Problem 26 - A Complete Resource Book in Chemistry for JEE Main

Question

If for the reaction given below \(2 \mathrm{PQ} \rightleftharpoons \mathrm{P}_{2}+\mathrm{Q}_{2} \mathrm{~K}_{1}=2.5 \times 10^{5}\) \(\mathrm{PQ}+\frac{1}{2} \mathrm{R}_{2} \rightleftharpoons \mathrm{PQR} \mathrm{K}_{2}=5 \times 10^{-3}\) find \(K_{3}\) for the reaction \(\frac{1}{2} P_{2}+\frac{1}{2} Q_{2}+\frac{1}{2} R_{2} \rightleftharpoons P Q R\) (a) \(2.5 \times 10^{-3}\) (b) \(2.5 \times 10^{3}\) (c) \(1 \times 10^{-3}\) (d) \(5 \times 10^{-3}\)

Step-by-Step Solution

Answer

Given reactions transform to desired reaction with \( K_3 = 2.5 \times 10^{-3} \).

Step 1: Understanding the Given Reactions

We are given two equilibrium reactions with their equilibrium constants:1. Reaction 1: \( 2 \text{PQ} \rightleftharpoons \text{P}_2 + \text{Q}_2 \) with \( K_1 = 2.5 \times 10^5 \).2. Reaction 2: \( \text{PQ} + \frac{1}{2} \text{R}_2 \rightleftharpoons \text{PQR} \) with \( K_2 = 5 \times 10^{-3} \).We are asked to find \( K_3 \) for the reaction \( \frac{1}{2} \text{P}_2 + \frac{1}{2} \text{Q}_2 + \frac{1}{2} \text{R}_2 \rightleftharpoons \text{PQR} \).

Step 2: Express Desired Reaction with Given Reactions

The target reaction: \( \frac{1}{2} \text{P}_2 + \frac{1}{2} \text{Q}_2 + \frac{1}{2} \text{R}_2 \rightleftharpoons \text{PQR} \) can be obtained from the given reactions by:- Reversing and halving the first reaction to have \( \frac{1}{2} \text{P}_2 + \frac{1}{2} \text{Q}_2 \rightleftharpoons \text{PQ} \).- Keeping the second reaction as is.Then, add these reactions.

Step 3: Calculate Equilibrium Constant for Intermediate Steps

When reactions are reversed, the equilibrium constant is inverted. The halved Reaction 1 becomes \( \frac{1}{2} \text{P}_2 + \frac{1}{2} \text{Q}_2 \rightleftharpoons \text{PQ} \) with equilibrium constant \( K_1' = (K_1)^{-1/2} = (2.5 \times 10^5)^{-1/2} \).

Step 4: Determine Final Equilibrium Constant for Desired Reaction

The combined reaction equilibrium constant is the product of the constants for the intermediate steps. Thus, \( K_3 = K_1' \times K_2 \).Calculate: \[ K_1' = \frac{1}{\sqrt{2.5 \times 10^5}} = \frac{1}{500} = 2 \times 10^{-3} \] Then,\[ K_3 = 2 \times 10^{-3} \times 5 \times 10^{-3} = 10 \times 10^{-6} = 1 \times 10^{-5} \].

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction is equal to the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant. This state is dynamic because the reactions continue to occur, but there is no net change in the concentrations of the substances involved. In essence, chemical equilibrium is like a well-balanced tug of war where neither side is winning, yet both are continuously pulling. It’s crucial to understand that equilibrium does not mean the reactants and products are in equal amounts; instead, it means their rates of formation are equal. When studying reactions, the equilibrium constant (K) is used to express the ratio of the concentrations of the products to the reactants, each raised to the power of their respective coefficients in the balanced equation.

Reaction Kinetics

Reaction kinetics is the study of the rate at which chemical reactions occur. It helps us understand how different conditions such as temperature, concentration, and pressure affect the speed of a reaction. In the context of the given problem, reaction kinetics is important because it influences the equilibrium position. When a system reaches chemical equilibrium, the forward and reverse reactions occur at the same rate.

• The rate of a reaction is determined by the frequency of the reacting particles colliding with enough energy and in the correct orientation.
• Catalysts can significantly speed up reactions by providing an alternate pathway with a lower activation energy.
• The specific rate constant and the reaction order provide further insight into how reactant concentration affects the rate.

Understanding these kinetics aspects is essential for manipulating and predicting reaction behavior in various conditions.

Equilibrium Calculations

Equilibrium calculations involve determining the equilibrium constants for reactions, which helps us predict the concentrations of reactants and products at equilibrium. In the exercise provided, we must find the equilibrium constant (K_3) for a specific reaction by using the constants given for other reactions. This requires a clear understanding of how to manipulate equations and their constants. When reactions are reversed or multiplied, the equilibrium constant changes accordingly.

• Reversing a reaction means you take the reciprocal of the equilibrium constant.
• If a reaction is multiplied by a factor, the equilibrium constant is raised to the power of that factor.
• Adding reactions results in multiplying the equilibrium constants of the individual reactions.

In this problem, the reactions were manipulated and added accordingly to derive the equilibrium constant for the desired reaction, demonstrating a practical application of equilibrium calculations.