Problem 127 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The ratio of total pressures at equilibrium is (a) 1:36.

Step 1: Write the Expression for Kp

For the reaction \( \mathrm{X}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Y}(\mathrm{g}) \), the expression for \( K_{p_1} \) is \( K_{p_1} = \frac{{(P_Y)^2}}{{P_X}} \), where \( P_Y \) is the partial pressure of \( Y \) and \( P_X \) is the partial pressure of \( X \). Similarly, for the reaction \( \mathrm{Z}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g})+\mathrm{Q}(\mathrm{g}) \), the expression for \( K_{p_2} \) is \( K_{p_2} = \frac{{P_P \times P_Q}}{{P_Z}} \), where \( P_P \) and \( P_Q \) are the partial pressures of \( P \) and \( Q \), and \( P_Z \) is the partial pressure of \( Z \).

Step 2: Understand Degree of Dissociation

Let the degree of dissociation of \( X \) and \( Z \) be \( \alpha \). For \( \mathrm{X} \rightleftharpoons 2 \mathrm{Y} \), initially 1 mole of \( X \) dissociates to produce \( 2\alpha \) moles of \( Y \). Thus, total moles at equilibrium are \( 1+\alpha \).

Step 3: Calculate Partial Pressures

Let the initial pressure of \( X \) be \( P_1 \). Then the partial pressures are given by: \( P_X = P_1(1-\alpha) \) and \( P_Y = 2P_1 \alpha \). Total pressure \( = P_1(1+\alpha) \). Similarly, for \( Z \), with initial pressure \( P_2 \), partial pressures are \( P_Z(1-\alpha) \), \( P_P = \alpha P_2 \), and \( P_Q = \alpha P_2 \). Total pressure \( = P_2(1+\alpha) \).

Step 4: Express Kp in Terms of α and Total Pressure

For \( \mathrm{X} \), \[ K_{p_1} = \frac{{(2\alpha P_1)^2}}{{P_1(1-\alpha)}} = \frac{{4\alpha^2 P_1}}{{1-\alpha}} \]. For \( \mathrm{Z} \), \[ K_{p_2} = \frac{{\alpha^2 P_2^2}}{{P_2(1-\alpha)}} = \frac{{\alpha^2 P_2}}{{1-\alpha}} \].

Step 5: Using Given Kp Ratio

Given \( \frac{{K_{p_1}}}{{K_{p_2}}} = \frac{1}{9} \), we set up the equation: \[ \frac{{4\alpha^2 P_1}}{{1-\alpha}} \div \frac{{\alpha^2 P_2}}{{1-\alpha}} = \frac{1}{9} \]. Simplifying, we get: \[ \frac{{4P_1}}{{P_2}} = \frac{1}{9} \], leading to \( \frac{P_2}{P_1} = 36 \).

Step 6: Calculate Ratio of Total Pressures

The total pressures at equilibrium are \( P_{\text{total}_1} = P_1(1+\alpha) \) and \( P_{\text{total}_2} = P_2(1+\alpha) \). Therefore, \( \frac{P_{\text{total}_1}}{P_{\text{total}_2}} = \frac{P_1}{P_2} = \frac{1}{36} \), implying \( \frac{P_{\text{total}_2}}{P_{\text{total}_1}} = 36 \).

Step 7: Choose the Correct Answer

The correct ratio of total pressures is \( 1: 36 \), which matches option (a).

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state of a chemical reaction where the concentrations of reactants and products remain constant over time.
This occurs because the rate at which the reactants turn into products equals the rate at which products revert to reactants. In simple terms, it's a balanced state where neither the reactants nor products have any tendency to change on their own.

For students studying chemistry, understanding equilibrium is essential for predicting the amounts of substances in a reaction at a given point.

• An equilibrium position can shift due to changes in conditions such as pressure and temperature.
• The equilibrium constant, denoted as \(K_{P}\), is a vital concept as it mathematically represents the state of balance in a gaseous reaction.
• This constant is specific for each reaction at a given temperature and helps to compare the relative amounts of reactants and products present.

Recognizing that equilibrium is dynamic, even though no net change is observed, enriches a student's ability to predict how different factors influence chemical reactions.

Dissociation

Dissociation is a process where a compound breaks down into its constituent parts, often leading to simplification into ions or simpler molecules.
In chemical equations, dissociation fundamentally changes the nature of the compounds involved by splitting them. For example, in the exercise given, the dissociation of \(X\) into \(Y\) is considered.

It's important to understand that dissociation impacts the equilibrium of a reaction since it alters the amounts of different species present.

• When considering dissociation in chemical equations, keep in mind the concept of the "degree of dissociation," which is shown by the symbol 0041lpha.
• This value indicates the fraction of a substance that has dissociated and can widely impact equilibrium calculations.

Reducing confusion around dissociation will allow students to accurately predict the outcomes of reactions and understand how changing conditions affect chemical balance.

Partial Pressure

Partial pressure is a crucial concept when dealing with gases, especially in reactions involving equilibrium. It is defined as the pressure that individual gas components in a mixture would exert if it alone occupied the entire volume.

This is key to calculating and understanding equilibria involving gases, such as those worked with in the problem set.

• Each gas in a mixture at equilibrium exerts a partial pressure, contributing to the total pressure of the system.
• The equilibrium constant \(K_{P}\) is often expressed using partial pressures of gases, making it essential to grasp the relationships between them.

A student aware of partial pressures is better equipped to navigate challenges posed in chemical equilibria and will adeptly apply this knowledge, enabling clear calculations and accurate predictions. Mastery of this concept strengthens the foundational knowledge required for advanced studies in physical chemistry.