Problem 74 - A Complete Resource Book in Chemistry for JEE Main

Question

For a gaseous reaction \(2 \mathrm{~A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\), the partial pressures of \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) at equilibrium are \(0.5\), \(0.8,0.7\) and \(1.2 \mathrm{~atm}\). The value of \(\mathrm{K}_{\mathrm{p}}\) for this reaction is (a) \(2.4 \mathrm{~atm}\) (b) \(6.2 \mathrm{~atm}^{-2}\) (c) \(4.2 \mathrm{~atm}^{-1}\) (d) \(8.4 \mathrm{~atm}^{-3}\)

Step-by-Step Solution

Answer

The value of \( K_p \) is \( 4.2 \, \text{atm}^{-1} \) (option c).

Step 1: Write the expression for Kp

The expression for the equilibrium constant in terms of partial pressures, \( K_p \), for the given reaction \( 2 \mathrm{~A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D} \) is given by:\[K_p = \frac{P_C \cdot P_D}{(P_A)^2 \cdot P_B}\] Where \( P_A, P_B, P_C, \) and \( P_D \) are the partial pressures of \( A, B, C, \) and \( D \) respectively.

Step 2: Substitute the partial pressures

Substitute the given values of partial pressures into the expression:\[K_p = \frac{0.7 \times 1.2}{(0.5)^2 \times 0.8}\]

Step 3: Simplify the expression

First, calculate the multiplication in the numerator and the powers in the denominator:- Numerator: \( 0.7 \times 1.2 = 0.84 \)- Denominator: \( (0.5)^2 \times 0.8 = 0.25 \times 0.8 = 0.2 \)Substitute these values back:\[K_p = \frac{0.84}{0.2}\]

Step 4: Calculate Kp

Now, divide the numbers to find \( K_p \):\[K_p = \frac{0.84}{0.2} = 4.2\]The units of \( K_p \) will be \( \text{atm}^{-1} \) given the stoichiometry changes between reactants \((2+1)\) and products \((1+1)\) which results in \(2-2=0\) net change in coefficients, keeping denominator unchanged as atm^{-1}.

Key Concepts

Equilibrium Constant

In chemical equilibrium, the equilibrium constant is a crucial concept that helps describe how far a reaction proceeds before reaching a state of balance. For gaseous reactions, we often use the equilibrium constant in terms of partial pressures, denoted as \(K_p\). In simple terms, \(K_p\) is a numeric value calculated by the ratio of the products' concentrations or pressures raised to the power of their coefficients, divided by the reactants' concentrations or pressures raised to their respective coefficients.
In a balanced chemical equation like \(2 \mathrm{~A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}\), the \(K_p\) expression is formulated as follows:

• Numerator: \(P_C \times P_D\)
• Denominator: \((P_A)^2 \times P_B\)
• \(K_p = \frac{P_C \cdot P_D}{(P_A)^2 \cdot P_B}\)

\(K_p\) helps chemists predict how changes in pressure or concentration might shift the equilibrium position. A large \(K_p\) value indicates that the products are favored at equilibrium, while a small \(K_p\) suggests that the reactants are favored.

Partial Pressure

Partial pressure refers to the pressure exerted by a single gas in a mixture of gases. Each component of a gas mixture has its own partial pressure, and the sum of these partial pressures equals the total pressure of the system.
In the context of our reaction \(2 \mathrm{~A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}\), partial pressures for gases \(A\), \(B\), \(C\), and \(D\) are given respectively at \(0.5\), \(0.8\), \(0.7\), and \(1.2\) atm. Each gas's partial pressure depends on both the total pressure and the mole fraction of that particular gas in the mixture (which itself is based on the amount of the substance relative to the total amount of all substances present).
When calculating \(K_p\), these partial pressures are substituted into the \(K_p\) expression. This usage highlights the importance of partial pressures in understanding and calculating equilibria in gas-phase reactions.

Reaction Stoichiometry

Reaction stoichiometry involves balancing chemical equations and understanding the relationships between reactants and products as dictated by the equation. In our exercise, the reaction is \(2 \mathrm{~A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}\).
Stoichiometry tells us the molar ratio of each reactant and product. Here, it takes 2 moles of \(A\) and 1 mole of \(B\) to produce 1 mole each of \(C\) and \(D\). This balanced equation is important because it helps us write the expression for \(K_p\) correctly and understand how changes in concentration or pressure will affect each component's ratio.
When calculating \(K_p\), stoichiometry ensures that we raise the pressures to the proper power, reflecting these ratios: \((P_A)^2\) because there are two \(A\) molecules. Understanding how stoichiometry affects \(K_p\) is key to predicting how varying conditions will move the system towards or away from equilibrium.