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 \(\mathrm{O}_{2}\). As soon as \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) are formed they react to form \(\mathrm{N}_{2} \mathrm{O}_{5}\). Two equilibria $$ \begin{aligned} \mathrm{N}_{2} \mathrm{O}_{4} & \rightleftharpoons 2 \mathrm{NO}+\mathrm{O}_{2} \\ 2 \mathrm{NO}+\frac{3}{2} \mathrm{O}_{2} &=\mathrm{N}_{2} \mathrm{O}_{3} \end{aligned} $$ 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) \([\mathrm{NO}]+\left[\mathrm{O}_{2}\right]=\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]+\left[\mathrm{N}_{2} \mathrm{O}_{3}\right]\) (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 (d) is correct: \(\frac{1}{2}[\mathrm{N}_2\mathrm{O}_5]+[\mathrm{O}_2]=\frac{1}{2}[\mathrm{NO}]\)."

Step 1: Initial Conditions and Degree of Dissociation

We are given that the degree of dissociation of \(\mathrm{N}_2\mathrm{O}_4\) is 50%. This means that if we start with 2 moles of \(\mathrm{N}_2\mathrm{O}_4\), then 1 mole dissociates according to the first reaction. Thus, at equilibrium, 1 mole of \(\mathrm{N}_2\mathrm{O}_4\) remains, and 1 mole forms 2 moles of \(\mathrm{NO}\) and 0.5 moles of \(\mathrm{O}_2\).

Step 2: Determine Concentrations from First Equilibrium

From the dissociated \(1\) mole of \(\mathrm{N}_2\mathrm{O}_4\), \(2\) moles of \(\mathrm{NO}\) and \(0.5\) moles of \(\mathrm{O}_2\) are produced. Therefore, at equilibrium:- \(x = 2\) moles of \(\mathrm{NO}\).- \(y = 0.5\) moles of \(\mathrm{O}_2\).

Step 3: Analyze Second Reaction

According to the second equation, \(2\mathrm{NO} + \frac{3}{2}\mathrm{O}_2 \rightarrow \mathrm{N}_2\mathrm{O}_5\), the \(\mathrm{NO}\) and \(\mathrm{O}_2\) formed will react further to establish the second equilibrium. The equilibrium ratio consumes \(\mathrm{NO}\) and \(\mathrm{O}_2\) to form \(\mathrm{N}_2\mathrm{O}_5\), but we are not given exact concentrations for each but know they both contribute towards \(\mathrm{N}_2\mathrm{O}_5\).

Step 4: Evaluate Provided Options

Now we will test each option based on the calculated concentrations:- **Option (a):** \(\frac{1}{2}[\mathrm{NO}]=\frac{3}{2}[\mathrm{O}_2]\). This implies \(\frac{1}{2}\times 2 = \frac{3}{2}\times 0.5\Rightarrow 1 = 0.75\). This is not correct.- **Option (b):** \(2[\mathrm{N}_2\mathrm{O}_4]=[\mathrm{NO}]+\frac{3}{2}[\mathrm{O}_2]+[\mathrm{N}_2\mathrm{O}_5]\) implies \(2=2+0.75+z\), where \(z\) is the concentration of \(\mathrm{N}_2\mathrm{O}_5\). This is not possible without proper values for \([\mathrm{N}_2\mathrm{O}_5]\).- **Option (c):** \([\mathrm{NO}]+[\mathrm{O}_2]=[\mathrm{N}_2\mathrm{O}_4]+[\mathrm{N}_2\mathrm{O}_3]\) gives \(2 + 0.5 = 1 + 2\), also incorrect.- **Option (d):** \(\frac{1}{2}[\mathrm{N}_2\mathrm{O}_5]+[\mathrm{O}_2]=\frac{1}{2}[\mathrm{NO}]\) implies \(\frac{1}{2}z + 0.5 = 1\), meaning \([\mathrm{N}_2\mathrm{O}_5] = 1\). This is valid given the conditions.

Key Concepts

Degree of Dissociation

In chemical equilibrium, the degree of dissociation refers to the fraction of a substance that dissociates into its components. Consider when you start with a specific amount of a compound in a reaction. A part of it will break down or dissociate into different substances.
For example, in the exercise, \(\mathrm{N}_2\mathrm{O}_4\) initially undergoes dissociation. Starting with 2 moles of \(\mathrm{N}_2\mathrm{O}_4\), and given a 50% degree of dissociation, 1 mole breaks down into the products, leaving 1 mole of \(\mathrm{N}_2\mathrm{O}_4\) remaining.
The dissociation process can be expressed as:

• 1 mole of \(\mathrm{N}_2\mathrm{O}_4\) forms 2 moles of \({NO}\) and 0.5 moles of \({O}_2\).

Understanding this concept is crucial for calculating equilibrium concentrations, as the degree of dissociation affects the overall balance of substances in a reaction mixture.

Reaction Mechanism

A reaction mechanism provides a detailed step-by-step description of how reactants transform into products. It tells us what happens in between the start and finish of a chemical reaction.
In the given scenario, the mechanism involves two reactions. The first reaction involves the dissociation of \(\mathrm{N}_2\mathrm{O}_4\) into \({NO}\) and \({O}_2\):

• \(\mathrm{N}_2\mathrm{O}_4 \rightleftharpoons 2\mathrm{NO} + \mathrm{O}_2 \)

Then, the produced \({NO}\) and \({O}_2\) react further in a second equilibrium to form \(\mathrm{N}_2\mathrm{O}_5\):

• \(2\mathrm{NO} + \frac{3}{2}\mathrm{O}_2 \rightarrow \mathrm{N}_2\mathrm{O}_3 \)

These mechanisms show how initial reactants can transform into various intermediates before reaching a final product, emphasizing the complexity of chemical reactions.

Equilibrium Concentrations

Equilibrium concentrations represent the amount of each substance present at equilibrium in a reversible reaction. At equilibrium, rates of forward and backward reactions are equal, so concentrations remain constant.
To determine equilibrium concentrations, consider the initial amounts and changes due to reactions. Using the degree of dissociation, you determine how much \(\mathrm{N}_2\mathrm{O}_4\), \({NO}\), and \({O}_2\) remain or form at equilibrium.

• After the dissociation of 1 mole of \(\mathrm{N}_2\mathrm{O}_4\), we have 2 moles of \({NO}\) and 0.5 moles of \({O}_2\).

Any further transformations, like the formation of \(\mathrm{N}_2\mathrm{O}_5\), need to account for these mole amounts. Determining and understanding the concentrations at equilibrium are key for predicting the behavior and yield of a chemical reaction.