Problem 127 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The ratio of total pressures is 1:36, matching option (a).

Step 1: Identify the Given Information

We are given two expressions for equilibrium constants, \(K_{p_1}\) for the reaction \(\mathrm{X}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Y}(\mathrm{g})\) and \(K_p\) for the reaction \(\mathrm{Z}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g})+\mathrm{Q}(\mathrm{g})\). We know that the ratio of these equilibrium constants \(K_{p_1}:K_p = 1:9\) and that the degrees of dissociation for \(\mathrm{X}\) and \(\mathrm{Z}\) are equal.

Step 2: Write the Expression for Equilibrium Constants

The equilibrium constant \(K_{p_1}\) for \(\mathrm{X}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Y}(\mathrm{g})\) is \(K_{p_1} = \left(\frac{P_Y^2}{P_X}\right)\) and for \(\mathrm{Z}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g}) + \mathrm{Q}(\mathrm{g})\), it is \(K_p = \left(\frac{P_P \cdot P_Q}{P_Z}\right)\).

Step 3: Use Degree of Dissociation

Assuming the degree of dissociation is \(\alpha\), for \(\mathrm{X}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Y}(\mathrm{g})\) initial pressure is \(P_1\), the partial pressures become \(P_X = P_1(1-\alpha)\) and \(P_Y = 2P_1\alpha\). Similarly, for \(\mathrm{Z}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g}) + \mathrm{Q}(\mathrm{g})\), \(P_P = P_2\alpha\) and \(P_Q = P_2\alpha\) when initial pressure is \(P_2\).

Step 4: Substitute Partial Pressures into \(K_{p_1}\) and \(K_p\)

Substitute the expressions from Step 3 into the formulas for \(K_{p_1}\) and \(K_p\): For \(K_{p_1}\), we have:\[K_{p_1} = \frac{(2P_1\alpha)^2}{P_1(1-\alpha)}\]For \(K_p\), we have:\[K_p = \frac{(P_2\alpha)(P_2\alpha)}{P_2(1-\alpha)}\]

Step 5: Simplify to Find Pressure Ratio

The expressions simplify to \[K_{p_1} = \frac{4P_1^2\alpha^2}{P_1(1-\alpha)}\] which further simplifies to \[K_{p_1} = \frac{4P_1\alpha^2}{1-\alpha}\]Similarly, \[K_p = \frac{P_2^2\alpha^2}{P_2(1-\alpha)} = \frac{P_2\alpha^2}{1-\alpha}\]Using the given ratio \(K_{p_1}:K_p = 1:9\), solve \[\frac{1}{9} = \frac{4P_1\alpha^2}{P_2\alpha^2} = \frac{4P_1}{P_2}\]Thus, \[\frac{P_1}{P_2} = \frac{1}{36}\].

Step 6: Identify the Correct Answer

Comparing our result to the options given, \(\frac{P_1}{P_2} = \frac{1}{36}\) which matches with option (a).

Key Concepts

Degree of Dissociation

The degree of dissociation refers to the fraction of a mole of reactant that breaks down into its products in a chemical reaction. It is denoted by the symbol \(\alpha\). Understanding the degree of dissociation is important in gaseous reactions since it indicates how much of the original substance has dissociated.

In the reaction \(\mathrm{X}(\mathrm{g}) \rightleftharpoons 2\mathrm{Y}(\mathrm{g})\), if \(\alpha\) is the degree of dissociation, then initially there might be a mole of \(\mathrm{X}\). As the reaction proceeds, \(\alpha\) moles of \(\mathrm{X}\) dissociate, creating \(2\alpha\) moles of \(\mathrm{Y}\), considering the stoichiometry that doubles. The initial pressure \(P_1\) of \(\mathrm{X}\) then leads to partial pressures:

• \(P_X = P_1(1-\alpha)\) for the undissociated \(\mathrm{X}\).
• \(P_Y = 2P_1\alpha\) for the formed \(\mathrm{Y}\).

If two reactions have the same degree of dissociation, it implies they are at a similar stage of progress in terms of conversion of reactants to products.

With \(\mathrm{Z}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g})+\mathrm{Q}(\mathrm{g})\), the dissociation also leads to equal moles of \(\mathrm{P}\) and \(\mathrm{Q}\), calculated as \(\alpha P_2\) each, thus controlling how we evaluate changes in total pressure.

Pressure Ratio

Understanding the pressure ratio is essential in evaluating the behavior of gases at equilibrium. Pressure ratio in the context of these chemical reactions refers to the ratio of the total pressures at equilibrium states for two different reactions.

In our provided reactions, the pressure ratio \(\frac{P_1}{P_2} = \frac{1}{36}\) determines the relative scale of pressures exerted by each reaction at equilibrium. In the reaction \(\mathrm{X}(\mathrm{g}) \rightleftharpoons 2\mathrm{Y}(\mathrm{g})\), the initial pressure \(P_1\) must be measured against \(P_2\) of the reaction \(\mathrm{Z}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g})+\mathrm{Q}(\mathrm{g})\), taking into account their differing stoichiometries.This relationship informs us that despite having the same degree of dissociation \(\alpha\), their equilibrium pressures differ due to the stoichiometric coefficients and other intrinsic reaction properties. The equilibrium constant dependencies on pressures (partial or total) directly influence the ratio, found through simplifying equilibrium expressions:

• For \(K_{p_1}\), it's expressed as \( \frac{4P_1\alpha^2}{1-\alpha} \).
• For \(K_{p}\), it's expressed as \( \frac{P_2\alpha^2}{1-\alpha} \).
• Since the given \(K_{p_1}:K_{p} = 1:9\), the pressures align accordingly \(\frac{1}{36}\).

Gaseous Reactions

Gaseous reactions involve reactions between substances in the gaseous state. They are particularly interesting because of their dependence on parameters like pressure and temperature, which can significantly alter the behavior of the reaction equilibrium.In our example reactions:- \(\mathrm{X}(\mathrm{g}) \rightleftharpoons 2\mathrm{Y}(\mathrm{g})\)- \(\mathrm{Z}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g})+\mathrm{Q}(\mathrm{g})\)we observe how these reactions reach a balance between the forward and backward steps. Equilibrium in gaseous reactions is influenced by the balance between products formed and reactants consumed, captured through their equilibrium constants \(K_{p_1}\) and \(K_{p}\).

The study of gaseous reactions is crucial when assessing how changing conditions like the pressure affect the state of equilibrium. The equilibrium constant expression depends on the partial pressures of reactants and products, revealing a direct link between the reaction's stoichiometry and the equilibrium state:

• The reaction \(\mathrm{X} \rightleftharpoons 2\mathrm{Y}\) indicates a significant increase in gas moles \((1 \rightarrow 2)\), affecting pressure dynamics.
• The reaction \(\mathrm{Z} \rightleftharpoons \mathrm{P} + \mathrm{Q}\) doubles the initial moles, maintaining simpler pressure dynamics by a 1:1 formation ratio in products.

Understanding these dynamics explains the complexity and outcomes seen in systems like our given reactions, highlighting pressure as a pivotal factor.