Problem 104 - A Complete Resource Book in Chemistry for JEE Main

Question

For a gaseous equilibrium \(2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}_{(\mathrm{g})}, \mathrm{K}_{\mathrm{p}}\) has a value of \(1.8 \mathrm{at}\) \(700 \mathrm{~K}\). What is the value of \(\mathrm{K}_{\text {for the equilibrium }}\) \(2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}_{(\mathrm{g})} \rightleftharpoons 2 \mathrm{~A}\) at the same pressure? (a) \(0.031\) (b) \(1.3 \times 10^{-3}\) (c) \(44.4\) (d) 38

Step-by-Step Solution

Answer

None of the options match the expected reciprocal of 1.8 (0.556).

Step 1: Understand the Relationship between Reactions

The given problem presents a gaseous equilibrium reaction: \(2 \mathrm{~A} \rightleftharpoons 2 \mathrm{~B} + \mathrm{C}\) with a known equilibrium constant \(K_{\mathrm{p}} = 1.8\). It asks for the equilibrium constant for the reverse reaction: \(2 \mathrm{~B} + \mathrm{C} \rightleftharpoons 2 \mathrm{~A}\). In a chemical reaction, the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction.

Step 2: Calculate the Equilibrium Constant for the Reverse Reaction

The equilibrium constant for the reverse reaction is given by the formula:\[K_{\text{reverse}} = \frac{1}{K_{\text{forward}}}\]Substituting the given value:\[K_{\text{reverse}} = \frac{1}{1.8} \approx 0.5556\]This value will help us determine which of the provided options is correct.

Step 3: Match to Given Options

Now, compare your calculated \(K_{\text{reverse}} = 0.5556\) with the provided options:- (a) 0.031- (b) 1.3 \times 10^{-3}- (c) 44.4- (d) 38The calculated \(K_{\text{reverse}} = 0.5556\) does not directly match any of the options, indicating there might have been an oversight. Review the calculation, assumptions, or options provided.

Key Concepts

Gaseous Equilibria

Gaseous equilibria involve chemical reactions where all participating substances are in the gaseous state.
These reactions reach a balance where the rates of the forward and reverse reactions are equal.
As a result, the concentrations of reactants and products remain constant over time.

In gaseous equilibria, pressure and temperature significantly impact the position of equilibrium.

• Higher pressure tends to favor the side with fewer moles of gas due to reduced volume.
• Conversely, lower pressure favors the side with more moles.

Temperature also affects gaseous equilibria, with exothermic reactions being more favorable at lower temperatures.
Gaseous equilibria are governed by principles such as Le Chatelier's, which predicts how changes in conditions impact equilibrium shifts.
Understanding these principles is essential for predicting how a change in conditions will affect a given reaction.

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), measures the ratio of product concentrations to reactant concentrations at equilibrium.
It is crucial to note that each gaseous species is raised to the power of its coefficient in the balanced equation.

The form of the equilibrium constant can vary depending on whether the reaction involves gases or solutions. For gaseous reactions, it is typically represented by \( K_p \), based on partial pressures.
\[ K_p = \frac{{(P_B)^2 \cdot P_C}}{{(P_A)^2}} \] where \(P_B\) and \(P_C\) represent the partial pressures of the products, and \(P_A\) represents the partial pressures of the reactants.

A large \( K \) value means the position of equilibrium favors products, while a small \( K \) value indicates a preference for reactants.

• A \( K \) value of 1 suggests neither side is favored significantly.

It is vital to remember that changing the temperature can alter the value of \( K \), but changes in pressure or concentration do not.

Reverse Reaction Calculation

Calculating the equilibrium constant for a reverse reaction is a simple yet crucial concept.
To find the equilibrium constant of the reverse reaction, you take the reciprocal of the equilibrium constant of the forward reaction.

This relationship is explained by the fact that reversing a reaction essentially swaps its products and reactants, thus inverting the equilibrium expression.
Mathematically, for a reaction with a forward equilibrium constant \( K_{forward} \), the reverse reaction equilibrium constant \( K_{reverse} \) is:\[ K_{reverse} = \frac{1}{K_{forward}} \]
Applying this to the exercise provided, if the forward \( K_p \) is 1.8, then:\[ K_{reverse} = \frac{1}{1.8} \approx 0.556 \]
This simple calculation underscores the importance of understanding chemical equilibria and their reversibility.
It forms the basis for more complex equilibrium calculations in chemistry and is essential for problem-solving in chemical reactions.