Problem 110 - A Complete Resource Book in Chemistry for JEE Main

Question

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at 300 is kept in a closed container under one atmosphere. It is heated to 600 when \(20 \%\) by mass of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) decomposes to \(\mathrm{NO}_{2}\) (g). The resultant pressure is (a) \(1.2 \mathrm{~atm}\) (b) \(2.4 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.0 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The resultant pressure is 1.2 atm.

Step 1: Understand the Decomposition Reaction

The chemical reaction is \[ \mathrm{N_2O_4}(g) \rightarrow 2 \mathrm{NO_2}(g) \]. Initially, we have 1 mole of \(\mathrm{N_2O_4}\) at 1 atm in a closed container.

Step 2: Calculate the Amount of Decomposed \(\mathrm{N_2O_4}\)

Since 20% of \(\mathrm{N_2O_4}\) by mass decomposes, \[ \text{Moles of decomposed } \mathrm{N_2O_4} = 1 \times 0.2 = 0.2 \text{ moles} \].

Step 3: Determine Moles of \(\mathrm{NO_2}\) Produced

According to the stoichiometry of the reaction, 1 mole of \(\mathrm{N_2O_4}\) gives 2 moles of \(\mathrm{NO_2}\). Hence, 0.2 moles of \(\mathrm{N_2O_4}\) will form\[ 0.2 \times 2 = 0.4 \text{ moles of } \mathrm{NO_2} \].

Step 4: Calculate Remaining Moles of \(\mathrm{N_2O_4}\)

The remaining moles of \(\mathrm{N_2O_4}\) are \[ 1 - 0.2 = 0.8 \text{ moles} \].

Step 5: Determine Total Moles of Gas at Final Condition

The total moles of gas after decomposition and recombination are \[ 0.8 + 0.4 = 1.2 \text{ moles} \].

Step 6: Calculate Final Pressure Using Ideal Gas Law

According to the ideal gas law, if the volume and temperature are constant, pressure is proportional to the number of moles. The initial moles of gas were 1 mole corresponding to 1 atm. Thus, \[ P_{final} = \left(\frac{1.2}{1}\right) \times 1 \text{ atm} = 1.2 \text{ atm} \].

Key Concepts

Decomposition Reaction

A decomposition reaction is a chemical process where a single compound breaks down into two or more simpler substances. In the case of \(\mathrm{N_2O_4}(g)\), it decomposes into \(\mathrm{NO_2}(g)\) when heated. This type of reaction usually requires energy input, such as heat, for the compound's bonds to break.
The reaction can be represented as:\[\mathrm{N_2O_4}(g) \rightarrow 2\mathrm{NO_2}(g)\]In this reaction, one mole of \(\mathrm{N_2O_4}\) decomposes to produce two moles of \(\mathrm{NO_2}\). Decomposition reactions are crucial in various industrial and laboratory processes. They help synthesize complex molecules, produce gases, and break down waste products.
When performing exercises involving decomposition reactions, it's essential to keep track of

• the reactants and products,
• the conditions required (such as temperature),
• and the stoichiometry involved.

These factors help determine the amount of products formed from a given amount of reactant.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in physical chemistry that describes the behavior of gases. The equation is expressed as \(PV = nRT\), where

• \(P\) stands for pressure, light how much force the gas exerts within the container.
• \(V\) is the volume that the gas occupies.
• \(n\) is the number of moles of gas.
• \(R\) is the ideal gas constant, a value that makes the units work.
• \(T\) is the temperature, measured in Kelvin.

In this exercise, the volume and temperature remain constant when the pressure changes, resulting from decomposition.
If you increase the number of moles of gas, pressure has to increase to keep everything balanced, assuming volume and temperature are constant.In the given scenario, the initial 1 mole of gas at 1 atm increased to 1.2 moles due to decomposition. Thus, the pressure increased proportionately to 1.2 atm. This principle is essential for solving problems that involve changes in gas quantities under controlled conditions.

Stoichiometry

Stoichiometry refers to the calculation of reactants and products in chemical reactions. It is fundamentally rooted in the conservation of mass, meaning matter is neither created nor destroyed.
To solve problems with stoichiometry, like in this decomposition reaction of \(\mathrm{N_2O_4} \), you need to understand

• the balanced chemical equation,
• the mole ratio of reactants to products,
• and the law of conservation of mass.

For the given exercise:- The reaction \(\mathrm{N_2O_4}(g) \rightarrow 2 \mathrm{NO_2}(g)\) is balanced.- 1 mole of \(\mathrm{N_2O_4}\) produces 2 moles of \(\mathrm{NO_2}\).Understanding the mole ratio is critical. In the problem, 0.2 moles of \(\mathrm{N_2O_4}\) decomposed, forming 0.4 moles of \(\mathrm{NO_2}\). By following these ratios and using stoichiometry, you can predict how much of each product forms from a given amount of reactant, and determine the total moles after the reaction.