Problem 48 - A Complete Resource Book in Chemistry for JEE Main

Question

In which of the following case, the value of \(K_{p}\) is less than \(\mathrm{K}_{c}\) ? (a) \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\) (b) \(\mathrm{H}_{2}+\mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{HCl}\) (c) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) (d) \(\mathrm{PCl}_{5}^{2} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}^{3}\)

Step-by-Step Solution

Answer

(c) \( 2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3} \)

Step 1: Understand the Relationship Between Kc and Kp

The relationship between the equilibrium constant in terms of concentration \( K_c \) and pressure \( K_p \) is given by:\[ K_p = K_c (RT)^{\Delta n}\]where \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas (moles of products - moles of reactants).

Step 2: Calculate Δn for Each Reaction

(a) \( \mathrm{N}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO} \):\( \Delta n = 2 - (1+1) = 0 \)(b) \( \mathrm{H}_{2} + \mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{HCl} \):\( \Delta n = 2 - (1+1) = 0 \)(c) \( 2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3} \):\( \Delta n = 2 - (2+1) = -1 \)(d) \( \mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3} + \mathrm{Cl}_{2} \):\( \Delta n = (1+1) - 1 = 1 \)

Step 3: Analyze the Effect of Δn on Kp and Kc

If \( \Delta n = 0 \), then \( K_p = K_c \) because \( (RT)^{\Delta n} = 1 \).If \( \Delta n < 0 \), then \( (RT)^{\Delta n} < 1 \), so \( K_p < K_c \).If \( \Delta n > 0 \), then \( (RT)^{\Delta n} > 1 \), so \( K_p > K_c \).

Step 4: Determine the Correct Option

Based on the analysis:(a) and (b) have \( \Delta n = 0 \), so \( K_p = K_c \).(d) has \( \Delta n = 1 \), so \( K_p > K_c \).(c) has \( \Delta n = -1 \), so \( K_p < K_c \).Therefore, the correct option is (c) \( 2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3} \).

Key Concepts

Kp and Kc relationship

When it comes to chemical reactions in equilibrium, there are two main ways to express the equilibrium constant: in terms of concentration \( K_c \) and in terms of pressure \( K_p \). These constants allow you to predict the behavior of a reaction under different conditions. It is essential to understand their relationship to determine how they vary with temperature and changes in the number of moles of gases. The relationship between \( K_c \) and \( K_p \) is expressed by the equation: \[ K_p = K_c (RT)^{\Delta n} \] Here,

• \( R \) is the universal gas constant.
• \( T \) is the absolute temperature in Kelvin.
• \( \Delta n \) represents the change in moles of gases.

This equation tells us that the values of \( K_p \) and \( K_c \) are related by the change in moles of gases along with the temperature factor. Understanding this equation is crucial when analyzing equilibrium reactions involving gases.

Delta n - change in moles

In the context of chemical equilibrium, \( \Delta n \) serves as a crucial factor in understanding the shift between concentration and pressure-based equilibrium constants. \( \Delta n \) represents the change in moles of gaseous products minus the moles of gaseous reactants in a balanced equation:

• If \( \Delta n = 0 \), it means there is no net change in the number of gas molecules. In this case, \( K_p \) and \( K_c \) will be equal, because \( (RT)^{\Delta n} \) simplifies to 1.
• If \( \Delta n > 0 \), an increase in the number of moles of gas takes place, meaning \( K_p \) will be greater than \( K_c \), since \( (RT)^{\Delta n} \) becomes a value greater than 1.
• Conversely, if \( \Delta n < 0 \), a decrease in the moles of gas occurs, leading to \( K_p \) being less than \( K_c \), as \( (RT)^{\Delta n} \) becomes a fractional value less than 1.

Understanding \( \Delta n \) helps in identifying how reaction conditions will affect the equilibrium state, providing insights into whether pressure or concentration predominates in determining the constant value.

Equilibrium reaction analysis

Analyzing equilibrium reactions involves understanding how different factors influence the state of equilibrium. In a reaction at equilibrium, the rate of the forward reaction equals the rate of the backward reaction, and concentrations of the reactants and products remain constant over time. To determine how these constants, \( K_p \) and \( K_c \), compare, examine the individual reaction's changes in moles and consider the temperature.For instance, consider the reaction: \( 2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3} \). Here, \( \Delta n = -1 \), indicating a decrease in the number of gas moles. As a result, \( K_p < K_c \) because the pressure-based constant is reduced by the \((RT)^{-1}\) term being a factor of less than 1.Such an analysis is crucial:

• It helps predict the effect of pressure changes on equilibrium.
• Provides insights into how equilibrium will shift if temperature changes.
• Assists in determining which products or reactants will be favored under given conditions.

Employing these analyses will enhance understanding of chemical reactions and aid in predicting the behavior of equilibrium systems under different scenarios.