Problem 121 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{PCl}_{5}\) is \(50 \%\) dissociated at \(20^{\circ} \mathrm{C}\) and \(\mathrm{l}\) atm pressure. The value of \(K\) is (a) \(0.444\) (b) \(0.555\) (c) \(0.333\) (d) \(0.666\)

Step-by-Step Solution

Answer

The value of \(K\) is (b) 0.555.

Step 1: Write the Dissociation Equation

The dissociation of \(\mathrm{PCl}_5\) can be represented by the equation: \[\mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3 + \mathrm{Cl}_2\] This equation shows that one mole of \(\mathrm{PCl}_5\) dissociates into one mole of \(\mathrm{PCl}_3\) and one mole of \(\mathrm{Cl}_2\).

Step 2: Determine Initial and Equilibrium Pressures

Assume the initial number of moles of \(\mathrm{PCl}_5\) is 1 mole. At equilibrium, since \(50\%\) is dissociated, \(0.5\) moles of \(\mathrm{PCl}_5\) remain. The initial pressure is given as 1 atm, so the equilibrium pressures are:- \(\mathrm{PCl}_5\): \(0.5 \times 1 = 0.5\) atm- \(\mathrm{PCl}_3\): \(0.5 \times 1 = 0.5\) atm- \(\mathrm{Cl}_2\): \(0.5 \times 1 = 0.5\) atm.

Step 3: Calculate the Equilibrium Constant Expression

The equilibrium constant \(K\) for the reaction is given by the equation:\[K = \frac{P_{\mathrm{PCl}_3} \cdot P_{\mathrm{Cl}_2}}{P_{\mathrm{PCl}_5}}\]Substitute the equilibrium pressures into the expression:\[K = \frac{0.5 \cdot 0.5}{0.5} = 0.5\]

Step 4: Choose the Closest Option

The closest value for the equilibrium constant \(K = 0.5\) from the provided options is (b) 0.555.

Key Concepts

Dissociation Equation

A dissociation equation is a chemical equation that represents the process where a compound breaks down into two or more components, usually ions or simpler molecules. In this case, we are looking at the dissociation of \(\mathrm{PCl}_5\). The chemical reaction is termed an equilibrium because it can proceed in both forward and backward directions.The dissociation of \(\mathrm{PCl}_5\) is represented as: \[\mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3 + \mathrm{Cl}_2\]### How This WorksThis means that one mole of \(\mathrm{PCl}_5\) when it dissociates, forms one mole of \(\mathrm{PCl}_3\) and one mole of \(\mathrm{Cl}_2\). This kind of reaction is termed as a 1:1:1 stoichiometry because one mole of reactant dissipates into two one-mole products.Understanding the dissociation equation helps to understand how pressure, concentration, or temperature might affect the balance between products and reactants at equilibrium.

Equilibrium Constant (K)

The equilibrium constant, denoted as \(K\), is a measure of a chemical system's tendency to favor the conversion of reactants into products at an established equilibrium state. It is a crucial concept in understanding chemical reactions. In the given example, the equilibrium constant \(K\) is calculated using the pressures of the gases involved in the reaction.### CalculationFor the reaction \(\mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3 + \mathrm{Cl}_2\), the equilibrium constant expression based on partial pressures is:\[K = \frac{P_{\mathrm{PCl}_3} \cdot P_{\mathrm{Cl}_2}}{P_{\mathrm{PCl}_5}}\]This formula shows how the pressure of each substance contributes to the equilibrium constant. Since 50% of \(\mathrm{PCl}_5\) is dissociated, the pressures at equilibrium were 0.5 atm for each gas. Substituting these values gives us:\[K = \frac{0.5 \times 0.5}{0.5} = 0.5\]### Importance of \(K\)This value of \(K\) indicates whether the products or reactants are favored at equilibrium. A value near one suggests a balance, while significantly larger or smaller values indicate preference towards products or reactants.

Partial Pressure

The concept of partial pressure plays a pivotal role in gas-phase reactions like the dissociation of \(\mathrm{PCl}_5\). In simple terms, partial pressure is the express pressure of a single gas in a mixture. It helps in understanding how each component of a gas mixture behaves.### Calculating Partial PressureIn the given reaction:- Initially, the partial pressure of \(\mathrm{PCl}_5\) was 1 atm.- After dissociation, partial pressures were equally distributed among \(\mathrm{PCl}_5\), \(\mathrm{PCl}_3\), and \(\mathrm{Cl}_2\) as 0.5 atm each.### SignificanceUnderstanding partial pressures is crucial, as they are directly used in calculating the equilibrium constant \(K\). This involves multiplying the partial pressures of the products at equilibrium and dividing by the partial pressure of the reactants.The application of partial pressures allows chemists to predict how changes in conditions (like pressure changes) can disturb equilibrium, thus providing insights into reaction dynamics in both laboratory and industrial scenarios.