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 '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

Option (c): \( \mathrm{P} \frac{\alpha^{3}}{3} \) is correct.

Step 1: Understand Initial Conditions

In this reaction, the compound \( \mathrm{AB}_2 \) dissociates into \( \mathrm{AB} \) and \( \mathrm{B}_2 \). Start by assuming the initial moles of \( \mathrm{AB}_2 \) to be \( a \). So, initially for \( \mathrm{AB}_2 \), we have \( 2a \) moles.

Step 2: Express Change in Terms of Dissociation

When \( \mathrm{AB}_2 \) dissociates, the change in moles can be expressed using \( \alpha \), the degree of dissociation. Thus, \( 2a \cdot \alpha \) moles of \( \mathrm{AB}_2 \) react to form \( 2a \cdot \alpha \) moles of \( \mathrm{AB} \) and \( a \cdot \alpha \) moles of \( \mathrm{B}_2 \).

Step 3: Calculate Total Moles at Equilibrium

At equilibrium, the moles are \( 2a (1-\alpha) \) for \( \mathrm{AB}_2 \), \( 2a \alpha \) for \( \mathrm{AB} \), and \( a \alpha \) for \( \mathrm{B}_2 \). The total number of moles is \( 2a(1-\alpha) + 2a\alpha + a\alpha = 2a + a\alpha \).

Step 4: Determine Partial Pressures

Using total pressure \( P \) and the total number of moles, the partial pressures are derived as follows: \( P_{\mathrm{AB}_2} = P \times \frac{2a(1-\alpha)}{2a+a\alpha} \), \( P_{\mathrm{AB}} = P \times \frac{2a\alpha}{2a+a\alpha} \), and \( P_{\mathrm{B}_2} = P \times \frac{a\alpha}{2a+a\alpha} \).

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

The equilibrium constant \( K_p \) is given by \( K_p = \frac{(P_{\mathrm{AB}})^2 \cdot P_{\mathrm{B}_2}}{(P_{\mathrm{AB}_2})^2} \). Substitute the expressions from step 4 into this equation, which simplifies to \[ K_p = P \frac{\alpha^3}{3} \].

Step 6: Select the Correct Option

Based on the simplified expression for \( K_p \), the correct option is (c) \( \mathrm{P} \frac{\alpha^{3}}{3} \).

Key Concepts

Degree of Dissociation

The degree of dissociation, represented by the symbol \( \alpha \), is an important concept in chemistry. It measures the fraction of the original substance that breaks down (or dissociates) in a chemical reaction. When dealing with gaseous reactions, \( \alpha \) helps in evaluating how much a compound like \( \mathrm{AB}_2 \) has turned into its gaseous components. In the given exercise, \( \alpha \) represents the portion of \( \mathrm{AB}_2 \) that has dissociated into \( \mathrm{AB} \) and \( \mathrm{B}_2 \). This helps in calculating changes in the number of moles during a chemical reaction.
To consider this step-by-step:

• Initially, assume a number of moles of \( \mathrm{AB}_2 \), let's say \( 2a \).
• As the reaction proceeds, \( 2a\alpha \) moles dissociate into \( 2a\alpha \) moles of \( \mathrm{AB} \) and \( a\alpha \) moles of \( \mathrm{B}_2 \).

Degree of dissociation impacts the total pressure and is essential for computing the equilibrium state of a reaction.

Equilibrium Constant (Kp)

The equilibrium constant, specifically \( K_p \) for reactions involving gases, is a key concept in understanding the balance point of a chemical reaction. \( K_p \) quantifies the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients.
In the context of gaseous reactions, partial pressures replace concentrations when calculating \( K_p \). The solution to the exercise involved finding \( K_p \) as

• A combination of the partial pressures of \( \mathrm{AB} \) and \( \mathrm{B}_2 \) at equilibrium,
• divided by the partial pressure of \( \mathrm{AB}_2 \) raised to its stoichiometric coefficient.

The formula for \( K_p \) is given by \[ K_p = \frac{(P_{\mathrm{AB}})^2 \cdot P_{\mathrm{B}_2}}{(P_{\mathrm{AB}_2})^2} \]After substituting the partial pressure values derived from the initial moles, total moles, and degree of dissociation, it was simplified to \( K_p = P \frac{\alpha^3}{3} \), as shown in the solution.

Gaseous Reactions

Gaseous reactions involve reactants and products in the gas phase, and are greatly influenced by changes in conditions such as pressure and temperature. In the given example involving the dissociation of \( \mathrm{AB}_2 \), we observe how gaseous reactants form gaseous products, unaffected by external influences other than the stated pressure.
Key points about gaseous reactions:

• The behavior of gases follows the ideal gas law, where pressure, volume, and temperature are related.
• The partial pressures of gases are crucial in determining the extent of a reaction's progress to equilibrium.

During the dissociation of \( \mathrm{AB}_2 \), a careful consideration of moles and their conversion to partial pressures highlights the dynamic nature of gaseous reactions.
This helps to explain why understanding reactions in the gas phase is vital for predicting reaction behavior under various conditions.