Problem 181 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant \(\mathrm{K}_{\mathrm{P}}\) and \(\mathrm{K}_{\mathrm{P}_{0}}\) for the reactions \(\mathrm{X}=2 \mathrm{Y}\) and \(\mathrm{Z} \rightleftharpoons{\mathrm{P}}+\mathrm{Q} ;\) respectively are in the ratio of \(1: 9\). If the degree of dissociation of \(\mathrm{X}\) and \(\mathrm{Z}\) be equal then calculate the value of \(\sqrt{\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}}\).

Step-by-Step Solution

Answer

\(\sqrt{2}\)

Step 1: Write the expressions for equilibrium constants

For the reaction \(\mathrm{X} = 2\mathrm{Y}\), the equilibrium constant \(\mathrm{K}_{\mathrm{P}}\) is given by \(\mathrm{K}_{\mathrm{P}} = \frac{p_Y^2}{p_X}\). For the reaction \(\mathrm{Z} \rightleftharpoons \mathrm{P} + \mathrm{Q}\), the equilibrium constant \(\mathrm{K}_{\mathrm{P0}}\) is given by \(\mathrm{K}_{\mathrm{P0}} = \frac{p_P \cdot p_Q}{p_Z}\).

Step 2: Use relation between equilibrium constants

We know that \(\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{K}_{\mathrm{P}_{0}}} = \frac{1}{9}\). Thus, \(\mathrm{K}_{\mathrm{P}} = \frac{\mathrm{K}_{\mathrm{P}_{0}}}{9}\).

Step 3: Define the degree of dissociation

Let \(\alpha\) be the degree of dissociation for both \(\mathrm{X}\) and \(\mathrm{Z}\). This means \(\mathrm{X} = 1 - \alpha\) and \(\mathrm{Z} = 1 - \alpha\). Substitute these into the expressions for \(p_Y, p_X, p_P, p_Q,\) and \(p_Z\).

Step 4: Calculate partial pressures using \(\alpha\)

For \(\mathrm{X} = 2\mathrm{Y}\), \(p_X = 1 - \alpha\) and \(p_Y = 2\alpha\). Similarly, for \(\mathrm{Z} \rightleftharpoons \mathrm{P} + \mathrm{Q}\), \(p_Z = 1 - \alpha\) and \(p_P = p_Q = \alpha\).

Step 5: Substitute and simplify

Substitute \(\alpha\) values into the expressions: \(\mathrm{K}_{\mathrm{P}} = \frac{(2\alpha)^2}{1-\alpha}\) and \(\mathrm{K}_{\mathrm{P}_{0}} = \frac{\alpha \cdot \alpha}{1-\alpha}\) to obtain \(4\alpha^2 = \frac{\alpha^2}{9}\). Simplify to find \(\alpha = \frac{1}{2}\).

Step 6: Calculate \(\sqrt{\frac{p_2}{p_1}}\)

Given that \(\alpha = \frac{1}{2}\), then \(p_Y = 1\), \(p_X = \frac{1}{2}\), and similarly \(p_P = p_Q = \frac{1}{2}\). Compute \(\frac{p_2}{p_1} = \frac{1}{\frac{1}{2}} = 2\), and therefore, \(\sqrt{\frac{p_2}{p_1}} = \sqrt{2}\).

Key Concepts

Degree of Dissociation

The degree of dissociation, commonly represented by the symbol \( \alpha \), is a measure of how much a substance has dissociated into its constituent particles in a chemical reaction. In simpler terms, it shows the extent to which a chemical compound breaks apart during a reaction at equilibrium. For instance, if a compound dissociates completely, \( \alpha \) would be 1, indicating all molecules have dissociated. Conversely, if no dissociation occurs, \( \alpha \) would be 0, signaling no breakdown occurred.

In our exercise, \( \alpha \) is used as the degree of dissociation for both reactions X = 2Y and Z \( \rightleftharpoons \) P + Q. We have calculated \( \alpha \) using given relationships to determine partial pressures and the value of \( \sqrt{\frac{p_2}{p_1}} \). Understanding \( \alpha \) helps us predict the behavior of reactants and products under equilibrium conditions.

Partial Pressure

Partial pressure refers to the pressure that a gas in a mixture would exert if it alone occupied the entire volume. In a mix of gases, each gas exerts a partial pressure, which contributes to the total pressure. This concept is crucial when dealing with gaseous equilibrium reactions because each gaseous participant in a reaction will contribute a specific partial pressure.

For the equilibrium reaction \( \mathrm{X} = 2\mathrm{Y} \), we assign partial pressures to \( \mathrm{X} \) and \( \mathrm{Y} \) as \( p_X \) and \( p_Y \) respectively. Similarly, for \( \mathrm{Z} \rightleftharpoons \mathrm{P} + \mathrm{Q} \), we denote the partial pressures as \( p_Z \), \( p_P \), and \( p_Q \). By substituting the degree of dissociation, \( \alpha \), into these partial pressures, we can easily plug them into the equilibrium constant expressions. This highlights how changes in \( \alpha \) affect the relative amounts of reactants and products and hence their respective partial pressures.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction goes in both directions at an equal rate, leading to stable concentrations of reactants and products. At this point, the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of all species involved remain constant.

In our example problems, we're dealing with two equilibrium reactions: \( \mathrm{X} = 2\mathrm{Y} \) and \( \mathrm{Z} \rightleftharpoons \mathrm{P} + \mathrm{Q} \). The system is at equilibrium when the specified concentrations and partial pressures of the reactants and products remain unchanged over time. Knowing this allows us to use the equilibrium constants (\( \mathrm{K}_\mathrm{P} \) and \( \mathrm{K}_{\mathrm{P}_0} \)) to relate the concentrations and calculate values like \( \sqrt{\frac{p_2}{p_1}} \) under equilibrium conditions.

Reaction Quotient

The reaction quotient, often represented by \( Q \), is a handy tool in chemistry to predict the direction a reaction at any point (not necessarily at equilibrium) will proceed to attain equilibrium. It is calculated using the same expression as the equilibrium constant, but with the current concentrations or partial pressures of the reactants and products.

In our reactions, calculating \( Q \) helps us determine whether the reaction mixture needs to shift towards products or reactants to reach equilibrium. If \( Q < K \), the forward reaction is favored; if \( Q > K \), the reverse reaction is favored. In our solved problem, using \( K \) allows a comprehensive understanding of reaction direction and degree of dissociation. Although \( Q \) is not explicitly calculated in the exercise, the fundamental concept underlines the importance of equilibrium constants in predicting reaction behavior.