Question

The equilibrium constants \(\mathrm{K}_{\mathrm{P}_{1}}\) and \(\mathrm{K}_{\mathrm{P}_{2}}\) 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) \(1: 9\)

Short answer

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

Step-by-step solution

Understand the Reaction Equilibria

The given reactions are:- \( \text{X} \rightleftharpoons 2 \text{Y} \) with equilibrium constant \( K_{P_1} \).- \( \text{Z} \rightleftharpoons \text{P} + \text{Q} \) with equilibrium constant \( K_{P_2} \).The ratio of \( K_{P_1} \) to \( K_{P_2} \) is \( 1:9 \).

Define Degree of Dissociation

Let \( \alpha \) be the degree of dissociation for both reactions, which is given to be equal.

Express Equilibrium Constants

For reaction \( \text{X} \rightleftharpoons 2 \text{Y} \):- At dissociation \( \alpha \), the moles at equilibrium are \( 1 - \alpha \) for \( \text{X} \) and \( 2\alpha \) for \( \text{Y} \).- Total moles = \( 1 + \alpha \).- \( K_{P_1} = \frac{(P \cdot 2\alpha)^2}{(P \cdot (1 - \alpha))} = \frac{4\alpha^2 P}{1 - \alpha} \).

Express the Second Equilibrium Constant

For reaction \( \text{Z} \rightleftharpoons \text{P} + \text{Q} \):- At dissociation \( \alpha \), the moles are \( 1 - \alpha \) for \( \text{Z} \), \( \alpha \) for \( \text{P} \), \( \alpha \) for \( \text{Q} \).- Total moles = \( 1 + \alpha \).- \( K_{P_2} = \frac{(P \cdot \alpha)(P \cdot \alpha)}{(P \cdot (1 - \alpha))} = \frac{\alpha^2 P}{1 - \alpha} \).

Calculate Pressure Ratio

Given \( \frac{K_{P_1}}{K_{P_2}} = \frac{1}{9} \), substitute the expressions:\[ \frac{\frac{4\alpha^2 P_1}{1 - \alpha}}{\frac{\alpha^2 P_2}{1 - \alpha}} = \frac{1}{9} \]Simplify:\[ \frac{4P_1}{P_2} = \frac{1}{9} \]This yields \( \frac{P_1}{P_2} = \frac{1}{36} \).

Select the Correct Option

From \( \frac{P_1}{P_2} = \frac{1}{36} \), the ratio matches option (a).

Key concepts

Equilibrium Constant

The concept of an equilibrium constant, often denoted as \( K \), is a central idea in the study of chemical equilibria. It quantifies the balance between reactants and products in a reversible chemical reaction.
An equilibrium constant is specific to a particular reaction and varies with temperature. It is expressed in terms of the partial pressures of the gaseous reactants and products, known as \( K_p \), or in terms of concentration, known as \( K_c \).

• A large \( K \) value indicates that the reaction favors the formation of products.
• A small \( K \) value suggests that the reactants are favored at equilibrium.

For the reactions in the exercise, \( K_{P_1} \) and \( K_{P_2} \) are the equilibrium constants for two distinct reactions. Their given ratio of \( 1:9 \) plays a crucial role in determining equilibrium pressures, which will be further explored. Remember, the value of \( K \) is imperative in predicting how a reaction system behaves under equilibrium conditions.

Degree of Dissociation

The degree of dissociation, represented as \( \alpha \), is an important concept in understanding how much of a reactant has turned into products at equilibrium. It is a measure of the fraction of the initial amount of a substance that dissociates.
In the context of the exercise, the same degree of dissociation \( \alpha \) is applied to both reactions. It helps simplify calculations since the extent to which \( X \) becomes \( 2Y \) and \( Z \) becomes \( P + Q \) is identical.

• Let's say initially you have 1 mole of the reactant, and after dissociation, \( \alpha \) moles have transformed.
• The expression becomes \( 1 - \alpha \) for leftover reactants, and \( \alpha \) for each product formed.

This degree is crucial for setting up equations that relate to equilibrium constants in a mathematical form, ultimately helping us link these equations to experimental or theoretical pressure data.

Pressure Ratio

Pressure ratio plays a significant role in chemical equilibria as it indicates how pressure affects the equilibrium state between reactants and products. In reactions involving gases, total pressure at equilibrium can be influenced by changes in equilibrium constants and the degree of dissociation.
In the provided exercise, the equilibriums involve gas-phase reactions, where total pressure is a key factor. The comparison of \( P_1 \) to \( P_2 \) in the given equilibrium conditions reveals how the equilibrium constant ratio \( 1:9 \) affects this pressure ratio.
From the steps:

• The formula \( \frac{K_{P_1}}{K_{P_2}} = \frac{1}{9} \) is used to derive \( \frac{P_1}{P_2} = \frac{1}{36} \).
• This calculation shows how a change in \( K_p \) directly affects pressure, aligning with the concept that pressure and equilibrium conditions are deeply interconnected.

Understanding this interplay provides insight into the significance of pressure in maintaining and analyzing reaction equilibria.

Reaction Equilibria

Reaction equilibria refer to the state in which the forward and reverse reactions occur at equal rates, resulting in constant concentrations of reactants and products. This state is governed by the equilibrium constant, and various factors like concentration, temperature, and pressure can affect it.
In chemical equilibria, we often need to consider different forms such as gaseous equilibria or those occurring in solutions. This encompasses the delicate balance achieved between ongoing conversion of reactants into products and vice versa.

• For the reactions \( X \rightleftharpoons 2Y \) and \( Z \rightleftharpoons P + Q \), the equilibrium state is represented by \( K_{P_1} \) and \( K_{P_2} \) which are specific to each reaction's conditions.
• The constant value for each reaction serves as a tool for predicting the concentrations or pressures under equilibrium.

Understanding reaction equilibria involves not only recognizing this balance but also knowing that at equilibrium, the system is settled until external conditions force a shift, as per Le Chatelier's Principle.