Problem 34 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the reaction, \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})+0.5 \mathrm{O}_{2}(\mathrm{~g})\), calculate the mole fraction of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})\) decomposed at a constant volume and temperature, if the initial pressure is \(600 \mathrm{~mm} \mathrm{Hg}\) and the pressure at any time is 960 \(\mathrm{mm}\) Hg. Assume ideal gas behaviour.

Step-by-Step Solution

Answer

The mole fraction of decomposed \( \mathrm{N}_{2} \mathrm{O}_{5} \) is approximately 0.24.

Step 1: Understand the Reaction and Variables

The reaction is \( \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})+0.5 \mathrm{O}_{2}(\mathrm{~g}) \). Initially, the pressure is 600 mm Hg, representing \( P_0 \) for \( \mathrm{N}_{2} \mathrm{O}_{5} \). At any given time, the total pressure is 960 mm Hg. We need to calculate the mole fraction of \( \mathrm{N}_{2} \mathrm{O}_{5} \) decomposed.

Step 2: Define the Changes in Pressure

Initially, only \( \mathrm{N}_{2} \mathrm{O}_{5} \) is present, so its partial pressure is 600 mm Hg. Let \( x \) be the amount of \( \mathrm{N}_{2} \mathrm{O}_{5} \) decomposed. Then, 2x of \( \mathrm{NO}_{2} \) and 0.5x of \( \mathrm{O}_{2} \) are produced. These contribute to the increase in pressure: \( 2x + 0.5x = 2.5x \).

Step 3: Calculate Total Pressure Change

The initial pressure of \( \mathrm{N}_{2} \mathrm{O}_{5}\) is 600 mm Hg. After decomposition, the pressure is 960 mm Hg, indicating the pressure due to produced gases is equal to the pressure change: \( \Delta P = 960 - 600 = 360 \) mm Hg.

Step 4: Relate the Pressure Change to Decomposition

Using the equation from Step 2, where \( 2.5x = \Delta P \), solve for \( x \):\[ 2.5x = 360 \] \[ x = \frac{360}{2.5} \] \[ x = 144 \] mm Hg

Step 5: Calculate Mole Fraction of Decomposed \( \mathrm{N}_{2} \mathrm{O}_{5} \)

The mole fraction of \( \mathrm{N}_{2} \mathrm{O}_{5} \) decomposed is the ratio of decomposed to initial moles. Since x represents decomposed pressure and initial pressure was 600 mm Hg:\[ \text{Mole fraction} = \frac{x}{P_0} = \frac{144}{600} \approx 0.24 \]

Key Concepts

Ideal Gas Law

The Ideal Gas Law is an essential principle in chemistry that helps us understand how gases behave. It relates pressure, volume, temperature, and the number of moles of a gas. The law is expressed by the formula: \[ PV = nRT \] Where:

• \( P \) represents the pressure of the gas.
• \( V \) is the volume it occupies.
• \( n \) stands for the number of moles of the gas.
• \( R \) is the ideal gas constant.
• \( T \) indicates the temperature in Kelvin.

This equation is extremely useful to calculate how changes in one variable, like pressure, will affect another, such as volume, assuming the gas behaves ideally.
In our original exercise, the Ideal Gas Law assumption enables treating pressure changes due to chemical reactions directly in terms of mole ratios. This simplifies calculations and is generally applicable when dealing with gases at standard conditions of temperature and pressure.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It represents the ratio of moles of one component to the total moles of all components present. For a component \( A \) in a mixture, the mole fraction is given by:\[ X_A = \frac{n_A}{n_{total}} \] Where:

• \( X_A \) is the mole fraction of component \( A \).
• \( n_A \) is the number of moles of \( A \).
• \( n_{total} \) is the total number of moles in the mixture.

In our example, the mole fraction of decomposed \( \mathrm{N_2O_5} \) helps to understand how much of it has transformed into its products: \( \mathrm{NO_2} \) and \( \mathrm{O_2} \). A mole fraction close to 1 means nearly complete decomposition, while a fraction close to 0 means little decomposition has occurred. Calculating the mole fraction allows us to quantify this change in a simple way.

Partial Pressure

Partial pressure reflects the pressure exerted by a single type of gas in a mixture of gases. It's a crucial concept in understanding gas reactions, like the decomposition in our exercise. The total pressure exerted by a mixture of gases is the sum of the partial pressures of each gas present. For a gas component \( i \) in a mixture, its partial pressure can be expressed as:\[ P_i = X_i \times P_{total} \] Where:

• \( P_i \) is the partial pressure of component \( i \).
• \( X_i \) is the mole fraction of \( i \).
• \( P_{total} \) is the total pressure of the gas mixture.

In the earlier exercise, the initial pressure provided for \( \mathrm{N_2O_5} \) eventually changes due to reaction, resulting in different partial pressures for its products. Using partial pressures helps in predicting how the concentration of each component will influence the overall reaction, which is a foundation for understanding chemical kinetics and equilibrium. Understanding these subtle changes in gas pressure during reactions offers important insights into both theoretical and practical applications.