Problem 199 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction \(\mathrm{AB}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{A}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{~g})\) The degree of dissociation ' \(\alpha\) ' is negligible as compared to 1 (unity); the degree of dissociation may be expressed as: (a) \(\alpha \propto \frac{1}{\sqrt{\mathrm{P}}}\) (b) \(\alpha \propto \sqrt{\mathrm{V}}\) (c) \(\alpha \propto \frac{1}{\mathrm{P}}\) (d) \(\alpha \propto \frac{1}{\mathrm{~V}}\)

Step-by-Step Solution

Answer

Option (a): \( \alpha \propto \frac{1}{\sqrt{P}} \).

Step 1: Understanding the Reaction

The given chemical reaction is \( \mathrm{AB}_2(\mathrm{~g}) \rightleftharpoons \mathrm{A}(\mathrm{g})+\mathrm{B}_2(\mathrm{~g}) \). This is a decomposition reaction where one mole of \( \mathrm{AB}_2 \) dissociates into one mole of \( \mathrm{A} \) and one mole of \( \mathrm{B}_2 \). The degree of dissociation, \( \alpha \), represents the fraction of \( \mathrm{AB}_2 \) that has dissociated.

Step 2: Expression of Equilibrium Constant

At equilibrium, the equilibrium constant \( K_p \) for the reaction can be expressed as:\[ K_p = \frac{P_{A} \cdot P_{B_2}}{P_{AB_2}} \]Where \( P_A \) and \( P_{B_2} \) are the partial pressures of \( \mathrm{A} \) and \( \mathrm{B}_2 \), and \( P_{AB_2} \) is the partial pressure of \( \mathrm{AB}_2 \).

Step 3: Initial and Equilibrium Pressures

Assume the initial pressure of \( \mathrm{AB}_2 \) is \( P \). At equilibrium, \( \mathrm{AB}_2 \) will have a pressure of \( P(1-\alpha) \), and \( \mathrm{A} \) and \( \mathrm{B}_2 \) will each have a pressure of \( P \alpha \), assuming the total pressure remains constant. Thus, the pressures at equilibrium are:- \( P_{AB_2} = P(1-\alpha) \)- \( P_A = P_{B_2} = P \alpha \)

Step 4: Substitute into Equilibrium Expression

Substitute the equilibrium pressures into the expression for \( K_p \):\[ K_p = \frac{(P \alpha) \cdot (P \alpha)}{P(1-\alpha)} = \frac{P^2 \alpha^2}{P(1-\alpha)} \approx P \alpha^2 \text{ (since \( \alpha \) is negligible, \( 1-\alpha \approx 1 \))}\]

Step 5: Solving for \( \alpha \)

Rearrange the above equation to solve for \( \alpha \):\[ \alpha^2 = \frac{K_p}{P} \]\[ \alpha = \sqrt{\frac{K_p}{P}} \]This indicates that \( \alpha \) is inversely proportional to the square root of pressure \( P \), or \( \alpha \propto \frac{1}{\sqrt{P}} \).

Step 6: Conclusion

Based on the relationship derived, the correct expression for the degree of dissociation \( \alpha \) is \( \alpha \propto \frac{1}{\sqrt{P}} \), corresponding to option (a).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time. This does not mean that the reactions have stopped; rather, the rate of the forward reaction equals the rate of the reverse reaction. Think of it like a busy office where email is constantly sent and received at an equal rate, resulting in a consistent inbox size.
At equilibrium, the system has reached a balance. For our example with the reaction \( \mathrm{AB}_2(\mathrm{~g}) \rightleftharpoons \mathrm{A}(\mathrm{g})+\mathrm{B}_2(\mathrm{~g}) \), once equilibrium is established, the amounts of \( \mathrm{A} \), \( \mathrm{B}_2 \), and \( \mathrm{AB}_2 \) will not change as long as conditions such as temperature and pressure remain constant.

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a value that expresses the relationship between the concentrations or pressures (for reactions involving gases) of the products and reactants at equilibrium. It is specific to a particular reaction at a given temperature.
For our reaction, the equilibrium constant in terms of pressure, \( K_p \), is given by:\[ K_p = \frac{P_{A} \cdot P_{B_2}}{P_{AB_2}} \]
This equation tells us that higher equilibrium constant values indicate a greater extent of reaction towards the products, while lower values suggest the reactants are favored. The equilibrium constant is crucial because it helps predict how a change in conditions (like pressure or temperature) might affect the position of equilibrium.

Partial Pressure

Partial pressure is the contribution each gas in a mixture makes to the total pressure. It is an essential concept in understanding chemical equilibrium in gaseous reactions. Each gas in a mixture exerts a pressure as if it were the only gas present.
In the case of the decomposition of \( \mathrm{AB}_2 \), the partial pressures \( P_A \) and \( P_{B_2} \) correspond to the pressures that \( \mathrm{A} \) and \( \mathrm{B}_2 \) exert independently in the system. At equilibrium, these can be calculated from the degree of dissociation \( \alpha \), as shown:

• \( P_{AB_2} = P(1-\alpha) \)
• \( P_A = P_{B_2} = P \alpha \)

This concept helps in deriving the equilibrium constant expression where these partial pressures are a crucial part of the equation.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. It's like a big Lego block breaking into smaller pieces. In the given reaction \( \mathrm{AB}_2(\mathrm{~g}) \rightleftharpoons \mathrm{A}(\mathrm{g})+\mathrm{B}_2(\mathrm{~g}) \), \( \mathrm{AB}_2 \) decomposes into \( \mathrm{A} \) and \( \mathrm{B}_2 \).
Decomposition reactions often require energy input in the form of heat, light, or electricity. They are crucial in understanding equilibrium because the extent to which a compound decomposes can be quantitatively described by the degree of dissociation, \( \alpha \). In equilibrium systems, the decomposition is balanced by the reverse reaction, where \( \mathrm{A} \) and \( \mathrm{B}_2 \) combine to reform \( \mathrm{AB}_2 \), thus maintaining equilibrium conditions.