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 \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) is dissociated. What is the equilibrium constant (in \(\mathrm{mol} \mathrm{L}^{-1}\) )? (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 \( K_c \) is 0.2. The correct answer is (b) 0.2.

Step 1: Calculate the initial moles of N2O4

First, we need to find the initial moles of \(\mathrm{N}_{2}\mathrm{O}_{4}\). Use the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Given mass is 9.2 g and the molar mass is 92 g/mol, the initial moles are \( \frac{9.2}{92} = 0.1 \text{ moles} \).

Step 2: Determine moles of dissociated N2O4

At equilibrium, 50% of \(\mathrm{N}_{2}\mathrm{O}_{4}\) is dissociated. Thus, the moles dissociated is \(0.1 \times 0.5 = 0.05 \text{ moles} \). This amount dissociates to form \(2 \times 0.05 = 0.1 \text{ moles of } \mathrm{NO}_{2}\).

Step 3: Calculate the equilibrium moles

At equilibrium, the moles of \(\mathrm{N}_{2}\mathrm{O}_{4}\) is \(0.1 - 0.05 = 0.05 \text{ moles}\). The moles of \(\mathrm{NO}_{2}\) is 0.1 moles (calculated from dissociation).

Step 4: Calculate equilibrium concentrations

The equilibrium concentrations in a 1 L vessel correspond directly to the moles. Thus, [\(\mathrm{N}_{2}\mathrm{O}_{4}\)] = 0.05 M and [\(\mathrm{NO}_{2}\)] = 0.1 M.

Step 5: Use the equilibrium expression to calculate Kc

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

Key Concepts

Equilibrium Constant (Kc)

The equilibrium constant, often denoted as \( K_c \), provides insight into the position of equilibrium in a chemical reaction. It mathematically describes the ratio of the concentrations of products to reactants at the point of equilibrium. For the given reaction, \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2 \text{NO}_2(g) \), the formula to find \( K_c \) is:

• The concentration of \( \text{NO}_2 \), squared, because there are two moles produced.
• Divided by the concentration of \( \text{N}_2\text{O}_4 \).

At equilibrium, each concentration value is a measure of how the substances distribute when the forward and reverse reaction rates are equal. In this example, the calculated \( K_c \) value is 0.2, indicating the extent to which the reactants have turned into products.

Molecular Weight

Molecular weight, also known as molar mass, is a measure used to compare the mass of molecules. It is expressed in grams per mole (g/mol) and is crucial for converting masses into moles, which are needed in calculations involving chemical reactions. For \( \text{N}_2\text{O}_4 \), the molecular weight is given as 92 g/mol. By knowing this, we can compute the initial moles using the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).This allows the conversion of the provided 9.2 grams of \( \text{N}_2\text{O}_4 \) into moles, which is necessary for further calculations to determine the extent of dissociation and equilibrium concentrations.

Dissociation

Dissociation is a process where a compound breaks down into two or more components. In our problem, \( \text{N}_2\text{O}_4 \) dissociates into \( \text{NO}_2 \). It is crucial to understand the stoichiometry of the reaction:

• For every mole of \( \text{N}_2\text{O}_4 \) that dissociates, two moles of \( \text{NO}_2 \) are formed.

In this scenario, we know that 50% of the original \( \text{N}_2\text{O}_4 \) molecules dissociate, providing us with a direct method to calculate how many moles of each substance are present at equilibrium. By understanding dissociation, we can accurately predict the dynamics at play when reaching equilibrium.

Initial Moles Calculation

Initial moles calculation is a fundamental step in solving chemical equilibrium problems. It involves converting the given mass of a compound into moles using its molecular weight.For our \( \text{N}_2\text{O}_4 \) scenario, we begin with 9.2 grams. Using the given molar mass of 92 g/mol, we find the initial moles by applying:

• \( \text{moles} = \frac{9.2 \text{ grams}}{92 \text{ g/mol}} = 0.1 \text{ moles} \)

This initial moles calculation sets the stage for determining the system's behavior as it approaches equilibrium. It is the foundation upon which further dissociation and concentration assessments are based, ultimately leading to the calculation of the equilibrium constant \( K_c \).