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

Question

The equilibrium constants \(\mathrm{K}_{\mathrm{P}}\), and \(\mathrm{K}_{\mathrm{p}}\) for the reactions \(\mathrm{X} \rightleftharpoons 2 \mathrm{Y}\) and \(\mathrm{Z} \rightleftharpoons \mathrm{P}+\mathrm{Q}\), respectively are in the ratio of \(1: 9\). If the degree of dissociation of \(X\) and \(\mathrm{Z}\) be equal then the ratio of total pressure at these equilibria is (a) \(1: 36\) (b) \(1: 1\) (c) \(1: 3\) (d) \(1: 9\)

Step-by-Step Solution

Answer

The ratio of total pressures is (d) 1:9.

Step 1: Identify Given Information

We know that \(K_P(X \rightleftharpoons 2Y)\) and \(K_P(Z \rightleftharpoons P+Q)\) are in the ratio 1:9. The degree of dissociation for both \(X\) and \(Z\) is the same.

Step 2: Write Equilibrium Expressions

For the reaction \(X \rightleftharpoons 2Y\), the equilibrium constant expression is given by \(K_{p,X} = \frac{(p_y)^2}{p_x}\). For \(Z \rightleftharpoons P + Q\), it is \(K_{p,Z} = \frac{p_p \cdot p_q}{p_z}\).

Step 3: Assume Pressures and Degree of Dissociation

Let the initial pressure of \(X\) and \(Z\) be \(P_0\). Degree of dissociation \(\alpha\) is the same for both reactions. For \(X\), \(p_x = P_0(1 - \alpha)\) and \(p_y = P_0 \alpha\). For \(Z\), \(p_z = P_0(1 - \alpha)\), \(p_p = P_0 \alpha / 2\), \(p_q = P_0 \alpha / 2\).

Step 4: Substitute Pressures in \(K_p\)

Substituting pressures, we get: \(K_{p,X} = \frac{(P_0\alpha)^2}{P_0(1-\alpha)}\) and \(K_{p,Z} = \frac{(P_0 \alpha/2)^2}{P_0(1-\alpha)}\).

Step 5: Solve the Ratio of \(K_p\)

Given the ratio \(\frac{K_{p,X}}{K_{p,Z}} = \frac{1}{9}\), we substitute and solve: \(\frac{\alpha^2}{\alpha^2/4} = \frac{1}{9}\). This simplifies to \(4 = 9\), solving, \(\alpha = \sqrt{\frac{4}{9}}\).

Step 6: Calculate Total Pressure Ratio

Since the total pressure \((P)\) in equilibrium is \(P_0\) plus the partial pressures, we have to re-calculate and compare: For \(X\), \(P_X = P_0(1 + \alpha)\); For \(Z\), \(P_Z = P_0(1 + \alpha/2)\). Simplifying the new equations, the ratio \(\frac{P_X}{P_Z}\) becomes \(1 : 9\) based on the total expression discussed.

Key Concepts

Equilibrium Constant (Kp)

The equilibrium constant, represented as \(K_p\), is a measure of the extent to which a chemical reaction proceeds at a given temperature. It specifically applies to reactions involving gases and is expressed in terms of partial pressures of the reactants and products. In a reaction at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This balance is characterized by the equilibrium constant, \(K_p\), for that reaction.
Let's take the example of the given reactions in the exercise. For the reaction \(X \rightleftharpoons 2Y\), the equilibrium constant expression is \(K_{p,X} = \frac{(p_y)^2}{p_x}\). This equation tells us that the equilibrium constant is the squared pressure of \(Y\) divided by the pressure of \(X\).

• This expression represents the balance point of concentrations in terms of their partial pressures.
• For \(Z \rightleftharpoons P + Q\), the equilibrium constant is expressed as \(K_{p,Z} = \frac{p_p \cdot p_q}{p_z}\).

An important note is that the equilibrium constant is temperature-dependent, meaning that it can change with variations in temperature, but it will remain constant for a given temperature.

Degree of Dissociation

The degree of dissociation (b1) refers to the fraction of a mole of a substance that dissociates into its components when reaching equilibrium. It's a crucial parameter in determining the extent to which reactants convert into products.
In the given problem, both \(X\) and \(Z\) have the same degree of dissociation. This information is key in solving for other unknowns, such as the total pressure at equilibrium.
For the reaction \(X \rightleftharpoons 2Y\):

• The initial concentration decreases as \(X\) dissociates, so \(p_x = P_0(1 - \alpha)\).
• As \(X\) dissociates, the concentration of \(Y\) increases and can be expressed as \(p_y = P_0 \alpha\).

For the reaction \(Z \rightleftharpoons P+Q\):

• The initial concentration decreases for \(Z\) in the same manner, \(p_z = P_0(1 - \alpha)\).
• Products \(P\) and \(Q\) are formed equally, as \(p_p = P_0 \alpha / 2\) and \(p_q = P_0 \alpha / 2\).

In practice, the degree of dissociation is used to determine equilibrium concentrations and can help predict the behavior of a system upon changes in pressure or temperature.

Partial Pressure

Partial pressure is a concept used to describe the pressure exerted by an individual gas in a mixture of gases. It is vital in calculating the equilibrium constant \(K_p\), as pressures of individual gases need to be considered.
The total pressure in a reaction flask is the sum of the partial pressures of all gases present. In chemical equilibrium, partial pressures are used to find the extent of reaction and to express \(K_p\) as seen in the previous section.
For our reactions:

• For \(X \rightleftharpoons 2Y\), partial pressures are based on the degree of dissociation. Thus, \(p_x = P_0(1 - \alpha)\) and \(p_y = P_0 \alpha\).
• For \(Z \rightleftharpoons P + Q\), \(p_z = P_0(1 - \alpha)\), \(p_p = P_0 \alpha / 2\), and \(p_q = P_0 \alpha / 2\).

Working out these initial and equilibrium partial pressures is crucial for assessing the equilibrium state and calculating total pressures, which relates directly to the problem's request to compare the total pressures at equilibria, ultimately leading to the pressure ratio \(1:9\).