Problem 129 - A Complete Resource Book in Chemistry for JEE Main

Question

At certain temperature compound \(\mathrm{AB}_{2}(\mathrm{~g})\) dissociates according to the reaction \(2 \mathrm{AB}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{~g})\) The value of \(K_{p}\) in terms of degree of dissociation ' \(\alpha\) ' and total pressure ' \(\mathrm{P}\) ' is (a) \(\mathrm{P} \frac{\alpha^{3}}{2}\) (b) \(\mathrm{P} \frac{\alpha^{2}}{3}\) (c) \(\mathrm{P} \frac{\alpha^{3}}{3}\) (d) \(\mathrm{P} \frac{\alpha^{2}}{2}\)

Step-by-Step Solution

Answer

The correct option is (a) \( \text{P} \frac{\alpha^{3}}{2} \).

Step 1: Initial Mole Calculation

Consider 1 mole of \( \mathrm{AB}_2 \) at the start. Since the compound dissociates into \( 2 \mathrm{AB} \) and \( \mathrm{B}_2 \), the reaction can be represented as: \[2 \mathrm{AB}_2 \rightleftharpoons 2 \mathrm{AB} + \mathrm{B}_2.\]

Step 2: Relate Degree of Dissociation in Moles

If \( \alpha \) is the degree of dissociation, then for 1 mole of \( \mathrm{AB}_2 \), \( 2\alpha \) moles of \( \mathrm{AB} \) and \( \alpha \) moles of \( \mathrm{B}_2 \) are formed. This means at equilibrium, moles of \( \mathrm{AB}_2 \) = \( 1 - \alpha \).

Step 3: Calculate Total Moles at Equilibrium

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

Step 4: Partial Pressures in Terms of Total Pressure

Total pressure \( P \) implies partial pressures will be equivalent to mole fractions times \( P \). So, \( P_{\mathrm{AB}_2} = P \frac{1 - \alpha}{1 + 2\alpha} \), \( P_{\mathrm{AB}} = P \frac{2\alpha}{1 + 2\alpha} \), \( P_{\mathrm{B}_2} = P \frac{\alpha}{1 + 2\alpha} \).

Step 5: Express \( K_p \) in Terms of \( \alpha \) and \( P \)

\[ K_p = \frac{(P_{\mathrm{AB}})^2 \cdot P_{\mathrm{B}_2}}{(P_{\mathrm{AB}_2})^2} = \frac{\left(P \frac{2\alpha}{1 + 2\alpha}\right)^2 \left(P \frac{\alpha}{1 + 2\alpha}\right)}{\left(P \frac{1 - \alpha}{1 + 2\alpha}\right)^2}. \]

Step 6: Simplify the \( K_p \) Expression

After simplifying, \[ K_p = P \frac{8 \alpha^3}{(1 + 2\alpha)^2} \cdot \frac{(1 + 2\alpha)^2}{(1 - \alpha)^2} = P \frac{8 \alpha^3}{(1 - \alpha)^2}. \] Assumes \( \alpha \) is small, then \( 1 - \alpha \approx1 \). Thus, \[ K_p \approx P \frac{\alpha^3}{2}. \]

Step 7: Correct Option Identification

After solving, the suitable option that matches the simplified expression is (a) \( \text{P} \frac{\alpha^{3}}{2} \).

Key Concepts

Degree of Dissociation

The degree of dissociation, typically denoted by the symbol \( \alpha \), is a measure of the extent to which a substance dissociates into its components at equilibrium. It generally describes how much of the initial substance has split into simpler fragments or products in a chemical reaction.
For instance, in the dissociation of \( \mathrm{AB}_2 \) into \( 2 \mathrm{AB} \) and \( \mathrm{B}_2 \), the degree of dissociation tells us what fraction of the original \( \mathrm{AB}_2 \) molecules have decomposed. If one mole of \( \mathrm{AB}_2 \) starts the reaction, at equilibrium, the produced moles would be related to \( \alpha \) like this:

• \( 2\alpha \) moles of \( \mathrm{AB} \)
• \( \alpha \) moles of \( \mathrm{B}_2 \)

Moreover, the moles of \( \mathrm{AB}_2 \) remaining would be \( 1 - \alpha \).
This approach provides a clearer understanding of what is happening in the system as it approaches equilibrium.

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \( K_p \), plays a crucial role in understanding the behavior of gaseous reactions at equilibrium. It is determined by the partial pressures of the reactants and products. When dealing with gaseous equilibria, \( K_p \) is especially significant because it is influenced by the pressure changes during the reaction.
For the given reaction:\[ 2 \mathrm{AB}_2 \rightleftharpoons 2 \mathrm{AB} + \mathrm{B}_2 \]We express \( K_p \) in terms of the partial pressures of products and reactants:\[ K_p = \frac{(P_{\mathrm{AB}})^2 \cdot P_{\mathrm{B}_2}}{(P_{\mathrm{AB}_2})^2} \]Here, the partial pressures are derived from the total pressure \( P \) and depend on the mole fractions, which involve \( \alpha \).
When simplified, especially when assuming small degrees of dissociation (\( \alpha \) is small, hence \( 1 - \alpha \approx 1 \)), \( K_p \) approximates to \( P \frac{\alpha^{3}}{2} \). This approximation helps in arriving at the correct options offered in the multiple-choice format generally encountered in chemistry problems.

Partial Pressure

Partial pressure is the pressure contributed by an individual gas component in a mixture of gases. It is highly significant in analyzing chemical equilibria involving gases. Each gas in a mixture reacts independently, and their partial pressures help in calculating equilibrium expressions like \( K_p \).

• In the given reaction, partial pressure of \( \mathrm{AB}_2 \) is \( P_{\mathrm{AB}_2} = P \frac{1 - \alpha}{1 + 2\alpha} \).
• For \( \mathrm{AB} \), it is \( P_{\mathrm{AB}} = P \frac{2\alpha}{1 + 2\alpha} \).
• For \( \mathrm{B}_2 \), it is \( P_{\mathrm{B}_2} = P \frac{\alpha}{1 + 2\alpha} \).

The total pressure \( P \) acts as a reference point, and the individual mole fractions dictate each component's partial pressure. These calculations are integral to determining the equilibrium constant \( K_p \), as they provide the necessary values for expressing products and reactants in terms of pressure. Understanding partial pressure ensures that you can correctly solve for \( K_p \) and better comprehend the balance of reactions at equilibrium.