibrium.

To calculate the reaction quotient, you use the same formula as the equilibrium constant, substituting the current concentrations of reactants and products instead of equilibrium concentrations. Once you have \( Q \), it can tell you how far a system is from equilibrium:

• **If \( Q = K \):** The system is at equilibrium.
• **If \( Q < K \):** The forward reaction is favored; the system will proceed to form more products.
• **If \( Q > K \):** The reverse reaction is favored; the system will form more reactants to reach equilibrium.

The concept of the reaction quotient helps students and scientists predict how shifts in concentration, pressure, or temperature might impact a reaction's progress towards equilibrium.