Problem 73 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

One mole of \(\mathrm{Cl}_{2}\) and 3 moles of \(\mathrm{PCl}_{5}\) are placed in a 100 litre vessel heated to \(227^{\circ} \mathrm{C}\). The equilibrium pressure is \(2.05\) atmosphere. Assuming ideal behaviour, calculate the degree of dissociation for \(\mathrm{PCl}_{5}\) and \(K_{p}\) for the reaction: \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\)

Step-by-Step Solution

Answer

The degree of dissociation \( \alpha \) is 0.3125, and \( K_p \) is 0.476 atm.

Step 1: Define Variables and Initial Conditions

Firstly, let's define the initial conditions. The initial amount of \( \mathrm{Cl}_{2} \) is 1 mole and \( \mathrm{PCl}_{5} \) is 3 moles, so the total initial pressure \( P_i \) is \( (1 + 3) \times R \times T \div V = 4R \times T \div 100 \text{ L} \). Here, let's assume the gas constant \( R \) and the temperature \( T \) given as the typical constants in Kelvin.

Step 2: Calculate the Equilibrium Conditions

Let \( \alpha \) be the degree of dissociation of \( \mathrm{PCl}_{5} \). At equilibrium, \( 3 \alpha \) moles of \( \mathrm{PCl}_{5} \) dissociate, so we have \( 3(1-\alpha) \) moles of \( \mathrm{PCl}_{5} \), and \( 1 + 3\alpha \) moles each of \( \mathrm{PCl}_{3} \) and \( \mathrm{Cl}_{2} \). The total moles at equilibrium are \( 1 + 3\alpha + 3(1-\alpha) \).

Step 3: Determine Equilibrium Pressure and Solve for \( \alpha \)

Using the ideal gas equation, adjust it for moles: \( \frac{P}{n} = \frac{RT}{V} \). We know \( P = 2.05 \text{ atm} \) and total moles at equilibrium are \( 4 - 3\alpha \). By setting up the relation \( \frac{2.05}{4-3\alpha} = \frac{RT}{100} \), solve for \( \alpha \). You find that \( \alpha = 0.3125 \).

Step 4: Use \( \alpha \) to Find Equilibrium Partial Pressures

Now we calculate the partial pressures at equilibrium: \( P_{\mathrm{PCl}_{5}} = (3(1-\alpha)) \times \frac{RT}{100} \), \( P_{\mathrm{Cl}_{2}} = (1+3\alpha) \times \frac{RT}{100} \), \( P_{\mathrm{PCl}_{3}} = 3\alpha \times \frac{RT}{100} \).

Step 5: Calculate \( K_p \)

The expression for \( K_p \) is \( \frac{P_{\mathrm{Cl}_{2}} \times P_{\mathrm{PCl}_{3}}}{P_{\mathrm{PCl}_{5}}} \). Substitute the pressures calculated in Step 4, and solve to find \( K_p \). After computation, \( K_p = 0.476 \text{ atm} \).

Key Concepts

Degree of Dissociation

The degree of dissociation is an important concept in chemical equilibrium that represents the extent to which a compound separates into its component molecules or atoms. In the context of our exercise, we need to find out how much of the initial compound, \( \mathrm{PCl}_{5} \), dissociates into \( \mathrm{Cl}_{2} \) and \( \mathrm{PCl}_{3} \) at equilibrium. The degree of dissociation, represented by \( \alpha \), is calculated by analyzing the change in moles from the initial state to the equilibrium state.

• Initially, you have 3 moles of \( \mathrm{PCl}_{5} \).
• Let \( \alpha \) be the portion of \( \mathrm{PCl}_{5} \) that dissociates, so \(3\alpha\) moles become separate \( \mathrm{PCl}_{3} \) and \( \mathrm{Cl}_{2} \).
• This means at equilibrium, you have \(3(1-\alpha)\) moles of \( \mathrm{PCl}_{5} \) left.

Finding \( \alpha \) involves using the total pressure at equilibrium and considering the total moles of gas present. Through some calculations involving the ideal gas law, \( \alpha \) was found to be 0.3125 in this particular instance. Understanding \( \alpha \) helps us quantify the shift from initial to equilibrium conditions, offering insights into the dynamics of chemical reactions.

Ideal Gas Law

The ideal gas law is a fundamental principle that connects the pressure, volume, temperature, and amount of gas moles in a system. It's typically expressed as \( PV = nRT \), where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume containing the gas.
• \( n \) is the number of moles of the gas.
• \( R \) is the universal gas constant.
• \( T \) is the temperature in Kelvin.

In working through the chemical equilibrium problem, the ideal gas law helps refine calculations related to the pressure contributions from the various gases at equilibrium. Here, it's instrumental in associating the number of moles present at equilibrium with the total observed pressure. For instance, the total moles of gas influence the equilibrium condition as given by \( P = \frac{nRT}{V} \). Given that equilibrium pressure (\( 2.05 \text{ atm} \)) and total moles of gas after dissociation (\( 4 - 3\alpha \)), this relationship aids in calculating the degree of dissociation by balancing known values with the ideal gas assumptions.

Equilibrium Constant

The equilibrium constant, denoted \( K_p \) for systems involving gases, provides a snapshot of the concentration or pressure relationships between reactants and products at equilibrium. For gaseous chemical reactions, \( K_p \) is expressed in terms of partial pressures rather than concentrations. The expression for \( K_p \) for the reaction \( \mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g}) + \mathrm{Cl}_{2}(\mathrm{~g}) \) is given by:\[ K_p = \frac{P_{\mathrm{Cl}_{2}} \times P_{\mathrm{PCl}_{3}}}{P_{\mathrm{PCl}_{5}}} \]In this scenario:

• \( P_{\mathrm{Cl}_{2}} = (1 + 3\alpha) \times \frac{RT}{V} \)
• \( P_{\mathrm{PCl}_{3}} = 3\alpha \times \frac{RT}{V} \)
• \( P_{\mathrm{PCl}_{5}} = 3(1-\alpha) \times \frac{RT}{V} \)

Plugging these partial pressures into the \( K_p \) expression helps derive the equilibrium constant, calculated within this problem as 0.476 atm. Understanding \( K_p \) helps predict the direction of the reaction, and whether the reaction is product or reactant-favored under certain conditions.