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 \(1.2\) atm. The value of \(K_{p}\) 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 \(K_p\) for the reaction is (c) \(4.2 \text{ atm}^{-1}\).

Step 1: Write the expression for Kp

The expression for the equilibrium constant in terms of partial pressures, \(K_p\), for the 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 \(P_C\), \(P_D\), \(P_A\), and \(P_B\) are the partial pressures of \(\mathrm{C}, \mathrm{D}, \mathrm{A},\) and \(\mathrm{B}\) respectively.

Step 2: Substitute the given partial pressures

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

Step 3: Calculate the numerator

Calculate the numerator of the expression:\[ 0.7 \cdot 1.2 = 0.84 \]

Step 4: Calculate the denominator

Calculate the denominator of the expression:\[ (0.5)^2 \cdot 0.8 = 0.25 \cdot 0.8 = 0.2 \]

Step 5: Solve for Kp

Now, solve for the equilibrium constant \(K_p\) by dividing the numerator by the denominator:\[ K_p = \frac{0.84}{0.2} = 4.2 \]

Step 6: Determine the correct unit for Kp

Since \(K_p\) is calculated from the ratio of pressures, its units will be in pressures as well. In this case, since the denominator involves an extra factor of pressure (\(P_A^2 \cdot P_B\) instead of just \(P_A \cdot P_B\)), the unit is \(\text{atm}^{-1}\).

Key Concepts

Gaseous Reactions

Gaseous reactions are chemical reactions that occur between substances in the gaseous phase. These reactions are significant because the reactants and products are all gases, meaning their concentrations can be expressed in terms of partial pressures. This allows for easier manipulation and measurement of the reaction’s progress compared to reactions involving solids or liquids.

In a gaseous reaction like the one given: \(2 \mathrm{~A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}\), gases A and B combine to form gases C and D. Such reactions can be reversible, meaning they can reach a point where the forward and reverse reactions happen at the same rate. This state is referred to as equilibrium, where the concentrations of reactants and products remain constant over time, although the reactions are still proceeding.

At equilibrium, it is crucial to understand that the partial pressures of the reactants and products will determine the equilibrium constant \(K_p\), which provides insight about the position of the equilibrium and the proportions of products to reactants.

Partial Pressures

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. It is an essential concept in dealing with gaseous reactions as it represents the effective concentration of the gas in the mixture.

In a mixture of gases, each gas behaves independently and exerts pressure as if it were the only gas present. For a reaction such as \(2 \mathrm{~A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}\), the partial pressures are given by \(P_A, P_B, P_C,\) and \(P_D\). These values are used directly in calculations involving the equilibrium constant \(K_p\).

The individual partial pressures can be summed to get the total pressure in the system. The relationship is described by Dalton's law of partial pressures, which states that the total pressure is equal to the sum of the partial pressures of the individual gases. Thus, knowing each gas's partial pressure enables chemists to calculate the overall behavior of gaseous reactions.

Equilibrium Expressions

Equilibrium expressions are mathematical equations that describe the equilibrium state of a chemical reaction. They relate the concentrations or partial pressures of the reactants and products to the equilibrium constant.

For gaseous reactions, the equilibrium expression is written in terms of partial pressures and is denoted by \(K_p\). In the example given, the expression for \(K_p\) is \[ K_p = \frac{P_C \cdot P_D}{P_A^2 \cdot P_B} \] where the numerator represents the product of the partial pressures of the products \(\mathrm{C}\) and \(\mathrm{D}\), and the denominator represents the partial pressures of reactants \(\mathrm{A}\) and \(\mathrm{B}\) raised to the power of their stoichiometric coefficients.

Writing the equilibrium expression correctly is key to determining \(K_p\). It reflects the balance point of the reaction under given conditions. By substituting the actual measured partial pressures into this expression, we can calculate \(K_p\) and gain insight into how the system is behaving at equilibrium.

Pressure Units

Pressure units are an important aspect when dealing with equilibrium calculations, particularly involving gases. In most chemistry problems, we use the atmosphere (atm) as a standard unit of pressure to express partial pressures and equilibrium constants.

For the reaction \(2 \mathrm{~A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}\), each of the reactants and products is measured in atmospheres to calculate their contributions to the equilibrium state. When calculating \(K_p\), it is crucial to understand that these units affect the expression.

• The equilibrium constant \(K_p\) itself carries units derived from the powers of pressure involved in the expression.
• In this case, the units of \(K_p\) are \(\text{atm}^{-1}\) because the denominator includes an extra factor of pressure due to \(P_A^2\).

When performing calculations with \(K_p\), recognizing these units ensures accuracy and allows us to correctly interpret the behavior of the system at equilibrium.