Question

For the reaction (1) and (2) \(\mathrm{A} \rightleftharpoons(\mathrm{g}) \quad \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\) \(\mathrm{X} \rightleftharpoons(\mathrm{g}) \quad 2 \mathrm{Y}(\mathrm{g})\) Given, \(K_{p_{1}}: K_{p_{2}}=9: 1 .\) If degree of dissociation of \(A\) (g) and \(\mathrm{X}\) (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) \(3: 1\)

Short answer

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

Step-by-step solution

Write the Expression for Equilibrium Constants

For reaction (1), the equilibrium constant in terms of partial pressures is given by: \[ K_{p_1} = \frac{P_B P_C}{P_A} \] For reaction (2), the equilibrium constant is: \[ K_{p_2} = \frac{(P_Y)^2}{P_X} \] Given, \( K_{p_1} : K_{p_2} = 9 : 1 \).

Relate Degree of Dissociation to Partial Pressures

Let \( \alpha \) be the degree of dissociation, \( P \) be the total pressure for each system. For reaction (1), initial pressure of \( A \) is \( P \), and at equilibrium, \( P_B = P_C = \alpha P \), and \( P_A = (1-\alpha)P \). For reaction (2), the initial pressure of \( X \) is \( P \), and at equilibrium, \( P_Y = 2\alpha P \) and \( P_X = (1-\alpha)P \).

Substitute in Equilibrium Expressions

Using the relation between the derivative of the elements: For reaction (1): \[ K_{p_1} = \frac{(\alpha P)^2}{(1-\alpha)P} = \frac{\alpha^2 P}{1-\alpha} \] For reaction (2): \[ K_{p_2} = \frac{(2\alpha P)^2}{(1-\alpha)P} = \frac{4\alpha^2 P}{1-\alpha} \]

Equate the Ratios with Given Values

Given \( \frac{K_{p_1}}{K_{p_2}} = \frac{9}{1} \), therefore equate: \[ \frac{\alpha^2 P_1}{1-\alpha} \div \frac{4\alpha^2 P_2}{1-\alpha} = \frac{9}{1} \] This simplifies to the equation: \[ \frac{P_1}{4P_2} = \frac{9}{1} \]

Solve for the Pressure Ratio

Solving \( \frac{P_1}{4P_2} = \frac{9}{1} \) gives: \[ P_1 = 36P_2 \] Therefore, the ratio of total pressures \( \frac{P_1}{P_2} \) is \( 36:1 \).

Key concepts

Degree of Dissociation

The degree of dissociation, often represented by the Greek letter \( \alpha \), is a key concept in understanding chemical reactions in equilibrium. It describes the fraction of a mole of reactant that breaks apart into its products in a reaction. For example, in the reaction \( \mathrm{A} \rightleftharpoons \mathrm{B} + \mathrm{C} \), if the initial concentration of \( \mathrm{A} \) is \( P \) and at equilibrium, a portion \( \alpha \) of \( P \) dissociates into \( \mathrm{B} \) and \( \mathrm{C} \), then the concentrations of \( \mathrm{B} \) and \( \mathrm{C} \) each become \( \alpha P \), and the remaining concentration of \( \mathrm{A} \) is \( (1-\alpha)P \).
Understanding \( \alpha \) is crucial because it directly impacts the partial pressures of species in gaseous reactions and plays a significant role in calculating equilibrium constants. The degree of dissociation can vary with changes in pressure, temperature, and the nature of the reaction itself.

Partial Pressure

Partial pressure refers to the pressure exerted by a particular component of a mixture of gases. It is an essential concept when discussing chemical equilibria involving gases, as it helps determine the equilibrium constant, \( K_p \). In a mixture, the partial pressure of a gas is proportional to its mole fraction and the total pressure of the system.
For the reaction \( \mathrm{A} \rightleftharpoons \mathrm{B} + \mathrm{C} \), the total pressure \( P \) is distributed among \( \mathrm{A} \), \( \mathrm{B} \), and \( \mathrm{C} \) based on their equilibrium concentrations. At equilibrium, the partial pressures are \( P_A = (1-\alpha)P \), \( P_B = \alpha P \), and \( P_C = \alpha P \). This relationship is crucial for calculating the equilibrium constant and understanding how conditions such as total pressure affect the equilibrium state.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction reaches a state where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products. This state can be described by the equilibrium constant (\( K \)), which is a ratio of the concentrations of the products to reactants, each raised to the power of their respective coefficients in the balanced chemical equation.
In terms of partial pressures (for gaseous reactions), the equilibrium constant \( K_p \) provides insight into the extent to which the reaction will proceed to products. A larger \( K_p \) indicates a greater tendency for the formation of products at equilibrium. In the provided exercises, \( K_{p_1} \) and \( K_{p_2} \) are essential for determining how different reactions behave under the same conditions when the degree of dissociation is similar.

Reaction Kinetics

Reaction kinetics is the study of the rates at which chemical reactions occur and the factors that influence these rates. While chemical equilibrium focuses on the state at which a reaction becomes stable, reaction kinetics explores how long it takes to reach equilibrium and how different variables, such as temperature and concentration, affect this process.
In the context of the given reactions, understanding reaction kinetics can shed light on how quickly \( \mathrm{A} \) and \( \mathrm{X} \) dissociate into their products. Knowing the degree of dissociation (\( \alpha \)) does not just help in calculating equilibrium constants but also influences how we perceive the speed of the reaction. Factors such as catalysts, temperature, and pressure can significantly impact reaction rates and therefore change how quickly equilibrium is achieved.