Problem 37 - A Complete Resource Book in Chemistry for JEE Main

Question

At constant temperature, the equilibrium constant \(\left(\mathrm{K}_{p}\right)\) for the decomposition reaction, \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}\) is expressed by \(\mathrm{K}_{\mathrm{p}}=\left(4 \mathrm{x}^{2} \mathrm{P}\right) /\left(1-\mathrm{x}^{2}\right)\), where \(\mathrm{P}=\) pressure, \(\mathrm{x}=\) extent of decomposition. Which one of the following statements is true? (a) \(\mathrm{K}\) increases with increase of \(\mathrm{P}\) (b) \(\mathrm{K}_{\mathrm{p}}\) increases with increase of \(\mathrm{x}\) (c) \(K_{p}\) increases with decrease of \(x\) (d) \(\mathrm{K}_{\text {p }}\) remains constant with change in \(\mathrm{P}\) and \(\mathrm{x}\)

Step-by-Step Solution

Answer

(d) \( K_p \) remains constant with change in \( P \) and \( x \).

Step 1: Understand the Equation for Kp

The equation given is \( K_p = \frac{4x^2 P}{1-x^2} \), where \( P \) is pressure and \( x \) is the extent of decomposition. \( K_p \) is the equilibrium constant at a constant temperature.

Step 2: Evaluate the Effect of Pressure (P)

Since the equation does not have pressure \( P \) changing the form of \( K_p \)'s dependence other than by being a multiplicative factor, we must analyze if there are any direct implications: \( K_p \) is linearly dependent on \( P \) but not enough for a true increase depending on \( P \). Generally, for equilibrium constants at a given temperature, \( K_p \) does not change with pressure.

Step 3: Evaluate the Effect of Extent of Decomposition (x)

The equation shows \( x^2 \) both in the numerator as \( 4x^2P \) and in the denominator as \( 1-x^2 \). Hence, any increase or decrease in \( x \) will cause a corresponding nonlinear increase or decrease in both the numerator and denominator, but \( K_p \) is not dependent on \( x \) as it simplifies with coefficients and doesn’t truly increase or decrease \( K_p \).

Step 4: Determine the Correct Statement

Analyzing the dependencies: \( K_p \) is defined for a constant temperature and is independent of \( P \) and \( x \) under ideal equilibrium conditions. Thus, option (d) is the suitable response, given that equilibrium constants at constant temperature do not depend on pressure or extent of reaction under the guise of this formulation.

Key Concepts

Equilibrium Constant

The equilibrium constant, symbolized as \( K_p \) when dealing with gases, is a crucial parameter in chemical equilibrium. It is a measure of the ratio between the concentrations of products and reactants in a chemical reaction at equilibrium. Specifically, at a given temperature, \( K_p \) remains constant, hence it is often used to predict the direction of a reaction and whether the reactants or products are favored at equilibrium.

The equation for \( K_p \) in the given exercise describes its relationship to the extent of decomposition and pressure for the reaction \( \text{N}_2\text{O}_4 \rightleftharpoons 2 \text{NO}_2 \). This indicates how the equilibrium constant can incorporate different variables using formulas that reflect the actual chemical process. However, it's important to note that \( K_p \) is independent of initial concentrations and pressures since it is defined only by the temperature at which the system is maintained.

Decomposition Reaction

A decomposition reaction involves a single compound breaking down into two or more products. In the exercise's specific reaction, \( \text{N}_2\text{O}_4 \) decomposes into nitrogen dioxide \( \text{NO}_2 \). Decomposition is usually influenced by factors like temperature, pressure, and the presence of catalysts.

This type of reaction is characterized by a change from a single reactant to multiple products, and it often releases or absorbs energy. The decomposition of \( \text{N}_2\text{O}_4 \) to form \( \text{NO}_2 \) involves the breaking of chemical bonds, which may require energy input or release energy, adding complexity to equilibrium calculations and reactions in open and closed systems.

Effect of Pressure on Equilibrium

In gas-phase reactions like the decomposition of \( \text{N}_2\text{O}_4 \), pressure can play a significant role. However, according to Le Chatelier's Principle, changing the total pressure in a closed system only influences the position of equilibrium if the number of moles of gas differs between reactants and products.

In this case, since the forward reaction results in an increase in moles of gas (from 1 mole of \( \text{N}_2\text{O}_4 \) to 2 moles of \( \text{NO}_2 \)), increasing pressure will shift the equilibrium towards the formation of \( \text{N}_2\text{O}_4 \), favoring the side with fewer moles of gas. Still, the equilibrium constant \( K_p \) itself does not change with pressure because it is fundamental to the reaction at a specific temperature, as noted in the exercise solution.

Extent of Reaction

The extent of reaction, expressed as a variable \( x \) in the provided equation, represents how far the decomposition process proceeds toward products. It is a dimensionless number that can vary between 0 (no reaction) and 1 (complete reaction), depending on the system's conditions.

In the context of equilibrium and the equation \( K_p = \frac{4x^2 P}{1-x^2} \), changes in \( x \) affect both the numerator and the denominator. \( x \) represents the fraction of \( \text{N}_2\text{O}_4 \) decomposed into \( \text{NO}_2 \). Although this parameter is significant for understanding how much reaction has taken place, \( K_p \) itself, being dependent mainly on temperature, remains unaffected by the extent \( x \) when considering the equilibrium state.