Problem 119 - A Complete Resource Book in Chemistry for JEE Main

Question

Two moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is heated to form \(\mathrm{NO}\) and \(\mathrm{O}_{4}\). As soon as \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) are formed they react to form \(\mathrm{N}_{2} \mathrm{O}_{5}\). Two equilibria \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}+\mathrm{O}_{2}\) \(2 \mathrm{NO}+\frac{3}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{3}\) Are simultaneously established. At equilibrium, the degree of dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) was found to \(50 \%\). Which of the following is correct at equilibrium? (a) \(\frac{1}{2}[\mathrm{NO}]=\frac{3}{2}\left[\mathrm{O}_{2}\right]\) (b) \(2\left[\mathrm{~N}_{2} \mathrm{O}_{4}\right]=[\mathrm{NO}]+\frac{3}{2}\left[\mathrm{O}_{2}\right]+\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\) (c) \([\mathrm{NO}]+\left[\mathrm{O}_{2}\right]=\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]+\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\) (d) \(\frac{1}{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]+\left[\mathrm{O}_{2}\right]=\frac{1}{2}[\mathrm{NO}]\)

Step-by-Step Solution

Answer

Option (b) is correct: \(2\left[\mathrm{N}_{2}\mathrm{O}_{4}\right]=[\mathrm{NO}]+\frac{3}{2}\left[\mathrm{O}_{2}\right]+\left[\mathrm{N}_{2}\mathrm{O}_{5}\right]\)."

Step 1: Understand the Reaction Process

First, acknowledge the reactions involved. The first equilibrium is \(_2\mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO} + \mathrm{O}_{2}\), and the second equilibrium is \(2 \mathrm{NO} + \frac{3}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{5}\). This involves decomposition of \(\mathrm{N}_{2}\mathrm{O}_{4}\) and subsequent formation of \(\mathrm{N}_{2} \mathrm{O}_{5}\).

Step 2: Calculate Moles at Equilibrium

There are 2 moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\), with a dissociation of \(50\%\). This means \(1\) mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) dissociates to form \(2\) moles of \(\mathrm{NO}\) and \(1\) mole of \(\mathrm{O}_{2}\) at equilibrium.

Step 3: Analyze the Second Reaction

The \(2 \mathrm{NO} + \frac{3}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{5}\) reaction uses \(2\) moles of \(\mathrm{NO}\) and \(\frac{3}{2}\) moles of \(\mathrm{O}_{2}\) to form \(\mathrm{N}_{2} \mathrm{O}_{5}\), meaning that all \(\mathrm{NO}\) produced initially is consumed.

Step 4: Determine Final Concentrations

Finally, note that at equilibrium, all \(\mathrm{NO}\) formed initially is used to form \(\mathrm{N}_{2} \mathrm{O}_{5}\). Thus, \(\mathrm{N}_{2} \mathrm{O}_{4} = 1\) mole, \(\mathrm{NO} = 0\) moles (after reacting), \(\mathrm{O}_{2} = 0.5\) moles, and \(\mathrm{N}_{2} \mathrm{O}_{5} = 1\) mole from the stoichiometry of the reactions.

Step 5: Check Given Statements

(a) and (b) can be assessed: (a) \(\frac{1}{2}[\mathrm{NO}]=\frac{3}{2}\left[\mathrm{O}_{2}\right]\) - not true, as [NO] becomes 0. (b) \(2\left[\mathrm{N}_{2}\mathrm{O}_{4}\right]=[\mathrm{NO}]+\frac{3}{2}\left[\mathrm{O}_{2}\right]+\left[\mathrm{N}_{2}\mathrm{O}_{5}\right]\) - holds true with the stoichiometry, as both sides equal 2. For (c) and (d), they do not match balance.

Key Concepts

Degree of Dissociation

The degree of dissociation refers to the fraction of molecules that dissociate into simpler molecules or atoms in a chemical reaction. In our given chemical equilibrium, the degree of dissociation of \(\text{N}_2\text{O}_4\) is 50%.
This indicates that half of the \(\text{N}_2\text{O}_4\) molecules have broken down into \(\text{NO}\) and \(\text{O}_2\). In numerical terms, if we start with 2 moles of \(\text{N}_2\text{O}_4\), then 1 mole of it dissociates, leading to the formation of 2 moles of \(\text{NO}\) and 1 mole of \(\text{O}_2\).
Understanding the degree of dissociation allows us to predict the amounts of products and reactants at equilibrium by recognizing what proportion of the original reactants has broken down or reformed into other chemicals. This is crucial for calculating the outcome of reversible chemical reactions.

Equilibrium Constant

The equilibrium constant, denoted as \(K_c\), measures the relative concentrations of the products and reactants of a reversible reaction at equilibrium. In the reactions involving \(\text{N}_2\text{O}_4\) dissociating into \(\text{NO}\) and \(\text{O}_2\) and the subsequent formation of \(\text{N}_2\text{O}_5\), the equilibrium constants provide vital information.
They determine the extent to which reactants are converted into products at equilibrium. For a chemical equilibrium expressed as \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant is written as:
\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
At equilibrium, changes in the chemical atmosphere, such as temperature adjustments, can affect \(K_c\). A high \(K_c\) value signals that the formation of products at equilibrium is favored, leading to more products than reactants. Conversely, a low \(K_c\) indicates that the reactants are more favored.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In our problem, stoichiometry helps us determine the exact amounts of \(\text{N}_2\text{O}_4\), \(\text{NO}\), \(\text{O}_2\), and \(\text{N}_2\text{O}_5\) at equilibrium.
This is achieved by balancing the number of atoms of each element in the reactants and products, ensuring that matter is conserved. For instance, when 1 mole of \(\text{N}_2\text{O}_4\) dissociates, it forms 2 moles of \(\text{NO}\) and 1 mole of \(\text{O}_2\). Then, these products react further to create \(\text{N}_2\text{O}_5\).
Understanding the stoichiometry of this problem confirms that chemical equilibrium has been achieved when the rate of the forward reaction equals the rate of the reverse reaction. This equilibrium is reflected in the stability of the concentrations of all reactants and products involved, within the constraints of stoichiometric balance.