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

Question

In which of the following case, the value of \(\mathrm{K}_{\mathrm{p}}\) is less than \(\mathrm{K}_{\mathrm{c}} ?\) (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) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\)

Step-by-Step Solution

Answer

In case (c), the value of \(K_{p}\) is less than \(K_{c}\).

Step 1: Understanding the Relationship Between Kp and Kc

The relationship between the equilibrium constants at constant temperature for reactions involving gases can be understood using the formula: \ \( K_{p} = K_{c} (RT)^{\Delta n} \) \ where \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in number of moles of gas, calculated as the moles of gaseous products minus the moles of gaseous reactants.

Step 2: Evaluate Delta n for Case (a)

The reaction is: \( N_{2} + O_{2} \rightleftharpoons 2 NO \). Calculate \( \Delta n \) as follows: \ \( \Delta n = (\text{moles of NO}) - (\text{moles of } N_{2} + \text{moles of } O_{2}) = 2 - (1+1) = 0 \). With \( \Delta n = 0 \), \( K_{p} = K_{c} \).

Step 3: Evaluate Delta n for Case (b)

The reaction is: \( H_{2} + Cl_{2} \rightleftharpoons 2 HCl \). Calculate \( \Delta n \) as follows: \ \( \Delta n = (\text{moles of HCl}) - (\text{moles of } H_{2} + \text{moles of } Cl_{2}) = 2 - (1+1) = 0 \). With \( \Delta n = 0 \), \( K_{p} = K_{c} \).

Step 4: Evaluate Delta n for Case (c)

The reaction is: \( 2 SO_{2} + O_{2} \rightleftharpoons 2 SO_{3} \). Calculate \( \Delta n \) as follows: \ \( \Delta n = (\text{moles of } SO_{3}) - (\text{moles of } SO_{2} + \text{moles of } O_{2}) = 2 - (2+1) = -1 \). With \( \Delta n = -1 \), \( K_{p} = K_{c} (RT)^{-1} \). Since \((RT)^{-1}\) is less than 1, \( K_{p} < K_{c} \).

Step 5: Evaluate Delta n for Case (d)

The reaction is: \( PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2} \). Calculate \( \Delta n \) as follows: \ \( \Delta n = (\text{moles of } PCl_{3} + \text{moles of } Cl_{2}) - (\text{moles of } PCl_{5}) = (1+1) - 1 = 1 \). With \( \Delta n = 1 \), \( K_{p} = K_{c} (RT)^{1} \). Since \((RT)^{1}\) is greater than 1, \( K_{p} > K_{c} \).

Key Concepts

Equilibrium Constant

The equilibrium constant is a crucial concept in chemistry that helps in understanding how a chemical reaction proceeds at a given temperature. In essence, it provides a ratio of the concentrations of products to reactants at equilibrium for a given reaction. This balance point, known as equilibrium, indicates how much of the initial substances remain and how much has been converted into products.

• At equilibrium, the reaction appears to stop, but in reality, it reaches a state where the rates of the forward and reverse reactions are equal.
• For reactions in the gaseous phase, the equilibrium constant can be expressed both in terms of concentration \( K_{c} \) and in terms of pressure \( K_{p} \).

Equilibrium constants are essential because they give us insight into the favorability of reactions. A high equilibrium constant indicates that, at equilibrium, the reaction mixture contains more products than reactants. Conversely, a low equilibrium constant indicates more reactants than products.
Understanding the equilibrium constant guides chemists in determining how a reaction behaves under different conditions and assists in predicting the outcome of reactions.

Gaseous Reactions

In gaseous reactions, substances in the gas phase undergo transformations that are described by equilibrium constants. Gaseous reactions are unique due to the compressible nature of gases, which brings pressure into consideration.

• For gaseous reactions, particularly those happening in closed vessels, the concentration of gases can be expressed as partial pressures, making \( K_{p} \) a valuable expression.
• Unlike solid and liquid reactions, changes in volume or pressure can significantly affect gaseous reactions.

Consider the general reaction:\( aA + bB \rightleftharpoons cC + dD \)For this reaction, the equilibrium constant expression in terms of pressure is:\[K_{p} = \frac{(P_C)^c \cdot (P_D)^d}{(P_A)^a \cdot (P_B)^b}\]The partial pressures (\( P_A \), \( P_B \), etc.) reflect the number of moles of each gas and the temperature and volume of the system. Understanding these reactions involves calculating changes in various states like volume and pressure, affecting the equilibrium state.

Kp and Kc Relationship

The relationship between \( K_{p} \) and \( K_{c} \) is fundamental in relating pressure-based and concentration-based equilibrium expressions. Understanding this connection allows predicting how changes in conditions affect these constants.

• The relationship is given by the equation:\[K_{p} = K_{c} (RT)^{\Delta n}\]where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas (products minus reactants).
• \( \Delta n \) is crucial since it determines how \( K_{p} \) and \( K_{c} \) are related. When \( \Delta n \) is zero, these two constants are equal, but when it's not, \( K_{p} \) can be either greater or lesser than \( K_{c} \) depending on the sign of \( \Delta n \).

For example, in the reaction involving sulfur dioxide and oxygen forming sulfur trioxide:
\( 2 SO_2 + O_2 \rightleftharpoons 2 SO_3 \)
The \( \Delta n = -1 \), leading to \( K_{p} = K_{c} (RT)^{-1} \), which results in \( K_{p} < K_{c} \). Comprehending these relationships is key to measuring reaction tendencies and making accurate predictions about the behavior of gaseous systems.