Problem 199 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constants \(\mathrm{K}_{\mathrm{P}}\) and \(\mathrm{K}_{\mathrm{P}}\) for the reactions \(\mathrm{X} \longrightarrow 2 \mathrm{Y}\) and \(\mathrm{Z} \rightleftharpoons \mathrm{P}+\mathrm{Q}\), respectively are in the ratio of \(1: 9 .\) If the degree of dissociation of \(\mathrm{X}\) and \(Z\) be equal then the ratio of total pressure at these equilibria is (a) \(1: 36\) (b) \(1: 1\) (c) \(1: 3\) (d) \(1: 9\)

Step-by-Step Solution

Answer

The ratio of total pressure at these equilibria is (a) 1:36.

Step 1: Understand the Equilibrium Constants

The equilibrium constant, \( K_p \), is a measure of the concentration of products to reactants at equilibrium, specifically for gaseous reactions. Given, \( K_p(X \rightarrow 2Y) : K_p(Z \rightleftharpoons P + Q) = 1 : 9 \), so \( K_{p1} = K \) and \( K_{p2} = 9K \).

Step 2: Degree of Dissociation Equation

The degree of dissociation \( \alpha \) for a generic reaction impacts the equilibrium expression. Given that the degree of dissociation for both reactions is equal, let \( \alpha_1 = \alpha_2 = \alpha \). We need to determine how this affects the total pressure ratios.

Step 3: Reaction 1: Calculate Shift Due to Dissociation

For reaction \( X \rightarrow 2Y \), initially, assume the pressure of X is \( P \). At equilibrium, the pressure will be \((1-\alpha)P\) for X and \(2\alpha P\) for 2Y. Therefore,\[ K_{p1} = \frac{(2\alpha P)^2}{(1-\alpha)P} = \frac{4\alpha^2 P^2}{(1-\alpha)P} \].

Step 4: Simplifying Reaction 1

Simplify the expression for \( K_{p1} \): \[ K_{p1} = \frac{4\alpha^2 P}{(1-\alpha)} \].

Step 5: Reaction 2: Calculate Shift Due to Dissociation

For reaction \( Z \rightleftharpoons P + Q \), similarly, assume initial pressure is \( P \). At equilibrium, the pressure becomes \((1-\alpha)P\) for Z and \(\alpha P\) each for P and Q. Thus,\[ K_{p2} = \frac{(\alpha P)^2}{(1-\alpha)P} = \frac{\alpha^2 P^2}{(1-\alpha)P} \].

Step 6: Simplifying Reaction 2

Simplify the expression for \( K_{p2} \): \[ K_{p2} = \frac{\alpha^2 P}{(1-\alpha)} \].

Step 7: Solve for Pressure Ratio

With \( K_{p1} = K \) and \( K_{p2} = 9K \), and comparing the two expressions:- \( \frac{4\alpha^2 P_1}{(1-\alpha)} = K \)- \( \frac{\alpha^2 P_2}{(1-\alpha)} = 9K \).Dividing the two equations gives:\[ \frac{4\alpha^2 P_1}{\alpha^2 P_2} = \frac{K}{9K} \]Thus, \[ \frac{4P_1}{P_2} = \frac{1}{9} \], implying \( P_2 = 36P_1 \).

Step 8: Conclude with Ratio of Total Pressure

The ratio of total pressure \( P_1 : P_2 \) is \( 1 : 36 \).

Key Concepts

Degree of Dissociation

The degree of dissociation is an essential concept when discussing equilibrium in chemical reactions, particularly in gaseous systems. It refers to the fraction of the original substance that decomposes into products at equilibrium. In simpler terms, if a substance partially breaks down into simpler atoms or molecules, we need a way to measure this transformation.
For a reaction of the form \(X \rightarrow 2Y\), the degree of dissociation, \(\alpha\), can be understood as the proportion of \(X\) that converts into \(Y\) over time. If initially, X had a pressure \(P\), and it dissociates to form \((1-\alpha)P\) and \(2\alpha P\), this shows how \(\alpha\) affects equilibrium concentration.
Knowing \(\alpha\) is crucial because it also connects directly to equilibrium constants, such as \(K_p\), which involve ratios of concentrations of products and reactants raised to certain powers, indicative of their stoichiometric coefficients. With this understanding, students can more easily compute equilibrium concentrations, facilitating a better grasp of reaction dynamics.

Reaction Equilibria

Reaction equilibria describe the state of a chemical reaction where the rates of the forward and backward reactions are equal, leading to constant concentrations of reactants and products. This equilibrium concept is pivotal because it determines the final composition of substances involved in a reaction.
For distinct reactions such as \(X \rightarrow 2Y\) and \(Z \rightleftharpoons P + Q\), their respective equilibrium constants, \(K_{p1}\) and \(K_{p2}\), offer insights into how far each reaction goes. In the problem, these constants are directly proportional to the dissociation and pressures involved, as given by their specific conditions and stoichiometry.
This equilibrium also links to Le Chatelier's Principle, which predicts how changes in conditions such as pressure will shift a reaction's equilibrium position. Thus, understanding equilibria is about recognizing these dynamic balances and predicting how reactions adjust when disturbed.

Pressure Ratios

Pressure ratios in reaction equilibria derive from understanding how changes in pressure can affect the amounts and partial pressures of substances in reactions. Considering equilibrium reactions, pressure plays a crucial part in shifting equilibria according to Le Chatelier’s principle.
Take as an example our reactions: \(X \rightarrow 2Y\), \(Z \rightleftharpoons P +Q\). As we inferred from equilibrium reactions, the pressure of gases changes as they interact and transform at equilibrium. Initially, we assume some pressure \(P\); dissociation and resulting pressure changes are calculated using relationships based on \(\alpha\), the degree of dissociation.
For the listed problem, the calculations for pressure ratios between two points in equilibria relate closely to their equilibrium constants, \(K_{p1}\) and \(K_{p2}\). They show that the total pressure alignment, like the \(1:36\) ratio derived, demonstrates how markedly pressures adjust during equilibrium—deepening our appreciation of dynamic equilibrium systems in chemistry.