Question

In the case of \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) the degree of dissociation " \(x^{\prime \prime}\) is related to the equilibrium pressure \(P\) as (1) \(x \propto \frac{1}{P}\) (2) \(x \alpha \frac{1}{\sqrt{P}}\) (3) \(x \alpha P\) (4) \(x \alpha P^{2}\)

Short answer

Option (2) \( x \propto \frac{1}{\sqrt{P}} \).

Step-by-step solution

Write the equilibrium expression

For the given reaction, first express the equilibrium condition using the equilibrium constant expression. The reaction is: \[ \text{PCl}_5 (g) \rightleftharpoons \text{PCl}_3 (g) + \text{Cl}_2 (g) \]

Define the equilibrium constant (K)

The equilibrium constant for the reaction can be defined as: \[ K = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]

Express concentrations in terms of degree of dissociation

Let the initial concentration of \( \text{PCl}_5 \) be \([A]_0\). If the degree of dissociation is \( x \), then:\[ [\text{PCl}_5] = [A]_0 (1 - x) \]\[ [\text{PCl}_3] = [A]_0 x \]\[ [\text{Cl}_2] = [A]_0 x \]

Substitute into the equilibrium expression

Substituting these into the equilibrium expression yields:\[ K = \frac{[A]_0 x [A]_0 x}{[A]_0 (1-x)} = \frac{[A]_0^2 x^2}{[A]_0 (1-x)} = \frac{[A]_0 x^2}{(1-x)} \]

Relate to pressure

Assume the total pressure at equilibrium is P and the initial pressure is \( P_0 \). The degree of dissociation can be related to the partial pressures via:\[ P = P_0 (1 + x) \]Therefore, \( x \propto \frac{1}{\sqrt{P}} \)

Find the correct option

Comparing with the given options: \[ x \propto \frac{1}{\sqrt{P}} \]The correct option is (2).

Key concepts

Degree of Dissociation

The degree of dissociation, often represented by the symbol \( x \), measures how much a substance breaks down or dissociates into simpler forms when it reaches equilibrium. For the decomposition of \( \text{PCl}_5 \), the degree of dissociation represents the fraction of \( \text{PCl}_5 \) molecules that have dissociated into \( \text{PCl}_3 \) and \( \text{Cl}_2 \) molecules. For instance, if we start with an initial amount of substance \( [A]_0 \), and at equilibrium, a fraction \( x \) has dissociated, then:

• The concentration of \( \text{PCl}_5 \) remaining is given by: \( [\text{PCl}_5] = [A]_0 (1 - x) \)
• The concentration of \( \text{PCl}_3 \) and \( \text{Cl}_2 \) formed is: \( [\text{PCl}_3] = [A]_0 x \) and \( [\text{Cl}_2] = [A]_0 x \)

The degree of dissociation is vital in deriving relationships involving equilibrium constants and partial pressures in chemical reactions. The degree of dissociation is directly influenced by the conditions of the system, mostly temperature and pressure for gas-phase reactions.

Equilibrium Constant

The equilibrium constant, denoted by \( K \), is a dimensionless number that provides a ratio of the concentrations or partial pressures of the products and reactants at equilibrium. For the decomposition of \( \text{PCl}_5 \), the equilibrium expression is:

\[ K = \frac{[\text{PCl}_3] [\text{Cl}_2]}{[\text{PCl}_5]} \]

Insert the values of the concentrations in terms of the degree of dissociation to obtain:

\[ K = \frac{[A]_0 x [A]_0 x}{[A]_0 (1-x)} = \frac{[A]_0 x^2}{1-x} \]

The equilibrium constant's value depends on the temperature and nature of the reactants and products but remains unchanged for a fixed temperature. The equilibrium expression links the reaction's degree of dissociation with the initial concentration of the reactant and the equilibrium constant itself. This relationship serves as the foundation for understanding how changes in conditions, such as pressure, will affect the system's degree of dissociation.

Partial Pressure

Partial pressure refers to the pressure exerted by an individual gas in a mixture of gases. According to Dalton's law, the total pressure of a gas mixture is the sum of the partial pressures of each individual gas. For our reaction involving \( \text{PCl}_5 \) decomposing into \( \text{PCl}_3 \) and \( \text{Cl}_2 \):

• The partial pressures of \( \text{PCl}_3 \) and \( \text{Cl}_2 \) are equal, both proportional to the degree of dissociation \( x \).
• The total pressure at equilibrium, \( P \), is given by: \( P = P_0 (1 + x) \)

Hence, the degree of dissociation can be related to the partial pressure with the expression:

\[ x \propto \frac{1}{\sqrt{P}} \]

This indicates that as the total pressure increases, the degree of dissociation decreases. The relationship between partial pressure and degree of dissociation is crucial for predicting how a chemical system will respond to changes in pressure, enabling chemists to manipulate reaction conditions to favor desired outcomes.