Problem 126 - A Complete Resource Book in Chemistry for JEE Main

Question

For the dissociation reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\) the degree of dissociation \((\alpha)\) in terms of \(\mathrm{K}_{\mathrm{p}}\) and total equilibrium pressure \(\mathrm{P}\) is: (a) \(\alpha=\sqrt{\frac{4 P+K_{p}}{K_{p}}}\) (b) \(\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{P}}}{4 \mathrm{P}+\mathrm{K}_{\mathrm{p}}}}\) (c) \(\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{4 \mathrm{P}}}\) (d) None of these

Step-by-Step Solution

Answer

The correct answer is (b) \( \alpha = \sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{4 \mathrm{P} + \mathrm{K}_{\mathrm{p}}}} \).

Step 1: Write the Balanced Reaction Equation

The given reaction is \( \mathrm{N}_2\mathrm{O}_4(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{g}) \). Here, one mole of \( \mathrm{N}_2\mathrm{O}_4 \) dissociates into two moles of \( \mathrm{NO}_2 \).

Step 2: Define the Initial and Change in Moles at Equilibrium

Assume initially there is 1 mole of \( \mathrm{N}_2\mathrm{O}_4 \). Let \( \alpha \) be the degree of dissociation. At equilibrium, \( 1 - \alpha \) moles of \( \mathrm{N}_2\mathrm{O}_4 \) remain and \( 2\alpha \) moles of \( \mathrm{NO}_2 \) are formed.

Step 3: Write the Expression for Total Moles at Equilibrium

Total moles at equilibrium are \( (1 - \alpha) + 2\alpha = 1 + \alpha \).

Step 4: Express Partial Pressures at Equilibrium

Partial pressure of \( \mathrm{N}_2\mathrm{O}_4 \) is \( P(1-\alpha)/(1+\alpha) \) and of \( \mathrm{NO}_2 \) is \( 2P\alpha/(1+\alpha) \), where \( P \) is the total pressure.

Step 5: Write the Expression for Equilibrium Constant \( K_p \)

From the equation: \[ K_p = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2\mathrm{O}_4}} = \frac{\left( \frac{2P\alpha}{1+\alpha} \right)^2}{\frac{P(1-\alpha)}{1+\alpha}} = \frac{4P^2\alpha^2}{P(1-\alpha)(1+\alpha)} \]

Step 6: Simplify the Expression and Solve for \( \alpha \)

Simplify the equation: \[ K_p = \frac{4P^2\alpha^2}{P(1 - \alpha^2)} = \frac{4P\alpha^2}{1 - \alpha^2} \] Rearrange to solve for \( \alpha^2 \): \[ K_p(1 - \alpha^2) = 4P\alpha^2 \] This gives: \[ \alpha^2 = \frac{K_p}{4P + K_p} \]

Step 7: Final Step: Compare to Given Options

Compare \( \alpha = \sqrt{\frac{K_p}{4P+K_p}} \) to the options. It matches option (b).

Key Concepts

Degree of Dissociation

The degree of dissociation, commonly represented by the Greek letter \( \alpha \), is a measure of the fraction of a mole of a substance that has dissociated into its components at equilibrium. This is particularly important in equilibrium reactions where dissociation occurs, such as in the decomposition of \( \mathrm{N}_2\mathrm{O}_4 \) gas into \( \mathrm{NO}_2 \) gas.
Initially, we assume 1 mole of \( \mathrm{N}_2\mathrm{O}_4 \). During dissociation, \( \alpha \) represents the part of this mole that has dissociated. Hence, at equilibrium, we have \( 1-\alpha \) moles of \( \mathrm{N}_2\mathrm{O}_4 \) and \( 2\alpha \) moles of \( \mathrm{NO}_2 \).
Understanding \( \alpha \) is crucial because it affects the concentration and therefore the partial pressures of the gases involved. The degree of dissociation is linked to other concepts like the equilibrium constant and the initial conditions of the reaction, furthering our understanding of how systems approach equilibrium.

Equilibrium Constant

The equilibrium constant, denoted as \( K_p \) when dealing with gaseous reactions, represents the ratio of the concentrations (or partial pressures) of products to reactants at equilibrium, raised to the power of their stoichiometric coefficients. For the reaction \( \mathrm{N}_2\mathrm{O}_4(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}_2(\mathrm{g}) \), the expression for \( K_p \) is given as:
\[ K_p = \frac{(P_{\mathrm{NO}_2})^2}{P_{\mathrm{N}_2\mathrm{O}_4}} \]
This formula reflects how \( K_p \) depends on the partial pressures of the reactants and products at equilibrium. The value of \( K_p \) informs us about the position of equilibrium. A high \( K_p \) value indicates that products are favored at equilibrium, while a low \( K_p \) suggests that reactants are preferred.
In solving problems related to equilibrium, understanding how to manipulate \( K_p \) with respect to total pressure (\( P \)) and the degree of dissociation (\( \alpha \)) is crucial. The interplay between these variables helps predict how changes in conditions might shift the equilibrium position.

Partial Pressure

Partial pressure is a vital concept in understanding gaseous equilibria. It is the pressure that an individual gas within a mixture of gases would exert if it occupied the entire volume alone. In the dissociation of \( \mathrm{N}_2\mathrm{O}_4 \), calculating the partial pressures of \( \mathrm{N}_2\mathrm{O}_4 \) and \( \mathrm{NO}_2 \) at equilibrium is key.
The partial pressure of a gas can be calculated using its mole fraction and the total pressure \( P \). For \( \mathrm{N}_2\mathrm{O}_4 \), it is:
\[ P(1-\alpha)/(1+\alpha) \]
For \( \mathrm{NO}_2 \), with \( 2\alpha \) moles generated:
\[ 2P\alpha/(1+\alpha) \]
These expressions show how the total pressure \( P \) influences the pressure each gas exerts at equilibrium. Knowing the partial pressures helps in deriving \( K_p \) and determining how changes in the reaction conditions, such as pressure or volume, might affect the system's equilibrium state.