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 \(\left.\operatorname{mol} \mathrm{L}^{-1}\right)\) ? (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 (option b).

Step 1: Calculate Initial Moles

First, determine the initial moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) in the vessel. Given that the weight of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is 9.2 grams and its molecular weight is 92 g/mol, the initial moles can be calculated as follows: \[\text{Moles of } \mathrm{N}_{2} \mathrm{O}_{4} = \frac{9.2}{92} = 0.1 \text{ moles}\].

Step 2: Determine Moles at Equilibrium

At equilibrium, 50% of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is dissociated. This means \(0.05\) moles (half of 0.1 moles) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) have dissociated to form \(\mathrm{NO}_{2}\). Thus, the amount of \(\mathrm{N}_{2} \mathrm{O}_{4}\) remaining is \(0.1 - 0.05 = 0.05\) moles, and \(\mathrm{NO}_{2}\) produced is \(2 \times 0.05 = 0.10\) moles because each mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) gives two moles of \(\mathrm{NO}_{2}\) upon dissociation.

Step 3: Write the Expression for Equilibrium Constant

The equilibrium constant (\(K_c\)) for the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\) can be expressed in terms of concentrations:\[K_c = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{N}_{2} \mathrm{O}_{4}]}\]where \([\mathrm{NO}_{2}] = 0.10 \text{ M}\) and \([\mathrm{N}_{2} \mathrm{O}_{4}] = 0.05 \text{ M}\).

Step 4: Calculate the Equilibrium Constant

Substitute the equilibrium concentrations into the equilibrium expression:\[K_c = \frac{(0.10)^2}{0.05} = \frac{0.01}{0.05} = 0.2\]

Key Concepts

Equilibrium Constant Calculation

The equilibrium constant, often denoted as \( K_c \), is a crucial concept that helps us understand how concentrations of reactants and products relate to each other in a chemical equilibrium.
Imagine a reaction reaching a point where the rates of the forward and backward reactions are equal. At this point, the concentrations of the involved substances remain constant over time. The equilibrium constant quantifies the ratio of concentrations of products to reactants when this equilibrium is achieved.
For the dissociation of \( \mathrm{N_2O_4} \) to \( \mathrm{NO_2} \), the equation is expressed as:

• \( K_c = \frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} \)

By substituting the known equilibrium concentrations into this formula, we can calculate \( K_c \). This value provides insight into the extent to which the reaction proceeds before reaching equilibrium. A larger \( K_c \) indicates a higher concentration of products, suggesting the reaction favors product formation.

Dissociation Reactions

Dissociation reactions are when a compound breaks down into two or more simpler substances. This plays a significant role in chemical equilibria and affects the overall balance of reactions.
In the case of \( \mathrm{N_2O_4} \) dissociation, the molecule decomposes into two \( \mathrm{NO_2} \) molecules.
This process of splitting can be described as:

• \( \mathrm{N_2O_4(g)} \rightarrow 2 \mathrm{NO_2(g)} \)

Understanding and calculating the extent of dissociation is essential. Here, given that \(50\%\) of \( \mathrm{N_2O_4} \) dissociates, we know exactly how many moles of the original compound are left and how many have turned into \( \mathrm{NO_2} \). Grasping this process allows for a more accurate description of chemical equilibria, as it directly impacts concentration calculations.

Equilibrium Concentrations

Equilibrium concentrations refer to the amounts of different substances in a reaction mixture once equilibrium is established.
They are crucial for calculating the equilibrium constant and understanding the state of the reaction.
In the given example with \( \mathrm{N_2O_4} \) and \( \mathrm{NO_2} \), equilibrium concentrations were determined by observing that \(50\%\) of \( \mathrm{N_2O_4} \) dissociates.

• The remaining \( \mathrm{N_2O_4} \) concentration is \(0.05\) M.
• The \( \mathrm{NO_2} \) concentration produced is \(0.10\) M.

Measuring these concentrations accurately is fundamental.
Using this data, we substitute the values into the equilibrium expression to find \( K_c \). Understanding how equilibrium concentrations determine the direction and completion of a reaction gives insight into the chemical dynamics at play.