Problem 83 - A Complete Resource Book in Chemistry for JEE Main

Question

The reaction \(\mathrm{PCl}_{5}(\mathrm{~s}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) is in equilibrium. If the equilibrium concentration of \(\mathrm{PCI}_{3}\) (g) is doubled, then concentration of \(\mathrm{Cl}_{2}(\mathrm{~g})\) would become (a) \(1 / 2\) of its initial value (b) \(1 / 4\) of its initial value (c) four times of its initial value (d) two times of initial value

Step-by-Step Solution

Answer

(a) 1/2 of its initial value

Step 1: Understanding the Equilibrium

Consider the equilibrium reaction given: \( \mathrm{PCl}_{5}(\mathrm{~s}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g}) + \mathrm{Cl}_{2}(\mathrm{~g}) \). In an equilibrium reaction, concentrations of the reactants and products do not change with time as long as the conditions remain constant. The concentration of solids like \( \mathrm{PCl}_{5} \) does not appear in the expression for the equilibrium constant \( K_c \).

Step 2: Equilibrium Constant Expression

For the given reaction, the equilibrium constant \( K_c \) can be expressed as: \[ K_c = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]} = [\mathrm{PCl}_{3}][\mathrm{Cl}_{2}] \] because \( \mathrm{PCl}_{5}(\mathrm{~s}) \) is a solid and not included in the expression.

Step 3: Relation on Increasing Concentration of \(\mathrm{PCl}_{3}(g)\)

Given that the equilibrium concentration of \(\mathrm{PCl}_{3}(\mathrm{g})\) is doubled. Let \([\mathrm{PCl}_{3}] = a\) initially, thus \([\mathrm{Cl}_{2}] = a\) initially (since they are produced in a 1:1 ratio). When \( [\mathrm{PCl}_{3}]=2a \), find the new concentration \([\mathrm{Cl}_{2}]\) that maintains the \( K_c \) constant.

Step 4: Solve for New \([\mathrm{Cl}_{2}]\)

Using the equilibrium expression: \[ K_c = [\mathrm{PCl}_{3}][\mathrm{Cl}_{2}] = 2a \times \mathrm{New}\, [\mathrm{Cl}_{2}] \] The new value of \([\mathrm{Cl}_{2}]\) must satisfy \(2a \times \mathrm{Cl}_{2,new} = a^2\). Thus, \( \mathrm{Cl}_{2,new} = \frac{a^2}{2a} = \frac{a}{2} \). It shows that the concentration of \(\mathrm{Cl}_{2}(\mathrm{~g})\) becomes half of its initial value.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \) for reactions in a chemical system, serves as a measure of the relationship between the concentration of reactants and products at equilibrium. It tells us how far a reaction proceeds and is crucial for predicting the direction of the reaction under different conditions.

In essence, \( K_c \) is calculated using the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their respective coefficients in the balanced chemical equation. For the given reaction, \( \mathrm{PCl}_{5}(s) \rightleftharpoons \mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g) \), the equilibrium constant is expressed only using gases or aqueous species. Therefore, the equation becomes:

• \( K_c = [\mathrm{PCl}_{3}][\mathrm{Cl}_{2}] \)

Notice that \( \mathrm{PCl}_{5} \) is a solid and does not appear in the equation for \( K_c \). This is because the concentration of solids and liquids are considered constant and do not affect the equilibrium constant in the same way as gases and solutions.

Le Chatelier's Principle

Le Chatelier's Principle is an essential concept in understanding how chemical reactions at equilibrium respond to changes in conditions. This principle states that if an external change is applied to a system at equilibrium, the system adjusts to counteract that change and restore a new equilibrium state.

When applied to the exercise at hand, doubling the concentration of \( \mathrm{PCl}_{3} \) disturbs the original equilibrium. According to Le Chatelier's Principle, the system will attempt to minimize this disturbance by adjusting the concentration of \( \mathrm{Cl}_{2} \).

This is observed as the system seeks to re-establish equilibrium by decreasing the concentration of \( \mathrm{Cl}_{2} \). The equilibrium shifting towards reactants, or reducing \( \mathrm{Cl}_{2} \), is a classic application of Le Chatelier's Principle in action.

Concentration Changes

Concentration changes play a pivotal role in determining and shifting the state of equilibrium in a chemical reaction. When the concentration of any component in a reaction mixture is altered, the system responds by attempting to restore equilibrium.

• Doubling the concentration of \( \mathrm{PCl}_{3} \) means an increase in the amount of this product.
• As the concentration of \( \mathrm{PCl}_{3} \) increases, according to the equation \( K_c = [\mathrm{PCl}_{3}][\mathrm{Cl}_{2}] \), the concentration of \( \mathrm{Cl}_{2} \) must adjust to maintain a constant \( K_c \).

The solution to our original problem revealed that as \( [\mathrm{PCl}_{3}] \) is doubled, \( [\mathrm{Cl}_{2}] \) reduces to half to maintain equilibrium. This illustrates the dynamic nature of equilibrium systems, where changes in one part trigger compensatory changes in another to sustain constancy in \( K_c \).