Problem 140 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The resultant pressure is (b) 2.4 atm.

Step 1: Understand the Problem

We start with 1 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\). When heated to 600 K, 20% of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) decomposes into \(\mathrm{NO}_{2}(\mathrm{~g})\). We need to find the new pressure of the system after this decomposition.

Step 2: Determine the Amount of Decomposition

20% of 1 mole means \(0.2\) moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) decomposes. This leaves \(0.8\) moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) unchanged.

Step 3: Calculate Moles After Decomposition

The decomposition reaction is \( \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g}) \). Therefore, \(0.2\) moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\) decomposes into \(0.4\) moles of \(\mathrm{NO}_{2}\). Total moles of gas after decomposition are: \(0.8 + 0.4 = 1.2\) moles.

Step 4: Use the Ideal Gas Law

Assuming ideal gas behavior, we use the formula: \(PV = nRT\). For the same system at different conditions, we have: \(\frac{P_1}{T_1} =\frac{P_2}{T_2}\). Initially, at 300 K, \(P_1 = 1\) atm. After heating to 600 K, let the final pressure be \(P_2\). Ceiling moles after decomposition is \(1.2\) and initial moles is \(1\), thus, \(P_2=n_2/n_1 \times T_2/T_1 \times P_1 = 1.2/1 \times 600/300 \times 1 = 2.4\) atm.

Key Concepts

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. In the exercise, we start with dinitrogen tetroxide, \(\mathrm{N}_2\mathrm{O}_4\), which breaks down into nitrogen dioxide, \(\mathrm{NO}_2\). This is represented by the chemical equation:

\[ \mathrm{N}_2\mathrm{O}_4\rightarrow 2\,\mathrm{NO}_2 \]
This means one mole of \(\mathrm{N}_2\mathrm{O}_4\) decomposes to form two moles of \(\mathrm{NO}_2\). The process is initiated by heating the compound, increasing the kinetic energy of the molecules, and thereby favoring decomposition.

• The initial amount of \(\mathrm{N}_2\mathrm{O}_4\) is 1 mole.
• 20% of \(\mathrm{N}_2\mathrm{O}_4\) decomposes, equating to 0.2 moles.
• Each mole of \(\mathrm{N}_2\mathrm{O}_4\) yields 2 moles of \(\mathrm{NO}_2\).
• Thus, 0.2 moles of \(\mathrm{N}_2\mathrm{O}_4\) yields 0.4 moles of \(\mathrm{NO}_2\).

The importance of decomposition reactions lies in the ability to produce multiple products from a single reactant, which is essential in chemical manufacturing and synthesis.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is expressed as:

\[ PV = nRT \]
Where:

• \(P\) is the pressure of the gas,
• \(V\) is the volume of the gas,
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant, and
• \(T\) is the absolute temperature in Kelvin.

In the context of the original problem, the decomposition reaction occurs under conditions that allow us to assume the gases involved behave ideally. This means the changes in pressure and temperature directly influence the gas properties according to the Ideal Gas Law.

In practice, for the problem discussed:

• The initial condition was one mole of \(\mathrm{N}_2\mathrm{O}_4\) at 300 K and 1 atm.
• The volume remains constant as it was heated to 600 K.
• By calculating the total moles of gas after decomposition, the law helps to determine the final pressure.
• Using the simplifying relation as \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\), the problem is solved to find the new pressure as 2.4 atm.

Partial Pressure

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. It is a key concept in understanding how gases behave in different conditions and plays a crucial role when applying Dalton's Law of Partial Pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases.

In the decomposition reaction, both \(\mathrm{N}_2\mathrm{O}_4\) and \(\mathrm{NO}_2\) are present in the container after heating. Each gas contributes to the total pressure of the system.

• The partial pressure of \(\mathrm{N}_2\mathrm{O}_4\) refers to the 0.8 moles remaining unreacted.
• The partial pressure of \(\mathrm{NO}_2\) is related to the 0.4 moles formed from the decomposition.
• Both gases together contribute to the final pressure of 2.4 atm within the system.

When calculating total pressure after a decomposition reaction, understanding the concept of partial pressures allows us to properly account for the contributions of each gas present. It ensures accurate predictions of system behavior and aids in solving real-life chemistry problems efficiently.