Problem 19 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at \(300 \mathrm{~K}\) is kept in a closed container under one atmosphere. 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) \(1.2 \mathrm{~atm}\) (b) \(2.4 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.0 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The resultant pressure is 2.4 atm.

Step 1: Determine initial moles and conditions

Initially, you have 1 mole of \(\mathrm{N}_2\mathrm{O}_4\) at 300 K and 1 atm. When the temperature increases to 600 K, 20% by mass of \(\mathrm{N}_2\mathrm{O}_4\) decomposes into \(\mathrm{NO}_2\).

Step 2: Calculate moles decomposed

Since 20% by mass of \(\mathrm{N}_2\mathrm{O}_4\) decomposes, it means 0.2 moles of \(\mathrm{N}_2\mathrm{O}_4\) decomposes, since 1 mole initially was present. This leads to 0.8 moles remaining.

Step 3: Write balanced chemical equation

The balanced decomposition equation is: \[ \mathrm{N}_2\mathrm{O}_4(g) \rightarrow 2 \mathrm{NO}_2(g) \]. This implies each mole of \(\mathrm{N}_2\mathrm{O}_4\) decomposes to form 2 moles of \(\mathrm{NO}_2\).

Step 4: Calculate moles of gas after decomposition

For 0.2 moles of \(\mathrm{N}_2\mathrm{O}_4\) that decomposes, it forms \(0.2 \times 2 = 0.4\) moles of \(\mathrm{NO}_2\). Thus, the moles after decomposition are 0.8 moles of \(\mathrm{N}_2\mathrm{O}_4\) remaining and 0.4 moles of \(\mathrm{NO}_2\).

Step 5: Find total moles of gas post-reaction

Combine moles after decomposition: \(0.8 + 0.4 = 1.2\) moles of gas in total after the reaction.

Step 6: Use ideal gas equation for pressure

Assuming ideal behavior, using \(PV = nRT\), and noting that volume \(V\) and gas constant \(R\) stay the same, the new pressure \(P_2\) at 600 K can be calculated by \(P_2 = \frac{(1.2 \text{ moles})( ext{600 K})}{(1 \text{ mole})(300 \text{ K})} \times 1 \, \mathrm{atm}\).

Step 7: Calculate resultant pressure

The resultant pressure is \(P_2 = 2.4 \, \mathrm{atm}\) after substituting the number of moles and temperature into the pressure formula.

Key Concepts

Gas Laws

Gas laws help us understand the behavior of gases under varying conditions of temperature, pressure, and volume. These laws provide crucial insights into how gases expand, compress, and react with changes in their environment.

There are several fundamental gas laws that include Boyle's Law, Charles's Law, and Avogadro’s Law. These are all part of the ideal gas law which is often used in calculations.

• Boyle's Law: States that the pressure of a gas is inversely proportional to its volume, at constant temperature. Mathematically, it is represented as: \[ P_1V_1 = P_2V_2 \]
• Charles's Law: Indicates that the volume of a gas is directly proportional to its temperature, at constant pressure. It can be expressed by: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
• Avogadro’s Law: Argues that volume is directly proportional to the number of moles of gas, with constant temperature and pressure. Simplified as: \[ V \propto n \]

These principles form the foundation for understanding more complex gas behaviors like in the decomposition of a gas, as demonstrated in the given problem scenario.

Decomposition Reactions

Decomposition reactions involve breaking down a single substance into two or more simpler substances. In chemical terms, a larger molecule decomposes to form simpler molecules.

In the given problem, we observe the decomposition of dinitrogen tetroxide (\[\mathrm{N}_2\mathrm{O}_4\]) into nitrogen dioxide (\[\mathrm{NO}_2\]). This can be represented with the chemical equation:\[\mathrm{N}_2\mathrm{O}_4\rightarrow 2\mathrm{NO}_2\].

Such reactions are usually accompanied by a change in physical conditions, such as temperature or pressure. The process may also involve a phase shift, say from gas to another state, although our particular exercise remains in the gas phase.

• For every mole of \[\mathrm{N}_2\mathrm{O}_4\] decomposed, two moles of \[\mathrm{NO}_2\] are produced.
• This change significantly affects the number of moles present in the system post-reaction, vital in calculating resultant pressures.

Understanding decomposition helps predict outcomes like pressure changes when dealing with gas mixtures post-reaction.

Ideal Gas Equation

The ideal gas equation is a mathematical model used to describe the behavior of an ideal gas. This equation integrates multiple gas laws into one useful formula: \[ PV = nRT \]. Here:

• \( P \) represents pressure in atmospheres (atm).
• \( V \) stands for volume in liters (L).
• \( n \) is the amount of substance in moles (mol).
• \( R \) is the ideal gas constant (0.0821 Latm/molK).
• \( T \) represents the absolute temperature measured in Kelvin (K).

The ideal gas equation allows chemists to calculate unknown properties of a gas when other variables are known.

In this problem, the temperature increase altered the pressure due to changes in the number of moles of gas. As a result, we applied the equation \[ P_2 = \frac{n_2 \cdot T_2}{n_1 \cdot T_1} \cdot P_1 \]. This rearranged form helps in finding the new pressure after the reaction takes place at a different temperature.
This equation provides a simplified yet powerful tool to predict and analyze the behavior of gases as conditions vary within any given reaction scenario.