Problem 88 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\), the equilibrium concentrations of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) are \(0.4\) and \(0.2\) mole/litre respectively. If the value of \(\mathrm{K}_{\mathrm{c}}\) is \(0.5\), what is the concentration of \(\mathrm{Cl}_{2}\) in mole \(/\) litre? (a) \(2.0\) (b) \(1.5\) (c) \(1.0\) (d) \(0.5\)

Step-by-Step Solution

Answer

The concentration of \(\text{Cl}_2\) is 1.0 mol/L.

Step 1: Write the Equilibrium Expression

For the reaction \(\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g),\) the equilibrium constant \(K_c\) is given by the expression: \[K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}.\]

Step 2: Substitute the Known Values

Substitute the known equilibrium concentrations into the \(K_c\) expression. We have \([\text{PCl}_5] = 0.4\, \text{mol/L}\) and \([\text{PCl}_3] = 0.2\, \text{mol/L}\). The value of \(K_c\) is \(0.5\).Thus, \[0.5 = \frac{0.2 \cdot [\text{Cl}_2]}{0.4}.\]

Step 3: Solve for [Cl2]

Now, solve the equation for \([\text{Cl}_2]\):\[0.5 = \frac{0.2 \cdot [\text{Cl}_2]}{0.4}\]Multiply both sides by \(0.4\):\[0.5 \times 0.4 = 0.2 \cdot [\text{Cl}_2]\]\[0.2 = 0.2 \cdot [\text{Cl}_2]\]Finally, divide both sides by \(0.2\) to solve for \([\text{Cl}_2]\):\[[\text{Cl}_2] = \frac{0.2}{0.2} = 1.0 \, \text{mol/L}.\]

Key Concepts

Equilibrium Constant (Kc)

In chemical reactions, the equilibrium constant, denoted as \( K_c \), is crucial for understanding how a reaction behaves when it reaches equilibrium. Equilibrium is the state where the rate of the forward reaction equals the rate of the reverse reaction.
It does not mean the concentrations of reactants and products are equal, but that they remain constant over time.The equilibrium constant \( K_c \) is a ratio of the product concentrations to the reactant concentrations, each raised to their respective coefficients in the balanced chemical equation. For example, in the reaction \( \mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g}) + \mathrm{Cl}_2(\mathrm{~g}) \), the expression for \( K_c \) is:

• \( K_c = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]} \)

A higher \( K_c \) value indicates that, at equilibrium, products are favored, while a lower \( K_c \) suggests that reactants are more prevalent.

Concentration Calculation

Calculating concentrations at equilibrium allows us to predict how much of each substance is present in the mixture when the reaction stops changing biologically or chemically. The concentrations of the reactants and products are crucial for the calculating of \( K_c \) and for understanding the progress of the reaction.
In the given reaction \( \mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g}) + \mathrm{Cl}_2(\mathrm{~g}) \), you're given that \([\mathrm{PCl}_5] = 0.4\, \text{mol/L}\) and \([\mathrm{PCl}_3] = 0.2\, \text{mol/L}\). Using the equilibrium constant \( K_c = 0.5 \), you can solve for \([\mathrm{Cl}_2]\).First, substitute these values into the equilibrium expression:

• \( 0.5 = \frac{0.2 \cdot [\mathrm{Cl}_2]}{0.4} \)

Next, multiplying both sides by 0.4 isolates \([\mathrm{Cl}_2]\):

• \(0.5 \times 0.4 = 0.2 \cdot [\mathrm{Cl}_2] \)

Finally, solve for \([\mathrm{Cl}_2]\):

• \([\mathrm{Cl}_2] = \frac{0.2}{0.2} = 1.0 \text{ mol/L} \)

Le Chatelier’s Principle

Le Chatelier’s Principle helps predict how a change in conditions affects the equilibrium of a system. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and restore a new balance. Changes could include:

• Concentration adjustments of reactants or products
• Alterations in temperature
• Pressure changes in gaseous reactions

When a system in equilibrium experiences an increase in the concentration of \( \mathrm{PCl}_3 \) or \( \mathrm{Cl}_2 \), the system will shift to the left to form more \( \mathrm{PCl}_5 \), reducing the added substances.
Conversely, removing some \( \mathrm{PCl}_5 \) will also shift equilibrium to the left to partially restore the original system.
Le Chatelier's Principle is a guiding tool that finds applications in synthesis and industrial processes by predicting how a reaction can be driven forward or reversed to optimize production.