Problem 135 - A Complete Resource Book in Chemistry for JEE Main

Question

The degree of dissociation of \(\mathrm{PCl}_{5}\) at one atmosphere is \(0.3 .\) The pressure at which \(\mathrm{PCl}_{5}\) is dissociated to \(50 \%\) is (a) \(0.273 \mathrm{~atm}\) (b) \(0.3 \mathrm{~atm}\) (c) \(1.34 \mathrm{~atm}\) (d) \(2.73 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The pressure at 50% dissociation is approximately 1.34 atm.

Step 1: Understanding the Reaction

The dissociation of \(\mathrm{PCl}_5\) is represented by the equilibrium reaction: \(\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g) + \mathrm{Cl}_2(g)\). Initially, consider 1 mole of \(\mathrm{PCl}_5\). As it dissociates, the number of moles changes, affecting the total pressure.

Step 2: Setting Up Initial Conditions

At the beginning, you have 1 mole of \(\mathrm{PCl}_5\) without any \(\mathrm{PCl}_3\) or \(\mathrm{Cl}_2\). Thus, the initial total moles of the system is 1. The initial pressure is set at 1 atm, which is the reference state.

Step 3: Expression of Degree of Dissociation

The degree of dissociation (\(\alpha\)) is the fraction of \(\mathrm{PCl}_5\) that has dissociated. For 50% dissociation, \(\alpha = 0.5\). This means half of the initial \(\mathrm{PCl}_5\) dissociates to form \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\).

Step 4: Calculating Moles of Each Component

If \(\alpha = 0.5\), then 0.5 moles of \(\mathrm{PCl}_5\) dissociate, creating 0.5 moles of \(\mathrm{PCl}_3\) and 0.5 moles of \(\mathrm{Cl}_2\). Now, \(0.5\) moles of \(\mathrm{PCl}_5\), \(0.5\) \(\mathrm{PCl}_3\), and \(0.5\) \(\mathrm{Cl}_2\) remain.

Step 5: Calculate Total Moles at Equilibrium

At equilibrium, the total moles (n) in the container are the sum of all individual species: \(n = 0.5 + 0.5 + 0.5 = 1.5\) moles.

Step 6: Ideal Gas Law Application

Using the ideal gas law, \( P = \frac{{nRT}}{V} \). Since \(RT/V\) remains constant under the same conditions, pressure is proportional to the total moles. Therefore, pressure at equilibrium \(P = 1.5 \times 1\,\mathrm{atm} = 1.5\,\mathrm{atm}\).

Step 7: Identify Closest Option

The calculated pressure at which 50% of \(\mathrm{PCl}_5\) is dissociated is approximately \(1.5\,\mathrm{atm}\). Among the given options, the closest is (c) \(1.34\,\mathrm{atm}\).

Key Concepts

Degree of Dissociation

In a chemical reaction, the degree of dissociation is a key factor. It represents the fraction or percentage of the original compound that has dissociated into other components. Think of it as how much of the puzzle has been broken apart to reveal new pieces.

For example, if we start with 1 mole of \(\mathrm{PCl}_5\), and 0.3 moles dissociate, then the degree of dissociation \(\alpha\) is 0.3, or 30%. In simpler terms, every time we hear the term 'degree of dissociation', we're looking at how much of our original chemical has broken down.

• Degree of dissociation is represented by \(\alpha\)
• Calculated by dividing the dissociated amount by the initial amount
• Directly affects the composition of products and reactants

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates pressure, volume, temperature, and the number of moles of gas. It's like the rulebook for how gases behave under various conditions.

This law is written as \(PV = nRT\). Here, \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature. When you know three of these variables, you can easily find the fourth. Imagine it as having a toolbox that allows you to solve various gas problems by plugging in the known numbers.

• Widely used to calculate pressures or volumes in reactions
• Assumes gases are 'ideal', meaning they have no interactions
• Perfect for situations where temperature and volume are constants

PCl5 Dissociation Reaction

When we talk about the dissociation of \(\mathrm{PCl}_5\), we're diving into an example of chemical equilibrium. This particular reaction is expressed as:\[ \mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g) + \mathrm{Cl}_2(g) \]

In simple terms, \(\mathrm{PCl}_5\) breaks down into \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\). Initially, the reaction starts with just \(\mathrm{PCl}_5\). As the reaction progresses, \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\) start to form.

This system reaches equilibrium when the rate of the forward reaction equals the rate of the backward reaction. At this point, the composition of the mixture remains steady, although molecules continue to move between reactants and products.

• Illustrates the balance between forming and breaking compounds
• Equilibrium concept highlights unchanging concentrations despite ongoing reactions