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) \(\mathrm{K}_{\text {p increases with increase of } \mathrm{P}}\) (b) \(K_{p}\) increases with increase of \(x\) (c) \(\mathrm{K}_{\mathrm{p}}\) increases with decrease of \(\mathrm{x}\) (d) \(\mathrm{K}_{\text {p }}\) remains constant with change in \(\mathrm{P}\) and \(\mathrm{x}\)

Step-by-Step Solution

Answer

\(K_{p}\) increases with increase of \(x\) (option b).

Step 1: Understanding the Given Expression

The equilibrium expression given is \(K_{p}=\frac{4x^{2}P}{1-x^{2}}\), where \(P\) is pressure and \(x\) is the extent of decomposition. We are to analyze the impact of changes in \(P\) and \(x\) on \(K_{p}\).

Step 2: Analyzing Effect of Pressure on \(K_{p}\)

To determine if \(K_{p}\) changes with pressure \(P\), observe that \(K_{p}\) contains \(P\) directly in its expression. Therefore, \(K_{p}\) does change with changes in pressure due to its dependency on \(P\). Thus, option (d) is incorrect.

Step 3: Analyzing Effect of Extent of Decomposition on \(K_{p}\)

Considering the expression \(K_{p}=\frac{4x^{2}P}{1-x^{2}}\), note that if \(x\) changes, the terms \(4x^2\) and \(1-x^2\) both vary, influencing \(K_{p}\). An increase in \(x\) results in increased numerator value \(4x^2P\) and decreased denominator \(1-x^2\), leading to an increase in \(K_{p}\). Therefore, option (b) is true.

Key Concepts

Equilibrium Constant (Kp)

The equilibrium constant, represented as \(K_p\), is a crucial part of understanding chemical equilibria. It quantifies the ratio of the concentrations of products to reactants at equilibrium. Specifically, in gas-phase reactions like the decomposition of \(\text{N}_2\text{O}_4\) to \(2\text{NO}_2\), \(K_p\) is defined based on partial pressures. The given formula for \(K_p\) is:\[K_{p}=\frac{4x^{2}P}{1-x^{2}}\]Here, \(P\) represents the pressure, and \(x\) denotes the extent of decomposition.

• This equation suggests that \(K_p\) is dependent on both \(P\) and \(x\).
• It is not a constant that is immune to changes, but rather, it actively varies with system conditions.
• Understanding how \(K_p\) responds to different variables helps predict the behavior of the system under various conditions.

This understanding is fundamental for predicting how a chemical reaction will shift when external conditions change.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where a single compound breaks down into two or more simpler substances. In this context, we examine the chemical equation:\[\text{N}_2\text{O}_4 \rightleftharpoons 2 \text{NO}_2\]

• Here, dinitrogen tetroxide (\(\text{N}_2\text{O}_4\)) decomposes into nitrogen dioxide (\(\text{NO}_2\)).
• This is a reversible reaction, indicated by the equilibrium symbol (\(\rightleftharpoons\)).
• In decomposition reactions such as this, the forward reaction involves breaking chemical bonds, while the reverse reaction involves bond formation.

The decomposition process is heavily influenced by factors such as temperature, pressure, and the extent of decomposition, which directly impacts the equilibrium constant \(K_p\). Understanding the impact of these factors allows chemists to manipulate the conditions to favor either the forward or reverse reaction, depending on the desired outcome.

Effect of Pressure and Extent of Decomposition

Pressure and the extent of decomposition play significant roles in how a chemical equilibrium behaves. Let's take a closer look:

• Pressure (P): In the expression \(K_{p}=\frac{4x^{2}P}{1-x^{2}}\), pressure is clearly a variable within the formula. As pressure changes, \(K_p\) is directly affected because it multiplies the numerator, which results in a change in \(K_p\).
• Extent of Decomposition (x): The extent of decomposition, \(x\), also influences \(K_p\). As \(x\) increases, both the numerator \(4x^2P\) increases and the denominator \(1-x^2\) decreases, thus elevating \(K_p\). Conversely, a decrease in \(x\) would decrease the value of \(K_p\).

Recognizing the link between pressure and extent of decomposition means acknowledging that equilibrium is dynamic. Changes in pressure or the extent of decomposition can lead to a shift in equilibrium, affecting the concentrations of reactants and products. This knowledge is vital for anticipating how a reaction mixture will respond under varying experimental conditions.