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 \(K_{c}\) 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 \(\mathrm{Cl}_{2}\) is \(1.0\, \text{mol/L}\).

Step 1: Write the expression for the equilibrium constant

For the given chemical reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\), the equilibrium constant \(K_c\) is expressed as: \[ K_c = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]} \] We need to find \([\mathrm{Cl}_{2}]\) when \(K_c = 0.5\), \([\mathrm{PCl}_{5}] = 0.4\, \text{mol/L}\), and \([\mathrm{PCl}_{3}] = 0.2\, \text{mol/L}\).

Step 2: Rearrange the formula to find [Cl2]

To find the concentration of \([\mathrm{Cl}_{2}]\), rearrange the equilibrium constant expression: \[ [\mathrm{Cl}_{2}] = \frac{K_c \times [\mathrm{PCl}_{5}]}{[\mathrm{PCl}_{3}]} \] This will allow us to solve for \([\mathrm{Cl}_{2}]\) using the given values.

Step 3: Substitute the known values

Now, substitute \(K_c = 0.5\), \([\mathrm{PCl}_{5}] = 0.4\, \text{mol/L}\), and \([\mathrm{PCl}_{3}] = 0.2\, \text{mol/L}\) into the rearranged formula: \[ [\mathrm{Cl}_{2}] = \frac{0.5 \times 0.4}{0.2} \]

Step 4: Calculate the concentration of Cl2

Perform the calculation to find \([\mathrm{Cl}_{2}]\): \[ [\mathrm{Cl}_{2}] = \frac{0.2}{0.2} = 1.0\, \text{mol/L} \] Thus, the concentration of \(\mathrm{Cl}_{2}\) at equilibrium is \(1.0\, \text{mol/L}\).

Key Concepts

Understanding Chemical Equilibrium

Chemical equilibrium is a concept where reactants and products in a chemical reaction are in a state of balance. This means that the rate at which the reactants turn into products is equal to the rate at which the products revert back into reactants.
At this point, the concentration of the different components in the reaction remains constant over time.
This doesn't mean that the reaction has stopped, but rather that it proceeds at the same rate in both the forward and reverse directions.In our exercise involving \( \mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3} + \mathrm{Cl}_{2} \),we see this balance reflected in the equilibrium constant \( K_c \).
This constant is a crucial tool that helps us understand how far a reaction proceeds before reaching equilibrium.

• If \( K_c \) is large, the products are favored at equilibrium.
• If \( K_c \) is small, the reactants are favored.

By calculating \( K_c \), we can determine the concentrations of each component at equilibrium, essential for predicting the outcome of reactions.

Concentration Calculation in Equilibrium Reactions

Calculating concentrations in equilibrium reactions is all about using the equilibrium constant and known concentrations to solve for the unknown. In our reaction \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3} + \mathrm{Cl}_{2}\),we need to find the concentration of \(\mathrm{Cl}_{2}\).
The expression for the equilibrium constant \( K_c \) used is: \[K_c = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]}\]Here, the given values are \( [\mathrm{PCl}_{3}] = 0.2 \, \text{mol/L} \) and \( [\mathrm{PCl}_{5}] = 0.4 \, \text{mol/L} \), with \( K_c = 0.5 \).
We rearrange to solve for \([\mathrm{Cl}_{2}]\):\[[\mathrm{Cl}_{2}] = \frac{K_c \times [\mathrm{PCl}_{5}]}{[\mathrm{PCl}_{3}]}\]Substituting the known values gives us:\[[\mathrm{Cl}_{2}] = \frac{0.5 \times 0.4}{0.2} = 1.0 \, \text{mol/L} \]This calculation reveals the concentration of \(\mathrm{Cl}_{2}\) at equilibrium, indicating the system's balance between reactants and products.

Reaction Stoichiometry Basics

Reaction stoichiometry involves understanding the quantitative relationships between reactants and products in a chemical reaction.
It is vital to grasp stoichiometry to predict how much of each substance is needed or produced.
In the reaction \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3} + \mathrm{Cl}_{2}\), the stoichiometry is 1:1:1.
This implies that 1 mole of \(\mathrm{PCl}_{5}\) produces 1 mole of \(\mathrm{PCl}_{3}\) and 1 mole of \(\mathrm{Cl}_{2}\).Understanding stoichiometry helps to set up the equations used in equilibrium constant expressions.

• It ensures that equations are balanced, which is crucial for accurate calculations.
• Balancing tells us the correct ratios of the substances involved.

By applying stoichiometry, we ensure that all components are properly accounted for, which is crucial for solving equilibrium problems accurately.