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

Question

For the reaction equilibrium, \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})=2 \mathrm{NO}(\mathrm{g})\) 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 \(\mathrm{K}\) for the reaction is (b) \(3 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1}\) (a) \(3.3 \times 10^{2} \mathrm{~mol} \mathrm{~L}^{-1}\) (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

The correct value of \(K\) is (c) \(3 \times 10^{-3} \text{ mol L}^{-1}\).

Step 1: Write the Equilibrium Expression

The equilibrium constant expression for the reaction \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \) is given by: \[ K = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} \].

Step 2: Substitute the Known Values

Substitute the equilibrium concentrations \([\text{N}_2\text{O}_4] = 4.8 \times 10^{-2} \text{ mol L}^{-1} \) and \([\text{NO}_2] = 1.2 \times 10^{-2} \text{ mol L}^{-1} \) into the expression: \[ K = \frac{(1.2 \times 10^{-2})^2}{4.8 \times 10^{-2}} \].

Step 3: Calculate the Numerator

Calculate \((1.2 \times 10^{-2})^2\): \[ (1.2 \times 10^{-2})^2 = 1.44 \times 10^{-4} \text{ mol}^2 \text{ L}^{-2} \].

Step 4: Calculate the Equilibrium Constant

Divide the result from Step 3 by \(4.8 \times 10^{-2}\): \[ K = \frac{1.44 \times 10^{-4}}{4.8 \times 10^{-2}} = 3 \times 10^{-3} \text{ mol L}^{-1} \].

Step 5: Determine the Correct Answer from Options

Compare the calculated \(K\) value with the given options: it matches option (c) \(3 \times 10^{-3} \text{ mol L}^{-1} \).

Key Concepts

Equilibrium Constant

In any chemical reaction at equilibrium, the equilibrium constant (K) is a valuable tool that helps us understand the relative concentrations of products and reactants. For the reaction \[\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g),\]K can be expressed in terms of concentration. This is given by the formula:\[K = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}. \]This formula shows that the equilibrium constant depends on the concentrations of **NO** and **N\text{O}_2** at equilibrium.

• The square brackets \([\ ]\) denote the concentration of a substance.
• K is dimensionless in some cases or can have dimensions depending on the reaction order.

When calculations reveal that K is significantly larger than 1, it suggests the products are favored at equilibrium. Conversely, a K much less than 1 indicates the reactants are favored.

Concentration Calculation

Concentration calculations are crucial for finding out the equilibrium constant. At equilibrium, we already know the concentrations of the reactants and products. For this particular reaction:

• The initial concentration of **N\text{O}_2** is \(1.2 \times 10^{-2} \text{ mol L}^{-1} \).
• The initial concentration of **N\text{O}_4** is \(4.8 \times 10^{-2} \text{ mol L}^{-1} \).

By substituting these values into the equilibrium expression, we can calculate K:\[K = \frac{(1.2 \times 10^{-2})^2}{4.8 \times 10^{-2}}. \]Perform the calculations step by step:

• Calculate \((1.2 \times 10^{-2})^2 = 1.44 \times 10^{-4} \text{ mol}^2 \text{ L}^{-2} \).
• Divide by the concentration of **N\text{O}_4** to find K.

The final value is \(3 \times 10^{-3} \text{ mol L}^{-1} \).

Reaction Quotient

The reaction quotient, often represented as Q, is a tool to determine the direction the reaction will proceed to reach equilibrium. It uses the same expression as the equilibrium constant (K), but for current, non-equilibrium concentrations. For this exercise:

• Assume we calculate Q using concentrations found in a real-time scenario.
• If Q < K, the reaction will shift towards product formation to reach equilibrium.
• If Q > K, the reaction shifts toward reactants, trying to restore balance.
• If Q = K, the system is already at equilibrium.

For \[\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g),\]use the same formula:\[Q = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}.\]By checking Q against K, you can verify if a system is at equilibrium or predict which direction the reaction should move to achieve equilibrium.