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

Question

The equilibrium constants \(K_{p_{1}}\) and \(K_{p_{2}}\) for the reactions \(\mathrm{X} \rightleftharpoons 2 \mathrm{Y}\) and \(\mathrm{Z} \rightleftharpoons \mathrm{P}+\mathrm{Q}\), respectively are in the ratio of \(1: 9\). If the degree of dissociation of \(X\) 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 \(1: 3\) (option c).

Step 1: Understanding the Problem

The problem involves two gaseous equilibria with given equilibrium constants in a ratio. The goal is to find the ratio of total pressures at equilibrium given equal degree of dissociation for both reactions.

Step 2: Define Equilibrium Constants

For reaction 1 \( \text{X} \rightleftharpoons 2 \text{Y} \), the equilibrium constant \( K_{{p}_{1}} = (P_Y)^2 / P_X = \alpha^2 P^2 / (1-\alpha)P \), where \( \alpha \) is the degree of dissociation and \( P \) is the total pressure. For reaction 2 \( \text{Z} \rightleftharpoons \text{P} + \text{Q} \), the equilibrium constant \( K_{{p}_{2}} = (P_P \cdot P_Q) / P_Z = \alpha^2 P^2 / (1-\alpha)P \).

Step 3: Apply Given Ratio

Since \( K_{{p}_{1}} : K_{{p}_{2}} = 1:9 \), we have \( \frac{ \alpha^2 P_1^2 }{ (1-\alpha) P_1 } : \frac{ \alpha^2 P_2^2 }{ (1-\alpha) P_2 } = 1:9 \). This simplifies to \( \frac{P_1^2}{P_1} : \frac{P_2^2}{P_2} = 1:9 \), reducing to \( \frac{P_1}{P_1} : \frac{P_2}{P_2} = 1:9 \).

Step 4: Calculate Ratio of Pressures

The above simplifies further to: \( \frac{1}{P_2} = 1:9 \). Therefore, \( P_1 : P_2 = 1:3 \).

Step 5: Conclude the Solution

The ratio of total pressure at equilibria where the degree of dissociation of \( X \) and \( Z \) are equal is 1:3.

Key Concepts

Equilibrium Constants

Equilibrium constants are a fundamental concept in chemistry, crucial for understanding how reactions behave under stable conditions. In simple terms, the equilibrium constant ( K_p) is a number that provides insight into the ratio of products to reactants at equilibrium in a particular reaction. For gaseous reactions, we often use partial pressures instead of concentrations.

• For a general reaction like A ↔ B, K_p is determined using: K_p = (P_B) / (P_A).
• This reflects the way products and reactants balance under equilibrium conditions.

In our problem, we have two reactions, X ↔ 2Y and Z ↔ P + Q, where their equilibrium constants K_{p_{1}} and K_{p_{2}}, respectively, provide relative ratios of pressures involved.
According to the problem, these constants have a ratio of 1:9. This signifies that the reaction involving X and Y favors reactants more than the reaction involving Z, P, and Q when they reach equilibrium. Understanding these constants helps predict how much of each substance will be present when balance is achieved, critical for calculating pressures and dissociations.

Degree of Dissociation

The degree of dissociation refers to the fraction of a substance that splits into its components in a chemical reaction. It expresses how far a reaction has progressed towards completion under given conditions.

• It is typically represented by the variable α (alpha) indicating the proportion of the initial substance that has dissociated.
• The degree of dissociation is particularly important in reversible reactions where equilibrium is established.

In the given exercise, both substances, X and Z, have the same degree of dissociation denoted by α. This implies both reactions reach equilibrium while losing the same fraction of their original forms.
By maintaining the same degree of dissociation, it simplifies the ratio calculations for total pressures, as identical α affects both reactions in a similar way.
Therefore, it allows us to isolate pressure differences as the sole variable when applying the pressure ratio of two equilibrium constants. If the degree of dissociation changes, it directly influences the eventual pressure and concentration of each reactant and product.

Pressure in Equilibrium Reactions

The notion of pressure is pivotal in gas-phase reactions and is inherently tied to equilibrium constants and degree of dissociation.
The pressure in a system of reacting gases is the combined effect of the individual gas pressures or partial pressures and guides how the reaction achieves equilibrium.

• Total pressure is the sum of pressures of all gases in the reaction, influencing the equilibrium state.
• Changes in pressure can shift equilibria, as dictated by Le Chatelier’s principle.

Our problem focuses on finding the pressure ratio in two equilibrium systems.
The simplified expression for pressure reveals that, given identical α, the ratio of total pressures P_1 to P_2 is determined by the relation in their equilibrium constants.
The calculated ratio of 1:3 tells us how much more pressure is maintained by the second system involving Z, P, and Q compared to the first system of X and 2Y.
Understanding pressure changes and their impact helps in controlling and optimizing conditions in chemical processes to achieve desired outcomes.