Problem 118 - A Complete Resource Book in Chemistry for JEE Main

Question

A gaseous phase reaction is allowed to attain equilibrium as \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\) at constant pressure P. The partial pressure of \(\mathrm{A}\) at equilibrium is \(\mathrm{P} / 2\). The value of equilibrium constant \(K_{p}\) is (a) \(\frac{\mathrm{P}}{8}\) (b) \(\frac{\mathrm{P}}{6}\) (c) \(\frac{\mathrm{P}}{2}\) (d) \(\frac{\mathrm{P}}{4}\)

Step-by-Step Solution

Answer

The value of the equilibrium constant, \( K_p \), is \( \frac{P}{8} \).

Step 1: Initial Setup and Reaction Expression

Given the reaction \( \text{A}(\text{g}) \rightleftharpoons \text{B}(\text{g}) + \text{C}(\text{g}) \), we know the initial approach to equilibrium and can express the equilibrium pressures. The total pressure is \( P \). At equilibrium, \( P_A = \frac{P}{2} \), where \( P_A \) is the partial pressure of \( \text{A} \).

Step 2: Express Partial Pressures for B and C

Since \( \text{A} \) decomposes into \( \text{B} \) and \( \text{C} \), if \( \text{A} \) decreases by \( x \), both \( \text{B} \) and \( \text{C} \) will increase by \( x \). At equilibrium, \( x = \frac{P}{2} \). Since the total pressure remains \( P \), we solve for \( x \): \( P_B = \frac{P}{4} \) and \( P_C = \frac{P}{4} \).

Step 3: Plug Values into Equilibrium Constant Expression

The equilibrium constant \( K_p \) in terms of partial pressures is expressed as \( K_p = \frac{P_B \times P_C}{P_A} = \frac{\left(\frac{P}{4}\right) \times \left(\frac{P}{4}\right)}{\frac{P}{2}} \).

Step 4: Solve for K_p

Calculate \( K_p \): \( K_p = \frac{\frac{P^2}{16}}{\frac{P}{2}} = \frac{P^2}{16} \times \frac{2}{P} = \frac{P}{8} \). Simplify the expression to find \( K_p = \frac{P}{8} \).

Key Concepts

Gaseous Phase Reaction

In a gaseous phase reaction, reactants and products are all in the gas phase. This is important as it influences the behavior of the reaction, especially when considering factors like pressure and volume.
Gaseous reactions occur when gas molecules collide with enough energy to cause a chemical change. These reactions follow the principles of kinetic molecular theory. This theory also guides how we calculate and predict the rates of these reactions.
Since all participants in the reaction are gases, their behaviors follow the ideal gas law: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the gas constant, and \( T \) is temperature. Understanding this law is essential when calculating partial pressures in equilibrium reactions involving gases.

Partial Pressure

Partial pressure is the pressure that each gas in a mixture would exert if it alone occupied the entire volume. This concept comes from Dalton's Law of Partial Pressures.
Dalton's Law states that the total pressure of a gas mixture is the sum of the partial pressures of each individual component. In a gaseous equilibrium, each component contributes to the overall pressure. It's important in reactions like \( \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g}) \) because the partial pressures help determine the direction and extent of the reaction.
Calculations often involve subtracting or adding known pressures to find unknown partial pressures. These pressures then feed into equilibrium constant expressions like \( K_{p} \). In the example, since the partial pressure of A is known, we used it to deduce the pressures of B and C.

Chemical Equilibrium

Chemical equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction. This results in constant concentrations of reactants and products.
In the reaction \( \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g}) \), equilibrium establishes when the forward transformation of A into B and C matches the rate at which B and C transform back into A.
At chemical equilibrium, changes in pressure, temperature, or concentration could shift the equilibrium state according to Le Chatelier's principle. The equilibrium constant, \( K_{p} \) for gaseous systems, is specific at a given temperature and shows the ratio of product pressures to reactant pressures at equilibrium.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one reactant breaks down into two or more products. This process often requires energy, such as heat, light, or electricity, to occur.
In our example, the gaseous phase reaction \( \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g}) \) is a decomposition process. One molecule of A decomposes into one molecule of B and one molecule of C.
Decomposition reactions are critical in chemistry for understanding how complex molecules break into simpler ones. They also play a significant role in chemical equilibria, influencing how the reaction moves towards reaching a stable state. Understanding these reactions helps predict what conditions control the breakdown and formation of molecules.