Problem 92 - A Complete Resource Book in Chemistry for JEE Main

Question

At Kp for the following reaction is 1 atm \(\mathrm{X}(\mathrm{g}) \rightleftharpoons \mathrm{Y}(\mathrm{g})+\mathrm{Z}(\mathrm{g})\) At equilibrium, \(50 \%\) of \(X(g)\) is dissociated. The total pressure of the equilibrium system is 'P' atm. what is the partial pressure (in atm) of \(X(\mathrm{~g})\) at equilibrium? (a) 1 (b) 4 (c) 2 (d) \(0.5\)

Step-by-Step Solution

Answer

The partial pressure of \( X(g) \) at equilibrium is 1 atm.

Step 1: Understand the Reaction

The chemical reaction given is a decomposition reaction of \( X(g) \) into \( Y(g) \) and \( Z(g) \): \[ \mathrm{X}(\mathrm{g}) \rightleftharpoons \mathrm{Y}(\mathrm{g}) + \mathrm{Z}(\mathrm{g}) \] The equilibrium constant \( K_p \) is given as 1 atm. This means at equilibrium, the partial pressures of the products raise to maintain the same pressure as the reactant.

Step 2: Set up the Initial and Change in Pressure

Let's assume initially, the pressure of \( X(g) \) is \( P_0 \) atm. Since 50% of \( X(g) \) dissociates, the pressure change for \( X(g) \) is \(-0.5P_0\). Hence at equilibrium: - Pressure of \( X(g) \) = \( 0.5P_0 \) - Pressure of \( Y(g) \) = \( 0.5P_0 \) - Pressure of \( Z(g) \) = \( 0.5P_0 \)

Step 3: Express Total Pressure

The total pressure \( P \) of the system at equilibrium is the sum of partial pressures of all gases: \[ P = 0.5P_0 + 0.5P_0 + 0.5P_0 = 1.5P_0 \] This equation represents the total pressure of the system.

Step 4: Calculate Partial Pressure of X

We know from \( K_p \), the relation:\[ K_p = \frac{P_{Y} \cdot P_{Z}}{P_{X}} = \frac{(0.5P_0)(0.5P_0)}{0.5P_0} = 0.5P_0 \]At equilibrium, \( K_p \) is given as 1 atm, so:\[ 0.5P_0 = 1 \rightarrow P_0 = 2 \] Therefore, partial pressure of \( X(g) \) at equilibrium is:\[ 0.5P_0 = 1 \text{ atm} \]

Step 5: Verify the Solution

Verify by checking if the initial setup and calculations satisfy the given Kp and equilibrium conditions. With \( P_0 = 2 \) atm, partial pressures at equilibrium become:\( P_X = 1 \) atm, \( P_Y = 1 \) atm, and \( P_Z = 1 \) atm. Adding up gives \( 1.5 \times 2 = 3 \) atm total, which works according to our calculations. The equilibrium setup is consistent and valid.

Key Concepts

Equilibrium Constant (Kp)

The equilibrium constant, denoted by \( K_p \), is a crucial concept in understanding chemical equilibrium in gaseous reactions. Specifically, \( K_p \) involves the partial pressures of gases, unlike \( K_c \), which is based on concentrations. When dealing with a reversible reaction like the decomposition of \( \, X(g) \) into \( \, Y(g) + Z(g) \), \( K_p \) helps quantify the balance between reactants and products at equilibrium.

In this context, \( K_p \) is defined as the ratio of the product of partial pressures of the products over the reactants. For the reaction \( X(g) \rightleftharpoons Y(g) + Z(g) \), the equilibrium constant is expressed as:

• \( K_p = \frac{P_{Y} \cdot P_{Z}}{P_{X}} \)

Here, \( P_{Y} \), \( P_{Z} \), and \( P_{X} \) are the partial pressures of \( Y \), \( Z \), and \( X \) respectively. Understanding how these pressures interact allows you to predict the behavior of the system at equilibrium.

The given \( K_p \) of 1 atm reflects that at equilibrium, the pressures of the products are balanced with that of the reactant to maintain the pressure constant. This equilibrium constant plays a pivotal role in determining how much of a reactant dissociates and how the pressure shifts can occur in the system.

Partial Pressure Calculation

Partial pressure is the pressure that each gas in a mixture would exert if it occupied the entire volume alone. In the context of chemical equilibrium, calculating partial pressures is essential to understand how gases distribute themselves at a given temperature and total pressure.

For the given reaction: \( X(g) \rightleftharpoons Y(g) + Z(g) \), we can calculate partial pressures based on initial conditions and the changes during the reaction:

• Initially, assume \( X(g) \) has a pressure \( P_0 \).
• If 50% of \( X(g) \) dissociates, its pressure decreases by \( 0.5P_0 \), leaving \( 0.5P_0 \) at equilibrium.
• The pressures of \( Y(g) \) and \( Z(g) \) each become \( 0.5P_0 \) from the dissociation.

At equilibrium, the partial pressures of all gases need to be added to get the total pressure \( P \):
\[ P = 0.5P_0 + 0.5P_0 + 0.5P_0 = 1.5P_0 \]
This total pressure setup helps to visualize the actual distribution of gases, which is critical for using the given \( K_p \) value correctly.

Decomposition Reaction

A decomposition reaction, such as \( X(g) \rightleftharpoons Y(g) + Z(g) \), involves breaking down a single compound into simpler substances. It's essential to fully grasp how equilibrium in such reactions works to predict the behavior of the system.

In our exercise, the decomposition of \( X(g) \) yields two products, \( Y(g) \) and \( Z(g) \), and reaches equilibrium when 50% of the initial \( X(g) \) is converted. This type of reaction typically involves an energy change, frequently requiring heat or another form of energy to proceed.

Understanding decomposition reactions involves:

• Identifying the initial and final states.
• Recognizing how quantitatively the reactant converts to products (e.g., 50% dissociation).
• Applying the equilibrium constant to calculate partial and total pressures.

Such reactions are reversible, meaning the system can shift to favor either the reactants or the products under different conditions of pressure, temperature, or concentration. Recognizing when equilibrium is achieved and anticipating its effects on the system is central to mastering decomposition reactions.