Problem 118 - A Complete Resource Book in Chemistry for JEE Main

Question

A gaseous phase reaction is allowed to attain equilibrium as \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\) at constant pressure P. The partial pressure of A at equilibrium is \(\mathrm{P} / 2 .\) The value of equilibrium constant \(K_{p}\) is (a) \(\frac{\mathrm{P}}{8}\) (b) \(\frac{\mathrm{P}}{6}\) (c) \(\frac{\mathrm{P}}{2}\) (d) \(\frac{\mathrm{P}}{4}\)

Step-by-Step Solution

Answer

The equilibrium constant \( K_p \) is \( \frac{P}{8} \) (Option a).

Step 1: Write the Equilibrium Expression

Consider the reaction: \( \mathrm{A(g)} \rightleftharpoons \mathrm{B(g)} + \mathrm{C(g)} \). The equilibrium constant \( K_p \) is expressed in terms of partial pressures as \( K_p = \frac{P_B \cdot P_C}{P_A} \) where \( P_B \), \( P_C \), and \( P_A \) are the partial pressures of B, C, and A respectively at equilibrium.

Step 2: Define Initial and Equilibrium Partial Pressures

Assume that initially, only A is present with a pressure of \( P \). At equilibrium, \( P_A = \frac{P}{2} \). Let the change for A be \( -x \) and for B and C be \( +x \). Thus, \( P_B = x \) and \( P_C = x \) at equilibrium. Since total initial pressure is \( P \) and at equilibrium \( P_A = \frac{P}{2} \), it implies \( \frac{P}{2} = P - x \) leading to \( x = \frac{P}{2} \).

Step 3: Express Partial Pressures at Equilibrium

We found that \( x = \frac{P}{2} \). Thus, \( P_B = \frac{P}{2} \) and \( P_C = \frac{P}{2} \). Therefore, the partial pressures of B and C are both \( \frac{P}{2} \), and \( P_A = \frac{P}{2} \) at equilibrium.

Step 4: Substitute into Equilibrium Expression

Substitute the equilibrium partial pressures into the \( K_p \) expression: \( K_p = \frac{\left(\frac{P}{2}\right) \cdot \left(\frac{P}{2}\right)}{\frac{P}{2}} = \frac{\frac{P^2}{4}}{\frac{P}{2}} \). This simplifies to \( K_p = \frac{P}{8} \).

Key Concepts

Gaseous Phase Reaction

A gaseous phase reaction involves chemical substances that exist in the gas phase. This is different from reactions occurring in liquids or solids. The molecules in gaseous reactions are free to move and fill the container they are in, leading to a homogeneous mixture. An example of a gaseous phase reaction is the reaction \( \mathrm{A(g)} \rightleftharpoons \mathrm{B(g)} + \mathrm{C(g)} \). In this reaction, all components are gases, and they participate equally in reaching equilibrium, regardless of the total pressure. This characteristic allows for simplification when considering changes in concentration and pressure. Gaseous reactions are often characterized by their ease of observing changes in pressure and volume, which are instrumental in calculating reaction rates and equilibria.

Partial Pressures

In the context of gaseous reactions, understanding partial pressures is key. Partial pressure refers to the pressure exerted by a single type of gas in a mixture. For each component of a gaseous mixture, you can calculate its partial pressure using the equation \( P_i = \chi_i \cdot P_{\text{total}} \)where \( P_i \) is the partial pressure of the gas, \( \chi_i \) is the mole fraction, and \( P_{\text{total}} \) is the total pressure of the mixture.

• Partial pressures are useful because they allow us to focus on each reactant's contribution to the overall pressure independently.
• In our initial example, the equilibrium partial pressure of A is given as \( \frac{P}{2} \), meaning A contributes half of the total pressure at equilibrium.
• This data can help calculate equilibrium constants and predict the reaction's direction under changing conditions.

Equilibrium Expression

The equilibrium expression links the concentrations of reactants and products to the equilibrium constant, which is crucial in describing the state of a chemical reaction at equilibrium. For a gaseous reaction, the equilibrium constant is given in terms of partial pressures, symbolized as \( K_p \). For our specific reaction, the equilibrium expression is\( K_p = \frac{P_B \cdot P_C}{P_A} \), where \( P_B \), \( P_C \), and \( P_A \) are the partial pressures of B, C, and A respectively.

• The equilibrium expression provides a quantitative measure of the position of equilibrium.
• By substituting the values of partial pressures from the reaction, \( K_p \) is calculated as \( \frac{P}{8} \).
• This tells us how much of the reactants have been converted into products at equilibrium.

Knowing \( K_p \) can also show how changes in temperature and the presence of catalysts affect equilibrium.

Chemical Equilibrium

Chemical equilibrium is a dynamic state where the forward and reverse reactions occur at equal rates. At equilibrium, the concentrations of reactants and products remain constant over time, although the reactions themselves do not stop.

• For example, in our reaction \( \mathrm{A(g)} \rightleftharpoons \mathrm{B(g)} + \mathrm{C(g)} \), equilibrium is achieved when \( A \), \( B \), and \( C \) are present at stable concentrations.
• The condition of equilibrium reflects a balance but not necessarily equal amounts of reactants and products.
• The equilibrium is highly dependent on the reaction conditions like pressure, temperature, and concentration.

Understanding chemical equilibrium involves applying the equilibrium constant, \( K_p \), to predict how these conditions will shift the position of equilibrium, allowing for control over the extent of reactions in practical applications.