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 answer is (a) \(P \frac{\alpha^3}{2}\).

Step 1: Write the Balanced Reaction Equation

The given reaction is \(2 \mathrm{AB}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{~g})\). This is a reversible reaction, and as provided, \(\mathrm{AB}_2\) dissociates into \(\mathrm{AB}\) and \(\mathrm{B}_2\).

Step 2: Define Initial Pressure and Degree of Dissociation

Assume the initial pressure of \(2 \mathrm{AB}_{2}\) is \(P\). Let \(\alpha\) be the degree of dissociation. Initially, the pressure of \(2 \mathrm{AB}_{2}\) is \(P\), and the pressures of \(\mathrm{AB}\) and \(\mathrm{B}_2\) are 0.

Step 3: Express Pressure in Terms of Degree of Dissociation

After dissociation, the change in moles of \(2 \mathrm{AB}_{2}\) is \(-2\alpha\), forming \(+2\alpha\) moles of \(\mathrm{AB}\) and \(+\alpha\) moles of \(\mathrm{B}_2\). The partial pressures are:\[\text{For } 2 \mathrm{AB}_{2},\space P(1-\alpha)\]\[\text{For } \mathrm{AB},\space 2P\alpha\]\[\text{For } \mathrm{B}_{2},\space P\alpha\]

Step 4: Calculate Total Pressure

The total pressure \(P_{total}\) after dissociation is given by:\[P_{total} = P(1-\alpha) + 2P\alpha + P\alpha = P(1 + \alpha)\]

Step 5: Derive Expression for Equilibrium Constant \(K_p\)

The equilibrium constant \(K_p\) at the given temperature is given by:\[K_{p} = \frac{(P_{\text{AB}})^2 \cdot P_{\text{B}_{2}}}{(P_{\text{AB}_2})^2}\]Substitute the expressions:\[K_{p} = \frac{(2P\alpha)^2 \cdot (P\alpha)}{(P(1-\alpha))^2}\]Simplify:\[K_{p} = \frac{4P^3\alpha^3}{P^2(1-\alpha)^2} = 4P\alpha^3\]

Step 6: Set Final Expression and Compare

Since \(K_p = 4P\alpha^3\), and rearranging for small values of \(\alpha\), where \((1-\alpha) \approx 1\):\[K_{p} \approx P\frac{\alpha^3}{2}\]This matches option (a), which simplifies to \(K_p = P\frac{\alpha^3}{2}\).

Key Concepts

Degree of Dissociation

Understanding the degree of dissociation is crucial when dealing with chemical equilibria, especially in reversible reactions. The degree of dissociation, denoted by the symbol \(\alpha\), refers to the fraction of the original compound that breaks down during the reaction. Thus, it gives us a clear picture of how much of the compound has dissociated compared to its initial amount.

In the reaction \(2 \mathrm{AB}_{2}(\mathrm{g}) \rightleftharpoons 2\mathrm{AB}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{g})\), the compound \(\mathrm{AB}_2\) dissociates into products. Initially, you start with a certain pressure, which we denote with \(P\), from the undissociated \(\mathrm{AB}_2\). As the reaction progresses to reach equilibrium, a fraction \(\alpha\) of \(\mathrm{AB}_2\) dissociates. This fraction transforms into products \(\mathrm{AB}\) and \(\mathrm{B}_2\).

To calculate the number of moles that change as a result of dissociation: if \(2\alpha\) moles of \(\mathrm{AB}_2\) dissociate, they form \(2\alpha\) moles of \(\mathrm{AB}\) and \(\alpha\) moles of \(\mathrm{B}_2\). The degree of dissociation affects both the concentrations (or pressures in the gas phase) and the number of moles present at equilibrium. This factor plays a pivotal role in determining the equilibrium state of the system.

Equilibrium Constant (Kp)

The equilibrium constant, specifically \(K_p\) for gases, is a vital concept that reflects the ratio of the partial pressures of products to reactants at equilibrium. It helps in understanding how far a reaction proceeds and in predicting the position of equilibrium.

For gaseous reactions, \(K_p\) is used when dealing with partial pressures instead of concentrations. For the given reaction, \(K_p\) can be expressed as:

• \(K_{p} = \frac{(P_{\text{AB}})^2 \cdot P_{\text{B}_2}}{(P_{\text{AB}_2})^2}\)

This equation emerges from the balanced reaction, showing the dependence of the equilibrium constant on the pressures of all gaseous species involved.

In practice, by substituting the partial pressures, derived in terms of \(\alpha\) and \(P\), into this formula, we get:

• \(K_{p} = \frac{4P^3\alpha^3}{P^2(1-\alpha)^2}\)

This simplifies to \(K_p = 4P\alpha^3\). Such expressions reveal the intricate relationship between the degree of dissociation and the pressure conditions that maintain equilibrium.

Partial Pressure Calculations

In chemical reactions involving gases, understanding partial pressures is critical as they depict the fraction of total pressure exerted by each gas. It's important because these pressures are directly involved in the calculation of equilibrium constants like \(K_p\).

In our specific reaction, we start with an initial pressure \(P\) for the undissociated \(2 \mathrm{AB}_2\). As dissociation occurs, we calculate the partial pressures of each species at equilibrium by considering how much of \(\mathrm{AB}_2\) dissociates:

• For \(2 \mathrm{AB}_2\), the partial pressure becomes \(P(1-\alpha)\).
• For \(\mathrm{AB}\), formed from \(\mathrm{AB}_2\), it has \(2P\alpha\) pressure.
• For \(\mathrm{B}_2\), formed separately, it has \(P\alpha\) pressure.

By adding these pressures, the total pressure \(P_{total}\) at equilibrium is \(P(1 + \alpha)\).

Accurate partial pressure calculation is essential, not just for determining \(K_p\), but also for analyzing how reactions respond to pressure changes, known as Le Chatelier's Principle. By mastering partial pressure calculations, students can predict how changes in reaction conditions affect equilibrium.