Problem 133 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The pressure ratio at equilibrium is (b) 36:1.

Step 1: Understand the Reactions

Consider the reaction (i) \( A(g) \rightleftharpoons B(g) + C(g) \) and reaction (ii) \( X(g) \rightleftharpoons 2Y(g) \). For both reactions, we need to analyze the equilibrium in terms of the degree of dissociation, which is given to be the same for both reactions.

Step 2: Define Degree of Dissociation

Let \( \alpha \) be the degree of dissociation for both reactions. For reaction (i), the change in moles due to dissociation is \( \alpha \), resulting in \( (1-\alpha)A, \alpha B, \alpha C \). For reaction (ii), the change in moles due to \( X \) dissociating is also \( \alpha \), resulting in \( (1-\alpha)X, 2\alpha Y \).

Step 3: Calculate Total Moles at Equilibrium

For reaction (i), total moles at equilibrium \( = 1 + \alpha \). For reaction (ii), total moles at equilibrium \( = 1 + \alpha \).

Step 4: Relate Mole Fractions to Kp

For a reaction \( aA \rightleftharpoons bB + cC \), we have \( K_p = \left(\frac{{P_B}}{{P^\circ}}\right)^b \left(\frac{{P_C}}{{P^\circ}}\right)^c / \left(\frac{{P_A}}{{P^\circ}}\right)^a \). Applying it to reactions (i) and (ii) leads to equations where the equilibrium constant relates mole fractions ascribed to pressures.

Step 5: Simplify the Kp Equations

For reaction (i), using \( K_{p1} = \frac{{\alpha^2}}{{(1-\alpha)P}} \). For reaction (ii), \( K_{p2} = \frac{{(2\alpha)^2}}{{(1-\alpha)P/4}} \). Given \( \frac{{K_{p1}}}{{K_{p2}}} = 9 \), relate the expressions.

Step 6: Solve for Pressure Ratio

Substitute equations from step 5 into the relationship to derive \( (1-\alpha)^2: (1-\alpha)^2/4 = 9:1 \). Simplifying gives \( P = 4 \cdot P_{total}\) for reaction (ii). Thus the pressure ratios become \( 36: 1 \).

Step 7: Conclusion

Combining the information from all the steps, we conclude that the ratio of the total pressure at equilibrium for the reactions is \( 36:1 \).

Key Concepts

Degree of Dissociation Explained

In chemical reactions, the degree of dissociation is a measure of how much a compound breaks down into its components. It is represented by the Greek letter \( \alpha \). If a substance partially dissociates in a reaction, \( \alpha \) ranges between 0 and 1. A value of 0 indicates no dissociation, while a value of 1 signifies complete dissociation into products.

For a given reaction, if we start with 1 mole of the reactant, like \( A(g) \) in this context, and \( \alpha \) portions of it dissociate, it means at equilibrium, we have \( (1-\alpha) \) moles of \( A \) remaining, while \( \alpha \) moles each of the products are formed. This concept is crucial because it allows us to determine how far a reaction has progressed under specific conditions.

• The degree of dissociation is helpful in calculating equilibrium concentrations of reactants and products.
• It's used to predict the effectiveness and yield of chemical transformations.

Determining \( \alpha \) can also reveal information about the stability of products and reactants in the system at equilibrium.

Understanding Pressure at Equilibrium

In the context of chemical equilibria, pressure can significantly influence the state of the system, particularly when gas-phase reactions are involved. This exercise involves reactions where pressure plays a crucial role in determining equilibrium.

At equilibrium, the total pressure of the system is the sum of partial pressures of all gaseous substances present. Partial pressure is essentially the pressure that a gas would exert if it were alone in the container. In practice, knowing the total moles of gases helps calculate pressure as outlined by the ideal gas law.

• With \( P_{total} = P_A + P_B + P_C \), the partial pressures relate to mole fractions derived from dissociation levels and starting materials.
• The condition of same degrees of dissociation (\( \alpha \)) for two reactions ensures that relative changes in pressure can be efficiently compared.

In analyzing the pressures at equilibrium, especially using the equilibrium constant \( K_p \), fundamental insights into how state variables like pressure drive reactions towards or away from equilibrium can be gained.

Chemical Equilibrium Analysis in Practice

Chemical equilibrium analysis involves understanding how different factors like concentration, pressure, and temperature influence the position of equilibrium in a reaction. A state of chemical equilibrium is characterized by constant concentrations of reactants and products over time, resulting in no net change.

The key to analyzing these changes lies in the equilibrium constant, \( K_p \), which links the pressures (or concentrations) of reactants and products under given conditions. For reactions in gaseous phases, \( K_p \) is particularly useful as it directly relates equilibrium concentrations to partial pressures:

• \( K_p = \frac{{(P_B)(P_C)}}{(P_A)} \) for reaction (i), clearly indicating product-to-reactant pressure relationships.
• This ratio signifies how the balance of reaction species must adjust under equilibrium conditions.

Distinguishing changes brought about by pressure alterations is vital, especially in industrial applications where reaction conditions are tuned for optimal yield. Chemical equilibrium analysis facilitates a comprehensive understanding of such tuning, mediated through parameters like \( \alpha \) and \( K_p \), in response to changes in system variables.