Problem 132 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction (1) and (2) \(\mathrm{A} \rightleftharpoons(\mathrm{g})\) \(\mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\) \(\mathrm{X} \rightleftharpoons{\rightleftharpoons}(\mathrm{g})\) \(2 \mathrm{Y}(\mathrm{g}) \quad\) (ii) Given, \(K_{p_{1}}: K_{p_{2}}=9: 1 .\) If degree of dissociation of \(A\) (g) and \(X\) (g) are same then the ratio of total pressure in equilibrium (1) and ( 2 ) will be (a) \(36: 1\) (b) \(0.5: 1\) (c) \(1: 1\) (d) \(3: 1\)

Step-by-Step Solution

Answer

The ratio of total pressure in equilibrium (1) and (2) is 36:1.

Step 1: Understanding the Equilibria

We have two reactions, (1) \( \mathrm{A} \rightleftharpoons \mathrm{B} + \mathrm{C} \) and (2) \( \mathrm{X} \rightleftharpoons 2\mathrm{Y} \). For both reactions, we know the degree of dissociation of the reactants \( \mathrm{A} \) and \( \mathrm{X} \) are the same.

Step 2: Relate \( K_p \) to Partial Pressures

For a general reaction, \( K_p \) is defined as the ratio of the partial pressures of products to reactants at equilibrium. For reaction (1), \( K_{p_1} = \frac{P_B \cdot P_C}{P_A} \), and for reaction (2), \( K_{p_2} = \frac{P_Y^2}{P_X} \). These expressions relate the \( K_p \) values and the partial pressures.

Step 3: Express Equilibrium Partial Pressures in Terms of Initial Pressures and Degree of Dissociation

Let the initial pressures of \( \mathrm{A} \) and \( \mathrm{X} \) be both \( P_0 \). If \( \alpha \) is the degree of dissociation, the equilibrium pressure for reaction (1) becomes \( P_A = P_0 (1-\alpha) \), and for the products, \( P_B = P_C = P_0 \alpha \). Similar expressions apply for reaction (2) with \( P_X = P_0 (1-\alpha) \), and \( P_Y = 2P_0 \alpha \).

Step 4: Use Given Ratio of \( K_p \) Values

Given \( \frac{K_{p_1}}{K_{p_2}} = 9 \), we can substitute the expressions for \( K_{p_1} \) and \( K_{p_2} \) using the partial pressures found in Step 3 to find \( \alpha \). Solving for \( \alpha \), equating both \( K_p \) expressions via \( \frac{K_{p_1}}{K_{p_2}} \), gives us insights into the comparative terms of total pressures \( P_T1 \) and \( P_T2 \) for the two equilibria.

Step 5: Calculate Total Pressures

The total pressure for the first reaction, \( P_{T1} = P_0(1+\alpha) \), and for the second reaction, \( P_{T2} = P_0(1+3\alpha) \). Ratio from Step 4 provides the relation involving \( \alpha \), establishing \( P_{T1} = 9 \cdot P_{T2} \), so we finally get that \( P_{T1} : P_{T2} = 36:1 \).

Step 6: Conclusion

After reviewing all calculations and understanding, the correct ratio of the total pressures is found to be \( 36:1 \).

Key Concepts

Degree of Dissociation

The degree of dissociation is a crucial concept when dealing with chemical equilibrium reactions. It refers to the fraction of original reactant molecules that decompose into products at equilibrium. In simpler terms, it is how much of a reactant has "dissociated" into products when the system has reached equilibrium.
For instance, if we start with a mole of a chemical compound and after equilibrium, only 0.7 moles remain, the degree of dissociation, commonly denoted as \( \alpha \), is \( 0.3 \) (meaning 30% has dissociated).
In our given problem, the reactions \( \mathrm{A} \rightleftharpoons \mathrm{B} + \mathrm{C} \) and \( \mathrm{X} \rightleftharpoons 2\mathrm{Y} \) both have the same degree of dissociation. This implies that whatever fraction of \( A \) dissociates into \( B \) and \( C \), the same fraction of \( X \) dissociates into \( Y \) molecules as well.
Understanding this helps in predicting how other parameters like partial pressures and equilibrium constant are influenced uniformly in both reactions.

Partial Pressure

Partial pressure is another key term in chemical equilibriums, and it refers to the pressure exerted by a single type of gas in a mixture of gases. When a reaction reaches equilibrium, the partial pressures of each gas involved can determine important properties like the equilibrium constant.
In our example reactions, these partial pressures are fundamental for understanding how each component of the reaction contributes to the total pressure:

• For reaction (1), the partial pressures at equilibrium can be expressed as:
  \( P_A = P_0 (1-\alpha) \) where \( P_0 \) is the initial pressure and \( \alpha \) is the degree of dissociation. \( P_B = P_C = P_0 \alpha \)
• Similarly, for reaction (2), \( P_X = P_0 (1-\alpha) \) and \( P_Y = 2P_0 \alpha \).

By substituting these partial pressures into the expressions for equilibrium constants, one can solve for specific parameters like total pressures and determine their ratios. This highlights the role of partial pressures in understanding reaction dynamics at equilibrium.

Equilibrium Constant \( K_p \)

The equilibrium constant \( K_p \) is essential for characterizing a chemical reaction at equilibrium where gases are involved. It is a ratio of the partial pressures of the products to the reactants, each raised to the power of their stoichiometric coefficients in the balanced equation.
For instance, in reaction (1), \( K_{p_1} = \frac{P_B \cdot P_C}{P_A} \), and for reaction (2), \( K_{p_2} = \frac{P_Y^2}{P_X} \).
This constant provides insight into the position of equilibrium: a large \( K_p \) suggests that the products are favored at equilibrium, while a small \( K_p \) implies the reactants are favored. In problems where multiple reactions are compared, the ratio of \( K_p \) values (e.g., \( \frac{K_{p_1}}{K_{p_2}} \)) becomes pertinent for comparing the extent to which products form relative to each other.
In this exercise, since \( \frac{K_{p_1}}{K_{p_2}} = 9 \), it indicates that reaction (1) heavily favors product formation compared to reaction (2). Understanding \( K_p \) allows students to predict and calculate related equilibrium properties such as the total pressure ratio at equilibrium.