Problem 184 - A Complete Resource Book in Chemistry for JEE Main

Question

For the equilibrium \(\mathrm{AB}(\mathrm{g}) \rightleftharpoons \mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g})\), at a given temperature \(\frac{1}{3}\) rd of \(\mathrm{AB}\) is dissociated then \(\frac{\mathrm{P}}{\mathrm{K}_{\mathrm{p}}}\) will be numerically equal to

Step-by-Step Solution

Answer

The value of \( \frac{P}{K_\text{p}} \) is 6.

Step 1: Understand the Equilibrium Expression

The equilibrium is given by \( \mathrm{AB}(\mathrm{g}) \rightleftharpoons \mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \). For this equilibrium, the equilibrium constant \( K_\text{p} \) is expressed in terms of partial pressures as \( K_\text{p} = \frac{{P_A \cdot P_B}}{{P_\text{AB}}} \).

Step 2: Define the Initial and Change in Concentration

Initially, assume the concentration of \( \mathrm{AB} \) is \( P \). Since \( \frac{1}{3} \) rd of \( \mathrm{AB} \) dissociates, \( \frac{1}{3}P \) moles of \( \mathrm{AB} \) dissociate to form \( \frac{1}{3}P \) moles each of \( \mathrm{A} \) and \( \mathrm{B} \).

Step 3: Calculate Equilibrium Concentrations

At equilibrium, the concentration of \( \mathrm{AB} \) is \( P - \frac{1}{3}P = \frac{2}{3}P \). The concentrations of \( \mathrm{A} \) and \( \mathrm{B} \) are both \( \frac{1}{3}P \).

Step 4: Substitute into the Equilibrium Expression

Substitute the equilibrium concentrations into the equation for \( K_\text{p} \): \[ K_\text{p} = \frac{(\frac{1}{3}P)(\frac{1}{3}P)}{\frac{2}{3}P} = \frac{\frac{1}{9}P^2}{\frac{2}{3}P} \]. Simplifying, \[ K_\text{p} = \frac{1}{6}P \].

Step 5: Calculate \( \frac{P}{K_\text{p}} \)

Substitute \( K_\text{p} = \frac{1}{6}P \) into \( \frac{P}{K_\text{p}} \): \[ \frac{P}{K_\text{p}} = \frac{P}{\frac{1}{6}P} = 6 \].

Key Concepts

Equilibrium Expression

Understanding the equilibrium expression is vital in chemical equilibrium. It helps us express the relationship between the concentrations of reactants and products when a chemical reaction has reached equilibrium.
For a gas-phase reaction like \( \mathrm{AB}(\mathrm{g}) \rightleftharpoons \mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \), the equilibrium constant \( K_p \) is expressed in terms of the partial pressures of the participants.

The equilibrium expression for this reaction is structured as:

• The products are on the numerator: \( P_A \cdot P_B \).
• The reactant is on the denominator: \( P_\text{AB} \).

Thus, \( K_p = \frac{P_A \cdot P_B}{P_\text{AB}} \).
This expression helps us calculate the ratio of the products to the reactants needed to reach equilibrium under certain pressures.

Partial Pressures

Partial pressures play a crucial role in understanding gaseous equilibria. They represent the pressure exerted by an individual gas in a mixture of gases.
In the context of chemical equilibrium, the partial pressure of each component is used in the equilibrium expression to understand the state of the reaction.

Let's break it down:

• When a gas like \( \mathrm{AB} \) partially dissociates, it forms \( \mathrm{A} \) and \( \mathrm{B} \). The pressures that \( \mathrm{A} \) and \( \mathrm{B} \) exert separately are their respective partial pressures.
• Initially, if the partial pressure of \( \mathrm{AB} \) is \( P \), then after one-third dissociation, each \( \mathrm{A} \) and \( \mathrm{B} \) would exert a partial pressure of \( \frac{1}{3}P \).
• At equilibrium, the remaining \( \mathrm{AB} \) exerts a pressure of \( \frac{2}{3}P \).

This concept is fundamental in calculating the equilibrium constant, \( K_p \), which determines the extent of a reaction at equilibrium.

Equilibrium Constant (Kp)

The equilibrium constant, \( K_p \), is a dimensionless value that provides insight into the balance between the products and reactants at equilibrium in gaseous reactions.
\( K_p \) is derived from the ratio of the partial pressures of products to reactants, indicating the position of equilibrium.

Here's how it's calculated using the example reaction:

• From the expression \( K_p = \frac{P_A \cdot P_B}{P_\text{AB}} \), substitute the equilibrium partial pressures.
• For \( \mathrm{A} \) and \( \mathrm{B} \), each have \( \frac{1}{3}P \), while \( \mathrm{AB} \) has \( \frac{2}{3}P \).
• Substituting into the expression gives: \[K_p = \frac{(\frac{1}{3}P)(\frac{1}{3}P)}{\frac{2}{3}P} = \frac{\frac{1}{9}P^2}{\frac{2}{3}P} = \frac{1}{6}P\]

This calculation informs us that at this specific point, for every unit of \( P \) in equilibrium, the reaction is balanced such that the ratio becomes \( \frac{1}{6}P \). Understanding \( K_p \) is essential for predicting how changes in conditions like temperature or pressure might shift the equilibrium.