Problem 76 - A Complete Resource Book in Chemistry for JEE Main

Question

In which of the following gaseous reaction, \(\mathrm{K}_{0}\) and \(\mathrm{K}_{\mathrm{c}}\) have the same values? (a) \(2 \mathrm{H} 1 \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}\) (b) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (c) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) (d) \(\mathrm{PCI}_{5}=\mathrm{PCI}_{3}+\mathrm{Cl}_{2}^{3}\)

Step-by-Step Solution

Answer

In reaction (a), \( K_c \) and \( K_p \) have the same values.

Step 1: Understand the Concepts of Kc and Kp

In chemical equilibrium, \( K_c \) is the equilibrium constant in terms of concentration, while \( K_p \) is the equilibrium constant in terms of partial pressures. These two constants are related by the equation \( K_p = K_c (RT)^{\Delta n} \), where \( \Delta n \) is the change in moles of gas during the reaction, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

Step 2: Identify Reactions with Equal Mole Change

For \( K_c \) and \( K_p \) to be equal, \( \Delta n \) must be zero, meaning the number of moles of gas does not change from reactants to products. Therefore, check each reaction to determine \( \Delta n \): - (a) \( 2 \text{ moles of } \, \text{HI} \rightarrow 0 \text{ moles of } \) change- (b) \( 1 + 3 = 4 \rightarrow 2 \text{ moles of } \) change, so \( \Delta n = -2 \)- (c) \( 2 + 1 = 3 \rightarrow 2 \text{ moles of } \) change, so \( \Delta n = -1 \)- (d) \( 1 \rightarrow 1 + 1 = 2 \text{ moles of } \) change, so \( \Delta n = +1 \).

Step 3: Solution Analysis

By analyzing each option, only reaction (a) has no net change in the number of gas moles from reactants to products (\( \Delta n = 0 \)). Therefore, for this reaction, \( K_c = K_p \) since \( (RT)^0 = 1 \), meaning the effect of \( RT \) is eliminated, maintaining equal \( K_c \) and \( K_p \).

Key Concepts

Equilibrium Constant (Kc)

In the realm of chemical equilibrium, the equilibrium constant denoted as \( K_c \) describes the ratio of the concentrations of products to reactants at equilibrium. It is calculated using the molar concentrations of substances involved in a chemical reaction. The general expression for \( K_c \) is given by the formula:\[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]where \([A]^a\), \([B]^b\), \([C]^c\), and \([D]^d\) denote the molar concentrations of the reactants and products respectively.
This equation is applicable for a reversible reaction of the form:\[aA + bB \rightleftharpoons cC + dD\]It is important to note that \( K_c \) only changes with temperature, and not with the concentrations of reactants or products once equilibrium is established.
Therefore, understanding \( K_c \) is crucial for predicting the mix of chemicals at equilibrium, ensuring we can anticipate how the system responds to changes.

Equilibrium Constant (Kp)

The equilibrium constant in terms of partial pressure is represented as \( K_p \). It is primarily used for gaseous reactions. In such reactions, the equilibrium constant is based on the partial pressures of the gases involved. A general formula for \( K_p \) can be written as:\[K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}\]where \(P_A, P_B, P_C,\) and \(P_D\) are the partial pressures of the gases.
The conversion between \( K_c \) and \( K_p \) is linked through the equation:\[K_p = K_c(RT)^{\Delta n}\]Here, \( R \) is the ideal gas constant, \( T \) represents temperature in Kelvin, and \( \Delta n \) is the difference in moles of gas between products and reactants.
Understanding \( K_p \) is vital for reactions involving gases, as it conveys how changes like pressure adjustments impact equilibria.

Mole Change in Reactions

Mole change, denoted as \( \Delta n \), in a chemical reaction is the difference in the number of moles of gaseous products and reactants. Calculating \( \Delta n \) helps determine if there is a net change in gas moles which influences whether \( K_c \) and \( K_p \) are equal.
The formula for calculating \( \Delta n \) is:\[\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}\]

• If \( \Delta n = 0 \), then \( K_c = K_p \) since any factor related to pressure volume adjustments becomes nullified.
• If \( \Delta n eq 0 \), \( K_c \) and \( K_p \) differ depending on the temperature and gas constant values.

In reaction analysis, correctly computing \( \Delta n \) can simplify understanding if a change in pressure will influence the equilibrium state of the reaction.