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 \(\mathrm{K}_{\mathrm{p}}\) 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 \( \frac{P}{2} \) (Option c).

Step 1: Setup the Initial Equilibrium Equations

To determine the equilibrium constant \( K_p \), first set up the reaction equation: \[ \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g}) + \mathrm{C}(\mathrm{g}) \]. Initially, assume the concentration of \( A \) is \( P \) and both \( B \) and \( C \) are \( 0 \).

Step 2: Determine Changes in Pressure

At equilibrium, the partial pressure of \( A \) is \( \frac{P}{2} \). This means \( \frac{P}{2} \) of \( A \) has reacted, forming \( \frac{P}{2} \) each of \( B \) and \( C \) because for every mole of \( A \) that decomposes, 1 mole of \( B \) and 1 mole of \( C \) are formed.

Step 3: Calculate Total Pressure at Equilibrium

The total pressure at equilibrium is the sum of the partial pressures of the gases: \[ P = \frac{P}{2} + \frac{P}{2} + \frac{P}{2} = P \]. Everything checks out since the total pressure stays constant.

Step 4: Write the Expression for \( K_p \)

The equilibrium constant \( K_p \) is given by the expression: \[ K_p = \frac{(P_{B})(P_{C})}{P_{A}} \]. Substituting the equilibrium pressures: \( P_B = \frac{P}{2} \), \( P_C = \frac{P}{2} \), and \( P_A = \frac{P}{2} \).

Step 5: Simplify the Expression for \( K_p \)

Substitute the pressures into the expression for \( K_p \): \[ K_p = \frac{(\frac{P}{2})(\frac{P}{2})}{\frac{P}{2}} \]. Simplify the expression: \[ K_p = \frac{\frac{P^2}{4}}{\frac{P}{2}} = \frac{P}{2} \].

Key Concepts

Gaseous Phase Reactions

In chemistry, reactions taking place in the gaseous phase often involve substances in a dynamic environment where gases can freely mix and react. Gaseous phase reactions are common in both industrial and natural processes. A distinguishing feature of these reactions is the behavior of gases under different conditions, such as pressure and temperature.

Taking the reaction \( ext{A(g)} \rightleftharpoons \text{B(g) + C(g)}\) as an example, we see how the exchange between reactants and products occurs until a state of balance—referred to as equilibrium—is reached. At equilibrium, the rate at which \(A\) is converted into \(B\) and \(C\) is equal to the rate at which \(B\) and \(C\) revert to \(A\). This unique interplay is crucial for understanding the equilibrium state in gaseous reactions.

Partial Pressure

Partial pressure plays a significant role in gaseous phase reactions. It refers to the pressure exerted by an individual gas in a mixture of gases. Imagine each gas acting independently inside the same container—each contributes to the total pressure according to its mole fraction in the mixture.

In the equilibrium scenario \( ext{A(g)} \rightarrow \text{B(g) + C(g)}\), where each gaseous component exists at a specific partial pressure, let's consider that at equilibrium, the partial pressure of \(A\) has reduced to \(\frac{P}{2}\). This change signifies that half of the initial \(A\) pressure has transformed into products \(B\) and \(C\) each also constituting \(\frac{P}{2}\).

• Overall, the total pressure stays constant, maintaining the environment for equilibrium.
• Understanding partial pressures is pivotal to applying the equilibrium constant \(K_p\).

Chemical Equilibrium

In the chemical reactions involving gases, equilibrium is that magical state where the forward and reverse reactions occur at the same rate. The concentrations of reactants and products remain stable over time. Consider our reaction: \( ext{A(g)} \rightleftharpoons \text{B(g) + C(g)}\).

At equilibrium, the reaction doesn't stop but continues to manifest dynamically at a molecular level. The equilibrium constant \(K_p\) gives us a quantifiable means to express this state using the partial pressures of the participating gases:
\[ K_p = \frac{(P_B)(P_C)}{P_A} \]

This equation shows the ratio of the products' pressures to the reactants', providing insights into the reaction's tendency towards forming products or staying as reactants.

Reaction Stoichiometry

Stoichiometry plays a critical role in understanding how substances interact in chemical reactions, particularly in calculating equilibrium constant expressions. It examines the quantitative relationships based on balanced chemical equations.

For the reaction \(\text{A(g)} \rightleftharpoons \text{B(g) + C(g)}\), stoichiometry guides us to understand that when \(\frac{P}{2}\) of gas \(A\) reacts, it forms \(\frac{P}{2}\) of each \(B\) and \(C\). This 1:1:1 ratio derived from the balanced equation is crucial for calculating the equilibrium constant \(K_p\):
\[ K_p = \frac{(\frac{P}{2})(\frac{P}{2})}{\frac{P}{2}} \]

Simplifying gives us \(K_p = \frac{P}{2}\), demonstrating how stoichiometry helps us to navigate through reaction dynamics and equilibrium calculations.