Problem 198 - A Complete Resource Book in Chemistry for JEE Main

Question

A reaction, \(\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{C}(\mathrm{g})+\mathrm{D}(\mathrm{g})\) was studied using an initial concentration of B which was \(1.5\) times that of \(\mathrm{A}\). But the equilibrium concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) were found to be equal. What is the value of \(K_{p}\) for above equilibrium?

Step-by-Step Solution

Answer

The equilibrium constant \( K_p \) is 16.

Step 1: Initial Concentrations

The initial concentration of \( \text{A} \) is \( x \). Given that the concentration of \( \text{B} \) is \( 1.5 \) times that of \( \text{A} \), it is \( 1.5x \).

Step 2: Equilibrium Condition

It is given that the equilibrium concentrations of \( \text{A} \) and \( \text{B} \) are equal. Let's call this equilibrium concentration \( y \). Therefore, \( x - a = 1.5x - 2a = y \).

Step 3: Express Changes in Terms of x

From \( x - a = 1.5x - 2a \), solving yields \( a = x/2 \). Thus, the changes are: \( A \) decreases by \( x/2 \), \( B \) decreases by \( x \), \( C \) increases by \( x \), and \( D \) increases by \( x/2 \).

Step 4: Write Equilibrium Concentrations

At equilibrium:- \( [A] = x - x/2 = x/2 \)- \( [B] = 1.5x - x = 0.5x \)- \( [C] = 2( x - x/2 ) = x \)- \( [D] = x/2 \)

Step 5: Set Up Expression for Kp

The equilibrium constant \( K_p \) is given by: \[ K_p = \frac{(P_C)^2 \cdot P_D}{P_A \cdot (P_B)^2} \] We substitute the equilibrium pressures (equal to the concentrations for this exercise) into the expression.

Step 6: Calculate Kp

Substitute the equilibrium values: \[ K_p = \frac{x^2 \cdot (x/2)}{(x/2) \cdot (0.5x)^2} \]Simplify the expression:\[ K_p = \frac{2x^3}{0.125x^3} = \frac{2}{0.125} = 16 \]

Step 7: Verify the Calculation

Review each step to ensure that the values were substituted correctly and calculations were performed accurately to confirm that \( K_p = 16 \).

Key Concepts

Equilibrium Constant (Kp)

In a reversible chemical reaction like \( \mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{C}(\mathrm{g})+\mathrm{D}(\mathrm{g}) \), the equilibrium constant \( K_p \) is a reflection of the relative concentrations of reactants and products once the reaction has reached equilibrium.
This constant is extremely useful in predicting the direction of the reaction at any given point.
Essentially, if \( K_p \) is significantly greater than one, the equilibrium position favors the products. If it is much less than one, it largely favors the reactants.
For our specific reaction, calculating \( K_p \) involved substituting the equilibrium concentrations of gases into a mathematical equation. The equation used is: \[K_p = \frac{(P_C)^2 \cdot P_D}{P_A \cdot (P_B)^2}\] Where \( P \) stands for the partial pressure (or concentration) of each species at equilibrium. By substituting these appropriate values, you can determine \( K_p \) as a ratio of product concentrations to reactant concentrations.

Initial Concentrations

Initial concentrations refer to the amounts of reactants present before the reaction reaches equilibrium. In this exercise, the initial concentrations are expressed in terms of a variable \( x \).
The problem statement tells us that the amount of substance B started at 1.5 times that of A. So, initially:

• \( [A]_0 = x \)
• \( [B]_0 = 1.5x \)

These initial concentrations are crucial because they help determine how the concentrations will change once equilibrium is reached. By understanding these initial values, you can predict and calculate how much of each reactant will remain and how much of the products will form.

Equilibrium Concentrations

Once a reaction reaches chemical equilibrium, the concentrations of all reactants and products remain constant over time. This doesn't mean the reaction stops; instead, it means the rates of the forward and reverse reactions are equal.
In our case, it is given that the equilibrium concentrations of A and B are equal. This equality becomes a clue in determining the specific change in concentration values as the reaction proceeds.When solving for equilibrium concentrations, we used the relation \( x - a = 1.5x - 2a = y \), leading to a change \( a \) as \( x/2 \). So, the equilibrium concentrations were found to be:

• \( [A] = x/2 \)
• \( [B] = 0.5x \)
• \( [C] = x \)
• \( [D] = x/2 \)

These equilibrium values are necessary for calculating \( K_p \) as they show the actual concentrations used in the equilibrium constant equation.