Problem 96 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(\mathrm{N}_{2} \mathrm{O}_{4}\) is dissociation to \(33 \%\) and \(40 \%\) at total pressure \(\mathrm{P}_{1}\) and \(\mathrm{P}_{2}\) atm respectively. The ratio of \(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\) is (a) \(\frac{8}{5}\) (b) \(\frac{5}{8}\) (c) \(\frac{9}{5}\) (d) \(\frac{5}{9}\)

Step-by-Step Solution

Answer

The ratio \( \frac{P_1}{P_2} \) is \( \frac{5}{8} \).

Step 1: Understand the Problem

We need to find the ratio \( \frac{P_1}{P_2} \) given that \( \text{N}_2\text{O}_4 \) dissociates into \( 2\text{NO}_2 \) at different dissociation levels: 33% at pressure \( P_1 \) and 40% at pressure \( P_2 \).

Step 2: Write the Dissociation Reaction

The dissociation reaction is \( \text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2 \). The degree of dissociation (\( \alpha \)) helps determine the moles at equilibrium.

Step 3: Express Initial and Equilibrium Moles

Initially, assume 1 mole of \( \text{N}_2\text{O}_4 \). Moles at equilibrium for dissociation \( \alpha \): \( 1 - \alpha \) for \( \text{N}_2\text{O}_4 \) and \( 2\alpha \) for \( \text{NO}_2 \).

Step 4: Use Degree of Dissociation for Pressures

For \( P_1 \): \( \alpha_1 = 0.33 \), so total moles = \( 1 - 0.33 + 2 \times 0.33 = 1.33 \). For \( P_2 \): \( \alpha_2 = 0.4 \), so total moles = \( 1 - 0.4 + 2 \times 0.4 = 1.4 \).

Key Concepts

Dissociation Reaction

A dissociation reaction is a chemical process where a compound breaks down into two or more components. In this case, the compound involved is dinitrogen tetroxide (\(\text{N}_2\text{O}_4\)), which dissociates into nitrogen dioxide (\(\text{NO}_2\)). The reaction can be represented as:

• \(\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2\)

This equation indicates that one molecule of \(\text{N}_2\text{O}_4\) breaks apart to form two molecules of \(\text{NO}_2\). During such reactions, the equilibrium can shift based on various conditions such as temperature, pressure, and concentration of reactants and products.
The key is to understand that dissociation leads to an increase in the number of gas molecules, affecting the equilibrium position.

Degree of Dissociation

The degree of dissociation (\(\alpha\)) quantifies how much of a substance has dissociated into its constituent parts. It is represented as a fraction or percentage of the total number of initial moles that have dissociated. For the reaction \(\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2\), the degree of dissociation is calculated by observing how many moles of \(\text{N}_2\text{O}_4\) have converted into \(\text{NO}_2\).Here's how degree of dissociation is determined:

• For dissociation of 33% at pressure \(P_1\), it means \(\alpha_1 = 0.33\).
• For 40% dissociation at pressure \(P_2\), it implies \(\alpha_2 = 0.40\).

Knowing the degree of dissociation is crucial for calculating equilibrium moles and understanding how reactants and products distribute themselves at equilibrium.

Equilibrium Pressure

Equilibrium pressure refers to the pressure exerted by the gaseous components of a chemical system when it reaches equilibrium. In a gas-phase reaction such as \(\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2\), equilibrium pressure is important for determining how far the reaction has proceeded.
When the degrees of dissociation are known, they help calculate how the total pressure is distributed between the initial and resulting gas molecules.
Popping one can understand why equilibrium pressure will change as dissociation increases or decreases, affecting moles in the system.

• At \(P_1\), with a degree of dissociation of 33%, total moles = \(1 + 2(0.33) - 0.33 = 1.33\) moles.
• At \(P_2\), with 40% dissociation, total moles = \(1 + 2(0.40) - 0.40 = 1.40\) moles.

This tells us how deviations in pressure relate to the amount of dissociated products and provides a means to compare effects of different pressures.

Moles at Equilibrium

Moles at equilibrium refer to the quantity of reactants and products present once the system has reached equilibrium. For each portion of the \(\text{N}_2\text{O}_4\) that dissociates, a certain quantity of \(\text{NO}_2\) is formed, following the stoichiometry:

• Initially, start with 1 mole of \(\text{N}_2\text{O}_4\).
• Equilibrium moles of \(\text{N}_2\text{O}_4\) = \(1 - \alpha\).
• Equilibrium moles of \(\text{NO}_2\) = \(2\alpha\).

The process of calculating the moles at equilibrium highlights the distribution of particles in a reaction once a dynamic balance is achieved. Knowing these quantities allows chemists to ascertain the direction and extent of reactions.