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 value of \(K_{p}\) is 4.

Step 1: Define Initial Concentrations

Start by defining the initial concentrations of A and B. Let the initial concentration of A be \([A] = x\). Given that the initial concentration of B is 1.5 times that of A, its concentration would be \([B] = 1.5x\).

Step 2: Determine Changes in Concentration

As the reaction proceeds to equilibrium, let the change in the concentration of A be \(-y\). Thus, the change for B is \(-2y\), for C is \(+2y\), and for D is \(+y\).

Step 3: Express Equilibrium Concentrations

The equilibrium concentration for A is \([A]_e = x - y\). For B, it is \([B]_e = 1.5x - 2y\). Given the problem states that the equilibrium concentrations of A and B are equal, set these equations equal: \(x - y = 1.5x - 2y\).

Step 4: Solve for y

Rearrange the equation \(x - y = 1.5x - 2y\) to solve for y: \(x - 1.5x = -2y + y\). Simplifying, we get \(-0.5x = -y\), so \(y = 0.5x\).

Step 5: Calculate Equilibrium Concentrations for All Species

Substitute \(y = 0.5x\) back to found the equilibrium concentrations: - \([A]_e = x - 0.5x = 0.5x\)- \([B]_e = 1.5x - 2(0.5x) = 0.5x\)- \([C]_e = 2(0.5x) = x\)- \([D]_e = 0.5x\).

Step 6: Write the Equilibrium Expression for Kp

For the given reaction, the equilibrium constant expression in terms of partial pressure is: \[ K_p = \frac{[C]^2[D]}{[A][B]^2} \].

Step 7: Calculate Kp

Substitute the equilibrium concentrations: \[ K_p = \frac{(x)^2(0.5x)}{(0.5x)(0.5x)^2} \].This simplifies to \(K_p = \frac{0.5x^3}{0.125x^3} = 4\).

Key Concepts

Equilibrium Constant

In a chemical reaction at equilibrium, the equilibrium constant, denoted as \(K_p\) for reactions involving gases, is a crucial concept. It is an expression that reflects the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced equation. For our reaction, \( \mathrm{A} + 2 \mathrm{B} \rightleftharpoons 2 \mathrm{C} + \mathrm{D} \), the equilibrium constant in terms of partial pressures is expressed as:\[ K_p = \frac{[C]^2[D]}{[A][B]^2} \]This expression tells us how the concentrations of the reactants and products relate at equilibrium. For instance, in this particular case, the simplification of the expression \( K_p = \frac{0.5x^3}{0.125x^3} = 4 \) gives insight into the proportionate levels of each species. It's calculated by substituting the equilibrium concentrations into the equilibrium expression. Understanding \( K_p \) is essential because it beautifully captures the reaction's intrinsic properties at equilibrium state. This simplification assumes that the ratio of product concentrations to reactant concentrations stays constant at a given temperature, emphasizing the balanced nature of chemical equilibrium.

Stoichiometry

Stoichiometry is a fundamental aspect of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. This concept helps us predict the quantities of substances consumed and produced in a reaction. In our reaction \(\mathrm{A} + 2 \mathrm{B} \rightleftharpoons 2 \mathrm{C} + \mathrm{D} \), the coefficients indicate the stoichiometric ratios: one mole of \(\mathrm{A}\) reacts with two moles of \(\mathrm{B}\) to form two moles of \(\mathrm{C}\) and one mole of \(\mathrm{D}\).When calculating equilibrium concentrations, stoichiometry is essential. For every change \(-y\) in \(\mathrm{A}\), there is a corresponding change of \(-2y\) in \(\mathrm{B}\), and a gain of \(+2y\) and \(+y\) in \(\mathrm{C}\) and \(\mathrm{D}\), respectively. Understanding this allows us to correctly express initial concentrations and compute changes to reach equilibrium accurately. Stoichiometry provides the framework to ensure that equations are balanced and the law of conservation of mass is adhered to across reactions.

Le Chatelier's Principle

Le Chatelier's Principle is a key concept in understanding how systems at equilibrium respond to external changes. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and restore a new balance.In our reaction setup, where the initial concentration of \(\mathrm{B}\) is \(1.5\) times that of \(\mathrm{A}\), if any change occurs—such as a variation in pressure, concentration, or temperature—the system will adjust. For example, if the concentration of \(\mathrm{A}\) were increased, the equilibrium would shift toward producing more \(\mathrm{C}\) and \(\mathrm{D}\) to balance the change.Understandably, decreases or increases in any one of the reactants or products can cause the system to adjust in a way that minimizes the impact of the change. This principle not only guides us in predicting the direction of shift but is also crucial in industrial processes where maximizing product yield is essential. Recognizing how Le Chatelier's Principle affects reactions provides valuable insights into how reactions can be controlled and utilized effectively.