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

Question

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

Step-by-Step Solution

Answer

Option (a) \(\mathrm{K}_{p}\) increases with increase of \(\mathrm{P}\).

Step 1: Understand the Formula

The equilibrium constant Kp for the decomposition reaction \(\mathrm{N}_2\mathrm{O}_4 \rightarrow 2 \mathrm{NO}_2\) is given by \(\mathrm{K}_{p}=\frac{4 x^2 P}{1-x^2}\). This formula shows how Kp is dependent on both the extent of decomposition \(x\) and the pressure \(P\).

Step 2: Analyze the Effect of Pressure \(P\) on \(\mathrm{K}_{p}\)

Examine the formula \(\mathrm{K}_{p}=\frac{4 x^2 P}{1-x^2}\). The pressure \(P\) is in the numerator, meaning it directly multiplies the entire term. This implies that as \(P\) increases, \(\mathrm{K}_{p}\) also increases.

Step 3: Analyze the Effect of Extent \(x\) on \(\mathrm{K}_{p}\)

Look at the components involving \(x\) in the formula. Both \(x^2\) terms appear in the numerator and denominator. These \(x^2\) terms will have a balancing effect as they are a part of both the numerator and denominator.

Step 4: Compare \(\mathrm{K}_{p}\)'s Dependencies

From the analysis, we see that \(\mathrm{K}_{p}\) increases directly with pressure \(P\) but is not dependent on \(x\) in a straightforward manner since the terms including \(x\) balance out. Therefore, \(\mathrm{K}_{p}\) is influenced by pressure \(P\), not just \(x\).

Step 5: Determine the Correct Statement

Given the analysis, \(\mathrm{K}_{p}\) increases with \(P\) but cannot be simply affected by changes in \(x\) without considering \(P\). So, the statement (a) "\(\mathrm{K}_{\mathrm{p}}\) increases with increase of \(\mathrm{P}\)" is true.

Key Concepts

Decomposition Reaction

In chemistry, a decomposition reaction plays an essential role in understanding how complex substances break down into simpler components. Specifically, in the context of equilibrium reactions, a decomposition reaction refers to a type of chemical reaction where a single compound breaks down into two or more simpler substances. For example, the decomposition reaction of \(_2o_4\) transforming into \(2 ext{NO}_2\) is a classic example. Here, a single molecule of dinitrogen tetraoxide (_2o_4) decomposes into two molecules of nitrogen dioxide ( ext{NO}_2).

The simplicity of these products supports the concept of chemical equilibrium, where not all of the reactant converts back to the product; hence, some of it exists as a mixture. For students, this foundational understanding helps in predicting and analyzing the behavior of chemicals over time, especially in a closed system where equilibrium is reached.

Pressure Dependency

The dependency of the equilibrium constant on pressure is a crucial concept when studying reactions involving gases. For gaseous reactions like the decomposition of \(n_2o_4\) to \(2 ext{NO}_2\), the pressure directly influences the position of the equilibrium. In our specific exercise, the equilibrium constant ( ext{K}_p) formulation \(\frac{4 x^2 P}{1-x^2}\) clearly suggests that pressure \(P\) is a crucial factor influencing the equilibrium constant. This means that increasing the pressure will lead to an increase in \( ext{K}_p\).

It's important to differentiate between changes in pressure that affect \( ext{K}_p\) and those that change the amounts of reactants and products temporarily.

• If pressure increases due to the addition of an inert gas at constant volume, the equilibrium position does not change.
• With constant temperature and a change in total system pressure, reactions involving gases will shift towards fewer moles of gas if there's an increase in pressure or towards more moles if pressure decreases.

Understanding these nuances can help students build clear expectations of how real-world reactions might respond to pressure changes.

Extent of Decomposition

The extent of decomposition is a measure of how much a substance has broken down in a decomposition reaction. In our example of \(n_2o_4\) decomposing into \(2 ext{NO}_2\), the extent of decomposition is symbolized by \( ext{x}\). It represents the fraction or percentage of the initial substance that has decomposed at equilibrium.

In the equation for \( ext{K}_p\), \(x\) is critical because it accounts for how much reactant is converted into product. However, its effect on \(K_p\) is nuanced, as seen in the formula \(\frac{4 x^2 P}{1-x^2}\).

• The numerator depends on the square of \( ext{x}\) times \(P\), highlighting how decomposing more material impacts pressure and ultimately \(K_p\).
• The denominator having \(1-x^2\) indicates a limit on how \(x\) influences \(K_p\), representing a balance between decomposed and undecomposed material.

Hence, understanding the extent of decomposition aids in predicting the equilibrating behavior of a reaction mixture.

Chemical Equilibrium

Chemical equilibrium refers to the state in which the concentrations of reactants and products no longer change with time. It occurs when the forward and reverse reactions occur at the same rate. This state is essential in understanding reversible reactions, such as decomposition, as it defines the maximum extent a reaction can reach under given conditions before the rates equalize.

For reactions like the breakdown of \(n_2o_4\) to \(2 ext{NO}_2\), equilibrium is determined by the equilibrium constant \(K_p\), which embodies the relationship between pressure (in cases of gases) and the concentration of reactants and products. A constant \(K_p\) indicates that the reaction has reached equilibrium at a specific temperature.

Interestingly, changes such as increasing pressure or temperature are external factors that might shift this equilibrium, but at a given temperature, \(K_p\) remains consistent. This guiding principle helps predict and manipulate the conditions needed to favor a forward or backward reaction, a key application in chemical manufacturing and synthesis.