Problem 84 - A Complete Resource Book in Chemistry for JEE Main

Question

\(9.2\) grams of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) is taken in a closed one litre vessel and heated till the following equilibrium is reached \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) .\) At equilibrium, \(50 \%\) of \(\mathrm{N}, \mathrm{O}_{4}(\mathrm{~g})\) is dissociated. What is the equilibrium constant (in \(\left.\mathrm{mol} \mathrm{L}^{-1}\right)\) ? (molecular weight of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is 92 ) (a) \(0.1\) (b) \(0.2\) (c) \(0.4\) (d) 2

Step-by-Step Solution

Answer

The equilibrium constant is 0.2, so the answer is (b).

Step 1: Calculate Initial Moles of \( \mathrm{N}_{2} \mathrm{O}_{4} \)

We have 9.2 g of \( \mathrm{N}_{2} \mathrm{O}_{4} \) and its molecular weight is 92 g/mol. To find the initial number of moles, use the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molecular weight}} = \frac{9.2}{92} = 0.1 \text{ mol} \]

Step 2: Determine Moles of \( \mathrm{N}_{2} \mathrm{O}_{4} \) at Equilibrium

Since 50% of \( \mathrm{N}_{2} \mathrm{O}_{4} \) dissociates at equilibrium, half of the initial moles remain undissociated. Therefore, the moles of \( \mathrm{N}_{2} \mathrm{O}_{4} \) at equilibrium are: \[ 0.1 \times (1 - 0.5) = 0.05 \text{ mol} \]

Step 3: Determine Moles of \( \mathrm{NO}_{2} \) at Equilibrium

From the dissociation, 0.05 mol of \( \mathrm{N}_{2} \mathrm{O}_{4} \) has dissociated to form \( \mathrm{NO}_{2} \). Since 1 mole of \( \mathrm{N}_{2} \mathrm{O}_{4} \) forms 2 moles of \( \mathrm{NO}_{2} \), the moles of \( \mathrm{NO}_{2} \) are: \[ 2 \times 0.05 = 0.1 \text{ mol} \]

Step 4: Calculate Equilibrium Concentrations

The volume of the vessel is 1 liter, so the concentration of \( \mathrm{N}_{2} \mathrm{O}_{4} \) at equilibrium is: \[ \left[ \mathrm{N}_{2} \mathrm{O}_{4} \right] = \frac{0.05}{1} = 0.05 \text{ mol/L} \]The concentration of \( \mathrm{NO}_{2} \) at equilibrium is: \[ [\mathrm{NO}_{2}] = \frac{0.1}{1} = 0.1 \text{ mol/L} \]

Step 5: Apply Equilibrium Constant Expression

The equilibrium constant expression for the reaction \( \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2} \) is: \[ K_c = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2} \mathrm{O}_{4}]} \]Substituting the equilibrium concentrations: \[ K_c = \frac{(0.1)^2}{0.05} = \frac{0.01}{0.05} = 0.2 \]

Step 6: Choose the Correct Answer

From the calculated equilibrium constant, the correct option that matches \( K_c = 0.2 \) is (b).

Key Concepts

N2O4 Dissociation

Dissociation is a crucial concept in chemical equilibrium, especially when dealing with gases like dinitrogen tetroxide \(\mathrm{N}_2\mathrm{O}_4\). When \(\mathrm{N}_2\mathrm{O}_4\) is heated in a closed vessel, it partially dissociates into nitrogen dioxide \(\mathrm{NO}_2\). This process can be represented by the chemical equation \(\mathrm{N}_2\mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})\). The double arrow indicates that the reaction reaches a state of equilibrium where both the forward and reverse reactions occur at the same rate.
This balance of formation and dissociation leads to a dynamic state where the concentrations of \(\mathrm{N}_2\mathrm{O}_4\) and \(\mathrm{NO}_2\) become constant. Understanding this dissociation is essential to solving problems related to equilibrium because it determines the concentrations of reactants and products at equilibrium. By knowing the percentage of dissociation, you can calculate the amounts of each compound present when the system is at equilibrium.

Equilibrium Constant

In chemical equilibrium, the equilibrium constant \(K_c\) is a valuable mathematical representation that helps determine the relationship between the concentrations of reactants and products at equilibrium. For the reaction \(\mathrm{N}_2\mathrm{O}_4 \rightleftharpoons 2 \mathrm{NO}_2\), the equilibrium constant is given by the equation: \[ K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]} \]Where \([\mathrm{NO}_2]\) and \([\mathrm{N}_2\mathrm{O}_4]\) represent the molar concentrations of nitrogen dioxide and dinitrogen tetroxide, respectively.
This constant is specific to a given reaction at a particular temperature. It reflects how far the reaction proceeds to reach equilibrium. A higher \(K_c\) value implies a greater concentration of products at equilibrium, indicating the reaction favors the forward process. Conversely, a lower \(K_c\) suggests the equilibrium lies toward the reactants. Understanding and calculating \(K_c\) accurately is essential for predicting the behavior of chemical systems under different conditions.

Mole Calculations

Mole calculations form the backbone of quantitative chemistry. They allow us to convert between mass and the number of moles, providing a bridge between the macroscopic and molecular worlds. The mole is a standard unit in chemistry that quantifies the amount of substance, where one mole contains Avogadro's number of particles, typically atoms or molecules.
In this exercise, we started with 9.2 grams of \(\mathrm{N}_2\mathrm{O}_4\), and with the molecular weight given as 92 g/mol, we calculated the initial moles as: \[ \text{moles of } \mathrm{N}_2\mathrm{O}_4 = \frac{9.2 \text{ g}}{92 \text{ g/mol}} = 0.1 \text{ mol} \]These calculations provide the initial quantities necessary to understand how much substance is available to participate in equilibrium reactions. Accurate mole calculations ensure that we consider the precise amounts of reactants and products when predicting and analyzing chemical equilibria.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It is dictated by the balanced chemical equation and is crucial for determining how much of each substance will be consumed or produced.
In the dissociation of \(\mathrm{N}_2\mathrm{O}_4\), one mole dissociates to form two moles of \(\mathrm{NO}_2\). This 1:2 ratio is significant: it means that if 0.05 mol of \(\mathrm{N}_2\mathrm{O}_4\) dissociates, it produces: \[ 2 \times 0.05 = 0.1 \text{ mol of } \mathrm{NO}_2 \]Knowing the stoichiometric coefficients helps in predicting the amount of products formed from given quantities of reactants. It also aids in calculating equilibrium concentrations, essential for determining the equilibrium constant. Mastering stoichiometry provides the necessary tools to fully understand and predict the outcome of chemical processes.