Problem 33 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The degree of dissociation is \(0.4\) at \(400 \mathrm{~K}\) and \(1.0 \mathrm{~atm}\) for the gaseous reaction \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\). Assuming ideal behaviour of all gases, calculate the density of equilibrium mixture at \(400 \mathrm{~K}\) and \(1.0\) atmosphere. (Relative atomic mass of \(\mathrm{P}=31.0\) and \(\mathrm{Cl}=35.5\) ) \([1998-3\)

Step-by-Step Solution

Answer

The density of the equilibrium mixture is 4.54 g/L.

Step 1: Determine Molar Masses

Calculate the molar mass of each compound involved in the reaction. The molar mass of \( \mathrm{PCl}_5 \) is \(M(\mathrm{PCl}_5) = 31.0 + 5 \times 35.5 = 208.5 \) g/mol and for \( \mathrm{PCl}_3 \), \(M(\mathrm{PCl}_3) = 31.0 + 3 \times 35.5 = 137.5 \) g/mol, and for \( \mathrm{Cl}_2 \), \(M(\mathrm{Cl}_2) = 2 \times 35.5 = 71.0 \) g/mol.

Step 2: Establish Initial and Equilibrium Moles

Assume you start with 1 mole of \( \mathrm{PCl}_5 \). At equilibrium, due to the degree of dissociation \( \alpha = 0.4 \), \( 0.4 \) moles of \( \mathrm{PCl}_5 \) dissociate to form \( 0.4 \) moles of \( \mathrm{PCl}_3 \) and \( 0.4 \) moles of \( \mathrm{Cl}_2 \). Thus, the equilibrium moles of \( \mathrm{PCl}_5 = 1 - 0.4 = 0.6 \), \( \mathrm{PCl}_3 = 0.4 \), and \( \mathrm{Cl}_2 = 0.4 \).

Step 3: Calculate Total Moles at Equilibrium

Total moles at equilibrium = \( 0.6 + 0.4 + 0.4 = 1.4 \text{ moles} \).

Step 4: Calculate Total Molar Mass of Mixture

Calculate the mass of each component: \( \mathrm{PCl}_5 \): \(0.6 \times 208.5 = 125.1 \) g, \( \mathrm{PCl}_3 \): \(0.4 \times 137.5 = 55 \) g, and \( \mathrm{Cl}_2 \): \(0.4 \times 71.0 = 28.4 \) g. Sum gives total mass of mixture = \(125.1 + 55 + 28.4 = 208.5 \) g.

Step 5: Calculate Density of Mixture

Density \( \rho \) is defined as \( \rho = \frac{\text{mass}}{\text{volume}} \). Using the ideal gas law, \(PV = nRT\), where at \(1.0 \text{ atm}\) and \(400 \text{ K}\), \(V = \frac{1.4 \times 0.0821 \times 400}{1} = 45.96 \text{ L} \). So, density \( \rho = \frac{208.5 \text{ g}}{45.96 \text{ L}} = 4.54 \text{ g/L} \).

Key Concepts

Degree of Dissociation

The degree of dissociation is a concept used in chemistry to describe how much of a compound breaks down into its components. It's crucial for understanding how substances interact in a reaction, especially when dealing with reversible reactions, as in the case of \( \mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3 + \mathrm{Cl}_2 \).
In this exercise, the degree of dissociation (\( \alpha \)) is given as 0.4, meaning that 40% of the original \( \mathrm{PCl}_5 \) has dissociated into \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \).
This can be visualized as follows: if you start with 1 mole of \( \mathrm{PCl}_5 \), at equilibrium, 0.4 moles dissociate. This process affects how we calculate quantities at equilibrium.

• Initial moles of \( \mathrm{PCl}_5 \): 1 mole
• Equilibrium moles of \( \mathrm{PCl}_5 \): \(1 - 0.4 = 0.6 \text{ moles}\)
• Produced moles of \( \mathrm{PCl}_3 \): 0.4 moles
• Produced moles of \( \mathrm{Cl}_2 \): 0.4 moles

Overall, understanding the degree of dissociation helps predict the concentration of different species in a chemical equilibrium, which is essential for further calculations like density or reaction dynamics.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry and physics used to describe the behavior of gases. It is expressed as \( PV = nRT \), where \( P \) stands for pressure, \( V \) for volume, \( n \) for number of moles, \( R \) for the ideal gas constant, and \( T \) for temperature in Kelvin.
In this problem, the Ideal Gas Law helps in calculating the volume of the gas mixture at equilibrium. We know the following:

• Pressure (\( P \)) = 1.0 atm
• Total moles at equilibrium (\( n \)) = 1.4 moles
• Temperature (\( T \)) = 400 K
• Gas constant (\( R \)) = 0.0821 L·atm/mol·K

Using the Ideal Gas Law, we find: \[ V = \frac{nRT}{P} = \frac{1.4 \times 0.0821 \times 400}{1} = 45.96 \text{ L} \]So, all components combined occupy 45.96 liters in an ideal condition. By using the Ideal Gas Law, we assume no interactions between gas particles, making calculations straightforward when the system conditions are close to ideal.

Molar Mass Calculation

Molar mass is defined as the mass of one mole of a substance (usually in g/mol). It is a critical figure in stoichiometry, allowing you to convert between moles and grams.
In the given reaction, we calculate the molar masses for each participating gas:

• \( \mathrm{PCl}_5 \): \[ M(\mathrm{PCl}_5) = 31.0 + 5 \times 35.5 = 208.5 \text{ g/mol} \]
• \( \mathrm{PCl}_3 \): \[ M(\mathrm{PCl}_3) = 31.0 + 3 \times 35.5 = 137.5 \text{ g/mol} \]
• \( \mathrm{Cl}_2 \): \[ M(\mathrm{Cl}_2) = 2 \times 35.5 = 71.0 \text{ g/mol} \]

For a practical example, at equilibrium, we determine the mass contributions from each component. This involves multiplying the number of moles by their respective molar masses:

• \( \mathrm{PCl}_5 \): \[ 0.6 \times 208.5 = 125.1 \text{ g} \]
• \( \mathrm{PCl}_3 \): \[ 0.4 \times 137.5 = 55.0 \text{ g} \]
• \( \mathrm{Cl}_2 \): \[ 0.4 \times 71.0 = 28.4 \text{ g} \]

Summing these gives a total mass of 208.5 grams for the mixture, which can then be used to calculate the density or other properties. Understanding molar mass is foundational to translating theoretical chemistry into practical results.