Problem 189 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction equilibrium, \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) the concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) at equilibrium are \(4.8 \times 10^{-2}\) and \(1.2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}\) respectively. The value of \(\mathrm{K}_{\mathrm{c}}\) for the reaction is (a) \(3.3 \times 10^{2} \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(3 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(3 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\) (d) \(3 \times 10^{3} \mathrm{~mol} \mathrm{~L}^{-1}\)

Step-by-Step Solution

Answer

(c) \(3 \times 10^{-3} \text{ mol L}^{-1}\)

Step 1: Write the Equilibrium Expression

The equilibrium constant expression for the reaction \( \mathrm{N}_2\mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{g}) \) can be written as: \( K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]} \). This equation shows that the equilibrium constant \( K_c \) depends on the concentration of \( \mathrm{NO}_2 \) squared divided by the concentration of \( \mathrm{N}_2\mathrm{O}_4 \).

Step 2: Substitute the Equilibrium Concentrations

Substitute the given equilibrium concentrations into the equilibrium expression. Given: \([\mathrm{N}_2\mathrm{O}_4] = 4.8 \times 10^{-2} \) mol/L and \([\mathrm{NO}_2] = 1.2 \times 10^{-2} \) mol/L. The equilibrium constant expression becomes: \( K_c = \frac{(1.2 \times 10^{-2})^2}{4.8 \times 10^{-2}} \).

Step 3: Calculate the Value of K_c

First, calculate \((1.2 \times 10^{-2})^2 = 1.44 \times 10^{-4} \). Then divide by \(4.8 \times 10^{-2}\) to find \( K_c \): \( K_c = \frac{1.44 \times 10^{-4}}{4.8 \times 10^{-2}} = 3 \times 10^{-3} \).

Step 4: Compare with Given Options

The calculated \( K_c = 3 \times 10^{-3} \text{ mol L}^{-1} \). Compare it with the options provided: (a) \(3.3 \times 10^{2} \text{ mol L}^{-1}\), (b) \(3 \times 10^{-1} \text{ mol L}^{-1}\), (c) \(3 \times 10^{-3} \text{ mol L}^{-1}\), (d) \(3 \times 10^{3} \text{ mol L}^{-1}\). The correct option is (c).

Key Concepts

Equilibrium Constant (Kc)

In chemical reactions, particularly reversible reactions, the equilibrium constant, symbolized as \( K_c \), plays a crucial role. It is a numerical value that indicates the ratio of the concentration of products to reactants at equilibrium. For the reaction \( \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \), the expression for \( K_c \) can be written as:

\[ K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]} \]
This equation tells us how to relate the concentrations of products and reactants. The product \([\mathrm{NO}_2]^2\) is squared in the equation because there are two \(\mathrm{NO}_2\) molecules formed for every molecule of \(\mathrm{N}_2\mathrm{O}_4\) consumed. By substituting the equilibrium concentrations, we calculate \(K_c\).

• If \( K_c \) is large, it suggests that at equilibrium, products dominate over reactants.
• A small \( K_c \) value implies more reactants than products at equilibrium.

This powerful tool helps chemists understand how far a reaction will proceed before reaching equilibrium.

Equilibrium Concentration

Equilibrium concentration refers to the concentrations of all reactants and products in a reversible chemical reaction once equilibrium is established. Equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction.

For the reaction \( \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \), we are given the equilibrium concentrations:

• \([\mathrm{N}_2\mathrm{O}_4] = 4.8 \times 10^{-2} \text{ mol L}^{-1}\)
• \([\mathrm{NO}_2] = 1.2 \times 10^{-2} \text{ mol L}^{-1}\)

To plug these values into the equilibrium expression, we use them to calculate the equilibrium constant. The equilibrium concentration is key for calculating \( K_c \) and determining if a reaction at a certain point in time has reached equilibrium. Understanding equilibrium concentrations helps predict reaction behavior and product formation.

Reversible Reactions

A reversible reaction is one where the reactants form products that react together to give the reactants back in a chemical system. The double arrow symbol \( \rightleftharpoons \) in reactions like \( \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2\,\mathrm{NO}_{2}(\mathrm{g}) \) indicates its reversible nature.

- **Forward vs Reverse:** In a forward reaction, reactants convert to products. In reverse, products convert back to reactants.- **Dynamic Equilibrium:** It occurs when the rate of the forward reaction equals the rate of the reverse, meaning concentrations of reactants and products remain constant over time.Reversible reactions are fundamental in both industrial processes and biological systems. Understanding them allows control over chemical processes by adjusting conditions like pressure, temperature, or concentration to favor certain outcomes or maximize yield.